Problem Set 6, Fall 2015
Name:
8 points total
1. You suspect that two genes, d and e, might be linked. You cross a heterozygote with
homozygous recessive and get the results shown below (Where D = the dominant
phenotype, d = the recessive phenotype, etc. Assume that d+ and e+ are dominant.).
De
512 d+ e–/d– e–
1035
dE
523 d– e+/d– e–
(1035 + 940)/2 = 987.5
D E 471 d+ e+/d– e–
940
de
469 d– e–/d– e–
a. Use the chi square test to determine whether d and e are linked.
(1035 – 987.5)2/987.5 + (940 – 987.5)2/987.5 = 2256.25/987.5 + 2256.25/987.5
= 2.28 + 2.28 = 4.56
If you look at the chi square table you’ll see that the value of 4.56 with 1 degree of
freedom corresponds to a probability < 0.05 that the results are due to chance.
Therefore the two genes appear to be linked.
b. What was the genotype of the original heterozygous parent that was crossed to give the results
shown above?
d+ e–/d– e+ (often abbreviated d+ e/d e+ or even + e/d +)
2. While on a quest, you come across two pure-breeding strains of dragons, one with green scales
with jagged edges and the other with purple scales with smooth edges. You notice that the
dragons have offspring, all but one of which have green scales and jagged edges. The one
unique offspring is mostly green with jagged scales but also has a small spot where the scales are
purple and smooth.
a. Please provide the genotypes of the parents and the F1 offspring
g+ j+/g+ j+ X g– j–/g– j– => g+ j+/g– j–
b. Assuming the genes are linked, are the mutations in cis or trans in the F1 offspring?
cis
c. Draw out a mitotic recombination event that could account for the anomalous offspring
with the green, jagged scales and a spot with purple smooth scales.
1
Problem Set 6, Fall 2015
Name:
8 points total
3. You’ve discovered a new species of purple-eyed tree frogs, J. mannae. Some of the
individuals have blue rather than purple eyes. Eye color is determined by a single locus, blue
(bl). To study the genetic basis of eye color in J. mannae, you perform genetic crosses between
mutant and wild type and get the following results.
+/+ males X bl/bl females bl/bl males X +/+ females
bl/+ X bl/+
all purple-eyed
all blue-eyed
½ purple- and ½ blue-eyed
How do you explain these results? What mode of inheritance does this represent (10 pts)
bl appears to be subject to genomic imprinting. If bl/bl = blue then it is maternally
imprinted.
4. After determining pink skin (p) and fuzzy fur (f) to be recessive in a population of duckbill
platypus on the remote island of Mushkibibble, you notice an individual with a single pink spot.
a. Diagram 3 possible explanations of how this came to be.
2
Problem Set 6, Fall 2015
8 points total
Name:
b. You then find a platypus with a pink spot next to a fuzzy spot. Diagram how this could
have happened.
3
Problem Set 6, Fall 2015
Name:
8 points total
5. You have identified three genes on the third chromosome in the newly discovered singing
firefly S. myersae. Mutations in the first gene, dim prevent fluorescence of the fireflies.
Mutations in the second gene, grn, causes the animals to develop green wings. Mutations in the
third gene, voiceless (vls for short), cause the normally musical fireflies to be silent. You cross
triply heterozygous individuals by homozygous recessive individuals and obtain the results
shown below.
106
592
157
27
18
171
615
131
1817
wild type
green
dim
voiceless
green, dim
green, voiceless
dim, voiceless
green, dim, voiceless
total
A. What is the order of the three linked genes?
grn dim vls
B. Calcualte the map distances between each of the genes:
dim to grn
(131+106+27+18)/1817 X100% = 15.5 cM
grn to vls
15.5 + 20.5 = 36.0
vls to dim
(171+157+27+18)/1817 X 100 = 20.5
C. What is the genotype of the two parents used in this cross?
grn + +/+ dim vls X grn dim vls/grn dim vls
D. You have mapped two other genes, glo and hop to the same chromosome. In separate
mapping experiments, you find that glo is 5.5 mu from grn. hop is 10.5 mu from grn, and glo is 5
mu from hop. Please draw a map showing the relative location of the dim, grn, vls, glo and hop.
