BIOL 308 - Study Questionss3.amazonaws.com/prealliance_oneclass_sample/BMRXYo8daw.pdf · BIOL 308 - Study Questions Lectures 5-7 1. Thinking question re DNA/RNA structure: Certain

BIOL 308 - Study Questions

Lectures 1-4

1. What information(s) could you obtain from a genetic approach of studying mutants defective in a particular process? The potential function and importance of the gene/protein can be determined through identification, isolation and survival of mutants. The number of genes, the order of the genes (using complementation analysis), and the interactions between different genes (using genetic suppression) can also be determined.

2. How would you define permissive conditions in respect to temperature sensitive mutants? An environment where the mutant can survive, since it displays the wild type phenotype. Under nonpermissive conditions (too cold) mutant phenotype is displayed and mutants can’t survive.

3. Define (or compare and contrast): a) gene expression; transcription; replication; translation; b) gene; allele a) transcription: an RNA copy of the DNA is made, translation: RNA copy is used to create protein, gene expression: gene has undergone transcription and translation for form a product, replication: DNA makes a copy of itself before undergoing mitosis b) gene: contains code to produce RNA or a protein, allele: variations of the gene

4. Explain by using your own words the meaning/significance of gene expression. Gene expression occurs when a gene is “turned on,” and forms a product such as a protein. The gene will be expressed depending on the environment, developmental stage, or tissue type.

5. What are the roles of model organisms in molecular biology studies? Choose two model organisms and explain your reasoning. Model organisms can be easily studied and are used in place of species that can’t be study in the same manor. Ex. Arabidopsis and Drosophila are simple, easy to come by in bulk, cheap, easy to maintain, lots of data is already available, and reproduce quickly allowing scientists to observe future generations.

6. What are three main functions of DNA? Explain the importance of each of them. 1. Stores information: coding for proteins and RNA, and contains regulatory signals 2. Replicates faithfully (preservation of information): semi-conservative, complementary base pairing, preserves genes important to organism’s survival 3. Has ability to mutate (variability of information): allows for adaptation, may be harmful or have no effect

7. What is (are) the role(s) of phospho-diester bonds in DNA structure? What is (are) the role(s) of hydrogen bonds in DNA structure? What is (are) the role(s) of hydrophobic interactions in DNA structure? Phosphodiester bonds hold the sugar to the phosphate backbone. Hydrogen bonds assist with complementary base pairing. Hydrophobic interactions occur due to the highly negative phosphate backbone on the outside and the nonpolar bases inside.

8. What noncovalent interactions are involved in maintaining the double-helical conformation of DNA? Van der Waals (stacking interactions), hydrophobic, ionic (salts stabilize phosphate backbone), and hydrogen bonding

9. Learn to recognize nitrogenous bases (A,T,G,C,U).Purine Pyrimidine

Adenine Guanine Cytosine Thyamine Uracil

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10. Describe Meselson-Stahl experiment and explain how it showed that DNA replication is semiconservative? E.coli was grown in N15, and then moved to N14. In the first replication semiconservative material contained equal amounts of N15 and N14 in each strand. In the second replication two of the strands contained N15 and N14, while the other two contained N14 only.

11. What is meant by saying that a DNA strand has polarity? That two strands of DNA are antiparallel? That the strands are complementary to one another? The DNA strand is synthesized in a particular direction, 5’ to 3’, 5’ end is negatively charged. The two strands of DNA go in opposite directions of one another (there 3’ and 5’ ends are opposite). The DNA strands have matching pairs (AT and CG), that always line up together.

12. If a C content of a preparation of double-stranded DNA is 20%, what is the T content? 30%

13. What is the difference between nucleoside and nucleotide? What does dNTP stand for? Nucleotides contain a phosphate group, along with the nitrogenous base and the sugar. Nucleosides do not include a phosphate group. Deoxynucleotide Triphosphate.

14. Describe the conformational characteristics of B DNA (or A DNA, Z DNA, triple-helical DNA). When does this (any of the above) form of DNA occur? B DNA is a right handed, double helix, and has a diameter of 80A, a major groove of 22A, a minor groove of 12A, is 34A per helical turn, and represents most in vivo DNA. A DNA is also a right-handed double helix with 28A per helical turn, and is found in dormant spores of bacillus. Z DNA is a left handed, double helix with 45A per helical turn, and is found in chromosomal breakages and Alzheimer’s disease. Triple helical DNA is formed when purines make one strand and pyrimidines the other, the third strand can be accommodated, and they are found in vitro and likely in vivo during DNA recombination or repair. 1A = 0.1nm, 1nm = 1.0X10-9m

15. How does high salt concentration influence denaturation kinetics of DNA? Explain your reasoning. Salt stabilizes DNA since it neutralizes the negatively charged backbone, and therefore slows down denatuation kinetics of DNA.

16. What are the classes of DNA sequences in genomic DNA (based on renaturation kinetics)? Nonrepeating, where the complexity = number of nucleotides, and repetitive sequences where complexity = the number of unique nucleotides and the total number of nucleotides from one copy of each repetitive sequence.

17. What is Cot analysis? Used to determine the rate of renaturation or a measure of complexity of DNA. Co is the starting concentration (nucleotides/L) and t is the reaction time (s). Units of complexity are measured in terms of nucleotides.

18. Who received a Nobel Prize for 3D DNA structure? Watson and Crick.

19. If you had two solutions of DNA, one single-stranded and one double-stranded, with equivalent absorbance at 260 nm, how would the concentrations of DNA compare in these two solutions? (You can use a diagram if it makes it easier for you to explain.) There would be more double-stranded DNA because single stranded DNA takes up light easier, since purines and pyrimidines exposed.

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Lectures 5-7

1. Thinking question re DNA/RNA structure: Certain chemical agents acting on DNA could convert cytosine to uracil through the process of deamination (chopping off the amino group). This mutation is routinely repaired by the existing repair mechanism (uracil is removed and it gets replaced by cytosine). Knowing this, how would you explain why DNA contains thymine and NOT uracil. If the DNA contained uracil instead of thymine the enzyme would remove all uracils, including those intended to be present. The DNA would end up with adenine-cytosine pairs and be greatly mutated.

2. How does complexity of bacterial genome differ from that of eukaryotic (calf) genome? Bacterial genome is smaller, but contains few/no repeats. In eukaryotic genomes there are many repeats, but the length of the gene is substantially longer, and even without the repeats would be greater in length than the bacterial genome. Eukaryotic genes contain multiple isoforms of a protein, which can be produced through alternative slicing of mRNA.

3. Explain C value paradox. There is no correlation between the amount of DNA (size of genome) and the apparent complexity of the organism.

4. List and briefly explain factors that influence DNA renaturation kinetics. DNA concentration: complementary single strands have a better chance at finding another single strand in greater concentrations, Salt concentration: ionic conditions mask repulsion forces of phosphate backbones, Temperature: needs to be 20-25C below Tm, time: enough time need for reaction to occur, size: larger fragments have more difficulty lining up to anneal, complexity: simple sequences re-nature faster than complex.

5. You have found a new species of insects. To evaluate the complexity of the genome of this species, you isolate genomic DNA from, fragment the DNA to uniform 500 base pair pieces, denature the DNA and measure the rate of reassociation. Your data is represented in the curve below (sorry for the bad drawing):

(a) How many classes of DNA (in respect to sequence complexity) are found in this organism? Nonrepetative, moderately repetitive, and repetitive.

(b) What can you say about the relative complexity of each class? What fraction of the genome falls into each class? Those with repeats will be less complex.

6. List three (3) differences between prokaryotic Topoisomerase I and Gyrase. Topoisomerase 1 relaxes negative supercoiling by adding positive supercoil, but gyrase introduces negative supercoils by reducing positive supercoiling. Topoisomerase 1 makes transient cut in one strand, but gyrase being a toposiomerase 2 makes a double-stranded cut and passes a duplex DNA through it and reseals it. Gyrase is ATP hydrolysiszed.

7. What are topological isomers of DNA? Topological isomers or isomers are used in DNA to reduce stress (in the form of supercoiling) in DNA specifically during replication.

