Problem Book, 20th ed., revised  (2013) Corrections Listed in Order of Problems

This is the page you need if you have the 20th edition, 2012.  Your book already contains all the corrections made up to 2012. The additional (significant) corrections made in 2013 are listed below. The additions or corrections of a few words are highlighted in blue. Where a paragraphs was added, the addition is noted in green. (Minor changes and typos are not listed.)


If you have an earlier edition, you will need to look at previous corrections as well as the corrections below. See the current corrections page or the old-probbook corrections page for links to previous corrections.


Since last year, only two questions (11-4 & 13-4) were changed significantly, and only one answer (to 13R-4) has been significantly altered. Here are the details:


Changes to Questions:

Problem 11-14. The phenotypes used in this problem have been changed from the ability to make large (L) vs small (S) plaques to the ability (or inability) to kill and lyse cells of E. coli, strain K. (This is actually the phenotype that was measured by the investigators.)  The results described in the older edition of the problem book are not realistic, as complementation between mutants that form small plaques is not likely to lead to formation of large plaques. However, complementation between mutants that cannot lyse strain K is likely to lead to ability to lyse cells of strain K. The basic set up, and the interpretation of the results, is the same in both versions of the problem. (As long as you assume, in the older version, that complementation between mutants that form small plaques will lead to formation of large plaques.)  If you want to see the revised question and answer, borrow a 2013 edition from a friend. (The principles tested are not changed.)


Problem 13-4, Part C. This question has been broken up into C-1 (similar to before) and C-2 (added): Part C should read:


C. Suppose the enzyme you chose in A (= EatI, II, III or IV) cuts the site AT/GGCCAT (cuts at the slash).

The restriction enzyme BugI cuts at the site GG/CC. BugII cuts at TA/GGCCTA.

            C-1. To make a recombinant plasmid, you could cut up human DNA containing gene X with
(the correct Eat) (BugI) (BugII) (none of these). Show how you would connect the plasmid DNA cut with Eat and the piece of human DNA containing gene X to make a recombinant plasmid.
            C-2. Suppose you make your recombinant plasmid. Which enzyme(s) could you cut it up with? Same choices as in C-1. (Note: Answer will depend on your choice(s) in C-1.)


The answer to C-1 is unchanged. The answer to C-2 is below.


Changes to Answers:

Answer to problem 13-14, part C-2:

C-2. If you used Eat X to cut the human DNA in C-1, you can cut the recombinant plasmid with EatX or BugI. If you used BugII in C-1, only BugI will cut the recombinant DNA.


Answer to problem 13R-4, part A.  The answer in the older version is what happens if you start with total DNA and use a probe. If you start with purified DNA, the answer is different. Here is the complete explanation:


A. Allele (haplotype) A has one restriction site within the gene, 13KB from one end and 2 KB from the other end of the gene. Allele (haplotype) B has an additional site in the middle of the region that generates the 13 KB fragment. This site is 7 KB from one end and 6 KB from the other end of the ends of the 13KB piece.  (If you start with whole DNA from each person, and not with the purified genes, you'd need a probe to detect the pieces from the gene of interest. To get the patterns described, using a probe, there would have to be restriction sites at both ends of the gene, or the area containing it, not just within the genes.)