Changes in the Problem Book to Update to the 18th edition (2007)

This is the information you need if you have the 17th edition, revised (2005). If you have a later edition (2007 or later), all these corrections & changes have been made, and you don't need this list.

If you have the 17th edition (2004), click here. If you have the 16th edition (2003 or 2002), click here. If you have an earlier edition, 14th or 15th, click here.
 

There are two things on this page -- first a list of corrections (typos mostly) and then a list of changes (mostly additions) made in converting the 17th ed, revised, to the 18th ed.

Corrections -- see below for  updates (additions) made to convert to the 18th edition.

Corrections to Problems:

Problem 1-24. There are bonds missing in the picture. Alpha glucosamine has an NH₂ group, not an NH on position 2 of the ring. NSG has an amide bond between the NH at position 2 and the side chain of C=0 etc.

Problem 4-13. Reaction X: products should be 2PGAlde + ADP. (ADP is omitted in book.)

Problem 9-9. A. Assume male has a normal karyotype.

Problem 11-6.  The 'chromosomal mutation' in this case = a mutation on a chromosome, not on a plasmid.

Problem 11-16. 'Assume his is not closely linked to leu and lys' should read 'assume the his gene is not close to the leu and lys genes.'

Corrections to Answers

Answer to Prob. 1-9. Reference to Becker is p. 184 in the 6th ed. (This is helpful for part B, not for part A.)

Answer to Prob. 2-13, D. If you got +2, did you remember the ends?

Answer to Prob. 3-8. Part A. The answer reads: "This ratio (k2/k1) = dissociation constant for ES and is an indication of the affinity of E for S and the position of equilibrium for ES <--> E + P." It should be "ES <--> E + S." This mistake is repeated two more times in the body of the long paragraph. The correct answer should read:

3-8.       A.  ii, iv. K_m = (k₂ + k₃)/k₁. If k₃ is relatively small, as it usually is, this reduces to k₂/k₁. This ratio = disassociation constant for ES and is an indication of the affinity of E for S and the position of equilibrium for ES ↔ E + S.

Note that K_m is a measure of the position of equilibrium for ES ↔E + S but NOT the position of equilibrium for S ↔ P. For S ↔ P, the enzyme is a catalyst. Presence or absence of enzyme (or state of enzyme) has no effect on the equilibrium of S ↔ P. (See answer to B.) For ES ↔ E + S, the enzyme is a reactant and/or product. When you change the structure of E, you change the reactants/products so you actually have a different reaction with a different equilibrium constant and therefore a different k₂/k₁. To put it another way, if you change the structure of E, you can change the free energy involved in ES ↔ E + S (& thus the position of equilibrium). But changing E has no effect on the free energy change involved in S ↔ P.

Answer to Prob. 3-10, part C. Last line should say "committed step is before C → A," not "before A → C."

Answer to Prob. 5-4, part (iii). Cell can live on ATP from Krebs if you don't need ATP for the transport of reduced NAD into mitochondria. (Should work in prokaryotes since there are no mitochondria; may or may not work in yeast, since they are eukaryotes.)

Answer to Prob. 5-14, part B. Reference to Becker is to 5th edition. There is no corresponding fig. in the 6th ed.

Answer to Prob. 8-2, at *. How to count centromeres. 'Centromere' can refer to the DNA itself or to the structure (including proteins & DNA) that holds the sister chromatids together. (The protein structure that connects to the spindle fibers forms at the centromere, but it is a separate structure called the kinetochore.) The centromeric DNA replicates in S, like all the DNA, but the two copies remain stuck together at the centromere region until anaphase. The protein structure associated with the DNA probably doubles in size in G-2, and splits in half at anaphase.

Answer to Prob. 9-7, Part B-4. It has a '6' instead of an arrow at 3 places. First sentence should read: A allele produces active enzyme; a allele produces inactive enzyme; A' allele produces partially active enzyme.

Answer to Prob. 14-2. Assume p = frequency of the Rh^- allele; q = frequency of the Rh⁺ allele.

Additions & Changes

Below is a list of the significant changes made when updating the 17^th edition, revised (2005), to the 18^th ed. (2007). There is no change in the total number of pages. The number of pages in each problem set, and the number in each answer key, are the same as in the previous edition. However a few problems or parts have been added, and some problems have been renumbered and/or appear on different pages. There are no changes in problem sets 3R, 5R, 7R, 10R, 12R or 15, and no significant changes in problem set 14. The problem numbers are the same, except for problem sets 3 & 13 (see below for details).