8. Explain the importance of DNA supercoiling for the cell survival? Supercoiling allows the DNA to be store in the cell9. What are the differences between primary (or secondary, or tertiary) structures of RNA and DNA? Primary RNA is single

stranded and contains uracil, and primary DNA contains thyamine instead of uracil. Secondary structures of RNA involve the folding over of RNA on itself causing hairpins and stem loops., while DNA takes on a double helix structure. Tertiary structures of RNA involve possible formation of U:A:U base triple and pseudoknots can form between nonadjacent sequences, and DNA forms supercoils/complexes with proteins.

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10. You have a small 4800 bp long circular DNA. It has a linking number of 450 (L=450). What are the twist (T) and the writhe (W) of this DNA? What assumptions about the structure of the DNA have you made in your answer? (Hint: what is the definition of the twist #?) Assuming the DNA is double stranded:

Twist = total # of base pairs/# of base pairs or turns = 4800/450 = 10.7

Writhe = sign of link number = positive

11. Thinking question 1: Many different mutations have been observed in almost all genes. However, only few have been isolated in histones. How would you explain this finding? Histones have a critical function in the cell, and those with mutations would not survive.

12. Thinking question 2: would it be an easy task to produce polyclonal antibodies to histones? Explain your reasoning. It would be difficult because histones are wrapped in DNA, therefore difficult to access.

13. What are some of the distinctive features of eukaryotic chromosomes? (note: I expect you to first define chromosomes and after that you have to briefly explain nucleosomes/histone proteins/octet +H1/wrapped DNA, different levels of chromosome condensation, centromere and telomere regions) Chromosomes are organized structures of DNA. DNA is wrapped onto histone octects or cores, this is referred to as a nucleosome. The octet + H1 is a 25 to 100 fold compression, H1 associates with DNA and octet in linker region binding two distinct regions of the DNA duplex. The nucelosomes can be condensed into a 10nm fiber where H1 is not required or a 30nm fiber. Centromeres are specific sequences where sister chromatids meet. Telomeres occur at the end of the chromosome.

14. What is unusual about the amino acid composition of histones? How is the function of histones related to their amino acid composition? Histones contain large amounts of arginine and lysine, which contain amine groups. Histones can use the positively charged amine groups to bind to negatively charged phosphate backbone of DNA.

15. Name few nonhistone proteins which are a part of chromatin structure and explain why you would expect them to be found there. Matrix attachment regions (MARs) or scaffold attachment regions (SARs) help bind DNA to histone core. DNA replication proteins, transcription factors and chaperone proteins would also be found to help with DNA replication which occurs before DNA is bound to histone. Topoisomerase determines if coiling is too tight.

16. What are heterochromatin and euchromatin? What is their importance in DNA replication and transcription? Heterochromatin is tightly packed DNA and is less susceptible to DNase digestion and transcriptionally inactive. Euchromatin is lightly packed since it contains transcriptionally active DNA, but is susceptible to DNase digestion.

17. Does the degree of chromosomal condensation play a role in controlling replication and transcription? How (explain briefly; use your own words)? More condensed DNA is less available for replication.

18. Would you expect there to be more histones per kilobase in euchromatin or heterochromatin? Explain your reasoning. There would be more histones per kilobase in heterochromatin because the DNA needs to be tightly wound.

19. Thinking question: the sequences of a particular set of genes are found by in situ hybridization (note: methodology is irrelevant for this question, so do not think about it) to be heterochromatic in some cells and euchromatic in cells at different stages of development. How would these sequences be categorized (what would be your conclusion about these sequences in respect to gene expression)? These genes can be turned on and off, and it is better to have them turned off, if not required, and tightly wound to prevent DNase digestion. However, the genes are being translated at certain points of development due to the cells requirments.

Lectures 8-9

1. The human gametes have about 3 billion bp of DNA in their chromosomes.a. If the entire DNA was in relaxed B-DNA form, what would be the average length of a chromosome in the cell? (3 billion/10)*3.4nm = 1,020,000,000nm or 1.02mb. On average, how many complete turns would be in each chromosome? 3 billion/10 = 300,000,000

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c. If there are around 30-40,000 genes in a human gamete, how many genes are there in an average chromosome? 40,000 genes/gamete = 40,000 genes/46chromosome = 870 genes/chromosome80,000genes/somatic cell = 80,000 gene/46 chromosomes = 1749 genes/chromosome

2. Define homologous chromosomes. Pairs of chromosomes, one maternal and one paternal, that are the same length, contain the same genes, and have the centromere in the same position.

3. Define non-homologous chromosomes. Chromosomes that do not match.

4. How many homologous chromosomes are there in a germ cell of a woman? 23, men only have 22 because XY is not homologous.

5. Distinguish between homologous chromosomes and sister chromatides. Sister chromatids make up a chromosome and come from the same parent. One of each homologous chromosome comes from a parent.

6. What is the purpose of cell division in Prokaryotes? In Eukaryotes? The purpose of cell division in prokaryotes is to produce “offspring.” In eukaryotes the purpose is to increase mass, or replace dead cells.

7. Distinguish between DNA replication and cell division. DNA replication duplicates the DNA in preparation for cell division, where the cell splits in two, one copy of each DNA going to each daughter cell.

8. Distinguish between reason/purpose for/of mitosis and reason/purpose for/of meiosis? Mitosis increases mass and replaces dead cells, where meiosis creates haploid cells to be used in reproduction.

9. How is variability of genetic information attained by meiosis and fertilization? Eggs and sperm are haploid and when eggs are fertilized the DNA from the father and mother come together and crossing over can occur.

10. What is a cell cycle? What are the stages of cell cycle? The cell cycle is the process the cell undergoes consistently. Interphase – G1: growth occurs as organelles double and the cell carries out normal metabolism, S (synthesis): DNA replication occurs as chromosomes duplicate, and G2: growth occurs as cell prepares to divide. Mitosis: cell division, however some cells never leave interphase (G0).

11. List and briefly describe the checkpoints in cell cycle. What is their purpose? Checkpoints determine if the cell is ready to move to the next stage – if not error is corrected or apoptosis occurs.

Start 1: G1 checkpoint, determines if cell is large enough and is the environment is favourable.S phase: Determines if replication is proceedingEnter Metaphase: G2 checkpoint, determines if cell is large enough, if the environment is favourable and if all DNA is

replicated.Exit from Metaphase: metaphase checkpoint, determines if chromosomes are aligned on spindle.

12. What can trigger arrest during the cell cycle? Errors such as DNA damage, and cell size must be adequate and the environment must be favourable.

13. How many chromosomes are there in a somatic cell of a person with Down syndrome (trisomy of chromosome 21) 47

a) How many autosomes does this person have in a somatic cell? 45b) How many sex chromosomes does this person have in a somatic cell? 2 In a germ cell? 2 In a gamete? 1 In a

spermatozoid? 1 In an ovum? 1 In a zygote? 2c) How many DNA molecules does this person have in mitotic metaphase? 47 chromosomes each containing 2 sister

chromatids or DNA molecules, 94 In G1 phase? Chromosomes are unipartite, or one of the sister chromotids, 47d) How many telomeres are there in a person’s somatic cell during G2 phase? 4 per chromosome, 4x47=188e) There are 4 alleles for a certain gene carried by chromosome 21 in human population. How many alleles does a

person with Down syndrome have in a somatic cell in G1 phase? Each chromatid has 1 allele for the gene, are 6 chromatids for individuals with Trisomy 21, 6. How many different alleles for this gene could the same person have? 4

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f) There are 13 alleles for a certain gene carried by chromosome 21 in human population. How many alleles does a person with Down syndrome have in a somatic cell in G1 phase? 6 How many different alleles for this gene could the same person have? 6, since only contains 6 alleles.

