Name ________________________________
BIO 2110 – Genetics
Exam 2
4 November 2016
calculators are permitted
see last two pages for potentially useless information
1. In cuttlefish, allelic variation in a single gene results in variation in swimming speed.
Allelic variation in a linked gene results in variation in the ability of cuttlefish to change
their skin color. Based on the segregation data presented below, determine the
recombination frequency between these genes. [P0s are homozygous.] SHOW YOUR
WORK
P0 Fast, Polychrome x Slow, Monochrome
↓
F1 Fast, Monochrome x Fast, Monochrome
↓
F2 Fast, Monochrome
Fast, Polychrome
Slow, Monochrome
Slow, Polychrome
4,125
1,875
1,875
125
2. Below are shown segregation from reciprocal monohybrid crosses for two genes, the
fat head gene in grasshoppers and the tetrodotoxin gene pufferfish. For each of these
genes, explain the segregation patterns of F1 and F2 phenotypes. [All P0s are from
populations that are fixed for either the dominant or recessive allele.]
reciprocal crosses for the fat head gene
cross 1 cross 2
P0 ♀ fat head ♂ thin head ♀ thin head ♂ fat head
↓ ↓
F1 ♀ fat head ♂ fat head ♀ thin head ♂ thin head
↓ ↓
F2 ♀ thin head
♀ fat head
♂ thin head
♂ fat head
156
0
153
0
♀ thin head
♀ fat head
♂ thin head
♂ fat head
129
0
134
0
reciprocal crosses for the tetrodotoxin
cross 1 cross 2
P0 ♀ toxic ♂ nontoxic ♀ nontoxic ♂ toxic
↓ ↓
F1 ♀ toxic ♂ toxic ♀ toxic ♂ nontoxic
↓ ↓
F2 ♀ toxic
♀ nontoxic
♂ toxic
♂ nontoxic
137
0
68
69
♀ toxic
♀ nontoxic
♂ toxic
♂ nontoxic
53
47
49
51
3. In octopi, allelic variation in the tententacle gene results in variation in tentacle number.
Allelic variation in the fuscluster gene results in frequent fusions of clusters of suckers
on the tentacles. Based on the segregation data presented below, are these two genes
linked? [P0s are homozygous.] SHOW YOUR WORK
P0 ten, unfused x eight, fused
↓
F1 eight, unfused X eight, unfused
↓
F2 eight, unfused
ten, unfused
eight, fused
ten, fused
435
163
173
29
4. Meiotic recombination (crossover) is precise to the base pair. Explain, based on the
molecular mechanism of recombination, how such precision is possible.
5. Based on the test cross data presented below, place the evrs, to, tnkr and chnc genes on
a recombination map in their proper order. IT IS NOT NECESSARY to calculate
recombination frequencies between these genes. Progeny genotypes are shown only
for the chromosomes derived from the heterozygous parent. Among progeny,
reciprocal genotypes are grouped together.
cross 1
evrs tnkr chnc
EVRS TNKR CHNC
x evrs tnkr chnc
evrs tnkr chnc
↓
EVRS TNKR CHNC
evrs tnkr chnc
315
315
EVRS TNKR chnc
evrs tnkr CHNC
35
35
EVRS tnkr chnc
evrs TNKR CHNC
15
15
EVRS tnkr CHNC
evrs TNKR chnc
135
135
cross 2
evrs to tnkr
EVRS TO TNKR
x evrs to tnkr
evrs to tnkr
↓
EVRS TO TNKR
evrs to tnkr
360
360
EVRS TO tnkr
evrs to TNKR
40
40
EVRS to tnkr
evrs TO TNKR
90
90
EVRS to TNKR
evrs TO tnkr
10
10
6. Based on the epistasis data presented below, construct a genetic pathway for the
regulation of eye number in Brachypelma.
genotype phenotype
single mutants
sxorb six eyes
hexocular six eyes
decocular ten eyes
xyes ten eyes
teneye ten eyes
double mutants
sxorb; decocular six eyes
sxorb; xyes ten eyes
sxorb; teneye six eyes
hexocular; decocular ten eyes
hexocular; xyes ten eyes
hexocular; teneye ten eyes
7. Allelic variation in a single gene results in flower color
variation in Laqueum diaboli. Based on the segregation data
presented below, describe the dominance relationships for
the alleles of this gene. [P0s are homozygous.]
