Problem Book, 19th ed., (2009) Corrections Listed in Order of Problems
This is the page you need if you have the 19th edition, 2009. If you have an earlier edition, see the corrections page for links to corrections for older editions. If you have the 19th edition, revised (2010) or re-revised (2011) all these corrections have already been made.
Significant Corrections made for the 19th edition, revised, 2010.
The words highlighted in blue below were accidentally omitted. The words below in italics are replacements for the original wording. (Changes that were made to correct typos or fix the formatting are not listed.)
Corrections to the Questions
p. 35, Prob. 5-10. Note: O= = O2-
p. 37, Prob. 5-14. Assume there is an enzyme that can catalyze hydrolysis of acetyl-CoA to acetate + CoA.
p. 45, Prob. 6-15, part C should read: How many cycles of PCR will it take to make ds molecules that have 2 newly synthesized strands in the same molecule?
p. 48, Prob. 7-7. For parts B &C, if you can’t figure out what to do, see hints in answer key.
Part C should read: Something to think about: How would you actually do this experiment, i.e., how would you measure the amount of RNA or DNA in hybrid?
p. 51. Prob. 7-17, part B.
Assume you denature the met-tRNA before you hybridize it to the 2 separated
strands of DNA.
Prob. 7-18, part B. Add the following:
#Note: tRNAs are extensively modified after transcription. One of the modifications is the addition of the nucleotides C-C-A to the 3’ end, so that the standard cloverleaf structure has an extension on the 3’ end. For this problem, assume no modifications have occurred after transcription (no CCA end added yet).
p. 60, Prob. 8-6. In the note, it should read: thymidine = thymine + deoxyribose.
p. 95. Prob. 12-0. Part A should read: Which of the following components are necessary for the proper functioning of the greasy operon?
p. 96, Prob. 12-4, part A. The following hint was added: Can’t see how to do this problem? Draw out the components of each operon and see which parts of the constitutive operon (promoter? operator? repressor?) will affect the inducible operon, and vice versa.
p. 98, Prob. 12-9, parts D & E. The following hint was added: What happened when you put the two operons together? Did either one affect the other?
p. 106, Prob. 13-8, part D. In D-3-(i) the first choice should be (the same size as in D-2).
p. 115, Prob. 13R-3, part C. This part of the problem has been rewritten to make it clearer. The revised part C reads as follows:
C. Suppose the hybrid plasmid has the operator and promoter of the lactose operon (whose function is to breakdown lactose) hooked up correctly to the human gene for enzyme A (which has no operator or promoter of its own). You use the hybrid plasmid to transform bacterial cells that have a complete, normal lactose operon. You select transformed bacteria that have received the hybrid plasmid. If you want to isolate mRNA for human enzyme A, you should incubate the transformed bacteria (w/ cys) (w/o cys) (w/ lac) (w/o lac) (with both) (without either) (w/ cys but w/o lac) (w/ lac but w/o cys) (any of these). Note: lac = lactose; cys = cysteine.
Corrections to the Answers
Short explanations were added to the following answers: 1-12, 1-13, 1-15, 1-16, 1-20, 2-4, 2-11 (A), 2-13, 3-8 (F), 4-1 , 4-15 (A & D), 4-17 (A), 5-3, 6-2, 6-6 (A). See the 2010 edition if you are having trouble with any of these questions.
p. 146, answer to 2-20, part C was rewritten. New answer is below. Highlighted sentence is the only sig. difference.
C. I. Both oligopeptides could form a secondary structure like an alpha helix. If they did, the 6 leucine side chains of oligopeptide I (case A below) would all be sticking out along the same side, approximately above and below each other, since there are about 3-4 residues per turn of an alpha helix (3.6 to be exact). If the side of the helix with these 6 long leucine side chains faced the exterior of the subunit, then it could interact with the same face presented by an identical subunit via hydrophobic forces, thus contributing to the association of the 2 subunits in forming the homodimer. (Note: Peptides that interact like this usually have leu or another hydrophobic residue every 7 amino acids, not every 3 or 4. See **) In oligopeptide II (case B below) 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.
