Biochemistry Exam 1

1. Identify. Examine the following four amino acids (A-D): A: 5-ring connected to alpha carbon and amino group B: CH2-6Ring-OH C: CH2-CH-CH3CH3 D: (CH2)4NH3+

(A) Proline, Pro, P; (B) tyrosine, Tyr, Y; (C) leucine, Leu, L; (D) lysine, Lys, K

3. Match 'em. Match each amino acid in the left-hand col- umn with the appropriate side-chain type in the right-hand column. (a) Leu (1) hydroxyl-containing (b) Glu (2) acidic (c) Lys (3) basic (d) Ser (4) sulfur-containing (e) Cys (5) nonpolar aromatic (f) Trp (6) nonpolar aliphatic

(a) 6; (b) 2; (c) 3; (d) 1; (e) 4; (f ) 5

4. Solubility. In each of the following pairs of amino acids, identify which amino acid would be more soluble in water: (a) Ala, Leu; (b) Tyr, Phe; (c) Ser, Ala; (d) Trp, His.

(a) Ala; (b) Tyr; (c) Ser; (d) His.

6. Name those components. Examine the segment of a pro- tein shown here. (NTerm-CH3-CTerm-NTerm-H-Cterm-NTerm-CH2OH-Cterm) (a) What three amino acids are present? (b) Of the three, which is the N-terminal amino acid? (c) Identify the peptide bonds. (d) Identify the a -carbon atoms.

(a) Alanine-glycine-serine; (b) Alanine; (c and d): Peptide bonds between CO and NH. Alpha carbons connected to H and R group.

2. Properties. In reference to the amino acids shown in Problem 1, which are associated with the following characteristics? (a) Hydrophobic side chain ______________ (b) Basic side chain ______________ (c) Three ionizable groups ______________ (d) p K a of approximately 10 in proteins ______________ (e) Modified form of phenylalanine ______________

(a) C, B, A; (b) D; (c) D, B; (d) B, D; (e) B.

20. Part of the mix. Your frustrated colleague hands you a mixture of four proteins with the following properties: Isoelectric point (pI) Molecular weight (in kDa) Protein A 4.1 80 Protein B 9.0 81 Protein C 8.8 37 Protein D 3.9 172 (a) Propose a method for the isolation of Protein B from the other proteins. (b) If Protein B also carried a His tag at its N-terminus, how could you revise your method?

(a) Ion exchange chromatography will remove Proteins A and D, which have substantially lower isoelectric point; then gel filtra- tion chromatography will remove Protein C, which has a lower molecular weight. (b) If Protein B carries a His tag, a single affinity chromatography step with an immobilized nickel(II) column may be sufficient to isolate the desired protein from the others.

1. Valuable reagents. The following reagents are often used in protein chemistry: CNBr Trypsin Urea Performic acid Mercaptoethanol 6 N HCl Chymotrypsin Phenyl isothiocyanate Which one is the best suited for accomplishing each of the following tasks? (a) Determination of the amino acid sequence of a small peptide. (b) Reversible denaturation of a protein devoid of disulfide bonds. Which additional reagent would you need if disul- fide bonds were present?

(a) Phenyl isothiocyanate; (b) urea; b -mercaptoethanol to reduce disulfides;

5. Complements. Write the complementary sequence (in the standard 5 9 S 3 9 notation) for (a) GATCAA, (b) TCGAAC, (c) ACGCGT, and (d) TACCAT.

(a) TTGATC; (b) GTTCGA; (c) ACGCGT; (d) ATGGTA.

6. Compositional constraint. The composition (in mole- fraction units) of one of the strands of a double-helical DNA molecule is [A] 5 0.30 and [G] 5 0.24. (a) What can you say about [T] and [C] for the same strand? (b) What can you say about [A], [G], [T], and [C] of the complemen- tary strand?

(a) [T] 1 [C] 5 0.46. (b) [T] 5 0.30, [C] 5 0.24, and [A] 1 [G] 5 0.46.

26. Charge! Suppose two phosphate groups in DNA (each with a charge of 2 1) are separated by 12 Å. What is the energy of the ionic interaction between these two phos- phates assuming a dielectric constant of 80? Repeat the calculation assuming a dielectric constant of 2.

1 1.45 kJ mol 2 1 ( 1 0.35 kcal mol 2 1 ); 1 57.9 kJ mol 2 1 ( 1 13.8 kcal mol 2 1 )

What's the ratio? An acid with a p K a of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the proton- ated to the deprotonated form of the acid?

