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1. What is the amino acid sequence encoded by the DNA sequence?
The DNA sequence given is: GCATGCTGCGAAACTTTGGCTGA
The 3-letter codons are: ATG CTG CGA AAC TTT GGC TGA
The amino acid sequence is: Met Leu Arg Asn Phe Gly Stop
2. What does a missense mutation look like?
Here's one example: GCATGGTGCGAAACTTTGGCTGA
The 3-letter codons are: ATG GTG CGA AAC TTT GGC TGA
The amino acid sequence is: Met Val Arg Asn Phe Gly Stop
3. What does a nonsense mutation look like?
Here's one example: GCATGCTGTGAAACTTTGGCTGA
The 3-letter codons are: ATG CTG TGA AAC TTT GGC TGA
The amino acid sequence is: Met Leu Stop
4. You can make more than one frameshift mutation. What do they look like? How do they change the
amino acid sequence? Here are some examples of different frameshift mutations.
Insertion: GCATGCTGATCGAAACTTTGGCTGA
The 3-letter codons are: ATG CTGATC GAA ACT TTG GCT GA
The amino acid sequence is: Met Leu Ile Glu Thr Leu Ala
Deletion: GCATGCT-CGAAACTTTGGCTGA
The 3-letter codons are: ATG CTC GAA ACT TTG GCT GA
The amino acid sequence is: Met Leu Glu Thr Leu Ala
Insertion: GCATGCTGAAAGATAGTTGTCGAAACTTTGGCTGA
The 3-letter codons are: ATG CTG AAA GAT AGT TGT CGA AAC TTT GGC TGA
The amino acid sequence is: Met Leu Lys Asp Ser Cys Arg Asn Phe Gly
Deletion: GCATGCTGCG-----TTGGCTGA
The 3-letter codons are:ATG CTG CGT TGG CTG
The amino acid sequence is: Met Leu Arg Trp Leu
5. Notice how a single amino acid can be encoded by a number of different codons. What are some
examples of silent mutations in the DNA sequence?
Here's one example: GCATGCTGCGTAACTTTGGCTGA
The 3-letter codons are: ATG CTG CGT AAC TTT GGC TGA
The amino acid sequence is: Met Leu Arg Asn Phe Gly Stop
A silent mutation changes a nucleotide without changing the codon. For many amino acids the third
nucleotide of the codon can be variable. For example, CCT, CCC, CCA and CCG all code for Proline.
However, for some amino acids, both the first and the third nucleotide of the codon can vary. For example,
CTT, CTC, CTA, TTA, CTG, AND TTG all code for Leucine.
Multiple Choice Questions
1. Which of the following is not known to be involved in initiation by eukaryotic RNA
polymerase II?
A) DNA helicase activity
B) DNA polymerase activity
C) Formation of an open complex
D) Protein binding to specific DNA sequences
E) Protein phosphorylation
2. Processing of a primary mRNA transcript in a eukaryotic cell does not normally involve:
A) attachment of a long poly(A) sequence at the 3' end.
B) conversion of normal bases to modified bases, such as inosine and pseudouridine.
C) excision of intervening sequences (introns).
D) joining of exons.
E) methylation of one or more guanine nucleotides at the 5' end.
3. The 5'-terminal cap structure of eukaryotic mRNAs is a(n):
A) 7-methylcytosine joined to the mRNA via a 2',3'-cyclic linkage.
B) 7-methylguanosine joined to the mRNA via a 5' 3' diphosphate linkage.
C) 7-methylguanosine joined to the mRNA via a 5' 5' triphosphate linkage.
D) N
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-methyladenosine joined to the mRNA via a 5' 5' phosphodiester bond.
E) O
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-methylguanosine joined to the mRNA via a 5' 5' triphosphate linkage.
4. The excision (splicing) of many group I introns requires, in addition to the primary
transcript RNA:
A) a cytosine nucleoside or nucleotide and a protein enzyme.
B) a guanine nucleoside or nucleotide (only).
C) a protein enzyme only.
D) a small nuclear RNA and a protein enzyme.
E) ATP, NAD, and a protein enzyme.
5. A branched (“lariat”) structure is formed during:
A) attachment of a 5' cap to mRNA.
B) attachment of poly(A) tails to mRNA.
C) processing of preribosomal RNA.
D) splicing of all classes of introns.
E) splicing of group II introns.
6. Splicing of introns in nuclear mRNA primary transcripts requires:
A) a guanine nucleoside or nucleotide.
B) endoribonucleases.
C) polynucleotide phosphorylase.
D) RNA polymerase II. 暨南大學應用化學系 992 生化小考(4)試題 2011
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E) small nuclear ribonucleoproteins (snurps).
7. Which one of the following is not true of the mRNA for ovalbumin?
A) Exons are used for polypeptide synthesis.
B) Introns are complementary to their adjacent exons and will form hybrids with them.
C) The mature mRNA is substantially shorter than the corresponding region on the
DNA.
D) The mRNA is originally synthesized in the nucleus, but ends up in the cytoplasm.
E) The splicing that yields a mature mRNA occurs at very specific sites in the RNA
primary transcript.
8. Differential RNA processing may result in:
A) a shift in the ratio of mRNA produced from two adjacent genes.
B) attachment of the poly(A) tail to the 5' end of an mRNA.
C) inversion of certain exons in the final mRNA.
D) the production of the same protein from two different genes.
E) the production of two distinct proteins from a single gene.
9. Which of these polymerases does not require a template?
A) RNA pol I
B) RNA pol II
C) Reverse transcriptase
D) Polyadenylate polymerase
E) RNA replicase
10. RNA polymerase:
A) binds tightly to a region of DNA thousands of base pairs away from the DNA to be
transcribed.
B) can synthesize RNA chains de novo (without a primer).
C) has a subunit called (lambda), which acts as a proofreading ribonuclease.
D) separates DNA strands throughout a long region of DNA (up to thousands of base
pairs), then copies one of them.
E) synthesizes RNA chains in the 3' 5' direction.
11. Which of the following statements about E. coli RNA polymerase (core enzyme) is
false?
A) In the absence of the subunit, core polymerase has little specificity for where
initiation begins.
B) The core enzyme contains several different subunits.
C) The core enzyme has no polymerizing activity until the subunit is bound.
D) The RNA chain grows in a 5' 3' direction.
E) The RNA product is complementary to the DNA template. 暨南大學應用化學系 992 生化小考(4)試題 2011
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12. RNA polymerase from E. coli (core enzyme alone) has all of the following properties
except that it:
A) can extend an RNA chain and initiate a new chain.
B) is required for the synthesis of mRNA, rRNA, and tRNA in E. coli.
C) produces an RNA polymer that begins with a 5'-triphosphate.
D) recognizes specific start signals in DNA.
E) requires all four ribonucleoside triphosphates and a DNA template.
Short Answer Questions
13. Write the sequence of the messenger RNA molecule synthesized from a DNA template
strand having the sequence:
(5')ATCGTACCGTTA(3')
Ans: (3')UAGCAUGGCAAU(5'). Also acceptable is (5')UAACGGUACGAU(3').
14. For each of the following statements, indicate with a P if the statement applies only to
prokaryotes, an E if the statement applies only to eukaryotes, and an E and a P if the
statement applies to both eukaryotes and prokaryotes.
A.___ RNA synthesis is blocked by actinomycin D.
B. ___ A single RNA polymerase transcribes genes that encode mRNAs, tRNAs, and
rRNA.
C. ___ Transcription of mRNA is blocked by -amanitin.
D. ___ Sigma () subunit detaches from RNA polymerase shortly after transcription has
initiated.
E. ___ The 5' end of the mature mRNA begins with a triphosphate.
F. ___ Termination of transcription requires the protein factor.
Ans: E and P; P; E; P; P; P
15. Indicate whether each of the following statements about eukaryotic cells is true (T) or
false (F).
A. ___ They have three distinct RNA polymerases.
B. ___ Their mRNAs are generally synthesized by RNA polymerase I.
C. ___ RNA polymerase III synthesizes only rRNAs.
D. ___ The 5S rRNA is synthesized by RNA polymerase I
E. ___ Their RNA polymerases initiate transcription at specific promoter sites on the
DNA
Ans: T; F; F; F; T
16. Name four general types of postsynthetic processing reactions that are observed in RNA.
Briefly (one sentence or less) point out an example of each type. In your example,
identify the type of RNA molecule involved (tRNA, mRNA, rRNA, etc.), the type of
“processing” involved, and whether the example is characteristic of eukaryotes or 暨南大學應用化學系992 生化小考(4)試題 2011
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prokaryotes, or both. Do not describe specific genes, sequences, complicated
structures, or enzymes.
17. What is a telomere? Describe the key features of its structure. What is unusual about
the structure and/or mechanism of action of telomerase?
18. List one basic property that distinguishes RNA polymerases from DNA polymerases,
and list one basic property they share.
BIOCHEMISTRY TEST
You have 546 questions in this exam.
1. Increasing the concentration of dGTP directly causes ribonucleotide reductase toA. increase the rate of production of dGDP.B. increase the rate of production of dADP.C. decrease the rate of production of rCDP.D. decrease the rate of production of dADP.E. increase the rate of production of dTTP.
Show answer
Correct Answer: B
Feedback A: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
Feedback C: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
Feedback D: This rate increases
Feedback E: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
2. Glycogen phosphorylase is activated most directly byA. epinephrine.B. phosphorylase kinase.C. phosphorylase phosphatase.D. cAMP.E. glucagon.
Show answer
Correct Answer: B
Feedback A: See B
Feedback B: glucagon or epinephrine ---> increased cAMP ---> active PKA ---> active phosphorylase kinase ---> active glycogen phosporylase
Feedback C: See B
Feedback D: See B
Feedback E: See B
3. Which of the following cofactors is most directly concerned with a CO2 fixation reaction?
A. biotinB. vitamin DC. coenzyme AD. thiamine pyrophosphateE. lipoic acid
Show answer
Correct Answer: A
Feedback A: Biotin is a carrier of activated CO2 and is a component of acetyl carboxylase, propionyl CoA carboxylase, and pyruvate carboxylase.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
4. Which statement about reactions catalyzed by transaminases (aminotransferases) is false?A. The equilibrium constant is close to 1.B. The reaction involves a Schiff base intermediate.C. Pyridoxal phosphate is covalently bound to the enzyme protein through the epsilon amino group of a lysine residue.D. The reactions are important for the biosynthesis of non-essential amino acids.E. Ammonia is liberated.
Show answer
Correct Answer: E
Feedback A: Aminotransferases can participate in degradation or biosynthesis depending on substrate concentrations.
Feedback B:
Feedback C:
Feedback D: Aminotransferases most commonly transfer amino groups to alpha-ketoglutarate forming glutamate. Glutamate can then be the amino group donor for amino acid biosynthesis.
Feedback E: False, an amino group is transferred to an alpha-keto acid.
5. The enzyme in muscle which is most directly stimulated by epinephrine isA. glycogen synthase.B. phosphofructokinase.C. isocitrate dehydrogenaseD. glucose-6-phosphatase.E. adenylate cyclase.
Show answer
Correct Answer: E
Feedback A: inactivated due to epinephrine.
Feedback B: decreased stimulation of PFK-1
Feedback C: very indirect modification
Feedback D: Changes in its activity are determined by synthesis and degradation.
Feedback E: increased intracellular cAMP.
6. Acetyl-CoA regulates gluconeogenesis by activation ofA. phosphoenolpyruvate carboxykinase.B. pyruvate kinase.C. pyruvate carboxylase.D. lactate dehydrogenase.E. fructose 1,6-bisphosphatase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Which catalyzes: pyruvate + CO2 ---> oxaloacetate.
Feedback D:
Feedback E:
7. Which compound does NOT cross the inner mitochondrial membrane due to lack of a specific transport protein?A. malateB. glutamineC. NADHD. citrateE. ATP
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: NADH reducing potential must be shuttled across the membrane by another molecule.
Feedback D:
Feedback E:
8. All of the following statements are correct with regard to chromatin EXCEPTA. Core nucleosomes consist of 8 histone protein molecules and about 146 base pairs of DNA.B. Histones have been highly conserved during evolution.C. Almost all of the DNA in a eukaryotic cell is assembled into nucleosomes.D. Assembly of DNA into nucleosomes compacts it about 100-fold in length.E. Nucleosome cores are separated by variable amounts of linker DNA.
Show answer
Correct Answer: D
Feedback A: see D
Feedback B: see D
Feedback C: see D
Feedback D: Nucleosomes effect about a 5-7 fold compaction of the length of the DNA
Feedback E: see D
9. Which is LEAST likely to be characteristic of a regulated enzyme in a metabolic pathway?A. Catalyzes the slowest step of the pathwayB. Has a short half-life in vivoC. Concentration is hormonally regulatedD. Activity under allosteric controlE. Catalyzes a physiologically reversible reaction
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Irreversible reactions are usually regulated, especially those that involve ATP.
10. Which of the following are examples of tertiary (3°) structure? 1. the sequence of amino acids in a protein 2. the alpha-helix of hemoglobin 3. the folding of alpha-helix onto beta-sheet of triose phosphate isomerase 4. hydrogen bonding of one beta -sheet to another beta-sheet 5. juxtaposition of amino acids that are separated by a "long distance" in primary (1°) structure 6. the un-ordered portions of carboxy peptidase A 7. juxtaposition of hemoglobin beta-7 Glu with alpha-68 AsnA. all the even-numbered itemsB. all the odd-numbered itemsC. 3, 4, and 5D. 3, 4, 5, and 7E. 3, 4, 5, 6, and 7
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: 1-primary, 2-secondary, 6-hard to classify--primary or secondary, 7-quaternary
Feedback D:
Feedback E:
11. The H1 histoneA. is one of the four histones in the nucleosome core.B. helps form the octamer nucleosome complex.C. is rich in acidic amino acid residues.D. is present at a 1:1 weight ratio with DNA.E. helps fold or stabilize the "beads-on-a-string" structure into higher order structures.
Show answer
Correct Answer: E
Feedback A: No, the core histones are H2A, H2B, H3, and H4
Feedback B: No, the octamer contains only H2A, H2B, H3, and H4
Feedback C: It is rich in basic amino acids
Feedback D: Overall, all histones (H1 and H2A, H2B, H3, and H4) are about 1:1 with DNA
Feedback E:
12. Which mitochondrial enzyme requires acetyl-CoA as a substrate?A. Citrate synthaseB. Succinyl-CoA synthaseC. Succinic dehydrogenaseD. Pyruvate dehydrogenaseE. Isocitrate dehydrogenase
Show answer
Correct Answer: A
Feedback A: Catalyzes: acetyl CoA + oxaloacetate ---> citrate.
Feedback B: CoA is required coenzyme, but not acetyl CoA.
Feedback C:
Feedback D: Produces acetyl CoA from pyruvate.
Feedback E:
13. Transfer of the methyl group from 5-methyl tetrahydrofolate to homocysteine to form methionine requiresA. NADPHB. vitamin B12C. FADD. pyridoxal phosphateE. none of the above
Show answer
Correct Answer: B
Feedback A:
Feedback B: Methycobalamin, a derivative of B12, is the coenzyme that mediates the transfer of methyl groups in methionine biosynthesis.
Feedback C:
Feedback D: Aminotransferases require this coenzyme.
Feedback E:
14. All of the following are true of bacterial plasmids or episomes EXCEPTA. Episomes like the F factor can be transferred from one cell to another.B. Plasmids or episomes can confer resistance to many antibiotics simultaneously.C. Plasmids or episomes can be transferred from one species of bacteria to another.D. Since plasmids must be replicated, they confer a disadvantage to their hosts and can therefore be expected to disappear from human pathogens soon.E. Genes can be transferred from bacterial genomes to plasmids at a low frequency.
Show answer
Correct Answer: D
Feedback A: This is true, they can pass through a pilus into another cell.
Feedback B: This is true.
Feedback C: This is true
Feedback D: The disadvantages caused by “carrying†plasmids are outweighed by the advantages �they provide; this advantage increases as we use antibiotics.
Feedback E: Strange but true.
15. If one mole of glucose is metabolized to carbon dioxide and water via glycolysis and the tricarboxylic acid cycle, and the glycerol phosphate shuttle is operative, the net conversion of ADP (or equivalent) to ATP (or equivalent) would theoretically beA. 12 moles.B. 24 moles.C. 32 moles.D. 36 moles.E. 38 moles.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The 2NADH produced by glycolysis are converted to 2FADH2 in the mitochondria by the glycerol phosphate shuttle, so they are worth 2ATP each.
Feedback E: With the malate-aspartate shuttle the 2NADH from glycolysis are converted to 2NADH in the mitochondria, and are worth 3ATP each.
16. The cofactor for transamination isA. nicotinamide.B. biotin.C. thiamine pyrophosphate.D. pyridoxal phosphate.E. vitamin B12.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: A vitamin B6 derrivative that is a prosthetic group for each transaminase.
Feedback E:
17. Which enzyme catalyzes a reaction in which carbon dioxide (or bicarbonate) is neither a substrate nor
a product?A. isocitrate dehydrogenaseB. pyruvate carboxylaseC. carboxypeptidaseD. phosphoenolpyruvate carboxykinaseE. alpha-ketoglutarate dehydrogenase
Show answer
Correct Answer: C
Feedback A: CO2 produced.
Feedback B: CO2 is a substrate.
Feedback C: Cleaves peptide bonds, no CO2 involved.
Feedback D: CO2 produced.
Feedback E: CO2 produced.
18. 5-phosphoribosyl-1-pyrophosphate (PRPP) is an intermediate inA. the de novo synthesis of purine nucleotides.B. the de novo synthesis of pyrimidine nucleotides.C. the salvage pathway for the synthesis of purine nucleotides.D. A and C.E. A, B and C.
Show answer
Correct Answer: E
Feedback A: True (it is activated as the committed step for purine synthesis), but so are B and C
Feedback B: True (orotidylate is formed by reacting the free base with PRPP), but so are A and C
Feedback C: True (PRPP reacts with free bases like guanine and hypoxanthine through HGPRTase), but so are A and B.
Feedback D: True, but B is also correct
Feedback E:
19. Which statements are FALSE? In mitochondria, tricarboxylic acid cycle reactions generally proceed more(1) rapidly as the ADP concentration rises.(2) slowly as the ADP concentration rises.(3) slowly as the NADH concentration rises.(4) rapidly as the NADH concentration rises.(5) rapidly as the oxaloacetate concentration increases.
A. 1 and 3B. 2 and 4C. 3 and 5D. 1 and 4E. 2 and 5
Show answer
Correct Answer: B
Feedback A:
Feedback B: High levels of ATP and NADH signal that the cell’s energy level is high, high ADP and NAD+ signal that more energy is needed.
Feedback C:
Feedback D:
Feedback E:
20. In Lesch-Nyhan syndrome, lack of HGPRTase activity should result in higher than normal tissue concentrations of all of the following EXCEPTA. adenine.B. guanine.C. uric acid.D. hypoxanthine.E. PRPP.
Show answer
Correct Answer: A
Feedback A: Adenine levels are not directly affected by loss of this enzyme.
Feedback B: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
Feedback C: Increased hypoxanthine leads to increased xanthine and therefore increased uric acid.
Feedback D: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
Feedback E: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
21. Which of the following represents the correct number of moles of products formed as one mole of acetate (in the form of acetyl-CoA) moves through one turn of the citric acid cycle? (respectively)GTPCO2NADHFADH2
A. 1 1 3 1B. 2 2 3 1C. 1 2 3 1D. 1 2 2 1E. 1 1 2 2
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Memorize the cycle.
Feedback D:
Feedback E:
22. How many net moles of ATP are produced per glucose equivalent when glycogen is the substrate for anaerobic glycolysis?A. 0B. 1
C. 2D. 3E. -2
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: glycogen ---> glucose-1-phosphate ---> glucose-6-phosphate ---> glycolysis. Because glycose-6-phosphate enters glycolysis the reaction catalyzed by glucokinase or hexokinase is not carried out and one less ATP is used.
Feedback E:
23. One of the strands of a double-stranded, 10 kbp DNA duplex has the following numbers of base residues: adenine(a) - 3800, thymine(t) - 2600. The base composition of the whole double-stranded molecule will beA. a - 5200, t - 5200, g - 4800, c - 4800B. a - 7600, t - 7600, g - 2400, c - 2400C. a - 6400, t - 6400, g - 3600, c - 3600D. a - 6400, t - 3600, g - 6400, c - 3600E. There is not enough information to determine the overall base composition
Show answer
Correct Answer: C
Feedback A: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback B: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback C:
Feedback D: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback E: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
24. A major control point of the citric acid cycle is exerted at the level ofA. succinate dehydrogenase.B. citrate synthase.C. succinyl CoA synthase.D. malate dehydrogenase.E. aconitase.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Citrate synthase, isocitrate dehydrogenase, and alpha-ketoglutarate dehydrogenase are the regulated enzymes of TCA cycle.
Feedback C:
Feedback D:
Feedback E:
25. Secondary structure of proteinsA. is maintained by hydrogen bonds.B. refers to proteins consisting of two (or more) subunits held together by noncovalent forces.C. encompasses any hydrogen-bonded interaction found in proteins.D. refers to proteins consisting of one or more polypeptide chains plus a nonprotein moiety.E. is found only in fibrous proteins.
Show answer
Correct Answer: A
Feedback A: alpha-helix and beta-sheet.
Feedback B: This is quaternary.
Feedback C: No, H-bond interactions between residues far apart linearly would be tertiary interactions.
Feedback D: This is quaternary.
Feedback E: Any protein.
26. Each of the following statements concerning DNA is true EXCEPT which one?A. C does not have to equal G in single stranded DNA.B. DNA can be found in either circular or linear forms.C. The two DNA strands of the double helix are antiparallel.D. Hydrogen bonds alone hold the DNA helix together.E. Base pairs lie in a plane perpendicular to the long axis of the helix.
Show answer
Correct Answer: D
Feedback A: This is true. For example, a single-stranded DNA could be all C and have no G.
Feedback B: This is true; most bacterial plasmids are circles while most mammalian chromosomes are linear.
Feedback C: Please don’t tell me you picked this one, I just don’t want to know.
Feedback D: No, they contribute but they are only one part of the picture.
Feedback E: This is true.
27. The reduction potential for a reactionA. is an oxidation potential.B. is unrelated to the free energy of the reaction.C. is negative for a spontaneous process.D. can be used to decide if one substance will reduce another.E. is not dependent on reactant and/or product concentration.
Show answer
Correct Answer: D
Feedback A: Has the opposite sign of the reduction potential.
Feedback B:
Feedback C: Positive for a spontaneous redox reaction.
Feedback D: More negative (smaller) reduction potentials donate electrons to more positive (larger) reduction potentials.
Feedback E:
28. GlutamineA. is not incorporated into proteins because there is no codon or tRNA for glutamine.B. is the principal donor of amino groups in transamination reactions.C. donates its alpha-amino group in the de novo biosynthesis of both purine and pyrimidine nucleotides.D. is synthesized from glutamate primarily in the liver.E. represents an important non-toxic transport form of ammonia.
Show answer
Correct Answer: E
Feedback A: There are codons and tRNA’s for both glutamine and glutamate.
Feedback B: Glutamine is the ultimate product of tissue transamination. Glutamine is transported to the liver where nitrogen metabolism to urea takes place.
Feedback C: Donates its gamma-amide in both syntheses.
Feedback D: Synthesis occurs primarily in the tissues and then glutamine is broken down into glutamate and ammonia in the liver.
Feedback E: See B.
29. As a cell initiates DNA replication, it contains sufficient pools of dNTPs to completeA. about 1% or less of DNA replication without making new dNTPs.B. about 10% of DNA replication without making new dNTPs.C. 100% of DNA replication without making new dNTPs.D. about 10 times more dNTPs than it needs to replicate the genome once.E. no replication at all.
Show answer
Correct Answer: A
Feedback A: About 0.1% in mammals, 1% in bacteria
Feedback B: About 0.1% in mammals, 1% in bacteria
Feedback C: About 0.1% in mammals, 1% in bacteria
Feedback D: About 0.1% in mammals, 1% in bacteria
Feedback E: About 0.1% in mammals, 1% in bacteria
30. The uncoupling of oxidative phosphorylation in the human might be of physiological importance because itA. allows storage of nutrients.B. produces water.C. increases carbon dioxide level in the blood.D. produces heat.E. raises the oxygen level in the blood.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The energy produced by electron transport is released as heat rather than being used to synthesize ATP.
Feedback E:
31. The condition of LACTOSE INTOLERANCE can be caused by defective or deficientA. lactoseB. UDP-galactose epimeraseC. galactose-1-phosphate uridyl transferaseD. lactaseE. galactase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
32. During the second day of a fast, after liver glycogen is depleted, blood glucoseA. is derived mainly from fatty acids of adipose tissue.B. is derived mainly from muscle glycogen.C. is derived mainly from the amino acids of liver proteins.D. falls to low levels (e.g., 20 mg/100 ml) until eating is resumed.E. is derived mainly from the amino acids of muscle proteins.
Show answer
Correct Answer: E
Feedback A: After approximately 3 weeks of starvation muscle almost exclusively oxidizes fatty acids for energy.
Feedback B:
Feedback C:
Feedback D:
Feedback E: Rapid breakdown of muslce protein takes place the first few days of starvation to provide amino acids for gluconeogenesis.
33. A complete lack of adenosine deaminase causes SCID: severe combined immuno-deficiency. Which of the following is LEAST true?A. Loss of the enzyme causes increased levels of dATP because there is less turnover of adenosine nucleosides in general.
B. Increased dATP decreases the concentration of all rNTPs, blocking RNA synthesis.C. Increased dATP inhibits ribonucleotide reductase, such that de novo production of all dNDPs is inhibited.D. Adenosine deaminase loss causes SCID because T cells are particularly sensitive to DNA replication inhibition.E. It is not clear why adenosine deaminase loss affects T cells so specifically since other cells have the same kind of nucleotide synthesis regulation and might be expected to be equally affected.
Show answer
Correct Answer: B
Feedback A: This is true, failure to degrade adenosine forms causes their concentrations to rise.
Feedback B: Lack of dNTPs blocks DNA synthesis, not RNA synthesis
Feedback C: This is true, dATP levels rise (see A) which shuts down all RNR activities
Feedback D: This is true; the reason is not fully known but it is observed to be the case.
Feedback E: This is true, see D.
34. In the oxidation of NADH by the electron transport system, energy released is coupled to the formation of ATP from ADP plus inorganic phosphate. For every mole of NADH oxidized, how many moles of ATP formed are formed?A. 1B. 5C. 3D. 6E. None of the above
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: This is for NADH in the mitochondria. NADH outside of the mitochondria produce different numbers of ATP depending on the shuttle system used to transfer the reducing potential from the cytosol to the mitochondria.
Feedback D:
Feedback E:
35. In analysis of isoenzymes of serum CPK for diagnosis of an AMIA. CPK above six times normal total CPK indicates an AMI.B. CPK-MM above four times normal total CPK indicates an AMI.C. CPK-BB above 1% of total normal, and total CPK of six times normal CPK indicates an AMI.D. CPK-MB above 3% of total normal CPK indicates an AMI.E. Total CPK above six times normal, with CPK BB in the normal range indicates an AMI.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: CPK-MB is only 1-3% in other muscle, but 15-20% in heart muscle.
Feedback E:
36. Which of the following is true about the structure and properties of ATP?A. The structure contains 3 phosphoanhydride bonds and one glycosidic bond.B. At physiological pH, the molecule has a net charge of approximately -6.C. The transfer of a phosphoryl group (phosphate residue) is the only biological reaction of ATP that cleaves a high-energy bond.D. The high free energy of hydrolysis results, in part, from electrical repulsion between phosphate residues.E. The standard free energy of hydrolysis of ATP is highter than that of other all phosphate derivatives in the glycolytic pathway.
Show answer
Correct Answer: D
Feedback A: 2 phosphoanhydride bonds and one glycosidic bond.
Feedback B:
Feedback C:
Feedback D:
Feedback E: No, hydrolysis of PEP gives more energy than hydrolysis of ATP.
37. Which of the following species of denatured DNA will renature most rapidly in solution, under appropriate conditions of ionic strength, pH, and temperature?A. Human liver nuclear DNAB. Vaccinia (viral) DNAC. E. coli DNAD. Yeast nuclear DNAE. Mouse neuronal DNA
Show answer
Correct Answer: B
Feedback A: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback B: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback C: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback D: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the
slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback E: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
38. When oxidative phosphorylation in mitochondria is uncoupledA. AMP is formed and O2 consumption stops.B. the oxidation of acetyl CoA in the tricarboxylic acid cycle stops.C. mitochondrial metabolism stops.D. ATP synthesis stops but O2 consumption continues.E. the cytochromes lose their heme groups into the cytoplasm.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Electron transport continues to reduce O2, but the proton gradient is not maintained so ATP is not synthesized.
Feedback E:
39. Oxaloacetate moves through the mitochondrial membraneA. after conversion to glycerol phosphate.B. by passive diffusion.C. after reaction with carnitine.D. after oxidation to pyruvate.E. after reduction to malate.
Show answer
Correct Answer: E
Feedback A: DHAP + NADH ---> glycerol-3-phosphate + NAD in the glycerolphosphate shuttle.
Feedback B:
Feedback C: Long-chain fatty acids are transported into the mitochondria for oxidation by carnitine.
Feedback D:
Feedback E: The Malate-Aspartate shuttle for NADH.
40. The absence of which enzyme results in PKU (phenylketonuria)?A. Tyrosine hydroxylaseB. Amino acid oxidaseC. Phenylalanine hydroxylaseD. Tyrosine transaminaseE. Phenylalanine transaminase
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Phenylalanine is not hydroxylated to tyrosine, and builds up to toxic levels.
Feedback D:
Feedback E:
41. All of the following statements are consistent with the chemiosmotic hypothesis of oxidative phosphorylation EXCEPT which one?A. Agents which uncouple oxidative phosphorylation from electron transport disrupt permeability barriers such that mitochondrial membranes become permeable to protons.B. The term electrochemical gradient or proton gradient operationally describes the free energy generated by electron transport which drives phosphorylation of ADP.C. Protons are transferred into the mitochondria during electron transport establishing a pH gradient.D. Intact mitochondrial membranes are necessary for oxidative phosphorylation to occur.E. Electron carriers of the electron transport chain are vectorially oriented.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Protons are transferred from the matrix to the intermembrane space of the mitochondria.
Feedback D: The proton gradient must be set up across the inner membrane.
Feedback E:
42. Which statement about the pyruvate dehydrogenase complex is true?A. Lipoamide functions in the decarboxylation of pyruvate.B. Coenzyme A is covalently bound to a lysine residue in one of the proteins of the complex.C. Phosphorylation catalyzed by a kinase results in decreased activity.D. NADH is an allosteric activator of pyruvate oxidation.E. The function of FAD is to oxidize NADH.
Show answer
Correct Answer: C
Feedback A: TPP functions in decarboxylation; lipoamide transfers the acetyl group to CoA.
Feedback B: Lipoamide is bound to lysine.
Feedback C: Increased acetyl CoA/CoA or NADH/NAD ratios activate the kinase which phosphorylates and deactivates PDH.
Feedback D: Increased NADH causes indirect inhibition of PDH.
Feedback E: FADH2 reduces NAD to NADH.
43. Which of the following is the single most important factor accounting for the high rates of enzyme catalysis?A. Enzymes bring the substrate [S] into close proximity of the coenzyme.B. Certain enzyme amino acid side chains interact with the substrate and stabilizing reactive intermediate(s).C. The "free energy" of activation is increased during enzyme catalysis.
D. Enzymes contain prosthetic groups which promote catalysis.E. Enzymes are globular proteins which contain hydrophobic and hydrophilic amino acids.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Enzymes stablaize the transition state.
Feedback C: Acitvation energy is lowered by enzymes.
Feedback D: True, but not the best explanation.
Feedback E: Does not explain catalysis.
44. Oxidative phosphorylation in mitochondria is carried out by enzymesA. on the outer membrane.B. on the inner membrane.C. in the matrix.D. in the intermembrane space.E. all of the above are correct..
Show answer
Correct Answer: B
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
45. What change from normal is likely to result from the ingestion of ethanol?A. increased reduction of pyruvateB. increased conversion of beta-hydroxybutyrate to acetoacetateC. decreased ketogenesis due to lack of acetyl-CoAD. increased production of oxaloacetate from malateE. increased production of dihydroxyacetone phosphate from glycerol phosphate
Show answer
Correct Answer: A
Feedback A: Ethanol metabolism requires reduction of NAD to NADH , and as in anaerobic glycolysis, NAD is regenerated by reduction of pyruvate to lactate.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
46. GlucagonA. is released from the pancreas in response to lowered blood glucose.
B. inactivates liver phosphorylase.C. increases glucose release from muscle glycogen.D. utilizes cAMP as a second messenger in muscle cells.E. increases glucose uptake into hepatocytes.
Show answer
Correct Answer: A
Feedback A:
Feedback B: Activates the phosphorylase for glycogen degradation.
Feedback C: Glucagon does not affect muscle.
Feedback D: See C.
Feedback E: Insulin increased glucose uptake when blood glucose is high.
47. Muscle can carry out all of the following processes EXCEPTA. transamination of branched chain amino acids.B. synthesis of glycogen.C. synthesis of glucose from alanine.D. oxidation of ketone bodies.E. phosphorylation of creatine.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Muscle cannot carry out gluconeogenesis because pyruvate caboxylase is found only in liver and kidney cells.
Feedback D:
Feedback E:
48. All of the following are likely to occur following the ingestion of a very large amount of carbohydrate EXCEPTA. the synthesis of triacylglycerol in the liver.B. the synthesis of glycogen in the liver.C. the oxidation of glucose by the pentose phosphate pathway in liver.D. lipolysis in adipose tissue.E. the synthesis and release of VLDL from the liver.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: High glucose levels lead to high insulin which supports synthesis of storage molecules, not breakdown.
Feedback E:
49. Catalysis by chymotrypsin depends on the concerted participation of three amino acid side chains (catalytic triad) which are contributed by which of the following amino acids?A. Arginine, histidine, and serineB. Histidine, serine, and glutamateC. Serine, glutamate, and aspartateD. Aspartate, serine, and histidineE. Serine, histidine, and arginine
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Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Memorize it. Know the part each plays in the mechanism.
Feedback E:
50. Which of the following is the most accurate description of phosphofructokinase-1?A. This enzyme uses fructose-6-phosphate as a substrate and converts it to fructose-2,6- diphosphate.B. This enzyme is inhibited by ATP, citrate, and fructose-2,6-biphosphate.C. This enzyme catalyzes a fully reversible reaction under physiological conditions.D. This activity of this enzyme is indirectly increased by cyclic AMP.E. No statement above is accurate.
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Correct Answer: E
Feedback A: PFK-2 makes fructose-2,6-bisphosphate, PFK-1 produces fructose-1,6-bisphosphate.
Feedback B: Inhibited by ATP and citrate, stimulated by fructose-2,6-bisphosphate and AMP.
Feedback C: The committed step of glycolysis, irreversible.
Feedback D: Indirectly decreased because increased cAMP causes PFK-2/FBP-2 complex to act as a phosphatase.
Feedback E:
51. All of the following amino acids are required as precursors for collagen (procollagen) synthesis EXCEPTA. lysine.B. glycine.C. proline.D. hydroxyproline.E. phenylalanine.
Show answer
Correct Answer: D
Feedback A: Commonly hydroxylated to hydroxylysine in collagen.
Feedback B: Found every third residue.
Feedback C: Commonly hydroxylated to hydroxyproline.
Feedback D: Proline is the precursor which undergoes posttranslational modification to hydroxyproline.
Feedback E:
52. Which of the following statements is an accurate comparison between chymotrypsinogen and chymotrypsin?A. Chymotrypsinogen has a single amino terminus and is inactive.B. Chymotrypsinogen is a storage form of chymotrypsin and is active.C. Chymotrysinogen and chymotrypsin are readily interconvertable forms of this hydrolase.D. More than one of the above is an accurate statement.E. None of the above is an accurate statement.
Show answer
Correct Answer: A
Feedback A: Chymotrypsinogen is a single polypeptide chain that is cleaved to give chymotrypsin (3 chains that are linked by 2 disulfide bonds),
Feedback B: Chymotrypsinogen is inactive.
Feedback C: No, cleavage of chymotrypsinogen to chymotrypsin is irreversible.
