Problem Set 05, Fall 2015
Name:
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 purple
A-B- and A-bb
62 purple and yellow
aaB19 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.
The probability of AA is 1/3
c) What is the likelihood that purple individuals will be
because aa never has the purple
phenotype. By contrast the
homozygous for all genes involved?
probability of BB and bb each is
Purple homozygous for both genes are AABB or AAbb
Probability{AABB} or P{AAbb} = 1/3(1/4) + 1/3(1/4) = 1/6 ¼ because purple includes all
possible alleles of B (i.e. purple
genotype
BB,
2. In a new strain of dog, you find two new recessive mutations. can
Onehave
mutation
results
inBb
theor bb.
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,
To produce the homozygous recessive
assuming no double crossovers?
individuals, two st po gametes must fuse.
st +/+ po can produce
Note that the st po and + + gametes are the
the following gametes:
st +
products of recombination. Recombinants
+ po
will be 22% of the gametes if the genes are
st po
22 mu apart. Therefore the st po gametes will
++
be half of these or 11% of total.
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:
rl
76% +
+l
+
rl
r+
24% + +
+l
Answer: 0.12(0.38) = 0.0456 or 4.56% = 1.2%
1
Problem Set 05, Fall 2015
8 points total
Name:
4. In studying skin spot color in the brilliant tropical frog A. millerae, you find that there are two
true breeding strains. You cross a true breeding white-spotted strain with a true breeding redspotted 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
orange spots
red spots
O-ww and ooww
ooWO-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: ooWW X ooww => ooWw
ooWw X ooww => ½ ooWw and ½ ooww
of orange spotted F2’s, 1/3 are ooWW 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%
2
Problem Set 05, Fall 2015
8 points total
Name:
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%
y
15
cv
ct can be either 16.1 cM to the left or right of
m. You should indicate both possibilities on
your map.
21.1
c. In map units, how far apart are y and f?
15 + 21.1 + 20.6 = 56.7
3
m
20.6
f
Problem Set 05, Fall 2015
8 points total
Name:
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.
4
Problem Set 05, Fall 2015
8 points total
Name:
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:
278
Pink, dull, odorless
53
Blue, sparkly, odorless
305
Blue, sparkly, smelly
5
Problem Set 05, Fall 2015
8 points total
8
57
6
63
60
Name:
Pink, sparkly, odorless
Pink, sparkly, smelly
Blue, dull, smelly
Blue, dull, odorless
Pink, 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
l to p 16.1 + 15.3 = 31.4 cM or mu
c. Draw the genetic map, including the distances between the genes.
16.1
15.3
p
r
l
____________________________________________
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