Revisions to Problem Book, 16th Edition = New Things in 16th ed, revised.
Use this list to update the 16th edition to the revised version. The changes in the problems are listed first, followed by all the changes to the answers. Some typos and minor clarifications are not listed.
Changes in Problems
2-16. Neither hydrolysate contains any cysteine or cystine.
2-17. Assume the two proteins have no quaternary structure. You get a single band after electrophoresis of your mixture on a SDS gel.
3-7. There should be an arrow, not a number nine, going from B to F in the diagram.
3-10. Assume each mutant has only one mutation.
5-1, part B. Add (v): Suppose you run out of glucose and lactate, and you still want to make wine. What can you feed the yeast?
5-11. Whenever it says 'NAD is oxidized,' it should say 'NADH2 is oxidized'.
5-14. Coenzyme A should be included in the equation; the fatty acids and the acetate in the equation should each be attached to CoA.
6-12. Assume DNA shown is linear. In part D, nucleotides = nucleoside triphosphates.
7-26, part C. GC, AT etc. should be written G/C, A/T, and so on. Each pair refers to a base pair in double stranded DNA, with G (or A) on one strand and C (or T) on the other.
8-9. Add the following hint: When dealing with meiosis, be careful not to confuse chromosomes (which may be single or double) with chromatids. When the DNA is doubled before meiosis, the chromatid # changes, not the chromosome #. (See also the note to answer of 9-16.)
9-1. Ignore the term "trait."
10-2 & 10-15. Change the term "trait" to "character."
10-11. This problem has been significantly changed. The new version is as follows:
10-11. Genes A-F are arranged in alphabetical order on a chromosome. The distances are as follows:
Map is not drawn to scale. Numbers refer to distance in map units. Fill in each blank below and explain each answer briefly.
A. The recombination frequency between genes D and E should be __________ %.
B. The map distance between genes D and F should be __________ map units.
C. The recombination frequency between genes A and F should be __________ %.
D. The map distance between genes A and F should be ______________ map units.
E. Suppose you cross a homozygous recessive for genes D and E to a homozygous dominant for both genes. You get a heterozygote that produces many gametes. What proportion of the gametes should carry the dominant alleles of both genes?
11-14. This problem has been rewritten so that experiments #1 and #2 are switched. Therefore part A now asks about the experiment that generated the data in table 2, and part B asks about the experiment that generated the data in table 1.
12-2. Part C of his problem has been significantly altered. The new version of C is:
C. Suppose the lac operon is next to the his operon, and both are transcribed off the same DNA strand; assume the lac operon is to the left of the his operon, and transcription goes left to right. The lac operon is induced by lactose.
Now suppose a deletion occurs that removes all the structural genes of the lac operon and both the operator and promotor of the his operon, so the promotor and operator of the lac operon are joined to the structural genes of the his operon. (What does the DNA in the deletion mutant look like? Draw in all the promotors, structural genes etc.)
(i) What will happen to the items listed in A if you add histidine to a culture of the deletion mutant growing in minimal medium containing lactose as a carbon source? (Minimal means salts plus lactose; no glucose or his.)
(ii). What will happen to the items listed in A if you transfer the bacteria to minimal medium containing glucose instead of lactose? (Still no his). Will the bacteria grow?
13-11. A new part A has been inserted; what used to be part A is now B and so on.
A. This DNA is probably from a (prokaryote) (eukaryote) (can’t predict) AND what is the usual symbol for “start,” meaning the start of transcription? How should it be indicated on the diagram?
Changes in Answers
1-15. Amino acids may be found in tube 5. This is because some membrane phospholipids contain serine as the alcohol linked to the phosphate groups. Most phospholipids contain alcohols that are not amino acids. See Becker fig. 7-7.
2-17. Answer has been expanded.
5-1, part B. Answer to (iv) includes the answer to the new part (v). Note that yeast may have the enzymes to convert lactate to pyruvate and then into carbon dioxide and ethanol. But yeast cannot grow by using this process -- no ATP is generated by these reactions.
5-6 & 5-7. Answers have been slightly expanded.
