Exam 1 Answers

BCH4024 SU2015 E1A 90 minutes

1. D 26. D 2. D 27. B 3. D 28. C 4. E 29. B 5. C 30. B 6. C 31. C 7. B 32. D 8. C 33. C 9. C 34. D 10. C 35. E 11. B 36. B 12. B 37. B 13. B 38. C 14. B 39. A 15. D 40. B 16. C 41. D 17. C 42. C 18. D 43. C 19. B 44. C 20. B 45. A 21. E 46. B 22. B 47. D 23. A 48. C 24. C 49. D 25. E 50. A (B) B


1. Saccharomyces cerevisiae, a yeast, would be most closely related to which of the following species?

D. Homo sapiens L1/S4

Yeasts, which are fungi, are closer in phylogeny to animals than to molds or plants. 2. Which of the following statements regarding the cytoskeleton is correct?

D. Each filament monomer interacts by noncovalent interactions. L1/S7,10

Actin filaments are the smallest; it is dynamic; it helps cells and organelles move; it also aids in cell division (mitosis). “Each filament is compose of protein monomers bound noncovalently to form a long polymer.”

3. Which of the following functional groups is not represented in the molecule below?

D. Sulfhydryl L1/S19

Don’t miss these problems. 4. Which molecule is the heaviest, on average, and how many of its molecules exist in a typical

human cell?

E. DNA; 46 L1/S16 DNA is the heaviest molecule at 1 × 1011 g mol-1 (or Da), and there are 46 molecules (46 chromosomes) in a typical diploid cell.


5. The arrangement of water molecules in ice is:

C. A tetrahedral lattice due to 4 hydrogen bonds. L2/S4 Solids typically have lattice structures. Ice is stated to use all 4 of the possible hydrogen bonds. Gases are the ones with no fixed arrangement. Liquid water has the “flickering” structure. Sheets of molecules that can slide are actually a characteristic of graphite, not ice.

6. An unknown solution has a pH of 12.33 due to a compound. Given that ["

#][%"]

is 24.24, what

is the 𝐾' of the chemical in solution?

C. 1.12 × 10-11 L2/S22 Use the Henderson-Hasselbalch equation:

𝑝𝐻 = 𝑝𝐾' + 𝐿𝑜𝑔

𝐴0

𝐻𝐴

12.33 = 𝑝𝐾' + 𝐿𝑜𝑔 24.24 12.33 − 1.39 = 10.95 = 𝑝𝐾' 𝐾' = 100;<= = 100>?.@A

𝐾' = 1.13×100>>

7. A researcher discovers a new acid, X. At pH 6.4, there is one thousand more times XH than

X-. What is the pKa of X?

B. 9.4 L2/S22

Use the Henderson-Hasselbalch equation:

𝑝𝐻 = 𝑝𝐾' + 𝐿𝑜𝑔 𝐴0

𝐻𝐴

6.4 = 𝑝𝐾' + 𝐿𝑜𝑔 1

1000

6.4 + 3 = 9.4 = 𝑝𝐾'

Remember that it’s + Log, not – Log. This makes sense because if you increase [A–], you increase the amount of base, so pH should increase because the Log expression increases.

8. Which of the following statements is true?


C. Protein structure determines phenotype. L3

DNA is transcribed into RNA, not translated. mRNA is translated into protein, not transcribed. The minor groove is a “palindrome”, so to speak, in that the forward and reverse base sequences are read in the same way, so it can’t be more specific than the major groove. Nucleosides lack the phosphate, nucleotides have the phosphate. It is true that protein expression is what determines one’s phenotype. Even if you were unsure about whether this was true, you could eliminate the other definitely false answers to arrive here.

9. The molecules below are the monophosphate versions of (d)NTPs. Pyrophosphorylation of

which of these molecules forms the dominant energy “currency” for all living cells?

C.

L3/S8 Pyrophosphate is 2 inorganic phosphates. The molecule in question is adenylate, or adenosine 5’—monophosphate, AMP. Addition of 2 inorganic phosphates to AMP forms ATP, the main form of energy in living cells. Note that it is not the deoxy form, or else we would call it dATP, so the 2’—OH is present.

