Category: Answers

Correct answers in Bold.

#1. (5 points) When population size is large and mutation rates are low (~10^-8), as an evolutionary force mutation is generally characterized as:

A. Strong

B. Weak

C. Balancing

D. Beneficial

 

#2. (5 points) You sequenced a RNA polymerase gene from Caenorhabditis elegans, and a RNA polymerase gene from its sister species Caenorhabditis briggsae. You performed a Ka/Ks analysis on these genes and the calculated value was 0.005. What kind of selection is mostly likely influencing the evolution of these genes?

A. Negative (or, purifying) selection

B. Positive (or, directional) selection

C. Balancing selection

D. Neutrality (no selection)

 

#3. You are studying the population genetics of Mendel’s pea plants. You remember that they are diploid, and that the G allele is fully dominant and results in yellow peas; the g allele is recessive and results in green peas (when homozygous, of course). Upon visiting a field, you discover 200 pea plants. 160 of the plants produce yellow peas, and 40 of the plants produce green peas. Assume this population is in Hardy-Weinberg equilibrium.

 

Part A. (3 points) What is the genotype frequency for gg plants?

40/200 = 0.20

 

Part B. (3 points) What is the allele frequency for g?

q = 0.447

 

Part C. (4 points) What is the expected number of heterozygous plants in this population?

98.9 (or round to 99) plants

correct answers in bold

#1 (5 points) How can the abundance and function of a gene’s products (RNA, protein) be regulated inside the cell?

A. At the level of transcription, synthesis of a RNA molecule from coding DNA

B. At the level of translation, synthesis of a protein product from a template mRNA

C. RNA stability and structure, post-transcriptional regulation of RNA

D. All of the above

#2 (5 points) You have a lacI^S (‘super repressor’) mutant of coli. The LacI protein of this mutant cannot bind lactose, but can bind the operator. The lac operon structural genes (lacZ, lacY, lacA) will be expressed:

A. Only when there is lactose in the growth media.

B. Only when there is not lactose in the growth media

C. Always, regardless of whether or not there is lactose in the growth media

D. Never, regardless of whether or not there is lactose in the growth media

 

#3 (5 points) You are a population geneticist studying a diploid species of salamander. The gene B influences salamander coloration. Two alleles of this gene exist, allele B (dominant) results in yellow salamanders, allele b (recessive) results in brown salamanders. The frequency of allele B is 0.07. What is the frequency of the b allele?

A. 0.93

B. 0.14

C. 0.07

D. 0.01

 

#4. (5 points) In which of the following populations is genetic drift expected to be strongest, in terms of its effects on change in allele frequencies over time?

 

A. 10,000 rabbits

B. 1,000 rabbits

C. 100 rabbits

D. 10 rabbits

Chapter 9:

9.3: (a) One mutation that might cause the constitutive phenotype is a mutation in an operator region of the enzyme-coding gene that makes it insensitive to repression. The second is a mutation that impairs the structure of the repressor. Such mutations can range from deletion of the repressor gene to subtler mutations that impair binding of the repressor to arginine, to the DNA, or to both. (b) A mutation in an arginine biosynthetic enzyme, not sufficient to cause a requirement for arginine but enough to reduce the amount of arginine in the cell, could activate a regulatory response by a normal regulatory system and induce constitutive synthesis.

9.6: The mutant gene should bind more of the activator protein, or have more efficient binding of the activator. Thus, the mutant gene should be induced with lower levels of activator protein or expressed at higher levels, compared to wild-type.

9.10: (a) The phenotype of cells with mutant Gal4p would be non-inducible; the mutant gene would be recessive because the wild-type Gal4p would still function normally. (b) The phenotype of cells with mutant Gal80p would be constitutive; the mutant gene would be recessive because wildtype Gal80p would bind Gal4p in the normal way.

9.18: (a): Yes, the repressor is functional, and the presence of lactose activates transcription of the lac genes; (b) and (d): Yes, at 42C, the repressor cannot bind the operator, which means that the lac operon is transcribed whether or not the inducer is present. (c): No, at this temperature, the repressor functions normally – because lactose is absent, the lac operon is in a repressed state.