Changes to Questions:

Changes to Problem Set #1:
 

p. 10. Prob. 1-24. Part C-3. Added:   

            C-3.  The formation of glycosidic bonds in cells is enzyme-catalyzed. In the alternating polymer, all linkages are β 1→6. What is the minimum number of enzyme types required for the formation of the glycosidic bonds in this alternating polymer?  (1)  (2)  (3) (>3) (can’t predict).

Changes to Problem Set #2
 

p.11. Prob. 2-0 -- Warm up (and/or review) questions on chemistry -- added:

2-0.      A. Why is a carboxyl group (or a molecule containing one) considered to be an acid but a hydroxyl   group is not?

            B. Why is an amino group considered a base?

            C. How can atoms that are covalently bonded, like the atoms in a peptide bond, form hydrogen bonds to    other atoms?

            D. Is an amino acid every totally un-ionized? At any pH?

p. 14. Prob. 2-11. Part C. added:

C.    Should HbA and HbX be separable by SDS-PAGE? (SDS-PAGE = polyacrylamide gel electrophoresis in the presence of SDS). Explain why or why not.

p. 16. Problem 2-20 added:

*2-20.   Consider the two hypothetical 24-amino-acid oligopeptides below. Each has the same overall amino acid composition.

            I.   leu-ser-ala-ala-leu-ala-ser-leu ala-ala-ser-leu-ala-ser-ala-leu-ala-ala-leu-ala-ser-ala-leu-ser

            II.  leu-leu-leu-ala-ala-ser-ser-ala-ser-ala-ala-ala-leu-ser-leu-leu-ala-leu-ala-ser-ala-ser-ala-ala

            A.  The 2 oligopeptides are mixed in equal amounts and subjected to hydrolysis by strong (hydrochloric) acid. This treatment breaks (hydrolyzes) all the peptide bonds.   How many spots of amino acids do you expect to find after ascending paper chromatography of the hydrolysate in 70% isopropanol-30% water buffered at pH7?  Circle the best answer and indicate the composition and relative mobility of all spot(s). (1)  (2)  (3)  (4)  (5)  (6)  (>6)

            B.  If the paper chromatography were carried out at pH 2 instead of pH7, the migration rates would be (faster for all spots) (slower for all spots) (faster for some spots) (slower for some spots) (faster for some spots and slower for others)   (unchanged for all spots).  Circle the best answer and explain that choice.

            C.  Suppose each of these 2 oligopeptides exist as alpha-helical regions within a larger homodimeric protein and  both are found on the surface of the monomer subunit. One is found to contribute to the quaternary structure.  Which is the better candidate? (I)    (II)    (I and II are equally good candidates)  Hint: This is not easy.  It will be important to take into account that the peptides are in an alpha-helical conformation at the surface, and that there are 3.6 amino acids per turn of an alpha helix. Also remember that the protein is homodimeric.

p. 17. Questions for Prob. 3-1 are changed slightly. (Data is the same.) Questions are now:
3-1.     A.    What are the V_max and K_m values for the enzyme-catalyzed reaction without inhibitor? Please give units and explain your reasoning. (See note.)

B.    (i). What are the apparent V_max and K_m values plus inhibitor? 

      (ii). Is this a case of competitive or noncompetitive inhibition? How did you know?

All the other problems in set 3 are the same, but are renumbered because one new problem (see below) was added. Here is a table of the changes for problem set 3:

Here is the new problem:

3-5. Succinic dehydrogenase (SDH) catalyzes the interconversion of fumaric acid and succinic acid by adding or removing two hydrogen atoms from the central two carbons:

COOH-CH=CH-COOH ↔ COOH-CH₂-CH₂-COOH
              Fumaric acid                      Succinic acid

 (This reaction is part of an important metabolic pathway called the Krebs cycle, which will be discussed later.) Consider the SDH purified from mouse liver (SDH-M) and SDH purified from E. coli (SDH-C). Both enzymes catalyze the same chemical reaction indicated above. However, they are distinct proteins as they have a somewhat different primary structure. Suppose you are given a tube of each enzyme; you do not know the concentration of each enzyme in each tube. Using 1 ml. of the enzyme solution each time, you run 2 reactions for each enzyme, starting with fumarate as the substrate plus a source of hydrogen atoms, and you measure the rate of succinic acid production. The results are shown in the table.  As you can see there is some degree of uncertainty about the measured values for V_o. This real-life situation is included here to preclude solving this problem using the Michaelis-Menton equation .