100

19. Protein purification problem. Complete the following table. Purification Procedure Total protein (mg) Total activity (units) Specific activity (units mg ?1 ) Purification level Yield (%) Crude extract 20,000 4,000,000 __ 1 100 (NH 4 ) 2 SO 4 precipitation 5,000 3,000,000 __ __ __ DEAE-cellulose chromatography 1,500 1,000,000 __ __ __ Gel-filtration chromatography 500 750,000 __ __ __ Affinity chromatography 45 675,000 __ __ __

200 600 3 75 667 3.3 25 1,500 7.5 19 15,000 75 17

15. Find the pK a . For an acid HA, the concentrations of HA and A ? are 0.075 and 0.025, respectively, at pH 6.0. What is the p K a value for HA?

6.48

8. Alphabet soup . How many different polypeptides of 50 amino acids in length can be made from the 20 common amino acids?

8. There are 20 choices for each of the 50 amino acids: 20 50 , or 1 3 10 65

13) Acetic acid has a pKa of ~4.5, while carbonic acid has a pKa of ~6.5. Which statement below best describes their expected status at pH 5.5? A) Acetic acid is ~90% deprotonated, while carbonic acid is ~90% protonated B) Acetic acid and carbonic acid are both ~90% protonated C) Acetic acid is ~10% deprotonated, while carbonic acid is ~10% protonated D) Acetic acid and carbonic acid are both more than 99% deprotonated E) Acetic acid and carbonic acid are both ~90% deprotonated

A

18) ProtParam also tells me that the pI of citrate synthase is 6.86. Imagine that I had a mixture that included both citrate synthase and mystery 'protein X', and performed a 2D gel separation. Imagine that with the particular conditions that I used (exact type of gradient gel that I purchased, length of time that I applied the current, etc.), citrate synthase ended up in the center of the gel and 'protein X' ended up in the top right of the gel (the gel is oriented in the typical way - the same way as we saw in class and in the textbook). Which one of the following best indicates how we could describe 'protein X' compared with citrate synthase? A) Protein X is larger and has a higher pI. B) Protein X is smaller and has a higher pI. C) Protein X is larger and has a lower pI. D) Protein X is smaller and has a lower pI.

A

23) It is clear in Jmol (as well as in the supporting documentation, of course) that the citrate synthase protein is a homodimer; this is a way of describing its ______ structure. A) quaternary B) primary C) secondary D) tertiary E) domain

A

26) Citrate synthase is expressed in most cell types. Imagine that you had a population of cells, homogenized them, and then were going to make your first attempt to purify them by molecular exclusion chromatography them so that you could try to recover only the citrate synthase. Imagine that you had the following data in your lab notebook (units are the same as we saw in class and in the textbook; units are omitted in this question and the answer options to avoid giving hints about how these calculations are performed!): Step Total protein Total activity Homogenization 10,000 400,000 Chromatography 2,000 40,000 What is the purification level of the sample after the chromatography step? A) 0.5 B) 40 C) 0.1 D) 20 E) 10

A

27) A quick glance at the amino acid sequence of citrate synthase shows that it has an eye-catching string of six histidines in a row near the N-terminal end. Think about what we covered regarding histidine. Which one of the following is true regarding histidine? A) At pH 5 its expected protonated:deprotonated ratio is approximately 10:1 B) Its alpha-amino N is directly covalently linked to its R-group. C) It is the only amino acid that contains more oxygen atoms than nitrogen atoms. D) We (and our textbook) classified it as an acidic R-group. E) Its deprotonated form is negatively charged.

A

4) Peptide bond synthesis from free amino acids _________ a water molecule. This reaction has a _________ deltaG value. A) produces; positive B) produces; negative C) consumes; negative D) consumes; positive E) none of the above; there is no production or consumption of water during peptide bond synthesis.

A

At what pH is a functional group expected to be ~90% deprotonated if its pKa is 4? A 5 B 3 C 2 D 6 E 4

A

Which one of the following best indicates the nucleophilic 'attacker' in the DNA polymerase mechanism of action? A. A hydroxyl group B. A phosphate group C. An amino group D. A carboxyl group E. A guanidinium group

A

1. A t instead of an s? Differentiate between a nucleoside and a nucleotide.

A nucleoside is a base attached to a ribose or deoxyribose sugar. A nucleotide is a nucleoside with one or more phosphoryl groups attached to the ribose or deoxyribose

14. Power( ful ) tools. Monoclonal antibodies can be conju- gated to an insoluble support by chemical methods. Explain how these antibody-bound beads can be exploited for pro- tein purification.