10. In the year 2013, a deceased alien life form was found in a crater. Scientists analyzed its

genetic code, and found that it was similar to Earth-based life. They discovered that the percent of thymine was 37%, and its genetic code was 3.2 × 109 base pairs long, similar to humans. Given that the GC content in humans is about 41%, how would the thermal stability of the alien’s genome compare with that of a human’s?

C. It would be less stable, because genome stability depends linearly on GC content L3/S10

%T = 37% = %A, so %AT = 74%, and %GC = 26%. This is lower than the human %GC of 41%, so its genome would be less stable. From graph 2 on S20, Tm is seen to depend linearly on %GC. A higher Tm inplies a greater stability.


11. Which of the following diagrams correctly indicates the pattern of hydrogen bonding in an AT base pair?

B.

L3/S13 Just draw these bases out a bunch of times. One way to remember it might be to pair up the numbers. Ex. N1—NH’3 and NH6—O’4. Try to see these from multiple perspectives, because, like here, they may not be in the same arrangement as in the slides.

12. β-mercaptoethanol is commonly used to break disulfide bonds. Which of the following

molecules would react with β-mercaptoethanol?

B. Cystine L4/S20

Only cystine and methionine have sulfurs, so eliminate A and D. Recall that the pairing of 2 cysteine molecules yields cystine, which has a disulfide bond.

13. Which of the following sequences is not a valid sequence of amino acids?

B. ADMIRALACKBAR L4/S29

If you know your one-letter codes, you should realize that B is not an amino acid you are responsible for, so treat it as an invalid entry.

14. Which of the following amino acids can be synthesized de novo in the human body?

B.

L4/S30 If the body can synthesize it, it’s not an essential amino acid. Y is not an essential amino acid. Some people like to use PVT TIM HaLL to remember the essential ones, but that’s a little confusing since those aren’t actually the one-letter abbreviations. I like WTF MILK H(I)V better.

15. What is the pI of the following amino acid, given that it has buffer points at 2.18, 8.95, and

10.53?


D. 9.74 L4/S10

“Buffer points” just means those are its pKa’s. pI occurs when the net charge is 0, and that will happen when the C terminus (2.18) is deprotonated, the N terminus (8.95) is protonated, and the side chain (10.53) is neutral. So, between 8.95 and 10.53, Lys (this amino acid) will have a neutral charge. Average 8.95 and 10.53 to obtain 9.74. If you had trouble determining which was the N terminus and which was the R group, most N termini are around 9 and most C termini are near 2. 8.95 is closer to 9 than 10.53 is.

16. What is the approximate length of a peptide bond?

C. 1.32 Å L5/S7

Peptide bonds are 1.32 Å long, notably shorter than the C—N bond adjacent.


17. Find the isoelectric point of the following peptide:

Given this information:

Amino acid pKa1 pKa2 pKa3 His 1.82 6.00 (3) 9.17 (4) Glu 2.19 4.25 (2) 9.67 Leu 2.36 (1) 9.60 Lys 2.18 8.95 10.53 (5) Ser 2.21 9.15

C. 7.59 L5/S16-17

First, work out the relevant pKa’s. The ones to consider are highlighted in red, and are ranked numerically. Start at the fully protonated form, at pH = 1: pH = 1: +3 charge pH = 2.36: +2.5 charge pH = 3: +2 charge pH = 4.25: +1.5 charge pH = 5: +1 charge pH = 6.00: +0.5 charge pH = 7: 0 charge pH = 9.17: -0.5 charge pH = 10: -1 charge Average the pKa’s on either side of the 0 charge: 𝑝𝐼 = E.??F@.>G

H= 𝟕. 𝟓𝟗

If you weren’t sure which pKa’s to use for Lysine and got 7.48, try to remember that the N terminus is usually around 9, and the C terminus is around 2. 8.95 is closer to 9 than 10.53 is, so 10.53 should be the R group and 8.95 the N terminus.


18. A dodecapeptide is capable of having how many possible sequences?

D. 4.10 × 1015 L5/S19

Dodeca = 12, so a peptide with 12 amino acids. To find the number of sequences of a dodecapeptide (12), do 2012.

19. A homodimerized protein weighs 213 kDa. About how many amino acids are in the

undimerized protein?

