Chapter 7 Practice Problems

Question 1

an A-T base pair wild-type gene is changed to a G-C pair

Answer 1

transition
base substitution
deamination

Question 2

an A-T base pair changed to a T-A base pair

Answer 2

base substitution

transversion

Question 3

the sequence AAGCTTATCG is changed to AAGCTATCG

Answer 3

deletion
X-irradiation
intercalator

Question 4

the sequence CAGCAGCAGCAGCAGCAG is changed to CAGCAGCAGCAGCAGCAGCAGCAGCAG

Answer 4

insertion

slipped mispairing

Question 5

the sequence AACGTTATCG is changed to AATGTTATCG

Answer 5

transition
base substitution
deamination

Question 6

the sequence AACGTCACACACACATCG is changed to AACGTCACATG

Answer 6

deletion
x-ray irradiation
slipped mispairing

Question 7

the sequence AAGCTTATCG is changed to AAGCTTTATCG

Answer 7

insertion

intercalator

Question 8

the DNA sequence of one strand of a gene from three independently isolated mutants is given here (5’ ends are at left). Using this information, what is the sequence of the wild-type gene in this region?
mutant 1: ACCGTAATCGACTGGTAAACTTTGCGCG
mutant 2: ACCGTAGTCGACCGGTAAACTTTGCGCG
mutant 3: ACCGTAGTCGACTGGTTAACTTTGCGCG

Answer 8

5’ ACCGTAGTCGACTGGTAAACTTTGCGCG 3’

Question 9

among mammals, measurements of the rate of generation of autosomal recessive mutations have been made almost exclusively in mice, while many. measurements of the rate of generation of dominant mutations have been made both in mice and in humans. What do you think is the reason for this difference?

Answer 9

dominant mutations can be detected immediately in the heterozygous progeny who receive the mutant gamete. Recessive mutations can be detected only when they are homozygous. To detect the appearance of new recessive alleles, you must test cross with a recessive homozygote. Testcrosses can be done in mice, where the researcher can control the mating, but it cannot be done in humans

Question 10

Over a period of several years, a large hospital kept track of the number of births of babies displaying the trait achondroplasia. Achondroplasia is a very rare autosomal dominant condition resulting in dwarfism with abnormal body properties. After 120,000 births, it was noted that 27 babies had been born with achondroplasia. One physician was interested in determining how many of these dwarf babies resulted from new mutations and whether the apparent mutation rate in this geographical area was higher than normal. He looked up the families of the 27 dwarf babies and discovered four of the dwarf babies had a dwarf parent. What is the apparent mutation rate of the achondroplasia gene in this population? is it unusually high or low?

Answer 10

the mutation rate = 23 mutant gametes/240,000 gametes = 9.5 x 10^-5. This rate is somewhat higher than 2 to 12x10^-6 mutations per gene per generation which is the average mutation rate for humans

Question 11

Suppose you wanted to study genes controlling the structure of bacterial cell surfaces. You decide to start by isolating bacterial mutants resistant to infection by bacteriophage that binds to the cell surface. The selection procedure is simple: spread cells from a culture of sensitive bacteria on a petri plate, expose them to a high concentration of phages, and pick the bacterial colonies that grow. To set up the selection you could 1) spread cells from a singe liquid culture of sensitive bacteria on many different plats and pick every resistant colony; or 2) start many different cultures, each grown from a single colony of sensitive bacteria, spread one plate from each culture, and then pick a single mutant from each plate. which method would ensure that you are isolating many independent mutants?

Answer 11

2

Question 12

So-called two-way mutagens can induce both a particular mutation and a reversion of the mutation that restores the original DNA sequence. In contrast. one-way mutagens can induce mutations but not exact reversions of these mutations. Which of the following mutagens can be classified as one-way and which as two-way?

a. 5-bromouracil
b. hydroxylamine
c. ethyl methane sulfonate
d. nitrous acid
e. proflavin

Answer 12

a. two-way
b. one-way
c. one-way
d. two-way
e. two-way

Question 13

when a particular mutagen identified by the Ames test in injected into mice, it causes the appearance of many tumors, showing that this substance is carcinogenic. When cells from these tumors are injected into other mice not exposed to the mutagen, almost all of the new mice develop tumors. However, when mice carrying mutagen-induced tumors are mated to unexposed mice, virtually all of the progeny are tumor free. Why can the tumor be transferred horizontally but not vertically?