Antibody molecules bound to a solid support can be used for affinity purification of proteins for which a ligand molecule is not known or unavailable.

16. Making more enzyme ? In the course of purifying an enzyme, a researcher performs a purification step that results in an increase in the total activity to a value greater than that present in the original crude extract. Explain how the amount of total activity might increase

An inhibitor of the enzyme being purified might have been present and subsequently removed by a purification step. This removal would lead to an apparent increase in the total amount of enzyme present.

11. All for one . If the forces—hydrogen bonds and stacking forces—holding a helix together are weak, why is it difficult to disrupt a double helix?

Although the individual bonds are weak, the population of thousands to millions of such bonds provides much stability. There is strength in numbers.

12. One from many. Differentiate between amino acid com- position and amino acid sequence .

Amino acid composition refers simply to the amino acids that make up the protein. The order is not specified. Amino acid sequence is the same as the primary structure—the sequence of amino acids from the amino terminal to the carboxyl terminal of the protein. Different proteins may have the same amino acid composition, but amino acid sequence identifies a unique protein.

7. Who's charged? Draw the structure of the dipeptide Gly- His. What is the charge on the peptide at pH 5.5? pH 7.5?

At pH 5.5, the net charge is 1 1: (H3N+...H...ch2Ring with NH+...COO-) At pH 7.5, the net charge is 0: (H3N+...H...ch2Ring neutral...COO-)

11) I hope that you've been getting enough retinol in your diet recently - it's a form of vitamin A. {It's needed in several physiological processes, and is on the World Health Organization's list of essential medicines, because those who don't receive enough in their diet should be provided access to it.} Take a look at its structure below (One 6-ring, 3 branches. Then 11-C branch) Determine how many carbons it has, and then choose the option with the range that contains your answer. Retinol contains: A) 23-24 C's. B) 19-20 C's. C) 21-22 C's. D) 25 or more C's. E) 18 or fewer C's.

B

14) Visualize this short DNA sequence. I've given you the sequence of only one strand, so write out the sequence of the opposite strand before proceeding. 5'- A G C G G G -3' 3'- -5' Which one of the following is true? When you consider the double-stranded DNA molecule, there are a total of: A) four guanines B) six purines C) twelve 2' OH's D) twelve phosphodiester bonds E) thirteen H-bonds

B

Which amino acid has an expected net charge of +1 at pH 7? A all four of the listed amino acids B Lys C Asn D Trp E Glu

B

21) Looking at the 3D structure of citrate synthase, it is notable immediately that it has many alpha-helices. Imagine any H-bond in any of these alpha-helices. Which one of the following is true? A) The H-bond acceptor is a water molecule. B) The distance between the H and the donor is shorter than the distance between the H and the acceptor. C) The H-bond is more than 20 Angstroms in length. D) The H-bond acceptor is a N. E) None of the above because there are no H-bonds associated with an alpha-helical structure in a protein.

B

29) I looked in one of my favorite databases, OMIM (Online Mendelian Inheritance in Man) to see what citrate synthase mutants have been described. It is clear that researchers believe that several specific lysines in citrate synthase are critical (there are 26 lysines in total in the polypeptide). Researchers have tried to establish the importance of certain lysines by studying the enzyme in the lab; one strategy involved mutating key lysines to arginines. Which one of the following best summarizes this Lys -> Arg mutation strategy (assuming physiological pH of ~7)? A) Lysine has a charge; the mutation introduces a residue that is expected to have the opposite charge. B) Lysine has a charge; the mutation introduces a residue that is expected to have the same charge. C) Lysine has a charge; the mutation introduces a residue that is expected to be polar uncharged. D) Lysine has a charge; the mutation introduces a residue that is expected to be hydrophobic. E) None of the above because lysine is not expected to be charged at physiological pH.

B

5) Think about a solution with a proton concentration of 0.0004M. This solution has a pH with a value ________. A) higher than 4, but lower than 5 B) higher than 3, but lower than 4 C) of 4 D) of 3 or lower E) of 5 or higher

B

What is the antilog (base 10) of -4? A 4 B .0001 C 0.4 D None of the others is correct E 10000

B

What is the log (base 10) of 0? A 0 B Undefined C None of the others is correct D 1 E 10

B

Which one of these is an 'activated precursor' for DNA synthesis? A. the DNA template B. dGTP C. the primer D. ATP E. pyrophosphate