Chapter 14:

14.1. The frequency of A₁ equals 0.35.

14.2. The expected frequency of A₂ in the next generation is the same as its frequency in the current generation: 0.65.

14.4. (A) no; (B) yes.

14.8. The frequency of homozygous recessives is q² = 0.16; this implies that q=0.4. The frequency of the dominant allele is therefore p=0.6.

14.9. Two issues need to be considered. First, recessive alleles are maintained in heterozygous individuals and so are not exposed to selection. Second, new mutations in each generation replenish the number eliminated by selection in the homozygous recessives.

14.16. The numbers of alleles are as follows: A, 8+10+2 = 20; B, 10+48+20=78; C, 20+20+2 = 42. The total number of alleles is 140, so the allele frequencies are as follows: A, 20/140 = 0.14; B, 78/140 = 0.56; C, 42/140 = 0.30. The expected numbers among the 70 plants are:

AA: 1.42, AB: 11.14, BB: 21.72, BC: 23.40, CC: 6.30, AC: 6.00.

#7:

D, dosage compensation

#13:

D, 0.0001

#20:

Cyclins tether specific target proteins and bring them to the complex; CDKs phosophorylate those target proteins to change their structure/function.

#21:

(0.10 x 0.005) x 10,000 = 5; coeff. coinc. = 2/5 = 0.4

i = [1-(0.4)] = 0.6

12.1: (b) Formation of crosslinked thymine dimers.

12.2: Such people are somatic mosaics, a condition that can be explained by somatic mutations in the pigmented cells of the iris of the eye or in their precursor cells.

12.5: There are 18 depurinations per minute and 1440 minutes per day, so the total number of depurinations per cell per day is 18 x 1440 = 25,920.

 

13.2: It is a “genetic disease” in that genetic mutations take place in cells and make them undergo uncontrolled growth and division. However, most cancers are caused by somatic mutations in the cells of affected individuals; hence, they are not inherited from the parents (familial) but rather are due to new somatic mutations.

13.11: A defect in the G1/S checkpoint allows cells with damaged chromosomes to enter DNA synthesis, when replication and repair can cause chromosome rearrangements or replication errors leading to aneuploidy.

13.19. Effectively, HPV removes all the brakes on cellular proliferation, turns on all the stimuli, and removes the infected cell’s ability to induce apoptosis to kill itself. The cell will effectively be p53−, because E6 is present to ensure that levels of p53 do not rise. This means that synthesis of Bax will not be induced, and cells will not undergo apoptosis. In addition, synthesis of p21, GADD45, and 14-3-3s will not be accelerated, so cells will not arrest in the cell cycle. Because E7 disables RB, E2F will be active at all times, resulting in new rounds of DNA synthesis and continued cellular proliferation.

13.20. For generations I and II, the band at position c is associated with the disease. In generation III, individuals 2 and 7 are at risk because of parent 5, and the individuals 21 and 22 are at risk because of parent 20. However, individual 13 is not at risk because the parent 14 with a band at position c is not affected; band c in this individual evidently originates from a nonmutant allele.

Extra Question #1:

Endogenous mutagens originate from sources inside the cell or organism; exogenous mutagens come from outside (or, environmental) sources.

Endogenous mutagen sources include nucleotide tautomerism (enol/keto), replication errors, and metabolic byproducts (such as 8-oxoguanine).

Exogenous mutagen examples include UV light, chemical base analogs (such as 5-bromouracil), and diesel fumes.

Correct answers in bold

1. (5 points) The figure to the right shows a chromosome during mitotic metaphase. Which term best describes the structure of this chromosome? (Note: see bottom left panel of textbook Figure 5.5 for image)

a) metacentric

b) acrocentric

c) chiasma

d) compensatory

 

2. (5 points) The Y chromosome in humans is thought to have evolved from a homologous partner to which other chromosome?

a) Chromosome 21

b) Chromosome 22

c) The Z Chromosome

d) The X Chromosome

 

3. You are studying the rb gene (rb indicates recessive mutant allele, + indicates wild-type) and cv gene (cv indicates recessive mutant allele, + indicates wild-type) in Drosophila. Both genes are on the X chromosome. You set up a testcross between a doubly heterozygous female and a male tester fly. You infer the following genotypes from the phenotypes observed in the male progeny of this cross:

 

Genotype                                Number Observed

+   + / Y                                              90

+ cv / Y                                              11

rb + / Y                                             9

rb cv / Y                                             90

 

Part A (5 points): Does this data suggest that the rb and cv genes are linked?

a) Yes

b) No

c) Not Enough Information Provided

 

Part B (5 points): What is the genetic distance (in cM) between rb and cv?

