Problem Book, 19th ed., re-revised  (2011) Corrections Listed in Order of Problems

This is the page you need if you have the 19th edition, re-revised, 2011.  Your book already contains all the corrections made in 2010 & 2011. The additional (significant) corrections made in 2012 are listed below. All additions or corrections of a few words are highlighted in blue. Where long notes or paragraphs were added, the addition is noted in green. (Minor changes and typos are not listed.)

If you have the 19th edition (2009) or the 19th edition, revised (2010), or an earlier edition, you will need to look at previous corrections as well as the corrections below. See the current corrections page for links to previous corrections.

Corrections to Questions

p5. The note at the bottom of the page has been changed because we don't want to imply that you should do the problems over and over. (See the link below for more detailed problem solving advice.) The new version of the note reads as follows:

[1] Note: The starred questions are practice problems from old exams. You are encouraged to treat these problems differently from the others. We suggest that you postpone doing the starred problems until after you have done all the other problems (on that particular topic) and you feel confident that you understand the material. Then use the starred problems to test yourself. For more advice on problem solving, go to See esp. tip # 7. If you run into difficulties figuring out the problems, do not hesitate to contact your TA or one of the instructors. Reminder: On actual exams, the questions usually require both a short answer and a short explanation, so it's not enough to get the right answers -- you also have to be able to explain how you got your answers. For advice on 'how to explain' see the acknowledgements page.

p. 16, Problem. 2-20. Sequences are changed to: 



In the one letter code for the amino acids, L = leu; S = ser; A = ala. The questions remain unchanged, but the answer to C has been updated to match the new sequences.

p. 18-22. Replaced 3-6 with a new question, and added a new questions, now numbered 3-7. The remaining questions in problem set 3 are the same in the 19th & 20th editions, but are numbered 3-8 to 3-14 in the 20th and 3-7 to 3-13 in the 19th.

New Problem 3-6:

3-6. Consider the ribbon diagram of a protein shown below. In the diagram:

A. The number of polypeptides is                        ______

B. The number of subunits is                              ______

C. The number of alpha helices is about  ______

D. The number of strands per alpha helix is         ______

E. The number of beta sheets is                          ______

F. The total number of strands in beta sheets is ______

G. The number of bound ligands is                      ______

H. The number of amino terminals is                   ______

I. The number of carboxyl terminals is                 ______

J. The strands in the beta sheets are (parallel) (antiparallel) (both) (neither).

Picture for Problem 3-6

This diagram can be seen more clearly online at:

This diagram is in the public domain:

Added Problem:

3-7. Suppose enzyme X has a single substrate and follows Michaelis-Menton kinetic behavior.


A. At very low substrate concentrations (i.e., <Km/100) the shape of the Michaelis-Menton curve (as usually plotted) looks the most like:

      B. At very high substrate concentrations (i.e., > 100 X Km) the shape of the Michaelis-Menton curve (as usually plotted) looks the most like ___________. (Same choices as in A).


C. Suppose the Km of enzyme X is 10-4M and its turnover number is 100. From these data, the best estimate for the association constant (estimated Ka) of the enzyme for its substrate is:

(10-6)  (10-4 )  (10-2 )  (0)   (1)  (100)   (10,000)   (106 )


D. Suppose an independent measurement of the real association constant (real Ka) turns out to be different from the estimate in C above. Compared to the estimate based on the enzyme kinetics, the real association constant could be (higher) (lower) (either) (neither).


E. Suppose you have an inhibitor of Enzyme X. When you add inhibitor, the Vmax of the reaction is changed, but the Km is not.

E-1. The value of [S] needed to reach Vmax/2 should (increase) (decrease) (stay the same) (can't predict).

E-2. The inhibitor described here is (competitive) (noncompetitive) (either way) (neither).
If you knew the structure of the substrate for this reaction could you predict anything about the structure of the inhibitor in this problem or in problem 1?

p. 24 Add the following note, which was omitted accidentally from the previous edition:

            *Note: a residue is that part of a monomer structure that remains after it is incorporated into a polymer, i.e. most of it.

p. 30. Problem 4-14 part A: 'How many net molecules of ATP will be generated...'

p. 34. Problem 5-6. Add the following:

Hint for part F: In the triose phosphate isomerase reaction (rxn 5) does the phosphate group move?

p. 36. Problem 5-11, part B. The third sentence reads: 'This time the same amount of NADH2 is oxidized and the same amount of oxygen is used up, but the concentration of ADP and Pi does not change.' Cross out the indicted section and substitute instead: 'but you do not measure the changes in ADP and Pi.' (In the revised version of B-5, you have to decide what will happen to the concentrations of ADP and Pi .)