B

15) Later in the semester we'll discuss the citric acid cycle (also known as the Krebs cycle or the TCA cycle). The first enzyme that acts in this cyclic pathway is citrate synthase, and I decided to look up a bit of information about this protein in some common databases. The next questions are about this protein. The structure of citrate synthase has been solved by several research groups, and so it appears in the Protein DataBank several times. One of its structures reportedly has a resolution of '2.3 Angstroms', which helps us to understand how confident these researchers are in their assignments of the atoms relative to each other in 3D space. It's standard to report structure resolution in Angstroms, but what if they had for some reason chosen nanometers instead? 2.3 Angstroms is: A) 0.023 nm B) 230 nm C) 0.23 nm D) 23 nm E) None of the above is correct in representing the conversion between Angstroms and nm

C

19) Citrate synthase does not contain any disulfide bonds. This indicates that the cysteines are in the ______ form, which is the form in which they are relatively ______ in electrons. A) oxidized; deficient B) reduced; deficient C) reduced; enriched D) oxidized; enriched E) None of the above because cysteines are not the amino acids that are involved in disulfide bonds

C

20) In the previous question, it was just stated that citrate synthase doesn't contain any disulfide bonds. Imagine that you were performing SDS-PAGE only on citrate synthase (no other proteins). Which one of the following would therefore be fairly unimportant (i.e. imagine that you could omit only one of the following)? A) polyacrylamide B) SDS C) beta-mercaptoethanol D) the cathode E) the anode

C

3) Ala, Ile, and Phe share a key feature: A) they all lack chirality at the alpha-carbon. B) their R-groups all have pKa's in the 3-4 range. C) their R-groups all are composed exclusively of carbon and hydrogen atoms. D) their R-groups all have an expected positive charge at physiological pH. E) their R-groups display resonance.

C

6) Look at this schematic of ATP: Which one of the following best explains why it is not a nucleoside? (Adenine connected to Ribose connected to three phosphates) A) It has three phosphates instead of one. B) It has a nitrogenous base. C) It has phosphates. D) The carbohydrate is ribose. E) None of the above; according to the definition that we (and the textbook) discussed, ATP is a nucleoside.

C

9) Visualize one deoxyribose within a DNA strand. Focus specifically on the 4' carbon. Which one of the following best describes what is covalently attached to that 4' C? A) One C, one H, one O, and one nitrogenous base B) One C, two H's, and one O C) Two C's, one H, and one O D) Two C's, one H, and one P E) Two C's and two H's

C

The reading specifically noted that DNA polymerases require a primer with a free hydroxyl group at the _______ position. A. 1' B. 2' C. 3' D. 4' E. 5'

C

What is the log (base 10) of 100,000,000? A. 4 B. 10 C. 8 D. -8 None of the others is correct

C

What is the net charge of the tripeptide Cys-Arg-Gln at pH 2? A -1 B None of the others is correct C +2 D +1 E 0

C

Which amino acid has the molecular formula C3H7NO3? A Ala B Gln C Ser D Met E Lys

C

Which one of the following is a product of the reaction that is catalyzed by DNA polymerases? A. water B. ADP C. PPi D. CO2 E. a polypeptide chain

C

10) Visualize the structures of the amino acids mentioned below, and recall the definitions of chiral and achiral. Choose the one option below that creates an achiral molecule from a chiral starting molecule. A) Begin with Gly; remove its R-group and replace it with an H. B) Begin with Met; remove its R-group and replace it with a methyl (-CH3) group. C) Begin with Thr; phosphorylate the hydroxyl group. D) Begin with Phe; remove its R-group and replace it with an amino group. E) Begin with Lys; alter the pH so that the R-group is in deprotonated form.

D

12) Green Fluorescent Protein can be purified by 'anion exchange' chromatography, which relies on the attraction of GFP to ______ charged beads. The GFP can then be eluted using NaCl, with the ______ competing with the GFP for binding to the beads. A) negatively; Na+ B) positively; Na+ C) negatively; Cl- D) positively; Cl

D

) During the conversion of glucose to carbon dioxide, entropy ________ . Therefore, during this reaction deltaS is ________ . A) decreases; positive B) increases; negative C) decreases; negative D) increases; positive E) None of the above options are suitable because according to our coverage (and the textbook) S indicates a different variable, not entropy.

D

16) Citrate synthase catalyzes the first step in the citric acid cycle, and so we can predict that it must have excellent binding sites for the first two reactants in the citric acid cycle. One of the first two reactants in the citric acid cycle is oxaloacetate (OAA), a four-carbon molecule that has two carboxyl groups. Think about the expected charge of carboxyl groups at typical cellular pH. Now imagine OAA binding to citrate synthase; we can predict that there might be favorable electrostatic interactions between OAA and a few ________ residues in the binding site. We would expect those residues to remain charged, unless the surrounding pH became ________. A) acidic; very low pH (e.g. pH 2). B) acidic; very high pH (e.g. pH 13). C) basic; very low pH (e.g. pH 2). D) basic; very high pH (e.g. pH 13).

