Genetics 371B, Autumn 1999

Answer Key, Exam 1: 20 October, 1999

Open book, open notes, 100 points total. Answer Questions 1-4 and either 5 or 6, but NOT BOTH. Show your work!

1.

(20 pts total)

Eyes in the Great Northwestern Squirrel can be red or black. A red-eyed female was crossed with a black-eyed male. The F1 progeny consisted of equal numbers of red-eyed females, red-eyed males, black-eyed females, and black-eyed males.

(a)

State if each of the following statements is consistent with the data you have been given so far. If so, show the parental genotypes (specify alleles!); if not, explain in ONE sentence why not. (16 pts)

(i) black eye is autosomal recessive
Consistent. If B = black, R= red, the parents would have to be BR (female) and BB (male) to give 1:1 ratio of the two phenotypes in F1.
 
(ii) black eye is autosomal dominant
Consistent. If B = black, R = red, the parents would have to be RR (female) and BR (male) to give 1:1 ratio of the two phenotypes in F1.
 
(iii) black eye is X-linked recessive
Consistent. If Xb = black, XR = red, the parents would have to be XRXb (female) and XbY (male) to give 1:1 ratio of the two phenotypes in both sexes in F1.
 
(iv) black eye is X-linked dominant
Not consistent! If XB = black, Xr = red, the male parent would XBY -- and therefore all his daughters should inherit XB and be black-eyed.
(b)

Black-eyed F1 females were crossed to black-eyed F1 males. The progeny were 3/8 black-eyed females, 3/8 black-eyed males, 1/8 red-eyed females, and 1/8 red-eyed males. Which of the above four is the correct mode of inheritance of black eye, and why? (One or two sentences!) (3 pts)

No difference between males and females => autosomal trait; black : red ratio is 3:1, so black must be dominant. So black eye must be autosomal dominant.

(c)

Which would be easier to eliminate from your stock of squirrels, the allele for red eyes or the allele for black eyes? Why? (ONE sentence!) (1 pts)

Black eyes being dominant, that phenotype will always be seen if the B allele is present, whereas the red eye allele could be present but masked -- therefore, it'll be easier to eliminate the black eye allele.

2.

(18 pts total)

Below is a pedigree that shows the inheritance of green hair in a human family (individuals with green hair are shown as affected ). Assume complete expressivity and penetrance.
(a)

Which one of the following modes of inheritance best fits this pedigree? (Circle one.) (8 pts)

(i) autosomal recessive (not consistent -- affected couple II-4 and II-5 have unaffected child)
(ii) autosomal dominant <-- this is the best fit (all others are not consistent)!
(iii) X-linked recessive (not consistent -- affected woman II-4 has unaffected father I-2)
(iv) X-linked dominant (not consistent -- affected man II-3 has unaffected daughter III-3)
(b)

Based on your answer in (a), write the genotypes of all individuals in the pedigree in the space beneath each individual. Use G for the dominant allele and g for recessive. (4 pts)
(c)

A second trait, purple eyes, is determined by a second gene, and segregates perfectly with the green hair trait. In other words, all individuals in the pedigree with green hair also have purple eyes, while individuals without green hair do not have purple eyes. What can you say about the positions of the loci for these two traits? (ONE sentence!) (2 pts)

Since the traits segregate together, the two loci must be linked.

(d)

Individuals III-1 and III-2 had another child after this pedigree was drawn. This youngest child (not included above) has green hair, but DOES NOT have purple eyes. Given your answer in part (b), what event could explain this outcome, and which person was involved? (4 pts)

Recombination between the two loci must have occurred in the heterozygous parent (III-2).

3. (22 pts total)
(a)

Hemophilia is an X-linked trait in humans. Using long lines for X chromosomes and short lines for Y, show the possible sex chromosome configurations that might be seen at the end of Anaphase I of Meiosis in a Klinefelter male who is heterozygous for hemophilia (XHXhY). Your diagrams should be accurate with respect to the number of chromatids , and show the appropriate alleles of the hemophilia locus. (For this part of the question, ignore the temporary checkpoint delay. Also ignore the autosomes. And don't worry about his sterility or fertility!) (16 pts)

Remember that this is meiosis, so homologs pair up in Metaphase I, and are separated in Anaphase I. Because there's an extra chromosome, three possible pairings can occur: XH with Y (in which case XH and Y would go to opposite poles in Anaphase I), Xh with Y, or XH with Xh. Chromatids have not separated yet, so there should be two chromatids per chromosome.

(b)

A tissue culture cell line was established from a normal (XY) male. Tissue culture cells that were just about to enter mitosis were incubated with radioactively labeled DNA precursors. Cell samples were withdrawn at regular intervals, and the amount of radioactive DNA in each sample was plotted:

(i) In the plot above, circle and label every S phase that you can detect. (2 pts)

S phase is when DNA precursors are incorporated into DNA -- so the periods when there's in increase in the amount of radioactive DNA correspond to S phases.

(ii) Approximately how many hours does one complete cell cycle last in the XY cells? (Plus or minus ~2 hours is good enough.) (2 pts)

About 24 hours (start of one S phase to start of next S phase -- see Figure above).

(iii) Do you think the presence of the extra X chromosome in XXY cells will cause a mitotic checkpoint delay? Explain in ONE sentence. (2 pts)

No -- homologs don't pair up in mitosis; the extra X chromosome will have two chromatids, so everything will be okay.

4.