Feedback D:
Feedback E:
53. Reactions occurring during anaerobic GLYCOLYSIS in liver which require the input (use of) of ATP includeA. pyruvate kinase and glucokinaseB. pyruvate kinase and phosphofructokinaseC. glyceraldehyde-3-phosphate dehydrogenase and pyruvate kinaseD. glucokinase and phosphofructokinaseE. glucokinase and glyceraldehyde-3-phosphodehydrogenase
Show answer
Correct Answer: D
Feedback A: See C.
Feedback B: See C.
Feedback C: Glyceraldehyde-3-phosphate dehydrogenase produces NADH, pyruvate kinase produces ATP.
Feedback D:
Feedback E: See C.
54. The conversion of the two nitrogen atoms of glutamine to the two nitrogen atoms of urea requires the participation of urea cycle enzymes andA. glutaminase only.B. glutamate dehydrogenase only.C. glutaminase and aspartate aminotransferase (transaminase).D. glutaminase and glutamate dehydrogenase.E. glutamate dehydrogenase and aspartate aminotransferase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: One urea nitrogen comes from glutamine and enters the urea cycle as carbamoyl phosphate, the second nitrogen enters the cycle as aspartate.
Feedback D:
Feedback E:
55. Which molecule can donate a one-carbon unit most directly to tetrahydrofolate?A. serineB. methionineC. glutamateD. histidineE. choline
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
56. "Induced fit" is observed during catalysis byA. adenylate cyclase.B. glycogen phosphorylase "b".C. chymotrypsin.D. carboxypeptidase A.E. cyclic AMP dependent protein kinase.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Binding of substrate alters the conformation of the enzyme.
Feedback E:
57. Overall totally anaerobic glycolysis provides less ATP than aerobic glycolysis becauseA. less or no NADH is recovered during anaerobic glycolysis.B. less NAD is available under anaerobic conditions, so glycolysis cannot continue.C. lactate produced anaerobically inhibits phosphofructokinase-1.D. lactate is not a substrate for pyruvate dehydrogenase.E. more than one of the above is correct.
Show answer
Correct Answer: A
Feedback A: The NADH is oxidized to NAD+ by the conversion of pyruvate to lactate, so that the NAD+ can be used for another round of glycolysis. The NADH is thus not available for oxidative phosphorylation.
Feedback B: See A.
Feedback C: Citrate and ATP inhibit PFK-1.
Feedback D: True, but it is the NADH that is important because it and FADH2 transfer reducing potential to electron transport. The TCA cycle only makes one GTP without electron transport.
Feedback E:
58. The protective mechanism preventing accumulation of ammonia in most tissues resides in which enzyme?A. aspartate aminotransferase (GOT)B. glutaminaseC. gamma-glutamyl transpeptidaseD. glutamine synthetaseE. argininosuccinate synthetase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Glutamine removes ammonia from tissues and transports it to the liver for further metabolism to urea.
Feedback E:
59. Changes in enzyme activities IN SITU often involve "allosteric effects." This regulation most commonly is seen through changes inA. Vmax.B. Km.C. enzyme concentration.D. phosphorylation of reactive serines.E. Vmax and Km.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Allosteric modifiers most often change an enzyme’s affinity for substrate; high Km indicates low affinity.
Feedback C:
Feedback D:
Feedback E:
60. Phosphofructokinase-2 was described as a "tandem" or bifunctional enzyme. This means that
A. this enzyme can catalyze more than one reaction.B. this enzyme hydrolyzes fructose-1,6-diphosphate and may also catalyze the formation of this same compound.C. this enzyme has "group" or "class" specificity.D. this enzyme catalyzes the formation of fructose-1,6-diphosphate.E. this enzyme catalyzed the hydrolysis of fructose-2,6-diphosphate to give fructose-2-phosphate.
Show answer
Correct Answer: A
Feedback A: It has two parts: the kinase that catalyzes phosphorylation and the phosphatase that catalyzes dephosphorylation.
Feedback B: Hydrolysis and phosphorylation that converts between fructose-2,6-bisphosphate and fructose-6-phosphate.
Feedback C:
Feedback D: See B.
Feedback E: See B.
61. An important source of reducing power for biosynthesis is the reaction catalyzed byA. malate dehydrogenase.B. pyruvate dehydrogenase.C. acyl-CoA dehydrogenase.D. glycerol-3-phosphate dehydrogenase.E. glucose-6-phosphate dehydrogenase.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Part of the HMP pathway that produces NADPH which is needed for biosynthesis of fatty acids, steroids, etc.
62. Argininosuccinate is an intermediate in the de novo biosynthesis ofA. heme.B. adenosine monophosphate.C. uridine monophosphate.D. creatine.E. urea.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: arininosuccinate ---> fumarate + arginine ---> urea + ornithine
63. What is the net yield of NADH when glucose 6-phosphate is converted to lactate by anaerobic glycolysis?A. 0B. 1C. 2D. 3E. 4
Show answer
Correct Answer: A
Feedback A: Glyceraldehyde-3-phosphate dehydrogenase produces NADH and lactate dehydrogenase oxidizes NADH to NAD+.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
64. An International Unit (IU) is that quantity of enzyme which can convert substrate to product at a rate ofA. 1 micromole/min.B. 1 micromole/sec.C. 1 millimole/min.D. 1 millimole/sec.E. 1 nanomole/sec.
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
65. You have just finished a meal containing 100 grams of carbohydrate, 50 grams of fat and 25 grams of protein. How many kilocalories did you consume?A. 350B. 650C. 950D. 1250E. 1550
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: 100g carb (4 Cal/g) + 50g fat (9 Cal/g) + 25g protein (4 Cal/g) = 950 Cal or kcal.
Feedback D:
Feedback E:
66. All of the following can increase the activity of muscle glycogen phosphorylase "b" (dependent) except forA. epinephrine.B. cyclic AMP.C. glycogen phosphorylase kinase.D. 5'-adenosine monophosphate (AMP).E. glucagon.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glucagon does not effect muscle.
67. The major rate-limiting step of glycolysis in liver cells isA. the conversion of glucose to glucose 6-phosphate.B. the conversion of glucose 6-phosphate to fructose 6-phosphate.C. the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate.D. the aldolase reaction.E. none of the above.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: The committed step catalyzed by PFK-1.
Feedback D:
Feedback E:
68. A new enzyme has been isolated and shown to have a native molecular weight of 240,00. In the presence of 8M urea the molecular weight of the enzyme was estimated to be 120,000. When beta- mercaptoethanol was added to the urea system, the molecular weight was found to be 60,000. Which of the following best describes the subunit or polypeptide structure of the enzyme?A. The enzyme consists of two subunits with no interchain disulfide bridges.B. The enzyme consists of four polypeptide chains with no interchain disulfide bridges.C. The enzyme consists of two polypeptide chains with disulfide bridges.D. The enzyme consists of four polypeptide chains interconnected by disulfide bridges.E. The enzyme consists of two pairs of polypeptide chains with each pair interconnected by a disulfide bridge.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Urea disrupts the noncovalent interactions between the 2 subunits and beta-mercaptoethanol disrupts the disulfide bonds in each subunit.
69. Assume that the STANDARD FREE ENERGY change for the reaction:ATP ---> ADP + Piis -7.0 Cal/mole. What is the best estimate of the "actual" free energy change when the following concentrations exist:[ATP = 1x10-4], [ADP = 1x10-5], and [Pi = 1x10-4]?Use R = 2.0 cal/mole K°, and T = 300 K°.A. -7.3 CalB. -10.0 CalC. -10.3 CalD. -12.3 CalE. -13.9 Cal
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: G = Go + 2.3RT log K G = -7000 cal + 2.3(2)(300) log(10-4 X 10-5/10-4) G = -13900 cal or -13.9 Cal
70. Which citric acid cycle intermediate helps to regulate the rate of glycolysis by directly influencing the activity of phosphofructokinase?A. CitrateB. SuccinateC. Alpha-ketoglutarateD. OxaloacetateE. Malate
Show answer
Correct Answer: A
Feedback A: Inhibits PFK-1.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
71. Which primers will amplify my favorite gene if it is in the context below? 5'-GGTACAGTC-MY FAVORITE GENE-CTAGATCAT-3'A. 5'-GACTGTACC and 5'-CTAGATCATB. 5'-GGTACAGTC and 5'-ATGATCTAGC. 5'-GGTACAGTC and 5'-CTAGATCATD. 5'-GACTGTACC and 5'-ATGATCTAGE. 5'-GGTACAGTC and 5'-GACTGTACC
Show answer
Correct Answer: B
Feedback A: Both primers point away from the target.
Feedback B: Yes, these anneal to the sequences flanking the target and are oriented with their 3’ ends towards the target.
Feedback C: The first primer is correct, but the second one points away from the target.
Feedback D: Both primers point the same way, with one pointing away from the target.
Feedback E: The first primer is correct, but the second one is nonsense, it is complementary to the second flank but parallel.
72. The carbon atom in urea synthesized in the liver arises most directly fromA. aspartate.B. CO2.C. fumarate.D. ornithine.E. glutamate.
Show answer
Correct Answer: B
Feedback B: The CO2 comes into the urea cycle through the synthesis of carbamoyl phosphate.
Feedback C:
Feedback D:
Feedback E:
73. Phosphocreatine is important in metabolism because itA. is a compound formed by oxidative phosphorylation which can transfer a phosphate group to ADP.B. serves in synthetic reactions as the source of the phosphate groups of CDP-choline.C. is formed by substrate-level phosphorylation and irreversibly transfers a phosphate group to ADP and forms creatinine.D. reacts with ADP to give ATP and creatine.E. is the source of a nitrogen and the carbon atoms of urea.
Show answer
Correct Answer: D
Feedback A: Phosphocreatine is substrate level phosporylation not oxidative phosphorylation.
Feedback B:
Feedback C: The reaction is reversible.
Feedback D: Phosphocreatine is a temporary storage form of ATP in muscle.
Feedback E:
74. DNA sequencing depends on generating a set of molecules that terminate at every possible position. This is done by stopping the synthesisA. using modified nucleotides lacking the 3' hydroxyl group.B. using a DNA polymerase purified from a thermophilic organism.C. using modified nucleotides phosphorylated at the 5' hydroxyl.D. using modified nucleotides with deaminated bases.E. using a non-processive DNA polymerase.
Show answer
Correct Answer: A
Feedback A: Yes, the 3’ hydroxyl is needed for elongation so removing it causes termination
Feedback B: This can be done, but does not contribute to the mechanism of chain termination
Feedback C: No, normal nucleotides have 5’ phosphates
Feedback D: No, this would just make damaged DNA, it would not terminate the elongation.
Feedback E: No, this would just make short products, it would not allow specific termination.
75. Each of the following may be used to estimate the molecular weight of a protein EXCEPTA. sucrose density gradient centrifugation.B. ion exchange chromatography.C. SDS (sodium dodecyl sulfate) gel electrophoresis.D. gel filtration chromatography.E. sedimentation and diffusion measurements.
Show answer
Correct Answer: B
Feedback B: Separates molecules based on charge differences.
Feedback C:
Feedback D:
Feedback E:
76. AB arises from a 770 amino acid membrane protein called APP (amyloid precursor protein). APP is mostly processed by alpha and gamma secretases to yield, among other fragments, a small peptide p3 that causes little harm. Enhanced processing by beta-secretase and gamma-secretase generates the 40 or 42 residue AB peptide that results in Alzheimer amyloid deposits. APP spans the lipid bilayer between residues 671 and 770, and the last few residues of AB are in the bilayer. From what you know about membrane proteins which one of the following sequences identifies the last 5 amino acids in the 42 residue AB peptide.A. Asp-Ala-Glu-Phe-ArgB. Gly-Val-Val-Ile-AlaC. His-Asp-Ser-Gly-TyrD. Lys-Val-His-His-GlnE. Glu-Asp-Val-Gly-Ser
Show answer
Correct Answer: B
Feedback B: Because the last few residues of the AB peptide are in the membrane they need to by hydrophobic (non-polar).
Feedback C:
Feedback D:
Feedback E:
77. The glycerol phosphate shuttleA. results in production of dihydroxyacetone phosphate in the matrix of mitochondria.B. transfers electrons from cytosolic NADH to the respiratory chain of mitochondria.C. inhibits glycolysis.D. supports phosphorylation of three moles of ADP per mole of cytosolic NADH oxidized.E. uses 3-phosphoglyceric acid as a substrate.
Show answer
Correct Answer: B
Feedback A: DHAP is transferred out of the mitochondria.
Feedback B:
Feedback C:
Feedback D: Transfers NADH reducing potential into the mitochondria as FADH2, which supports the phosphorylation of 2ATP.
Feedback E:
78. When modifying the Southern blot procedure to do a Northern blot, you would want to avoid the step in whichA. the nucleic acid molecules are separated by electrophoresis in agarose.B. the gel is soaked in sodium hydroxide to denature the nucleic acids.C. the nucleic acids are transferred to a solid matrix by capillary action.D. a denatured probe is added to the solid matrix and incubated at 65° C.E. probe that has hybridized to the immobilized nucleic acids is detected.
Show answer
Correct Answer: B
Feedback A: No, this is necessary to gather size information
Feedback B: Yes, RNA is sensitive to hydrolysis in solutions with high pH values, so this would tend to destroy the RNA that is being tested in a Northern.
Feedback C: No, this is the same in Southerns and Northerns.
Feedback D: No, this is the same in Southerns and Northerns.
Feedback E: No, this is the same in Southerns and Northerns.
79. Urea is the product of an enzyme in liver which catalyzes the hydrolysis ofA. carbamoyl phosphate.B. citrulline.C. ornithine.D. argininosuccinate.E. arginine.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: carbamoyl phosphate + ornithine ---> citrulline + aspartate ---> argininosuccinate ---> arginine ---> urea
80. From what you know about protein chemistry, which peptide migrates fastest to the cathode (or negative charged electrode) at pH 6.0.A. Asp-Ala-Glu-Phe-ArgB. Gly-Val-Val-Ile-AlaC. His-Asp-Ser-Gly-TyrD. Lys-Val-His-His-GlnE. Glu-Asp-Val-Gly-Ser
Show answer
Correct Answer: D
Feedback A: -1, 0, -1, 0, +1 = -1
Feedback B: 0, 0, 0, 0, 0
Feedback C: +1, -1, 0, 0, 0 = 0
Feedback D: +1, 0, +1, +1, 0 = +3
Feedback E: -1, -1, 0, 0, 0 = -2 This peptide would migrate fastest toward the anode.
81. Which of the following is a substrate for RNA synthesis, but not DNA synthesis?A. adenosine triphosphate (ATP)B. deoxyuridine triphosphate (dUTP)C. guanosine-5', 5'-guanosine triphosphate (GpppG)D. cyclic adenosine monophosphate (cAMP)E. pseudouridine triphosphate (psiTP)
Show answer
Correct Answer: A
Feedback A: Yes, this a ribonucleoside triphosphate suitable for synthesis of RNA.
Feedback B: No, dNTPs are not found in RNA
Feedback C: No, this is not a substrate for an RNA polymerase.
Feedback D: No, this has no free hydroxyl to participate in the polymerization reaction.
Feedback E: No, this is not a standard NTP used by RNA polymerase.
82. Ammonium ion in the urine arises primarily fromA. the action of glutamate dehydrogenase in the kidney.B. the action of glutaminase in the kidney on the amide nitrogen of glutamine.C. the action of glutamine aminotransferase in the kidney.D. the action of glutaminase in the kidney on the alpha-amino group of glutamine.
E. the action of asparaginase in the kidney.
Show answer
Correct Answer: B
Feedback A:
Feedback B: In the proximal tubules glutamine is deaminated to glutamate and ammonia.
Feedback C:
Feedback D:
Feedback E:
83. The compound which accepts a two-carbon fragment from acetyl-CoA to initiate the citric acid cycle isA. oxalosuccinate.B. oxalate.C. oxaloacetate.D. oxytocin.E. oxybutyrate.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: acetyl CoA + oxaloacetate ---> citrate
Feedback D:
Feedback E:
84. Ketosis may develop as a result of which of the circumstances below?A. glucose synthesis from fatty acids during glucose deprivationB. acute attack of goutC. carnitine transferase I deficiencyD. hyperglycemiaE. fat mobilization to support gluconeogenesis
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Ketone bodies are produced because more acetyl CoA is derived from fatty acid oxidation than can be used in TCA cycle.
85. The alpha-amino groups on most mammalian proteins are acetylated. That is they have the structure: CH3-C(=O)-NH-C-peptide However, the alpha-amino groups on alpha- and beta- hemoglobin subunits are not acetylated. Their acetylation would likely:
A. prevent binding of protons to HbB. prevent BPG binding to HbC. Reduce CO2 elimination by the lungsD. A + CE. B + C
Show answer
Correct Answer: E
Feedback A: Protons bind to alpha-amino groups of the alpha chains and to carboxyl terminal His residues of the beta chains. The beta chains would still bind protons.
Feedback B: Yes, the terminal amino groups of the beta chains aid in bonding BPG.
Feedback C: Alpha-amino groups react with CO2: -NH2 + CO2 ---> NHCOO- + H+
Feedback D:
Feedback E:
86. A person with Xeroderma pigmentosumA. is missing the enzyme that reverses thymine dimersB. is sensitive to ultraviolet radiation (UV)C. is sensitive primarily to ionizing radiation (x-rays)D. is prone to cancers and should be treated prophylactically with radiation therapyE. is missing one of the 7 genes that encode type II topoisomerases
Show answer
Correct Answer: B
Feedback A: No, that is called photolyase, and has not been found in humans.
Feedback C: No, they are sensitive to UV
Feedback D: No, due to the defect in damage repair, radiation therapy is not indicated.
Feedback E: There are 7 classes of XP, but none are known to have a defect in a type II topoisomerase.
87. A person regularly eating a high protein diet (diet 1) changes to diet 2 which contains adequate but substantially less protein. After a period of several days on diet 2, what will occur?A. Nitrogen balance will be positive.B. Nitrogen balance will be negative.C. Daily output of urea will increase over that while on diet 1.D. Daily output of urea will decrease below that while on diet 1.E. Two of the above are true.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Consumed amino acids are used for protein replacement and excess amino acids are catabolized for energy. There is no real storage form of amino acids. Because diet 2 has less protein
intake, there is less excess protein to catabolize.
Feedback E:
88. The three stranded collagen superhelix is characterized by the following properties EXCEPT :A. Glycine is present every third residueB. H-bonds are parallel to the helix axisC. Some lysines and prolines are hydroxylatedD. Sugars can be attached to hydroxylated lysine residuesE. No exceptions
Show answer
Correct Answer: B
Feedback A:
Feedback B: The H-bonds occur btween the three strands and are more perpendicular to the axis.
Feedback C:
Feedback D:
Feedback E:
89. The DNA that is copied by the two replication forks traveling in opposite directions from a common site of initiation is called:A. a duplexB. an originC. a batteryD. a repliconE. a transposon
Show answer
Correct Answer: D
Feedback A: No, a duplex is the name for the two stranded form of DNA
Feedback B: No, the origin is the site of initiation itself.
Feedback C: No, a battery is a group of adjacent origins that fire simultaneously
Feedback D:
Feedback E: No, this is a piece of DNA capable of moving to other sites in the genome.
90. Lysosomal enzymes are often maximally active at pH 5.0 or less. In many cases, they are inactive at pH 7.0 which makes some biological sense in that lysosomal rupture would be lethal if the released proteases and nucleases were significantly active at cytosolic pH (~ 7.0). The amino acid R group most affected by a change in pH from 5 to 7 is:A. arginineB. aspartateC. cysteineD. histidineE. lysine
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Histidine’s imidazole ring readily accepts or donates protons. pKa = 6.5.
Feedback E:
91. The processivity of DNA polymerasesA. increases in the presence of ligasesB. decreases in the presence of helicasesC. increases in the presence of single strand binding proteinsD. increases in the presence of Okazaki factorsE. decreases in the presence of clamp proteins
Show answer
Correct Answer: C
Feedback A: No, ligases seal nicks in DNA, they do not influence the tendency to move forward or fall off.
Feedback B: Some helicases affect polymerase processivity, but they increase this property.
Feedback C: Yes, SSBs straighten the template and increase both the rate and processivity of DNA polymerases
Feedback D: No, there is no such thing as an Okazaki factor, and the size of Okazaki fragments are the result, rather than the cause, of processivity changes.
Feedback E: Clamp proteins increase the processivity of DNA polymerases.
92. McArdle's disease is due to a muscleA. debrancher enzyme defect.B. glucosidase defect.C. phosphofructokinase defect.D. phosphorylase absence.E. phosphohexose isomerase deficiency.
Show answer
Correct Answer: D
Feedback A: Cori’s Disease of Andersen’s Disease
Feedback B: Pompe’s Disease
Feedback C: Glycogen storage disease VII
Feedback D: Cannot degrade glycogen; limited ability to perform strenuous exercise.
Feedback E:
93. The following secondary structures can be classified according to the length of, say ten residues, in the specific conformation. From longest to shortest they are:A. beta strand, collagen strand, alpha-helix, pi-helixB. Collagen strand, beta strand, alpha-helix, pi-helixC. alpha-helix, beta strand, pi-helix, collagen strandD. alpha-helix, pi-helix, beta strand, collagen strandE. Collagen strand, alpha-helix, beta strand, pi-helix
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Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
94. A bacterial culture grown for many generations in a "heavy" (15N)medium was transferred to a "light" (15N) medium. After the DNA had been copied 3 times in light medium, what would be the relative distribution of DNA containing two light strands (LL), one heavy and one light strand (HL), and two heavy strands (HH)?A. 2 HH:2 LLB. 1 HH:1HL:6 LLC. 0 HH:1 HL: 7LLD. 0 HH:1 HL: 3LLE. 1 HH:2 HL: 1LL
Show answer
Correct Answer: D
Feedback A: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
Feedback B: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
Feedback C: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. This ratio would result after 4 rounds
Feedback D: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL.
Feedback E: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
95. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to alter the absorbance of a protein at 280 nm?A. 2B. 3C. 4D. 5E. 6
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glycine is unique in being optically inactive, and tryptophan is aromatic.
96. In methylation-directed mismatch repair in bacteriaA. MutL binds to mismatches in duplex DNAB. The methylated strand is judged to be the older strandC. MutL binds to methylated DNA strandsD. One mismatched base is excised leaving a single nucleotide gap to repairE. RecA makes an incision only at hemimethylated sites
Show answer
Correct Answer: B
Feedback A: No, MutS has this activity, the activity is MutL remains poorly understood.
Feedback B: Yes, since newly synthesized DNA does not have this post-synthetic modification, the methylated strand is older.
Feedback C: No, MutH seems to have this property, the activity is MutL remains poorly understood.
Feedback D: No, this would be base-excision repair.
Feedback E: An incision is made at hemi-methylated sites, but by MutH, not RecA
97. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to disrupt the folding of a globular protein?A. 1B. 2C. 3D. 4E. 5
Show answer
Correct Answer: C
Feedback A: Both are aromatic.
Feedback B: Only one group is different and both are polar.
Feedback C: Ile is non-polar, Arg is very polar and its R group is approximately twice as long as Ile.
Feedback D: Both have long R groups and both are bases.
Feedback E: Both are polar.
98. All of the following statements concerning eukaryotic DNA replication are true EXCEPTA. single-strand binding proteins coat some DNA intermediates in replicationB. replication of half of the DNA chains occurs in short discontinuous piecesC. helicases unwind the parental strands for the polymerasesD. a single type of DNA polymerase acts on both leading and lagging strandsE. ring-shaped clamp proteins promote high processivity of DNA polymerase
Show answer
Correct Answer: D
Feedback A: SSBs are needed for all phases of DNA metabolism, including replication.
Feedback B: This is true, describing the synthesis of Okazaki fragments on the lagging strand.
Feedback C: While the precise identity of the eukaryotic replicative helicase is not known, this is very
likely to be true given the rate of replication fork movement.
Feedback D: This is unlikely to be true since both Pol alpha and Pol delta seem to be needed for eukaryotic replication, although their exact roles remain unknown.
Feedback E: This is true, the clamp is called PCNA.
99. Substrate Km #1 2.5 #2 12.0 #3 32.0 Comparison of the respective Km values for the hydrolysis of the peptide bond of substrates 1-3 by chymotrypsin shows thatA. The Vmax of substrate #1 must be greater than that for #2 or #3.B. The Vmax for substrate #3 must be greater than that for #1 or #2.C. A lower concentration of #3 will be needed to attain Vmax.D. A lower concentration of #1 will be needed to attain Vmax.E. Substrate #3 has the highest affinity for the active stie of chymotrypsin.
Show answer
Correct Answer: D
Feedback A: The Km tells us the concentration of substrate needed to reach 1/2 Vmax for a particular enzyme, but provides no comparison for the actual velocities.
Feedback B: See A.
Feedback C: Since Km, which tells us the concentration needed to reach 1/2 Vmax, is highest for #3, #3 must need the highest concentration to reach full Vmax.
Feedback D: See C.
Feedback E: Higher Km indicates less affinity of the enzyme for the substrate.
100. Ionizing radiation is capable of producing genetic mutations byA. Altering bases so that they form different base pair interactionsB. Causing deletions of DNA sequencesC. Causing inversions of DNA sequencesD. Causing translocations of one part of a chromosome onto anotherE. A-D are all correct.
Show answer
Correct Answer: E
Feedback A: True, but so are all of the choices, so the answer is E
Feedback B: True, but so are all of the choices, so the answer is E
Feedback C: True, but so are all of the choices, so the answer is E
Feedback D: True, but so are all of the choices, so the answer is E
Feedback E:
101. Which of the following cases represents a metabolically important control?A. Substrate oxidation in mitochondria is enhanced by elevated ATP levels.B. The activity of liver hexokinase is stimulated by high levels of its product, glucose 6-phosphate.C. Acetyl-CoA inhibits pyruvate carboxylase in the liver.D. Phosphofructokinase (PFK) is inhibited by high levels of its substrate, MgATP.E. Glucagon stimulates hepatic glycogen synthase.
Show answer
Correct Answer: D
Feedback A: Increased ATP shuts down enzymes of glycolysis and TCA cycle, so oxidation is decreased.
Feedback B: Hexokinase is inhibited by glucose-6-phosphate.
Feedback C: The activity of pyruvate carboxylase requires acetyl CoA.
Feedback D: ATP and citrate inhibit PFK-1. Kinases require divalent metal ions such as Mg2+ or Mn2+ for activity. The divalent ions complex with ATP.
Feedback E: Glucagon stops glycogen synthesis and promotes glycogen degradation.
102. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to inactivate an enzyme without greatly altering its conformation?A. 2B. 3C. 4D. 5E. 6
Show answer
Correct Answer: A
Feedback A: Asp and Asn are structurally very similar, but Asp is an acid, and usually charged at physiological pH.
Feedback B: Would greatly affect conformation.
Feedback C: Both are basic and have fairly large R groups, so there would likely be little affect.
Feedback D: Less likely than answer A (2), but this was accepted as correct when this exam was administered. His is a base and Asn is not, and they are structurally quite different.
Feedback E: Structrually very different, would affect conformation.
103. Cells lacking p53 are abnormal because theyA. lack enzymes needed for base excision repairB. fail to enter S phase if their DNA is damagedC. arrest in mitosis in the presence of high levels of caffeineD. fail to die after exposure to low levels of DNA damageE. A and C are both correct
Show answer
Correct Answer: D
Feedback A: No, BER is normal in p53 mutants
Feedback B: No, this would be the normal response, not an abnormal one.
Feedback C: No, this would describe an overactive checkpoint, not the underactive one found in p53 mutants
Feedback D: Yes, cells with DNA damage undergo apoptosis, but this response requires p53.
Feedback E: No, neither A nor C are correct.
104. The following are general properties of fibrous proteins EXCEPT:A. They serve structural rolesB. They involve extended conformations of polypeptide chainsC. They are located outside cells
D. They are often composed of intertwined polypeptide chainsE. No exceptions
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Proteins make up cellular components; fibrous proteins make up cytoskeletal elements.
Feedback D:
Feedback E:
105. Telomeres get shorter in somatic tissues as the individual ages, thereforeA. mice have long telomeres because they have a short life spanB. aging correlates negatively with telomere lengthC. telomerase must be present for cells to ageD. blocking telomerase would stop cells from agingE. enhancing telomerase would stop cells from aging
Show answer
Correct Answer: B
Feedback A: No, this does not follow from the observation, and is in fact the opposite of what would be expected.
Feedback B: Yes, this is just a restatement of the observation.
Feedback C: No, this is not a logical conclusion to the observation
Feedback D: No, this is not a logical conclusion, and in fact is opposite of the expectation.
Feedback E: No, this might be true, but is not necessarily so, and is unlikely to be correct in any complete sense.
106. All of the following statements apply to the electron transport chain EXCEPT which one?A. The cytochromes of the system are embedded in the inner mitochondrial membrane.B. An uncoupler stops the flow of electrons down the respiratory chain.C. A maximum of 3 ATP is generated by the reoxidation of one NADH through oxidative phosphorylation.D. Cytochrome oxidase is a key component of this system.E. The rate of electron transport is normally controlled by the concentration of mitochondrial ADP.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Electron transport still frunctions and protons are pumped to the intermembrane space, but the uncoupler allows the protons to bypass the ATP synthase.
Feedback C:
Feedback D:
Feedback E: Respiratory control.
107. Given this information, what technique(s) would be least useful for determining whether an individual's huntingtin proteins contained extensions of 60 additional glutamines.A. Ion exchange chromatographyB. Native gel electrophoresisC. SDS-PAGED. Amino acid analysisE. A and B
Show answer
Correct Answer: E
Feedback A: Separates based on charge, but 60 glutamines do not alter charge.
Feedback B: Proteins are separated based on shape and charges, not by size differences.
Feedback C: SDS creates uniform charge density and allows separtation based on size difference.
Feedback D: Would show a higher percentage of glutamine, and the complete sequence could be determined.
Feedback E:
108. Bacteria use visible light to reverse pyrimidine dimer damage. Which of the following is an accurate statement regarding this process?A. A pyrimidine dimer absorbs a photon of the same wavelength that caused the damage and reverses the reaction.B. An enzyme called photolyase makes two incisions in the DNA near a pyrimidine dimer, releasing the damaged strandC. During repair, the damaged bases are removed and replaced with newly synthesized materialD. Photolyase removes the damaged bases only, leaving abasic sites that are repaired by other enzymes later.E. Photolyase absorbs a photon of visible light and converts the energy to specifically reverse the pyrimidine dimerization reaction
Show answer
Correct Answer: E
Feedback A: No, the enzyme absorbs the photon, and it is always a different (longer, visible) wavelength.
Feedback B: No, this describes nucleotide excision repair, not reversal.
Feedback C: No, reversal involves repair without resynthesis
Feedback D: No, the bases are not removed, they are chemically returned to their original form
Feedback E:
109. The chemiosmotic hypothesis suggests that the potential energy of the electrons moving down the mitochondrial electron transport chain from a negative to a positive oxidation potential is initially conserved in the form of aA. proton gradient across the membrane.B. a high energy phosphate bond.C. a different conformational form of the electron carriers.D. a protonated form of coenzyme Q.E. a reduced non-heme iron protein.
Show answer
Correct Answer: A
Feedback A: The chemiosmotic hypothesis explains the link between electron transport and ATP synthesis, which is the proton gradient.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
110. Which technique would be the best for detecting the additional glutamines?A. Ion exchange chromatographyB. Native gel electrophoresisC. SDS-PAGED. Amino acid analysisE. A and B
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C: SDS-PAGE would tell us that the abnormal protein is larger, but not what amino acids were added.
Feedback D: With techniques such as Edman Degradation we can break off and analyze one amino acid at a time.
Feedback E:
111. Mutations caused by one of the following agents almost always inactivate genes specifying protein products, but often do not affect the function of genes specifying only stable RNA products. Which agent is it?A. UV lightB. cis-platinC. intercalating agentsD. "mustard" gasesE. 5-bromouracil
Show answer
Correct Answer: C
Feedback A: No, UV light typically causes single base changes, which are as bad for proteins as for RNAs
Feedback B: No, this is typically a crosslinker leading to double-strand breaks.
Feedback C: Yes, these cause insertions and deletions leading to frame-shifts that are much worse for ORFs than for RNAs
Feedback D: No, these are typically a crosslinkers leading to double-strand breaks.
Feedback E: No, this causes an enforced tautomeric shift leading to single base changes.
112. When epinephrine binds to its receptor on skeletal muscle, glycogenolysis is stimulated. The hormone signal influences the activity of each of these enzymes EXCEPT:A. phosphorylase.B. adenylate cyclase.C. phosphorylase kinase a.
D. cyclic AMP-dependent protein kinase.E. debranching enzyme.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: ephinephrine binds ---> active adenylate cyclase ---> increased cAMP ---> active PKA ---> active phosporylase kinase ---> active glycogen phosphorylase
113. Although theoretically possible based on acceptable angles, the pi-helix has not been observed in real proteins because:A. Carboxyl-amide H-bonds cannot form.B. R groups lie above one another.C. H-bonds are too long.D. Lack of favorable Van der Waals interactions in the helix axis.E. The pi-helix has been found!
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The alpha-helix has a radius that allows excellent van der Waals attractions across the axis.
Feedback E:
114. In which disorder is mismatch repair found to be defective?A. Ataxia telangiectasiaB. Cockayne's SyndromeC. Bloom's syndromeD. Hereditary non-polyposis colon cancer (HNPCC)E. Xeroderma pigmentosum (XP)
Show answer
Correct Answer: D
Feedback A: No, the ATM gene defective in this disorder appears to be a checkpoint gene.
Feedback B: This is associated with repair defects, but not mismatch repair.
Feedback C: This is associated with repair defects, but not mismatch repair.
Feedback D: Yes, these disorders are associated with mutations in homologs of the bacterial MutS and MutL genes.
Feedback E: This is associated with repair defects, but not mismatch repair.
115. Which statement about gluconeogenesis is correct?A. The acetate group of acetyl-CoA is used for the net synthesis of glucose.B. It occurs primarily in skeletal muscle.C. It occurs through reversal of the reactions of glycolysis.D. Lactate and alanine can both serve as substrates.E. ATP is not required.
Show answer
Correct Answer: D
Feedback A:
Feedback B: Primarily in the liver and some in the kidney.
Feedback C: The irreversible reactions of glycolysis must be reversed by enzymes unique to gluconeogenesis.
Feedback D:
Feedback E: 4 ATP are required for each glucose produced.
116. An enzyme affects the rate of a chemical reaction byA. decreasing the free energy of the reaction.B. increasing the free energy of the reaction.C. lowering the energy of activation of the reaction.D. raising the energy of activation of the reaction.E. displacing the equilibrium constant.
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Correct Answer: C
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
117. Concerning beta sheets the following statements are correct EXCEPT:A. The beta strands can be parallel or anti-parallel.B. The sheet can have an even or odd number of strands.C. The R groups of two consecutive amino acids are on opposite sides of the sheetD. Gly and Pro are not present.E. All are correct.
Show answer
Correct Answer: D
Feedback A: Parallel are usually found interiorly and antiparallel on the surface of globular proteins.
Feedback B:
Feedback C: R groups are staggered.
Feedback D: Turns are enriched with gly and pro, but they can be found in a sheet, especially gly.
Feedback E:
118. The inhibition caused by a non-competitive inhibitor causes which of the following alteration(s) in a Lineweaver-Burk plot?A. no change in the y-intercept, but a decrease in the x-interceptB. no change in the y-intercept, but an increase in the x-interceptC. no change in the y-intercept, but a decrease in the y-interceptD. no change in the x-intercept, but an increase in the y-interceptE. increases in both the x- and y-intercepts
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Non-competative inhibitors change Vmax, but not Km. The x-intercept = -1/Km and the y-intercept = 1/Vmax.
Feedback E:
119. What is the chemical difference between NAD and NADP?A. NADP has an additional phosphoryl group at position #2' on the pentose adjacent to adenine.B. NAD has a nicotinic acid moiety rather than nicotinamide.C. NADP has an additional phosphoryl group at position #3' on the pentose adjacent to nicotinamide.D. NADP has an additional phosphoryl grouping as an acyl phosphate on the nicotinic acid moiety.E. Chemically, they are identical.
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Correct Answer: A
Feedback A:
Feedback B: Both have nicotinamide.
Feedback C:
Feedback D:
Feedback E:
120. Gluconeogenesis from lactate does not require activity ofA. aldolase.B. phosphofructokinase.C. glyceraldehyde 3-phosphate dehydrogenase.D. triose phosphate isomerase.E. phosphoglycerate kinase.