B. 968 L5/S18

213 kDa = 213,000 Da, H>L??? M'>>? M' ""#N

= 1936 𝐴𝐴 in the full protein, so in a homodimer (dimer = 2 units, homo = same → a 2 unit protein with each unit having the same sequence), each subunit will have >@LE

H = 968 amino acids

20. The protein PMP–22 is a structural protein of the myelin surrounding certain neurons. Mutations in PMP–22 can lead to Charcot–Marie–Tooth (CMT) disease, which occurs in several types. Type 1E CMT disease is caused by amino acid substitutions, and hearing loss is a common symptom of type 1E CMT disease. PMP–22 consists partly of four transmembrane α–helices embedded within the myelin membrane. A researcher theorizes that type 1E CMT disease may be caused by a substitution of alanine by proline. Is this a likely assessment?

B. Yes, because proline substitution is implicated in destruction of α–helices. L6/S9

The term “missense mutation” may not mean much to you, but you should know that a single changed amino acid won’t always destroy a protein’s structure. While C might be partially correct, it isn’t in this case because Pro is known to destroy α–helices.


21. For the following arrangement of β–sheets, indicate the direction in which each strand runs:

E. None of these.

Without an initial condition given, this problem is a little more tricky. But you should realize that none of the answers would have worked.


22. The structure of PMP—22, described in question 20, is shown below:

Where N and C represent the respective termini of the protein. In which of the regions of the associated Ramachandran plot would one expect a high density of allowed conformation angles?

I. A II. B

III. E IV. D

B. I and III only L6/S26-28

PMP—22 is seen to contain RH helices and antiparallel strands. Thus, one would expect torsion plots to have greatest density in regions A and E of the Ramachandran plot above.

N

C


23. Which of the following sequences is most likely to form 𝛽 sheets?

A. YRVIWYTWV L6/S23

Know those three mnemonics. Use “IVY For The Win”. This is a little on the simpler side of what could be asked.

24. What is the value of ΔG for the following reaction?

C. ΔG = 0 L7/S12

Keq for the reaction is 1. You could use 2 solutions to this problem: Either use 𝛥𝐺° = −𝑅𝑇 ln 𝐾, plugging in 1 for K to obtain 0; or realize that hydrogen bonding does not drive protein folding forward, and that therefore it would not effect a change in ΔG.

25. Which of the following statements is true of Van der Waals’ interactions?

I. They are a type of non-covalent interaction.

II. They contribute to protein stability in hydrophobic molecules. III. The attractive force (𝑉) of Van der Waals’ forces follows 𝑉 ∝ 𝑟0E, with 𝑟 being the

dipole length.

E. I, II, and III L8/S11 Non-covalent interactions are just that: Interactions that are not covalent bonds, which still contribute to protein stability and structure. Clearly, VDW forces are non-covalent. They contribute to protein structure. Individually, they are very weak, but the summation of all these forces makes a great difference. “Attractive force is proportional to (distance)-6”.

OH2N

O

OH

+

HH3N O

O

OH

HOHH

OH

OH2N

O

OH

+

H

NH3O

O

HO

HOH

HOH


26. The GroEL–GroES complex is an example of a chaperonin that assists certain proteins in folding. What is the correct sequence of steps in the chaperonin–assisted folding of a protein by GroEL–GroES?

D. Intermediate delivered by Hsp70–ADP → ATP binds → GroES binds → ADP dissociates → ATP is hydrolyzed

L7/S19 27. Amyloid–β is a transmembrane protein, the misfolding of which has been implicated in the

formation of plaques in Alzheimer patient brain tissue. Which of the following statements regarding Alzheimer’s Disease involving amyloid–β would be observed?

B. A shift in density from (𝜙 = – 90°, 𝜓 = – 60°) to (𝜙 = – 170°, 𝜓 = 150°) on a Ramachandran plot (refer to the plot in question 22).

This one is difficult. First, you must realize that amyloid–β involves the change from an α–helix to a β strand. Therefore, the distance between adjacent residues would increase, not decrease (L6/S22). Self–association does occur in prion diseases (L7/S22,23), but this happens in the unwinding of α to β. Choice D implies a shift from α to π helices, which is not a characteristic of amyloid–β. Denaturation of helices due to Pro substitution is likewise not applicable for this disease, although it is responsible for others. You could answer this by process of elimination or by referring back to where the four main 2º structures are located on the Ramachandran plot, in terms of their ϕ/ψ angles.

