Problem Book, 19th ed., revised (2010) Corrections Listed in Order of Problems
This is the page you need if you have the 19th edition, revised, 2010. Your book already contains all the corrections made in 2010. The additional (significant) corrections made in 2011 are listed below. All additions of a few words are highlighted in blue. Where notes or paragraphs were added, the addition is noted in blue. (Minor changes and typos are not listed.)
If you have the 19th edition, 2009 (not revised) you need the corrections page for the 19th edition as well as the corrections below. If you have an earlier edition, see the corrections page for links to corrections for older edition.
Corrections to the
p. 1 Note to Students. The following paragraphs have been added:
Every effort has been made to match the terms used in class with the terms in this book. However, there may be some technical terms used in these problems that were not (or not yet) covered in class. Don’t panic if you hit a term you don’t recognize – look the term up in your text or go online and ‘Google’ it. You don’t have to ask the TA or instructor.
These problems are intended to help you learn the material, not just to allow you to test yourself. Therefore it pays to go through them carefully and be sure that you understand the point of each problem, even if you think the answer is obvious. See note at bottom of page 5.
Corrections to the Questions
Note that almost all
of the following corrections are clarifications, not additional questions or
changes in meaning. The only 'new question' is an addition to 13-11.
p. 13, prob. 2-8 B. Change 'gel electrophoresis' to 'PAGE'. Electrophoresis in this case is NOT with very large pore gels (see next correction).
p. 13 & 15 -- probs. 2-9E & 2-17. Change 'starch' to 'agarose', and 'large pores' to 'very large pores.' Starch was used historically, but today agarose is used to make gels with very large pores. (If pores are very large, as with starch or agarose, there is no sieving effect & the separation depends only on charge.)
p. 17, prob. 3-1. At end of first line, change 'presence and absence' to presence or absence.'
p. 24, prob. 3R-3 -- see headings for the diagrams (underlined). Change 'large pores' to 'agarose' and '+SDS' to 'PAGE + SDS'. (For explanations, see above for corrections to pp. 13 & 15.)
p. 27, prob. 4-7. The symbol 'Q' was replaced with the symbol 'M.' This is to avoid confusion with the use of Q in chemistry to mean [products]/[reactants].
p. 37, prob. 5-13. In choices (2) and (4) change 'electrons crossing' to 'net electron flow across'.
p. 45, prob. 6-15. Add: Note that the DNA includes sections to the right and left of the sequences shown, as indicated by the .....
p. 49, prob. 7-10. Assume cells used are E. coli.
p. 49, prob. 7-12.
A. The question should read: Suppose mRNA is transcribed from this section of the DNA, using the complementary strand ....
B. Assume no start codon is required for B-F.
G. Suppose the stretch of DNA shown is found in a real E. coli, and the reading frame used in vivo is not know.
p. 50, prob. 7-14. Make the polylysine 150 amino acids long, not 50. (This is to be sure it is long enough to accommodate more than one ribosome. One ribosome covers about 30 codons.)
p. 77, prob. 10-15.
B. 1/16 are dark brown, same color as brown parent or grandparent.
Note at * (at end of problem) should read:
* Having trouble with this problem? Can’t come up with any reasonable solution? What type of cross will give you 1/16 in the F2 that look like each original parental strain? For the other colors, don’t forget the possibility of intermediate dominance, and that the effects of different genes can be additive. For more details, see the explanation in the key.
p. 89, prob. 11-6. Problem should not be starred.
p. 92, prob. 11-12, part B. Part B should read as follows, and the note at ** was added:
B. Suppose this polypeptide is part of the
of a spherical virus. If the
mutant virus gets into a cell somehow, it will probably NOT synthesize or
make (DNA) (protein) (infectious virus)
(any of these).
** Coat proteins = surface proteins that make up the external shell or capsid surrounding the internal nucleic acid. (Simple viruses contain only nucleic acid inside the capsid; some complex viruses contain internal proteins along with the nucleic acid).
p. 101, prob. 12R-4, part C. Change 'should' to 'is most likely to.'
p. 109, prob. 13-11, part B. B was changed to B-1. The following B-2 was added:
B-2. The length of the 5’ UTR is _____________ and it is encoded by (an exon) (an intron)
(either one) (neither).