Show answer
Correct Answer: B
Feedback A:
Feedback B: PFK-1 catalyzes an irreversible reaction in glycolysis; fructose-1,6-bisphosphatase is needed in gluconeogenesis.
Feedback C:
Feedback D:
Feedback E:
121. During gluconeogenesis, which enzyme must be present in addition to the glycolytic enzymes?A. fructose 1,6-bisphosphate phosphataseB. phosphofructokinaseC. pyruvate kinaseD. pyruvate dehydrogenaseE. glyceraldehyde 3-phosphate dehydrogenase
Show answer
Correct Answer: A
Feedback A: Carries out the reversal of the reaction catalyzed by PFK-1 in glycolysis.
Feedback B: Only used by glycolysis
Feedback C: Only used by glycolysis
Feedback D: Only used by glycolysis to TCA.
Feedback E: reversible.
122. In oxidative phosphorylation, an uncoupling agent causes which of the following?A. Both respiration and phosphorylation to decreaseB. Phosphorylation to remain constant and respiration to decreaseC. Respiration to increase and phosphorylation to decreaseD. Respiration and phosphorylation to increaseE. None of the above
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Electron transport is functioning but ATP synthesis is decreased. The decreased levels of ATP stimulate glycolysis and TCA cycle to produce more NADH and FADH2, which increase electron transport.
Feedback D:
Feedback E:
123. The NET products of anaerobic glycolysis areA. pyruvate, NAD, ATP.B. lactate, NAD, ATP.C. lactate, ATP.D. acetyl-CoA, NADH, ATP.E. pyruvate, ATP.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: In aerobic glycolysis pyruvate, ATP, and NADH are produced, but in anaerobic glycolysis pyruvate + NADH ---> lactate + NAD.
Feedback D:
Feedback E:
124. Glycolysis is only partially reversible because of energy barriers at the reactions catalyzed byA. hexokinase, triose phosphate isomerase, and pyruvate kinase.B. phosphofructokinase, aldolase, and lactate dehydrogenase.C. hexokinase, pyruvate dehydrogenase, and phosphoenolpyruvate carboxykinase.D. hexokinase, phosphofructokinase, and pyruvate kinase.E. hexokinase, glyceraldehyde 3-phosphate dehydrogenase, pyruvate kinase.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The 3 irreversible reactions of glycolysis which all involve ATP.
Feedback E:
125. The following properties are characteristic of all natural mammalian proteins EXCEPT forA. planar peptide bonds.B. defined amino acid sequence.C. origins on ribosomes.D. composed of L-amino acids.E. all are correct.
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Correct Answer: E
Feedback A: Double bond character.
Feedback B:
Feedback C: Ribosomes synthesize proteins.
Feedback D:
Feedback E:
126. An "allosteric effector" of an enzyme was defined asA. a compound which increased the catalytic rate by binding to a site removed from the active site.B. a compound which decreased the catalytic rate by binding to a site removed from the active site.C. a compound which decreased the Km for a substrate.D. a compound which changed the catalytic rate by binding to a site removed from the active site.E. a compound which increased Vmax for an enzyme.
Show answer
Correct Answer: D
Feedback A: The rate can be increased or decreased.
Feedback B: See A
Feedback C: Changes Km by increasing or decreasing it.
Feedback D:
Feedback E: Usually changes Km.
127. The cause of the most common form of galactosemia isA. loss of lactase activity in the intestinal mucosa.B. defective fructose 1-phosphate aldolase.C. inability to convert UDP-galactose to UDP-glucose.D. milk allergy.E. defective uridylyltransferase.
Show answer
Correct Answer: E
Feedback A: This can be a cause, but is not the most common. The major dietary source of galactose is lactose, which is broken down by lactase.
Feedback B: Causes fructose intolerance.
Feedback C: See E.
Feedback D: See A
Feedback E: Classical galactosemia; the following reaction does not occur: UDP-glucose + glactose-1-phosphate ---> UDP-galactose + glucose-1-phosphate.
128. Cellular DNA and RNA synthesis are distinguished by which of the following?A. release of the product polynucleotide from the templateB. use of a DNA templateC. selection of incoming nucleotides by Watson-Crick base pairing rulesD. release of pyrophosphate in the chemical step of synthesisE. polarity of synthesis of the new strand
Show answer
Correct Answer: A
Feedback A: In DNA synthesis the template and the newly synthesized strand become dsDNA. RNA is released from the DNA template.
Feedback B: Both RNA synthesis and DNA synthesis use a DNA template
Feedback C: While the nucleotides are different (T in DNA, U in RNA), the next correct nucleotide is chosen by base-pairing in both cases.
Feedback D: Both RNA synthesis and DNA synthesis use nucleoside triphosphates and release pyrophosphate
Feedback E: Both DNA and RNA are synthesized 5’ to 3’.
129. The most common male human germ-line mutation isA. A to T transversionsB. A to G transitionsC. C to A transversionsD. C to T transitionsE. G to C transversions
Show answer
Correct Answer: D
Feedback A: The most common mutation in the male germ line is the C to T transition caused by deamination of 5-methyl cytosine.
Feedback B: The most common mutation in the male germ line is the C to T transition caused by deamination of 5-methyl cytosine.
Feedback C: The most common mutation in the male germ line is the C to T transition caused by deamination of 5-methyl cytosine.
Feedback D: The most common mutation in the male germ line is the C to T transition caused by deamination of 5-methyl cytosine.
Feedback E: The most common mutation in the male germ line is the C to T transition caused by deamination of 5-methyl cytosine.
130. Which of following is NOT true concerning B form nucleic acids?A. The DNA duplex is destabilized by repulsion of the phosphates in the backbones.B. There are about 10.5 bp per turn of the DNA helix.C. The DNA duplex is stabilized by base stacking interactions.D. Hydrogen bonds form between bases on the two strands of the duplex.E. RNA does not adopt the B form because of the 3' hydroxyl on the ribose group.
Show answer
Correct Answer: E
Feedback A: This is true, the charge repulsion is a significant force pushing the strands apart, and is the basis of stabilization by salts.
Feedback B: This is true for B-forms
Feedback C: This is true, and provides the main force stabilizing the duplex
Feedback D: This is true; base pairing is largely due to these cross-strand interactions
Feedback E: The 2’ hydroxyl is important for A form, so DNA cannot adopt this form, but RNA has a 3’ hydroxyl so this response is incorrect (making it the right answer!)
131. Which of the following properties of nucleotides does NOT affect the structure of nucleic acids?A. The bases are hydrophobic.B. The phosphates are negatively charged at neutral pH.C. The bases can each form multiple hydrogen bondsD. The bases absorb ultraviolet light at a wavelength near 260 nm.E. The sugar group is highly water soluble.
Show answer
Correct Answer: D
Feedback A: This provides the main driving force for formation of the duplex
Feedback B: The repulsion of the backbones is a significant force affecting the structure and stability of duplexes
Feedback C: The ability to form hydrogen bonds is an important determinant of the specificity of the interaction between strands (the complementarity).
Feedback D: This is true but does not affect the structure of the duplex
Feedback E: This provides solubility to nucleic acids
132. Acylovir is an analog of guanosine in which the ribose ring is disrupted. Which of the following reasons explains why this drug is useful for treating herpesvirus infections?A. Normal cells are not permeable to the drug, but infected cells are.B. Cellular DNA polymerase will not bind the analog, but viral polymerase will.C. Ribonucleotide reductase acts on acyclovir to form a suicide inhibitor of only the viral DNA polymerase.D. Cellular thymidine kinase will not act on the drug, but the viral enzyme will.E. Viral replication is faster than cellular replication, so the virus is more sensitive to any change in the level of dNTPs.
Show answer
Correct Answer: D
Feedback A: Both kinds of cell are permeable to this drug.
Feedback B: The polymerase doesn’t use acyclovir because it is not a triphosphate, not because it doesn’t bind.
Feedback C: RNR does not act on acyclovir; it does not have a sugar moiety so it cannot have a 2’ hydroxyl.
Feedback D:
Feedback E: No, this is just confusing.
133. Adding high levels of thymidine to mammalian cells blocks their progression through the cell cycle becauseA. these cells lack thymidine kinase.B. ribonucleotide reductase is inhibited by thymidine so no dTTP is produced.C. DNA polymerase is inhibited by thymidine.D. the levels of dCTP are reduced due to the regulation of ribonucleotide reductase.E. this blocks type II topoisomerases in a covalent intermediate with DNA.
Show answer
Correct Answer: D
Feedback A: No, this is just made up
Feedback B: No, but this is closer, see D
Feedback C: No, this is just made up
Feedback D:
Feedback E: No, this is just made up
134. Histones are usuallyA. basic, and therefore negatively charged at neutral pH.B. basic, and therefore positively charged at neutral pH.C. acidic, and therefore negatively charged at neutral pH.D. acidic, and therefore positively charged at neutral pH.E. close to neutral, and therefore uncharged at neutral pH.
Show answer
Correct Answer: B
Feedback A: Basic residues are positively charged at neutral pH
Feedback B:
Feedback C: Acidic residues are negatively charged at neutral pH, but histones are rich in basic residues
Feedback D: Acidic residues are negatively charged at neutral pH, and histones are rich in basic residues
Feedback E: If neutral, there would be little charge, but histones are rich in basic residues.
135. As a result of X chromosome inactivationA. Males do not express the genes on the X chromosome.B. One copy of the X chromosome is inactivated no matter how many copies are present.C. Expression of the genes on the Y chromosome is increased.D. A heterochromatic Barr body forms and this structure is propagated in subsequent rounds of replication.E. The same X chromosome is inactivated in all tissues, revealing recessive "sex-linked" mutations.
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Correct Answer: D
Feedback A: Males have only one X chromosome, inactivation affects one of the two copies found in female cells
Feedback B: The opposite is true; only one copy is left active no matter how many are present
Feedback C: The Y chromosome has few genes and their expression is unaffected by X inactivation, even in cases where a Y chromosome is found in a cell with two X chromosomes where inactivation occurs
Feedback D:
Feedback E: Females are mosaics because different X chromosomes are inactivated in different cell lineages.
136. The processivity of leading strand DNA polymerasesA. increases with the addition of ligases.B. decreases with the addition of helicases.C. decreases with the addition of single strand binding proteins.D. increases with the addition of Okazaki factors.E. increases with the addition of clamp proteins.
Show answer
Correct Answer: E
Feedback A: Ligase does not affect processivity, it seals nicks after replication or repair is otherwise complete
Feedback B: Helicases unwind the template and enhance processivity
Feedback C: SSBs straighten the template and therefore enhance processivity
Feedback D: There is no such thing.
Feedback E: Clamps prevent polymerases from leaving the primer/template and so increase processivity
137. Knots formed in DNA molecules are disentangled in eukaryotic cells byA. type I topoisomerases.B. type II topoisomerases.C. type III topoisomerases.D. restriction endonucleases and DNA ligase.E. DNA helicases and DNA ligase.
Show answer
Correct Answer: B
Feedback A: Type I topo only breaks one DNA strand and therefore cannot perform the strand passage reaction required to untie knots
Feedback B:
Feedback C: There is no such thing
Feedback D: Restriction enzymes are only found in prokaryotes, and they don’t perform this role their either.
Feedback E: This would not work; helicases just unwind they don’t break the DNA backbone
138. All of the following statements are true with regard to DNA replication EXCEPT which one?A. It involves a transient covalent link between RNA and DNA chains.B. Replication forks move away in BOTH directions from origins.C. Chain growth occurs by addition at the 3' end of new strands.D. Ribonucleoside triphosphates are not used during replication.E. Chain growth on one side of the replication fork does not require continual action of primase.
Show answer
Correct Answer: D
Feedback A: This is true; the primers are RNA and are extended as DNA, producing the transient link.
Feedback B: This is true, origins give rise to two forks that travel apart.
Feedback C: This is true, nucleic acids are synthesized 5’-3’.
Feedback D: Yes, they are used in the synthesis of RNA primers, and also as an energy source for clamp loading and helicase action.
Feedback E: This is true, the leading strand is synthesized continuously and does not require repeated priming (unlike the lagging strand).
139. Since the genetic code is redundant, an important gene that encodes a protein will have itsA. DNA sequence accumulate changes more slowly than its amino acid sequence.B. DNA sequence accumulate changes more rapidly than its amino acid sequence.C. DNA sequence accumulate changes at the same rate as its amino acid sequence.D. mRNA accumulate changes more slowly than its amino acid sequence.E. exons accumulate changes more rapidly than introns.
Show answer
Correct Answer: B
Feedback A: Since many DNA sequences can encode the same protein, the DNA sequence is under less constraints and mutates more rapidly than the protein sequence it encodes.
Feedback B: Since many DNA sequences can encode the same protein, the DNA sequence is under less constraints and mutates more rapidly than the protein sequence it encodes.
Feedback C: Since many DNA sequences can encode the same protein, the DNA sequence is under less constraints and mutates more rapidly than the protein sequence it encodes.
Feedback D: mRNA sequence matches the DNA and is only a temporary copy, not a source of heritable change.
Feedback E: Exons are under greater selective pressure and therefore change more slowly than introns
140. Mutations in DNAA. are not important if they are in intron sequences.B. are not important unless they occur in open reading frames.C. always have a negative impact on the gene product affected.D. only affect single-copy genes.E. can change the translational reading frame of proteins.
Show answer
Correct Answer: E
Feedback A: This is often but not always true. In particular, mutations in introns that affect the ability for the introns to be spliced affect gene function.
Feedback B: ORFs are the most obvious target for mutations that affect function, but since flanking regions alter the expression of the ORF and the splicing patterns of messages, these sequences are also important targets.
Feedback C: Evolution is the process of the slow accumulation of beneficial mutations
Feedback D: Mutations occur at random and affect all genes. Phenotypes are more likely to be observed in single copy genes, but that wasn’t the question.
Feedback E: These are called frameshifts.
141. Ultraviolet light damages cells most seriously byA. causing double strand breaks in DNA.B. altering the secondary structure of RNA molecules.C. forming dimers between pyrimidines stacked adjacent to one another.D. crosslinking pyrimidines on opposing strands that are base paired to one another.E. accelerating the production of abasic sites.
Show answer
Correct Answer: C
Feedback A: ds DNA breaks are rare with UV light
Feedback B: RNA molecules are temporary copies of information and tend to be abundant, so RNA damage is not considered to be a serious problem
Feedback C: Yes, the most abundant class is T-T linkages or thymine dimers
Feedback D: Due to geometrical constraints, this is less likely than same strand dimerization.
Feedback E: Possible, but rare compared to C
142. During agarose gel electrophoresis, DNA moleculesA. all migrate at the same rate because they have the same charge to mass ratio.B. migrate according to size because they all have the same charge to mass ratio.C. migrate differently depending on nucleotide sequence because the different bases change the number of charges found in a sequence.D. migrate at rates independent of their shape.E. fail to migrate into the gel at all if they are larger than about 25,000 base pairs.
Show answer
Correct Answer: B
Feedback A: The regular repeating structure of DNA leads to a uniform charge to mass ratio, but the total mass is not the same
Feedback B: The regular repeating structure of DNA leads to a uniform charge to mass ratio, but the total mass is not the same, leading to the basis for separation
Feedback C: The regular repeating structure of DNA leads to a uniform charge to mass ratio, but the total mass is not the same
Feedback D: Two molecules of the same size but different shape will migrate differently in an agarose gel; see the 2-D gel method for an example.
Feedback E: They fail to separate well due to “snakingâ€, but they enter the gel.�
143. E. coli DNA is not degraded by its own restriction enzymes becauseA. the restriction enzymes are not located in the E. coli nucleus.B. there are not enough restriction enzyme molecules in the cell.C. DNA methylases have modified the restriction sites.D. the enzymes are made as inactive precursors before they are secreted.E. the E. coli genome does not contain DNA sequences which are recognized by its own restriction-modification system.
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Correct Answer: C
Feedback A: Bacteria have no nuclei, and restriction enzymes are active throughout the cell
Feedback B: Restriction enzymes are abundant, but are paired with modification enzymes; see C
Feedback C: yes, restriction enzymes are always paired with a modification enzyme with the same recognition site.
Feedback D: They are neither made as inactive precursors, nor are they secreted; they function intracellularly, see C.
Feedback E: See C, EcoRI recognizes a 6 bp site, which occurs over 1000 times in the E. coli genome.
144. Which of the following are examples of secondary structure? 1. the sequence of amino acids in a protein 2. the alpha-helix of myoglobin 3. the folding of alpha-helix onto beta-sheet of triose phosphate isomerase 4. the beta-sheet of an immunoglobulin 5. the subunit nature of hemoglobin 6. the un-ordered portions of follitropin 7. the beta-turn in any polypeptideA. all of theseB. only the even-numbered itemsC. only the odd-numbered itemsD. 2, 4 and 7E. 2, 4, 6, and 7
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Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: 1-primary, 3-tertiary, 5-quaternary, 6-primary or secondary.
Feedback E: Also scored as correct (along with D)
145. Cytochrome a, a3 (cytochrome oxidase)
A. contains copper which is required for the reduction of oxygen.B. is a terminal electron acceptor which forms hydrogen peroxide upon reduciton of oxygen.C. can be reduced directly by cytochrome b.D. is not inhibited by carbon monoxide.E. is inhibited directly by rotenone and amytal.
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Correct Answer: A
Feedback A:
Feedback B: O2 is the terminal electron acceptor and most often forms H2O.
Feedback C: Cytochrome c carries electrons from cytochrome b (part of cytochrome reductase) to cytochrome a/a3 (part of cytochrome oxidase).
Feedback D: Is inhibited by carbon monoxide, cyanide, and azide.
Feedback E: These drugs block electron transfer at NADH Q reductase.
146. Which statement about glucose metabolism is false?A. Phosphofructokinase is a key enzyme which can be regulated to control the metabolism of glucose.B. The pentose phosphate pathway must be operative in cells which synthesize significant amounts of fatty acids.C. A five-carbon compound required for the biosynthesis of RNA can be formed by the oxidation of glucose.D. In the cytoplasm, glucose can be metabolized to pyruvate.E. Glucose derived from glycogen in muscle serves as a source of blood glucose during fasting.
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Correct Answer: E
Feedback A: PFK-1 catalyzes the committed step of glycolysis.
Feedback B: HMP pathway provides the NADPH needed for fattty acid synthesis and ribose-5-phosphate for nucleotide synthesis.
Feedback C: The HMP pathway produces ribose-5-phosphate.
Feedback D: Glycolysis takes place in the cytoplasm.
Feedback E: This is false; muscle lacks glucose-6-phosphatase.
147. The substrate normally utilized as the major source of energy by the brain isA. triglycerides.B. fatty acids.C. ketone bodies.D. amino acids.E. glucose.
Show answer
Correct Answer: E
Feedback A:
Feedback B: Fatty acid degredation does not take place in the brain.
Feedback C: Some of the brain’s dependence on glucose shifts to ketone bodies during starvation, but some glucose is always required.
Feedback D:
Feedback E: The brain normally accounts for 60% of the glucose used by the body. The brain has an absolute requirement for some glucose even during starvation.
148. Consider a single-stranded DNA fragment 1000 bp long in a solution hybridization experiment. Which of the following parameters has the largest effect on the rate of reannealing of this fragment with its complementary sequence?A. The species from which the DNA was derivedB. The concentration of the complementary strand in the solutionC. The fraction of Gs and Cs in the sequence of the strandD. The amount of helicase added to the solutionE. The concentration of detergent in the solution
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Correct Answer: B
Feedback A: The RATE of reannealing is strictly determined by the probability of collisions between complementary strands, and therefore by the concentration of complementary strands. This is affected by the species, but is not diagnostic (while the concentration is). For example, two different species of bacteria will give similar reannealing rates because their genomes are similarly complex.
Feedback B: The RATE of reannealing is strictly determined by the probability of collisions between complementary strands, and therefore by the concentration of complementary strands. The concentration of complementary strands is determined by the complexity of the genome of the organism the DNA was extracted from.
Feedback C: The G+C content strongly affects the stability of duplex DNA, but not the RATE of reannealing.
Feedback D: Helicases are not added to reannealing experiments, and if they were would promote denaturation of all double-stranded DNAs
Feedback E: The stability of DNA duplexes is not strongly affected by detergents.
149. Which is NOT true of the side chain of cysteine?A. forms disulfide bondsB. pKa is near 8C. is positively charged at physiological pHD. is hydrophilicE. has a non-asymmetric beta carbon
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: False. Cysteine has no net charge close to pH=7, but the side chain can lose a proton at alkaline pH.
Feedback D: Cysteine is polar like water so it is relatively hydrophilic, but highly hydrophobic when cystine(forming a disulfide bond).
Feedback E: The beta carbon is not asymmetric because of its two bound hydrogens.
150. Which primers will lead to amplification of MY FAVORITE GENE in a PCR if one strand of the duplex template has the following sequence? 5'-GGTACCGATC-MY FAVORITE GENE-GCATGAGATC-3'A. 5'-GGTACCGATC and 5'-GCATGAGATC
B. 5'-GGTACCGATC and 5'-GATCTCATGCC. 5-GCATGAGATC and 5'-CGTACTCTAGD. 5'-GATCGGTACC-and 5'-CTAGAGTACGE. None of these primer pairs will amplify MY FAVORITE GENE
Show answer
Correct Answer: B
Feedback A: first primer is right, second primer points away from target (anneals to the same strand as the first primer)
Feedback B:
Feedback C: No, these both anneal to the right flank of the target (and to each other)
Feedback D: No, the first primer points away from the target (anneals to the left flank), and the second is nonsense: it is the reverse of the second site, but not its complement.
Feedback E: No, see B
151. Which is NOT true about protein folding?A. disulfide bond formation stabilizes a folded structureB. cooperative hydrogen bond formation is a driving forceC. protein folding is driven in large part by solvent entropyD. hydrogen bonding of protein side chains to water drives protein foldingE. van der Waals forces stabilize the folded state
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C: Hydrophobic residues disrupt hydrogen bonding between water molecules. Energetically it is more favorable for the hydrophobic groups to be folded to the inside of the protein.
Feedback D: Hydrophobic interaction with water is the major driving force for folding. See C.
Feedback E:
152. The sequence of the complementary DNA strand for DNA with the sequence 5'-ATCGTACCGTTA 3 is:A. 5'-TAGCATGGCAAT-3'B. 5'-ATCGTACCGTTA-3'C. 5'-TAACGGTACGAT-3'D. 5'- ATTGCCATGCTA-3'E. None of the above
Show answer
Correct Answer: C
Feedback A: This is complementary, but not antiparallel.
Feedback B: No, this is the same, not the antiparallel complement
Feedback D: This is reverse (antiparallel) but not complementary
Feedback E: No, see C
153. Deamination of 5-methylcytosine in DNAA. Does not alter base pairing and so has no effect on the information encodedB. Creates uracil in the DNA which must be removed by uracil-N-glycosylaseC. Can be repaired by dihydrofolate reductase (DHFR)D. Leads pnicipally to frameshift mutationsE. Forms thymine which leads to mismatches
Show answer
Correct Answer: E
Feedback A: No, this changes the base-pairing specificity (paired with G, now pairs with A)
Feedback B: No, deamination of C creates U, but 5-methyl U is T.
Feedback C: No, DHFR is a enzyme needed during synthesis of nucleotides.
Feedback D: No, this creates mismatches that lead principally to missense mutations
154. Glycine is conserved in nature for the following reasonA. lack of side chain for steric packingB. the f dihedral angle confined to -60 쳌C. hydrogen binding potential of the R = H side chainD. often found at positions where no side chain is permittedE. limiting amount of glycyl-tRNAs
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Correct Answer: D
Feedback A: See D
Feedback B:
Feedback C:
Feedback D: This includes packing and turns.
Feedback E:
155. Which of the following is NOT true regarding ribonucleotide reductase?A. RNR converts rNDPs to dNDPs by reducing the 2' carbon from -CHOH- to - CH 2B. RNR is subject to a complicated system of regulation in which the product dNTPs alter its substrate specificity.C. RNR can be inhibited by the free radical scavenger hydroxyurea.D. RNR's reaction mechanism makes use of a stable free radical of a tyrosine generated by an iron atom.E. RNR requires regeneration after each catalytic event through a cycle catalyzed by dihydrofolate reductase.
Show answer
Correct Answer: E
Feedback A: This is true
Feedback B: This is true
Feedback C: This is true
Feedback D: This is true
Feedback E: No, RNR does not use a folate derivative to maintain its oxidation state, so DHFR is not involved.
156. Which modification reaction does not occur at a polypeptide chain terminus?A. MyristylationB. ADP-ribosylationC. AcetylationD. Glycosylphosphatidylinositol additionE. amidation
Show answer
Correct Answer: B
Feedback A: Amino terminus
Feedback B: Attachment to a side-chain
Feedback C: Amino terminus
Feedback D: Carboxyl terminus
Feedback E: Carboxyl terminus
157. Acylovir is an analog of guanosine in which the ribose ring is disrupted. Which of the following reasons explains why this drug is useful for treating herpesvirus infections?A. Normal cells are not permeable to the drug, but infected cells are.B. Cellular DNA polymerase rejects this substrate, but the viral polymerase uses it.C. Ribonucleotide reductase acts on acyclovir to form a suicide inhibitor that only affects the viral DNA polymerase.D. Only the viral version of thymidine kinase will act on acyclovir, so only infected cells convert it to an active form.E. Viral replication is faster than cellular replication, so the virus is more sensitive to any change in the level of dNTPs.
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Correct Answer: D
Feedback A: Both kinds of cell are permeable to this drug.
Feedback B: The polymerase doesn’t use acyclovir because it is not a triphosphate, not because it is rejected.
Feedback C: RNR does not act on acyclovir; it does not have a sugar moiety so it cannot have a 2’ hydroxyl.
Feedback D:
Feedback E: No, this is just confusing.
158. Which is NOT true about the structure of an alpha helix?A. Amide hydrogen of every peptide bond is hydrogen bondedB. Side chain alternate between a surface location and buried within the helixC. The rise per residue in the helix is 1.5 angstroms.D. Prolines tend to be excluded from helicesE. There are 13 atoms in a hydrogen bonded loop
Show answer
Correct Answer: B
Feedback A: All amide hydrogens and carbonyl oxygens are hydrogen bonded.
Feedback B: Side-chains are staggered on the surface. There are no interior side-chains.
Feedback C:
Feedback D: Proline is rigid, which reduces structural flexibility, and it has no amide hydrogen for hydrogen bonding.
Feedback E:
159. A child is found to have a disorder indicating a mutation in a blood clotting factor gene. When the relevant gene from the child is compared to the same sequence from the parents, an insertion of 300 bp is found in one copy from the child that is not present in either copy from either parent. Which is the likeliest explanation?A. The child can not be the progeny of both of the supposed parents since he or she has a different configuration of genes from the parents.B. The child has probably suffered a translocation between two chromosomesC. This is the result of the transposition of a SINED. This is the result of the transposition of a LINEE. This is the result of the movement of an intron
Show answer
Correct Answer: C
Feedback A: This would be the first thing to check, but the conclusion is not warranted; see C
Feedback B: A translocation would not cause the observed change.
Feedback C: Yes, SINEs are about 300 bp long and are transposons capable of relocating.
Feedback D: LINEs are typically larger than this, usually about 6 kbp.
Feedback E: Introns rarely move since few are transposons, so C is more likely.
160. Polypeptides that traverse phospholipid bilayers are often helical for which reason?A. Helix formation minimizes solvent contactB. Helices have all amides and carbonyl groups hydrogen bondedC. Helices have side chains packed against one anotherD. Helix formation is the one conformation that minimizes the number of residues being in contact with the hydrophobic bilayer.E. Beta strands are rare in nature
Show answer
Correct Answer: B
Feedback A: See D
Feedback B: Thus they are more lipid soluble.
Feedback C: No, side-chains are staggered to reduce steric hindrance.
Feedback D: No, all side-chains are on the surface.
Feedback E: They are common.
161. Which secondary structural element is not exposed at the surface of a globular protein?A. alpha helixB. Parallel beta conformersC. Antiparallel beta confomers
D. Type I tight turnE. 3/10 helix
Show answer
Correct Answer: B
Feedback A: Alpha helices often have hydrophobic and hydrophilic sides.
Feedback B: Both sides of parallel beta sheets are hydrophobic, and thus interior structures.
Feedback C: Antiparallel beta sheets have a hydrophobic and hydrophilic sides.
Feedback D: Occur at the surface.
Feedback E:
162. Which of the statements is CORRECT regarding the composition of CORE nucleosome particles?A. They contain 5 different kinds of histonesB. They compact the length of the DNA 20-30 foldC. They contain 2 molecules each of H2A, H2B, H3, and H4D. They contain about 400 bp of DNAE. They wrap DNA around themselves about 3 times
Show answer
Correct Answer: C
Feedback A: Core histones contain only H2A, H2B, H3, and H4.
Feedback B: Nucleosomes compact the DNA 5-7 fold.
Feedback C:
Feedback D: A core nucleosome has about 146 bp of DNA and a 0-60 bp linker.
Feedback E: The DNA wraps just less than twice around the nucleosome.
163. Silk fibers formed by the silk worm exhibit the following features. 1. adjacent chains are hydrogen bonded 2. the fiber chains are largely extended 3. the silk protein shows a repeating motif of (Ser-Gly-Ala-Gly) Which of the following secondary structural element is likely to be he major type of element?A. alpha helixB. beta conformersC. reverse turnsD. omega loopsE. 3/10 helix
Show answer
Correct Answer: B
Feedback A: alpha helices hydrogen bond within the helix, not between chains.
Feedback B: The repeating motif rich in glycine is similar to a collagen triple helix, but that is not a choice. Hydrogen bonding between chains suggests a beta conformer.
Feedback C:
Feedback D:
Feedback E:
164. In human somatic tissues, telomeresA. Get longer with age, so inhibiting telomerase will make cells younger.B. Get shorter with age, so activating telomerase will make cells younger.C. Must be elongated to allow immortalization, so inhibiting telomerase will prevent tumors.D. Must be removed to allow immortalization, so inhiting telomerase will prevent tumors.E. Must be elongated to allow immortalization, so activating telomerase will prevent tumors.
Show answer
Correct Answer: C
Feedback A: They get shorter, and telomerase is not causally related to age.
Feedback B: Telomerase lengthens telomeres, but this does not mean the cells get younger.
Feedback C: “Will†is probably too strong, we are still in the “might†stage.� �
Feedback D: Immortalization requires a way to maintain telomeres, not to remove them.
Feedback E: Activating telomerase would cause lengthening of telomeres, which would aid tumor growth.
165. Which is true about the structure of collagen?A. Amide/carbonyl hydrogen bonding occurs between adjacent tropocollagen unitsB. Gly residues are located in the interior the helixC. Hydroxyproline residues are necessary for crosslinking of adjacent helicesD. The rise per residue of the collagen helix is smaller than that of the a helixE. Desmosine crosslinks stabilize fibrils
Show answer
Correct Answer: B
Feedback A: A bridge involving one hydroxypyridinium and two hydroxylysines froms between tropocollagen molecules.
Feedback B: Every 3rd residue faces the crowded interior and only glycine will fit.
Feedback C: See A
Feedback D: alpha helix rise per residue = 1.5 Collagen helix = 2,9 angstroms
Feedback E: Desmosine crosslinks are found in elastin, not collagen.
166. In error-prone (SOS) repair in bacteria,A. RecA binds to single stranded DNAB. RecA assists in the proteolysis of lexAC. RecA acts to assist single stranded DNA search for homologous sequencesD. Transcripts encoding enzymes required for repair are inducedE. All of the above are correct
Show answer
Correct Answer: E
Feedback A: True, but so are the other choices, so E is correct.
Feedback B: True, but so are the other choices, so E is correct.
Feedback C: True, but so are the other choices, so E is correct.
Feedback D: True, but so are the other choices, so E is correct.
Feedback E: True, but so are the other choices, so E is correct.
167. What type of crosslink does NOT stabilize procollagen helices?A. LysinonorleucineB. Disulfide bridgesC. Hydroxypridinium structureD. DesmosineE. Hydrogen bonding of certain side chains
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Found in elastin.
Feedback E:
168. Which statement is NOT TRUE regarding cell cycle checkpoints?A. Checkpoint proteins participate catalytically in the repair of DNA damageB. Checkpoints can be suppressed by chemicals like caffeine, allowing inappropriate passage through the cell cycle.C. Checkpoints delay progression of the cell cycle to allow time for repairs to occurD. Loss of checkpoints leads to genomic instabilityE. Checkpoints can distinguish accidental double-strand DNA breaks from telomeres
Show answer
Correct Answer: A
Feedback A: Checkpoints are defined to be agents that only ALLOW repair by delaying the cell cycle, they do not participate in the repair.
Feedback B: This is true
Feedback C: This is true
Feedback D: This is true, inappropriate passage through the cycle often leads to loss of chromosomes or rearrangement.
Feedback E: Yes they can, ds breaks trigger checkpoints but telomeres do not.
169. Which is true about the cooperative oxygen binding by hemoglobin?A. Oxygenated hemoglobin has the heme iron atom pushed out of the plane of the heme.B. Oxygen binding induces a series of salt links formed between the four chains.C. Oxygenation induces movement of the two alpha/beta subunit pairs.D. Oxygen binding alters the number of contacts between the two alpha subunits.E. Oyxgenation of the 4 subunits occurs in an precise order
Show answer
Correct Answer: C
Feedback A: False, In deoxyhemoglobin the iron is pulled out of the plane by a histidine. Binding of oxygen counteracts the histidine’s pull.
Feedback B: False, oxygen binding requires that salt links be broken.
Feedback C: .
Feedback D: There is always little contact between the two alpha subunits or the two beta subunits
Feedback E: No, binding of O2 to one heme affects the affinity of the other hemes, but there is no precise order of oxygenation.
170. Sequencing DNA molecules by dideoxy nucleoside triphosphate termination requiresA. a mixture of normal dNTPs and modified forms with a 2'triphosphateB. a mixture of normal dNTPs and modified forms with a 3'triphosphateC. a mixture of normal dNTPs and modified forms lacking the 2'hydroxylD. a mixture of normal dNTPs and modified forms lacking the 3'hydroxylE. a mixture of normal dNTPs and modified forms lacking the 5'hydroxyl
Show answer
Correct Answer: D
Feedback A: dNTPs have no 2’ hydroxyl
Feedback B: terminators have no 3’ hydroxyl
Feedback C: normal dNTPs already lack the 2’ hydroxyl
Feedback D: Yes, the 3’ hydroxyl is the nucleophile required for polymerization, so loss of the group creates a chain terminator.
Feedback E: Polymerization requires a 5’ triphosphate, so a form lacking this would never be incorporated and could not act as a terminator.
171. A physician told a patient that she had no Bohr effect. What amino acids were missing in her hemoglobin?A. carboxyl-terminal residuesB. heme binding ligandsC. proximal F8 HisD. B8 GluE. beta chain DPG binding residues
Show answer
Correct Answer: A
Feedback A: H+ is bound to carboxy terminal His146 of the beta chain and to terminal alpha amino groups. CO2 binds to the terminal alpha amino groups.
Feedback B:
Feedback C: This His pulls the iron out of the porphyrin plane in deoxyhemoglobin; it is not associated with the Bohr effect.
Feedback D:
Feedback E:
172. E. coli DNA is not degraded by its own restriction enzymes becauseA. the restriction enzymes are activated only outside the nucleus.B. the restriction enzymes are transported out of the nucleus.C. DNA methylases modify the restriction sites.D. the restriction enzymes are activated only after they are secreted out of the cell.E. The E. coli genome does not contain DNA sequences which are recognized by its own restriction-modification system.
Show answer
Correct Answer: C
Feedback A: Bacteria have no nuclei, and restriction enzymes are active throughout the cell
Feedback B: Bacteria have no nuclei, and restriction enzymes are active throughout the cell
Feedback C: yes, restriction enzymes are always paired with a modification enzyme with the same recognition site.
Feedback D: Restriction enzymes act inside the cell.
Feedback E: See C, EcoRI recognizes a 6 bp site, which occurs over 1000 times in the E. coli genome.