D

17) It's straightforward from the Protein DataBank record to download the protein sequence, and then copy that to the web-based ProtParam tool, which allows the calculation of a variety of protein features. One of the things that ProtParam quickly tells us is that some residues are more common in citrate synthase than others. For example, there's a particular amino acid that occurs only 9 times even though this protein is 462 amino acids long. (Many other amino acids occur several dozen times in this protein.) I'll give you the chemical formula (at pH 7) of the amino acid that occurs 9 times: it's C11H12N2O2. Which amino acid has this chemical formula? (Remember - of course you weren't supposed to memorize amino acid formulas! Deduce which one it is by looking at all of the options and seeing how they differ!) A) Asn B) His C) Glu D) Trp E) Met

D

22) Using Jmol, I looked at the structure of citrate synthase, focusing on the first (most N-terminal) of its many alpha-helices. I wanted to see whether in 3-D there was evidence that the helix was amphipathic, and it was easy to confirm that it was. I'll give you five residues (in no particular order) that are found in that first alpha-helix; four are found on one face and one is found on the opposite face. You can deduce which one doesn't belong! The residue that is found on the opposite face of the amphipathic helix is: A) Thr B) Asp C) Gln D) Ile E) Lys

D

25) Imagine that you were looking at the citrate synthase structure, and focusing on the backbone. Imagine that I asked you to highlight only the pure single bonds along the backbone. Make sure to ignore any bond that has resonance-stabilization, and limit yourself to the backbone! What type of bond(s) have you highlighted? A) C-N bonds only. B) C-N bonds and C-O bonds. C) C-C bonds only. D) C-N bonds and C-C bonds. E) C-C bonds and C-O bonds.

D

28) Now think about the DNA that encodes the citrate synthase gene. Imagine a copy of the citrate synthase gene in the process of being copied in the genome. Which one of the following best indicates the thermodynamic driver of this copying reaction? A) The attack of the hydroxyl group on the innermost phosphate. B) The base-pairing between the incoming nucleotide and the template. C) The magnesium cofactor of the enzyme. D) The hydrolysis of pyrophosphate. E) The electrostatic repulsion of nucleophiles by the nucleic acid backbone.

D

30) An examination of the C-terminal end of the sequence of citrate synthase reveals that its final three residues are Lys-Ser-Gly. Let's use this information as our inspiration to 'randomly' create a 'net charge on a tripeptide' question. So now imagine a free molecule that is only comprised of the sequence Lys-Ser-Gly. Which one of the following indicates the most likely charge of the molecule Lys-Ser-Gly at pH 13, based on our coverage? A) +1 B) -2 C) 0 D) -1 E) None of the above because it is either lower than -2 or higher than +1.

D

8) There are plenty of sources for looking up pKa values besides our textbook. One source is AnaSpec.com, which estimates the pKa for threonine's alpha-carboxyl group to be approximately 2, and the pKa for threonine's alpha-amino group to be approximately 9. Given these values for use in only this question, what is the most likely net charge on a threonine molecule (not side chain) at pH 3? A) +2 B) -2 C) -1 D) 0 E) +1

D

If a functional group has a pKa of 9, then what is the deprotonated:protonated ratio for this functional group at pH 4? A 10,000:1 B 1:10,000 C 100,000:1 D 1:100,000 E None of the others is correct

D

What is the log (base 10) of 0.1? A None of the others is correct B Undefined C 1 D -1 E 10

D

What is the log (base 10) of 1? A None of the others is correct B Undefined C 10 D 0 E 1

D

What is the net charge of the dipeptide Asp-Glu at pH 10? A -2 B +1 C -1 D None of the others is correct E 0

D

7) At what pH could glutamine side chains be approximately 99% deprotonated? Choose the best option range below (and use process of elimination to get rid of the four deeply unsuitable options!). A) pH 2-4 B) pH 5-7 C) pH 8-10 D) pH 11-13 E) None of the above because we did not identify this as an amino acid with a side chain that could reasonably be expected to become deprotonated.