Answer 13

the mutagen induced tumor-causing mutations in somatic cells, not in gamete producing cells in the germline

Question 14

When the His- Salmonella strain used in the Ames test is exposed to substance X, no His+ revenants are seen. If, however, rat liver supernatant is added to the cells along with substance X, revenants do occur. Is substance X a potential carcinogen for human cells? explain.

Answer 14

yes the rat liver supernatant contains enzymes that convert substance X to a mutagen, and His+ revertants occur. Our livers contain similar enzymes

Question 15

The Ames test uses the reversion rate (His- to His+) to test compounds for mutagenicity.
a. is it possible that a known mutagen, like proflavin, would be unable to revert a particular His- mutant used in the Ames test? How do you think that the Ames test is designed to deal with this issue?
b. can you think of a way to use forward mutation (His+ to His-) to test a compound for mutageniity?
C. given the rate of forward mutation is so much higher than the rate of reversion, why does the Ames test use the reversion rate to test for mutagenicity?

Answer 15

a. Yes; specific mutagens can revert only particular types of mutations. Proflavin, for example, can revert only single-base insertions or deletions; it cannot revert nucleotide substitutions caused by mutagens such as base analogs. The Ames test deals with this issue by testing potentially mutagenic compounds for their ability to revert His− strains that have different types of mutations at the molecular level.
b. A wild-type (His+) strain could be grown in the presence of a potentially mutagenic compound (plus rat liver enzymes) and then plated for single colonies on minimal medium + histidine. Replica-plating on minimal medium (without a histidine supplement) would identify colonies that are His−; they would fail to grow
c. Identification of forward mutations as described in part (b) is through a screen, not a selection. Even though the forward mutation rate is higher than the reversion rate, this screening process is much more labor-intensive than the selection that is employed when testing for revertants.

Question 16

a. Seymour Benzer’s fine structure analysis of the rII region of bacteriophage T4 depended in large part on deletion analysis. But to perform such deletion analysis, Benzer had to know which rII- bacteriophage strains were deletions and which were point mutations. How do you think he was able to distinguish rII- deletions from point mutations?
b. A key feature of Benzer’s fine structure map of point mutation is the existence of hotspots, which Benzer interpreted as nucleotide pairs that were particularly susceptible to mutation. How could Benzer say that all of the independent mutations in a hotspot were due to mutations of the same nucleotide pair?

Answer 16

a. First, deletions are mutations that never revert to wild type (because so much information is lost that it can never be restored). Second, phage strains with rII− deletion mutations fail to recombine to produce wild-type progeny in separate co-infections (recombinations) with rII− point mutations that affect different base pairs in the rII region. This second method is based on the idea that if a point mutation is included in (deleted by) the deletion, it is impossible to recover rII+ phage by recombination because neither rII− mutant gene has the wild-type base pair to contribute to the recombinant.
b. rII− mutants in the same base pair cannot recombine with each other to produce rII+ phage.

Question 17

a. you have a test tube containing 5 mL of a solution of bacteriophages and you would like to estimate the number of bacteriophages in the tube. Assuming the tube actually contains a total of 15 billion bacteriophages, design a serial dilution experiment that would allow you to estimate this number. Ideally the final plague-containing plates you count should contain more than 10 ad fewer than 1000 phages
b. when you count bacteriophages by the serial dilution method as in part a you are assuming a plating efficiency of 100% that is the number of plaques on the petri plate represents exactly the number of bacteriophages you mixed with the plating bacteria. Is there any way to test the possibility that only a certain percentage of bacteriophage particles can form plaques (so that the plating efficiency would be less than 100%). why is it fair to assume that na plaques are initiated by one rather than multiple bacteriophage particles?

Answer 17

a. The starting tube (call it tube A) contains 5 ml of bacteriophage with 1.5 × 1010 phages. Take a 1 µl (0.001 ml) sample of tube A, corresponding to 3 × 106 phages, and add it to 999 µl (in practice, 1 ml) of diluent in tube B. This step is a 2 × 10-4 dilution (1 µl / 5000 µl). Repeat this step with 1 µl of tube B (3 × 103 phages) and mix it with 999 µl of diluent in tube C (2 × 10-7 dilution). Next take 10 µl of tube C (30 phages) and mix it with bacteria [about 100 × more cells than phage = a low multiplicity of infection (MOI)]. Allow the phages to infect the cells, then add cells to a top agar and pour on an agar plate. Repeat the infection/top agar step with 100 µl of tube C (300 phages) and plate. There should be 30 plaques on the first plate and about 300 plaques on the second plate.
b. To determine the total number of phage, you need to look at a specific dilution in the electron microscope and count all the phage particles. The ratio of plaques made from the dilution to total phage in the dilution is the plating efficiency. In part (a), it is fair to assume that only one phage initiated each plaque because of the very low MOI. That is, because many more bacterial cells than phage existed, the chances are very high that any individual bacterial cell would have been infected only by a single phage.