173. A patient had trouble with exercise and you as the physician found only beta chains in his hemoglobin. His blood was bright red on the arterial side and wasn't the usual blue on the venous side. It was red! Why was the hemoglobin reluctant to give up enough oxygen in the tissues for the man to exercise comfortably?A. Deficiency of 2,3-DPGB. Overexpression of fetal hemoglobin chainsC. Abnormally high oxygen binding affinity and no cooperativityD. Hemoglobin mutation prevented binding of CO2E. Tissues were lacking in myoglobin
Show answer
Correct Answer: C
Feedback A: This would cause hemoglobin to have a higher oxygen affinity, but this patient has a stated abnormality of only beta chains.
Feedback B:
Feedback C: This could be HbH disease. Accumulation of beta tetramers which have no heme-heme interactions.
Feedback D: Binding of CO2 normally lowers hemoblogin’s affinity for 02, so loss of C02 binding would increase oxygen affinity, but the stated abnormality is missing alpha chains.
Feedback E: In a study of mice who lacked myoglobin no significant effects were found.
174. Gout can be caused by all of the following EXCEPTA. Inhibition of ribonucleotide reductaseB. underexcretion of uric acidC. dietary and environmental factorsD. errors in the feedback regulation of purine biosynthesisE. overproduction of uric acid
Show answer
Correct Answer: A
Feedback A: This affects only DNA replication, not purine turnover.
Feedback B: This causes elevated uric acid levels, leading to gout
Feedback C: These can cause increased uric acid by altering the rate of synthesis or the rate of excretion
Feedback D: Excess purine synthesis causes excess turnover, leading to high urate and gout.
Feedback E: Excess uric acid is the cause of gout.
175. One of your patients was a mechanic who worked on automobile exhaust systems exclusively. He came in to the clinic drowsy and with pink cheeks. He claimed he often fainted when he exerted himself. What did his hemoglobin analysis show?A. Low oxygen binding due to high levels of carboxyhemoglobin (Hb CO)B. Abnormally low levels of beta chainsC. Abnormally low levels of gamma chainsD. Substitution of F8 His by TyrE. Monomeric hemoglobin molelcules
Show answer
Correct Answer: A
Feedback A: CO binds tightly to one heme shifting hemoglobin to the relaxed, high O2 affinity, state. Thus the affected Hb is unable to release O2 to the tissues.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
176. Identify the main type of mutation caused by type II topoisomerase inhibitors.A. Base pair substitutionsB. Frameshift mutationsC. DeletionsD. InsertionsE. Translocations
Show answer
Correct Answer: E
Feedback A: Failure of type II topo causes ds DNA breaks, leading to translocations
Feedback B: Failure of type II topo causes ds DNA breaks, leading to translocations
Feedback C: Failure of type II topo causes ds DNA breaks, leading to translocations
Feedback D: Failure of type II topo causes ds DNA breaks, leading to translocations
Feedback E: Failure of type II topo causes ds DNA breaks, leading to translocations
177. A patient had b chains that were colorless. Which part of the hemoglobin molecule may be defective?A. alpha-terminal residuesB. heme binding residuesC. DPG binding residuesD. the proximal F8 HisE. bilireubin binding site
Show answer
Correct Answer: B
Feedback A:
Feedback B: It is the heme that gives oxegenated blood its red color.
Feedback C:
Feedback D:
Feedback E:
178. A mutated virus is found to copy its genome faster than it used to. It is very likely that at the same time it also acquiredA. A much smaller genomeB. A much larger genomeC. A lower error frequencyD. A higher error frequencyE. A narrower host specificity
Show answer
Correct Answer: D
Feedback A: This is unlikely since it would probably incapacitate the virus completely
Feedback B: This would not cause faster replication
Feedback C: Faster copying is associated with higher error rates.
Feedback D: Yes, speed and accuracy are a tradeoff
Feedback E: This would not be linked to replication rate in any obvious straightforward way.
179. A witch doctor with sickle cell anemia claims his disease improved when he drank his own urine. What possible benefit may be occurring with sickle cell hemoglobin?A. Protease in urine degraded aggregated Hb S thereby restoring normal cell shapeB. Low pH of urine solubilized aggregated Hb S with restoration of normal cell shapeC. Cyanate formed from urea blocked the alpha amino residues precluding salt bridgesD. Urea in urine solubilized aggregated Hb SE. He obtained extra iron atoms for new hemoglobin synthesis
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Cyanate carbamoylates the terminal amino groups behaving as a reactive analog of CO2.
Feedback D:
Feedback E:
180. Which of the following statements is NOT reasonable. Hemoglobin would not be an effective oxygen carrier if it were a plasma protein rather than present within erythrocytes, because ... ?A. subunits may dissociate and be filtered in the kidney.B. there is no DPG in serum.C. the CO2 transported by hemoglobin requires an enzyme within the erythrocyte.D. there is no way to maintain the heme iron in the ferrous state.E. more difficult to maintain proper ratios of DPG and hemoglobin concentrations
Show answer
Correct Answer: D
Feedback A: True, large erythrocytes escape filtering.
Feedback B: O2 affinity of Hb within erythrocytes is lower than Hb in plasma because in erythrocytes DPG is maintained at approximately the same molar concentration as Hb.
Feedback C: True, most CO2 is transported as bicarbonate, which is fromed in erythrocytes by carbonic anhydrase.
Feedback D: Beware of blanket statements. There is probably a way.
Feedback E: See B.
181. What is needed to achieve Vmax in an enzyme reaction?A. enzyme concentration above the KmB. physiological pHC. a 1:1 ratio of enzyme and substrateD. great luckE. excess substrate
Show answer
Correct Answer: E
Feedback A: Km is the concentration of substrate needed to achieve 1/2 maximal velocity.
Feedback B:
Feedback C: A 1:1 ratio is not suffient because substrate would be converted to product, and then active sites would be unoccupied. There must be more substrate to keep the enzyme saturated.
Feedback D:
Feedback E: Vmax is achieved when all of the active sites are occupied by substrate. See C.
182. Suppose the His in the charge transfer network of chymotrypsin was mutated to Ala. What effect would that have on catalysis?A. all enzyme molecules would end up in a stable, acylated intermediate.B. peptide substrates would not bindC. no formation of acylated intermediateD. no change in activityE. enzyme would not fold
Show answer
Correct Answer: C
Feedback A: See C.
Feedback B:
Feedback C: The alanine would not take H+ from serine as histidine does. Thus the serine would not be able to attack and form the acylated intermediate.
Feedback D: Activity would change dramatically.
Feedback E: The folding my differ some from normal chymotrypsin, but the protein would still fold.
183. Increasing the concentration of dATP directly causes ribonucleotide reductase toA. Increase the rate of production of dCDPB. Increase the rate of production of dGDPC. Decrease the rate of production of rCDPD. Decrease the rate of production of dUDPE. Increase the rate of production of dTTP
Show answer
Correct Answer: D
Feedback A: Increased dATP decreases the activity of RNR for all substrates and products
Feedback B: Increased dATP decreases the activity of RNR for all substrates and products
Feedback C: Increased dATP decreases the activity of RNR for all substrates and products; but rCDP is not a product of RNR.
Feedback D: Increased dATP decreases the activity of RNR for all substrates and products
Feedback E: Increased dATP decreases the activity of RNR for all substrates and products
184. Which of the following conditions would promote denaturation of double-stranded DNA?A. Decreasing the temperatureB. Adding NaClC. Adding detergentD. Increasing the pH (above neutral)E. Decreasing the concentration of organic solvent (formamide, for example)
Show answer
Correct Answer: D
Feedback A: DNA denatures at high temperatures so this would promote the native form
Feedback B: NaCl contributes to charge shielding for the phosphodiester backbone, stabilizing the duplex
Feedback C: Detergents don’t appreciably alter the stability of DNA duplexes
Feedback D: Increased pH destabilizes base pairing interactions and therefore promotes denaturation
Feedback E: Increasing organic solvents decreases the stability of DNA duplexes due to the decrease in energy derived from base stacking, so decreasing the concentration would promote stabilization
185. Bacterial plasmidsA. are usually linear DNA molecules.B. can confer resistance to many antibiotics simultaneously.C. cannot be transferred from one cell to another.D. are a disadvantage to their hosts.E. are not found in human pathogens.
Show answer
Correct Answer: B
Feedback A: Plasmids are typically circular
Feedback B: Scary, and true
Feedback C: Many plasmids encode machinery for transferring themselves to other cells
Feedback D: Many plasmids confer resistance to antibiotics which provides a strong advantage to the host if that substance is encountered.
Feedback E: Unfortunately, this is not true.
186. 5-phosphoribosyl-1-pyrophosphate (PRPP) is an intermediate inA. the de novo synthesis of purine nucleotides
B. the de novo synthesis of pyrimidine nucleotidesC. the salvage pathway for the synthesis of purine nucleotidesD. A and CE. A, B and C
Show answer
Correct Answer: E
Feedback A: Yes, it is activated as the committed step for purine synthesis, but B and C are also true so the answer is E.
Feedback B: Yes, it is used to “scavenge†the preformed base or� otate, but A and C are also true, so the answer is E.
Feedback C: Yes, it is combined with hypoxanthine or guanine by HGPRTase, but A and B are also true so the answer is E
Feedback D: Yes, but B is also true, so the answer is E
Feedback E:
187. One of the strands of a double-stranded, 10 kbp DNA duplex has the following numbers of base residues: adenine (A) - 2300, thymine (T) - 2800. The base composition of the whole double-stranded molecule will beA. a - 5100, t - 5100, g - 4900, c - 4900B. a - 4600, t - 5400, g - 5600, c - 4400C. a - 5600, t - 5600, g - 4400, c - 4400D. a - 4600, t - 5600 , g - 5400, c - 4400E. There is not enough information to determine the overall base composition
Show answer
Correct Answer: A
Feedback A: Due to pairing rules, the other strand has 2300 T and 2800 A, for a total of 5100 A and 5100 T. The remainder must be equal numbers of G and C for the same reason, making 4900 of each.
Feedback B: Due to pairing rules, the other strand has 2300 T and 2800 A, for a total of 5100 A and 5100 T. The remainder must be equal numbers of G and C for the same reason, making 4900 of each.
Feedback C: Due to pairing rules, the other strand has 2300 T and 2800 A, for a total of 5100 A and 5100 T. The remainder must be equal numbers of G and C for the same reason, making 4900 of each.
Feedback D: Due to pairing rules, the other strand has 2300 T and 2800 A, for a total of 5100 A and 5100 T. The remainder must be equal numbers of G and C for the same reason, making 4900 of each.
Feedback E: Due to pairing rules, the other strand has 2300 T and 2800 A, for a total of 5100 A and 5100 T. The remainder must be equal numbers of G and C for the same reason, making 4900 of each.
188. If you denature similar amounts of the following 5 DNA samples then return them to conditions that favor renaturation, which will remain the most single-stranded the longest?A. Human liver nuclear DNAB. Vaccinia (viral) DNAC. E. coli DNAD. Yeast DNAE. eye of newt DNA
Show answer
Correct Answer: E
Feedback A: The rate of reannealing for any piece of DNA is determined by the concentration of the complementary strand. Given the same amount of total DNA, the sample with the simplest composition (the fewest different sequences) will have the highest concentration of each sequence and will anneal the fastest. In this case the simplest sample is the viral DNA.
Feedback B: The rate of reannealing for any piece of DNA is determined by the concentration of the complementary strand. Given the same amount of total DNA, the sample with the simplest composition (the fewest different sequences) will have the highest concentration of each sequence and will anneal the fastest. In this case the simplest sample is the viral DNA.
Feedback C: The rate of reannealing for any piece of DNA is determined by the concentration of the complementary strand. Given the same amount of total DNA, the sample with the simplest composition (the fewest different sequences) will have the highest concentration of each sequence and will anneal the fastest. In this case the simplest sample is the viral DNA.
Feedback D: The rate of reannealing for any piece of DNA is determined by the concentration of the complementary strand. Given the same amount of total DNA, the sample with the simplest composition (the fewest different sequences) will have the highest concentration of each sequence and will anneal the fastest. In this case the simplest sample is the viral DNA.
Feedback E: The rate of reannealing for any piece of DNA is determined by the concentration of the complementary strand. Given the same amount of total DNA, the sample with the simplest composition (the fewest different sequences) will have the highest concentration of each sequence and will anneal the fastest. In this case the simplest sample is the viral DNA.
189. The processivity of DNA polymerasesA. increases in the presence of ligasesB. decreases in the presence of helicasesC. decreases in the presence of single strand binding proteinsD. increases in the presence of Okazaki factorsE. increases in the presence of clamp proteins
Show answer
Correct Answer: E
Feedback A: Ligases seal nicks after polymerase is finished, they do not affect processivity
Feedback B: Helicases unwind the template, making progression easier and therefore increase the processivity of polymerases
Feedback C: SSBs straighten the template and therefore increase the processivity of polymerases.
Feedback D: There is no such thing
Feedback E: Clamp proteins prevent dissociation of the polymerase from its template and therefore increase the processivity
190. A bacterial culture grown for many generations in a "light" (14N) medium was transferred to a "heavy" (15N) medium. After the DNA had been copied 3 times under the new conditions, what would be the relative distribution of DNA containing two light strands (LL), one heavy and one light strand (HL), and two heavy strands (HH)?A. 4 HH:4 HL:0 LLB. 6 HH:1HL:1 LLC. 4 HH:2 HL: 2 LLD. 6 HH:2 HL: 0 LLE. 2 HH:4 HL: 2 LL
Show answer
Correct Answer: D
Feedback A: After one copy, the DNA would all be HL, after 2 it would be 1 HL:1 HH, after 3 it would be 1
HL:3 HH.
Feedback B: After one copy, the DNA would all be HL, after 2 it would be 1 HL:1 HH, after 3 it would be 1 HL:3 HH. There would be no LL after the first copy.
Feedback C: After one copy, the DNA would all be HL, after 2 it would be 1 HL:1 HH, after 3 it would be 1 HL:3 HH. There would be no LL after the first copy.
Feedback D: After one copy, the DNA would all be HL, after 2 it would be 1 HL:1 HH, after 3 it would be 1 HL:3 HH (=6 HH:2 HL)
Feedback E: After one copy, the DNA would all be HL, after 2 it would be 1 HL:1 HH, after 3 it would be 1 HL:3 HH. There would be no LL after the first copy.
191. In methylation-directed mismatch repair in bacteriaA. MutS binds to mismatches in duplex DNAB. The methylated strand is judged to be the newer strandC. MutL binds to methylated DNA strandsD. Methylation of cytosine leads to deamination to thymineE. RecA makes an incision only at hemimethylated sites
Show answer
Correct Answer: A
Feedback A:
Feedback B: Methylation indicates that this DNA is old since this modification must have occured before DNA replication
Feedback C: MutH recognizes the methylation status
Feedback D: This is true, but irrelevant to the question asked
Feedback E: RecA acts in SOS and in recombination, but not in mismatch repair
192. Muscle can carry out all of the following processes EXCEPTA. transamination of amino acids.B. synthesis of glycogen.C. synthesis of glucose from alanine.D. oxidation of ketone bodies.E. phosphorylation of creatine.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Gluconeogenesis occurs in the liver and kidney.
Feedback D:
Feedback E:
193. Telomeres are different from broken ends of DNA becauseA. They have a series of repeats of a particular 6-10 base pair sequenceB. They are stabilized by the presence of short RNA primers needed to complete replicationC. They are recognized and extended by telomeraseD. Only A and C are correctE. A, B, and C are all correct
Show answer
Correct Answer: D
Feedback A: This is true, but so is C
Feedback B: This just confusing
Feedback C: This is true, but so is A
Feedback D:
Feedback E: Bis incorrect
194. Mutations caused by one of the following agents almost always inactivate genes specifying protein products, but often do not affect the function of genes specifying only stable RNA products. Which agent is it?A. UV lightB. cis-platinC. intercalating agentsD. "mustard" gasesE. 5-bromouracil
Show answer
Correct Answer: C
Feedback A: No, UV light typically causes single base changes, which are as bad for proteins as for RNAs
Feedback B: No, this is typically a crosslinker leading to double-strand breaks.
Feedback C: Yes, these cause insertions and deletions leading to frame-shifts that are much worse for ORFs than for RNAs
Feedback D: No, these are typically a crosslinkers leading to double-strand breaks.
Feedback E: No, this causes an enforced tautomeric shift leading to single base changes.
195. Which primers will lead to amplification of MY FAVORITE GENE in a PCR if one strand of the duplex template has the sequence below? 5'-GGTACCGATC-MY FAVORITE GENE-GCATGAGATC-3'A. 5'-GGTACCGATC and 5'-GCATGAGATCB. 5'-GGTACCGATC and 5'-GATCTCATGCC. 5'-GCATGAGATC and 5'-CGTACTCTAGD. 5'-GATCGGTACC and 5'-CTAGAGTACGE. 5'-CCATGGCTAG and 5'-GCATGAGATC
Show answer
Correct Answer: B
Feedback A: The first primer is correct, but the second one anneals to the same strand and points away from the target
Feedback B:
Feedback C: The first primer points away from the target and the second is nonsense (complementary but not antiparallel)
Feedback D: The first primer is correct, but the second is nonsense (antiparallel, but not complementary)
Feedback E: The first primer is complementary but not antiparallel, and the second points away from the
target.
196. Pyruvate carboxylase of liver is activated most directly byA. acetyl CoA.B. phosphoenolpyruvate.C. a high ratio of ADP/ATP.D. calcium ions.E. cyclic AMP.
Show answer
Correct Answer: A
Feedback A: allosteric activation of the 1st step of gluconeogenesis.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
197. DNA sequencing depends on generating a set of molecules that terminate at every possible position. This is done by stopping the synthesisA. using modified nucleotides lacking the 3' hydroxyl groupB. using a DNA polymerase purified from a thermophilic organismC. using modified nucleotides phosphorylated at the 5' hydroxylD. using modified nucleotides with deaminated basesE. using a non-processive DNA polymerase
Show answer
Correct Answer: A
Feedback A: Yes, the 3’ hydroxyl is a necessary nucleophile for the polymerization reaction
Feedback B: Sometimes, but this is not a necessary feature contributing to chain termination
Feedback C: Normal nucleotides are phosphorylated at the 5’ position
Feedback D: The bases contribute to specificity but do not cause chain termination
Feedback E: This would not provide specific chain termination
198. Which of the following is a general feature of both DNA and RNA?A. in vivo they are usually double helical structures formed from two separate chains.B. Their phosphodiester backbones are stable in solutions above pH 12.C. They are made partly of complementary G-C base pairs.D. They both have phosphodiester bonds linking the 5' and 2' hydroxyls of the ribose groups.E. They interact with histones to form nucleosomes.
Show answer
Correct Answer: C
Feedback A: RNA is usually single-stranded, so any duplexes are formed from intra-strand interactions
Feedback B: RNA is unstable at high pH due to the nucleophilic attack by the 2’ hydroxyl on the phosphodiester backbone.
Feedback C:
Feedback D: This is true of the 5’ and 3’ groups
Feedback E: RNA is not found in nucleosomes
199. Which one of the following enzymes might be expected to be more active as a consequence of higher citrate concentration?A. pyruvate carboxylaseB. phosphofructokinaseC. citrate synthaseD. acetyl CoA carboxylaseE. pyruvate dehydrogenase
Show answer
Correct Answer: D
Feedback A: Stimulated by acetyl CoA.
Feedback B: Inhibited by citrate.
Feedback C: Makes citrate, inhibited by ATP.
Feedback D: Citrate induces polymerization of the subunits.
Feedback E:
200. The sequence of the complementary DNA strand for DNA with the sequence 5'-ATCGTACCGTTA 3' is:A. 5'-TAGCATGGCAAT-3'B. 5'-TGACAGAGTAGA-3'C. 5'-TAACGGTACGAT-3'D. 5'-ATTGCCATGCTA-3'E. 5'-ATCGTACCGTTA-3'
Show answer
Correct Answer: C
Feedback A: Complementary, but not antiparallel
Feedback B: Nonsense
Feedback C: This complementary and antiparallel
Feedback D: Antiparallel, but not complementary
Feedback E: Not complementary or antiparallel; this is the same as the sequence in the question
201. Out of the storage forms in humans, which is largest in terms of calorie yield isA. protein.B. glucose.C. glycogen.D. triglyceride.E. lipoprotein.
Show answer
Correct Answer: D
Feedback A: Not really a storage form, 4 Cal/g.
Feedback B: Not a storage form.
Feedback C: A carbohydrate, 4 Cal/g.
Feedback D: 9 Cal/g.
Feedback E: Not a storage form.
202. The melting temperature of double stranded DNAA. increases with increasing guanine contentB. decreases with increasing cytosine contentC. increases with decreasing guanine contentD. decreases with decreasing adenine contentE. increases with increasing thymine content
Show answer
Correct Answer: A
Feedback A: G+C base pairs are more stable than A+T so as G content rises so does the Tm
Feedback B: G+C base pairs are more stable than A+T so as C content rises so does the Tm
Feedback C: G+C base pairs are more stable than A+T so as G content rises so does the Tm
Feedback D: G+C base pairs are more stable than A+T so as A content drops the Tm rises
Feedback E: G+C base pairs are more stable than A+T so as T content increases the Tm drops
203. Reducing equivalents generated during glycolysis in a normal liver cellA. enter the mitochondria as protons.B. are used in the pentose phosphate pathway.C. are transported into mitochondria in the form of malate.D. enter the mitochondria as reduced glutathione.E. enter the mitochondria as NADH.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Malate-Aspartate shuttle.
Feedback D:
Feedback E: The mitochondrial membrane is impermeable to NADH.
204. What structural feature of DNA accounts for its stability to alkaline hydrolysis?A. The presence of hydrogen bonding between purines and pyrimidines.B. The lack of a 2'-hydroxyl in the pentoseC. The lack of a 3'-hydroxyl in the pentoseD. The presence of thymine in DNA.E. The absence of uracil from DNA.
Show answer
Correct Answer: B
Feedback A: The 2’ hydroxyl is responsible for the instability of RNA, and its removal is the basis for the increased stability of DNA
Feedback B: The 2’ hydroxyl is responsible for the instability of RNA, and its removal is the basis for the increased stability of DNA
Feedback C: The 2’ hydroxyl is responsible for the instability of RNA, and its removal is the basis for the increased stability of DNA
Feedback D: The 2’ hydroxyl is responsible for the instability of RNA, and its removal is the basis for the increased stability of DNA
Feedback E: The 2’ hydroxyl is responsible for the instability of RNA, and its removal is the basis for the increased stability of DNA
205. All of the following can stimulate the activity of muscle glycogen phosphorylase EXCEPTA. calcium ions.B. epinephrine.C. cyclic AMP.D. phosphorylase kinase.E. glucagon.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glucagon does not affect muscle.
206. Which of the following properties of nucleotides does NOT affect the structure of nucleic acids?A. The bases are hydrophobicB. The phosphates are negatively charged at neutral pHC. The bases can each form multiple hydrogen bondsD. The bases absorb ultraviolet light at a wavelength near 260 nmE. The sugar group is highly water soluble
Show answer
Correct Answer: D
Feedback A: This provides the main driving force for formation of the duplex
Feedback B: The repulsion of the backbones is a significant force affecting the structure and stability of duplexes
Feedback C: The ability to form hydrogen bonds is an important determinant of the specificity of the interaction between strands (the complementarity).
Feedback D: This is true but does not affect the structure of the duplex
Feedback E: This provides solubility to nucleic acids
207. Accelerated de novo purine biosynthesis in the Lesch-Nyhan syndrome is probably due toA. depression of 5-phosphoribosyl-1-pyrophosphate amidotransferaseB. increased availability of 5-phosphoribosyl-1-pyrophosphateC. defective feedback inhibition of ribonucleotide reductaseD. increased uric acid synthesisE. increased hypoxanthine-guanine phosphoribosyltransferase
Show answer
Correct Answer: B
Feedback A: Lack of HGPRTase causes the level of its substrate PRPP to increase, leading to higher rates of activation and purine synthesis.
Feedback B: Lack of HGPRTase causes the level of its substrate PRPP to increase, leading to higher rates of activation and purine synthesis.
Feedback C: Lack of HGPRTase causes the level of its substrate PRPP to increase, leading to higher rates of activation and purine synthesis.
Feedback D: Lack of HGPRTase causes the level of its substrate PRPP to increase, leading to higher rates of activation and purine synthesis.
Feedback E: Lack of HGPRTase in this syndrome causes the level of its substrate PRPP to increase, leading to higher rates of activation and purine synthesis.
208. A common source of mutations in humans starts with the deamination ofA. 5-methyl-thymineB. 5-fluoro-uracilC. 5-methyl-cytosineD. 5-methyl-uracilE. 5-methyl-adenine
Show answer
Correct Answer: C
Feedback A: Thymine already has a 5-methyl group
Feedback B: This is not usually found in normal cells
Feedback C: Yes, this leads to formation of T, which is then a simple mismatch with the G usually found opposite C.
Feedback D: This is thymine (which is already deaminated).
Feedback E: No such thing
209. In DNA replication in human cellsA. replication initiates at about 4000 sites called originsB. a DNA polymerase that incorporates bases in the 3' to 5' orientation is foundC. replication is continuous on both strandsD. the replication or S phase lasts about 8 hoursE. replication and segregation overlap to allow shorter division cycles
Show answer
Correct Answer: D
Feedback A: Humans have about 40,000 origins of replication
Feedback B: No polymerase with this polarity is known
Feedback C: Replication is continuous on the leading strand but discontinous on the lagging strand
Feedback D:
Feedback E: This is a trick that only prokaryotes with single replication origins per genome can pull off
210. The compound which accepts a two-carbon fragment from acetyl-CoA to initiate the citric acid cycle isA. oxalosuccinate.B. oxalate.C. oxaloacetate.D. oxytocin.E. oxybutyrate.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Citrate synthase catalyzes: oxaloacetate + acetyl CoA ---> CoA + citrate.
Feedback D: A pituitary hormone.
Feedback E:
211. DNA polymerase, a primed DNA template, and the four deoxynucleoside triphosphates (dNTPs) are mixed together. Which is the most important in determining the base composition (the number of As, Gs, Cs, and Ts) of the product DNA?A. the addition of a processivity clampB. the ratios of the four dNTPs availableC. the species from which the template DNA was purifiedD. the type of primer used in the reactionE. the species from which the DNA polymerase was purified.
Show answer
Correct Answer: C
Feedback A: This is important for the rate of synthesis, but only the template determines the composition of the product.
Feedback B: These are necessary for replication, but only the template determines the composition of the product.
Feedback C: The composition of the template determines the composition of the product, so the source of the DNA template determines the composition.
Feedback D: The primer allows replication, but only the template determines the composition of the product.
Feedback E: DNA polymerase just takes instructions from the template, it doesn’t matter where it came from
212. The increase in blood sugar level resulting from epinephrine injection can best be explained by which one of the following?A. Synthesis of carbohydrate from fatB. Absorption of carbohydrate from the digestive tractC. Liver glycogenesisD. Liver glycogenolysisE. Gluconeogenesis
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C: Phosphorylated glycogen synthase in inactive.
Feedback D: Phosphorylated glycogen phosphorylase is active and glycogen in broken down providing glucose.
Feedback E: Not directly affected by epinephrine.
213. Nucleotide excision repair is known to need all of the following enzymes EXCEPT which one?A. A nucleaseB. A type I topoisomeraseC. A DNA helicaseD. A DNA ligaseE. A DNA polymerase
Show answer
Correct Answer: B
Feedback A: The excision itself is performed by a nuclease
Feedback B: No role is known at this point for a type I topo in NER
Feedback C: A helicase is need to remove the damaged DNA strand after it has been clipped by a nuclease
Feedback D: ligase is needed to seal the nick produced by resynthesis of the gap produced by removing the damaged DNA
Feedback E: DNA polymerase resynthesizes the DNA to fill the gap produced by removal of the damaged DNA
214. Allosteric inhibitors of regulatory enzymes most likely influence enzyme activity byA. inducing a conformational change in the enzyme.B. promoting a covalent modification of the enzyme.C. binding to the active site of the enzyme.D. reacting with the substrate.E. promoting dephosphorylation of the enzyme.
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C: This is competative inhibition.
Feedback D:
Feedback E:
215. PhotolyasesA. Use incident light to break N-glycosidic bondsB. Nick DNA molecules near sites of damage to initiate repair synthesisC. Seal nicks in the phosphodiester backbone of DNA moleculesD. Reverse the formation of cyclobutane dimers caused by UV light.E. Cut DNA at hemi-methylated sites.
Show answer
Correct Answer: D
Feedback A: This would cause, not repair, damage
Feedback B: Photolyases reverse damage by undoing the formation of cyclobutane rings
Feedback C: This is the job of ligase, photolyases reverse damage by undoing the formation of cyclobutane rings
Feedback D:
Feedback E: This is the job of MutH, photolyases reverse damage by undoing the formation of cyclobutane rings
216. In the liver, 3',5'-AMP increases the activity of all the following enzymes EXCEPTA. glycogen phosphorylase.B. fructose 2,6 bisphosphatase.C. pyruvate kinase.D. phosphorylase kinase.E. cyclic AMP-dependent protein kinase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Phosphorylation inactivates pyruvate kinase.
Feedback D:
Feedback E:
217. A DNA molecule is labeled and used to locate complementary strands in a restriction enzyme digest of genomic DNA separated by electrophoresis. This is called aA. Northern blotB. Western blotC. Southern blotD. Eastern blotE. Ink blot
Show answer
Correct Answer: C
Feedback A: Northern blots contain immobilized RNA, not DNA
Feedback B: Western blots contain immobilized proteins and are probed with antibodies
Feedback C:
Feedback D: No such thing, for some strange reason
Feedback E: Of course not, I was just being silly
218. The statements below refer to discrete steps involved in the breakdown of muscle glycogen to glucose 1-phosphate. Select the appropriate sequence from the list below.
1. phosphorylase b--> phosphorylase a2. dissociation of the cAMP dependent kinase3. activation of adenylate cyclase4. phosphorylation of phosphorylase kinase
A. 1,2,3,4B. 2,3,1,4C. 3,2,4,1D. 2,3,4,1E. 3,4,1,2
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
219. Cells lacking p53 are abnormal because theyA. lack enzymes needed for base excision repairB. fail to enter S phase if their DNA is damagedC. arrest in mitosis in the presence of high levels of caffeineD. fail to die after exposure to DNA damageE. A and C are both correct
Show answer
Correct Answer: D
Feedback A: No, BER is normal in p53 mutants
Feedback B: No, this would be the normal response, not an abnormal one.
Feedback C: No, this would describe an overactive checkpoint, not the underactive one found in p53 mutants
Feedback D: Yes, cells with DNA damage undergo apoptosis, but this response requires p53.
Feedback E: No, neither A nor C are correct.
220. Which statement regarding the actions of glucagon and insulin is FALSE?A. Insulin is an anabolic hormone.B. Glucagon increases levels of cAMP in target cells.C. Insulin activates a tyrosine kinase.D. Both insulin and glucagon bind specifically to proteins in the plasma membrane of target cells.E. Glucagon stimulates glycolysis in target cells.
Show answer
Correct Answer: E
Feedback A: It promotes uptake and storage.
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glucagon inhibits glycolysis by phosphorylation of pyruvate kinase and PFK-2.
221. An enzyme found in the liver but not in skeletal muscle isA. glycogen phosphorylase.B. lactate dehydrogenase.C. pyruvate dehydrogenase.D. glucose-6- phosphatase.E. hexokinase.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Muscle cannot convert glycogen breakdown or gluconeogenesis products into glucose. Liver can release glucose into the circulation.
Feedback E:
222. Phosphorylase aA. contains phosphorylated serine residues.B. exists as a tetramer.C. is the less active form of the enzyme.D. is activated directly by AMP.E. none of the above
Show answer
Correct Answer: A
Feedback A: increased cAMP ---> active PKA ---> active phosphorylase kinase ---> active glycogen phosphorylase.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
223. Insulin will causeA. enhanced glucose uptake by the muscle.B. increased glycogen synthesis.C. increased oxidation of glucose to carbon dioxide and water.D. increased conversion of glucose to fatty acids.E. all of the above.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
224. All of the following statements about phosphofructokinase (PFK- 1) are true EXCEPTA. it is a major control enzyme in glycolysis.B. ATP is a substrate for PFK.C. AMP is a negative effector of PFK.D. ATP is a negative effector of PFK.E. it catalyzes a metabolically irreversible reaction; i.e., its equilibrium point lies far in one direction.
Show answer
Correct Answer: C
Feedback A: The committed step.
Feedback B:
Feedback C: AMP and fructose-2,6-bisphosphate stimulate PFK-1; ATP and citrate inhibit.
Feedback D:
Feedback E:
225. In comparison to the current American diet it is recommended thatA. carbohydrate consumption be reduced and fat consumption be increased.B. that carbohydrate consumption be increased and fats be decreased.C. that protein consumption be decreased.D. that protein consumption be increased and fats be increased.E. that both carbohydrate and fat consumption be decreased.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Up for debate
Feedback C:
Feedback D:
Feedback E:
226. Which is LEAST likely to be characteristic of a regulated enzyme in a metabolic pathway?A. catalyzes the slowest step of the pathwayB. has a short half-life in vivoC. concentration is hormonally regulatedD. activity under allosteric controlE. catalyzes a physiologically reversible reaction
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Irreversible reactions (especially those with ATP involvement) are usually regulated.
227. Pyridoxal phosphate acts as a coenzyme in mostA. transmethylation reactions.B. carbon dioxide fixation reactions.C. transamination reactions.D. kinase reactions.E. none of the above.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Each transaminase has a pyridoxal phosphate (B6) prosthetic group.
Feedback D:
Feedback E:
228. Ammonia enters the urea cycle asA. free ammonia.B. free ammonia and carbamoyl phosphate.C. glutamine and aspartateD. aspartate and carbamoyl phosphate.E. glutamate and aspartate.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C: Glutamine transports nitrogen in the blood mainly to the liver.
Feedback D: The two nitrogens that end up in urea come into the cycle as part of aspartate and carbamoyl phosphate.
Feedback E:
229. Oxidation of fatty acids in liver mitochondria may facilitate gluconeogenesis byA. supporting the phosphorylation of ADP.B. forming acetyl CoA which can contribute to the net synthesis of glucose.C. reducing FAD and NADP+.D. transporting camitine into mitochondria.E. decreasing the dependence of phosphorylation of ADP on oxygen consumption.
Show answer
Correct Answer: A
Feedback A: Metabolism of fatty acids and resulting acetyl CoA gives higher NADH and FADH2, which leads to increased ATP.
Feedback B: Humans cannot make glucose from acetyl CoA. Formation of acetyl CoA from pyruvate is a key irreversible reaction.
Feedback C:
Feedback D:
Feedback E: ATP synthesis and O2 consumption are coupled.
230. Which of the following is a FALSE statement about creatine phosphokinase (CPK)?A. CPK is present predominantly as three different isoenzymes.B. CPK can catalyze the formation of ATP using creatine phosphate and ADP as substrates.C. All isoenzymes of CPK are dimers.D. An increase in total serum CPK is a positive indicator of myocardial infarction.E. An increase in the serum concentration of one isoenzyme of CPK above a certain percentage(e.g. 9%) of the total serum CPK is one indicator of a myocardial infarction.
Show answer
Correct Answer: D
Feedback A: True, MM, MB, and BB isoenzymes.
Feedback B:
Feedback C:
Feedback D: Serum CPK increases after many types of muscle damage. Myocardial damage is indicated if 3% or more of the increase in CPK is the MB isoenzyme.
Feedback E:
231. The enzyme which catalyzes the entrance of acetyl CoA into the citric acid cycle isA. aconitase.B. citrate synthase.C. succinate thiokinase.D. pyruvate dehydrogenase.E. isocitric dehydrogenase.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Catalyzes: acetyl CoA + oxaloacetate ---> citrate
Feedback C:
Feedback D: The PDH complex produces acetyl CoA from pyruvate.
Feedback E:
232. Which of the following is a FALSE statement about gluconeogenesis.A. This anabolic pathway only occurs during prolonged starvation (days to weeks).B. This process depends on the use of fatty acids as an energy source.C. This process does not occur in skeletal muscle.D. This process is stimulated by citrate.
E. This process requires the use of mitochondrial and cytoplasmic enzymes.
Show answer
Correct Answer: A
Feedback A: False, although critical during starvation it can play a role in maintaining blood glucose between normal meals.
Feedback B:
Feedback C: It occurs in liver (90%) and kidney (10%).
Feedback D: Citrate stimulates fructose-1,6-bisphosphatase.