E

24) As mentioned in a previous question, citrate synthase is 462 amino acids in length. Which one of the following is true? A) This protein is therefore 46.2 kDa. B) This protein is therefore 4.62 kDa. C) This protein is therefore 462 kDa. D) This protein is therefore 4,620 kDa. E) None of the above is correct because a protein's kDa value is not calculated directly from its size in this manner

E

If a functional group has a pKa of 3, then what is the (approximate) likelihood that this functional group will be protonated at pH 5? A 50% B 99% C 90% D 10% E 1%

E

What is the antilog of 1? A 1 B -1 C .1 D Undefined E 10

E

What is the log (base 10) of 0.001? A 3 B .01 C Undefined D 1/3 E -3

E

What is the pH of a solution with proton concentration 0.001M? A 12 B None of the others is correct C 11 D 2 E 3

E

4. Don't break the law . Given the following values for the changes in enthalpy ( D H ) and entropy ( D S ), which of the following processes can take place at 298 K without violat- ing the Second Law of Thermodynamics? (a) dH = -84 kJ/mol, dS = +125 J/mol (b) dH = -84 kJ/mol, dS = -125 J/mol (c) dH = +84 kJ/mol, dS = +125 J/mol (d) dH = 84 kJ/mol, dS = -125 J/mol

Processes a and b.

50. 3 is greater than 2. The adjoining illustration graphs the relation between the percentage of GC base pairs in DNA and the melting temperature. Suggest a plausible explana- tion for these results.

GC base pairs have three hydrogen bonds compared with two for AT base pairs. Thus, the higher content of GC means more hydrogen bonds and greater helix stability.

21. Potential partners . Identify the groups in a protein that can form hydrogen bonds or electrostatic bonds with an arginine side chain at pH 7.

Glutamate, aspartate, and the terminal carboxylate can form salt bridges with the guanidinium group of arginine. In addition, this group can be a hydrogen-bond donor to the side chains of glutamine, asparagine, serine, threonine, aspartate, tyrosine, and glutamate and to the main-chain carbonyl group. Histidine can form hydrogen bonds with arginine at pH 7.

20. Often irreplaceable . Glycine is a highly conserved amino acid residue in the evolution of proteins. Why?

Glycine has the smallest side chain of any amino acid. Its size is often critical in allowing polypeptide chains to make tight turns or to approach one another closely.

9. Sweet tooth, but calorie conscious . Aspartame (NutraSweet), an artificial sweetener, is a dipeptide composed of Asp-Phe in which the carboxyl terminus is modified by the attachment of a methyl group. Draw the structure of Aspartame at pH 7.

H3N...CH2COO...CH2+6Ring...COO+Methyl

2. A lovely pair . What is a Watson-Crick base pair?

Hydrogen-bond pairing between the base A and the base T as well as hydrogen-bond pairing between the base G and the base C in DNA.

6. Too much or not enough. Why do proteins precipitate at high salt concentrations? Although many proteins precipitate at high salt concentrations, some proteins require salt to dissolve in water. Explain why some proteins require salt to dissolve.

If the salt concentration becomes too high, the salt ions interact with the water molecules. Eventually, there will not be enough water molecules to interact with the protein, and the protein will precipitate. If there is lack of salt in a protein solution, the proteins may interact with one another—the positive charges on one protein with the negative charges on another or several others. Such an aggregate becomes too large to be solubilized by water alone. If salt is added, the salt neutralizes the charges on the proteins, preventing protein-protein interactions.

2. Resonance structures . The structure of an amino acid, tyro- sine, is shown here. Draw an alternative resonance structure.

Interchange the positions of the single and double bonds in the six-membered ring.

20. Finding a template. A solution contains DNA poly- merase and the Mg 2 salts of dATP, dGTP, dCTP, and TTP. The following DNA molecules are added to aliquots of this solution. Which of them would lead to DNA synthesis? (a) A single-stranded closed circle containing 1000 nucleo- tide units. (b) A double-stranded closed circle containing 1000 nucle- otide pairs. (c) A single-stranded closed circle of 1000 nucleotides base-paired to a linear strand of 500 nucleotides with a free 3 9 -OH terminus. (d) A double-stranded linear molecule of 1000 nucleotide pairs with a free 3 9 -OH group at each end

Molecules in parts a and b would not lead to DNA synthesis, because they lack a 3 9 -OH group (a primer). The molecule in part d has a free 3 9 -OH group at one end of each strand but no template strand beyond. Only the molecule in part c would lead to DNA synthesis.

4. But not always . A single strand of RNA is 20% U. What can you predict about the percentages of the remaining bases?

Nothing, because the base-pair rules do not apply to single- stranded nucleic acids.

