Corrections for the 2008 edition of the problem book (18th, revised)
This is the page you need to update from the 18th edition revised (2008) to the 19th. If you have the 18th edition (2007), you need the additional corrections listed here.
Below is a list of the changes that were made to convert the 18th ed, revised, into the 19th edition. (Some additions are highlighted in blue to make it clearer what was changed.) Most of these changes are minor clarifications or additional explanations. The following problems were added: Prob. 2-8, part D; Prob. 12-0; and prob. 13-8, parts B & C. (The previous part B is now part D.) Problem 4-2, parts D & E, were revised.
See links for changes/additions to problems 12-0 & 13-8. All other changes are listed below.
Changes to Questions:
1-6. The current number of people on earth was updated -- it is about 6 billion, not 3 billion.
2-8. Part D was added.
D. Suppose P1 and P2 have the amino acid differences described above, but P2 sediments more rapidly than P1 in the ultracentrifuge. Why might this be so? Explain how this is possible.
4-2. Parts D & E should read as follows:
D. Inside a cell, this reaction can be used in either direction. When glucose must be synthesized, the reaction goes to the right, converting F-6-P to G-G-P; during glycolysis, when glucose is broken down to yield energy, the reaction is used to the left, to convert G-6-P to F-6-P. How is it possible for cells to use this reaction in both directions?
E. If Keq were 23,000 instead of 2.3, would it be possible for cells to use this reaction to convert G-6-P to F-6-P? Try to explain your answer in terms of DG and DG°.
6-14. Part B-4. The term 'editing' should be replaced with 'proof reading.'
6-17, part A. Add the following possible choices:
(primer(s) for the leading strand) (primers(s) for the lagging strand) (primase)
7-22. Part B (ii) -- Question should read:
What is the minimum # of bases between the beginning of the mutated gene (the first translated part) and the mutation?
8-6. Add the following to the note:
This question is based on a
real experiment, which is why it says thymidine instead of thymine. Thymidine is
the nucleoside of thymine = thymine + ribose. Eukaryotic cells will take up
thymidine but not thymine.
10-15. Add the following note to B:
* Having trouble with this problem? Can’t come up with any reasonable solution? What type of cross will give you 1/16 like each parent? For the other colors, don’t forget the possibility of intermediate inheritance, and that the effects of different genes can be additive. For more details, see the explanation in the key.
11-4. Add Note: Assume all E. Coli strains have the same modification and restriction enzymes.
11-8. Right column in table should be headed "Average # Progeny virus produced per infected cell."
Problem set # 12 -- Problem 12-0 was added at the beginning. As a result the pagination is different, and two of the problems are in a different order, but all the problems (except 12-0) are the same. Problems 12-8 and 12-9 in the 18th ed rev. were renumbered 12-9 and 12-8 in the 19th edition. Click here for Problem 12-0.
13-3 A. Add: There is only one base difference between the DNA sequence in normal people and the DNA sequence in people with sickle cell disease.
13-8. Added new part B and C. What is labeled 'B' in the 2008 edition is part D in the 19th ed. Click here for complete question 13-8.
15-7. Add the following: Note: This problem is extremely long; it made up an entire exam. Do not panic if it takes a long time!
Changes to Answers:
Answer to 1-6. You need 2 X 1026 bacteria instead of just 1026 . This difference is not enough to change the approximate time needed, but it does alter the calculation.
Answer to 2-8. Part D. Answer: Four interspersed proline residues in P2 could create hinges between 5 short alpha helices, which could fold into a compact tertiary structure with a smaller diameter than the single extended alpha helix of P1.
Answer to 4-2. Corrected & expanded answers to parts D to E:
D. When [G-6-P]/[F-6-P] is high, the reaction
goes to the left; when [F-6-P]/[G-6-P] is high, the reaction goes to the right.
Which way the reaction goes depends on the DG,
not the DGo.
The DG depends on the
ratio of [G-6-P] to [F-6-P]. (Which compound is the product and which is the
reactant depends on the direction.) The ratio can be altered, favoring a
particular direction, either by providing a lot of substrate &/or by removing
product as it forms. When DGo
is relatively small (Keq = 2.3), relatively small changes in reactant
or product concentrations can have a big effect on
E. It would be possible, but it is unlikely. The reaction will go to the left if DG is positive. Since Keq is very large, DGo looks very favorable for conversion of F-6-P to G-6-P but very unfavorable for the reverse (to the left). However the actual direction depends on the DG, not the DGo. A very large ratio of [G-6-P] to [F-6-P] would be sufficient to drive the reaction to the left. This would require either a tremendous excess of substrate (G-6-P) or constant removal of product (F-6-P) to keep the F6P concentration very low and therefore to keep the [G-6-P]/[F-6-P] high.
Answer to 4-13. No ATP is needed for reaction Y. ATP is only needed for reaction X. The two reactions together (X + Y) make up the top reaction shown in the box in the answer. It is this combined reaction, not reaction Y, that uses up ATP.
The answer to part B is correct, but the explanation is not.
To fix the explanation, add the words in blue to the first full sentence:
It takes about 100 ATP to break down the glycogen and convert it to 100 molecules of PGAlde. First the glycogen is broken down to glucose-P (Reaction Y). Each molecule of glucose -P is then converted to glyceraldehyde-P (Reaction X), yielding.......
Answer to 6-9, part B. Unchanged -- if none of the old nucleotides are reused.
Answer to 7-14. A. You need 7 molecules of RNA total, not 6.
Answer to 10-15. Explanation for part B:
Coat color cannot be controlled by one gene – that won’t give you enough categories. The number of categories and the proportions should clue you in that this is a cross of two unlinked genes, but that some categories (genotypes) are combined – more than one genotype has the same phenotype. If there are 2 genes, one parent is AABB (brown), the other aabb (white), and F1 is AaBb. If the genes are unlinked, you expect that 1/16 of F2 will be AABB (like brown parent) and 1/16 aabb (like white parent). This fits, but how do you get the other categories? The simplest solution is to assume intermediate dominance for both genes, and that the effects of the two genes (as well as the effects of the alleles of each gene) are additive. Then, for example, AaBb and AAbb are the same color. (Each A or B allele causes production of approximately the same amount of brown pigment, possibly by different pathways.)
Answer to 10R-3, part C. Note: There is no obvious biochemical explanation for the dominance of striped over solid.
Answer to 11-12, part A. Explanation: Note that ‘frameshift’ is not a choice. In this question you are asked about the primary change in the DNA, not the consequence of that change. The change in the DNA is a deletion or insertion. The effect of the insertion or deletion is a frameshift during translation of the mRNA. (Deletions and insertions do not always cause frameshifts. Deletions or insertions in coding regions do cause frameshifts, but those in noncoding regions do not.)
Answer to 12-0 was added. Click here for question 12-0 & answer.
Answer to 12-1, part A. Explanation: Inorganic phosphate is the substrate of other enzymes, but not of phosphatase.
Answer to 13-3, part C. Answer should read: C. Maybe. Different genes can
differ in sequence and still produce
the same protein because of degeneracy in the genetic code.
Key to Problem set 13. Pagination is different because answers to 13-8 (B & C) were added. Click here for question 13-8 & answer.