Feedback E: Pyruvate carboxylase -- mitochondria Glucose-6-phosphatase -- ER All other enzymes needed for gluconeogenesis -- cytosol
233. The predominant products of anaerobic glycolysis differ from those formed during aerobicglycolysis because...A. aerobically, lactate diffuses from muscle and is recycled to pyruvate in liver.B. under aerobic conditions, NADH reduces pyruvate to give lactate as the major product.C. NAD must be replenished for glycolysis to continue anaerobically.D. NADH must be replenished for glycolysis to continue aerobically.E. lactate produced aerobically will be used to reduce NAD.
Show answer
Correct Answer: C
Feedback A: Anaerobically, the liver recycles lactate to pyruvate.
Feedback B: Anaerobically, NADH reduces pyruvate to lactate.
Feedback C: When O2 is low NADH cannot be oxidized to NAD+ by oxidative phosphorylation. NAD+ is replenished by reducing pyruvate to lactate.
Feedback D: NAD+ must be replenished for glycolysis to continue aerobically or anaerobically.
Feedback E: Lactate is produced in anaerobic conditions to replenish NAD+.
234. The rate-limiting reaction in urea production is that catalyzed byA. argininosuccinase.B. argininosuccinate synthetase.C. arginase.D. ornithine transcarbamoylase.E. carbamoyl phosphate synthetase.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Just memorize it.
235. Which of the following is the most accurate comparison of glucokinases in muscle and liver?A. Glucokinase in both tissues have about the same Vmax.B. Glucokinases in both tissues have about the same Km.C. Muscle glucokinase has a lower Vmax and a lower Km than liver glucokinase.D. Liver glucokinase has a lower Km than the muscle glucokinaseE. Liver glucokinase has both a lower Vmax and lower Km than the glucokinase in muscle.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Muscle glucokinase(hexokinase) can catalyze the phosphorylation of glucose at lower glucose concentrations. Liver glucokinase responds to higher glucose loads.
Feedback D:
Feedback E:
236. Which of the following is a FALSE statement about oxidative phosphorylation?A. This process will stop after ADP is depleted.B. Ox-phos requires that protons be pumped out of the mitochondria matrix.C. ATP synthesis occurs as protons are pumped out of the mitochondrial matrix.D. Electrons transport will continue if 2,4-dinitrophenol is added to respiring mitochondria.E. ATP synthesis occurs as protons enter the mitochondrial matrix through a specific channel.
Show answer
Correct Answer: C
Feedback A: This is respiratory control.
Feedback B: H+ is pumped from the matrix to the intermembrane space.
Feedback C: False, ATP in synthesized as protons enter the matrix via ATP synthase.
Feedback D: True, but ATP synthesis will fall off because dinitrophenol is an uncoupler.
Feedback E: The F0 subunit of the ATP synthase is the channel.
237. The major immediate source of urinary ammonia isA. the hydrolysis of urea by urease.B. the oxidative dearnination of glutamic acid by glutamate dehydrogenase.C. the dearnination of histidine by histidine-ammonia lyase.D. the hydrolysis of glutamine by glutaminase.E. the oxidative dearnination of amino acids by kidney D-amino acid oxidase.
Show answer
Correct Answer: C
Feedback A: Cleaves urea to CO2 and NH3 in the gut.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
238. The first product formed during glycogenolysis (glycogen degradation at alpha 1,4 glycosidicbonds) is...A. glucose.B. glucose-6-phosphate.C. glucose-1-phosphate.D. galactose.E. UDP-glucose
Show answer
Correct Answer: C
Feedback A: See C
Feedback B: See C
Feedback C: glycogen ---> glucose-1-phosphate ---> glucose-6-phosphate In the liver glucose-6-phosphate ---> glucose ---> blood glucose
Feedback D:
Feedback E: UDP-glucose is the immediate substrate of glycogen synthesis.
239. Ammonia is not transported in the blood from peripheral tissues such as brain in the free state. Instead, a major route for transport of "ammonia" isA. Glutamine, formed from glutamate as a result of transamination between alpha-ketoglutarate and ammonia.B. Glutamate, fonned from ammonium ion and alpha-ketoglutarate via a reaction involving pyridoxal phosphate as cofactor.C. Glutamine, formed from glutamate and ammonia via a reaction which involves the cofactor NADH.D. Glutamine, formed from glutamate and ammonia via a reaction which involves the hydrolysis of ATP.E. None of the other answers is correct.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Catalyzed by glutamine synthase.
Feedback E:
240. The rate of glycolysis is "coupled" to the rate of the citric acid cycle. Which compound limits glycolysis so that it dose not exceed the "needs" of the citric acid cycle?A. Oxaloacetate.B. Pyruvate.C. Citrate.D. NADE. NADH
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Citrate is an inhibitor of phosphofructokinase-1, the committed step of glycolysis.
Feedback D:
Feedback E:
241. Cyclic AMP dependent protein kinases...A. are tetrameres composed of two regulatory and two catalytic subunits.B. are dimers composed of one regulatory and one catalytic subunits.C. phosphorylate and deactivate phosphorylase kinase.D. phosphorylate and activate glycogen synthase.E. phosphorylate the PDH complex.
Show answer
Correct Answer: A
Feedback A: Protein kinase A is the specific example.
Feedback B:
Feedback C: Phosphorylated phosphorylase kinase is active.
Feedback D: Phosphorylated glycogen synthse in inactive.
Feedback E: The pyruvate dehydrogenase complex is inactive when phosphorylated, but its phosphorylation is mediated by ratios of acetyl CoA/ CoA, NADH/NAD, and ADP/ATP, not cAMP.
242. The immediate substrate used by glycogen synthetase is...A. UMP-glucoseB. Glucose-1-phosphateC. Glucose-6-phosphateD. UDP-glucoseE. Glucose
Show answer
Correct Answer: D
Feedback A: See D
Feedback B: See D
Feedback C: See D
Feedback D: glucose ---> glucose-6-phosphate ---> glucose-1-phosphate ---> UDP-glucose ---> glycogen
Feedback E: See D
243. Superoxide anion is a free radical which can be converted to non-radical products by the action ofA. superoxide hydrolase.B. catalase.C. superoxide dismutase (SOD) and catalase.D. superoxide hydrolase and catalase.E. ferric iron.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: superoxide ---> hydrogen peroxide ---> water
Feedback D:
Feedback E:
244. Which of the following is NOT an accurate comparison between muscle and liver glycogen phosphorylase?A. Both are increased by cyclic AMP.B. Glucagon and epinephrine increase glycogen phosphorylase in both tissues.C. Calcium partially activates muscle glycogen phosphorylase, but not glycogen phosphorylase in liver.D. Muscle and liver glycogen phosphorylase are fully active after phosphorylation by cyclicAMP dependent protein kinases.E. Glycogen phosphorylase is a monomer in muscle but exists as a dimer in liver.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Glucagon acts on the liver, epinephrine on muscle.
Feedback C:
Feedback D: Increased cAMP ---> active protein kinase A ---> active phosphorylase kinase ---> active glycogen phosphorylase
Feedback E:
245. A simple definition of "oxidative stress" isA. it involves the oxidation of biological tissues by free radicals.B. it always causes cancer.C. it is the "stress" that follows extreme exercise.D. it describes the over saturation of hemoglobin.E. it results in the formation of methemoglobin.
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
246. During gluconeogenesis, oxaloacetate crosses the inner mitochondrial membrane...A. after it's converted to glycerol phosphate.B. by passive diffusion.C. after decarboxylation to pyruvate.D. after reduction to malate.E. by active transport using ATP as an energy source.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Malate dehydrogenase reduces oxaloacetate to malate in the mitochondria and an isoenzyme reforms oxaloacetate in the cytosol.
Feedback E:
247. Which of the following forces does NOT contribute substantially to the stabilization of the tertiary conformation of globular proteins?A. van der Waals interactionsB. hydrogen bondingC. stacking of aromatic ringsD. electrostatic interactionsE. hydrophobic effect
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
248. Which statement about the pyruvate dehydrogenase complex is FALSE?A. A product of PDH catalysis is acetyl-CoA.B. PDH is found in the matrix of the mitochondria.C. Kinase catalyzed phosphorylation decreases the activity of PDH.D. Acetyl CoA and NADH both stimulate PDH.E. PDH catalyzes a the first reaction where carbon dioxide is formed from glucose.
Show answer
Correct Answer: D
Feedback A: Pyruvate + CoA + NAD+ ---> acetyl CoA + CO2 + NADH
Feedback B:
Feedback C: A kinase phosphorylates PDH inactivating it; a phosphatase can remove the phosphate reactivating PDH.
Feedback D: Acetyl Coa and NADH are products of the reaction PDH catalyzes. They cause feedback inhibition.
Feedback E: No CO2 is produced by glycolysis.
249. Which is NOT a common secondary structural elements in globular proteins?A. alph helices
B. parallel Beta strandsC. reverse turnsD. triple helical bundlesE. antiparallel Beta sheets
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Collagen has a triple helix structure and is a fibrous protein.
Feedback E:
250. Reducing equivalents formed during totally aerobic glycolysis in a normal liver cell...A. enter the mitochondria as protons.B. are used in the pentose phosphate pathway.C. are transported into mitochondria in the form of glycerol-3-phosphate.D. enter the mitochondria as NADH.E. reduce pyruvate to yield lactate and NAD.
Show answer
Correct Answer: C
Feedback A: Reducing equivalents donate electrons to the electron transport chain.
Feedback B: NADPH is produced by the HMP pathway.
Feedback C: The glycerophosphate shunt.
Feedback D: NADH cannot cross the mitochondrial membrane.
Feedback E: This happens under anaerobic conditions.
251. Which of the following statements does NOT describe the structure of globular proteins?A. the surface is hydrophilicB. the interior is hydrophobicC. polar side chains are hydrogen bonded to other groups or waterD. side chains are densely packed within interiorE. reverse turns are buried to prevent solvent contact
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Most reverse turns are at the surface and consist of hydrophilic residues.
252. Bacteria like E. coli can grow with doubling times shorter than the amount of time it normally takes
to copy their genome becauseA. Their genomes are circular, so they do not need telomeraseB. They can reinitiate replication before completing a previously intiated replication cycleC. They can increase the rate of replication fork progression in response to nutrient availabilityD. Only a small part of the bacterial genome is essential so only part needs to be duplicatedE. They can activate multiple origins, decreasing the amount of time it takes to copy the genome
Show answer
Correct Answer: B
Feedback A: This is true, but irrelevant
Feedback B: Yes, by reinitiating before completing a round, bacterial cells can overlap replication cycles
Feedback C: This would work, but it isn’t true
Feedback D: The entire genome must be accurately duplicated for subsequent viability, so this is not true
Feedback E: Eukaryotes use multiple origins per chromosome, but prokaryotes (especially those with circular genomes) do not commonly do this, and E. coli can’t
253. The synthesis of glucose from pyruvate by gluconeogenesisA. occurs exclusively in the cytosol.B. is inhibited by an elevated level of glucagon.C. requires the participation of biotin.D. involves lactate as an intermediate.E. requires the oxidation/reduction of FAD.
Show answer
Correct Answer: C
Feedback A: No, in the mitochondria, cytosol, and ER.
Feedback B: No, gluconeogenesis is stimulated by glucagon due to decreased levels of fructose-2,6-bisphosphate.
Feedback C: Yes, biotin is a coenzyme that provides CO2 to carboxylate pyruvate, forming oxaloacetate.
Feedback D: Just plain no.
Feedback E: Requires NAD.
254. What is NOT true of the peptide bondA. limited rotation of the peptide bondB. planarC. partially charged groupsD. bond between alpha carbon and carbonyl carbonE. resonance stabilized
Show answer
Correct Answer: D
Feedback A: True due to partial double bond character.
Feedback B: True due to partial double bond character.
Feedback C: True due to charged contributing resonance structures and electronegativity differences.
Feedback D: False, peptide bonds are between the carbonyl carbon and the amino nitrogen.
Feedback E: True, two important resonance structures
255. Which one of the following is NOT a characteristic of gluconeogenesis?A. It requires energy in the form of ATP or GTP.B. It is important in maintaining blood glucose during the normal overnight fast.C. It uses carbon skeletons provided by degradation of amino acids.D. It consists of all the reactions of glycolysis functioning in the reverse direction.E. It involves the enzyme fructose 1,6-bisphosphatase.
Show answer
Correct Answer: D
Feedback A: True, 4 ATP and 2 GTP.
Feedback B:
Feedback C: Most notably alanine.
Feedback D: No, the irreversible reactions of glycolysis require unique reactions in gluconeogenesis.
Feedback E: This enzyme catalyzes the following: fructose 1,6-bisphosphate --> fructose 6-phosphate. Phosphofructokinase catalyzes the reverse reaction.
256. The sequence of the complementary DNA strand for the sequence 5'-GCTAGCATACGAC-3' is:A. 5'-CAGCATACGATCGB. 5'-CGATCGTATGCTGC. 5'-GTCGTATGCTAGCD. 5'-GCTAGCATACGACE. None of these are correct
Show answer
Correct Answer: C
Feedback A: This is antiparallel, but identical, not complementary
Feedback B: This is complementary, but not antiparallel
Feedback C: This is both complementary and antiparallel, and is correct
Feedback D: This is identical; it is neither complementary nor antiparallel
Feedback E:
257. All of the following can directly or indirectly stimulate the activity of muscle glycogen phosphorylase except…A. calcium ions.B. epinephrine.C. cyclic AMP.D. active phosphorylase kinase.E. glucagon.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glucagon acts primarily on liver and does not affect muscle.
258. What condition or compound does NOT modulate the extent of oxygen binding in hemoglobin?A. carbon dioxide concentrationB. oxygen tensionC. ascorbate concentrationD. concentration of 2,3-bisphosphoglycerateE. pH
Show answer
Correct Answer: C
Feedback A: Increased CO2, decreased oxygen affinity.
Feedback B: Increased PO2, increased saturation.
Feedback C:
Feedback D: Increased BPG (DPG), decreased oxygen affinity.
Feedback E: Increased H+, decreased oxygen affinity.
259. Fetal hemoglobin (Hb) binds oxygen with greater affinity than adult hemoglobin due primarily to which fact?A. the heme Fe in fetal Hb is ferric ironB. fetal Hb has a lower affinity for 2,3-DPGC. fetal Hb binds 2 hemes per subunitD. fetal Hb is monomericE. fetal Hb cannot form salt bridges
Show answer
Correct Answer: B
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
260. Which is most effective in inhibiting the citric acid cycle?A. fructose-2,6-bisphosphate in the mitochondria matrix..B. a high ratio of ADP/ATP in the mitochondria matrix.C. a high ratio of NADH/NAD in the mitochondria matrix.D. a high concentration of citrate in the mitochondria matrix.E. glycerol-3-phosphate in the matrix.
Show answer
Correct Answer: C
Feedback A: Fructose-2,6-bisphosphate in the cytosol stimulates phosphofructokinase-1 and inhibits fructose-1,6-bisphosphatase.
Feedback B: High ratios of ATP/ADP dampen TCA cycle.
Feedback C: High NADH/NAD inactivates PDH and alpha-ketoglutarate dehydrogenase.
Feedback D: High citrate in the cytosol inhibits PFK-1
Feedback E: Glycerol-3-phosphate is a shuttle molecule for reducing equilvalents.
261. Salt bridges are key components in the transition from oxyhemoglobin to deoxyhemoglobin. Salt bridges are important for which reason?A. NaCl disruption of hydrogen bonds in oxyhemoglobinB. Electrostatic interactions stabilize deoxyhemoglobinC. NaCl displacement of 2,3-DPG in oxyhemoglobinD. Ferrous salts in hemoglobin bind oxygenE. Electrostatic interactions enhance the affinity of oxygen and heme
Show answer
Correct Answer: B
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Enhanced affinity aids in transition from deoxy to oxyHb, and salt bridges are broken.
262. Which is a positive allosteric effector of pyruvate carboxylase?A. glucose-6-phosphateB. acetyl-CoAC. citrateD. ADPE. ATP
Show answer
Correct Answer: B
Feedback A:
Feedback B: Acetyl CoA is an absolute requirement for the catalytic activity of pyruvate carboxylase.
Feedback C:
Feedback D:
Feedback E:
263. The heme-heme cooperativity in oxygen binding in hemoglobin initiates from which event?A. oxygen binding and movement of Fe atom into the heme planeB. oxidation of three critical cysteine residues in the alpha chainsC. oxygen competitive binding to the 2,3-DPG binding siteD. oxygen bridging of two adjacent hemesE. covalent bond formation between the four hemes in hemoglobin
Show answer
Correct Answer: A
Feedback A: This conformational change is propagated to the other subunits.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
264. Which statement about phosphofructokinase-1 is FALSE?A. It is a major point of control over the rate of glycolysis.B. ATP is a substrate.C. AMP is a negative effector.D. ATP is a negative effector.E. It catalyzes a physiologically irreversible reaction.
Show answer
Correct Answer: C
Feedback A: It catalyzes the committed step of glycolysis.
Feedback B: ATP + fructose-6-phosphate ---> fructose-1,6-bisphosphate
Feedback C: AMP is a positive effector.
Feedback D:
Feedback E:
265. Beta-thalassemia hemoglobin is best characterized by which of the following?A. rapid oxygen-induced Beta chain degradationB. no alpha chains yielding Beta4 tetramersC. substitution of a Tyr for a critical proximal His in Beta chainsD. excess alpha chains cause pathologyE. low solubility of Beta chains yielding Beta fibers
Show answer
Correct Answer: D
Feedback A:
Feedback B: alpha-thalassemia with severe anemia (HbH disease).
Feedback C:
Feedback D: Little or no Beta-chains. Excess alpha chains precipitate.
Feedback E:
266. "Respiratory control" over oxidative phosphorylation states that electron flow and ATPformation stop when which compound is absent?A. ATPB. ADPC. NADHD. FADHE. oxygen
Show answer
Correct Answer: B
Feedback A:
Feedback B: Electrons do not typically flow through electron transport unless ADP is simultaneously phosphorylated to ATP.
Feedback C: NADH and oxygen are required for oxidative phosphorylation but respiratory control is the need for ADP.
Feedback D:
Feedback E: See C
267. What is NOT true about the structure of collagen fibers?A. tropocollagen is stabilizes as a helical bundle of 4 chainsB. hydrogen bonds from hydroxyl prolines are importantC. propeptides in each chain are cleaved to form the fibrilsD. glycines appears at nearly every third position in the sequenceE. Crosslinking of lysine and hydroxylysine residues stabilize the fibers
Show answer
Correct Answer: A
Feedback A: Superhelix of 3 parallel helical strands.
Feedback B:
Feedback C: Propeptides are removed from procollagen to form tropocollagen.
Feedback D:
Feedback E:
268. An enzyme absent in skeletal muscle is...A. lactate dehydrogenase.B. hexokinase.C. pyruvate dehydrogenase.D. glucose-6-phosphatase.E. glycogen synthetase.
Show answer
Correct Answer: D
Feedback A: Anaerobic glycolysis
Feedback B: glycolysis
Feedback C: TCA cycle
Feedback D: An enzyme for gluconeogenesis which is carried out in the liver and kidney.
Feedback E: Glycogen synthesis.
269. Which statement describes how enzymes catalyze reactions?A. enzymes shift the equilibrium constant favoring product formationB. enzymes lower the activation energy barrier separating reactants from productsC. enzymes destabilize the transition state by binding more strongly to productsD. enzymes transfer energy to reactants making them reactiveE. enzymes increase the temperature of the reaction
Show answer
Correct Answer: B
Feedback A: Enzymes do not alter equilibrium.
Feedback B: Usually by stabilizing intermediates.
Feedback C: They stabalize the transition state.
Feedback D:
Feedback E:
270. Major products formed during the pentose phosphate pathway are...A. ATP and NADH.B. Ribose and NADH.C. Ribose-5-phosphate and NADPH.D. Glucose-6-phosphate and NADPH.E. Ribose-5-phosphate and NADH.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Ribose-5-phosphate is used for nucleotide synthesis and NADPH for fatty acid synthesis.
Feedback D: Glucose-6-phosphate and NADP are reactants in the HMP pathway.
Feedback E:
271. Serine proteases contain a catalytic triad important in catalysis. Which statement is true about the role of the triad residues?A. Asp protonates the active SerB. Ser is the active nucleophileC. Tyrosine provides a hydrophobic pocketD. The covalent intermediate is formed with HisE. Asp and Tyr form the substrate binding site
Show answer
Correct Answer: B
Feedback A: Asp stabalizes the charged His.
Feedback B: Ser-OH + His ---> Ser-O- + His-H+
Feedback C: His, Asp, and Ser make up the catalytic triad.
Feedback D: Serine forms a covalent bond to the peptide.
Feedback E: See C.
272. The ultimate effect of cyclic AMP on liver phosphofructose kinase-2 (PFK-2) is...A. that it phosphorylates PFK-2 which increases its phosphatase activity and slows glycolysis.B. that it dephosphorylates PFK-2 which increases its phosphatase activity and slows glycolysis.C. that it increases the concentration of fructose-2,6-bisphosphate.
D. enhances ATP inhibition of PFK-1.E. decreases ATP inhibition of PFK-2.
Show answer
Correct Answer: A
Feedback A: increased glucagon ---> increased cAMP ---> phosphorylated PFK-2 ---> decreased fructose-2,6-bisphosphate ---> slows glycolysis.
Feedback B: Dephosphorylated PFK-2 is a kinase that catalyzes: fructose -6-phosphate ---> fructose -2,6-bisphosphate which stimulates PFK-1.
Feedback C: See A
Feedback D: High levels of fructose-2,6-bisphosphate can overcome inhibition of PFK-1 by ATP.
Feedback E: See D
273. In noncompetitive kinetics, what is the effect of excess substrate?A. prevents binding of the inhibitorB. falls to alter the effect of the inhibitorC. yields maximal velocityD. excess substrate maximizes the inhibitory effect of the inhibitorE. increases Km
Show answer
Correct Answer: B
Feedback A: See B.
Feedback B: Because the inhibitor binds to a site different from the substrate binding site, the level of substrate does not affect the inhibitors binding.
Feedback C: Vmax in decreased.
Feedback D:
Feedback E: Km is unchanged.
274. The "committed step" of the citric acid cycle involves the conversion of...A. oxaloacetate to citrate.B. the conversion of citrate to isocitrate.C. the conversion of isocitrate to alpha ketoglutarate.D. the conversion of fumarate to malate.E. the conversion of succinyl-CoA to succinate.
Show answer
Correct Answer: C
Feedback A: This reaction is regulated and essentially irreversible, but not the committed step.
Feedback B: A reversible reaction
Feedback C: This decarboxylation is essentially irreversible. The reaction is stimulated by ADP and inhibited by ATP.
Feedback D: A reversible reaction
Feedback E: A reversible reaction.
275. What best describes the Km term in Michaelis-Menton kinetics?A. substrate concentration that yields maximal velocityB. the reaction rate U of breakdown of the ES complex to productC. a ratio of rate constants (k2+k3)/ K1D. the concentration of active enzymeE. a thermodynamic constant unique for the given enzyme
Show answer
Correct Answer: C
Feedback A: Substrate concentration that yields 1/2 Vmax.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
276. Coenzyme Q is unique transporter of electrons during oxidative phosphorylation because...A. it is freely mobile in the hydrophobic inner mitochondrial membrane.B. it is the point where cyanide interupts electron tranport.C. it can directly accept electron from both FADH and NADH.D. it is the only part of the electron transport chain that is soluble.E. it is the final donor of electron to reduce molecular oxygen.
Show answer
Correct Answer: A
Feedback A: Also called ubiquinone, it transfers electrons from complexes I and II to complex III.
Feedback B: Cyanide, azide, and carbon monoxide all inhibit electron transport at the a3 of complex IV.
Feedback C: Electrons are transferred from NADH through complex I and from FADH2 through complex II to ubiquinone.
Feedback D:
Feedback E: Cytochrome a/a3 is the final electron donor.
277. Respiratory poisons like cyanide and carbon monoxide act by...A. disrupting the inner mitochondrial membrane.B. complexing the the irons of cytochromes.C. preventing ADP from interacting with inorganic phosphate.D. interfering with the transport of oxygen into the matrix of mitochondria.E. inhibiting ATP synthetase.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Cyanide and azide bind to the ferric form of a3, carbon monoxide binds to the ferrous form.
Feedback C:
Feedback D:
Feedback E:
278. Which two compounds will increase FIRST in liver following a rapid increase in circulating glucagon?A. pyruvate and oxaloacetateB. pyruvate and cAMPC. cAMP and glucose 1-phosphateD. oxaloacetate and cAMPE. glucose 1-phosphate and oxaloacetate
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Glucagon binding to its receptor increases cytosolic cAMP and causes glycogen breakdown to glucose-1-phosphate.
Feedback D:
Feedback E:
279. NADPH for fatty acid synthesis is directly supplied byA. isocitrate dehydrogenase.B. glycogen phosphorylase.C. alpha-ketoglutarate dehydrogenase.D. glucose-6-phosphate dehydrogenase.E. the transfer of reducing power from NADH to NADP in mitochondria.
Show answer
Correct Answer: D
Feedback A: A TCA cycle enzyme that does not use or produce NADPH.
Feedback B: Degradation of glycogen, does not use or produce NADPH.
Feedback C: Produces NADH.
Feedback D: HMP pathway, supplies NADPH and ribose-5-phosphate for synthesis of other biomolecules.
Feedback E: NADH is the major reducing equivalent in the mitochondria and NADPH predominates in the cytosol.
280. Increased levels of ATP in the fed state helps to channel acetyl CoA into fat synthesis by inhibitingA. citrate synthase.B. isocitrate dehydrogenase.C. citrate lyase.D. acetyl CoA carboxylase.E. palmitate thiokinase (CoA synthase).
Show answer
Correct Answer: B
Feedback A: Inhibited by high ATP but no a significant rate limiting step.
Feedback B: The committed step of TCA cycle is inhibited so acetyl CoA is not being used up by TCA cycle and citrate accumulates.
Feedback C: This enzyme is needed for fatty acid synthesis.
Feedback D: Needed for fatty acid synthesis.
Feedback E:
281. An important product of the pentose phosphate pathway isA. ribose phosphate.B. ATPC. NADPD. ADPE. NADH
Show answer
Correct Answer: A
Feedback A: Ribose-5-phosphate is uesed in nucleotide synthesis. NADPH is another important product.
Feedback B: ATP is not used or produced in this reaction.
Feedback C: NADPH
Feedback D:
Feedback E: NADPH
282. Substrates in the PENTOSE PHOSPHATE PATHWAY includeA. NADPH + glucose.B. glucose + NAD.C. glucose-6-phosphate + NAD.D. glucose-6-phosphate + NADP.E. ribose-5-phosphate + NAD.
Show answer
Correct Answer: D
Feedback A: glucose-6-phosphate + 2NADP + H2O ---> ribose-5-phosphate + 2NADPH + 2H+ + CO2.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
283. Which of the following is FALSE about a person whose skeletal muscle is unable to perform glycolysis?A. They will depend almost totally on fatty acid oxidation as an energy source during exercise.B. Gluconeogenesis will supply glucose to support active exercise.C. They will likely have hyperuricemia after a period of exercise.D. They will require an overload of glucose before and during exercise.E. They will exhibit considerable exercise intolerance.
Show answer
Correct Answer: B
Feedback A:
Feedback B: It is of no use to supply glucose because the muscle cannot perform glycolysis to obtain energy from the glucose. D is also false.
Feedback C:
Feedback D: You're right, but also see B.
Feedback E:
284. Elevated levels of circulating glucagon are associated with which one of the following?A. Increased activity of phosphofructokinase-2B. Decreased level of fructose 2,6-bisphosphatase activityC. Decreased fructose-1,6-bisphosphatase activityD. FastingE. Ingestion of a carbohydrate-rich meal
Show answer
Correct Answer: D
Feedback A: Incorrect, because in liver kinase activity is highest for dephospho-PFK-2, and glucagon leads to phosphorylation of the PFK-2.
Feedback B: Incorrect because phosphatase activity is highest for phospho-PFK-2, so glucagon would increase activity.
Feedback C: Increased activity because glucagon promotes gluconeogenesis.
Feedback D: Glucagon promotes glycogenolysis and gluconeogenesis to maintain adequate blood sugar levels between meals.
Feedback E: Glucose is absorbed and blood glucose levels rise, so insulin is needed to promote cell uptake of glucose.
285. Which amino acid is NOT ionized at physiological pH?A. glutamateB. arginineC. serineD. lysineE. aspartate
Show answer
Correct Answer: C
Feedback A: Conjugate base, negative charge
Feedback B: Base, positive charge
Feedback C: Polar and can be ionized, but not at physiological pH.
Feedback D: Base, positive charge
Feedback E: Conjugate base, negative charge
286. Which one of the following statements concerning glutamine is INCORRECT?A. Glutamine is responsible for the transport and storage of ammonia in a nontoxic form.B. The only fate of glutamine is hydrolysis to glutamate and ammonia.C. ATP is required for the reaction leading to the synthesis of glutamine from glutamate and ammonia.D. The kidneys can hydrolyze glutamine to glutamate and ammonia.E. Glutamine is found in tissue proteins.
Show answer
Correct Answer: B
Feedback A: Glutamine transports ammonia from tissues to the liver where urea is produced.
Feedback B: Incorrect because glutamine is also used in the sythesis of purines, pyrimidines, and proteins.
Feedback C: Glutamate + NH4+ + ATP --> glutamine + ADP
Feedback D: Glutaminase is found in kidney, although most glutamine is hydrolyzed in the liver.
Feedback E: Glutamine is one of the 20 basic amino acids.
287. Satellite sequences in human cellsA. are repeated sequences whose density is different from the bulk of the genomeB. are extrachromosomal elements found in cells that have amplified particular genesC. are repeated sequences that are interspersed throughout the genomeD. make up less than 1% of the total genomic DNA of humansE. are thought to be "junk" DNA without any obvious function
Show answer
Correct Answer: A
Feedback A: This is the definition of a satellite sequence
Feedback B: Such things exist but are called episomes or double minutes, not satellites
Feedback C: Satellites are repeated, but are clustered in tandem repeats, such as the alpha-satellites in humans found near centromeres
Feedback D: To be detected as a satellite, a sequence must be present enough times to create a noticeable alteration of the DNA profile in a cesium chloride gradient; 1% is not enough
Feedback E: The alpha satellite in humans has been associated with centromere function, and telomeric repeats have been found to be needed for telomere function.
288. Which one of the following statements about the urea cycle is correct?A. The two nitrogen atoms that are incorporated into urea enter the cycle as ammonia and alanine.B. Urea is produced directly by the hydrolysis of ornithine.C. ATP is required for the reaction in which argininosuccinate is cleaved to form arginine.D. Urinary urea is increased by a diet rich in protein.E. The urea cycle occurs exclusively in the cytosol.
Show answer
Correct Answer: D
Feedback A: The two nitrogen atoms come from ammonia and aspartate.
Feedback B: Hydrolysis of arginine.
Feedback C: Just plain no.
Feedback D: True because there is no real storage form for amino acids, so excess amino acids are metabolized, with the carbon chains being used for energy and the nitrogen removed through urea.
Feedback E: Mitochondria and cytosol.
289. Which is NOT true as to why glycine is a unique amino acid?
A. no carbon side chainB. commonly found in reverse turnsC. more diverse phi,psi angles possible than other residuesD. typically found every third residue in collagenE. prefers cis-peptide bond
Show answer
Correct Answer: E
Feedback A: True, the R group is a hydrogen.
Feedback B: True, no side-chain hindrance allows unusual dihedral angles.
Feedback C: True, no side-chain hindrance allows unusual dihedral angles.
Feedback D: True. In the collagen triple helix structure every 3rd residue faces the crowded center, only Gly will fit because it has no carbon side-chain.
Feedback E: False. In all cis peptides mutual hindrance between neighboring side-chains is severe.
290. Which one of the following statements about the synthesis of carbamoyl phosphate by carbamoyl phosphate synthetase I is INCORRECT?A. The enzyme catalyzes the rate-limiting reaction in the urea cycle.B. The reaction is allosterically activated by N-acetylglutamate.C. The reaction requires two high-energy phosphates for each carbamoyl phosphate molecule synthesized.D. The enzyme incorporates CO2 into carbamoyl phosphate.E. The reaction is reversible.
Show answer
Correct Answer: E
Feedback A: See E.
Feedback B: See E.
Feedback C: See E.
Feedback D: CO2 + NH4+ + 2ATP --> carbamoyl phosphate + 2ADP.
Feedback E: When ATP is used to drive a reaction the reaction is essentially irreversible and usually regulated.
291. Synthesis of dTMP in human cells is inhibited by administeringA. 5-fluorouracilB. 5-fluoroadenineC. 5-fluoroguanineD. 5-fluorocytosineE. 5-fluorothymidine
Show answer
Correct Answer: A
Feedback A: Yes, this is processed to F-dUMP which is a suicide inhibitor of thymidylate synthase
Feedback B: See the note for choice A
Feedback C: See the note for choice A
Feedback D: See the note for choice A
Feedback E: See the note for choice A
292. In which one of the following tissues is glucose transport into the cell enhanced most by insulin?A. BrainB. LensC. Red blood cellsD. Adipose tissueE. Liver
Show answer
Correct Answer: D
Feedback A: See D.
Feedback B: See D.
Feedback C: See D.
Feedback D: Skeletal muscle and adipocytes are dependent on insulin for glucose uptake. Nervous tissue, hepatocytes, RBC’s, cornea are insulin independent.
Feedback E: See D.
293. Which modification reaction is NOT important in directing membrane association?A. myristylationB. palmitolytionC. ADP-ribosylationD. isoprenylationE. phosphoinositol-glycan
Show answer
Correct Answer: C
Feedback A: Fatty acid addition, highly lipid soluble
Feedback B: Fatty ester addition, highly lipid soluble
Feedback C: Modification from NAD, not very lipid soluble or normally part of membranes.
Feedback D: Polyisoprene addition, highly lipid soluble
Feedback E: Glycolipid addition
294. Insulin does all of the following EXCEPTA. enhance glucose transport into muscle.B. enhance glycogen formation by liver.C. increase lipolysis in adipose tissue.D. inhibit gluconeogenesis in liver.E. enhance amino acid transport into muscle.
Show answer
Correct Answer: C
Feedback A: See C.
Feedback B: See C.
Feedback C: Insulin is high in the fed-state and promotes uptake and storage of nutrients. Lipolysis (fat breakdown) occurs when glucagon is high.
Feedback D: Gluconeogenesis is stimulated by glucagon.
Feedback E: See C.
295. LINEs are NOTA. About 6 kbp elementsB. Found typically in blocks of tandem repeatsC. Able to encode reverse transcriptaseD. Found in the human genomeE. Retrotransposons
Show answer
Correct Answer: B
Feedback A: The size of individual elements varies, but an intact element is about 6000 bp
Feedback B: LINEs are transposons that are distributed throughout the genome (the “I†is for �“interspersedâ€).�
Feedback C: Not all elements still encode active enzyme, but some do so the class of elements is able to encode RT
Feedback D: LINEs are not limited to humans, but are found in humans
Feedback E:
296. Which one of the following is NOT a direct action of insulin binding to the cell membrane?A. Activation of tyrosine kinase activity of the b-subunit of the insulin receptorB. Phosphorylation of the intracellular domain of the insulin receptorC. Phosphorylation of intracellular proteinsD. Activation of adenylate cyclaseE. Increased activity of and number of glucose transport molecules in the cell membrane
Show answer
Correct Answer: D
Feedback A:
Feedback B: Autophosphorylation.
Feedback C: The covalent modification changes the activity of enzymes.
Feedback D: Adenylate cyclase can be activated upon receptor binding of either glucagon or epinephrine, but the insulin receptor is a tyrosine kinase.
Feedback E: To bring about increased glucose uptake.
297. Side chain amino acid modifications commonly occur:A. to amino acids prior to charging of the tRNAB. to the amino acyl-tRNA complexC. to ribosome bound preprotein chainsD. to nascent or completed polypeptide chainsE. proteins pre-attached to the ER membrane
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Mostly proteins are made according to the mRNA template, then modified when completely translated (post-translational modification).
Feedback E:
298. Muscle glycogen cannot contribute directly to blood glucose levels becauseA. muscle glycogen cannot be converted to glucose 6-phosphate.B. muscle lacks glucose 6-phosphatase.C. muscle contains no glucokinase.D. muscle contains no glycogen phosphorylase.E. muscle lacks phosphoglucoisomerase.