There are 13 alleles for a certain gene carried by chromosome 21 in human population. g) How many alleles does a person with Down syndrome have in a cell which is in meiosis I anaphase? 6h) How many different alleles for this gene could the same person have in a cell which is in meiosis I anaphase? 6

14. When does chromosome segregation happen in mitosis? In meiosis? When does chromatide segregation happen in mitosis? In meiosis? Chromosome segregation does not occur in mitosis, but it occurs in anaphase 1 in meiosis. Chromatid segregation occurs in anaphase in mitosis, and in anaphase 2 in meiosis.

Lectures 10-11

1. List proteins (enzymes and other factors) involved in the process of DNA replication in E. coli. Explain the role of each of these proteins in replication.

DnaA: initiates replication by recognition of DnaA boxesDnaB: helicase, separates the strandsDnaC: escort to DnaA forming pre-priming complexSSB, single-strand binding proteins: binds to ssDNA, prevents double helix from reformingPrimase, DnaG: elongates existing primer strandsRNase H: removes RNA primers except last nucleotide, in lagging strandDNA polymerase I: removes ribonucleotides (last nucleotide), fills in deoxyribonucleotide gapsDNA polymerase III: adds nucleotides in 5’-3’ directionDNA ligase: links fragments on lagging strandTopoisomerase: recognize and regulate supercoiling

2. What is meant by replication being bidirectional? Semiconservative? Continuous and discontinuous? There are two replication fork that move in bother directions from the origin. Replication is semiconservative because one strand is the original and the other is produced from the original. Since DNA is anitparrallel and DNA polymerase III can only move 5’ to 3’ a continuous or leading strand and a discontinuous or lagging strand are produced.

3. Contrast the role of DNA polymerase I and III in E. coli DNA replication. DNA polymerase I removes the leftover ribonucelotide from each gap in the lagging strand, and fills in the gap with deoxyribonucleotides in the 5’ to 3’ direction. DNA pol I also proofreads in the 3’ to 5’ direction. DNA polymerase III adds nucleotides in the 5’ to 3’ direction on both the leading (continuous) and lagging (discontinuous) strands.

4. Which subunit of DNA polymerase III provides processivity? Which protein complex loads this subunit onto the DNA? The beta subunit functions as a clamp and increases the processiviy of polymerase. The gamma complex loads and unloads the beta complex.

5. How can discontinuos synthesis of the lagging strand keep up with continuous synthesis of the leading strand? The leading strand’s replication speed is slowed down by the presence of the leagging strand.

6. Why is decatenation required after replication of circular DNAs? Decatenation is required to unloop the two pieces of circular DNA from one another.

7. Why do eukaryotes need telomeres but prokaryotes do not? Eukaryotes have linear DNA, where after replication a 3’end overhang exists from lagging strand since not enough room to fit primer. This overhang is generally cleaved, therefore can’t have it contain important information.

8. What would be the components necessary to make DNA in vitro by using DNA polymerase I. RNA primers, DNA polymerase III, DNA, ligase, topoisomerase, helicase/DnaB, SSBP, primase, RNase H, DnaA, DnaC, and dNTPs.

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9. What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA ligase? Unsealed fragments on lagging strand. Only half the DNA would be viable in harmful temperatures. At correct temperature no problems should occur.

10. What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA polymerase I? Gaps in lagging strand at extreme temperatures, DNA not viable with lagging strand. At correct temperatures no problems should occur.

11. Compare and contrast major eukaryotic and prokaryotic DNA polymerases. Eukaryotes: Pol alpha makes the RNA primer and does some elongation, Pol sigma is responsible for elongation and maturation of Okazaki fragments and proofreads Pol alpha, and Pol epsilon is responsible for assembly of replisome (DNA replicator machinery). Prokaryotes: DNase I, III, and II which is involved in repair.

12. What is telomerase and why is it important? Telomerase carries an RNA template that is complementary to the telomeric repeats and can therefore cause extension of telomeres. This helps preserve the DNA.

13. What is the major difference between bacterial and eukaryotic replication that allows a eukaryotic cell to replicate its DNA in a reasonable amount of time? Eukaryotic cells have multiple sites of replication.

14. Describe the events that occur at an origin of replication during initiation of replication in E. colii? Specific proteins recognize the origin of replication, contains DnaA boxes that DnaA recognizes. The area is rich in A-T base pairs allowing for easier denaturation. Initiation depends on methylation of oriC, on both strands. Dam methylase methylates DNA at GATC sequence. This prevents the DNA from continuous replication since both strands must be methylated ad after one round of replication on one of each strands is methylated.

16. What are cis- elements? What are transfactors? Cis-elements: sites or sequences on DNA like DnaA box and OriC. Trans-factors: proteins that diffuse through cell/nucleous and recognize cis-elements and bind to them, like DnaA, DnaB and DnaC.

17 Is the following statement true or false: Regardless of whether a gene is expressed in a given cell type, it will replicate at the same, characteristic time during S phase. Explain your reasoning. True, all genes need to be present and are required in the replicated DNA, whether or not expressed.

18. Thinking question: Is making of RNA primers (by primase), which have to be subsequently removed and replaced with dNTPs (by DNA polymerase I) actually wasteful and energetically inefficient process? (Hint: think about fidelity of primer-making vs. proofreading capability of DNA pol I, in other words what is the only purpose of RNA primers – is the accuracy of this process very important at this point?) DNA polymerase can only elongate chains and therefore requires a primer. Need RNA primer to build off of. Also, RNA primer doesn’t need to be prefect, and therefore less energy required, since it will be removed.

19. You preformed Cot analysis using genomic DNA samples obtained from a 2 year-old child and a 76 year-old individual. Results of this analysis show that one of the most rapidly reassociating classes of DNA is substantially reduced in the older individual with respect to the 2 year-old. How can you explain this finding? (hint: think about termination of linear DNA replication.) Telomeres contain repeats, which anneal quickly together in Cot analysis. As we grow older and out cells replicate, our telomeres reduce, and therefore reduce the number of repeats.

Lectures 12-14

1. Distinguish between the terms “mutation”, “DNA repair” and “recombination”. Mutation: errors in the DNA, may be lethal, silent, or beneficial, caused by replication errors, spontaneous changes in DNA, and external factors like radiation. DNA repair: mechanisms like DNAP and mismatch repair, repair mutations. Recombination: occurs during replication, repairs lesions the fork encounters to continue replication, fills in one strand by retrieving corresponding single strand from another homologous duplex.

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2. List and briefly explain three major causes for mutation in DNA. Replication errors not fixed by proofreader mechanisms, spontaneous changes in DNA like deaminations and depurinations, and external factors like radiation, temperature, and mutagens.

3. Explain how errors in DNA replication can lead to mutations. Wrong base may be put into place, and go unnoticed by repair mechanisms.

4. Distinguish between the effects of mutations on the somatic and germ cells of multicellular organism. Somatic mutations may effect organism. Mutations in germ cells may effect offspring.

5. List the various types of DNA repair mechanism (we have mentioned seven). 1. Proof-reading or Editing by DNA Polymeras 2. Direct Reversal of Damage 3. Base Excision Repair 4. Nucleotide Excision Repair 5. Mismatch Repair 6. Recombination Repair 7. Error-Prone Repairs

6. Give the detailed description of the base excision (or nucleotide excision) repair process in bacteria. The incorrect base flips to the other side of the backbone. Glycosylase recognizes and removes base (ie. Uracil in DNA) by hydrolyzing glycosidic bond. AP endonuclease and phosphodiesterase removes sugar phosphate backbone. The nucleotide is replaced by DNA pol and DNA ligase seals the nick.

7. Describe the mismatch repair process of bacteria (pay attention at the ways in which the daughter and parent strand are recognized by repair system). MutS scans DNA for distortion and binds to mismatch ie. loop. MutL is recruited by MutS-mismatch-DNA complex. MutS ttranslocates along DNA until a GATC sequence is reached (used to determine which strand is parent). MutSL activates MutH (endouclease), which recognizes GATC and binds MutSL. MutH nicks the unmethylated or daughter strand of DNA. Strand is then progressively degraded from GATC to mismatch and a new DNA strand is synthesized by DNaseIII and sealed by ligase so the mismatch is removed and replaced with correctly base-paired nucleotide.