28. Which of the following statements about protein folding is false?

C. Protein folding is driven by hydrogen bonding L7/S12

Protein folding is not driven by hydrogen bonding. All of the other statements are true.


29. An overworked and unpaid undergraduate performs chromatography on a sample of proteins in order to separate them. The protein she is interested in has several α–helices which loop around in the cytosol, and consists of a series of repeating units with the following sequence: … (1) LHAGSMQHLK MLKISTRKAG DKRERRGKLA (30) … Based on this sequence, which of the following chromatography types might she want to use?

B. Cation exchange L7/S13

H, R, and K makes this peptide positively charged, so a cation exchanger is required. You can eliminate gas because that is not a topic we’ve covered in this class. Eliminate size-exclusion because no statement is made about the sizes of the other proteins in the sample. You might be tempted to default to affinity, but for one, you’re not given any information as to whether a specific antibody or coenzyme exists for the protein, and for another, the question asks you to select a chromatography type “based on this sequence”. The best answer is cation exchange, because the sequence doesn’t tell you anything with regards to affinity for specific molecules, but only its charge.

30. A cell extract is comprised of proteins with the following molecular weights:

I. Hemoglobin: 68 kDa

II. Myoglobin: 16.7 kDa III. Bovine serum albumin: 66.4 kDa IV. Fibrinogen: 340 kDa

Which of these proteins would elute last in size–exclusion chromatography?

B. II only L7/S15

The smaller a protein, the longer the elution time, because they get stuck in the pores. Eluting last means that it would take the longest to fall off the column, making myoglobin, the smallest of the batch, the last to elute.


31. In an experiment to isolate myoglobin from hemoglobin from a sample of human blood, a researcher used a salt to precipitate the proteins sequentually from solution. He obtained the following graph:

Where 𝐼𝑜𝑛𝑖𝑐 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ ∝ 𝑀. Based on the graph, which of the following would the researcher be able to conclude?

C. Myoglobin is more soluble than hemoglobin because a higher salt concentration is needed to precipitate it.

L7/S10 The more salt needed, the more soluble it is, and vice versa.

32. Which of the following statements regarding electrophoresis is false?

D. Proteins with a higher native charge travel farther in SDS-PAGE. L7/20-22

The point of applying SDS-PAGE is to flood a protein with a negative charge. This makes it so that the initial charge doesn’t matter, and only its Mr does.


33. Which of the following statements about myoglobin is false?

C. The 1° structure of myoglobin is similar to that of a single hemoglobin subunit. L8/S13

The 3° and 2° structures are highly similar, but there are few similarities in the 1° structures other than the positions of the His residues.

34. Hemoglobin A, adult hemoglobin, is a tetramer comprised of 𝛼 (141 AA) and 𝛽 (146 AA)

chains. Hemoglobin A most commonly uses Heme B. What is the estimated molecular weight of fully oxidated hemoglobin? (Heme B = 𝐶Lk𝐻LH𝑁k𝑂k𝐹𝑒)

D. 65,732 Da 2 β-strands: 146 × 110 × 2

2 α-strands: 141 × 110 × 2 Heme B = (34 × 12) + 32 + (14 × 4) + (16 × 4) + 56 = 616 4 Heme B per Hb = 616 × 4 4 O2 per Heme B = 32 × 4 ∴ (141 × 110 × 2) + (146 × 110 × 2) + [(616 + 32) × 4] = 65,732

35. Arrange the affinities for oxygen of the following globins:

α2β2, α2β2–BPG, myoglobin, α2γ2

E. Myoglobin > α2γ2 > α2β2 > α2β2–BPG L8/S22,32 Myoglobin has the highest affinity for O2. The γ subunit (in fetuses) is higher affinity than the β subunit (adult subunit). BPG lowers the affinity of Hb for O2.

36. Which of the following statements is false?

B. Enzymes change the equilibrium constant. C. L10/S3

Enzymes do not affect equilibria, only rate.


37. IgG is depicted in the image belo

At which region(s) would IgG bind: ○1 to a foreign organism; and ○2 to a macrophage?

B. 1 & 2; 5 L8/S33-35

The “Y”-shaped region bind the antigens (foreign organism) at both tips (1, 2), and the base of the “Y” (5) binds the white blood cell.