Show answer
Correct Answer: B
Feedback A: False, see B.
Feedback B: Muscle can breakdown stored glycogen to glucose-6-phosphate but lacks glucose-6-phosphatase, so it cannot produce glucose.
Feedback C: True, but glucokinase phosphorylates glucose to glucose-6-phosphate, aiding in removal of glucose from the blood.
Feedback D: Just plain no. Muscle does have this enzyme to breakdown stored glycogen.
Feedback E:
299. If one strand of a 10 kbp duplex DNA molecule has 3000 G residues, the overall G+C content of the duplex isA. 15%B. 30%C. 45%D. 60%E. There is not enough information to determine the answer
Show answer
Correct Answer: E
Feedback A: See the note for choice “E†�; this value is below the minimum possible for the information given.
Feedback B: See the note for choice “Eâ€; this is the minimum possible value, but it could be higher�
Feedback C: See the note for choice “E†�
Feedback D: See the note for choice “E†�
Feedback E: If one strand has 3000 Gs, the other strand has 3000 Cs, but the total G+C content is undetermined since the first strand could have 0-7000 Cs. So the G+C content is a minimum of 30%, but could be as high as 100%. Made you think, though, didn’t I?
300. Which one of the following is NOT characteristic of the hexose monophosphate pathway?A. It produces CO2.B. It is controlled by inhibition of glucose-6-phosphate dehydrogenase by NADPH.
C. In requires ATP for phosphorylation.D. It produces ribose 5-phosphate.E. It involves the breakage and formation of C-C bonds.
Show answer
Correct Answer: C
Feedback A: 6 carbon glucose-6-phosphate looses C1 to become ribulose-5-phosphate, a 5 carbon sugar.
Feedback B: Feedback inhibition.
Feedback C: No ATP is consumed or produced.
Feedback D: See A.
Feedback E: See A.
301. Scurvy arises from the failure of which post-translational modification?A. hydroxylation of ProB. hydroxylation of LysC. gamma-carboxylation of GluD. S-nitrosylation of CysE. N-terminal myristylation
Show answer
Correct Answer: A
Feedback A: Lack of vitamin C leads to an inability to hydroxylate Pro, and thus collagen cannot form the appropriate H-bonds to make fibrils.
Feedback B: HydroxyLys is common in collagen, but not invloved with scurvy.
Feedback C:
Feedback D:
Feedback E:
302. Which one of the following reactions does NOT consume or produce NADPH?A. Reduction of oxidized glutathioneB. Synthesis of steroidsC. Conversion of glucose 6-phosphate to 6-phosphogluconolactoneD. Microsomal hydroxylation with the cytochrome P-450 oxygenase systemE. Oxidative phosphorylation
Show answer
Correct Answer: E
Feedback A: Glutathione reductase catalyzes the following reaction: G-S-S-H + NADPH + H+ ---> 2G-SH + NADP+
Feedback B: Desmolase catalyzes: cholesterol + NADPH + O2 ---> pregnenolone + NADP+
Feedback C: First step in the hexose monophosphate pathway; produces NADPH.
Feedback D: Cytochrome P-450 catalyzes: R-H + O2 + NADPH + H+ ---> R-OH + H2O + NADP+
Feedback E: NADH and FADH2(not NADPH) supply electrons to the electron transport chain,
303. Nucleosome coresA. contain about 300 bp of DNAB. contain DNA that is more sensitive to nuclease digestion than linkers between nucleosomesC. are stabilized by histone H1D. contain 2 copies of the histone H2AE. are found predominantly near centromeres and telomeres
Show answer
Correct Answer: D
Feedback A: A nucleosome core has 146 bp of DNA
Feedback B: The DNA between nucleosome cores, the linker DNA, is more sensitive to nuclease digestion than the DNA in the core.
Feedback C: H1 stabilizes the 30 nm fibers that are formed by interactions among adjacent nucleosome cores, H1 is not part of the core nucleosome.
Feedback D: Yes, as well as 2 copies each of H2B, H3, and H4.
Feedback E: Virtually all DNA in a eukaryotic cell is incorporated into nucleosomes; if anything they are less prominent at specific structures like these because other protein complexes are found in these areas.
304. Which one of the following conditions decreases the oxidation of acetyl CoA by the citric acid cycle?A. A low ATP/ADP ratioB. Low NADH due to rapid oxidization to NAD+ through the respiratory chainC. A low NAD+/NADH ratioD. High concentration of AMPE. Low GTP/GDP ratio
Show answer
Correct Answer: C
Feedback A: When ATP is low, pyruvate dehydrogenase complex is activated, and more acetyl CoA is put into TCA cycle. Citrate synthase, isocitrate dehydrogenase, and alpha-ketoglutarate dehydrogenase are also activated when ATP is low.
Feedback B: See C.
Feedback C: Pyruvate dehydrogenase complex is in the inactive form when phosphorylated. The kinase, which phosphorylates the PDC, is active when CoA/acetyl CoA or NAD+/NADH ratios are low.
Feedback D:
Feedback E: High GTP inhibits alpha-ketoglutarate dehydrogenase.
305. Which is NOT true about structures of globular proteins?A. packing density of nearly 75%B. surface hydrated with water moleculesC. peptide bond carbonyl and amide groups hydrogen bondedD. tertiary structures consist of layers of secondary structuresE. strong hydrophobic bonds in reverse turns
Show answer
Correct Answer: E
Feedback A: True. See B
Feedback B: True. The inner portion has packed hydrophobic residues and the surface has more
hydrophilic residues.
Feedback C: True, making alpha helices and beta sheets
Feedback D: True
Feedback E: False, turns are usually near the surface and hydrophilic.
306. An uncoupler of oxidative phosphorylation such as dinitrophenolA. inhibits electron transport and ATP synthesis.B. allows electron transport to proceed without ATP synthesis.C. inhibits electron transport without impairment of ATP synthesis.D. specifically inhibits cytochrome b.E. acts as a competitive inhibitor of NAD+-requiring reactions in the mitochondrion.
Show answer
Correct Answer: B
Feedback A: See B.
Feedback B: H+ ions are being pumped from the matrix side to the cytosolic side of the inner mitochondrial membrane, but the uncouplers transfer H+ back to the matrix side bypassing the ATP synthase.
Feedback C: See B
Feedback D: See B
Feedback E: See B
307. A mutated virus is found to replicate its genome faster than it used to. It is likely that it now alsoA. has a faster rate of transcriptionB. has a larger genome, since it can copy more DNA in the same amount of timeC. has a higher mutation frequencyD. is more resistant to antibioticsE. has a narrower host specificity
Show answer
Correct Answer: C
Feedback A: The rates of transcription and replication are largely independent
Feedback B: A larger genome would take longer to copy. Since the mutation rate is probably higher now (see “Câ€) the mutational load is higher, and the genome is not likely to get larger.�
Feedback C: Yes; less accuracy is a common consequence of faster replication. Typically, the DNA polymerase spends less time determining whether the most recent catalytic cycle was accurate, so it goes faster but makes more mistakes.
Feedback D: If we interpret “antibiotics†broadly enough to include � “antiviralsâ€, a virus with a �higher mutation rate (see choice C) could acquire resistance at a higher frequency, but this would have to be an indirect consequence of the direct effect of higher mutation rates. So C is a better answer.
Feedback E: Who knows, I needed a fifth answer and this sounded like it could happen, or would at least make you think about it.
308. GlucokinaseA. is widely distributed and occurs in most mammalian tissues.B. has a high Km for glucose and, hence, is important in the phosphorylation of glucose primarily after ingestion of a carbohydrate-rich meal.C. is inhibited by glucose 6-phosphate.
D. levels are decreased by carbohydrate-rich diets.E. catalyzes a readily reversible reaction.
Show answer
Correct Answer: B
Feedback A: Glucokinase is found is liver and B-cells. Hexokinase is widely distrubuted.
Feedback B: Hexokinase has a low Km and phosphorylates glucose when glucose levels are much lower.
Feedback C: Hexokinase is inhibited by glucose- 6-phosphate, but not glucokinase.
Feedback D: High carbohydrate consumption increases the amount of glucokinase.
Feedback E: The reaction involves the transfer of a high energy phosphate from ATP and is regulated and essentially irreversible.
309. The alpha helix and the collagen helix differ in many ways. Which of the following statements in NOT true in how they differ?A. alpha helix is more elongated than the collagen helix implying a greater rise per residueB. the alpha helix consists of a single polypeptide chain unlike the triple stranded helix in collagenC. hydrogen bonding in the alpha helix is within a chain unlike hydogen bonding between chains in the collagen helixD. all amide and carbonyl groups are hydrogen bonded to other residues within the alpha helix instead of some hydrogen bonding to solvent in the collagen helixE. alpha helices are usually limited to three helical turns, whereas the collagen helix extends much longer
Show answer
Correct Answer: A
Feedback A: False, the collagen helix is more elongated with a rise per residue of 2.9 compared to a rise per residue of 1.5 for an alpha helix.
Feedback B: True
Feedback C: True
Feedback D: True
Feedback E: True
310. The reaction catalyzed by phosphofructokinaseA. is activated by high concentrations of ATP and citrate.B. uses fructose-1-phosphate as a substrate.C. is the rate-limiting reaction of the glycolytic pathway.D. is near equilibrium in most tissues.E. is inhibited by fructose 2,6-bisphosphate.
Show answer
Correct Answer: C
Feedback A: PFK-1 is inhibited by high ATP and citrate.
Feedback B: No, the reaction is: fructose-6-phosphate ---> fructose- 1,6-diphosphate.
Feedback C: Yes, it is the committed step of glysolysis.
Feedback D:
Feedback E: Fructose-2,6-bisphosphate and AMP activate PFK-1.
311. Before catalyzing the incorporation of a nucleotide into a growing DNA chain, DNA polymerase checks forA. the appropriate formation of hydrogen bonds between the incoming base and the template baseB. the formation of stacking interactions between the incoming base and the adjacent primer baseC. the absence of rare tautomeric shifts in the incoming base that can alter base pairingD. solvent access of the catalytic site to ensure the rapid escape of pyrophosphate after the reactionE. the proper distance between the incoming alpha phosphate and the template phosphodiester group
Show answer
Correct Answer: E
Feedback A: Sounds reasonable, and you may even have been told this before, but it just isn’t true and I tried to make this clear in my class.
Feedback B: Stacking or hydrophobic interactions are very important energetically, but have no specificity and are not a concern of DNA polymerase.
Feedback C: Polymerase is not capable of performing this check, which is why it makes most of the mistakes it makes.
Feedback D: Sounds important, made it up.
Feedback E: Odd, but true. Nucleotide analogs incapable of hydrogen bonding, but with the right shape, are incorporated by DNA polymerase.
312. Compared to the resting state, vigorously contracting muscle showsA. an increased conversion of pyruvate to lactate.B. decreased oxidation of pyruvate to CO2 and water.C. a decreased NADH/NAD+ ratio.D. decreased concentration of AMP.E. decreased levels of fructose 2,6-bisphosphate.
Show answer
Correct Answer: A
Feedback A: “Vigorously†is the key word here. The muscle has a high energy demand that is not �being met by oxidative phosphorylation so anaerobic glycolysis increases.
Feedback B: No, the cell is using glycolysis and TCA cycle to fuel oxidative phosphorylation, so pyruvate oxidation increases.
Feedback C: No, because glycolysis and TCA cycle (if aerobic) are producing more NADH.
Feedback D: Increased ADP and AMP because ATP is being dephosphorylated.
Feedback E:
313. Which does NOT contribute to stability of the alpha helix?A. hydrogen bonding across diameter of helixB. stagger of side chains pointing outward from helix axisC. van der waals contact across the helix diameterD. aligned hydrogen bonds of amides and carbonyl oxygensE. packing against neighboring secondary structural elements
Show answer
Correct Answer: A
Feedback A: False, H-bonding is parallel to the axis of the helix.
Feedback B: True, decreases side-chain hindrance.
Feedback C: True
Feedback D: True, See A
Feedback E: True
314. Which one of the following statements concerning gluconeogenesis is correct?A. It occurs in muscle.B. It is stimulated by fructose-2,6-bisphosphate.C. It is inhibited by elevated levels of acetyl CoA.D. It is important in maintaining blood glucose during the normal overnight fast.E. It uses carbon skeletons provided by degradation of fatty acids.
Show answer
Correct Answer: D
Feedback A: Gluconeogenesis occurs 90% in the liver and 10% in kidney.
Feedback B: PFK-1, and therefore glycolysis, is stimmulated by fructose-2,6-bisphosphate.
Feedback C: The pyruvate dehydrogenase complex is the enzyme inhibited by high acetyl CoA. Acetyl CoA is essential for pyruvate carboxylase in the first step of gluconeogenesis.
Feedback D:
Feedback E: The carbon skeletons come from amino acids.
315. During renaturation of separated nucleic acid strands, the slowest step in the process isA. Lowering the temperature to the proper rangeB. Lowering the pH to the proper rangeC. The initial contact between two small complementary patchesD. Extension of the initial contact to adjacent regions ("zippering")E. Establishment of the hypochromic shift.
Show answer
Correct Answer: C
Feedback A: This is very rapid
Feedback B: This very rapid
Feedback C: Yes, contact is slow, determined by concentration of the partners, and rate-limiting
Feedback D: This is fast, see C
Feedback E: The hypochromic shift is an instantaneous consequence of the base stacking that occurs during annealing
316. In human liver, the initial step in the utilization of fructose is its phosphorylation to fructose 1-phosphate. This is followed byA. phosphorylation to fructose 1,6-bisphosphate.B. cleavage of fructose 1-phosphate to form glyceraldehyde and dihydroxyacetone phosphate.C. conversion to fructose 6-phosphate by action of a phosphofructomutase.D. isomerization to glucose 1-phosphate.E. hydrolysis to fructose followed by isomerization to glucose.
Show answer
Correct Answer: B
Feedback A: This is the fate of fructose 6-phosphate in glycolysis.
Feedback B: Catalyzed by aldolase B.
Feedback C: Phosphoglucomutase isomerizes glucose 6-phosphate and glucose 1-phosphate in glycogen synthesis and degradation.
Feedback D: See C.
Feedback E: Just plain no.
317. Which is NOT an important allosteric regulators of hemoglobin oxygenation?A. hydrogen ion concentrationB. CO concentrationC. chlorideD. DPGE. CO2 concentration
Show answer
Correct Answer: B
Feedback A: increased [H+] ---> decreased O2 affinity
Feedback B: CO binds at the same site as O2, and is thus not an allosteric interaction but a competative interaction.
Feedback C: Stabilizes the alpha chains in deoxyhemoglobin.
Feedback D: increased DPG ---> decreased O2 affinity
Feedback E: increased [CO2] ---> decreased O2 affinity
318. The enzyme in muscle which is most directly stimulated by epinephrine isA. glycogen synthase.B. phosphofructokinase.C. isocitrate dehydrogenase.D. glucose-6-phosphatase.E. adenylate cyclase.
Show answer
Correct Answer: E
Feedback A: This enzyme is inactivated due to epinephrine-induced phosphorylation.
Feedback B: Loss of stimmulation due to ephinephrine- induced decreases in the level of fructose 2,6-bisphosphate.
Feedback C: Indirectly affected.
Feedback D: This enzyme is found in liver and kidney and is involved in gluconeogenesis.
Feedback E: Ephinephrine binding to its receptor activates adenylate cyclase.
319. The best evidence that DNA carries genetic information isA. it can transform the properties of cells in heritable waysB. it can be duplicated using the information in either strandC. it is a chemically stable polymer
D. it comprises most of the material injected into cells by some virusesE. information in DNA is found to be copied into RNA
Show answer
Correct Answer: A
Feedback A: This is not only the best experiment in terms of clean results, it is conceptually the real evidence that DNA carries genetic information, because it shows that genetic information is altered by changing the DNA.
Feedback B: True, and useful for genetic material, but not evidence that DNA carries information
Feedback C: True, and useful for genetic material, but not evidence that DNA carries information
Feedback D: Often cited, but tainted by the “most†t� erm; this is not as powerful a demonstration as choice “A†�
Feedback E: True, and an indication that information may be present, but not actually evidence that DNA carries information
320. All of the following can stimulate the activity of muscle glycogen phosphorylase EXCEPTA. calcium ions.B. epinephrine.C. cyclic AMP.D. phosphorylase kinase.E. glucagon.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Affects the liver and kidney, but not muscle.
321. What is NOT required for oxygenation of Hb?A. one heme bound per subunitB. iron atom in ferric valence stateC. rupture of at least eight salt bridgesD. oxygen partial pressure in excess of 5 mm HgE. decrease in the packing density between alpha1/beta2 chains
Show answer
Correct Answer: B
Feedback A: Full oxygenation is when all four of the subunits have bound one oxygen molecule.
Feedback B: Ferrous state (Fe 2+), not ferric.
Feedback C: Changes that facilitate cooperativity.
Feedback D: The first oxygen molecule is the most difficult to bind, and then Hb’s affinity for oxygen increases with each additional oxygen bound.
Feedback E: Conformational changes facilitating cooperativity.
322. Which mitochondrial enzyme requires acetyl-CoA as a substrate?A. Citrate synthaseB. Succinyl-CoA synthaseC. Succinic dehydrogenaseD. Pyruvate dehydrogenaseE. Isocitrate dehydrogenase
Show answer
Correct Answer: A
Feedback A: In the first step of TCA cycle acetyl CoA + oxaloacetate ---> citrate
Feedback B: Requires CoA, but not acetyl CoA.
Feedback C:
Feedback D: Produces acetyl CoA.
Feedback E:
323. The synthesis of dTMP from its immediate precursor involvesA. reduction of rTMPB. methylation of dUMPC. a reaction catalyzed by ribonucleotide reductaseD. deamination of dCMPE. salvage of thymidine with the activated sugar derivative PRPP
Show answer
Correct Answer: B
Feedback A: No such thing except as a post-transcriptional modification in existing RNA
Feedback B: Yes, the reaction catalyzed by thymidylate synthase
Feedback C: No, RNR acts earlier during conversion of rCDP or rUDP to dCDP or dUDP which eventually provide the dUMP that is the immediate precursor of dTMP.
Feedback D: This produces dUMP which is the immediate precursor for dTMP synthesis
Feedback E: This can happen at a very low frequency, but is not a major pathway for producing dTMP
324. Allosteric control most often occurs through changes inA. enzyme levels.B. substrate concentrations.C. Vmax.D. Km.E. phosphorylation/dephosphorylation.
Show answer
Correct Answer: D
Feedback A: This is not allosteric control of the enzyme. Enzyme levels are determined by control of transcription, translation, and/or degradation.
Feedback B: Allosteric control modifys the affinity or catalytic activity of the enzyme in response to levels of specific molecules. It is the change in affinity or activity that is the control.
Feedback C: Could occur but it is not the most common mode of control.
Feedback D: Allosteric control usually involves a change in the enzyme’s affinity for the substrate.
Feedback E: This is a covalent modification rather than an allosteric control.
325. An abnormal hemoglobin (Hb) was identified with reduced DPG binding. What is the physiological consequence?A. reduced oxygen binding to HbB. reduced Bohr effectC. enhanced methemoglobin formationD. increased turnover of hemoglobinE. less oxygen transfer to tissues
Show answer
Correct Answer: E
Feedback A: False, decreased DPG binding ---> increased O2 affinity of Hb
Feedback B: Irrelevant because the Bohr effect involves H+ and CO2.
Feedback C: Sounds impressive but see E.
Feedback D: Sounds impressive but see E.
Feedback E: decreased DPG binding ---> increased O2 affinity of Hb, so O2 is not as readily released to the tissues.
326. Which is a positive allosteric effector of pyruvate carboxylase?A. Glucose 6-phosphateB. Acetyl-CoAC. CitrateD. ADPE. ATP
Show answer
Correct Answer: B
Feedback A:
Feedback B: Pyruvate carboxylase catalyzes the 1st step of gluconeogenesis: pyruvate + biotin-CO2 ---> oxaloacetate where CO2 is added to biotin only when acetyl CoA is bound to pyruvate carboxylase.
Feedback C: Citrate promotes gluconeogenesis by stimulating fructose 1,6-bisphosphatase.
Feedback D:
Feedback E:
327. In human cellsA. the replication or S phase lasts about 8 hoursB. replication initiates at about 4000 sites per cell called originsC. replication and segregation overlap to allow shorter division cyclesD. origins fire in groups of about 1000 replicons called batteriesE. replication is continuous on both leading and lagging strands
Show answer
Correct Answer: A
Feedback A: This is a typical value for a typical human cell
Feedback B: Human cells activate about 40,000 origins of replication
Feedback C: This occurs in prokaryotes, but not in eukaryotes, including humans
Feedback D: Batteries represent 10-100 origins that fire coordinately
Feedback E: NO known replication system has continuous lagging strand synthesis.
328. What change from normal is likely to result from the ingestion of ethanol?A. Increased reduction of pyruvateB. Increased conversion of b-hydroxybutyrate to acetoacetateC. Decreased ketogenesis due to lack of acetyl-CoAD. Increased production of oxaloacetate from malateE. Increased production of dihydroxyacetone phosphate from glycerol phosphate
Show answer
Correct Answer: A
Feedback A: Metabolism of ethanol by the liver increases the cytosolic concentration of NADH which promotes the reactions: pyruvate + NADH ---> lactate + NAD+ oxaloacetate + NADH ---> malate + NAD+
Feedback B: Ethanol metabolism increases the cytosolic NADH levels and would promote the reverse reaction: acetoacetate + NADH ---> b-hydroxybutyrate + NAD+
Feedback C:
Feedback D: High NADH would promote the reverse reaction: Oxaloacetate + NADH ---> malate + NAD+.
Feedback E: Increased cytosolic NADH would promote the reverse reacion: dihydroxyacetone phosphate + NADH ---> glycerol phosphate + NAD+
329. Mutations that affect alpha1beta2 interface can increase the amount of dimeric hemoglobin. What is the consequence of dimeric Hb?A. reduced oxygenationB. increased DPG bindingC. increased oxygenationD. enhanced methemoglobin formationE. heme precipitation
Show answer
Correct Answer: C
Feedback A: See C
Feedback B: DPG cross-links beta-chains in the tetramer, does not effectively interact with a dimer.
Feedback C: Loose cooperativety and loss of DPG’s effect make the dimeric Hb more like myoglobin with high oxygen affinity.
Feedback D: Sounds important but see C.
Feedback E: Sounds important but see C.
330. Each of the following is generally true about NIDDM EXCEPT (BAD QUESTION, credit was given for 2 of the following)A. It is associated with polydypsia and polyuria.B. Insulin is not detected in serum.C. Obesity in NIDDM is common.
D. There is a strong genetic component.E. There are antibodies against normal insulin.
Show answer
Correct Answer: E
Feedback A: True, these are common symptoms of diabetes.
Feedback B: The answer expected was E, but credit was also given for this response. This is false because the pancreas retains B-cells and insulin levels are typically high in obese NIDDM patients.
Feedback C: True
Feedback D: True
Feedback E: This was the expected answer, but credit was also given for B. This is false because NIDDM does not involve viruses or autoimmune antibodies.
331. Direct reversal of thymine dimers using the energy obtained from incident light is catalyzed byA. a complex of UvrA, UvrB, and UvrCB. photolyaseC. the repairosomeD. thymidine deaminaseE. phytochromase
Show answer
Correct Answer: B
Feedback A: These are enzymes of Nucleotide Excision Repair (NER), not photoreversal
Feedback B:
Feedback C: I made this up
Feedback D: I made this up, thymidine doesn’t have anything to deaminate with.
Feedback E: I made this up
332. Catabolism of 100 g of carbohydrate, 25 g of fat, and 20 g of protein yields approximately how many Cals?A. 300B. 500C. 700D. 900E. 1100
Show answer
Correct Answer: C
Feedback A: See C
Feedback B: See C
Feedback C: 100g carb (4 Cals/g) + 25g fat (9 Cal/g) + 20 g protein (4 Cal/g) = 705 Cals
Feedback D: See C
Feedback E: See C
333. Most peptide bonds in naturally occurring proteins:A. preferentially adopt a trans configurationB. preferentially adopt a cis configurationC. show no preference for trans or cis configurationsD. never assume a cis configurationE. never assume a trans configuration
Show answer
Correct Answer: A
Feedback A: Double bond character of the peptide bond is more stable in a trans configuration.
Feedback B: In all cis peptides mutual hindrance between neighboring side-chains is severe.
Feedback C:
Feedback D:
Feedback E:
334. Which of the following is FALSE about glucagon?A. It decreases in blood as glucose increases.B. It promotes the breakdown of hepatic and muscle glycogen.C. It promotes gluconeogenesis in liver.D. It promotes lipase activity in adipocytes.E. It can stimulate adenylate cyclase.
Show answer
Correct Answer: B
Feedback A: True, as blood glucose rises insulin levels increase and glucagon levels decrease.
Feedback B: False, glucagon does not affect muscle.
Feedback C: True, in order to maintain blood glucose between meals.
Feedback D: True, to provide fuel molecules when glucose is low.
Feedback E: Glucagon binding to its receptor activates adenylate cyclase.
335. Lesch-Nyhan Syndrome is associated with the loss ofA. xanthine oxidaseB. adenylate cyclaseC. thymidylate synthaseD. hypoxanthine-guanine phosphoribosyl transferaseE. dihydrofolate reductase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
336. Cyclic GMP (unlike cyclic AMP) causes few notable metabolic changes, but cyclic GMP doesA. stimulate cyclic AMP phosphodiesterase(s).B. relax smooth muscle in vascular beds.C. cause the release of calcium from the sarcoplasmic reticulum.D. reverse effects of insulin.E. stimulate phosphatase-1.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Nitric oxide activates cytosolic guanylate cyclase. Increased cGMP then induces smooth muscle relaxation.
Feedback C:
Feedback D:
Feedback E:
337. Consider the peptide: Gly-Ser-Glu-Asp-Lys-Val-Pro At neutral pH, the overall charge on the peptide will be approximately:A. -3B. -1C. 0D. +1E. +3
Show answer
Correct Answer: B
Feedback A:
Feedback B: Side chain charges are: Gly (0)-Ser (0)-Glu (-1)-Asp (-1)-Lys (+1)-Val (0)-Pro(0)
Feedback C:
Feedback D:
Feedback E:
338. Glycogen degradation in muscle can be increased by each of the following EXCEPT forA. calcium ion.B. phosphorylated glycogen phosphorylase kinase.C. phosphorylated glycogen synthase.D. high epinephrine.E. increased cyclic AMP in the cell.
Show answer
Correct Answer: C
Feedback A: Fluxes of Ca2+ act as a mediator of contraction and energy use in muscle.
Feedback B: Epinephrine binds to receptor ---> cAMP levels increase ---> phosphorylation of glycogen phosphorylase kinase ---> phosphorylation of glycogen phosphorylase ---> glycogen degradation
Feedback C: Glycogen synthase is inactive when phosphorylated and thus not synthesizing glycogen, but
this does not increase degradation.
Feedback D: See B
Feedback E: See B
339. The stability of DNA duplexes is increased byA. increasing the temperature above 20 degrees CB. changing the pH to values near 7C. adding formamide or other non-aqueous solventsD. decreasing the concentration of saltsE. the hypochromic shift
Show answer
Correct Answer: B
Feedback A: Increasing the temperature decreases the stability of duplex DNA
Feedback B: Yes, DNA is most stable near neutral pHs because this optimizes the formation of hydrogen bonding interactions
Feedback C: DNA is stabilized by the hydrophobic effect which is less important as the solvent becomes less polar.
Feedback D: Higher ionic strengths stabilize DNA due to shielding of the repulsive interactions between the charged phosphodiester backbones. So decreasing the salt destabilizes DNA
Feedback E: The hypochromic shift is a consequence of formation of duplexes between two single strands. It does not influence the stability of the molecules, it is an effect of the formation of dsDNA.
340. The term "oxidative stress" refers toA. the fact that oxidative phosphorylation requires ADP (respiratory control).B. the formation of oxygen free radicals because of the incomplete reduction of molecular oxygen.C. the fact that oxidative phosophorylation requires molecular oxygen.D. the inability of aerobic glycolysis to meet energy demands during strenuous exercise.E. the uncoupling of oxidative phosphorylation.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Free radicals can initiate harmful chain reactions, and because an essential part of our metabolism is reducing O2 to H2O there is a constant stress on the cell to deal with oxygen free radicals.
Feedback C:
Feedback D:
Feedback E:
341. Enzymes use which of the following to lower the energy of activation?A. change in substrate polarityB. ATP hydrolysisC. substrate strainD. temperature variationE. solvation of substrate
Show answer
Correct Answer: C
Feedback A:
Feedback B: Can provide energy to reach activation energy.
Feedback C: Enzyme active sites are complementary to the transition states of the reacting species.
Feedback D: Increased temperature is a way to increase rates of reactions, but it is not how enzymes work.
Feedback E:
342. All of the following statements about the enzymic complex that carries out the synthesis of ATP during oxidative phosphorylation are correct EXCEPTA. It is located on the matrix side of the inner mitochondrial membrane.B. It is inhibited by oligomycin.C. It contains a proton channel.D. It can exhibit ATPase activity.E. It can bind molecular oxygen.
Show answer
Correct Answer: E
Feedback A: The proton conducting F0 unit is in the inner mitochondrial membrane and the ATP synthesizing F1 unit in on the matrix side.
Feedback B: Oligomycin binds to the F0 unit of the ATP-synthase preventing proton flux through the channel to the F1 unit.
Feedback C: See A
Feedback D: If the proper [H+] gradient is not present the enzyme can hydrolyze ATP instead of synthesizing it.
Feedback E: False. Cytochrome a/a3 of the electron transport chain contains a heme iron that binds O2.
343. Histones are usuallyA. Acidic, and therefore positively charged at neutral pHB. Acidic, and therefore negatively charged at neutral pHC. Basic, and therefore positively charged at neutral pHD. Basic, and therefore negatively charged at neutral pHE. Close to neutral, and therefore uncharged at neutral pH
Show answer
Correct Answer: C
Feedback A: See C
Feedback B: See C
Feedback C: Basic residues are protonated at neutral pHs and are therefore positively charged. Histones have a high proportion of basic residues.
Feedback D: See C
Feedback E: See C
344. The number of moles of substrate converted to product per minute per mol of enzyme is the:A. KmB. Vmax
C. turnover numberD. apparent KmE. Km/Vmax
Show answer
Correct Answer: C
Feedback A: Km is the concentration of substrate ([S]) which gives half maximal velocity (1/2 Vmax).
Feedback B: Vmax is the maximum velocity of product formation.
Feedback C: Can be stated several different ways but is basically: amount of substrate to product per enzyme amount per time.
Feedback D: Increase is Km in the presence of a competative inhibitor.
Feedback E: Slope of the Lineweaver-Burke plot.
345. Which of the following is NOT in the pyrimidine family?A. thymineB. cytosineC. uracilD. orotateE. hypoxanthine
Show answer
Correct Answer: E
Feedback A: This is a pyrimidine
Feedback B: This is a pyrimidine
Feedback C: This is a pyrimidine
Feedback D: Orotic acid is the base that is converted to a nucleotide during pyrimidine synthesis.
Feedback E: Hypoxanthine is a purine
346. Which of the following statements is NOT true? Allosteric enzymes ...A. show sigmoidal kineticsB. are often multisubunit structuresC. are often regulated by effector moleculesD. have at least two specific binding sitesE. subunits exhibit minimal interactions when the enzyme is oligomeric
Show answer
Correct Answer: E
Feedback A: True, Michaelis-Menten kinetics give hyperbolic curves, allosteric kinetics give sigmoid curves.
Feedback B: True, Hemoglobin (although not an enzyme) is a great example of allosteric control. Hb is a tetramer.
Feedback C: True. Hb oxygen affinity is affected by CO2, H+, and DPG binding at allosteric sites.
Feedback D: True, at least one allosteric site and an active site.
Feedback E: False, subunits can be highly interactive with each other, as in Hb.
347. Which of the following compounds would DIRECTLY serve as a substrate of the enzyme that converts ribonucleotides to deoxyribonucleotides?A. GDPB. UMPC. CMPD. dTMPE. ATP
Show answer
Correct Answer: A
Feedback A: RNR acts only on ribonucleoside diphosphates
Feedback B: RNR acts only on ribonucleoside diphosphates, this is a monophosphate
Feedback C: RNR acts only on ribonucleoside diphosphates, this is a monophosphate
Feedback D: RNR acts only on ribonucleoside diphosphates, this is a deoxy sugar and a monophosphate
Feedback E: RNR acts only on ribonucleoside diphosphates, this is a triphosphate
348. A noncompetitive inhibitor will affectA. KmB. VmaxC. Km and VmaxD. the apparent KmE. enzyme stability
Show answer
Correct Answer: B
Feedback A: Is not affected because the noncompetative inhibitor does not interfere with the binding of substrate to the enzyme.
Feedback B: Noncompetative inhibition cannot be overcome by an increase in [S], and thus Vmax is affected.
Feedback C: Only Vmax is affected. See A and B.
Feedback D: Km is higher in the presence of a competative inhibitor but unaffected by a noncompetative inhibitor.
Feedback E: Just plain no.
349. Which substance does NOT contribute to the ring atoms during the synthesis of purines?A. AspartateB. ArginineC. GlycineD. GlutamineE. N-10-tetrahydrofolate
Show answer
Correct Answer: B
Feedback A: See B
Feedback B: Sounds trivial, but I found this exact question on the 1991 board exam that I just received, so I threw it in.
Feedback C: See B
Feedback D: See B
Feedback E: See B
350. Enzymes catalyze reactions commonly byA. lowering the activation energyB. activating cofactorsC. shifting the equilibrium to a more favorable positionD. increasing the energy content of the substrateE. protecting substrates from solvent hydrolysis
Show answer
Correct Answer: A
Feedback A: Activation energy is lowered through binding energy and transition state stabilization.
Feedback B: No, many enzymes do not have cofactors.
Feedback C: Enzymes do not affect equilibrium.
Feedback D: No, they lower the energy of activation.
Feedback E: Just because there needs to be an E.
351. Purified ribonucleotide reductaseA. makes more dCDP in the presence of dTTPB. makes more dGDP in the presence of dATPC. makes less dUDP in the presence of ATPD. makes more dADP in the presence of dGTPE. makes less dGDP in the presence of dTTP
Show answer
Correct Answer: D
Feedback A: See the regulation chart in lecture 2, page 5
Feedback B: See the regulation chart in lecture 2, page 5
Feedback C: See the regulation chart in lecture 2, page 5
Feedback D: See the regulation chart in lecture 2, page 5
Feedback E: See the regulation chart in lecture 2, page 5
352. What is a possible defect in the Ehlers-Danlos syndrome?A. defective conversion of procollagen to tropocollagenB. failure to enzymatically form lysinonorleucine crosslinksC. defective hydroxylation of prolineD. defective disulfide bond formation in elastinE. inactivation of inhibitor of elastase
Show answer
Correct Answer: A
Feedback A: In some cases an abnormal procollagen peptidase does not remove the propeptides.
Feedback B: This is Lathyrism
Feedback C: This is Scurvy
Feedback D: Impressive sounding, but no.
Feedback E: Alpha1-antitrypsin deficiency can lead to emphysema due to destruction of alveolar walls by elastase.
353. Which of the following is a general feature of both DNA and RNA?A. Their secondary structures can include G-C base pairsB. They interact with histones to form nucleosomesC. they are usually found as helices formed from two separate chainsD. Phosphodiester bonds link the 5' and 2' hydroxyls of adjacent sugarsE. They are stable in high pH solutions
Show answer
Correct Answer: A
Feedback A:
Feedback B: Nucleosomes contain DNA, not RNA
Feedback C: dsRNA involving two separate strands does form and can be found in nature, but is not the usual case for RNA as it is for DNA
Feedback D: Both RNA and DNA contain phosphodiester backbones with 5’ and 3’ linkages, not 2’.
Feedback E: The secondary structure of DNA is unstable at high pH, and both the secondary and primary (phosphodiester backbone) structures of RNA are unstable at high pH
354. The histidine residue within the catalytic triad in chymotrypsin participates in the following:A. facilitates the protonation of Ser as the initial event in catalysisB. serves as a proton donor and acceptorC. serves only as a proton donorD. forms a salt bridge to the side chain carboxylate of the substrateE. hydrogen bonds to the carbonyl of the labile bond in the substrate
Show answer
Correct Answer: B
Feedback A: The initial event is the deprotonation of serine.