8. Describe briefly the mechanism of direct reversal of damage in bacteria. Direct reversal repairs thymine dimmers through photoreactivation by photolyase. The thymine dimmer is broken by light-dependent activity.

9. Distinguish between the two DSB repair mechanisms we talked about in class. Double stranded breaks can be repaired by nonhomologous end joining (NHEJ) and recombination repair. NHEJ can directly ligate the two ends of broken DNA by cleaving overhangs, which is error prone. Recombination repair references the sister chromosome to repair the other and is less prone to error. *Recombination repair may also be used when one strand is broken.

10. What is SOS repair mechanism, when is it used and why is it important? SOS translesion repair is higly error-prone because polymerases add nucleotides randomly, without proper base pairing. It introduces mutations by still enables complete chromosome replication. It’s used as a last attempt and is important because it introduces mutations – variability.

11. Describe the term “non-homologous end joining” and explain how this process results in the repair of double strand breaks in DNA molecule. Cleaving of overhangs in a damaged piece of DNA allows for joining of non-homologous ends to join.

12. What are biological roles of DNA recombination? DNA repair, formation of new genes and integration of a specific DNA element.

13. List and briefly describe the ways genomic DNA can be rearranged (there are three of them). General or homologous recombination: genetic exchange between pairs of homologous DNA sequences. Site specific recombination: occurs between sequences with a limited stretch of similarity, involves recombination sites. Transposition: mobile DNA element moves from one site to another, usually little sequence similarity involved.

14. What are the key steps in single stranded model of homologous DNA recombination? The two homologous DNA molecules align. A break is introduced into one strand of each homologous DNA molecule. A single strand region from one strand invades and pairs with the complementary strand from the homologous molecule. The two molecules become connected through crossing DNA strands, forming a Holliday junction. Branch migration occurs, moving the Holliday junction. Cleavage of the Holliday junction results in new combinations of DNA.

15. Explain the relationship between hybrid duplex and heteroduplex. (You can use diagram.) A homoduplex is the strand without homologous chromosomes DNA. A heteroduplex contains a hybrid of DNA.

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16. Describe the function of proteins involved in homologous recombination of E. coli. Used in repair since E.coli is haploid. RecBCD is a nuclease/helicase that binds to duplex DNA and processes it into a substrate for recombination by generating single strands for invasion. RecA starts recombination by facilitating pairing of homologous DNAs and is involved in strand invasion. RuvAB is a complex with helicase activity and recognizes Holliday junctions and catylases branch branch migration. RuvC is an endonuclease that catalyses the resolution of Holliday junction.

RecBCD complex enters the DNA at the site of the double strand break and unwinds and degrades both strands of DNA in the presence of ATP. Upon reaching a chi sequence activity changes and the chi sequences control its activity. Degradation of 3’ end stops and degradation of 5’end increase creating 3’ overhang. RecA binds to the overhang and initiates strand exchange. Branch migration is caused by RuvAB and the Holliday junction is resolved by RuvC.

17. Describe the role of chi sequences in homologous recombination of E. coli. Chi sequences enhance recombination frequency.

18. What is the major role of homologous recombination in prokaryotes? Repair

19. What are the roles of homologous recombination in eukaryotes? Repair and variability (in meiosis)

20. When does the programmed creation of DSBs occur in eukaryotes? In which type of cells? Briefly describe the process. Programmed double stranded breaks occur between homologous chromosomes in germ cells during prophase 1 of meiosis 1. Double stranded breaks are formed at recombination hotspots. Spo11 generates double stranded breaks. Mre11 is a nuclease that creates 3’ overhangs. Dmc and Rad1 cause strand invasion. End up with 2 Holliday junctions.

21. Define gene conversion (use your own words). What is the significance of gene conversion? Gene conversion is where DNA from one chromatid or chromosome is exchanged with a homologous pair. It increases variability in the genes.

22. What is the role of mismatch repair mechanism in gene conversion? If the new DNA is incorrectly placed into the DNA ie. Forms a loop, doesn’t match, the mismatch repair mechanism will recognize this and replace the new DNA to match the parent DNA.

23. Explain the role of site-specific recombination in infection of E. coli genome by lambda phage. Lambda phage can insert it’s DNA into the E.coli genome at a recombination site, entering latent prophage state.

24. What are potential effects of transposons on the genome? Transposons can cause genetic changes by inserting into genes/coding sequences and into regulatory sequences, possibly disrupting the cell.

25. List and briefly describe three mechanisms by which genetic elements are able to move from one site in the genome to another. DNA-only transposons: move as DNA either by cut and past or replicative pathways. Retroviral-like retrotransposons: moves via RNA intermediate produced by a promoter in the direct long terminal repeats vertical movement (between cells). Nonretroviral retrotransposon: moves via an RNA intermediate that is often produced from a neighboring promotor, horizontal movement (within a cell).

26. Draw a fully annotated diagram illustrating the transposition mechanism of a simple IS transposon (or retroviral-like retrotransposon, or non-retroviral retrotransposon).

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27. Who was Barbara McClintock and what was her major scientific contribution [no more then two (2) sentences]. Nobel prize winner for her discovery of mobile genetic elements – transposons!

Lecture 15

1. What kind of information can we obtain from a northern blot? RNA detection 2. What kind of information can we obtain from a Southern blot? DNA desction

3. Compare and contrast Southern and northern blotting technique. In Southern blotting DNA is denatured and broken into fragments that are separated by agarose gel electrophoresis, then blotted onto nitrocellulose paper, and a probe is added to label DNA to be visualized by autoradiography, In northern blotting mRNA, tRNA, and rRNA are isolated from cells and elecrophoresed, very similar to Southern blotting except RNA does not need to be denatured since already single stranded.

4. Compare the information obtained by northern analysis with the information obtained by microarray experiments. A northern blot determines the steady-state level of a specific transcript in a certain RNA mixture (the abundance of specific mRNA at a certain time and under certain conditions). A microarray experiment is similar to running hundreds of northern blots, since you can compare different mRNAs, or mRNAs from different environments, to see if they are expressed.

5. How do we treat a DNA gel prior to Southern blotting? Explain why.6. Thinking question: you have made a short probe (50 nucleotides) from a certain genomic DNA (note: genomic

DNAs include both exons and introns). Are you sure that you would be able to use this probe for northern hybridization? Explain your reasoning. The DNA is cut with restriction nucleases because then we can determine if the gene sequence occurs more then once, and we can estimate the positions o the gene copies. The same probe used in DNA could not by used in RNA because all the information found in DNA is not transcribed to RNA, and RNA removes introns. The probe used for DNA may contain part of this intron and therefore not bind to the RNA.

7. Distinguish between a template, primer and a probe. Would you be able to use primer as a probe? If yes, explain when/how. A template is a known sequence that is similar to the target, that we can base our probe off of. A probe binds to the unknown DNA if there is a match. A primer binds to DNA to start transcription. A primer could be used as a probe as long as you do not want the primer to act as a primer, because the primer is meant to fall off the DNA strand after replication..

8. Why is it important to know the exact start site of transcription? It is important in order to insert a gene afterwards, so it will be transcribed.

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9. What does SDS-PAGE stand for? Explain the roles of SDS in SDS-PAGE (keep in mind - two major roles). Sodium deoceyl sulfate polyacrylamide gel. SDS is a detergent that denatures proteins, and gives them a nigative charge proportional to their mass, allowing for size seperation..

10. Compare the information you could obtain by SDS-PAGE with the information you could obtain by using Western blotting. SDS-PAGE seperates proteins based on size. These separated proteins can be transferred to a membrane and incubabted with a specific antibody that will bind to a specific protein. If that protein is present, a secondary antibody will bind to the first and allow it to be visible.

11. General knowledge: distinguish between antibody and antibiotic. An antibody (immunoglobin) is found in the body and is used to identify and neutralize foreign objects like bacteria and viruses. An antibiotic is a compound or substance that kills or slows down the growth of bacteria.

