Problem Set 05, Fall 2018Name:
8 points total
Note: Please show your work. Define your symbols and don't assume linkage unless otherwise stated.
1. While observing a new species of butterfly, you notice that two true-breeding strains exist. One strain is yellow and purple, and the other strain is just purple. You decide to perform some genetic crosses and cross a yellow and purple butterfly with a purple butterfly. All the F1s are purple. When you cross two F1 individuals, you find:
240 purpleA-B- and A-bb
62 purple and yellowaaB-
19 yellow aabb
a)Give the genotypes for the F2s and the original parental strains.
Original parents were aaBB X AAbb
b)What ratio of F2s is observed?
12:3:1 This is a variation of 9:3:3:1, telling us there are two alleles of two genes producing the phenotypes that we see above.
c)What is the likelihood that purple individuals will be homozygous for all genes involved?
Purple homozygous for both genes are AABB or AAbb
Probability{AABB} or P{AAbb} = 1/3(1/4) + 1/3(1/4) = 1/6
2. In a new strain of dog, you find two new recessive mutations. One mutation results in the animal having a short tail (st), and the other mutation causes the dog to be spotted instead of a solid color (po). After more study, you discover that the two genes for the traits are 22 mu apart. If you mate individuals heterozygous for both traits (st + / + po), what percentage of their offspring will be both short-tailed and spotted, assuming no double crossovers?
st +/+ po can produce
the following gametes:st +
+ po
stpo
+ +
Answer: 0.11 X 0.11 = 1.2%
3. In humans, a recessive mutation results in lips as red as a rose. Another recessive mutation causes excessively long, curly lashes. The two genes r and l are on the same chromosome and are 24 mu apart. If two heterozygous individuals cross, what is the fraction of the offspring that will have both red lips and long lashes? Assume the alleles are in trans in the mother and in cis in the father, and no double crossovers occurred.
mom produces::r +dad produces:r l
+ l+ +
rlr +
+ ++ l
Answer: 0.12(0.38) = 0.0456 or 4.56%
4. In studying skin spot color in the brilliant tropical frog J. collinsae, you find that there are two true breeding strains. You cross a true breeding white-spotted strain with a true breeding red-spotted strain, and find all F1’s have red spots. When you perform crosses amongst the F1’s, you get the following results:
797 white spots
599 orange spots
1803 red spots
A. What are the genotypes of the F2 progeny?
white spots O-wwand ooww
orange spots ooW-
red spots O-W-
B. If you cross orange spotted F2’s chosen at random with homozygous recessive individuals, what phenotypes will be observed? What will the relative frequencies of these phenotypes be?
ooW- X ooww
Cross is: ooWWX ooww => ooWw
ooWwX ooww => ½ ooWw and ½ ooww
of orange spotted F2’s, 1/3 areooWW and 2/3 are ooWw
So, 1/3(1) + 2/3(1/2) = 2/3 of progeny are ooWw and 2/3(1/2) are ooww
5. C. elegans is a self-fertile hermaphrodite. unc-34 mutations are recessive and cause uncoordinated movement. him-7 mutations are recessive, and cause a high frequency of X-chromosome nondisjunction, resulting in many males (C. elegans males are XO). The genes are 16 mu apart. If unc-34 +/+ him-7 animals self-fertilize, what fraction of the progeny will be both uncoordinated and Him (i.e. producing many males)? (Assume no doubles)
16% of gametes will be recombinant. To produce the double mutant each parent must contribute a unc-34 him-7 gamete. 8% (half of 16%) will be of this genotype so the answer is
0.08 X 0.08 = 0.0064 or 0.64%
6. Consider two genes on the X chromosome in Drosophila melanogaster: cut (ct) which causes the wing edges to be clipped off, and miniature (m) which causes small wings. You perform recombination mapping of these genes by crossing a ct m / + + female by a ct m / Y male. You get the following four classes of progeny:
cut, mini wings 1769
mini wings 331
wild type 1755
cut 345
total 4200
a. What is the map distance in centimorgans (cM) between m and ct? Please show all calculations.
(331 + 345)/4200 X 100% = 16.1 cM or mu
b. Consider three additional genes on the X chromosome: yellow (y) which causes yellow body color, forked (f ) which causes bent bristles, and crossveinless (cv) which causes reduced wing crossveins. Given the following recombination frequencies, order m, y, f, cv and ct on the X chromosome and indicate the genetic distances between the genes on the chromosome drawn below:
y and cv 15.0%
y and m 36.1%
y and f 50.0%
cv and m 21.1%
cv and f 41.7%
m and f 20.6%
c. In map units, how far apart are y and f?
15 + 21.1 + 20.6 = 56.7
7. Consider a diploid cell with one pair of long chromosomes. The cell is heterozygous for the A and B genes. The alleles are in the cis orientation. In your diagram, show a single crossover between A and B, and the four products of this meiotic division. You do not need to show all steps of meiosis, but rather show the products of each meiotic division, and metaphase of meiosis I, with the crossover indicated.
8. A population of caterpillars has three mutations; pink (p) instead of blue, sparkly (r) instead of dull coats, and smelly (l) instead of odorless secretions. You cross triply heterozygous adult females with triply homozygous recessive adult males. The progeny are as follows:
278Pink, dull, odorless
53Blue, sparkly, odorless
305Blue, sparkly, smelly
8Pink, sparkly, odorless
57Pink, sparkly, smelly
6Blue, dull, smelly
63Blue, dull, odorless
60Pink, dull, smelly
a. What were the parental genotypes?
p + +/ + r l X p r l/p r l
b. Calculate the distances between the genes of interest.
p to r (57 + 63 +8 + 6)/830 X 100 = 16.1 cM or mu
r to l (60 + 53 + 8 + 6)/830 X 100 = 15.3 cM or mu
lto p 16.1 + 15.3 = 31.4 cM or mu
c. Draw the genetic map, including the distances between the genes.
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