grn 5.5
hop 5 glo 5.5 grn
glo
5
5
hop
15.5
4
dim
20.5
vls
dim
20.5
vls
Problem Set 6, Fall 2015
Name:
8 points total
6. In your studies of the rare squirrel S. ydeaed in Bloomington you noticed several interesting
phenotypes. Your studies reveal three recessive mutations that affect the squirrels’ tails. b
produces blue instead of brown; f causes extremely fuzzy tails instead of normal, and t causes
large teeth instead of normal teeth. You cross a triply heterozygous female with a homozygous
recessive male and find the following results:
323
10
69
87
73
13
342
83
blue tail color, extremely fuzzy tail and normal teeth
blue tail color, normal fuzzy tail and larger sized teeth
blue tail color, normal fuzzy tail and normal teeth
blue tail color, extremely fuzzy tail and larger sized teeth
brown tail color, extremely fuzzy tail and larger sized teeth
brown tail color, extremely fuzzy tail and normal teeth
brown tail color, normal fuzzy tail, and larger sized teeth
all wild type
A. Are the genes linked? Briefly explain why you think they are or are not.
Yes because blue, fuzzy and normal and brown, normal and large teeth all
cosegregate preferentially.
B. What is the genotype of each of the original parents (i.e. those that were crossed to
produce the offspring listed above?
f b +/+ + t X f b t/f b t
C. What is the distance between the genes:
b and f: (73 + 69 + 10 + 13)/1000 X 100 = 16.5 cM
b and t: (87 + 83 + 10 + 13)/1000 X 100 = 19.3 cM
f and t: 16.5 + 19.3 = 35.8 cM
7. You suspect that two genes, cb and bx, are linked. A test cross of cb bx/ + + produces the
results shown below.
cb bx/cb bx 312
609
+ + /cb bx 297
(609 + 553)/2 = 581
cb +/cb bx 272
553
+ bx/cb bx 281
Are you correct in supposing the two genes are linked? What is the probability that the two genes
are unlinked and the results shown are the result of chance? Have you written the genotypes
correctly?
chi2 = (609-581)2/581 + (553 – 581)2/581 = 784/581 + 784/581 = 1.35 + 1.35 = 2.70
p>0.10 so we can’t reject the null hypothesis; the genes may be unlinked
genotypes should have been written
cb–/cb–; bx–/bx–
cb+/cb–; bx+/bx–
cb+/cb–; bx+/bx–
cb–/cb–; bx+/bx+
5
Problem Set 6, Fall 2015
Name:
8 points total
8. In your studies of the mythical Farlow’s sparrow in Bloomington you discovered three
recessive mutations that affect the birds: (b) produces a pointed beak instead of blunt, t causes
turquoise colored feathers instead of black, and l causes abnormally long tail feathers. You cross
triply heterozygous females with homozygous recessive males and find the following results:
94
372
14
80
11
381
84
89
1125
pointed beaks, turquoise feathers and normal tail feathers
pointed beaks, black feathers and long tail feathers
pointed beaks, black feathers and normal tail feathers
pointed beaks, turquoise feathers and long tail feathers
normal beaks, turquoise feathers and long tail feathers
normal beaks, turquoise feathers and normal tail feathers
normal beaks, black feathers and long tail feathers
normal beaks, black feathers and normal tail feathers
A. Are the genes linked? Briefly explain why you think they are or are not.
Yes, because combination of three alleles segregate together (giving the parentals).
B. What is the distance between the genes:
b and t: 17.2 + 18.0 = 35.2 cM
t and l: (80 + 89 + 11 + 14)/1125 X 100% = 17.2 cM
b and l: (94 + 84 + 14 + 11)/1125 X 100% = 18.0 cM
C. Previous mapping experiments have shown that the gene for flight (gene f, mutations
produce flightless birds) is 4 map units from gene t. You cross f and l heterozygotes with
homozygous recessive and get the following results:
101 flightless
104 long tail feathers
14 flightless and long tail feathers
17 wild type
236
How far apart are f and l? (14 + 17)/236 X 100 = 13.1cM
D. Please draw a map of the four genes (b, t, l and f), including distances.
b
18.0
l
13.1
6
f
4.0
t