12. How does salt concentration influence hybridization process between target DNA or mRNA and probe during nucleic acid hybridization step performed in northern (or Southern) blotting method? Higher salt concentration neutralizes charged backbones allowing strands to come together, so probe more easily binds.

13. Change “salt concentration” in the above question to:a) temperature at too high of a temp the probe won’t annealb) size of the probe* small probe more likely/takes less time to annealc) probe’s % of identity with the target sequence* The closer in similarity the probe is the more likely of have a

correct match, reduces mismatch.

Lectures 16-191. You have discovered base changes in the promoter region of the operon in a bacterial chromosome. Would you expect

these changes to act in trans on another copy of the operon? Explain your reasoning. No, since the change is in the sequence or cis element and not the protein or trans-factor.

2. What are cis- elements? What are transfactors? Give an example from the Trp operon (or form the Ara-operon). Cis elements are regulatory sequences where trans factors bind. Transfactors are regulatory proteins that bind to cis elements. Two types of trans-factors are activators and repressors.

3. Draw the diagram of the lac operon that illustrates negative control (be careful here and think about the complete picture!!!!!). Negative regulation occurs when regulatory proteins in active state turn off the expression.

When there is no lactose present the repressor protein (orange) binds to the operator, and therefore no beta-galatosilase, permease, or transaceetylase (or very little since leaky) is required to escort lactose into the cell.

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Lactose (blue dot) is present and therefore prevents repressor protein from binding.

4. You have isolated a protein that binds to DNA in the region upstream of the promoter sequence of the gene of interest. If this is a positive regulator (activator) which would be true:

A) Loss of function mutation in the gene encoding this DNA binding protein would cause constitutive expressionB) Loss of function mutation in the gene encoding this DNA binding protein would result in lower or no expression.

Explain your reasoning. The protein or activator would no longer be able to bind and turn on the gene.

5. Discuss why are lac Oc mutants cis-acting. They prevent the binding of the repressor protein by changing the sequence (cis element).

6. Discuss why are lac I- mutants trans-acting. They alter the repressor protein (trans-factor) preventing it from binding.

7. Discuss positive and negative regulation of L-ara operon. AraC regulator as an activator upon binding of arabinose and binds at one set of cis elements upstream from the promoter. araC regulator as a represon bindsa at different cis elements upstream of the promoter.

Positve:

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Negative:

8. Regarding the regulation of Trp operon, what do we call the amino acid tryptophan? Why? Tryptophan is referred to as a co-repressor because it binds to an aporepressor to form an active repressor. Tryptophan controls production of itself.

9. What is meant by polycistronic mRNA? Give an example. Polycistronic mRNA has all structural genes in the operon transcribed as part of the same mRNA. Ex. Lactose operon, tryptophan operon, and arabnose operon

10. What is catabolite repression? What is the role of Catabolite Activator Protein? Explain its action (remember the # of operons it activates!). Catabolic repression allows bacteria to adapt quickly to a preferred carbon or energy source first, typically glucose. Catabolite Activator Protein of cAMP is a positive regulator of the lac operon when cell is starving, causes catabolism of lactose. cAMP binds to CAP and complex binds to CAP enhancer element found upstream from RNAP binding site where is stimulates transcription of Z, Y and A. Found in more than 100 genes/operons.

11. Define: repressor, co-repressor, aporepressor and inducer. Repressor: prevents RNAP (RNA polymerase) from bidning to the promoter. Co-repressor: binds to aporeprssor which then binds to mRNA to prevent binding of RNAP. Aporepressor: forms with co-repressor to from represser, always found whereas corepressor may not be present. Inducer: starts gene expression.

12. Define effector and inducer. Give examples. Effector: recognize changes in cell environment ex. Inducers and co-repressors. Inducer: inactivates repressor, induces transcription ex. Lac operon.

13. What are activators? What are enhancers? Activators: are regulatory proteins that turn on the expression. Enhancers: increases the expression

14. What is the role of auxiliary operators? Auxiliary operators are operator regions that contain multiple operators. Having multiple operators can better prevent binding of RNAP in Lac-operon, better suppressing transcription.

15. Discuss the type of regulation of gene expression by two-component regulatory systems in bacteria? In two component regulatory systems, one protein is the sensor-transmitter protein, which monitors specific changes in the environment. The sensor is an outer part, which detects specific changes in the environment, and the transmitter is an inner part, which usually acts as a kinase. The other protein is a response regulator protein that either stimulates or represses regulation of

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specific genes, changes in gene expression are necessary for bacterium to adapt to environmental change.

16. Glutamine and arginine in DNA-binding proteins tend to make what kind of bonds with DNA? Glutamine and arginine in DNA binding proteins make hydrogen bonds with DNA due to amine group.

17. List different ways of control of prokaryotic transcription initiation and give one example for each of them (this question could be separated in few smaller questions; make sure you understand review slides and that you have at least one example for each of the “control ways”). Multiple operators (auxiliary operators): ex. Lac-operon region contains 3 operators. Two component regulatory system: ex. E.coli osmoregualrity system.

18. Define constitutive and regulated proteins. What is the difference between expression of constitutive and expression of regulated proteins? Constitutive proteins: proteins produced in fixed amounts regardless of the organisms requirement for them, they are expressed all the time. Regulated proteins: are produced only when needed, and are not expressed all the time.

19. What is the most important characteristic of binding sites for prokaryotic regulatory proteins (for example lac operon operator)? How are those binding sites different from the RNAP binding site (promoter)? Binding to the operator is regulated – negative control, whereas the RNAP will always bind to the promoter if it can; may be blocked by operator.

20. Describe the most common structural motif found in a DNA binding domain of prokaryotic regulatory proteins. The most common structural motid found in a DNA binding domain is the helix-turn-helix motif. It’s about 20 amino acids long and contains 2 short alpha helices (7-9 aa long) connected with a short turn.

21. Apart from the DNA binding domain, we have mentioned other two domains found in prokaryotic binding proteins. What are they? The effector binding domain is where effectors bind to repressors ex. Inducers (lac operon) or co-repressors (trp operon), and the Oligomerization (dimerization) domain are two additional domains.

22. Thinking question: three adjacent genes are involved in arginine biosynthesis. (They are structural genes of arginine operon.) You have three DNA fragments, each containing coding sequence for one of these three structural genes, and

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you have used these fragments to make three different probes for northern blotting (one probe for each of the three genes).

You have isolated mRNA from bacteria and you have performed the agarose gel electrophoresis followed by northern blotting.

What result do you expect to get if you use each of these three probes as a probe in northern analysis? What would be the result if three genes do not make up an operon? You could use diagrams to explain your reasoning. The expected result would be that the 3 different probes show three separate bands. Each gene will be partially protected by the repressor/activator when RNA polymerase is added. If the gene is recognized by the probe than means it was protected makes up the operon. If three genes do not make up the operon, 3bands will not be seen on the film.

Lectures 20-231. What are the major differences between prokaryotic and eukaryotic transcription? In eukaryotes:

- transcription and translation occur in separate compartments- eukaryote pre-mRNAs are subject to extensive post-transcriptional modification (processing)- chromoatin structure in eukaryotes limits accessibility (transcription is tightly regulated)- eukaryotic RNAP does not recognize binding sire by itself (needs general transcription factors to help)- three RNA polymerases (may be more)

2. Which hypothesis regarding eukaryotic RNAPs was proven with a-amanitin and actinomycin D (be specific)? It was proven that three nuclear RNA polymerases had different roles in transcription. RNAP II is the most sensitive to a-amanitin, since it binds to the RNAP II Rpb1 subunit and stops translocation of the RNAP II along the DNA template. RNAP I is the most sensitive to actinomycin D which intercalates inot GC rich regions and inhibits transcription. By inhibiting RNAP I or II transcription was also inhibited.

3. Which genes are transcribed by RNAP I? RNAP II? RNAP III? RNAP I synthesizes large rRNA precursors. And 28S, 18S and 5.8s rRNAs. RNAP II synthesizes hnRNAs, snRNAs and mRNAs. RNAP III synthesizes 5S rRNA precurser, tRNA precursors, 5s rRNA, tRNAs, U6 snRNA, 7SL RNA, and 7SK RNA.