5-11, part B-4. There is a major mistake in this answer -- the answer is backwards. Protons are stuck on the inside, not the outside. The correct answer should read as follows:
B-4. ADP levels stay constant because there is no ox. phos. Either protons are stuck on the inside of the vesicles and cannot pass back out because they can't get through the membrane, or the protons are bypassing the ATP synthase and leaking back through the membrane without driving ATP synthesis. If the protons are leaking back out, they must be doing so slowly, because a proton gradient is established. If the protons were flowing back freely out of the vesicles as quickly as they were pumped in, there would be no pH gradient. In either case, the membranes have been damaged by the experimental procedure so that the protons do not pass normally through the ATP synthase to leave the vesicles. Either the membranes are leaky and/or ATP synthase no longer works.
5-14. Answer changed slightly to make it clearer, and include CoA.
7-5. Primase is not the same as regular RNA polymerase. Primase catalyzes synthesis of very short RNA chains (primers) only. No long transcripts.
8-9. Here is a clearer version
of the answer:
Note that the question asks only about the abnormal gametes resulting from ND (nondisjunction) in male meiosis. What abnormal gametes do you expect?
(a) ND at 1st division in the male produces 2 gametes with extra sex chromosomes (both XY) and 2 gametes with no sex chromosomes (“0”).
(b) ND at 2nd division produces 1 gamete with extra sex chromosomes (XX or YY) and 1 “0” gamete with no sex chromosomes. (The other 2 gametes are normal.)
If ND is equally likely at 1st or 2nd division, then outcomes (a) and (b) are equally likely. However ND at first division produces 4 abnormal gametes per ND event (2 “0” and 2 XY), while ND at 2nd division produce only 2 abnormal gametes per ND event (1 “0” and one XX or one YY).
Given the argument above, “0” should be the most common abnormal gamete, and YY & XX the least common abnormal gametes. (If you can’t figure this out, assume there are 4 ND events, and calculate what abnormal gametes are expected – you’d expect 2 mistakes at 1st div, 2 at 2nd div, etc.)
10-5. Here are the explanations of the answers.
A. 27 = 3 X 3 X 3 = # genotypes for the 1st gene X # for the 2nd gene X # for the 3rd.
F1 = GgBbDd = triple heterozygote.
F2 has 3 genotypes (GG, Gg & gg) for color, 3 (BB, Bb, & bb) for size, and 3 (DD, Dd & dd) for fur type. The simplest way to see all the possible combinations is to draw out a branching diagram. This will show that GG can go with BB, Bb, or bb, and each of these 3 combinations can go with any of the 3 fur type genotypes (DD, Dd & dd) and so on.
B. 12 = 2 X 3 X 2 = # phenotypes for the 1st gene X # for the 2nd gene X # for the 3rd. There is complete dominance for color and fur type and intermediate dominance for size. In other words, GG & Gg look the same and so do DD & Dd, but Bb looks different from BB. So there are two different phenotypes for color (corresponding to G_ & gg), 3 for size and 2 for fur type for a total of 2 X 3 X 2. (You can also figure this out by drawing a branching diagram. G_ can go with Bb, bb, or BB and so on.)
C. You need to calculate the fraction that are triple homozygotes (gg bb dd) from a cross of two triple heterogyzgotes (GgBbDd X the same). Simplest way is to consider each gene separately. Answer = 1/64 = 1/4 X 1/4 X 1/4 = fraction of offspring that are homozygous for 1st gene X fraction homozygous for 2nd gene X fraction homozygous for the 3rd gene.
D. 3/32 = 3/4 X 1/2 X 1/4 = (fraction that are G_) X (fraction that are Bb) X (fraction that are dd).
10-11. Here is the answer to the new version of the question:
A. 12% recombination frequency. One map unit = 1% recombination frequency.
B. 18 map units. Map distances are additive (within reasonable limits).
C. 50%. When genes are far enough apart, R.F. and map distance do not correspond because of multiple crossovers. As two genes get farther apart, the two genes act less and less linked (assort more and more randomly) and R.F. approaches a maximum of 50%. No matter how far apart the genes are, they never give a R.F. of more than 50%. To give a concrete example, the R.F. between A and E or A and F would be the same – 50% -- indicating random assortment between A and E or between A and F.