1 2

3

4

5


38. The following graph shows the oxygen dissociation curves for hemoglobin and myoglobin:

Where pO2 (mmHg) is plotted on the x–axis, and mole fraction of oxygen (𝑌tH) is plotted on the y–axis. In the presence of BPG, how would the curves for hemoglobin and myoglobin change?

C. The curve for hemoglobin will shift right; the curve for myoglobin will not shift. L8/S22

BPG lowers the affinity of Hb by binding in between β-subunits. Mb has no other subunits, so it does not bind BPG. A rightward shift on the graph indicates a lower affinity for O2 (more pO2 is needed to have the same partial pressure).


39. A researcher wants to identify the sequence of an unknown peptide. One of the steps she takes in doing so is to digest part of it with chymotrypsin. SDS-PAGE analysis of the original protein shows one band, and analysis after digestion shows 4 bands. Which of the following could be the sequence of the peptide?

A. … (N) PSLY THECRRECTANSERY NAW TMPA (C) … L10/S21

“3 new bands” forming means there was 1 to begin with and now there are 4. The correct peptide pieces are shown by spaces. Chymotrypsin cleaves at the C-end of the amino acids Y, W, F, unless there is a P immediately afterwards (ruling out choice B).

40. A researcher is studying an enzymatic reaction scheme, and believes he has come up

with the correct pathway. A very simplified version is reproduced below:

To prevent this reaction from occurring, a competitive inhibitor can be developed, which is an analog of the molecule for which the enzyme has the highest affinity. Which of the labeled molecules might this analog resemble?

B. B only L9/S20-22

Enzymes have the highest specificity for the transition state. A good competitive inhibitor would resemble the transition state, which is molecule B.


41. In one particular trial, a researcher begins recording his measurements for 𝑉? after starting an enzyme-catalyzed reaction using 10 µμM of enzyme and 1 µμM of substrate. The initial velocity for this trial is determined to be about 3.6 µμM per minute, as per the graph below, and he determines 𝑉v'w to be 4.0 µμM per minute. From this trial, and the following graph of his complete results, he calculates the rate-limiting step to have a rate constant of about:

D. 0.396 𝑚𝑖𝑛0> L12/S14

The Michaelis-Menten equation is 𝑉? =yz={ |{ [}]<~F[}]

. The rate-limiting step is 𝑘�'�. We’ve

been given a particular set of initial conditions which will make it easier to calculate this value.

𝑉? = 3.6 𝜇𝑀 𝑚𝑖𝑛0> 𝐸� = 10 𝜇𝑀 𝑆 = 1 𝜇𝑀

Leaving us to calculate 𝐾v. 𝐾v is the substrate concentration [𝑆�] at which 𝑉� =

>H 𝑉v'w.

𝑉v'w is given to be 4.0 𝜇𝑀 𝑚𝑖𝑛0>, so 𝐾v will be found at 2.0 𝜇𝑀 𝑚𝑖𝑛0>. Draw the lines over, and you’ll find that [𝑆] at this point is roughly 0.1 𝜇𝑀. Therefore:

3.6 𝜇𝑀𝑚𝑖𝑛

=𝑘�'�×10 𝜇𝑀×1 𝜇𝑀0.1 𝜇𝑀 + 1 𝜇𝑀

𝒌𝒄𝒂𝒕 = 𝟎. 𝟑𝟗𝟔 𝐦𝐢𝐧0𝟏


42. The Lineweaver-Burk plot for the enzyme (labeled E1) studied in question 41 is shown below, with that of 2 other enzymes (which also catalyze the same reaction) also superimposed onto the graph (E2, E3).

Which of the following assessments is valid regarding the activity of these enzymes with regards to E1?

C. E2 has the same affinity and a higher 𝑉v'w; E3 has a lower affinity and the same 𝑉v'w. L12/S19

If they catalyze the same reaction, they can be compared. E1 and E2 are seen to intersect the x-axis at the same point, so they have the same 𝐾v. E1 and E3 are seen to intersect the y-axis at the same point, so they have the same 𝑉v'w. This much was given to you. E1 intersects the y-axis higher than E2 does. This means that >

�~=� is greater for E1. This

means that 𝑉v'w is actually smaller for E1. E1 intersects the x-axis more leftward than E3 does. This means that − >

<~ is greater for E1,

meaning 𝐾v is actually smaller as well. Don’t let that negative throw you off. An easy way to remember this: Slower-acting enzymes have steeper lines. This is why inhibition steepens the line. Any change that steepens this line (higher y-intercept, or more rightward x-intercept) will result in a slower reaction, which can result from, respectively, a lower 𝑉v'w, or a higher 𝐾v. (Remembering of course that a higher 𝐾v means a lower affinity!)