    A. From these results, which enzyme has the larger V_max in this experiment? (SDH-M) (SDH-C) (both the same) (can't tell).
    B. From these results, which enzyme has the larger turnover number? (SDH-M) (SDH-C) (both the same) (can't tell).
    C. From these results, assuming these enzymes are typical such that k₃ << k₂, which enzyme probably binds the substrate fumaric acid with the greater affinity?(SDH-M) (SDH-C) (both bind the same) (can't tell).
   D. Propenoic acid (CH2=CH-COOH) has a structure similar to fumaric acid, but it lacks one of the carboxylic acid groups. Although the hydrogenation of propenoic acid can be catalyzed by SDH-M, and the K_m of the enzyme for this substrate is the same as the Km of the enzyme for fumaric acid, the turnover number using propenoic acid is only 1/10 the turnover number using fumaric acid. If you started the reaction under the same conditions described above, but with 2 mM propenoic acid instead of fumaric acid, how many nanomoles of the product (propanoic acid, CH3-CH2-COOH) would accumulate after 10 minutes? (0) (0.31) (0.4) (0.62) (3.1) (4) (6.2) (31) (40) (62)

Changes to Problem Set #4

p. 28. Prob. 4-9C. Clarification added: Assume that reactions 1-3 have no other function -- they lead only to production of lys.

p. 29 Prob. 4-12. Changed reaction numbers from 1 and 9 to 1 and 10.

p. 30. Prob. 4-13. Added part C, and updated terminology. (Old parts C and D are now D and E.)

Part C:  In both A and B you start with 100 glucose units and ferment the glucose. How come you don’t end up with the same amount of ATP at the end? If you get more ATP, where did the energy to make the extra ATP come from?    

Changes to Problem Set #5

p. 37. Prob. 5-14, part B. There is a typo in the formula. It should be: COOH-(CH₂)₄-CH₃

Changes to Problem Set #6

p. 42. Prob. 6-6, Part C is is now starred. Also a hint added: Draw the DNA as in problem 7-1.

Changes to Problem Set #7

p. 47. Prob. 7-1, part B. Questions are now:

B-1. Will you find ³²P incorporated when any of these labeled GTP compounds are used? Indicate the           position of ³²P, when present, in the complementary RNA sequence.

B-2. If you want to get the maximum amount of radioactivity incorporated into your RNA, which is the best form of labeled GTP to use?

p. 48. Prob. 7-7. Hint added: Can’t figure out what to do? See hint in answer key.

p. 50, prob. 7-16. Hint added: Having trouble solving this problem? See the hint in the answer key.

Changes to Problem Set #8

p. 59. Prob. 8-0. Added D: D. Do you think this cell is diploid (2N) or haploid (N)?

            Prob. 8-2. Added: There is more than one way to count centromeres, as explained in the answer key.

Changes to Problem Set #8R

p. 66, Prob. 8R-3. Call the radioactive precursor X*. Then C should read:

C.    Suppose you grow the cells in medium containing X* for 2 generations, as follows:

                                                   interphase + X^*                     interphase + X^*

cells (Non-radioactive)     ------------------>  mitosis #1 -------------------> mitosis #2

Changes to Problem Set #9

p. 69, Prob. 9-9. A. Assume male has a normal karyotype.

Changes to Problem Set #10

p. 74. Prob. 10-8, E. Add Hint: The rarest classes are double crossovers. If the order is color-arg-strep, does it take 1 or 2 crossovers to get green arg-strep^S?

Changes to Problem Set #11

p. 89, Prob. 11-6. In this question, ‘chromosomal mutation’ means ‘on the chromosome’ as vs. ‘on the plasmid.’

p. 93, Prob. 11-14. Questions A and B have been rephrased to read as follows:

A.    Consider the results of experiment #1 shown in table 1. In this experiment,

(i) What are you testing for? (induction) (lysogeny) (recombination) (complementation) (transduction) (conjugation) (transformation).