10. Coming and going . What does it mean to say that the DNA strands in a double helix have opposite directionality or polarity?

One end of a nucleic acid polymer ends with a free 5 9 -hydroxyl group (or a phosphoryl group esterified to the hydroxyl group), and the other end has a free 3 9 -hydroxyl group. Thus, the ends are differ- ent. Two strands of DNA can form a double helix only if the strands are running in different directions—that is, have opposite polarity.

5. Bonding is good. Which of the following amino acids have R groups that have hydrogen-bonding potential? Ala, Gly, Ser, Phe, Glu, Tyr, Ile, and Thr.

Ser, Glu, Tyr, Thr

33. A picture is worth a thousand words . Write a reaction sequence showing why RNA is more susceptible to nucleo- philic attack than DNA

Show oxy vs deoxy on 2'C

11. Not a sidecar. Define the term side chain in the context of amino acid or protein structure.

Side chain is the functional group attached to the a -carbon atom of an amino acid.

23. Location is everything 1. Most proteins have hydro- philic exteriors and hydrophobic interiors. Would you expect this structure to apply to proteins embedded in the hydrophobic interior of a membrane? Explain.

Some proteins that span biological membranes are "the excep- tions that prove the rule" because they have the reverse distribu- tion of hydrophobic and hydrophilic amino acids. For example, consider porins, proteins found in the outer membranes of many bacteria. Membranes are built largely of hydrophobic chains. Thus, porins are covered on the outside largely with hydrophobic residues that interact with the neighboring hydrophobic chains. In contrast, the center of the protein contains many charged and polar amino acids that surround a water-filled channel running through the middle of the protein. Thus, because porins function in hydro- phobic environments, they are "inside out" relative to proteins that function in aqueous solution.

7. Size matters . Why are GC and AT the only base pairs permissible in the double helix?

Stable hydrogen bonding occurs only between GC and AT pairs. Moreover, two purines are too large to fit inside the double helix, and two pyrimidines are too small to form base pairs with each other.

3. Chargaff rules! Biochemist Erwin Chargaff was the first to note that, in DNA, [A] 5 [T] and [G] 5 [C], equalities now called Chargaff's rule. Using this rule, determine the percentages of all the bases in DNA that is 20% thymine

T is always equal to A, and so these two nucleotides constitute 40% of the bases. G is always equal to C, and so the remaining 60% must be 30% G and 30% C.

10. Vertebrate proteins? What is meant by the term polypep- tide backbone?

The (nitrogen- a carbon-carbonyl carbon) repeating unit.

24. Location is everything 2 . Proteins that span biological membranes often contain a helices. Given that the insides of membranes are highly hydrophobic (Section 12.2), pre- dict what type of amino acids would be in such an a helix. Why is an a helix particularly suited to existence in the hydrophobic environment of the interior of a membrane?

The amino acids would be hydrophobic in nature. An a helix is especially suited to crossing a membrane because all of the amide hydrogen atoms and carbonyl oxygen atoms of the peptide back- bone take part in intrachain hydrogen bonds, thus stabilizing these polar atoms in a hydrophobic environment.

28. Issues of stability . Proteins are quite stable. The lifetime of a peptide bond in aqueous solution is nearly 1000 years. However, the free energy of hydrolysis of proteins is nega- tive and quite large. How can you account for the stability of the peptide bond in light of the fact that hydrolysis releases much energy?

The energy barrier that must be crossed to go from the polym- erized state to the hydrolyzed state is large even though the reac- tion is thermodynamically favorable.

16. Active again . A mutation that changes an alanine resi- due in the interior of a protein to valine is found to lead to a loss of activity. However, activity is regained when a second mutation at a different position changes an isoleucine resi- due to glycine. How might this second mutation lead to a restoration of activity?

The first mutation destroys activity because valine occupies more space than alanine does, and so the protein must take a different shape, assuming that this residue lies in the closely packed interior. The second mutation restores activity because of a compensatory reduction of volume; glycine is smaller than isoleucine.

1. Donors and acceptors . Identify the hydrogen-bond donors and acceptors in each of the four bases on page 4 .

The hydrogen-bond donors are the NH and NH 2 groups. The hydrogen-bond acceptors are the carbonyl oxygen atoms and those ring nitrogen atoms that are not bonded to hydrogen or to deoxyribose.

9. Frequently used in shampoos. The detergent sodium dodecyl sulfate (SDS) denatures proteins. Suggest how SDS destroys protein structure.