(9+11)/200 = 20/200 = 0.10 recombination freq. (x100) = 10 cM

4.1: 0.2.

4.2: Females produce gametes of types A B, a b, A b, and a B in the proportions (1-0.08)/2 = 0.46 and (1-0.08)/2 = 0.46 for each of the nonrecombinant types. It is 0.08/2 = 0.04 and 0.08/2 = 0.04 for each of the recombinant types. Since there is no crossing over in male Drosophila, males produce only two gamete types (A B, a b) in equal (0.50, 0.50) proportions.

4.7: The map distance between s and c is 18 cM, but there is only 16% recombination, and the map distance between p and c is 13 cM, but there is only 12% recombination. The fact that the frequency of recombination is smaller than the map distance results from double crossovers.

4.8: The data include 108 progeny with either both mutations or neither, and 92 progeny with one mutation or the other. The first of these groups consists of parental chromosomes; the second consists of recombinant chromosomes.   The chi-square test compares the observed ratio (108:92) to the expected (100:100) under the hypothesis of no linkage. The chi-square value equals 1.28 and there is one degree of freedom. P is approximately 0.26, so there is no evidence of linkage even though both loci are on the X.

4.10: (a) the mutant genes are not alleles because they do not show segregation; if they were alleles, the progeny would show resistance to one insecticide or the other. (b) the genes are not linked. The chi-square of 177 parental:205 recombinants compared to 191:191 (expected) equals 2.05. With one degree of freedom, P is 0.15 (not significant). (c) The two genes must be far apart on the chromosome – too far to be ‘linked’.

4.11:

(a) The gene in the middle has the alleles D and d.

(b) The triply heterozygous parent has genotype A d B / a D b or B d A / b D a.

(c) The closest genes are B and D.

(d) B−D distance equals (40 + 60 + 10)/1000 = 0.11 = 11 map units = 11 centimorgans (cM).

4.12: (a) parental types = v+ pr bm and v pr+ bm+. Double recombinants = v+ pr+ bm+ and v pr bm. This means that gene v is in the middle. The pr-v recombination frequency is 20.7% and the v-bm frequency is 41.5%. The expected number of double crossovers = 85.9, so the coincidence is 0.90. The interference is thus 0.10. (b) The true map distances are larger than 20.7cM and 41.5 cM because the frequency of recombination between these genes is sufficiently large that there must have been some double crossover events that went undetected in the analysis.

4.20.

(a)       To solve this type of problem, you should first deduce the genotype of the gamete contributed by the triply heterozygous parent to each class of progeny. In this case the gamete types are, from top to bottom, + + +, + + unc, + dor unc, dpy + +, dpy dor +, dpy dor unc. To determine the order of the genes, you must identify which gene is in the middle. You can do this by comparing the parental type gametes with the double crossovers. The two parental types are dpy dor + and + + unc, because these are the most frequent. In these data there are no observed double crossovers (which means that interference across the region is complete). The missing classes of progeny correspond to the double crossovers. The missing phenotypes of progeny are normal body, deep orange eye, coordinated, corresponding to the genotype + dor +, and dumpy body, red eye, uncoordinated, corresponding to the genotype dpy + unc. Comparing the parental types with the inferred double crossover types reveals that dpy+ and dpy are interchanged relative to the alleles of the other two genes. Hence, the genotype of the F1 heterozygous female is dor dpy unc+ / dor+ dpy+ unc or unc+ dpy dor/ unc dpy+ dor+.

(b)       Analysis of the single recombinants indicates (96 + 110)/1000 = 20.6% recombination between dor and dpy, and (75 + 65)/1000 = 14.0% recombination between dpy and unc.

(c)        Because there are no double recombinants observed, the interference is complete, which means interference = 1.

 

Chapter 5

5.1, 5.2, 5.12, 5.13

5.1: The term “inactive” is not quite correct because about 15% of the genes in this chromosome are transcriptionally active to some degree, producing 15−50% of the RNA transcripts as are produced from the active X chromosome.