** It was originally thought that the two helices were interdigitated (as in Becker fig. 23-25 (c)) but we now know the two helices line up side by side as shown here or in Sadava fig. 14.15. To see the difference between the two models more clearly, click here.
p. 163, answer to 5-14 C. Yes, as long as you can hydrolyze acetyl-CoA (so you can reuse the CoA) and you can get rid of the acetate.
p. 177, answer to 7-17. Answer has been rewritten for clarity. New version is as follows:
A. 2 types (1 for initiation, 1 for elongation; both have same anticodon). Note: There are probably more than two genes involved here. There are only 2 types of met tRNA, and only two kinds of met tRNA genes, but there are usually several copies of each type of tRNA gene.
B. 1 or both – depends on the direction of transcription. If all genes for met tRNA’s are transcribed in the same direction (as from P1 & P3 in prob. 7-5), all genes will use the same strand as template, and the answer is “one.” If some genes are transcribed in one direction, while some are transcribed in the other direction (as from P1 & P2 in prob. 7-5), then some genes will use one strand for template and some will use the other, and the answer is “both.” Note that the tRNA, as well as the DNA, is denatured before hybridization, so that the entire tRNA sequence is available to hybridize, not just the anticodon. (The 2nd paragraph of the answer to B is unchanged.)
p. 213. Answers to 12-8 and 12-9 are reversed. The answer labeled ’12-8’ is the answer to problem 12-9 and vice versa.
p. 224, answer to 13-16 part A2. The following was added:
Note: The polyA tail is usually considered part of the primary transcript even though it is not encoded in the DNA and is added after transcription proper. That’s because the ‘tail’ is added as part of the process of transcription termination, so that pre-mRNAs without tails are not found. A ‘primary transcript’ without a polyA tail can be envisioned, but it does not exist.
p. 235, answer to 15-1. Answers have been expanded. Revised answers are as follows:
A-1. WT = 4; A-2. MUT = 1. The subunit in the WT is 40,000 Daltons; the subunit in the mutant is 30,000. The mutant subunits cannot associate to form proper quaternary structure.
B-1. In a structural gene – regulation appears to be normal, as the same number of mRNA molecules are made in both cases.
B-2. Mutation is probably a deletion. A nonsense mutation would shorten the protein, but wouldn’t shorten the mRNA.
p. 237, answer to problem 15-7. The explanations to A & B have been revised. The highlighted sections have been added:
A. A+C. These proteins migrate the same during electrophoresis with SDS, so they have the same mol. wt; They are separable during electrophoresis without SDS, so they have different charges.
B. i) Group 1 – defect in synthesis; require lys added for growth. Group 2 – defect in breakdown – can’t derive energy from lysine (so can’t grow without another carbon source, such as glucose, present).
ii) 2, 8-10 – mutants in these genes can’t break down lysine; therefore products of normal alleles of these genes must be needed for breakdown.
iii) C – lacking in group 2 mutants, which are defective in breakdown. So protein C must be needed for breakdown.
iv) These genes code for enzymes that catalyze steps in the pathways for the breakdown or synthesis of lysine. Only the enzymes for the first and last steps should bind lysine, since these are the only steps which involve lysine as a substrate, product, feedback inhibitor, or activator.
v) Last step in lysine synthesis. Mutants that lack protein A require lys for growth, so protein A must be involved in lys synthesis, not breakdown. The protein binds lysine, so it must catalyze the last step or contain the regulatory site for the first step. A lack of the peptide with the regulatory site should not cause a growth requirement for lysine—it should just mess up regulation. So answer given is the only correct one if you assume the regulatory site and the catalytic site of the first enzyme in the pathway are on different subunits. (If both sites are on one subunit, and that subunit is missing, answer could be first step.)