Feedback B: True due to its weakly basic character His can accept and donate H+ at physiological pH. Its imidazole ring can quite easily accommodate either state.
Feedback C: See B.
Feedback D:
Feedback E:
355. Production of thymidylate in human cells can be inhibited by administeringA. methotrexateB. folic acidC. acyclovirD. leucovorinE. dUMP
Show answer
Correct Answer: A
Feedback A: This blocks DHFR activity, preventing renewal of the required activated folate cofactor for thymidylate synthase
Feedback B: Folate is required for thymidylate synthase, it is not an antagonist
Feedback C: This is a chain-terminator if phosphorylated and therefore blocks DNA replication, but only in cells infected with herpesviruses (and therefore contain a thymidylate kinase that will work on this analog), and not by blocking thymidylate synthesis.
Feedback D: Leucovorin is an activated folate that rescues cells from the effects of DHFR inhibitors; it is an activator of thymidylate synthesis
Feedback E: dUMP is the substrate for thymidylate synthase and so would not be expected to block dTMP synthesis; in fact, just the opposite if it could get into cells.
356. Treatment of hyperuricemia, or gout, involves administering allopurinol whichA. prevents isomerization of uric acid from the keto to the insoluble enol formB. competes with urate for secretion by the kidneysC. inhibits xanthine oxidaseD. decreases the serum concentration of xanthineE. inhibits HGPRTase
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Yes, this decreases the formation of uric acid.
Feedback D:
Feedback E:
357. Which pair of oligonucleotide primers would amplify the following "target" sequence in a PCR experiment: 5'-GTATACGACC...target...TAGCATAGAC-3'A. 5'-TAGCATAGAC and 5'-GGTCGTATACB. 5'-GTATACGACC and 5'-TAGCATAGACC. 5'-GGTCGTATAC and 5'-GTCTATGCTAD. 5'-GTATACGACC and 5'-GTCTATGCTAE. 5'-GTATACGACC and 3'-TAGCATAGAC
Show answer
Correct Answer: D
Feedback A: Both primers point away from the target.
Feedback B: The first primer is correct, but the second anneals to the same strand and points away from the target.
Feedback C: The second primer is correct, but the first anneals to the same strand and points away from the target.
Feedback D: The sequence at the 5’ end of the given strand and the sequence at the 5’ end of the complementary strand.
Feedback E: The first primer is correct, but the second anneals to the same strand and points away from the target.
358. If you were to reanneal 1 microgram of denatured yeast DNA and 1 microgram of denatured human DNA separately in equal total volumes under equal solution conditions, you should expectA. the yeast DNA to decrease in total ultraviolet light absorbance faster than the human DNAB. the human DNA to increase in total ultraviolet light absorbance slower than the yeast DNAC. the yeast DNA sample to have fewer copies of any given sequence since the genome is smallerD. the human DNA sample to contain more RNA because the human genome has more genesE. both samples to anneal at the same rate because the concentration of DNA is the same
Show answer
Correct Answer: A
Feedback A: Yes, yeast DNA is less complex than human DNA, so any given sequence is present more times in the same total mass of DNA and will be higher in concentration in equivalent samples, and will therefore anneal more quickly. Formation of double stranded helices leads to a lowering of UV absorbance (the hypochromic shift) which will therefore occur faster with the yeast DNA.
Feedback B: No, annealing will be slower with the human DNA (see answer A), but during annealing UV absorbance is decreased not increased.
Feedback C: The genome is smaller, but this means that an equivalent mass of DNA will have more copies of the yeast genome.
Feedback D: It is true that the human DNA has more genes, but the DNA samples shouldn’t have RNA in them at all.
Feedback E: This would be true if both samples had the same complexity, but they don’t. See answer A.
359. You have probed a Southern blot with a labeled DNA probe but find that you get hybridization to too many "target" sequences. You correctly infer that you are getting hybridization to sequences similar to, but not identical to, the intended target. How would you increase the specificity of the hybridization?A. Decrease the hybridization timeB. Increase the hybridization temperatureC. Decrease the amount of organic solvent in the hybridization solutionD. Increase the concentration of salt in the hybridization solutionE. Decrease the stringency of the hybridization solution
Show answer
Correct Answer: B
Feedback A: No, this will reduce the signal from both the correct and incorrect targets.
Feedback B: Yes, this will increase the stringency and only allow correctly matched duplexes to remain.
Feedback C: No, the opposite would help to increase the stringency
Feedback D: No, this would stabilize duplexes, which would reduce the stringency
Feedback E: No, increasing the stringency is called for.
360. Bacterial fertility factorsA. are transferred from one cell to another, leaving the original host cured of the factorB. are limited to transfers between cells of the same speciesC. are toxic to pathogenic bacteria and are being developed as antibioticsD. are plasmids capable of conferring resistance to multiple antibiotics simultaneouslyE. induce bacteria to undergo meiosis and spore formation
Show answer
Correct Answer: D
Feedback A: They are transferred, but they are copied at the same time leaving both cells “infected†�
Feedback B: No, species boundaries can be crossed by these elements
Feedback C: See D
Feedback D: Disturbing but true
Feedback E: Well, they’re called fertility factors...but no, bacteria don’t do meiosis
361. The synthesis of dTTP, and therefore also DNA replication, is inhibited clinically by administering a precursor of a suicide substrate of thymidylate synthase called:A. methotrexateB. 5-fluorouridineC. acyclovirD. AllopurinolE. Both a and b are correct
Show answer
Correct Answer: B
Feedback A: No, this works, but acts by preventing cycling of folate forms, not by targeting thymidylate synthase.
Feedback C: No, this is a chain terminator effective against cells infected with herpesviruses.
Feedback D: No, this is an inhibitor of xanthine oxidase
Feedback E: No, A is incorrect.
362. A duplex DNA moleculeA. is about 30 nm in diameter and has about 6 base pairs per helical turnB. is about 10 nm in diameter and has about 10 base pairs per helical turnC. is about 2 nm in diameter and has about 10 base pairs per helical turnD. is about 10 nm in diameter and has about 16 base pairs per helical turnE. is about 2 nm in diameter and has about 16 base pairs per helical turn
Show answer
Correct Answer: C
Feedback A: See C
Feedback B: See C
Feedback C: Just checking for a sense of scale. Had a Biochemistry Ph.D. candidate fail to answer this question once and haven’t quite gotten over it.
Feedback D: See C
Feedback E: See C
363. Type II topoisomerase inhibitors that are used clinically as antibiotics includeA. leucovorinB. VP16C. 5-fluorocytosine
D. fluoroquinolonesE. Both b and c are correct
Show answer
Correct Answer: D
Feedback A: No, this a form of folate used to rescue patients from the harmful effects of methotrexate treatment.
Feedback B: No, this is a type II topo inhibitor, but only in eukaryotes.
Feedback C: No, this is an antibiotic, but it targets thymidylate synthase, not topo
Feedback E: No, neither is correct.
364. Mutations caused by one of the following agents almost always inactivate genes specifying protein products, but often do not affect the function of genes specifying only stable RNA products. Which agent is it?A. UV lightB. cis-platinC. 5-bromouracilD. Methyl-methane sulfonic acid (MMS)E. benzo[a]pyrene
Show answer
Correct Answer: E
Feedback A: Forms pyrimidine dimers, not frameshifts; see E
Feedback B: crosslinker, not a frameshifter; see E
Feedback C: mismatches due to tautomer shift stabilization, not frameshifts; see E
Feedback D: mismatches due to alkylation of bases, not frameshifts; see E
Feedback E: an intercalater and notorious frameshifter, this is much worse for ORFs (proteins) than structural RNAs.
365. An inhibitor of telomerase would be expected toA. cause telomeres to get shorterB. cause cells to age fasterC. promote growth of tumorsD. inhibit growth of fibroblastsE. promote immortalization of cultured cells
Show answer
Correct Answer: A
Feedback A: Yes, a telomerase inhibitor would prevent telomere elongation.
Feedback B: No, the correlation between age and telomere length has not been shown to be causal.
Feedback C: No, loss of telomerase should help to prevent tumor growth, although this is still being tested.
Feedback D: Yes, cells need to elongate their telomeres to grow indefinitely
Feedback E: No, immortalization requires activation of telomerase, not inhibition.
366. cis-platinA. blocks synthesis of folate and therefore prevents DNA replicationB. promotes solubiliztion of urate and therefore prevents attacks of goutC. is an intercalating agent and therefore affects protein coding regions more than structural RNAsD. causes thymine dimers and therefore can be used to target tumors defective in mismatch repairE. is a DNA crosslinker used clinically to target rapidly growing tumors
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Correct Answer: E
Feedback A: No, this is a crosslinker used to selectively kill rapidly growing cells
Feedback B: No, this is a crosslinker used to selectively kill rapidly growing cells
Feedback C: No, this is a crosslinker used to selectively kill rapidly growing cells
Feedback D: No, this is a crosslinker used to selectively kill rapidly growing cells
Feedback E:
367. Which amino acid can commonly functions as an acid and base in catalysis?A. SerB. TyrC. IleD. HisE. Lys
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Correct Answer: D
Feedback A:
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Feedback D: Histidine has an imidazole ring that accommodates switches between protonated and unprotonated states, and its pKa is close to physiological pH.
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368. Zymogens are commonly activated by:A. Addition of proteolipidB. Addition of glycolipidC. ProteolysisD. PhosphorylationE. ADP-ribosylation
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Correct Answer: C
Feedback A:
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Feedback C: Zymogens are inactive enzyme precursors that are commonly activated by cleavage of peptide bonds.
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369. Which is NOT true of the stability of proteins that are predominantly alpha helical in structure?A. Packing of alpha helical segmentsB. Hydrogen bonding between backbone amides of one helix and carbonyl oxygens of a second helixC. Hydrogen bonding between side chains of different helicesD. Packing of hydrophobic residues in the protein interiorE. Stabilizing interactions within reverse turns connecting two helices
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Correct Answer: B
Feedback A:
Feedback B: Hydrogen bonding occurs between backbone amides and carbonyl carbons that are 4 residues apart within the helix.
Feedback C: The side chains are staggered on the surface of the helix and can interact with side chains of other helices.
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370. Sickle cell anemia results from:A. Stability of oxy-hemoglobin in venous bloodB. Reduced DPG bindingC. Lack of heme binding to hemoglobin chainsD. Myoglobin precipitationE. Polymerization of deoxy-hemoglobin
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Correct Answer: E
Feedback A:
Feedback B: Fetal Hb has a lower affinity for DPG, and thus a higher affinity for oxygen.
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Feedback E: Substitution of glutamate by valine places a nonpolar residue on the outside of HbS, which binds to a sticky patch that is exposed on another deoxy Hb.
371. Hypoglycemia can results from a deficiency of each of the following EXCEPTA. fructose-1,6-bisphosphataseB. glucose-6-phosphataseC. phosphoenolpyruvate carboxykinaseD. pyruvate carboxylaseE. pyruvate dehydrogenase
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Correct Answer: E
Feedback A: Gluconeogenesis
Feedback B: Gluconeogenesis
Feedback C: Gluconeogenesis
Feedback D: Gluconeogenesis
Feedback E: All of the other enzymes listed are needed for gluconeogenesis. Pyruvate dehydrogenase is needed to transfer from glycolysis to TCA cycle.
372. The source of most ammonia excreted into the urine from the kidneys of most vertebrate animals isA. alanineB. asparagineC. glutamic acidD. glutamineE. arginine
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Correct Answer: D
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Feedback D: Glutamine from all tissues is transported to the liver where it is hydrolyzed to glutamate and ammonia, and urea is then synthesized.
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373. A deficiency of glucose-6-phosphate dehydrogenase most often appears clinically asA. hypoglycemiaB. hyperuricemiaC. hyperglycemiaD. hemolytic anemiaE. hyporicemia
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Correct Answer: D
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Feedback D: RBC’s are highly affected because their only means of generating NADPH is the HMP pathway, for which glucose-6-phosphate is needed.
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374. Insulin causes each of the following changes EXCEPTA. An increase in tyrosine kinase activityB. A decrease in gluconeogenesisC. An increase in triglyceride breakdown to fatty acidsD. A relative decrease in circulating glucoseE. An increase in glycogen synthesis
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Correct Answer: C
Feedback A: The insulin receptor’s cytosolic domain is a tyrosine kinase
Feedback B: Glucose levels are high, so the body does not need to synthesize glucose.
Feedback C: No, glucose is high an can be used for an energy source and excess is coverted to fatty acids for storage.
Feedback D: Due to the uptake of glucose into cells.
Feedback E: For storage of excess glucose.
375. The "committed step" of glycolysis occurs at which enzyme?A. Glucose 6-phosphateB. PFK-1C. GlucokinaseD. Fructose 1,6-bisphosphateE. Hexokinase
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Correct Answer: B
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Feedback B: Just memorize it.
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376. The reaction which requires a continuous supply of NAD+ for glycolysis isA. Isomerization of glucose 6-phosphateB. Phosphorylation of glucoseC. Oxidation of glyceraldehyde 3-phosphateD. Cleavage of fructose 1,6-bisphosphateE. Phosphorylation of fructose 6-phosphate
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Correct Answer: C
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Feedback C: Glyceraldehyde-3-phosphate is oxidized and NAD+ is reduced.
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377. The reaction which is inhibited by citrate isA. Oxidation of glyceraldehyde 3-phosphateB. Phosphorylation of fructose 6-phosphateC. Cleavage of fructose 1,6-bisphosphateD. Isomerization of glucose 6-phosphateE. Phosphorylation of glucose
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Correct Answer: B
Feedback A:
Feedback B: Phosphofructokinase is the enzyme for the committed step, and thus is highly regulated. Inhibitors: ATP, citrate. Stimulators: AMP, fructose-2,6-bisphosphate.
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378. An immediate substrate used by the enzyme glycogen synthase isA. GlucoseB. Glucose-6-phosphateC. Glucose-1-phosphateD. UMP-glucoseE. UDP-glucose
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Correct Answer: E
Feedback A: See E
Feedback B: See E
Feedback C: See E
Feedback D: See E
Feedback E: Glucose ---> glucose-6-phosphate ---> glucose-1-phosphate ---> UDP-glucose (glycogen synthase) ---> glycogen
379. Which of the following is FALSE about oxidative phosphorylation (ox-phos) occurring in intact mitochondria?A. Ox-phos will stop when ADP is depletedB. Ox-phos requires that protons be pumped out of the mitochondria matrix.C. ATP synthesis takes place as protons enter the mitochondrial matrix through a specific channel.D. The addition of 2,4-dinitrophenol will abolish electron transport.E. The addition of cyanide will inhibit electron transport
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Correct Answer: D
Feedback A: True. This is respiratory control.
Feedback B: Electron transport pumps H+ from the matrix to the intermembrane space and ATP is synthesized as protons move in the opposite direction.
Feedback C: The ATP synthase has a F0 subunit that is a channel and a F1 subunit that has enzymatic activity.
Feedback D: False. 2,4-dinitrophenol is an uncoupler. Electron transport still pumps H+ but the dinitrophenol carries the H+ back into the matrix, bypassing the ATP synthase.
Feedback E: Cyanide complexes with the ferric form of cytochrome oxidase, and CO complexes with the ferrous form.
380. Which of the following is FALSE about gluconeogenesis in liver?A. Enzymes from the mitochondria matrix and the cytoplasm are required.B. This anabolic pathways only occurs during prolonged starvation.C. This process does not occur in skeletal muscle.D. This process is indirectly stimulated by increased glucagon.E. Fatty acid oxidation supplies energy for this process.
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Correct Answer: B
Feedback A: Pyruvate carboxylase - mitochondria glucose-6-phosphatase - ER All other enzymes needed for gluconeogenesis - cytosol
Feedback B: False. It is critical during prolonged starvation, but it can play a part in maintaining blood glucose even between meals.
Feedback C: Gluconeogenesis occurs almost entirely in liver and kidney cells.
Feedback D: True. Glucagon causes the release of fatty acids which are converted to acetyl CoA and increase NADH levels. Increased acetyl CoA stimulates pyruvate carboxylase and increased NADH levels inhibit pyruvate dehydrogenase.
Feedback E: See D.
381. Which of the following about pyruvate dehydrogenase (PDH) is FALSE?A. This enzyme is in the mitochondrial matrix.B. Carbon dioxide is a product of this reaction.C. Acetyl CoA and NADH both stimulate PDHD. Specific phosphorylation of PDH decreases its catalytic activityE. A product of the reaction catalyzed by PDH is Acetyl CoA.
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Correct Answer: C
Feedback A:
Feedback B: pyruvate + CoA + NAD+ ---> acetyl CoA + CO2 + NADH
Feedback C: ATP, acetyl CoA, and NADH all inhibit pyruvate dehydrogenase.
Feedback D: True. High levels of ATP cause the phosphorylation.
Feedback E: See B
382. The major immediate source of urea isA. The hydrolysis of glutamine by glutaminaseB. The oxidative deamination of glutamic acid by glutamate dehydrogenase.C. The oxidative deamination of amino acids by kidney D-amino acid oxidase.D. The cleavage of arginine by arginase.E. The hydrolysis of asparagine by asparaginase.
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Correct Answer: D
Feedback A: Nitrogen is transported from the tissues to the liver as glutamine, and ammonia is released when glutaminase hydrolyzes glutamine. Not technically part of the urea cycle.
Feedback B: Not part of the urea cycle.
Feedback C: Not part of the urea cycle.
Feedback D: The final step of the urea cycle is arginine ---> urea + ornithine
Feedback E: Asparginase removes NH3 from asparagine. It is not part of the urea cycle.
383. All of the following are true about insulin EXCEPT forA. It contains two peptide chains.B. It increases in serum immediately after a carbohydrate rich meal.C. It promotes triglyceride synthesis.D. It promotes glycogen synthesis.E. It promotes gluconeogenesis
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Correct Answer: E
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Feedback E: Gluconeogenesis maintains blood glucose levels between carbohydrate intakes, while Insulin release is stimulated by the intake of carbohydrates.
384. Which of the following is the best description of lactose intolerance?A. The genetic deficiency of galactose-1-Pi-uridyl transferase.B. The genetic deficiency of lactaseC. The genetic deficiency of lactokinaseD. The genetic deficiency of fructokinase.E. The depletion of ATP in hepatocytes.
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Correct Answer: B
Feedback A: Classical galactosemia
Feedback B: The undigested lactose is metabolized by colonic bacteria producing gas and an osmotic gradient moves fluid into the intestine producing diarrhea.
Feedback C: Never heard of this enzyme
Feedback D: Essential fructosemia
Feedback E: No way
385. NIDDM (type 2) has each of the following characteristics (symptoms) EXCEPT forA. HyperglycemiaB. ObesityC. Near normal insulin in serumD. KetosisE. Polyuria
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Correct Answer: D
Feedback A: Types 1 and 2
Feedback B: Type 2
Feedback C: Type 2
Feedback D: Type 1
Feedback E: Types 1 and 2
386. Which statement about reactions catalyzed by transaminases is FALSE?A. The equilibrium constants are about 1.0.B. The reaction involves a Schiff base intermediate.C. Pyridoxal phosphate is covalently bount to the transaminase enzyme.D. Ammonia is liberated during the reactionE. These reactions are important for the biosynthesis of some amino acids.
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Correct Answer: D
Feedback A: This allows the reaction to function in both amino acid degradation and synthesis.
Feedback B:
Feedback C: Pyridoxyl phosphate (a derivative of vitamin B6) is a coenzyme for aminotransferases.
Feedback D: False. The amino group is transfered from an amino acid to an alpha-keto acid.
Feedback E: See A
387. The synthesis of carbamoyl phosphate is regulated most directly byA. UreaB. GlutamineC. N-acetyl glutamateD. AsparagineE. Ornithine
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Correct Answer: C
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Feedback C: NAG stimulates carbomyl phosphate synthase I.
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388. Which of the following statements is TRUE about the condition of fructosuriaA. Can be lethal in newborns if untreated.B. Reflects the genetic absence of fructokinase.C. Results in depletion of ATP in hepatocytesD. Reflects the absence of fructaseE. Results in hypoglycemia
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Correct Answer: B
Feedback A: Fructose intolerance, not fructosuria, can be lethal, which is the genetic deficiency of aldolase B.
Feedback B: This is a benign, asymptomatic condition.
Feedback C:
Feedback D: Never heard of this enzyme
Feedback E: Associated with fructose intolerance.
389. Each of the following can stimulate the activity of glycogen phosphorylase in muscle EXCEPTA. Calcium ionsB. EpinephrineC. Cyclic AMPD. GlucagonE. Active phosphorylase kinase
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Correct Answer: D
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Feedback D: Glucagon does not affect muscle.
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390. "Respiratory control" over oxidative phosphorylation reflects that electron flow and ATP synthesis stops when which compound is absent?A. ATPB. ADPC. NADHD. OxygenE. FADH
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Correct Answer: B
Feedback A:
Feedback B: Electrons go not typically flow through electron transport unless ADP is simultaneously phosphorylated to ATP.
Feedback C: NADH and oxygen are required for oxidative phosphorylation, but respritory control is the need for ADP.
Feedback D: See C
Feedback E:
391. Each of the following is an expected metabolic state after three days of starvation EXCEPTA. The ratio of glucagon to insulin is higher than in the absorptive stateB. Fatty acids are released from adipose tissueC. Gluconeogenesis takes place in liverD. Brain uses ketones alone as an energy sourceE. Liver and muscle use fatty acids as an energy source
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Correct Answer: D
Feedback A: Glucagon increases stimulating gluconeogenesis and fatty acid release from adipocytes.
Feedback B: See A
Feedback C: See A
Feedback D: During prolonged starvation some of the brain’s dependence on glucose shifts to ketone bodies, but the brain never loses its need for glucose.
Feedback E:
392. Each of the following statements about the hexose monophosphate pathway is correct EXCEPTA. It provides for the synthesis of NADPHB. It neither requires or generates ATPC. It allows fats to be converted into glucoseD. It takes place in the cytoplasmE. It provides for the synthesis of ribose from glucose
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Correct Answer: C
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Feedback C: False. HMP pathway provides NADPH needed for fatty acid synthesis.
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393. The chemiosmotic theory of oxidative phosphorylation states each of the following EXCEPTA. Oxidation and the phosphorylation of ADP are coupled by a proton gradientB. Protons are pumped from the matrix during electron transportC. Molecular oxygen is reduced during oxidative phosphorylationD. A protonmotive force drives ATP synthesisE. The matrix side of the membrane develops a lower pH relative to the cytoplasm side of the membrane
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Correct Answer: E
Feedback A:
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Feedback C: It is reduced to water.
Feedback D: H+ (protons) gradient driving the movement protons from the cytosolic to matrix side through the ATP synthase.
Feedback E: False. H+ is pumped from the matrix to the intermembrane space, so the pH in the intermembrane space is lower.
394. Which of the following enzymes is LESS ACTIVE when in a phosphorylated state?A. Glycogen phosphrylase
B. Triglyceride lipaseC. Pyruvate kinaseD. GlucokinaseE. Glycogen phosphorylase kinase
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Correct Answer: C
Feedback A: Increased glucagon induces phosphorylation of glycogen phosphorylase (stimulating) and phosphorylation of glycogen synthase (inhibiting).
Feedback B: Glucagon increases cAMP in adipocytes which activates the lipase. Glucagon also induces phosphorylation of acetyl CoA carboxylase decreasing synthesis of fatty acids.
Feedback C: Glucagon induces the phosphorylation of pyruvate kinase which inhibits glycolysis.
Feedback D: Glucokinase activity is under transcriptional control. Synthesis of glucokinase is decreased when glucagon is high.
Feedback E: Glucagon induced phosphorylation increases its activity.
395. Lactate dehydrogenase (LDH) catalyzed the reversible reduction of pyruvate:Pyruvate + NADH + H+----> Lactate + NAD+How exergonic is this reaction in the direction written given the following E0' for the half reactions? (Use a Faraday = 23.0 Cal/mol Ko)(1) Pyruvate + 2 H+ + 2e- ---> Lactate -0.20(2) NADH + H+ + 2e- ---> NADH -0.30A. -0.5B. -2.3C. -4.6D. +0.5E. +4.6
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Correct Answer: C
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396. Which of the following is a FALSE statement about the metabolic process of glycolysis?A. Takes place in the cytosol of all mammalian cells.B. Glycolysis itself only provides about 20% of the energy obtained from glucose.C. May occur aerobically or anaerobically.D. This process converts a six carbon substrate into pyruvate.E. All reactions of glycolysis are reversed during gluconeogenesis
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Correct Answer: E
Feedback A:
Feedback B: The remaining 80% is obtained through TCA cycle and electron transport.
Feedback C: Aerobically the endproduct is pyruvate Anaerobically the endproduct is lactate.
Feedback D: See C
Feedback E: No, the three irreversible reactions of glycolysis are carried out in the reverse direction by unique enzymes..
397. Two enzymes required for gluconeogenesis areA. Glucokinase and hexokinaseB. Fructose-1,6-bisphosphatase and pyruvate carboxykinaseC. Pyruvate kinase and glucose-6-phosphataseD. Aldolase and ketolaseE. Phosphofructokinase and pyruvate kinase
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Correct Answer: B
Feedback A: Isoenzymes involved in glycolysis
Feedback B:
Feedback C: pyruvate kinase --glycolysis glucose-6-phosphatase --gluconeogenesis
Feedback D: Aldolase --glycolysis Ketolase--never heard of it
Feedback E: glycolysis
398. Which of the following is complementary to the DNA sequence 5'-CCATTAGACC?A. 5'-GGTAATCTGGB. 5'-CCATTAGACCC. 5'-CCAGATTACCD. 5'-GGTCTAATGGE. 5'-GCCGGTACCT
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Correct Answer: D
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399. Adenylate cyclase from all tissues isA. stimulated by glucagon.B. stimulated by norepinephrine.C. intrinsic to the plasma membrane.D. found in the cytosol.E. found in the mitochondrial matrix.
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Correct Answer: C
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400. Which is likeliest to have its structure altered by an SSB (single-strand binding protein)?A. double-stranded DNAB. single-stranded DNAC. double-stranded RNAD. single-stranded RNAE. an RNA:DNA hybrid
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Correct Answer: B
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401. Cyclic AMP causes metabolic effects in tissues byA. being converted to AMP.B. inhibiting protein phosphatase.C. stimulating protein kinases.D. binding to the allosteric site on phosphorylase.E. inhibiting tyrosine kinase.
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Correct Answer: C
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402. The "Hershey-Chase" experimentA. proved that only DNA can carry genetic informationB. proved that proteins never carry heritable genetic informationC. provided the backdrop for the later, influential, "Pulse-Chase" experimentD. showed that bacteria can obtain heritable information from dead cellsE. showed that bacteriophages injected mostly DNA into their target cells
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Correct Answer: E
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403. The most common genetic defect in humans is seen inA. Hb (Sickle cell anemia).B. glucose-6-phosphate dehydrogenase.C. liver glucose-6-phosphatase (Von Gierke's disease).D. muscle glycogen phosphorylase (McArdle's disease).E. muscle phosphofructokinase.
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Correct Answer: B
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404. You could encourage a sample of duplex DNA to denature byA. placing the sample at 90 degrees C.B. increasing the pH to 7C. raising the concentration of saltD. decreasing the concentration of formamideE. increasing the melting temperature
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Correct Answer: A
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405. Which of the following is a FALSE statement about the process of oxidative phosphorylation?A. It requires ADP.B. It requires a continuous supply of oxygen.C. The synthesis of ATP can be uncoupled from electron transport by 2,4-dinitrophenol.D. It requires NADPH and NADH to supply reducing equivalents.E. It takes place in the inner mitochondrial membrane.
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Correct Answer: D
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406. Allopurinol ameliorates the symptoms of gout by inhibiting which enzyme?A. xanthine oxidaseB. DNA polymeraseC. Thymidine kinaseD. HGPRTaseE. RNA polymerase
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Correct Answer: A
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407. Untreated maternal diabetes during pregnancy is generally associated with all of the following EXCEPT FORA. low infant birth weightB. maternal hyperglycemiaC. infant hypoglycemia at birthD. cesarean section deliveryE. above normal fetal insulin
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Correct Answer: A
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408. Which duplex DNA (only one strand is shown) would melt at the highest temperature?A. AATTAGCGCAATB. GCGCATTAGCGCC. GCATTAAATGCGD. GAGAGAGAGAGAE. CTCTCTCTCTCT
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Correct Answer: B
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409. Cyanide is lethal because itA. complexes to the iron of the heme in cytochrome oxidase.B. inhibits ATP-ADP translocase.C. complexes to the iron of the heme of methemoglobin.D. "uncouples" oxidative phosphorylation.E. complexes to ubiquinone.
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Correct Answer: A
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410. Defects in adenosine deaminase cause blockage of the function of which enzyme, leading to which disease?A. Thymidine synthase and HNPCCB. HGPRTase and GoutC. DNA polymerase and CancerD. RNR and SCIDE. Thymidylate kinase and Herpes
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Correct Answer: D
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411. Ubiquinone (Co-Q) is unique because it isA. the only carrier of single electrons.B. the only electron carrier soluble in the inner mitochondrial membrane.C. the only electron carrier directly inhibited by cyanide.D. the only water soluble electron carrier.E. the only carrier of two electrons.
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Correct Answer: B
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412. RNA is different from DNA, because RNAA. is destabilized at high concentrations of saltB. has thymine but no uracilC. is found in the A form when it is a duplexD. has no 2' hydroxylE. is stable in strongly alkaline solutions
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Correct Answer: C
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413. Following an overnight fast, during minutes after a meal of syrup and pancakes, serumA. insulin and glucagon increase.B. insulin increases and glucagon decreases.C. insulin decreases and glucagon increases.D. insulin increases and glucagon remains constant.E. insulin remains constant and glucagon increases.
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Correct Answer: B
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414. Fluorocytosine is useful as an antimicrobial becauseA. it is converted to F-dUMP which is a suicide inhibitor of thymidylate synthaseB. the thymidylate synthase in human cells is not affected by F-dUMPC. bacteria and fungi can convert the base to a nucleotide but human cells can'tD. A and C are both true, but not BE. A, B, and C are all true
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Correct Answer: D
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415. Which of the following is FALSE about the enzyme phosphofructokinase-2 (PFK-2)?A. This enzyme is in the cell cytosol.B. This is a bifunctional enzyme which can have either kinase or phosphatase activity.C. The product of this enzyme can alter the rate of glycolysis.D. This enzyme catalyzes the committed step of glycolysis.E. Fructose-6-phosphate is a substrate for this enzyme.
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Correct Answer: D
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416. Lesch-Nyhan syndromeA. is caused by a lack of hypoxanthine-guanine phosphoribosyl transferaseB. is associated with severe combined immune deficiency (SCID)C. is partially corrected by factors that increase the level of urateD. is treated with methotrexateE. is associated with increased rates of synthesis of pyrimidines
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Correct Answer: A
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417. A person with lactose intolerance likely has diminished activity ofA. lactase.B. fructokinase.C. aldolase-B.D. galactokinase.
E. phosphofructokinase.
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Correct Answer: A
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418. Assuming 8-mers are an adequate length for priming a PCR amplification, which primers will amplify the target below in a PCR? 5'-GACATCAG-target-CGTAAGCTA. 5'-GACATCAG and 5'-CGTAAGCTB. 5'-GACATCAG and 5'-GCATTCGAC. 5'-CTGTAGTC and 5'-GCATTCGAD. 5'-GACATCAG and 5'-AGCTTACGE. There is not enough information to answer this question
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Correct Answer: D
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419. NucleosomesA. are made of histones and about 72 bp of DNAB. are disrupted during the formation of the 30 nm fiberC. contain 2 copies each of H2A, H2B, H3, and H4D. cover about 70% of the DNA in a human cellE. contain 2 copies each of H1, H2, H3, and H4
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Correct Answer: C
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420. During glycogen metabolism, phosphatase activity is regulated in which way?A. Phosphatase is phosphorylated by protein kinase making it inactive.B. Inhibitor-1 is phosphorylated by protein kinase and the phosphorylated inhibitor inhibits phosphatase.
C. Cyclic AMP binds to an allosteric site on the phosphatase, and this inhibits phosphatase.D. AMP allosterically inhibits phosphatase.E. The phosphate on the inhibitor is hydrolyzed to give a active inhibitor.
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Correct Answer: B
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421. Which of following is NOT TRUE of histones?A. They are small, basic proteinsB. They are found in nucleosomes in a ratio of 1:1:1:1 H2A:H2B:H3:H4C. They are only associated with transcriptionally active DNAD. They form a core of proteins about which DNA wraps about 1.8 timesE. They are abundant in eukaryotic chromatin
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Correct Answer: C
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422. Which of the following is administered after treatment with dihydrofolate reductase inhibitors to prevent death due to the loss of active forms of folate?A. allopurinolB. folic acidC. methotrexateD. thymidylateE. leucovorin
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Correct Answer: E
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423. The citric acid cycle (TCA cycle) is regulated at which three steps?
A. Citrate synthase, aconitase, and isocitrate dehydrogenaseB. Citrate synthase, aconitase, and alpha ketoglutarate dehydrogenase.C. Aconitase, succinate dehydrogenase, and isocitrate dehydrogenase.D. Isocitrate dehydrogenase, alpha ketoglutarate dehydrogenase, and citrate synthetase.E. Fumarase, succinate dehydrogenase, aconitase.
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Correct Answer: D
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424. Leucovorin is administeredA. To inhibit xanthine oxidase and reduce uric acid productionB. To replace active forms of folate after treatment with dihydrofolate reductase inhibitorsC. To inhibit ribonucleotide reductase to block the growth of certain human pathogensD. To block DNA replication in cells infected with herpesvirusesE. To block the salvage of purine metabolites by human parasites
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Correct Answer: B
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425. The target of the antibiotics in the fluoroquinolone class isA. type II topoisomeraseB. the replicative helicaseC. ribonucleotide reductaseD. folate synthesisE. aspartate transcarbamoylase
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Correct Answer: A
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426. Which one of the following is a FALSE statement about the TCA cycle?A. It accounts for most of the carbon dioxide produced during the total catabolism of glucose.B. It depends on a continuous supply of NAD.C. GTP is formed during one reaction of the cycle.D. Oxygen is required at the committed step.E. ATP inhibits the committed step.
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Correct Answer: D
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427. BrdU is a nucleoside analog that is readily taken up by human cells and incorporated into DNA. If you add BrdU to human cells and let them replicate their DNA 3 times, what fraction of the DNA strands will have the analog in them (assuming no recombination)?A. 1 out of 2B. 3 out of 4C. 1 out of 4D. 7 out of 8E. 1 out of 8
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Correct Answer: D
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428. Which of the following is a FALSE statement about the pentose phosphate pathway?A. This takes place in the cytosol of most cells.B. Two products are NADPH and ribose phosphate.C. This alternative pathway generates more ATP than aerobic glycolysis.D. The flux through this pathway depends upon concentrations of NADP.E. This pathway is important in erythrocytes.
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Correct Answer: C
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429. If a bacterium is grown for many generations in the presence of 15-N ("heavy" nitrogen) then shifted to medium containing 14-N, after the DNA is copied four times, what fraction of the total DNA will be hybrid (H:L) in density?A. All of itB. 1/2C. 1/4D. 1/8E. 1/16
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Correct Answer: D
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430. Activating telomerase wouldA. make telomeres shorterB. maker telomeres longerC. make cells olderD. make cells youngerE. A and D are both correct
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Correct Answer: B
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431. Which of the following is FALSE about the metabolic intermediate citrate?A. Citrate is formed from Acetyl CoA and oxaloacetate in the mitochondrial matrix.B. Citrate activates pyruvate kinase to increase Acetyl CoA for the TCA cycle.C. Citrate enhances ATP inhibition of phosphofructokinase-1.D. The synthesis of citrate is allosterically inhibited by ATP.E. Citrate is permeable to the inner mitochondrial membrane
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Correct Answer: B
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432. How many POTENTIAL translational reading frames does a double-stranded DNA molecule have?A. 1B. 3C. 6D. 9E. 12
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Correct Answer: C
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433. A virus with a single-stranded RNA genome makes its way into a human cell and is taken directly to the translation machinery. How many potential reading frames does this nucleic acid have?A. 1B. 2C. 3D. 6E. 9
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Correct Answer: C
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434. Lactate is the final product of totally anaerobic glycolysis becauseA. the concentration of pyruvate becomes limiting.B. lactate is formed by lactic acid dehydrogenase.C. pyruvate kinase is inactive under anaerobic conditions so lactate is produced.D. the ratio of NADH/NAD becomes high so NADH is used to reduce pyruvate.E. the reaction catalyzed by pyruvate kinase requires oxygen.