(20 pts total)

In a Great Northwestern plant, slimy (sl) is recessive to wet (+); chartreuse (ch) is recessive to green (+); and bushy (b) is recessive to fuzzy (+). A plant fully heterozygous for all of these genes was crossed to a plant which was fully recessive. Phenotypes and number of progeny were:

    Progeny phenotype Number
    + + + 88
    ch sl b 80
    + sl + 372
    ch + b 360
    + sl b 43
    ch + + 45
    + + b 5
    ch sl + 7
    Total 1000

(a)

What are the genotypes of the non-crossover (NCO) gametes produced by the trihybrid plant? The order of the genes is unimportant for this part of the question. (2 pt)

The most abundant types -- + sl + and ch + b

(b)

Determine the order of the three loci on the chromosome. Show your work! (6 pts)

The double crossover (DCO) products are + + b and ch sl + (the least abundant classes). So a double crossover event between the parental gamete types should give the DCO type:

 
 

(c)

Calculate the map distances between each of the adjacent pairs of loci. Write the distances on the chromosome figure at the bottom of the page. Note that the figure is not necessarily to scale. Show your work (facing page) for full credit. (12 pts)

Recombinant frequency in sl-ch interval:
= Single crossovers (SCO) in this interval + DCO
(SCO between sl and ch should give sl ch b and + + +)
= (80 + 88 + 12)/1000
Map distance = 18 cM

 

Recombinant frequency in ch-b interval:
= SCO in this interval + DCO
(SCO between ch and b should give sl + b and + ch +)
= (43 + 45 + 12)/1000
Map distance = 10 cM
 
 
The map is:
 

sl        18 cM           ch    10 cM       b
|-------------------------|-----------------|

 

Answer Question 5 or Question 6, BUT NOT BOTH. If you answer both, we will only grade #5.

5.

(20 pts total)

B/b and E/e, two loci in the garden pea, are linked at a map distance of 20 cM.

(a)

A fully heterozygous pea plant has the dominant alleles linked in trans. What will be the genotypes of gametes produced by this plant, and in what frequencies (or percentages)? Show your work! (10 pts)

The dominant alleles are in trans, so the two homologs have the allele configuration
B e and b E
-- so these will be the predominant (non-crossover) gamete types. Recombination between the loci will produce gametes of genotypes B E and b e. Because the map distance = 20 cM, the recombinants together should add up to 20% of the total, while the NCO types will add up to 80%. So the frequencies of the four gamete types are:
B e:  (0.8)/2 = 0.4
b E:  (0.8)/2 = 0.4
B E:  (0.2)/2 = 0.1
b e:  (0.2)/2 = 0.1
(The frequencies add up to 1.0, as they should.)

(b)

If this plant is self-pollinated, what progeny phenotypes will you expect to see, and in what frequencies? Use a Punnett square to illustrate your answer. (10 pts)

When doing a Punnett square for a dihybrid with independently assorting genes, we assume that the four gamete types are in equal proportions, i.e., frequency = 0.25 each. Here, we know that the frequencies aren't equal -- so we can simply include the frequencies in the Punnett square, and multiply frequencies to get the frequency of the product. For example -- when we say that the frequency of gamete genotype B e is 0.4, we mean that if we randomly picked one gamete from the pool of gametes, the probability that it is B e = 0.4. Likewise, the probability of picking a gamete of genotype b e = 0.1. Thus the probability of picking both (or, the probability that a gamete of genotype B e will meet and form a zygote with a gamete of genotype b e) = (0.4)(0.1) = 0.04. And so on. The Punnett square below shows the phenotypes of the progeny of the cross.

gamete
frequency
B e
0.4
b E
0.4
B E
0.1
b e
0.1
B e
0.4
B e
0.16
B E
0.16
B E
0.04
B e
0.04
b E
0.4
B E
0.16
b E
0.16
B E
0.04
b E
0.04
B E
0.1
B E
0.04
B E
0.04
B E
0.01
B E
0.01
b e
0.1
B e
0.04
b E
0.04
B E
0.01
b e
0.01

So the overall frequencies are:
B E:  0.51
B e:  0.24
b E:  0.24
b e:  0.01
Again, the frequencies add up to 1.0, as they should.

Answer Question 5 or Question 6, BUT NOT BOTH. If you answer both, we will only grade #5.

6.

(20 pts total)

Consider a and b, two autosomal genes in Drosophila . Suppose the following crosses were made, starting with pure-breeding parental flies:

The F2 progeny of this testcross were counted, scored for phenotype, and chi square analysis was done. The degrees of freedom = 3. The p value was found to be exactly p = 0.025.

(a)

If a and b are recessive to a+ and b+, what phenotypes were expected in the F2? (4 pts)

a+ b+

a+ b

a b+

a b

(b)

If 1000 F2 flies were scored, how much, on average, did the observed number for any single F2 class vary from the expected value? Show all calculations (12 pts)

For degrees of freedom = 3, a P value of 0.025 corresponds to a chi-squared value of 9.348

Therefore, the sum of ((O - E)2/E) = 9.348; the average ((O - E)2/E) = 9.348/4 = 2.337.
The average expected per class = 1000/4 = 250;
the variation (i.e., O - E) = square root of ((2.337)(250)) = +/- 24
Thus, the average variation from expected for each class = 24.

(c)

What percent of the time would this variation be expected by chance alone? (2 pts)

2.5% (that's what the P value means)

(d)

If the experimenter was testing independent assortment of a and b, would you accept or reject the hypothesis of independent assortment based on the chi squared results? (2 pts)

Reject -- there's only a very low probability (2.5%) of getting this much deviation from expected values due to chance fluctuation.

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