Corrections to 17th edition of the Problem Book (2004).  08/20/2009 03:55 PM

If you have the 2004 book, you need to make the corrections listed below. Then you need to look at the updates to convert to the latest edition. 

If you have the 17th edition, revised (2005), all these corrections have been made, and you do not need this list.

Corrections to questions are listed first; then corrections to answers.

1-21. (p. 9) Question: One of the choices should be (several different amino acids) instead of (amino acids). See corrections to answers below.

1-24. ( p.10) There is a typo in the question -- choice labeled (1) should be labeled (2) and vice versa.

4-19. (p. 32) Parts A & B should say n = 6; total number of carbons = 8. (Not n = 8.) For part C, assume that the supply of CoA is not limiting.

5R-6. (p. 40) Part A has a typo. End of first line of part A should read 'there is a lack of...' not 'there is a lace of....'

6-8. Part B. (p. 42) New medium is NOT radioactive.

6-12. (p. 43) Part D should read, "Given your answer in part C (not part B), the best primer should be...."

8-0. (p. 59) Problem should say the cell is in division not 'in metaphase'. (The stage doesn't matter.)

8-13. Part C. (p. 62) Last choice in the first line is (any of these). The "close parentheses" at the end is missing.

9-1. Part G. (p. 67) Add, "How many different characters?" (As well as genes, alleles, etc.)

9-5. Part A. ( p. 68) Change "traits" to "characters." 

13-11 part G-2. The last sentence (on p. 108) is missing. It should read, "to select bacteria that are resistant to (penicillin) (streptomycin) (both) (neither). Explain."

13-14. (p. 110) In the revised edition the set up is explained more clearly by the addition of the italicized words. At the end of the first paragraph, it should say: 'You then put the genetically modified plasmid (carrying the human gene) into bacteria. When you are done, you take the bacteria (which now have the modified plasmid) and do experiments (1) to (4). In part (1), “Total bacterial DNA” = DNA from bacterium containing plasmid with inserted human gene. In part (3), RNA should be in bold.  In part (4), add after the sentence given: 'You test for the enzyme missing in the mutant bacteria.'

14-3. (p. 119) Part A. Choices should be (<4%) (4%) (>4%) (unpredictable). <4% should not be listed twice.
         Part B: The question is about the % of males in the next generation.

15-6, part C. (p. 129) There is a typo in each of the choices involving substitutions -- there should be an arrow, not a 6. It should read (a substitution --> a nonsense codon) (a substitution --> missense; etc.)

Corrections to Answers:

Answer to 1-21:  (p. 138) Additional explanation: There is only one amino acid (serine) found in phospholipids, not several. Serine, but not any other amino acid, can be the alcohol attached to the phosphate of the phospholipid. Hydrolysis could produce “amino acids” meaning many molecules of serine, but it could not release “amino acids” meaning “several different amino acids.” Note: The question has been altered in the revised edition (see above, p. 9) to make this clearer.

Answer to 3R-2, part B. (p. 151) The Km should be stay the same or decrease. The rationale is as follows:  If you look at the formula for Km, Km = (k2 + k3)/k1. If k3 is decreased, Km should decrease too. But we usually assume k3 is very small, relative to k1 and k2, since k3 is the rate constant for the limiting step (ES → P). That’s why we usually ignore k3, and say  Km is about equal to the disassociation constant of the ES complex (= k2/k1). As long as k3 is small, relative to k1 and k2, a difference in k3 is not likely to have a significant effect on Km.  

Answer to 4-8, part B. (p.154) Last three words should be "excess of A", not "excess of B." (In other words, it takes a huge excess of [A] to drive reaction to right.)

Answer to 4-17, part C. (p.157) Structures shown are correct. You will get a mixture of the two labeled molecules shown -- Half the molecules will be of each kind.

Answer to 4-19, part B . (p. 158)  Add the following:

Note: For every turn of the Krebs cycle, two carbons are fed into the cycle (from acetate) and two carbons are released as CO2.  However the actual CO2 released in one turn is not derived from the acetate added in that turn – the carbon atoms of the released CO2 are derived from the OAA generated in the previous turn.

Answer to 4-19, part C . (p. 158) Answer is acetate or acetyl CoA. Delete the sentence in parentheses at the end of part C and substitute the following:

Note: CoA is needed for breakdown of fatty acids, but in this problem we are assuming CoA is not limiting – in other words, there is enough CoA to break down all the fatty acids to acetyl CoA. How can we have enough CoA? This will happen in a test tube if there is excess CoA; it will happen in a cell (with low levels of CoA) if the acetyl CoA is hydrolyzed to acetate, and the CoA is reused. If CoA were limiting, undegraded fatty acids would pile up as soon as all the CoA was converted to acetyl CoA.

Answer to 5-9: ( p. 160) Here is a slightly longer and more thorough explanation:

You will get a buildup of oxidized components after the block (& reduced components before the block). For example, consider a particular component, say cytochrome c. The net amount of cyt c won’t change, but the ratio of oxidized cyt c to reduced cyt c (the % of cyt c that is oxidized) will change. If cyt c. is before the block, almost all the cyt c will become reduced; if cyt c is after the block, virtually all the cyt c will become oxidized. How does this work? The compounds of the electron transport chain do not move down the chain; only electrons do. Electrons enter the pipeline, from reduced NAD and FAD but can't get past the block. So everything before the block should pick up electrons and be reduced . Everything after the block should give its electrons away to the next component and be oxidized. Eventually, all the electrons after the block will be passed down the chain to oxygen, leaving all the components after the block oxidized. See the answer to 5-10-D.

Answer to 5R-1, part D. (p. 165) In second line, it should say "but more NADH2 is oxidized...."

Answer to 6-12, part E. (p. 169) Every time it says ΔGE it should say ΔGo. Also, in the first line of the 2nd paragraph,  it says ΔG instead of ΔGo. The second paragraph should read, "Even if the ΔGo is zero, the reaction will go to the right as long as the overall ΔG (not the ΔGo) is negative. ...."

Answer to 7-18, part B. (p. 177) Answer should read: A = U because that region of the RNA is double stranded....." (The problem reads "DNA" instead of "RNA.")

Answer to 7-21, part D. (p. 178) Add to explanation: tRNAser carries alanine, but anticodon will match the serine codon.

Answer to 7-22, part A (ii).  (p. 178) Add to explanation: The mutation is in the codon for asp, and that codon is part of the gene for enzyme A.

Answer to 8-4, part A & B: ( p. 184) Four chromosomes means two pairs. Call one pair of chromosomes the A's and the other the B's.

            In part A., diploid adult is, say, AABB; both A’s are the same and both B’s are the same. Gametes are all AB – all the same.

            In part B, diploid adult is AaBb; two A’s are different (A & a) and two B’s are different (B & b). Gametes can be AB, Ab, aB or ab – 4 different kinds.

Answer to 9-1, part G. (p. 191) One character (fur color).

Answer to 9-5, part A. (p. 191) One character (= one trait) = blood type. Note the term "trait" can be used in multiple ways, so the term "character" is used here to avoid confusion.

Answer to 14-7, part A. (p. 230) You can assume q is about 1 in this case because p is so small (<<1).

Answer to 14-17, part D. (p. 234) Answer is 0.255 or 25.5%. Not 24.5%.