Questions B-1 to B-4 are the same. Replace Question B-5 with the following:

B-5. In this experiment, you would expect ATP to be generated from ADP (in the buffer) (inside the vesicles) (both) (neither).

p. 46. Problem 6-16, part D. Set up is the same, but change choices of answers as follows:

This information indicates that the native nucleic acid of this virus is probably (a single strand doubled back on itself) (single stranded ) (double stranded) (any of these) AND the DNA is probably (circular) (linear) (either way).

p.50 and thereafter -- the proper spelling is promoter, not promotor.

p. 61. Problem 8-9 has been rephrased to make it clearer. Here is the revised version:

8-9.      Suppose nondisjunction (ND) occurs during meiosis in a human male. Assume that ND occurs only once per meiosis, at either 1st or 2nd division. Also assume that ND is equally likely at either meiotic division, and that every meiosis produces 4 gametes (sperm). Now consider the gametes that will end up with an abnormal number of sex chromosomes. (Ignore the normal gametes.)

A. Which should be the most common type of gamete with an abnormal number of sex chromosomes?

B. Which should be the least common type of gamete with an abnormal number of sex chromosomes?

The hint at the end is the same.

p. 91. Problem 11-11. Assume the bacteria have no plasmids.

p. 94. Problem 12-0, part C. In the list of possible answers, one choice has been changed. Replace the choice 'binding of RNA polymerase to promoter of greasy operon' with the choice: 'initiation of transcription from ........'

p. 102. Problem 12R-5, part D. The 2nd sentence should read: The replication fork(s) should move.....

p. 106. Problem 13-8, part B. You cut up your DNA & run gels as in A, not as on A.
            In part B-2, the last enzyme treatment should be EatIII + Eat IV, not EatII + EatIV.

Corrections to Answers

p. 146, Ans. to 2-20, part C. The answer has been changed to match the new sequences. The new answer reads as follows:

C. Both oligopeptides could form an alpha helical secondary structure. The amino acid side chains protrude from the alpha helix. In the case of polypeptide 4, there are 4 leucine residues that are spaced 7 apart (1, 8, 15, 22). Since there are ~3.5 AAs per turn, then these leucine residues, being 7 AAs apart, would all be sticking out along the same side of the helix, aligned one over (or beside) the other. (See left diagram below.) If the side of the alpha helix with these 4 leucines faces the exterior of the protein, then it could interact by hydrophobic forces with the same face presented by an identical subunit of the protein, thus contributing to the association of two subunits in forming the homodimer. (See right diagram below.) In oligopeptide II the leucine side groups cannot help connect subunits, because the side chains would be pointing in various directions out from the alpha helix. Serines (which could interact via hydrogen bonding) cannot help connect subunits in either case, because the serines are distributed on all sides in both oligopeptides.


Picture for Problem 2-20, part C.

p. 148. Ans. to new 3-6 & 3-7 are as follows:

3-6. A & B. Two polypeptides &. two subunits. (Each peptide = one subunit.)

C & D. 10 Alpha helices, 1 strand per helix.

E & F. Two beta sheets, with 10 strands total.
Each beta sheet is curved, so you might conclude that each curved sheet should be considered two flat sheets at slight angles to each other. (In that case, each subunit would contain one sheet with 3 strands and a second sheet with 2 strands, instead of one sheet with 5 strands.) That the 'two' beta sheets really form one curved structure is shown most clearly in the bottom subunit.

G, H, & I. Two each. Each peptide chain has a carboxyl end and an amino end, and binds one ligand molecule.

J. Both. Some strands are parallel and some are antiparallel. The parallel ones are seen most clearly in the top subunit.

3-7. A. A straight line, since if S << Km, Vo will be proportional to [S]. At very low [S], Vo ≈ k3 Eo [S]/Km
B. A straight (horizontal) line, since if S>>Km, Vo will be constant. At very large [S], Vo ≈ k3 Eo.


C. 10,000, since Kd ~ Km and Ka = 1/Kd      (Kd = Keq for disassociation of ES or the reaction ES E + S; Ka = Keq for the reverse reaction.)