The long hydrophobic tail on the SDS molecule (see p. 72) dis- rupts the hydrophobic interactions in the interior of the protein. The protein unfolds, with the hydrophobic R groups now interact- ing with SDS rather than with one another.

14. Contrasting isomers . Poly- L -leucine in an organic sol- vent such as dioxane is a helical, whereas poly- L -isoleucine is not. Why do these amino acids with the same number and kinds of atoms have different helix-forming tendencies?

The methyl group attached to the b -carbon atom of isoleucine sterically interferes with a -helix formation. In leucine, this methyl group is attached to the g -carbon atom, which is farther from the main chain and hence does not interfere.

9. Uniqueness. The human genome contains 3 billion nucle- otides arranged in a vast array of sequences. What is the minimum length of a DNA sequence that will, in all prob- ability, appear only once in the human genome? You need consider only one strand and may assume that all four nucleotides have the same probability of appearance

The probability that any sequence will appear is 1/4 n , where 4 is the number of nucleotides and n is the length of the sequence. The probability of any 15-base sequence appearing is 1/4 15 , or 1/1,073,741,824. Thus, a 15-nucleotide sequence would be likely to appear approximately three times (3 billion 3 probability of appearance). The probability of a 16-base sequence appearing is 1/4 16 , which is equal to 1/4,294,967,296. Such a sequence will be unlikely to appear more than once.

11. Unexpected migration. Some proteins migrate anoma- lously in SDS-PAGE gels. For instance, the molecular weight determined from an SDS-PAGE gel is sometimes very differ- ent from the molecular weight determined from the amino acid sequence. Suggest an explanation for this discrepancy.

The protein may be modified. For instance, asparagine resi- dues in the protein may be modified with carbohydrate units (Section 2.6).

5. It's in the bag. Suppose that you precipitate a protein with 1 M (NH 4 ) 2 SO 4 and that you wish to reduce the concentra- tion of the (NH 4 ) 2 SO 4 . You take 1 ml of your sample and dialyze it in 1000 ml of buffer. At the end of dialysis, what is the concentration of (NH 4 ) 2 SO 4 in your sample? How could you further lower the (NH 4 ) 2 SO 4 concentration?

The sample was diluted 1000-fold. The concentration after dialysis is thus 0.001 M, or 1 mM. You could reduce the salt con- centration by dialyzing your sample, now 1 mM, in more buffer free of (NH 4 ) 2 SO 4 .

8. Strong, but not strong enough . Why does heat denature, or melt, DNA in solution?

The thermal energy causes the chains to wiggle about, which disrupts the hydrogen bonds between base pairs and the stacking forces between bases and thereby causes the strands to separate.

12. Overcharged . DNA in the form of a double helix must be associated with cations, usually Mg 2 . Why is this requirement the case?

There would be too much charge repulsion from the negative charges on the phosphoryl groups. These charges must be coun- tered by the addition of cations.

25. Neighborhood peer pressure? Table 2.1 shows the typical p K a values for ionizable groups in proteins. However, more than 500 p K a values have been determined for individual groups in folded proteins. Account for this discrepancy.

This example demonstrates that the p K a values are affected by the environment. A given amino acid can have a variety of p K a val- ues, depending on the chemical environment inside the protein

22. Quaternary structure. A protein was purified to homo- geneity. Determination of the mass by gel-filtration chro- matography yields 60 kDa . Chromatography in the pres- ence of 6 M urea yields a 30-kDa species. When the chro- matography is repeated in the presence of 6 M urea and 10 mM b - mercaptoethanol , a single molecular species of 15 kDa results. Describe the structure of the molecule.

Treatment with urea will disrupt noncovalent bonds. Thus the original 60-kDa protein must be made of two 30-kDa subunits. When these subunits are treated with urea and b -mercaptoethanol, a single 15-kDa species results, suggesting that disulfide bonds link the 30-kDa subunits.

29. Minor species . For an amino acid such as alanine, the major species in solution at pH 7 is the zwitterionic form. Assume a p K a value of 8 for the amino group and a p K a value of 3 for the carboxylic acid. Estimate the ratio of the concentration of the neutral amino acid spe- cies (with the carboxylic acid protonated and the amino group neutral) to that of the zwitterionic species at pH 7 (Section 1.3).

Using the Henderson-Hasselbach equation, we find the ratio of alanine-COOH to alanine-COO 2 at pH 7 to be 10 2 4 . The ratio of alanine-NH 2 to alanine-NH 3 1 , determined in the same fashion, is 10 2 1 . Thus, the ratio of neutral alanine to the zwitterionic spe- cies is 10 24 3 10 21 5 10 25