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Correct Answer: D
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435. Which of the following mutagens is more likely to disrupt the function of a protein coding region than it is to disrupt the function of a region encoding a structural RNA?A. ultraviolet lightB. Bromo-deoxyuracil (BrdU)C. an alkylating agentD. ethidium bromideE. photolyase
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Correct Answer: D
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436. Deamination of cytosineA. forms uracilB. forms thymineC. is repaired by photolyaseD. is the most common germ line mutation in humansE. B and D are both correct
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Correct Answer: A
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437. Which of the following is a FALSE statement about cyclic AMP?A. This compound is formed from ATP.B. Cyclic AMP is converted to 5'-AMP by phosphodiesterase.C. Cyclic AMP activates protein kinase(s).D. Glucagon increases the concentration of cyclic AMP in hepatocytes.E. Insulin increases the concentration of cyclic AMP in muscle.
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Correct Answer: E
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438. When a human cell is damaged by ultraviolet light, the most likely mechanism of repair isA. by photolyaseB. by a mechanism using an AP endonucleaseC. by base excision repairD. by a mechanism using a MutH homologE. by nucleotide excision repair
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Correct Answer: E
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439. Which of the following mutagens is more likely to disrupt the function of a protein coding region than it is to disrupt the function of a region encoding a structural RNA?A. Ultraviolet lightB. Ethidium bromideC. An alkylating agentD. Bromo-deoxyuracil (BrdU)E. Photolyase
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Correct Answer: B
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440. Which one of the following is an ACCURATE comparison of the polypeptide hormones glucagon and insulin?A. Both increase cyclic AMP in their respective target tissues.B. Both bind to extracellular receptors in their respective target tissues.C. Each increases in blood in response to an increase in blood glucose.D. Insulin increases cyclic AMP in muscle and fat, glucagon only increases cyclic AMP in liver.E. Insulin, but not glucagon, increases cyclic AMP in fat.
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Correct Answer: B
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441. You could increase the stringency of a Southern blot procedure byA. Raising the temperatureB. Raising the concentration of NaClC. Lowering the concentration of formamideD. Increasing the concentration of detergentE. Increasing the concentration of agarose
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Correct Answer: A
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442. When a human cell is damaged by ultraviolet light, the most likely mechanism of repair isA. by photolyaseB. by a mechanism using an AP endonucleaseC. by base excision repairD. by a mechanism using a MutS homologE. by nucleotide excision repair
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Correct Answer: E
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443. Each of the following are accurate comparisons of NADPH and NADH EXCEPT FORA. Most NADPH is formed in the cytosol but most NADH in the mitochondrion.B. NADH is a direct substrate for oxidative phosphorylation, but NADPH is not.C. The reduction potentials for NADH is greater than that for NADPH.D. NADPH supplies reducing equivalents for mixed function oxidases, but NADH does not.E. NADPH has one more phosphate than NAD.
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Correct Answer: C
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444. Acyclovir is a chain terminator and is usefulA. to limit the size of PCR productsB. against herpesvirus infectionsC. for determining the sequence of a DNA moleculeD. to kill rapidly growing tumorsE. for treating psoriasis
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Correct Answer: B
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445. Which of the following is a FALSE statement about pyruvate?A. Pyruvate can be converted to lactate in a reaction requiring NADH.B. Pyruvate can be converted to glucose.C. Pyruvate dehydrogenase will convert pyruvate to Acetyl-CoA.D. Pyruvate reacts with oxaloacetate to yield citrate.E. Pyruvate contains 3 carbon atoms.
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Correct Answer: D
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446. Mutations in homologs of bacterial MutS and MutL in humans causeA. Xeroderma pigmentosumB. Hereditary non-polyposis colon cancerC. Ataxia telangiectasiaD. Bloom's SyndromeE. Fanconi's anemia
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Correct Answer: B
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447. Which one of the following enzymes is more active when in its dephosphorylated state?A. Pyruvate kinaseB. Glycogen synthaseC. Glycogen phosphorylaseD. A and B correctE. B and C correct
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Correct Answer: D
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448. Which primers will amplify the target in the sequence below in a PCR experiment? 5'-GATCCATA-target-ACCATACCA. 5'-GATCCATA and 5'-GGTATGGTB. 5'-TATGGATC and 5'-ACCATACCC. 5'-GATCCATA and 5'-ACCATACCD. 5'-ACCATACC and 5'-GGTATGGTE. 5'TATGGATC and 5'-GGTATGGT
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Correct Answer: A
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449. What is the function of a translocase in the inner mitochondrial membrane?A. To shuttle reducing equivalents from NADH formed during aerobic glycolysis.B. To transport ATP from the matrix to the cytosol.C. To transport ADP from the matrix to the cytosol.D. To allow the formation of a proton gradient.
E. To transport protons through ATPsynthase.
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Correct Answer: B
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450. If two homologous DNA molecules have sequences that can be represented as ABCDEF and abcdef participate in recombination,A. both products will be ABCDEFB. both products will be abcdefC. neither product will have any mismatches, only exchangesD. resolution in the patch configuration will produce ABCdEFE. resolution in the recombinant configuration will produce ABcdEF
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Correct Answer: D
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451. The chemiosmotic explanation for oxidative phosphorylation states thatA. ATP is synthesized as protons are pumped from the mitochondrial matrix.B. Protons are pumped from the inner to the outer side of the inner mitochondrial membrane during electron transport.C. The inner mitochondrial membrane is freely permeable to protons.D. Protons are pumped from the cytosolic side of the inner mitochondrial membrane into the matrix during electron transport.E. More that one of the above is correct.
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Correct Answer: B
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452. Which of the following is NOT true regarding mismatch repair in bacteria
A. the methylated strand is cut by MutHB. the mismatch is recognized by MutSC. the average repair patch is 100-500 basesD. the role of MutL is not currently knownE. the DNA strands are only hemi-methylated for a few minutes after replication
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Correct Answer: A
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453. Which of the following is TRUE about cytochrome c?A. This is the only soluble part of the electron transport chain.B. This is the only part of the electron transport chain which transfer single electrons.C. This is the "terminal" part of electron transport involved in the reduction of molecular oxygen.D. This is the only heme which can complex cyanide.E. This is the only cytochrome which doesn't have a protein component.
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Correct Answer: A
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454. All of the following methods can be used to diagnose an infection with HIV. Which will produce a reliable signal the soonest after infection?A. A Southern blotB. A Northern blotC. A Western blotD. PCRE. A microarray
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Correct Answer: D
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455. You are planning an expression construct to force the production of your new, improved version of human growth hormone (hGH) in E. coli. Which of the following steps would you NOT have to take in order to insure expression of this protein?A. Choose a vector with an origin of replication that functions in bacteria.B. Make sure that the splicing signals at the borders of all introns are intact.C. Provide a promoter for bacterial RNA polymerase in front of the hGH gene.D. Make sure that the coding sequence of hGH is fused in frame to bacterial translation initiation signals.E. All of the above are required.
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Correct Answer: B
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456. Which of the following is TRUE about creatine kinase-MB (CK-MB)?A. CK-MB only originates from cardiac muscle.B. CK-MB is one of three different isoenzymes of creatine kinase.C. This is the only dimeric isoenzyme of creatine kinase.D. In serum, CK-MB is converted into CK-BB.E. An increase in the concentration if CK-MB in serum always indicates a MI.
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Correct Answer: B
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457. Let's say you decided to get fancy and control expression of the hGH gene in this clone using the normal control sequences of the E. coli tryptophan (Trp) operon. What would happen to hGH expression if you added lots of free tryptophan to the growth medium of cells carrying this construct?A. hGH expression would go down.B. hGH expression would go up.C. Nothing would happen unless you restricted glucose in the medium as well.D. Transcription of hGH mRNA would proceed, but translation would be blocked.E. Nothing, since Trp operon expression is not affected by tryptophan levels.
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Correct Answer: A
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458. An above normal level of total serum creatine kinase (T-CK) could result from any of the following EXCEPTA. a fall off a ladderB. extreme exertionC. a good night's sleepD. blunt chest traumaE. a myocardial infarction
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Correct Answer: C
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459. Which of the following features is shared by both DNA synthesis and RNA synthesis?A. The polynucleotide product stays associated with the template strand after synthesis.B. Synthesis is usually bidirectional after initiation at a single site.C. A primer is required for the polymerase to initiate synthesis.D. Both strands are copied essentially simultaneously.E. Nucleotides are inserted stepwise according to the Watson-Crick pairing rules.
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Correct Answer: E
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460. CK-MB#1 (isoform #1) is described as an isoform of CK-MB. Where does CK-MB#1originate?A. It is formed from CK-BB. It is formed from CK-MMC. It is native to cardiac muscleD. It is native to skeletal muscleE. It originates from CK-MB#2, which is defined as the native form from muscle
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Correct Answer: E
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461. Which of the following modifications does NOT take place during the production of mature tRNAs?A. Rotation of a uracil base to produce pseudouridine.B. Methylation of a uracil base to produce (ribo)thymidine.C. PolyA addition at the 3' end.D. Addition of some complicated substituents, like an isopentenyl group to adenosine.E. Conversion of uridine to dihydrouridine (DHU).
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Correct Answer: C
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462. Rick weighs about 340 lbs, and his daily energy requirement is 3070 Cal. His nutritionist specifies a diet per day of 300 g protein, 200 g carbohydrate, and 30 g fat. The theoretical weight loss for Rick each day would be closest toA. 100 gB. 200gC. 50 gD. 800 gE. 454 g
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Correct Answer: A
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463. We looked at some electron microscope pictures of ribosomal RNA transcription in progress that looked sort of like Christmas trees. What feature of the underlying complex corresponds to the trunk that runs the length of the Christmas tree?A. The splicing machinery.B. The 45S rRNA precursor.C. The promoter for RNA polymerase I.D. The ribosomal proteins waiting to be assembled.E. The template DNA (or chromatin).
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Correct Answer: E
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464. Which metabolic pathway accounts for most oxygen consumption during catabolism of foods?A. Aerobic glycolysisB. Oxidative phosphorylationC. The citric acid cycleD. Beta oxidationE. Formation of methemoglobin
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Correct Answer: B
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465. The SECOND chemical step in the splicing of eukaryotic mRNA precursors is:A. addition of the 7-methylguanosine triphosphate cap.B. addition of the polyA tail.C. joining of two exons with the release of the lariat intron.D. attack of the branchpoint A on the upstream exon-intron junction.E. not a transesterification, unlike the first chemical step.
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Correct Answer: C
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466. Fats inherently contain more calories per gram than carbohydrates becauseA. they are easier to digestB. they are more reducedC. they are easier to catabolizeD. they are catabolized to carbon dioxideE. they are more oxidized
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Correct Answer: B
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467. Imagine that you have isolated a new gene that you suspect is involved in a human genetic disease. Upon examining the sequence of the gene and the protein you predict is encoded by it, you recognize three apparent Cys2His2 zinc finger motifs. What can you reasonably conclude from this?A. The protein is likely to be a transcription factor.B. There is probably a leucine zipper motif nearby in the protein sequence.C. The protein is probably a repressor of transcription.D. The protein is likely a receptor for a steroid hormone.E. All of the above.
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Correct Answer: A
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468. Which residue is an aromatic amino acid?A. ThrB. ProC. MetD. TyrE. His
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Correct Answer: D
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469. Which of the following is a ribozyme?A. RNA polymerase II.B. mRNA.C. Lac repressor.D. Self-splicing group I intron.
E. tRNA.
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Correct Answer: D
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470. High levels of blood glucagon activate gluconeogenesis and inactivate glycolysis byA. activating fructose 1, 6 bisphosphatase, inactivating glucokinase, and inducing the transcription of glucose 6-phosphatase.B. activating aldolase and lowering the concentration of fructose 2,6 bisphosphate.C. inactivating pyruvate kinase, lowering the concentration of fructose 2,6 bisphosphate and inducing the transcription of PEP carboxykinase.D. activating pyruvate carboxylase, inactivating phosphofructokinase 1, and inducing the transcription of PEP carboxykinase.E. none of the above.
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Correct Answer: C
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471. Which amino acid is NOT a hydrophobic residue?A. IleB. leuC. ArgD. ValE. Phe
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Correct Answer: C
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472. What is the most likely consequence of a mutation in an intron that does not change the consensus
sequences at the donor site, the acceptor site, or the branchpoint?A. Termination of transcription.B. Formation of a new splice donor site.C. Change of one amino acid in the corresponding protein.D. Increased level of expression of the mRNA.E. No effect.
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Correct Answer: E
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473. Gluconeogenesis occursA. only in liver mitochondria.B. only after extensive fasting.C. until muscle protein is exhausted.D. in the liver and, to a small extent, kidney.E. by glycolysis in reverse.
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Correct Answer: D
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474. Which residue is positively charged at physiological pH?A. SerB. CysC. LysD. AspE. Thr
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Correct Answer: C
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475. What feature is shared by promoters and enhancers in eukaryotic gene expression?A. There are TATA boxes in both promoters and enhancers.B. Both are binding sites for RNA polymerase II.C. It doesn't matter what their orientation is relative to the gene they control.D. Their effects are mediated by sequence-specific DNA-binding proteins.E. They always cause repression of transcription.
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Correct Answer: D
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476. Glucogenic amino acidsA. are often broken down to intermediates in the citric acid cycleB. include all amino acids except lysine and leucine.C. are released from muscle protein by low blood insulin levels.D. include essential and nonessential amino acidsE. all of the above.
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Correct Answer: E
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477. What factor is NOT an essential step in general protein folding?A. Cis-trans peptide bond isomerizationB. Protein disulfide exchangeC. Sequestration of hydrophobic groups away from solventD. Formation of inter-residue hydrogen bondsE. Formation of lysinenorleucine crosslinks
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Correct Answer: E
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478. Which of the following is NOT a characteristic of bacterial helix-turn-helix proteins that regulate gene expression?A. The recognition helix contacts DNA bases in the major groove.B. The helix-turn-helix motif is the dimerization domain of the proteins.C. The proteins typically bind to DNA as dimers, contacting consecutive major grooves on one face of the DNA duplex.D. Both the proteins and their DNA binding sites are typically two-fold (dyad) symmetric.E. Recognition involves interaction between amino acid side chains and specific base pairs.
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Correct Answer: B
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479. A deficiency in glucose 6-phosphataseA. is the only cause of type I glycogen storage disease (Von Gierke's disease).B. causes severe fasting hypoglycemia.C. blocks only gluconeogenesis.D. results in decreased lactate production.E. all of the above.
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Correct Answer: B
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480. What is important in the activation of zymogens?A. Proteolysis of a peptide bondB. ADP-ribosylationC. Glycolipid additionD. Formation of salt bridgesE. Activity of cis-trans proline isomerase
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Correct Answer: A
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481. Why does a hepatitis virus have a binding site for a liver-specific transcription factor?A. All human diseases arise in the liver.B. Transcription occurs only in the liver.C. The virus targets liver cells, and it needs to express its genes there.D. Liver-specific transcription factors are found in all cells.E. The virus is really an albumin gene gone bad.
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Correct Answer: C
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482. Why does the production of glucose require an energy input higher than that released during glycolysis?A. Because gluconeogenesis occurs both in the mitochondria and cytosol, and the shuttling of intermediates requires energy.B. This occurs so that the free energy change of the overall process remains negative.C. Because glycolysis has three completely irreversible steps.D. Because of the connection between gluconeogenesis and the citric acid cycle.E.
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Correct Answer: B
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483. What modification fails to occur in individuals afflicted with scurvy?A. Gamma- carboxylation of glutamatesB. ADP-ribosylationC. Hydroxylation of ProD. N-terminal myristylationE. Cleavage of signal sequence
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Correct Answer: C
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484. You are designing a new vector that is designed to allow regulated expression of genes in mammalian cells. Which of the following approaches has essentially NO hope of working?A. Put a bacterial terminator sequence in the middle of the gene, and control expression by modulating Lac repressor levels with IPTG.B. Put an iron response element from the transferrin receptor gene into the 3'-UTR of the expressed mRNA, and control mRNA stability with iron levels in the growth medium.C. Put a binding site for the glucocorticoid receptor in the promoter of the gene, and control transcription by applying or withholding glucocorticoid hormones.D. Put a binding site for the heat shock transcription factor in the promoter of the gene, and control transcription with temperature.E. All of the above will work.
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Correct Answer: A
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485. High levels of ethanol ingestion inhibit glucose biosynthesis byA. preventing the utilization of pyruvate, glycerol and intermediates of the citric acid cycle from being utilized as precursorsB. inhibiting the induction of PEP carboxykinase.C. increasing the NAD+/NADH ratio.D. producing acetate, from oxidation of ethanol in the liver, which inhibits pyruvate carboxylase.E. all of the above.
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Correct Answer: A
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486. Which is NOT true of protein secondary structures?A. Hydrogen bonding in an alpha helix occurs between residues within the helixB. Hydrogen bonding in beta sheets occurs primarily between side chains of alternating residuesC. Amphipathic helices have separate hydrophobic and hydrophilic surfacesD. Parallel beta sheets tend to have both surfaces hydrophobicE. Antiparallel beta sheets are often connected by short reverse turns
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Correct Answer: B
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487. What cellular component(s) is(are) responsible for recognizing splice sites in mRNA precursors?A. tRNAs.B. snRNPs.C. RNA polymerase II.D. PolyA polymerase.E. Steroid receptor proteins.
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Correct Answer: B
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488. Derivatives of vitamin B6 (pyridoxine) are essential for:A. the pyruvate carboxylase reaction.B. transaminase and decarboxylation reactions of amino acids.C. the hydroxylation of phenylalanine, tyrosine and tryptophan.D. all of the above.E. none of the above.
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Correct Answer: B
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489. In what way are amino acid side chains oriented in an alpha helix?A. Side chains project toward the interior of alpha helicesB. Side chains project outward from the helical axisC. Side chains alternate projecting inward and outwardD. Side chains show no preference in their orientationE. Side chains project toward only one face of the helix, the side facing the solvent
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Correct Answer: B
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490. Which of the following is TRUE?A. mRNA is read 3' to 5', and protein is synthesized C-terminus to N-terminus.B. mRNA is read 5' to 3', and protein is synthesized C-terminus to N-terminus.C. mRNA is read 3' to 5', and protein is synthesized N-terminus to C-terminus .D. mRNA is read 5' to 3', and protein is synthesized N-terminus to C-terminus.E. mRNA can be read either 3' to 5' or 5'to 3'.
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Correct Answer: D
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491. The purpose of the urea cycle is to convert ammonium into a soluble, non-toxic form.A. TrueB. FalseC.D.E.
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Correct Answer: A
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492. What is TRUE about chain turns (reverse turns) in polypeptides?A. Chain turns typically occur in the interior of proteinsB. Chain turns tend to have apolar residuesC. Chain turns are usually helicalD. Chain turns occur on the surface of proteins
E. Chain turns tend to have the important catalytic residues
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Correct Answer: D
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493. Which of the following is TRUE for both prokaryotes and eukaryotes?A. mRNA is modified by a 5' methylguanylate cap.B. mRNA is modified by addition of a polyA tail.C. The small ribosomal subunit binds to the Shine-Dalgarno sequence.D. The first codon is always AUG (i.e. codes for Met).E. Ribosomes are made of protein and DNA.
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Correct Answer: D
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494. Genetic defects in the urea cycle cause disease primarily because they result in liver damage.A. TrueB. FalseC.D.E.
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Correct Answer: B
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495. What is NOT true about the tertiary structure of proteins?A. Some proteins are predominantly alpha helicalB. Some proteins are formed primarily by antiparallel beta strandsC. Some proteins are a mixture of alpha helical and beta structures
D. Globular proteins tend to have hydrophobic interiorsE. Proteins interiors usually have a number of cavities for binding of glucose
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Correct Answer: E
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496. Which of the following is TRUE?A. Production a charged tRNAs requires hydrolysis of GTP by amino tRNA synthetases.B. The fidelity of some amino tRNA synthetases is enhanced by editing.C. Charged tRNA molecules are bound by EF-Tu, when EF-Tu is also bound to GDP.D. Inhibitors of bacterial ribosomes are never used as antibiotics because they also inhibit human ribosomes.E. tRNA molecules bind their activated amino acid residues through their anticodon loops.
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Correct Answer: B
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497. Neonatal jaundice can be dangerous becauseA. heme cannot be properly synthesized in these infants.B. bilirubin diglucuronide cannot be excreted.C. unconjugated bilirubin can enter the brain.D. these infants cannot produce enough vitamin D without phototherapy.E. none of the above.
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Correct Answer: C
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498. The binding of BPG to hemoglobin occurs primarily by which of the following?A. Heme iron
B. Gamma chainC. Alpha chainD. Beta chainE. Alpha amino ends of each chain
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Correct Answer: D
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499. Which of the following is TRUE?A. The genetic code of mammals if different from that of E. coli.B. Exons are segments of genes that are NOT expressed in the resultant protein molecule.C. Introns and Exons are common in bacteria.D. Ribosomal frameshifting allows one segment of the HIV genome to code for more than one protein molecule.E. Hydrolysis of GTP is required for the initiation and termination steps of protein translation, but not for elongation.
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Correct Answer: D
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500. Phenylketonuria results in the production of toxic phenylalanine metabolites that are not normally found in the body.A. TrueB. FalseC.D.E.
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Correct Answer: B
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501. What most accurately describes the function of the proximal His in hemoglobin?A. Proximal His is one of the important residue in hydrogen bonding to BPGB. Proximal His is derivatized with carbon dioxideC. Proximal His is the residue that confers the major Bohr effectD. Proximal His hydrogen bonds the incoming dioxygen moleculeE. Proximal His coordinates the heme iron ion
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Correct Answer: E
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502. Which of the following is the TRUE cause of I-cell disease?A. Failure of proton pumps to acidify lysosomes.B. Failure of clathrin to form transport vesicles.C. Eating too much cheese.D. Absence of a KDEL sequence on an ER proteinE. Failure to phosphorylate a mannose groups in the cis Golgi.
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Correct Answer: E
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503. Which of the following is NOT a neurotransmitter produced by amino acid decarboxylation:A. norepinephrine.B. g-aminobutyrate.C. epinephrineD. acetylcholineE. serotonin.
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Correct Answer: D
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504. What is NOT important in the cooperative binding of dioxygen to hemoglobin?A. Concentration of methemoglobin reductaseB. Concentration of BPGC. Concentration of hydrogen ionsD. Concentration of carbon dioxideE. Concentration of chloride ions
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Correct Answer: A
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505. Which of the following is TRUE?A. Iron-loaded transferrin is internalized by receptor-mediated endocytosis and targeted to the nucleus.B. Transferin and its receptor dissociate at the low pH of endosomes.C. Endocytosed proteins and receptors are sorted in mitochondria.D. LDL receptors are usually degraded following their endocytosis.E. Transferin receptors are usually returned to the cell surface following their endocytosis
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Correct Answer: E
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506. Argininosuccinate lyase deficient patients excrete argininosuccinate into the urine. Why are arginine supplements an effective part of their treatment?A. The arginine can be hydrolyzed by arginase to replace depleted ornithine.B. Arginine can serve as a "sink" for free ammonium.C. Because hydrolysis of arginine produces urea.D. all of the above.E. none of the above.
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Correct Answer: A
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507. What describes the sliding contacts in the hemoglobin tetramer?A. Interaction of alpha chainsB. Interaction of beta chainsC. Movement of proximal His along axis of F helixD. Interaction of alpha1,beta1 chainsE. Interaction of alpha1,beta2 chains
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Correct Answer: E
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508. Which of the following is TRUE?A. The cytosolic Hsp70 chaperone does not function in the import of proteins into mitochondria.B. N-terminal signal sequences are cleaved from large proteins following their import into the nucleus.C. The nuclear and ER membranes are contiguous with each other.D. Large proteins can diffuse freely between the nucleus and cytosol.E. All of the above are true.
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Correct Answer: C
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509. Frogs begin life as tadpoles in an aqueous environment. As they mature, they convert from being largely ammonotelic to being ureotelic. Which of the following enzyme activities would you expect to increase dramatically as frogs mature?A. GlutaminaseB. Alanine aminotransferaseC. Glutamine synthaseD. EnteropeptidaseE. Carbamoyl phosphate synthetase I
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Correct Answer: E
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510. Which is NOT true about mutant hemoglobin molecules in sickle cell anemia?A. Precipitation of deoxyhemoglobinB. Glu to Val substitution on the surface of beta chainsC. Can be compensated by increase expression of fetal gamma chainsD. Fiber formation of oxyhemoglobinE. Tendency to precipitate can be reversed by modification of alpha amino groups
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Correct Answer: D
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511. Which if the following is TRUE?A. Binding of signal recognition particle to a signal sequence results in arrest of translation.B. Core oligosaccharide is added to proteins in the cytosol.C. Calnexin is a mitochondrial chaperone.D. Vesicles destined for the cell surface are usually trafficked through lysosomes before going to the plasma membrane.E. Extracellular proteins are synthesized by ribosomes that are attached to the plasma membrane.
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Correct Answer: A
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512. What is NOT an important feature in the structure of tropocollagen?A. Hydroxylation of prolineB. Gly residue every third residueC. Disulfides links between chainsD. Lysinenorleucine crosslinksE. Hydroxylation of lysine
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Correct Answer: C
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513. Which of the following is TRUE?A. The S phase of the cell cycle is immediately followed by G1.B. DNA replication occurs during mitosis.C. The G1 and G2 checkpoints monitor external environments (e.g. growth factor signals) and internal signals such as cell size and integrity of the DNA.D. G0 is an accelerated form of the G1 phase that allows for accelerated progression through the cell cycle.E. Chromosomes are condensed throughout interphase, but become decondensed in order to pass the metaphase-anaphase checkpoint.
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Correct Answer: C
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514. What is NOT an important feature of enzyme catalysts?A. Enzymes lower the activation energy barrierB. Enzymes shift the equilibrium of a reaction to product formationC. Enzymes are complementary to substrate transition statesD. Enzymes bind substrates by weak noncovalent interactionsE. Enzymes are active within a narrow pH range
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Correct Answer: B
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515. Which of the following is TRUE?A. Cdk enzymes are activated by cyclins.B. The activity of CDK enzymes is regulated by phosphorylation.C. Cyclins are degraded by the ubiquitin-proteosome pathway.D. Cdks phosphorylate, and thereby activate, some proteins that function to further stimulate Cdk activity.E. All of the above are true.
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Correct Answer: E
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516. What is TRUE of the function of the catalytic triad in chymotrypsin?A. Asp102 and His57 activate Ser195B. Asp102, His57 and Ser195 form the substrate binding siteC. His57 and Ser195 activate Asp102 for nucleophilic attackD. Asp102, His57 and Ser195 activate a water molecule for hydrolysis of the peptide bondE. Asp102, His57 and Ser195 stabilize the structure of the chymotrypsinogen zymogen enabling zymogen activation
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Correct Answer: A
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517. Which of the following is TRUE?A. Actin often functions to shape the cell surface.B. Kinesin and Dyneins are motor proteins that move along actin filaments.C. Actin filaments radiate out from the centrosome.D. Polymerization and depolymerization of actin involves a cycle of GTP binding and hydrolysis.E. Polymerization of actin requires rapid synthesis of new actin molecules.
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Correct Answer: A
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518. The inhibitory effects of a noncompetitive inhibitor on an enzyme are overcome by which of the following treatments?A. Increasing the substrate concentrationB. Increasing the temperature of the reactionC. Cross-linking the enzyme side chains
D. All of the aboveE. None of the above
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Correct Answer: E
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519. Which of the following is TRUE?A. Once formed, microtubules are generally stable for the lifetime of the cell.B. Centrosomes replicate and divide to form the poles of the spindle body.C. The nuclear membrane remains intact throughout the mammalian cell cycle.D. Cytokinesis only starts after telophase is finished.E. In mitosis, one cell divides to create two gametes.
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Correct Answer: B
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520. The citric acid cycle (TCA cycle) is initiated by the condensation of two molecules which areA. NAD and oxaloacetate.B. pyruvate and carbon dioxide.C. acetyl CoA and carbon dioxide.D. acetyl CoA and oxaloacetate.E. pyruvate and ATP.
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Correct Answer: D
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521. This is the organ or tissue which only uses glucose in the fed state or during starvation.A. adipose tissueB. brain
C. skeletal muscleD. liverE. erythrocytes
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Correct Answer: E
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522. This is the main site of synthesis for ketone bodies during starvation.A. adipose tissueB. brainC. skeletal muscleD. liverE. erythrocytes
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Correct Answer: D
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523. These tissues require insulin to facilitate glucose transport into cells. A. adipose tissue B. brain C. skeletal muscle D. liver E. erythrocytesA. A & BB. B & CC. C & DD. D & EE. A & C
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Correct Answer: E
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524. In these tissues, adenylate cycle will be stimulated by both glucagon and epinephrine. A. adipose
tissue B. brain C. skeletal muscle D. liver E. erythrocytesA. A & BB. B & CC. A & DD. D & EE. B & D
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Correct Answer: C
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525. The glucocorticoidsA. stimulate adenylate cyclase in liver.B. alter rates of protein synthesis.C. inhibit gluconeogenesis.D. stimulate glycolysis.E. inhibit oxidative phosphorylation.
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Correct Answer: B
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526. Which one of the following is NOT a characteristic of the hexose monophosphate pathway?A. Carbon dioxide is produced.B. It forms 2 ATP per glucose oxidized.C. It produces ribose-5-phosphate.D. It produces NADPH.E. The first reaction is controlled by the concentration of NADPH.
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Correct Answer: B
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527. Which one of the following conditions in the matrix of a mitochondrion would decrease the oxidation of Acetyl CoA in the TCA cycle?A. A low ratio of ATP/ADP.B. A high concentration of AMP.C. A high ratio of NADH/NAD.D. A low ratio of GTP/GDP.E. A high concentration of oxaloacetate.
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Correct Answer: C
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528. In human liver, fructose is phosphorylated to give fructose-1-phosphate. This is followed byA. phosphorylation to give fructose-1,6-bisphosphate.B. conversion to fructose-6-phosphate.C. cleavage to form glyceraldehyde and dihydroxyacetone phosphate.D. isomerization to glucose-1-phosphate.E. hydrolysis to give fructose.
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Correct Answer: C
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529. The enzyme catalyzing the rate limiting reaction of glycolysisA. is activated by higher ATP.B. catalyzes a reaction which is reversible.C. is inhibited by fructose-2,6-bisphosphate.D. is activated by higher citrate.E. uses fructose-6-phosphate as a substrate.
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Correct Answer: E
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530. The surest treatment for a patient who accidentally ingested antifreeze would be toA. infuse sodium nitrite and then use amyl nitrate inhalation.B. give a fructose rich drink like Ultra Fuel.C. give a couple shots of scotch whisky.D. give two cups of strong coffee.E. give a drink which is enriched in citrate.
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Correct Answer: C
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531. Each of the following is capable of increasing the activity of muscle glycogen phosphorylase b EXCEPT FORA. epinephrine.B. calcium ions.C. active phosphorylase kinase.D. glucagon.E. cyclic AMP.
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Correct Answer: D
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532. Each of the following is generally true about NIDDM EXCEPT thatA. it is accompanied by polydipsia and polyuria.B. it is associated with obesity.C. no insulin is detected in serum.D. there is moderate to high glucose in serum.E. there is a strong genetic component.
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Correct Answer: C
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533. "Oxidative stress" refers toA. the fact that oxidative phosphorylation requires ADP.B. the fact that oxidative phosphorylation requires molecular oxygen.C. the failure of aerobic glycolysis to meet the energy demands of muscle.D. the uncoupling of oxidative phosphorylation.E. the formation of free radicals because of the imperfect reduction of oxygen to give water.
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Correct Answer: E
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534. Viagra owes its efficacy to the fact that itA. activate adenylate cyclase in smooth muscle.B. stimulates glycolysis.C. stimulates guanylate cyclase.D. inhibits cyclic GMP phosphodiesterase.E. stimulates oxidative phosphorylation.
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Correct Answer: D
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535. The binding of insulin to receptors on hepatocyte causes each of the following EXCEPTA. increased activity of tyrosine kinase.B. increased activity of glycogen phosphorylase.C. increased synthesis of glycogen.D. increased synthesis of fatty acids.E. increased glycolysis.
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Correct Answer: B
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536. Each of the following statements about the protein complex that carries out the synthesis of ATP during oxidative phosphorylation is correct EXCEPTA. it contains a proton channel.B. it is inhibited by oligomycin.C. it binds molecular oxygen.D. it is embedded in the inner mitochondrial membrane.E. ATP synthesized by this complex is released to the matrix.
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Correct Answer: C
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537. After 5 days of starvation all of the following would be true EXCEPTA. the brain (nervous system) would only use ketone bodies for energy.B. glycogen in liver and muscle would be totally depleted.C. most circulating glucose would come from hepatic gluconeogenesis.D. the ratio of insulin/glucagon would be much lower than before the fast.E. glucagon would increase the release of fatty acids from adipocytes.
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Correct Answer: A
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538. Acetyl CoA is described by all of the following EXCEPTA. it's formed by the pyruvate dehydrogenase complex.B. it condenses with oxaloacetate to give citrate.C. the acetate carbons of acetyl CoA are incorporated into fatty acids.D. it enhances the phosphorylation of pyruvate dehydrogenase.E. it is the end product of glycolysis.
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Correct Answer: E
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539. The drug Antabuse (disulfiram)A. inhibits the complete oxidation of ethyl alcohol.B. inhibit isocitrate dehydrogenase.C. uncouples oxidative phosphorylation.D. inhibits electron transport.E. inhibits pyruvate dehydrogenase.
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Correct Answer: A
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540. What is NOT true about the stability of a globular protein?A. Hydrophobic group are shielded away from solventB. Hydrogen bonding between residues spatially separated in sequenceC. Loop structure connecting secondary structural elements are on the surface and hydrogen bonded to waterD. All peptide bond carbonyl oxygens and amides are hydrogen bonded to solvent or to other polypeptide donors/acceptorsE. All aromatic rings are stacked giving pi electron resonance
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Correct Answer: E
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541. What is NOT true about the ability of hemoglobin to bind oxygen?A. Each subunit binds one heme groupB. Proximal methionine binds heme ferrous ion activating it for oxygen binding.C. Ferrous iron in heme group binds to oxygenD. Heme iron is shielded from solvent keeping iron as ferrousE. Hemoglobin is present in erythrocytes which pass through lung for oxygenation
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Correct Answer: B
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542. What is NOT important for the deoxygenation of hemoglobin?A. Binding of four bisphosphoglycerates per hemoglobin tetramerB. Protonation induced salt bridge formation between subunitsC. Binding of carbon dioxide and induced formation of salt bridge between subunitsD. Increase in the number of subunit contacts between a2b1 a1b2 subunitsE. Cooperativity between hemoglobin subunits
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Correct Answer: A
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543. What is NOT true of an enzyme's ability to be a catalyst?A. Enzymes can specifically bind substrate molecules.B. Binding of substrate induces a strain in the substrate molecule.C. Enzymes lower the energy barrier in the conversion of substrates to a transition stateD. Enzymes shift the equilibrium between substrates and productsE. Enzymes have specific residues in the active site capable of carrying out acid/base chemistry or covalent catalysis.
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Correct Answer: D
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544. What is NOT a way proteins are modified post-translationally?A. Cleavage of a polypeptide bondB. Modification of the N-terminal residueC. Acetylation of peptide bond carbonyl oxygensD. Addition of components to the C-terminusE. Crosslinking of certain residues
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Correct Answer: C
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545. What is NOT a common way an enzyme can be inhibited?A. by compounds that increase the Vmax of the enzymeB. by compounds that increase the Km of the enzyme for a substrateC. by compounds that destabilize the enzymeD. by compounds that perturb both the Vmax and Km parameters of an enzymeE. by other enzymes that covalently modify the particular enzyme
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Correct Answer: A
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546. Which of the follow does NOT describe how mutations often attenuate the function of proteins?A. mislocalization of proteinB. destabilization of proteinC. misfolding of the proteinD. increasing the number of disulfide linkagesE. removal of nucleophilic residues in enzymes
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Correct Answer: D
19. Describe briefly the process of initiation by eukaryotic RNA polymerase.