D. Higher. Remember we are asking about how the real Ka relates to the Ka calculated using the Km , not how Kd relates to Km. Kd can be lower than Km , so the real Ka or 1/Kd can be higher than 1/Km. 'The reasoning is as follows: Km = (k2+k3)/k1 = k2/k1 + k3/k1 = Kd+k3/k1. Therefore Kd = Km-(k3/k1), and Kd can only be lower than Km (usually it is insignificantly lower). Since Ka = 1/Kd , if Kd is lower than than expected, then Ka can only be higher than expected (higher > 1/Km).

E-1. Stay the same. The Vmax/2 has changed, but the amount of S it takes to get there is the same.
E-2. Non-competitive (because Km stays the same, but the Vmax is affected).
E-3. No for this problem; yes for problem 1. A competitive inhibitor (as in problem 1) should look like the substrate; a non-competitive inhibitor (as in this problem) does not bind to the substrate binding site and should look different from the substrate.

p. 154, Ans to Problem 4-7, part B. Add the following:           

Note that starting concentrations of K and M are as in part A, not as described at the start of the problem.
p. 157, Ans to Problem 4-14, part A. Add: You put in one ATP to convert glycerol to glyceraldehyde, but you get two ATP back per glyceraldehyde phosphate broken down.

p. 158, Answer to Problem 4-18, part A (i). Add:

The radioactive acetate combines with OAA to give citric acid (citrate) before any CO2 is lost or other radioactive compounds are formed.

p. 160, Answer to Problem 5-6, part F. Add:  

Note that in reaction 5, the conversion of DHAP to glyceraldehyde P, the P doesn't move, so you have to turn the molecule upside down.

p. 162, Answer to Problem 5-11, parts B-1 & B-5 are redone. (The other answers are unchanged.) Here are the revised answers to B-1 & B-5:

B-1.Electron transport is occurring (and a proton gradient is being generated; see below.) There is not enough info given to indicate if ox. phos. is occurring or not, but it should be okay if all the components in the membrane are working. If the membranes are not damaged, ATP can be made with the membranes in either orientation (if the appropriate substrates are on the appropriate sides). For an example, see problem 5-16, and also B-5 below.

In normal undamaged membranes, ox. phos. and electron transport are coupled; if you block one, the other backs up. Since electron transport is normal, we assume that ox phos is occurring too, and everything is working as it should. (Why does blocking ox. phos. block electron transport? If ox. phos. does not occur, the H+ gradient usually builds up to very high values and electron transport is inhibited as a result. This has not happened here.)

            If the membranes are damaged, as often happens in real experiments, then ox. phos. and electron transport can get uncoupled. Protons can leak out, the H+ gradient doesn't get too high, and electron transport can continue without ox. phos.

B-5. In the buffer, assuming the membranes are not damaged. Remember the membrane is reversed, relative to part A and to the usual orientation. Since the topology is reversed in this experiment (relative to the normal), the H+ are pumped into the vesicles, and flow out, through ATP synthetase, driving phosphorylation of ADP to ATP.
  If the membranes are damaged, then electron transport and ox. phos. can be uncoupled, as explained above. Even though electron transport and H+ pumping continue, H+ can leak back out, bypassing ATP synthetase, and no ATP will be made on either side.

p. 172, Answer to problem 6-16, part D. The revised answer is as follows:

D. Double stranded & linear. (Double stranded RNA is not found in normal cells, but is found in some viruses and infected cells.) The heat treatment breaks hydrogen bonds, not covalent bonds. Since the molecular weight is reduced to 1/2 by heat, the native molecule must consist of 2 complementary strands connected by H bonds that can separate completely when heated. When the nucleic acid is cooled down, the complementary strands match up and the H bonds are reformed. If the DNA were circular, the two strands would be wound around each other in a double helix, and would not separate when the DNA was denatured. (You would have two linked rings.)

p. 174, Answer to problem 7-7, part C. Replace 'One way' with 'One way to do the experiment in A.'


p. 175, Answer to problem 7-14. Replace 'lys-tRNA' and 'tRNA-lys' with 'tRNAlys.'

Answer to Part A: The last sentence of part A should read: 'Only two tRNAs are bound to the ribosome at any one time (either in A & P or in P & E), but you need a third tRNAlys because the E site will not empty until a lys-tRNAlys arrives in the A site.'

p. 198, Answer to problem 10-5, parts C & D. The answers have been rewritten slightly to make them clearer. The revised answers are as follows:

C. You need to calculate the fraction that are triple homozygotes (gg bb dd) from a cross of two triple heterogyzgotes (GgBbDd X the same). Simplest way is to consider each gene separately.