            (ii) What goes in table 1 where the question mark is? ______ Explain briefly.

 B. Consider the results of experiment #2, shown in table 2. In this experiment,

            (i) What are you testing for? (induction) (lysogeny) (recombination) (complementation) (transduction) (conjugation) (transformation).

            (ii) What goes in table 2 where the question mark is? ______. Explain briefly.

p. 94, Prob. 11-16. The sentence after the table should read: All four his^- mutations are in the same cistron.  Assume this gene (cistron) is not closely linked to the genes for synthesis of leu and/or lys.

Changes to Problem Set #12

p. 96, Prob. 12-5. Assume glucose is the only carbon source both times.

Changes to Problem Set #13

P. 106. There is a new problem 13-8 (see below), and some of the old problems have been renumbered. Here is a summary of the changes in problem set 13:

 Here is new problem 13-8:

13-8.  Consider the plasmid described in problem 13-4. Assume it is 3150 BP long, contains no repeated sequences, and the order of restriction sites is Eat1, EatII, EatIII, EatI, EatIV, EatIV. Assume the first 2 sites are 150 base pairs apart, and all other restriction sites are 600 BP apart.

            A. Suppose you digest the plasmid with one or more of the restriction enzymes listed.  (See the table below for the enzyme(s) used in each case, A-1 to A-4.) After digestion, you separate the resulting pieces on a gel, and stain the gels for DNA. (No probe – just a stain containing a dye that binds to DNA.)          For each of the cases, fill in the following table:

            (i). How many different pieces of DNA will you get? (Two pieces are different if they have different sequences.)

            (ii). How many bands will you see on the gel?

            (iii). How long is the DNA in the band that migrates the fastest on the gel?

B. Suppose you insert a fragment of human gene X into the plasmid in the EatII restriction site (whether this is the best way to do it or not). Assume the human DNA has one EatI restriction sites (& no EatII sites). You then cut up the chimeric plasmid with EatII, and separate the resulting pieces on a gel.  You have a labeled probe for gene X. Suppose the probe is 1 KB long and the human DNA is 2 KB long. You make a blot of your gel and hybridize to the probe.

B-1. How many labeled bands will you find on the blot?

B-2. The fastest migrating labeled band should correspond to a DNA fragment that is (<1 KB) (2 KB) (>1 but <2KB) (2KB) (>2KB). Explain briefly.

B-3. Suppose you repeat your experiment. You make your chimeric plasmid the same way, but this time you cut it up with EatI & EatII. Then you run a gel, make a blot and hybridize to the same probe as before.

            (i). Suppose you find one labeled band on your blot. This band will correspond to a DNA fragment that is (the same size as in B-2) (shorter) (longer) (can’t predict).

            (ii). Suppose you find two labeled bands on your blot. How will the intensity of the two bands compare? You expect that (one band will be darker than the other) (both bands will have the same intensity of label) (you can’t predict).

            For both (i) and (ii) explain how you could get the results given. What is different about the two cases?

            (iii) Is it possible to get a blot with more than 2 bands? Explain why or why not.

Changes to Problem Set #13

P. 117, prob. 13R-7. The first paragraph should read as follows:

13R-7.  Gene G codes for a protein. You have a probe to a region of the DNA that is part of gene G. (Assume the region in question is transcribed; it is not part of the promotor for gene G.) The region has no repeats, but it has one or more polymorphic restriction sites. Now suppose a deletion of one base pair occurs in this region 50 nucleotides upstream (5' on the sense strand) of the first polymorphic restriction site in the area of DNA under discussion. Assume this change does not affect hybridization to the probe or directly alter a restriction site. For each part, circle an answer and explain your reasoning.

            The questions are the same.
 

There are no significant changes to problem sets 14 & 15.

Changes to Answers

Answer to Prob. 1-9. Reference to Becker is p. 184 in the 6th ed. (This is helpful for part B, not for part A.)

Answer to 1-24, C.