D. 76 map units. Individual short distances can be summed to give accurate long distances, even if the two end points (A & F) are so far apart that they give an R.F. of 50%, and therefore act unlinked. In this case, the distance from A to F (or A to E) can’t be calculated directly from the corresponding R.F. between A and F (or A and E) but it can be calculated by adding up the smaller, shorter distances in between the two endpoints.
E. 44%. Since RF = 12%, heterozygote should à 12% recombinant gametes and 88% parental gametes; 1/2 the parental gametes should be double dominant and 1/2 double recessive.
|DE/DE X de/de||
|12% De + dE (recombinants)|
|88% de + DE (parentals) = 44% each of de & DE|
11-7. Note: the simplest known real viruses contain single stranded RNA and code for less than 5 proteins.
11-14. Remember that the experiments are switched in the revised edition.
12-2, part C. Here is the answer to the revised version of the question:
C. You have converted a
repressible operon (controlled by histidine) into an inducible one (controlled
by lactose). The operator that is specific to the lac operon (and binds lac
repressor, not his repressor) is now hooked up to (and controls) the structural
genes of the his operon. The switch mechanism of the lac operon (O & P) acts
exactly the same whether the Olac & Plac are next to the lac structural genes or
the his structural genes. It is as if the switch on the microwave were rewired
to the toaster, so that now when you turn on the switch on the microwave, the
toaster goes on.
(i) In medium containing lactose, the his operon will be induced. If you add histidine, it will have no effect because there is no his operator for his repressor to bind to. The cells will continue to make his. (1) will remain the same at its usual, constitutive level. (4) & (5) will remain the same at zero -- there is no his operator for repressor to bind to and no structural genes of the lac operon to be transcribed. (2) & (3) will remain the same, at high induced levels. (6) will increase but have no effect.
(ii) If you switch to medium containing glucose, you will no longer have any lactose to induce the his operon. Therefore (2) will decline to very low levels and no new enzymes will be made. Enzymes are stable, unlike mRNA, but the enzyme/cell will decline as the cells grow. The total amount of enzyme/culture will remain the same, but the amount per cell will decline as the cell number increases. (1) will remain constitutive, and (4) & (5) will remain at zero, as when cells are grown on lactose. (6) will decrease, as the cells run out of his, but the change will have no effect on the his operon.
The cells will grow until the enzymes for synthesis of his get too low per cell, and the his runs out. In a normal organism, the his operon would be turned on (de-repressed) when the cells run out of his, but in these mutants the his operon can’t be turned on (induced) without lactose. If you want the mutant bacteria to keep on growing in medium without lactose, you have to add his, because the cells can’t make it themselves (unless you add lac).
Answer to A (inserted in revised edition):. DNA has introns, so it should be
from a eukaryote. The start of transcription is usually indicated by a bent
arrow pointing in the direction of transcription.
Add to what is B in the 16th edition, but C in the revised edition: Poly A, if any, would be attached to the end of the primary transcript as part of the termination process. Texts and authorities differ on whether poly A should be considered part of the primary transcript or not.
15-4, part E. Explanation of "increase.":
The population will move from one state of equilibrium (with stronger selection against condition X) to a new state of equilibrium (with weaker selection against condition X). Equilibrium does not mean there is no selection – it just means that all factors, including selection, have reached a balance. In other words, the population can be in equilibrium, as described in this problem, even if there is selection. The assumption here is that condition X is not neutral, given the present human diet. There is selection against condition X. But, if there is more of the necessary amino acid available in the future, selection against condition X will decrease. Homozygous recessive individuals will be more likely to pass on their alleles, and the recessive allele will reach a higher steady state level in the population. (If the mutation causing condition X were completely neutral, and there was no selection, then you would expect the frequency of the allele to stay the same in a large population. But that is not the going assumption here.)
15-6, part A. On high levels of substrate, both enzymes would work equally well, and both WT and mutant would grow at the same rate.