Significant changes made to the Problem Book, 20th edition, revised (2013) to convert it to the 20th edition, re-revised (2014).

This is what you need if you have the 20th edition, revised (2013). If you have the 20th edition, re-revised (2014), all these changes have already been made.

Below are the significant changes. (Corrections to typos or formatting that do not change the meaning in any way are not listed.) Text that was changed is blocked in yellow.

Corrections to Problems:

Problem 2-1, p. 11. Add: 'Assume for the purposes of this problem that the peptide bonds are not hydrolyzed under any of the conditions used. '

Problem 2-13C, p. 14. Option iii) should read: 'partially within the lipid bilayer and partially protruding on one side or the other.'

Problem 2-20A, p. 16. Sentence two should read: 'Under the conditions used, all the peptide bonds are hydrolyzed (broken). ' This does not change the meaning, but should make it clearer.

Problem 3-1, p. 17. The units of velocity should be nanomoles/min. instead of mM/min. (It should be amount per minute, not concentration per minute.)

Problem 6-11, p. 43.  Parts B & C should read:
    B. Would dCTP be incorporated into DNA?
    C. Do you think cCTP would affect RNA synthesis?

Problem 6-16D, p. 46. The second part of the question should read: 'AND the native nucleic acid is probably (circular) (linear) (either way).'

Problem 8-6, p. 60. It should read: 'At t = 6 hours, but not before, all dividing cells contain radioactivity.

Problem 10-9. Part F has been added:
    F. How is it possible to have a person with hemophilia, but not color blindness (or vice versa) if the two genes involved -- for hemophilia and color blindness -- are linked?

Problem 13-16, p. 112. A. Add: 'Assume a polyA tail is 100 bases (nucleotides) long.' For the choices listed in A-2, change (950) to (900) and change (3950) to (3900).   For A-3 & A-4, change 'is' to 'should be.' (PolyA tails vary in length, but the average length seems to be closer to 100 bases than to 150, which was the value assumed in the previous editions of the problem book. If you assume the polyA tail is 150 bases long, the choices and answers in the '13 edition are correct. )

Corrections to Answers:

A few more details have been added to the answers to 1-19, 13-16C, 13R-4. These changes are minor, and are not listed below.

Ans. to 3-1, p.147. Units of Vmax have been changed; see corrections to problems above.

Ans to 3R-5, p. 152. The link given on that page no longer works. To find a picture of porin, go to Google or Google images, and search for 'porin.'

Ans. to 10-9F, p. 198. 

F. The two genes involved here are the genes that control color vision and  blood clotting. It is only the mutant alleles of these genes that result in color blindness and hemophilia.  The genes are linked, but not the two mutant alleles.  An individual chromosome can have the normal allele of one gene and the mutant allele of the other, or mutant alleles of both genes or normal alleles of both genes.  Genetic linkage means that the two alleles on any particular parental chromosome tend to be inherited together, not that the two mutant alleles tend to be inherited together in all cases.

Ans. to 13-16A, p. 224. Answers have been adjusted to fit the case if the polyA tail is 100 bases long, not 150. In that case:
    A-2. The primary transcript should be 3900 bases long.
    A-3. The mRNA should be 900 bases long.
    A-4. The answer given is correct, whatever the length of the polyA tail, but the explanation is not adequate. We now know that translation does not have to start in the 1st exon or end in the last exon. Here is the complete, revised answer and explanation:

A4. The length of the RNA that codes for polypeptide X is definitely <800 bases, and probably >200. (It canít be 200, since that isnít a multiple of 3, and it canít be 800 or more as explained below.) 
Why <800? Messenger RNA almost always contains sequences in addition to the protein coding part (& excluding the polyA).  Each mRNA usually contains a leader -- extra bases on the 5í end, before the ATG translational initiation codon -- and a trailer Ė extra bases on the 3í end, after the stop codon (but before the polyA). Since the total information containing part of the mRNA  (=  length of exons = leader + protein coding + trailer) is 800 bases, the protein coding part must be <800.
Why >200?  The protein coding part of the mRNA must contain 3 bases for each AA in the protein. In other words, the length of the protein coding part must be a multiple of 3. So it canít be 200 bases.  It could be shorter, if the protein is unusually small (but this is not a choice). Of the choices given, the only viable answer is >200 but <800. The coding part is indeed likely to be longer than 200 bases, since most proteins are at least 100 AA long.

Ans. to 13R-2, part F-4, p. 226 at the bottom. The last line of the answer, the second half of the parenthesis, is missing. The entire last sentence should read  '(It could be in a 3' regulatory element that affects transcription, but this is not a possibility that you were expected to think of.)'

Ans. to 15-2, part B. The last sentence should read: In induced cells, (with O2), the enzymes are synthesized at much higher rates.