Corrections to the Answers
Note that almost all of the following corrections to the answers are clarifications, usually in the form of notes. The only answer that is significantly different is the answer to 1-18 B.
p. 138, prob. 1-18, B. Answer should read:
B. Answer (yes or no) depends on how you look at it. Amino acids can be hooked in only one way to form a linear chain connected by peptide bonds, so both molecules are the same in the sense that they are both polyserine – linear chains of linked serines. However, the chains can be of different lengths, and in that sense they can be considered different -- if they are different enough in length, they will have different physical properties.
p. 143, prob. 2-9. Change 'starch gel has large pores' to 'agarose gel has very large pores'.
p. 144, prob. 2-15. Change % sign to µ .
p. 145, prob. 2-17. Second sentence should read: Electrophoresis on an agarose gel without SDS....
p. 146, prob. 2-18. The explanation for experiment 1 is missing. Add:
Experiment 1 – Protein is relatively spherical, less friction (than in #2), protein migrates faster (than in #2).
p. 146, prob.
2-20. For a current model of a 'leucine
zipper' go to
p. 154, probs. 4-6 & 4-7 B. Explanations were added as follows:
4-6. A. 0; B. 1; C. (iii); D. (v) . This reaction has an equilibrium constant of 1 and a DGo of zero. So you can’t predict the direction from the DGo – to predict the direction you must know the actual concentrations and/or the DG. Depending on the concentrations of reactants and products, the reaction could go either way. If [Reactants]/[Products] is high it will go to the right, and if [Products]/[Reactants] is high it will go to the left.
4-7. B. >2. Without any enzyme, rate should be negligible. See answers to 4-1D, 4-3, & 4-5C above.
p. 162, prob. 5-11. The following note was added after B-1:
Note: 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.
p. 168, prob. 6-6. The following note was added after B:
Note: When you replicate a ds molecule, you get two new ds molecules, but you do not have the original ds molecule any more. You have a total of 2 ds molecules, not 3. You still have the two single strands from the original molecule, but they aren’t together – each one is half of a different, ds molecule.
p. 169, prob. 6-12 F.
The first sentence of the answer should be replaced by the following: You
will have 8 ds (double stranded) molecules at the end, and 1/8 of them will be
radioactive. (Did you think there would
be 15 ds molecules at the end? See note at end of 6-6 B.) The radioactive
molecules will be hybrid (in terms of radioactive label) -- one strand
radioactive, and one not.
p. 183, prob. 8-2. Note added to case H:
# In anaphase of meiosis II, the cell contains two sets of haploid chromosomes, not one set of homologs. It is not diploid.
Note added to # chromosomes for cases F
During anaphase of mitosis or meiosis II, each doubled chromosome splits, and so the number of chromosomes per cell doubles and it equals the number of chromatids. You temporarily have two cells’ worth of single chromosomes (with one chromatid each) in one cell. However, in anaphase of meiosis I, the doubled chromosomes do not split, so the number of chromosomes (with two chromatids each) does not increase.
p. 206, prob. 11-6. The following was added:
Hint: What did you transfer, a fragment or a plasmid? Can you get colonies by complementation if you added a fragment? A plasmid?
A recipient cell (that gets a fragment) may be pro+, but it will not give rise to a colony of pro+ cells.
p. 208, prob. 11-12 B. Change 'head or heads' to 'coat or coats'.
p. 211, prob. 12-2. Added brief explanations to A, as follows:
A. (1) remains the same -- production of repressor protein is constitutive.
(2) decreases – binding of repressor-co-repressor complex will block transcription.
(3) decreases (in long run) – cell grows, but more enzymes to make his are not made.
(4) increases – co-repressor binds to repressor forming complex; complex binds to operator.
(5) same – repression by his does not affect the lac operon – his does not bind to lac repressor.
(6) increases – see (4). His is the co-repressor.
p. 212, prob. 12-5. Added the following to the explanation:
DNA is always the same. However different proteins (enzymes) are needed under different conditions, so which sections of the DNA are transcribed changes. In minimal medium, more DNA sections are transcribed in order to make more enzymes -- the enzymes needed to catalyze synthesis of amino acids (AAs), nucleotides, etc., from glucose. In rich medium, AAs, nucleotides, etc. are provided, so the enzymes for synthesis of AA etc. are not needed, and the corresponding sections of DNA are not transcribed. Therefore cells grown in rich medium would contain less different mRNAs, and the RNAs would hybridize to a smaller proportion of the DNA.
p. 216, prob. 12R-4, part C. Answer given is simplest case. It is possible, but less likely, for both Km and Vmax to increase, or both to decrease, so that the enzyme works worse (slower) at a given [S].
p. 221, prob. 13-11. Answer to new part (B-2): 2 KB, an exon. First exon goes from start of transcription (not start of translation) to first 5' splice site.
p. 228, prob. 13R-6.
The following note was added:
When you are
separating DNAs by gel electrophoresis, changes of a few base pairs in length
are not usually noticeable – the % change is too small.