C. Negative (if you link all NSG’s) or neutral (if you link alternating glucosamines and NSG’s.) At neutral pH, COOH loses a H+ and amino gains one. You would need 2 different enzymes to make the alternating polymer, because you need to form two different glycosidic bonds. Both are β 1 →6, but one is from the 1 position of glucosamine to the 6 position of NSG and the other is from the 1 position of NSG to the 6 position of glucosamine.  One enzyme would be needed for tight binding of, say, glucosamine on the left and NSG on the right.  Because enzyme-substrate binding is so specific, it is unlikely that the active site of that enzyme could accommodate NSG on the left and glucosamine on the right in its substrate binding site, as these two sugars are significantly different. Thus a second enzyme would be needed for the second type of glycosidic bond that is being formed. These two could suffice

Answer to Prob. 2-13, D. If you got +2, did you remember the ends?

Answer to (new) 2-0.

2-0.      A& B. According to the Brǿnsted-Lowry definition of acids and bases, an acid is a substance that donates a proton, and a base is a substance that accepts a proton. A COOH group ionizes, releasing a proton, so it is an acid.  An amino group ionizes, accepting a proton, so it is a base. A hydroxyl group does not ionize under normal circumstances. The O-H bond in a hydroxyl group is polar, but it does not break.

 

            C. In some covalent bonds, the sharing of electrons is uneven, so that the bonds are polar – there is an electron rich end (δ‑) and an electron poor end (δ+). Polar bonds (in biology) usually involve N or O atom attached to C or H atom. The N and O atoms are more electronegative and have a stronger pull on electrons, so they are at the electron rich (δ-) end; the C and H atoms are at the electron poor (δ+) end. Note that a polar covalent bond is neutral overall, even though the distribution of electrons is uneven. The H on the δ+ end of one polar covalent bond can form a weak attraction to an O or N on the δ- end of a second polar covalent bond. This weak attraction is what is known as a hydrogen bond.

 

            D. No. At neutral pH, both the carboxyl and amino ends are ionized. (The R group may be ionized or not.) The amino acid may be neutral, but that is not because it is un-ionized – if it is neutral, it is because the charges on the two ionized end groups cancel each other out. What about high and low pH values? Try adding protons at lower pH’s or removing protons at higher pH’s to see what happens to the end groups (carboxyl and amino).

            We are ignoring the R group, since its presence doesn’t change the result. The R group may be un-ionized at all pH values, or it may be ionized under some conditions and not others. However at least one group in the amino acid will be ionized at any particular pH.

 

Answer to 2-11, (new) part C.

        C. No. Any difference in the molecular weight caused by a change of one amino acid is too small to detect by standard SDS-PAGE. This is because proteins usually contain over 100 amino acids. A hemoglobin molecule contains 4 chains, each about 150 amino acids long. Two amino acids out of 600 would be different between HbA and HbX, since there are two identical α chains per molecule.

 

Answer to (new) 2-20.

2-20.    A. There are only 3 different amino acids present in these oligopeptides: ser, ala and leu.  These 3 should be resolvable by paper chromatography because they would be expected to have different solubilities in the organic phase solvent (leu > ala > ser).  The top spot would be leu, the middle spot would be alanine and the bottom spot would be serine.

            B. Faster for all spots.  At pH 2, the number of charged groups on all these amino acids will be reduced from 2 to 1 due to the protonation of the alpha carboxyl group to –COOH.  The reduction in charge will make the amino acids somewhat less hydrophilic and so will favor solubility in the mobile non-polar solvent compared to the stationary water phase.

            C.  Both oligopeptides could form a secondary structure like an alpha helix.  If they did, then since the 6 leucine residues of oligopeptide I (case A below) are spaced 3-4 amino acids apart from each other their hydrophobic side chains would all be sticking out along the same side, approximately above and below each other, since there are about 3-4 residues per turn of an alpha (3.6 to be exact). If the side of the helix with these 6 long leucine side chains faced the exterior of the subunit, then it could interact with the same face presented by an identical subunit via hydrophobic forces, thus contributing to the association of the 2 subunits in forming the homodimer.  Oligopeptide II (case B below) does not have this possibility, either for its leucines or for its serines (which can interacting via hydrogen bonding), since the potentially interacting side groups would be pointing in various directions out from the alpha helix.  Serines are also distributed on all sides in I.

There is a picture in the new edition, but it doesn't print properly in this document. See an 18th edition if you are interested.

Answer to New 3-5.