4. What does CTD stand for? Explain the role of CTD tail in eukaryotic gene expression? CTD stands for carboxyl tail domain. It is necessary for methyl cap addition and polyadenylation as well as splicing. Un-phosphorylated CTD tail is used to initiate transcription. It is not present in RNAP I or III, but is found in RNAP II.

5. How would you define enhancers? What are their characteristics? What is the difference(s) between enhancer and upstream control element? Enhancers are control elements that stimulate transcription. They can occur anywhere including downstream in an intron or in an exon and be thousands of bp away. As well the orientation of the enhancer does not matter. Upstream control elements are relatively close to the promoter (GC boxes, CCAAT boxes, promoter-proximal elements). Transcription requires the upstream control elements but can still transcribe without enhancers???

6. Explain the use of reporter genes for estimation of promoter strength. Reporter genes express characteristics that are easily identifiable and measurable. A reporter gene can be cloned after the promoter of interest. The level of expression is measurable and is proportional to the strength of the promoter.

7. Explain briefly 5’ deletion series. What kind of information do they reveal? By deleting/mutating a part of the promoter starting from the 5’ end and repeat the experiment and build a map of the control region of the promoter with indications of relative importance of sequence in that area.

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8. Explain the modular nature of RNAP II promoters. RNAP II promoters are broken up into sections. Many have a TATA box as a core element. They have proximal elements, enhancers, and cis elements. Expression of one gene can be induced or increases or decreased or shut off under/in different conditions due to the modularity of the gene’s promoter.

9. Draw a diagram of RNAP II promoter (show all types of elements that we have mentioned in class).

10. Explain the tissue (cell type) specificity of eukaryotic cis elements. The same cis elements are present in each tissue. Tissue specificity is due to either presence of absence of particular regulatory (binding) proteins (transcriptions factors).

11. Knowing that different genes may have the same promoter and enhancer elements and that different transcription factors contain the same structural features, how would you explain transcriptional specificity? There are multiple regulatory sites on genes which respond to different regulatory proteins. There are also gene specific transcription factors found that are required for transcription along with general transcription factors.

12. If you know the binding site for certain transcription factor (TF), outline experiments you would use to purify this TF and to assay its activity. After isolating nuclear proteins including TFs, the TFs can be isolated by ion exchange chromatography where by increasing salt concentrations, proteins are bound with different affinity and the highest salt concentration contains possibly highly purified TF. DNA-affinity chromatography can be used

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since beads attract DNA sequences from promoter region. Specific DNA-affinity chromatorgraphy can be used to

purify specify TF.

13. Distinguish between the function of promoters and enhancers in transcriptional regulation. Promoters are sequences that must be located in the vicinity of the start point and are required for initiation of transcription. Enhancers influence transcription rate.

14. Distinguish between the function of general transcription factors and transcription activators in transcriptional regulation. General transcription factors act on promoters and transcriptional activators act on enhancers.

15. List and briefly explain four major domains in eukaryotic transcription factors. DNA binding domain: Zn fingers, one alpha helix in homeodomain, basic alpha helix part in bZip and basic alpha helix in bHLHDimerization domain: one of the Zn fingers, leucine zipper part in bZip, and seond helix in bHLHTranscription activation domain: acidic, glutamine rich, and proline richLigand binding domain: in Zn finger – C4 factors (steroid hormone is ligand)

16. List most frequent structural motifs in eukaryotic DNA binding domains . Zinc finger (C2H2 zinc finger, C4 zinc finger, and C6 zinc finger-yeast), Homeodomain Proteins, Leucine Zipper proteins, and Helix loop helix.

17. List three classes of transcription activation domains in eukaryotic transcription factors. Acidic activation domains, glutamine rich domains, and proline rich domains.

18. What is achieved by the ability of some transcription factors to form heterodimers (basically two things/players in regulation)? Combinatorial control is achieved and is control by combination of different proteins. Co-activators or co-repressors come as a result of direct contact with activators of repressors through transcription activation domain.

19. What is meant by the independence of the DNA-binding and transcription-activating domain of a transcription factor? The activation domain is important only for activation and the DNA binding domain is only for binding, the two are independent. Ex. RNAP binds to promoter (DNA binding domain), where an activator would bind to an enhancer (activation domain).

20. What is the role of the TATA box? What happens when TATA box is removed from the RNAP II promoter? The TATA box is a core element where a TATA binding protein binds as the first stage in initiation for a positioning factor to bind and begin transcription. When the TAT box is removed transcription cannot occur unless the promoter can act as a TATA-less promoter.

21. What is combinatorial control of transcription? Multiple proteins are required to work together to control transcription. Ex. The transcription-initiation complex is made up of RNAP II and general transcription factors bound to the promoter region.

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22. What are the roles of TFIIH in transcription initiation by RNAP II? TFIIH contains 9 subunits and has heilcase activity allowing for unwinding of DNA downstream from the initiator site in the presence of ATP causing promoter clearance (relase of general transcription facfotrs). It also has proteon kinase activity and phosphorylated the CTD tail of RNAP II, which allows RNAP II to detach from TFIID.

23. What are the roles of TFIID in transcription initiation by RNAP II (be as specific as possible; have to talk about TBP and TAFs)? TFIID is the largest general transcriotion factors and consists of a TATA box-binding protein (TBP) and TBP-associated factors (TAFs). It is used in the foundation for the transcriptional complex since TBP has to be ound for transcription to start. It also prevents stabilization of the nucleosome.

24. What are the roles of TFIIB in transcription initiation by RNAP II? TFIIB is a monometic protein and binds one end to DNA and the other to TBP causing bending of DNA downstream from TBP. So, TFIIB orients RNAP on DNA.

25. What are TATA-less promoters? How could transcription be initiated at TATA-less promoters? TATA-less promoters are promoter regions lacking a TATA box but have other elements such as initiator elements and CG boxes CG boxes that allow for the entire TFIID to bind (TBP cannot bind by itself).

Lectures 24-281. Describe the role of histone acetylation/deacetylation in regulation of transcription. Transcription factors regulate

histone acetylation/deacteylation. When acetylated (by histone acetyltransferases – HATs) the positive charge of lysine (part of histone) is neutralized and eliminates interactions with DNA, causing the chromatin to be less condensed and the promoter regions accessible (euchromatin). Transcription can occur when the histone is acetylated.

2. Describe the role of chromatin remodelling complexes in regulation of transcription. Chromatin remodeling complexes are required since histone acetylation is not sufficient for activation since nucleosomes are still intact. Nucleosomes are repositioned (alter structure of nucleosome, move nucleosome on DNA) to expose the promoter elements by chromatin remodeling protein complexes.

3. Describe an influence of activators and repressors on assembly of initiation complexes. Activators are required to modify histones and bind to regulatory region to prepare the gene for formation of initiation complex. Repressors can inhibit gene activation by binding to the site that overlaps the binding site, sites adjacent to the activator site, sites upstream that interact through a mediator, or by forming a co-repressor. Basically, activators allow assembly of initiation complexes, whereas repressors restrict it.

4. Explain the role of enchancesomes and architectural proteins in regulation of transcription initiation? Enhancesomes are used for specific gene expression, they are made up of multiple enhancers and allows for fine control through different combinations and concentrations for transcription factors – combinatorial control. Architectural proteins change the shape of the DNA in a control region, usually help in causing DNA to bend or loop. In some cases if the distance between core promoter and enhancer is too short additional architectural TFs may be necessary to allow the DNA to bend or loop.

5. Explain the role of mediators (or insulators – different question) in regulation of transcription initiation. Mediators or co-activators are required by some transcription factors. These proteins do not come in direct contact with DNA but connect TFs with GTFs. Ex. CREB binding protein (CBP) connects cAMP response element binding protein (CREB) with the basal complex. Insulators set up boundaries between DNA domains, which prevent

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activation/repression of genes by close, but unrelated activatos and repressors and can also prevent gene silencing.