Answer = (fraction of offspring that are homozygous for 1st gene) X (fraction homozygous for 2nd gene) X (fraction homozygous for the 3rd gene) = fraction that are gg X fraction bb X fraction dd = 1/4 X 1/4 X 1/4 =1/64.

D. Ans. = (fraction that are G_) X (fraction that are Bb) X (fraction that are dd) = 3/4 X 1/2 X 1/4 = 3/32

p. 198, Answer to problem 10-8, part B. Add the following:

Note that 'linked' does not mean 100% linkage (no crossing over & no recombinants). It simply means there are more parentals than recombinants  -- no independent assortment. See explanation to 10-10, A. (2).

p. 202, Answer to problem 10-19, part A-1. Add the words in blue. 'If they look the same, the plants have the same phenotype. Plants with the dominant phenotype can have different genotypes (heterozygous  -- in F1, or homozygous -- in Parents).'

p. 207. Answer to problem 11-8, part B. Part ii should read:

ii) 4% = both normals & double mutants. The parentals were single mutants, with mutation 1, 2, or 3. Each crossing over event between 2 single mutants produced a double mutant and a 'mutant-less' normal.

The explanation to part iii had a serious typo in it. (The term in parentheses in line 4 should be 'mutant', not 'non-mutant.' ) Here is a corrected version: :

How to decide between complementation and recombination in part iii? If function is restored without any opportunity for crossing over, you know it was complementation. Otherwise, it is sometimes difficult to tell -- you have to look closely at the details in the question to see how to rule out one or the other in any particular case. (There is no quick foolproof method.) In this problem, most of the progeny are mutant and only a few are recombinant. This makes sense if complementation allowed infection to proceed to the point where some crossing over could occur. Only a few progeny DNA molecules engaged in crossing over, so the rest are mutants, produced as a result of complementation.

p. 209, Answer to problem 11-14, part B. The arrowheads in the diagram are not all properly aligned. However the rest of the diagram is correct.


p. 214, Answers to problem 12-0, part C.

For item (2): Initiation of RNA synthesis by RNA polymerase will increase, but RNA polymerase binding may not. RNA polymerase  may  already be bound, but if it is, it can't do anything because initiation of RNA synthesis is blocked by repressor. 

The note has been rewritten to make it clearer. Here is the latest version; in the first paragraph, only the blue words were added. The second paragraph is completely rewritten:

Note: In this problem, and in biology/biochemistry in general, 'enzyme activity' refers to the catalytic capability of the enzyme (if given enough substrate), not to how much catalyzing it is doing at this very minute (which depends on the supply of substrate). In other words, it refers to Vmax , not Vo.
            When the enzyme activity (V
max) changes, it can be because the number of enzyme molecules (Eo) changes, or because the capability of each enzyme molecule (the k3 or turnover number) changes. [Remember that Vmax depends on the product of both the k3 and Eo]. In this problem, the change in enzyme activity is due to a change in the amount of enzyme, but this is not always the case. For example, if inhibitors or activators are added, the enzyme activity can change (because of a change in k3) while the amount of enzyme (Eo) stays constant. When enzyme activity (Vmax) changes, you cannot assume that the amount of enzyme changes in parallel. You have to figure out which is more reasonable in each particular case -- a change in inhibition/activation, or a change in enzyme amount.

p. 219, answers to problem 13-7.

        In parts A, B & C, the lengths should be in BP (base pairs) not KB (kilobases).  In part D, the lengths should be in KB, as given.


p. 220, answers to problem 13-8.
        The numbers given in parts C and D are in BP.


p. 227, answer to problem 13R-5. The numbers are reversed.

The chromosome generating band #2 should come from dad and the chromosome generating band #3 should come from mom. The numbers are reversed. Any man with a #2 band (not a #3) could be the dad.

p. 229, answer to problem 14-1, part C. The answer has been updated. It should read:


C.  The answer depends on whether you think any of the following options are socially acceptable ways to avoid passing on the 'a' allele (if prospective parents are Aa or aa). The following options (at least) exist: amniocentesis plus therapeutic abortions of homozygous aa or Aa fetuses; use of pre-implantation genetic diagnosis (PGD) to implant a fetus who is Aa or AA; avoidance of reproduction; use of sperm, eggs, or both from a AA individual.  Which of these options do you consider to be socially acceptable?


p. 229, answer to problem 14-4, part B. For more recent editions of Purves/Sadava, look up Hardy-Weinberg equilibrium in the index.