3-5.               A.  SDH-M. At 2 mM fumaric acid, the V_o for SDH-M is greater than that of SDH-C and it is still rising almost linearly. (See graph below.) At 2 mM the V_o for SDH-C is lower than that of SDH-M and it is starting to reach a plateau. Thus SDH-M must have a larger V_max under these conditions (with the enzyme amount unknown).

              B. (can't tell) The turnover number is the maximum rate at which the enzyme can covert substrate to product. It is expressed per n mole of enzyme or per molecule of enzyme. Although we could guess at the V_max, we have no idea how much enzyme (E_o) is being used here, so we cannot divide V_max by E_o to get k3, the turnover number.

              C. (SDH-C). The V_o vs. S curve (the dependence of initial velocity on substrate concentration) for SDH-C is starting to reach a plateau at 2 mM, so its K_m (the concentration of substrate that results in 1/2 the V_max) should be < 2 mM. In the case of SD H-M, on the other hand, V_o is still rising nearly linearly at 2 mM substrate, so 2 mM should be below its K_m, so its K_m must be > 2 mM. Thus the K_m for SDH-C is less than the K_m exhibited by SDH-M. Since the K_m often is a good estimate of the dissociation of the enzyme substrate complex, a lower K_m for SDH-C indicates that this enzyme binds its substrate fumaric acid with less dissociation, i.e., with greater affinity.

              D. (6.2) Since the K_m's are the same, then the proportion of V_max achieved at 2 mM substrate is the same for both substrates. Since the same amount of enzyme is used in both cases, then the V_max (=k3E_o) for propenoic acid will be 1/10 the V_max for fumaric acids, since the k3 values differ by this amount. Thus propenoic acid will achieve 1/10 the V_o achieved by fumaric acid at 2 mM,or ).1 x 6.2 = 0.62 nanomoles/min. After 10 minutes, 10 x 0.62 = 6.2 nanomoles will be accumulated. The same result can be derived algebraically using the Michaelis-Menton equation and setting up the proportion: V_o(propenoic)/V_o(fumaric). The K_m terms cancel out, 10 X V_max (fumaric) can be substituted for V_max (propenoic), leading to the canceling out of V_max, and 0.62 can be substituted for V_o(fumaric) under these conditions.

Answer to 3-8. Part A. The answer reads: "This ratio (k2/k1) = dissociation constant for ES and is an indication of the affinity of E for S and the position of equilibrium for ES <--> E + P." It should be "ES <--> E + S." This mistake is repeated two more times in the body of the long paragraph. The correct answer should read:

3-8.       A.  ii, iv. K_m = (k₂ + k₃)/k₁. If k₃ is relatively small, as it usually is, this reduces to k₂/k₁. This ratio = disassociation constant for ES and is an indication of the affinity of E for S and the position of equilibrium for ES ↔ E + S.

Note that K_m is a measure of the position of equilibrium for ES ↔E + S but NOT the position of equilibrium for S ↔ P. For S ↔ P, the enzyme is a catalyst. Presence or absence of enzyme (or state of enzyme) has no effect on the equilibrium of S ↔ P. (See answer to B.) For ES ↔ E + S, the enzyme is a reactant and/or product. When you change the structure of E, you change the reactants/products so you actually have a different reaction with a different equilibrium constant and therefore a different k₂/k₁. To put it another way, if you change the structure of E, you can change the free energy involved in ES ↔ E + S (& thus the position of equilibrium). But changing E has no effect on the free energy change involved in S ↔ P.

Answer to 3-10, part C. Last line should say "committed step is before C → A," not "before A → C."

Answer to 3R-5. For a picture of porin, see http://bio.winona.edu/berg/ILLUST/porin2.GIF

 

Answer to  4-13, (new) part C.

            C. In A, you start with individual molecules of glucose. You need to ‘spend’ 1 molecules of ATP to phosphorylate each glucose molecule before fermentation can begin. In B, you start with glycogen. You get enough free energy from breaking the glycosidic bonds in the polymer to drive phosphorylation of the glucose using phosphate instead of ATP. So you don’t have to ‘spend’ any ATP to get to glucose-6-phosphate. To put it another way, lots of ATP or the equivalent was used to make the polymer, and you are getting back some of that free energy when you hydrolyze the polymer.