6. What is the role of DNA methylation in regulation of transcription initiation? Methylation of cytosine in CpG doublets prevents transcription. Methylation affects chromatin structure. Extensive methylation near (or in) a gene sequence generally prevents (or stops) transcription of that gene. Remember there are no methyl groups on the new DNA strand formed during replication, however the parent strand is methylatied.

7. Describe how TFs from nuclear receptors superfamily regulate transcription (one example is enough; remember: nuclear receptor could be in cytoplasm, bound to chaperone protein, or in nucleus, already bound to DNA). Transcription of many transcription factors is under regulation of extracellular signals such as hormones.

8. What is (are) the role(s) of transcription factors during development? During development genes are turned on and off (methylation) depending if needed.

9. Describe the role of TBP during transcription (think about promoters for all three eukaryotic RNAPs and TATA-less RNAPII promoters – how do RNAPs bind to them; also, think about coordination of activities of all three polymerases). TBP is a component of the positioning factor for any of the transcribing RNAPs, meaning it allows each types of polymerase to bind to its promoter. It is involved in coordination of activities of all three polymerases through binding to other polymerase-specific factors. Found within SL1 comples (RNAP I), TFII (RNAP II), TFIIIB (RNAP III), and in TATA-less II promoters.

10. What is the role of Sp1 protein? What is the role of SL1 protein? What do they have in common? Sp1 proteins bind to GC boxes (found in TATA less promoters) and allow entire TFIID to bind. SL1 is a complex responsible for ensuring RNAP I is properly positioned at the start point. Both are used in bind of TATA-binding protein to bind RNAP.

11. What is unusual about type 1 and 2 promoters for RNAPIII polymerase? Type 1 and 2 promoters are internal (not upstream), but the efficiency of transcription is altered by changes on region upstream from the start point.

12. Describe the mechanism of attenuation of the Trp operon. Explain the importance of this mechanism for a bacterium? Attenuation is the regulated premature termination of transcription. In negative regulation the presence

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of tryptophan the production of tryptophan will stop by forming a co-repressor. With attenuation RNAP with terminate transcription before reaching trp structural genes because regions 3&4 on mRNA from attenuator loop. Since translation starts right away loops 2 & 3 don’t have the chance to base pair since 2 is already being transcribed, but when Trp is low it takes longer to translate so 2&3 form a loop that can be taken apart easily to

translate.

13. Could you imagine a mechanism similar to the mechanism of attenuation of the Trp operon in Eukaryotes? Explain your reasoning. This would not occur in Eukaryotes becase transcription and translation do not occur at the same time since transcription occurs in the nucleus and translation occurs in the cytoplasm.

14. Describe two distinctly different ways in which the trp operon is controlled by the overall availability of tryptophan (one way is through the formation of repressor protein and the other through attenuation of transcription. Think about fine tuning of Trp production; the questions below might be helpful). A corepressor protein is formed when excess tryptophan is present or the peptide can be more quickly produced casuing the formation of a loop at 3&4 which prevents anymore transcription since RNAP falls off.

15. Describe the mechanism responsible for shutdown of the trp operon when a plentiful supply of free tryptophan is available. The formation of a co-repressor stops transcription of the trp operon by binding to the operator.

16. Describe the mechanism by which the leader-attenuator region fine tunes the extent of transcription of the structural genes in the trp operon when Trp is available (but not to the point to completely saturate the apo-repressor). Rho-independent termination of transcription causes the formation of a polyU (3&4), which stops transcription before trp operon is reached. Balance between leader and total mRNA synthesis is in accordance with the need for trp.

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17. Describe rho-dependant transcription termination. Rho is a hexameric RNA helicase that has an RNA-binding domain and ATPase activities. It binds to rut site on mRNA and tracks along RNA unitl it catched up to RNAP (since RNAP pauses at termination site it can catch up). Rho unwinds DNA-RNA hybrid (requires ATP). Allows for release of RNAP, mRNA, and rho.

18. Describe rho-independent transcription termination. These termination sites have GC rich self-complementary regions with several intervening nucleotides that when transcribed for stem loop structures and are followed by a series of U residues (polyU). The loop interacts with the surface of RNAP that has just synthesized the polyU region, which contains very weak bonds that break apart when RNAP pauses on the loop. Therefore RNAP dissociated from the DNA.

23. Describe the process of mRNA cleavage and polyadenilation. (Don’t forget the role of CTD tail) Polyadenylation signal is followed by a GU rich sequence and cleavage occurs in between. The polyadenylation signal (AAUAAA) codes for addition of poly-A-tail. CTD recruits enzymes necessary for polymerization. CPSF is a cleavage and polyadenylation specificity factor that binds to AAUAAA. CstF is a cleavage stimulatory factor that binds to GU region. CFI and II are cleavage factors that cleave mRNA when bent. PAP or poly A polymerase binds to create poly-A-tail.

24. How is 5’ cap added to the nascent RNA? 3 modifications are made: 7-met-guanosine is coupled to 5’end, there is a methylation of ribose, and an N6-methylation of adenine. Phosphorylated CTD recruits the following capping enzymes: RNA 5’Triphoshase (RTP) removes the 5’phosphate, guanylyl transferase attached GMP, and 7-methyltranserase modifies the terminal guanosine.

25. What is the relationship between hnRNA and mRNA? hnRNA is covered with proteins during transcription forming hnRNP. hnRNAs include pre-mRNA and snRNA. The binding of proteins may reduce secondary structure and facilitate other processing steps such as splicing and transport.

26. What are the general steps in processing of a pre-mRNA into an mRNA? Removal of introns, addition of 5’cap and

addition of poly-A-tail.

27. What is the role of snRNAs in the spliseosome? snRNAs involved in splicing include are U1, U2, U4, U5 and U6, which are all rich in urigine. snRNAs form snRNPs with Sm proteins. snRNPs are involved in splicing.

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28. What is the role of Sm proteins in the spliceosome? What other proteins (apart from Sm) are found to be associated with splicing? Sm proteins join with snRNAs to form snRNPs that are involved in splicing. The 7 Sm ptoteins are B/B’, D1, D2, D3, E, F, and G. They form an Sm RNA motif in snRNAs.

29. What is the difference between splicing of group I and group II introns? Between splicing of group II introns and spliceosomal splicing? Group I and II introns are self-splicing introns that don’t require any proteins (RNA is a ribozyme) and are removed as a lariat structure in two transesterfication reactions. However, group II introns mechanisms occur ar branch site A, while group I occurs at branch site G.

30. Describe the current model of spliceosomal splicing. U1 attaches to 5’splice site of intron, and U2 binds to branch point A with help from ATP. U4/U6 and U5 form a tri-complex and bind to the 3’intron splice site forming the spliceosome. U6 then dissociates from U4, displacing U1 at the 5’ splice site and U6 and U4 are released activatin the splicesome. The active splicesome contains only U2, U5 and U6 snRNPs. U5 is responsible for positioning and holding.

31. List the roles of 5’ methyl cap. List the roles of polyA tail. List the roles of CTD tail (yes, again). The 5’methyl cap protects the mRNA from degradation but also is important in splicing, nuclear transport and translation. The polyA tail is required to protect the mRNA from degradation. The CTD tail is required to recruit enzymes to form the 5’methyl cap and the polyA tail.

32. What are the two types of transcriptional units in Eukaryotes? Use diagrams. Simple transcription units (constitutive splicing) and complex transcriptional units (alternative splicing)

33. List and explain two means of control of gene expression that could happen during pre-mRNA processing. Alternative splicing: Selections of mRNA (introns) are removed based on cell type, developmental stage or environment. Trans-splicing: mRNAs are constructed by splicing of separate RNA molecules; splicesomes then joins exons of one gene with the exon of another.

34. Describe one case of control of gene expression at the level of mRNA processing by means of splicing. In drosophila regulated splicing of alternative exons results in differential expression in male and female. Therefore, alternative splicing has a role in sex determination in this case.