43. Which of the following is not a general example of covalent modification leading to regulation of an enzyme?

C. Allosteric interactions of molecular oxygen on hemoglobin. L13/S11

Hemoglobin, while being regulated by the presence of oxygen, is not an enzyme, but rather a binding protein. In addition, the binding of oxygen is allosteric. The rest of these are examples of covalent (as opposed to allosteric) modifications as per the slide given.

44. The structures of Saquinavir and Indinavir are shown below:

These two molecules are capable of inhibiting HIV proteases. What type of inhibition do both of these carry out?

C. Competitive non-covalent L12/S30

Even if you didn’t remember this outright, it is partially answerable by elimination. Both of these structures are fairly similar. Chances are, it’s competitive inhibition. Otherwise, covalent vs. non-covalent is more straight recall.

45. Coupling with the hydrolysis of the 𝛾 phosphate of ATP is one way enzymes overcome

energy barriers to unfavorable processes. Which law of thermodynamics does this exemplify?

A. First law L13/S12

“Energy may change form, and may be transported, but energy cannot be created or destroyed.” Energy from ATP’s hydrolysis contributing to overcoming an endergonic reaction’s activation energy is a great example of the first law of thermodynamics.


46.

Which term best describes the relationship between glucose and galactose?

B. Epimers L14/S7

Glucose and galactose, along with mannose and fructose, are the four carbohydrates whose names are boxed; take this to mean you should know their structures rote. Constitutional isomers are molecules that share the same molecular formula. Surely, glucose and galactose do so, but the question asks you to pick the best term. A more specific term for these exists. They are not enantiomers, because they do not differ in configuration at every carbon. Conformational isomers differ in their conformation. Examples are: Ring flips, Fischer projections, rotations about a single bond. Glucose and galactose differ in orientation at just 1 carbon, C4. The best term for these two molecules is “epimers”.

47.

You discover two new sugars, Tyrannose and Tricerose, named after the organisms from which they were found. Their structures are shown below:

Similarly to how 5 membered cyclic ethers

are called pyrans, 7 membered cyclic ethers are oxepanes and 4 membered ones are oxetanes. When mixed together, tyrannose and tricerose bond together to make a disaccharide thusly:

According to standard IUPAC carbohydrate nomenclature, characterize this disaccharide.

D. 𝛼-D-tyrannooxepanosyl-(1→2)-𝛽-D-tricerooxetanose L14/S13

The left sugar gets –osyl, the right sugar gets –ose. Tyrannose is 𝛼, because the anomeric OH and the CH2OH are trans. Tricerose is 𝛽, because the anomeric OH and the CH2OH are cis. The bond is from 1 to 2. Both molecules are D, because their Fischer projections have the next-to-last OH groups on the right.

Anomeric carbon

Anomeric carbon

1

2


48. The following molecule could be a part of what type of carbohydrate?

C. Dextran L14/S18

This molecule has a (1→2) linkage. Such linkages are a characteristic of dextrans. If you see a molecule that isn’t (1→6) or (1→4) there’s a good chance it’s a dextran.

49. Which of the following fatty acids would have the lowest melting point?

D. 16:1(∆9) cis-Hexadeceneoic acid L15/S5

Don’t memorize this table. But, do realize that even one unit of unsaturation will sharply decrease the melting point. 16:1(∆9) beats 18:1(∆9) because the latter has 2 more carbons, and increasing carbon count increases melting point. You might be tempted to pick 12:0 because it has 4 fewer carbons than 16:1(∆9), but it is important to see that the effects of adding carbon are much less pronounced than of adding a double bond.

50. What is the general structure of a phospholipid?

A. A polar head group attached to two fatty acid chains. L15/S11

Know the general structure of each type of fatty acid complex. Bonus1 T / F: The fatty acid below can be characterized as an omega-3 fatty acid:

B. F Omega-3 fatty acids have a double bond starting at the third carbon, when counting from the opposite side.

1 There are no bonuses on your exam

1 4 7

1 3

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Exam 1 Answers