Answer to 5-4, part (iii). Cell can live on ATP from Krebs if you don't need ATP for the transport of reduced NAD into mitochondria. (Should work in prokaryotes since there are no mitochondria; may or may not work in yeast, since they are eukaryotes.)

Answer to 5-14, part B. Reference to Becker is to 5th edition. There is no corresponding fig. in the 6th ed.

Answer to 8-0, (new) part D.

            D. The cell is probably diploid because there are 2 of each kind of chromosome. If it were haploid, all 4 chromosomes could be visibly different.

Answer to 8-2 at * -- Here is a longer, clearer version of the note.

*             Depending on how you count centromeres, the number is either the same as in the left & middle column or it is the same as in the right column (in which case all the numbers with a * are doubled).
              How to count centromeres is arbitrary, and different people do it differently. Here we are defining the centromere as a structure in a particular region of the chromosome, whether the chromosome is single or double. For a doubled chromosome, the centromere equals the constriction that forms where the two sister chromatids are attached.  In this way of reckoning, there is always one centromere per chromosome whether the centromeric DNA is replicated or not and whether the chromosome has 2 chromatids or one. (Therefore the middle column matches the left column.)
              Alternatively, you can think of a centromere as a specific DNA sequence (the centromeric DNA) on each chromatid. (The constriction/attachment point always forms at a specific sequence in the DNA.) In that case, there is one centromere per chromatid, and two per replicated chromosome. Special proteins stick to the centromeric DNA sequences and either glue the two sister chromatids together or form structures called kinetochores that attach to the spindle fibers. (Different sets of proteins are used for the two jobs. The state of the protein structures depends on the phase of the cell cycle. See a cell biology text for details if you are interested.) If you think of centromeres this way, the number of centromeres matches the # of chromatids (right column), not the number of chromosomes (left column); all the values with a * are doubled.  

Answer to Prob. 9-7, Part B-4. It has a '6' instead of an arrow at 3 places. First sentence should read: A allele produces active enzyme; a allele produces inactive enzyme; A' allele produces partially active enzyme.

Answer to (New) 13-8.

13-8.    A.

Case:	A-1	A-2	A-3	A-4
Enzyme(s) Used:	EatI	EatII	EatIII + EatIV	All 4 enzymes
(i) How many pieces?	2	1	3	6
(ii) How many bands?	2	1	3	2
(iii) How long is shortest?	1350 KB	3150 KB	600 KB	150 KB
# times cut?	2	1	3	6
All the pieces:	1350 & 1800 KB	Only 1; whole length	600, 1200, & 1350 KB	150 & 5 dif. Pieces all 600 KB long

 

            B.-1. One. You’ll cut out the inserted gene fragment, and it’s the only piece that hybridizes to the probe.

            B-2. 2 KB. The fragments are separated by size first, and then probe is added. It’s the size of the fragment that matters, not the size of the probe.

            B-3.      (i). Smaller.

                        (ii). Both bands will be equal in intensity.

In both cases, the fragment of gene X was cut out of the plasmid by EatII and cut into two ‘halves’ by EatI, each ‘half’ smaller than the original. In case (ii), one ‘half’ must be smaller than the other, but in case (i) they could be the same size.

            In case (i), one half was at least 1 KB long, and contained the entire sequence complementary to the probe. The molecules in the gel containing that sequence hybridized to the probe and formed a labeled band. The molecules containing the other ‘half’ wouldn’t hybridize. (It doesn’t matter if they were smaller, in a different band, or the same size, in the same band.) In this case, the sequence that hybridized to the probe was entirely on one side of the EatI site

            In (ii), the original fragment was cut into two unequal sized ‘halves’, and each part contained some of the sequence complementary to the probe. In this case, the probe overlapped the spot where the EatI restriction site cut. Therefore the probe overlapped with sequences in each ‘half’ and could hybridize to either one. Some molecules of probe hybridized to one ‘half’ (forming one labeled band) and some to the other ‘half’ (forming a second labeled band). Each molecule in the gel could trap one molecule of probe, so the two bands should be equally labeled. 

                        (iii). No. There is only one site for EatI in the insert, so the maximum # of pieces that can be generated is 2. One or both of them may hybridize to the probe, as explained above.

Answer to Prob. 14-2. Assume p = frequency of the Rh^- allele; q = frequency of the Rh⁺ allele.