Lectures 29-31

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35. List and briefly explain different mechanisms of post-transcriptional control of gene expression (think about examples)? Localization of mRNA to specific regions: move the mRNA close to the sites where the protein product is required. Transport to the specific region involves different elements that allow different mRNAs to be recognized as targets for transport, these elements are mapped within 3’UTR – zipcode sequence. Ex. Proposed model of ZBP1-mRNA localization: ZBP1 shuttles into and out of the nucleus. If it encounters the nascent zipcode sequences it binds there as the nascent RNA finishes transcription and processing before export to the cytoplasm. Once in the cytoplasm it associates with cytoskeletal elements and localizes to the leading edge.RNA editing: alteration in the sequence of the mRNA. Ex. Deamination of adenine to inosine (recognized as G) or cytosine to uracil. An extreme form of RNA editing occurs when many nucleotides are removed or added. In mitochondra of protazoa and the mitochondirai and chloroplast of plantthere is an addition or deletion of uracil residues, directed by guide RNAs (gRNAs). The 5’ of gRNA binds to 3’ of transcript mRNA and directs changes. Postranscriptional silencing by siRNA or miRNA: Stops the translation of mRNA. Ex. Only certain proteins needed at points of development, larval protein only needed in early development. Specific protein complex(es) – DICER will process short noncoding imperfect dsRNA hairpins into single stranded micro RNAs (miRNAs) or if long into double stranded sequences called small interfering RNAs (siRNAs or iRNAs).Translational control switch: coordinate and opposite translational regulation of ferritin and transferring receptor, switched between ferritin and transferring activation depending on presence of ironRNA stability: balance between mRNA degradation and synthesis determines the level of individual mRNAs in cells; short-lived mRNAs in eukaryotes have copies of AUUUA. Degradation rate of mRNAs can be regulated by interaction with multiple specific RNA binding proteins; there are different binding proteins for stabilization and degradation.

36. Explain the connection between pre m-RNA splicing and transport of mRNA from the nucleus. Before transported from the nucleus, pre-mRNA undergoes splicing, which allows for different protein products from a single gene using alternative splicing but also allows for addition of 5’cap and poly-A-tail need to protect mRNA in cytoplasm.

37. What is trans-splicing? Give an example. Generates unique mRNAs by association and linking of exons from different pre-mRNAs. Ex. In Trypanosoma mini exons will splice with 5’ end exons of all protein coding transcripts contains in polycistronic primary transcript (lack internal introns).

38. What does S in 16S stand for? What is the numerical value of this constant? S stands for Svedberg units which measure both size and shape.

39. What are the roles of three major RNAs in protein synthesis? mRNA: carries genetic information from DNA in the form of codons. tRNA: translates mRNA code; each amino acid has its own tRNA that binds to mRNA due to condon-anticodon complementarity. rRNA: associates with proteins to form ribosomes which catalyze the assembly of protein chains.

40. What is the name of the region of tRNA molecule which attaches to an amino acid? The acceptor stem attaches to an amino acid.

41. How many different tRNAs are there in an eukaryotic cell? How many different aminoacyl tRNA synthetasis are there in an eukaryotic cell? In eukaryotic cells there are around 50 tRNAs and 20 aminoacyl tRNA synthtasis (one enzyme will attach one of the 20 amino acids).

42. What are the roles of tRNA in translation? tRNA is linked with a particular amino acid, which it brings to a recognized codon in mRNA, building the protein.

43. What is the wobble position for an anticodon? For a codon? Anticodon: wobble position is on the 5’ end, or first position. Codon: wobble position is on 3’ end or last position. A wobble position has greater variability in recognition.

44. Explain what does it mean when we say that the code is degenerate? The code is degenerate because there is redundancy for the coding of amino acids (codons are synonymous).

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45. What is the role of Shine-Dalgarno sequence? The Shine-Dalgarno sequence is a conserved sequence found in prokaryotes 8-13 nucleotides upstream of the start codon. It is used to position the ribosome correctly with respect to the initiation codon, so it is involved in translation initiation.

46. What is the role of Kozak sequence? The Kozak sequence is found in eukaryotes and is contains the start codon the 40S ribosomal subunit to scans for.

47. What are the major differences in initiation of translation between eukaryotes and prokaryotes (remember: first AA and the way mRNA and tRNA bind to ribosomes)? Prokaryotes: The Shine-Delgarno sequence base pairs with the complementary sequence at the 3’ end of the 16S rRNA in the small 30S subunit. Steps: IF3 binds to free 30S subunit. IF1 binds and IF2 (GTPase) complexes with GTP and binds. mRNA binds to 30S subunit through interaction of Shine-Delgarno sequence with 16S rRNA. Initiator tRNA binds (anticodon-codon base pairing) to P site forming the 30S initiation complex. The 50S subunit binds displacing IF1, 2, and 3, forming the 70S initiation complex.

Eukaryotes: 40S ribosomal subunit first binds initiator tRNA; then 40S subunit-initiator tRNA complex binds mRNA and scans along mRNA until it reaches an appropriate AUG and positions initiator tRNA there. Steps: The free 40 subunit complexes with eIF3 and eIF1A, keeping it apart from the 60S subunit. The ternary complex forms from the initiator tRNA, eIF2 and GTP, which then binds to the 40S subunit creating a 43S preinitiation complex. Meanwhile, different eIF4 factors are involved in recognition of the 5’methyl cap and they keeo mRNA free of any secondary structures using energy from ATP. mRNA is scanned by 43S complex to find right AUG, inside kozak sequence. When 43S complex finds Kozak sequence, eIF5 binds and displaces all other factors and the 40S initiation complex is formed, allowing the 60S subunit to bind, forming the 80S initiation complex.

48. Briefly explain two cases of RNA having enzymatic capability (how many ribozymes did we mention in class)? rRNAs have enzymatic capability. 16S and 23S rRNAs have direct contact with tRNA. 16S base pairs with Shine-Delgarno sequence. 23S brings the P loop bound peptidyl-tRNA into an interaction with bases in central region; used in elongation of a protein chain.

49. What is a peptidyl transferase? What is catalyzed by peptidyl transferase? Peptidyl transferase forms the peptide bonds between previous and newly added amino acids. The 23S component in prokaryotes contains the peptidyl transferase component and acts as a ribozyme. Peptidyl trasnferase is rRNA or ribosomal proteins.

50. How is gene expression controlled at the level of translation (mRNA is now in contact with ribosomes or their subunits/translation factors; three things mentioned in the class)? Phosphorylation of translation initiation factors: Upstream AUG codons: IRES

51. Which two factors greatly influence efficiency of protein synthesis? 1. Simultaneous translation of a single mRNA by multiple ribosomes: formation of polysomes. 2. Rapid recycling of ribosomal subunits; circularization of eukaryotic mRNAs allows everything to stay close to the start site for repeated protein synthesis.

52. 52. What are the roles of PABI? What is the role of PABII? PolyA-binding protein I is involved in rapoid recycling of

ribosomal subunits (bound to poly-A-tail).

53. Explain what happens with polypeptides after translation. Polypeptides undergo posttranslation modification and transport after translation. This includes folding of polypeptide into protein.

54. How are protein modifications related to the control of gene expression? List possible protein modifications. Protein modifications are necessary for transport from the cytoplasm. Proteolytic Cleavage: most proteins undergo proteolytic cleavage following translation; the simplest form is removal of the first methionine. Acylation: in most cases an acetyl group is added to the N-terminal amino acid. Glycosilation: glycoproteins consist of proteins with covalently linked sugars. Methylation: post-translational methylation occus at lysine residues in some proteins. Prenylation: addition of the compounds derived from the cholesterol biosynthetic pathway. Phosphorylation: post-translational phosphorylation is one of the most common protein modifications; occurs as a mechanism to regulate the biological activity of a protein.

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Sulfation: sulfate modification of proteins occurs at tyrosine residues; since sulfate is added permanently it is necessary for the biological activity and not used as a regulatory modification.

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BIOL 308 - Study Questionss3.amazonaws.com/prealliance_oneclass_sample/BMRXYo8daw.pdf · BIOL 308 - Study Questions Lectures 5-7 1. Thinking question re DNA/RNA structure: Certain