5.2: The genetic consequence of the obligatory crossover is that alleles in the pseudoautosomal region in the X chromosome can be interchanged with their homologous alleles in the pseudoautosomal region in the Y chromosome, yielding a pattern of inheritance in pedigrees that is indistinguishable from that of ordinary autosomal inheritance.

5.12: The mother has a translocation that joins the long arm of chromosome 21 with the long arm of another acrocentric chromosome (a ‘Robertsonian’ translocation). The affected child has 46 chromosomes. This karyotype differs from the usual karyotype for trisomy 21 because the extra chromosome 21 is not a free chromosome.

5.13: Because the X chromosome in the 45, X daughter contains the color-blindness allele, the 45, X daughter must have received the X chromosome from her father by way of a normal X-bearing sperm. The nondisjunction must, therefore, have occurred in the mother, resulting in an egg cell lacking an X chromosome.

Additional Question not in your textbook:

#1. In Drosophila, waxy wings are inherited as a sex-linked recessive trait, and hairy body is inherited as an autosomal dominant trait. Indicate the F₁ and F₂ phenotypic ratios expected from a cross between a waxy wing male and a hairy body female.

Approach this problem by first denoting the genotypes at both genes of the parents:

P          waxy male                 X          hairy female

wY h⁺h⁺                                                    w⁺w⁺ HH             

 Then, proceed to the F₁ generation. Remember that males get their X chromosome from their mother and their Y chromosome from the father. Females get an X chromosome from each of the mother and father.

F₁         males are:                 w⁺Y Hh⁺

females are:                         w⁺w Hh⁺

 Next, indicate the F₂ generalized genotypes that will result in each phenotype. Calculate the probability of obtaining the genotype at each gene pair and multiply the individual probabilities. This give the expected ratio of the corresponding phenotype as indicated below:

F₂         w⁺_ H_                        probability = ½ X ¾ = 3/8 hairy females

w⁺_ h⁺h⁺         probability = ½ X ¼ = 1/8 wildtype females

w⁺Y H_                        probability = ¼ X ¾ = 3/16 hairy males

w⁺Y h⁺h⁺         probability = ¼ X ¼ = 1/16 wildtype males

wY H_             probability = ¼ X ¾ = 3/16 waxy, hairy males

wY h⁺h⁺           probability = ¼ X ¼ = 1/16 waxy males

 

 

#1: C

#6: A

#16: A) black scales, bubble eyes; B) 1/4 x 1/4 = 1/16; C) (1/2 x 1/2) x 100 = 25 fish

#20: A) two (from genes F and G); B) 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1 = 1/64

 

Correct Answers are in bold; note that there are two possible correct answers for #4.

#1. (5 points) An allele is:

a) the homozygous genotype

b) the heterozygous genotype

c) one of many possible forms of a gene

d) another word for a gene

#2. (5 points) You perform a testcross involving two genes, D and E, each with two alleles showing simple dominant/recessive relationships. D is dominant to d, and E is dominant to e. Which of the following crosses accurately describes a testcross involving these genes?

a) An individual with both dominant phenotypes, crossed with a DD, EE

b) An individual with one dominant and one recessive phenotype, crossed with a DD, EE

c) An individual with both dominant phenotypes, crossed with a dd, ee

d) An individual with one dominant and one recessive phenotype, crossed with a dd, ee individual

#3. (5 points) You are studying Mendel’s peas and have set up a cross between a true-breeding plant with round peas and a true-breeding plant with wrinkled peas. The round phenotype is dominant to the wrinkled phenotype. You then perform a F₁ x F₁ cross and analyze 400 F₂ How many F₂ plants do you expect to have the round pea phenotype?

a) 400

b) 300

c) 100

d) 0

#4. (5 points) Two genetic mutants of Neurospora crassa are defective in the Alanine biosynthesis pathway.   After complementation testing, the two mutants were placed into different complementation groups. Which of the following statements most accurately describes the two mutants:

a) they must have mutations in different specific genes

b) they must have mutations at different exact nucleotide positions

c) they must have mutations in the same specific gene

d) they must have mutations at the same exact nucleotide positions

e) both A and B are true

f) both C and D are true