Question 18

The rosy (r) gene of Drosophila encodes an enzyme called xanithe dehydrogenase. Flies homozygous for ry mutations exhibit a rosy eye color. Heterozygous females were made that had ry^41 Sb on one homolog and Ly ry^564 on the other homolog, where ry^41 ad ry^564 are two independently isolated alleles of ry. Ly (Lyra (narrow) wings) and Sb (stubble bristles) are dominant mutant alleles of genes to the left and right of ry, respectively. These females are now mated to males homozygous for ry^41. Out of 100,000 progeny, 8 have wild type eyes, Lyra wings, and stubble bristles, while the remainder of rosy eyes

a. what is the order of these two ry mutations relative to the flanking genes Ly and Sb?
b. What is the genetic distance separating ry^41 and ry^564?

Answer 18

a. When this diploid is sporulated, the PD asci are: 4 ARG-E− ARG-H+ (Arg−) : 4 ARG-E+ ARG-H− (Arg−), and the NPD asci are: 4 ARG-E+ ARG-H+ (Arg+) : 4 ARG-E– ARG-H− (Arg−). The frequency of PD = frequency of NPD. ARG-E− ARG-H+ × ARG-E+ ARG-H− → ARG-E− ARG-H+ / ARG-E+ ARG-H−
In this case, the PD ascus is the same as above: 4 ARG-E− ARG-H+ (Arg−) : 4 ARG-E+ ARG-H− (Arg−). The NPD asci are the same as above also: 4 ARG-E+ ARG-H+ (Arg+) : 4 ARG-E−ARG-H−(Arg−). PD and NPD ascus types can show either MI or MII segregation for both genes, whether they are linked or not.
b. The ARG-E− ARG-H+ spores will grow when you supplement the media with ornithine, citrulline, argininosuccinate or arginine. For the ARG-E+ ARG-H– or ARG-E− ARG-H− spores, only arginine itself in the media allows growth. The ARG-E+ ARG-H+ spores are prototrophs that grow on minimal medium without supplementation.

Question 19

In corn snakes, the wild-type color is brown. One autosomal recessive mutation causes the snake to be orange, and another causes the snake to be black. An orange snake was crossed to a black one and the F1 offspring were all brown. Assume that all relevant genes are unlinked.

a. indicate what phenotype and ratios you would expect in the F2 generation of this cross if there is one pigment pathway, with orange and black being different intermediates on the way to brown
b. indicate what phenotypes and ratios you would expect in the F2 generation if orange pigment is a product of another pathway and brown is the effect of mixing the two pigments in the skin of the snake

Answer 19

a. The phenotypic ratio depends on the order of orange and black in the pathway to brown, which is not specified. If the pathway is orange  black  brown, the F2 would have 9 brown (B– O–): 3 black (bb O–): 4 orange (B– oo or bb oo).the F2 would have a 9 brown (B– O–) : 3 orange (bb O–) : 4 black (B– oo or bb oo) ratio
b. 9 brown (B– O–) : 3 black (B– oo) : 3 orange (bb O–) : 1 nonpigmented (bb oo).

Question 20

In a certain species of flowering plants with a diploid genome, four enzymes are involved in the generation of flower color. The genes encoding these four enzymes are on different chromosomes. The biochemical pathway involved is as follows; the figure shows that either of two different enzymes is sufficient to convert a blue pigment into a purple pigment.
white -> green -> blue -> purple
a true breeding green-flowered plant is mated with a true-breeding blue flowered plant. All of the plants in the resultant F1 generation have purple flowers. F1 plants are allowed to self fertilize, yielding an F2 generation. Show genotypes for P, F1, and F2 plants, and indicate which genes specify genes specify which biochemical steps. Determine the fraction of F2 plants with the following phenotypes: white flowers, green flowers, blue flowers, and purple flowers. Assume the green flowered parent is mutant in only a single step of the pathway

Answer 20

The ratio is 0 white: 45 purple : 16 green : 3 blue.