P0 purple x white
↓
F1 pink x pink
↓
F2 purple 315
pink 627
white 312
8. In Gavialis gangeticus, recessive mutations in the pink gene result in pink teeth (the
wild-type tooth color is white). Recessive mutations in the unlinked toothless gene
result in animals without teeth. From the dihybrid cross shown below;
i. what phenotypes would be observed in the F2 generation?
ii. out of 800 F2s, how many of each phenotype would be expected?
iii. briefly, explain how you determined the number of expected animals in each
phenotypic class.
[P0s are homozygous]
P0 toothless x pink teeth
↓
F1 white teeth white teeth
↓
F2
phenotypes number expected
9. The huey, duey and louie genes all are on the same chromosome (see below). huey and
duey are separated by 10 cM. duey and louie are separated by 30 cM. Based on the
segregation data presented, how much, if any, crossover interference is present in this
region of this chromosome? Progeny genotypes are shown only for chromosomes
derived from the heterozygous parent. Chromosomes are shown in reciprocal pairs.
SHOW YOUR WORK
huey duey louie
HUEY DUEY LOUIE
x huey duey louie
huey duey louie
↓
HUEY DUEY LOUIE
huey duey louie
315
309
HUEY DUEY louie
huey duey LOUIE
37
39
HUEY duey louie
huey DUEY LOUIE
141
135
HUEY duey LOUIE
huey DUEY louie
11
13
huey
duey
louie
10 cM 30 cM
10. Null mutations are mutations that completely eliminate gene function. In some genes,
null mutations are dominant. Provide a biochemical explanation for the dominance of
loss-of-function mutations.
probability
df 0.90 0.50 0.20 0.05 0.01 0.001
1 0.02 0.46 1.64 3.84 6.64 10.83
2 0.21 1.39 3.22 5.99 9.21 13.82
3 0.58 2.37 4.64 7.82 11.35 16.27
4 1.06 3.36 5.99 9.49 13.28 18.47
5 1.61 4.35 7.29 11.07 15.09 20.52
χ2 values
Lod = Log10(odds) χ2 = Σ [(obs-exp)2/exp] p2 + 2pq + q2 = 1
p2 + q2 + r2 + 2pq + 2pr + 2qr = 1 df = n – 1
!!!!!
! ¼ p2
!!!!!
!
p + q = 1 #classes = 2n + 1 p + q + r = 1
parental gamete frequency = ½ (1 – p) recombinant gamete frequency = ½ p
Nt = N0 x e-rt r = ln[N(t2)/N(t1)]/(t2 – t1) θT = π prob12 = (prob1 x prob2)
__ __
W = p2WAA + 2pqWAa + q2Waa pn+1 = (pn2WAA + pnqnWAa)/W C = DCOobs/DCOexp
€
p =
1
2
(TT) + 3(NPD)
N
€
x =
−b ± b2 − 4ac
2a
€
p =
1
2
(2nd div)
N
par = ½ (1-p)
1 cM equals a recombination frequency of 0.01 a = 1 + ½ + … + 1/(n – 1)
__
qn+1 = (qn2Waa + pnqnWAa)/W θ = 4Neμ #classes = log10(freqext)/log10(¼)
1
θW = k/a I = 1 – C 1 1 Eco RI = GAATTC
1 2 1
odds = probx/prob0.5 1 3 3 1 (2Ne – 1)/2Ne
1 4 6 4 1
rec = ½ p 1 5 10 10 5 1 1/(2Ne)
1 6 15 20 15 6 1
pn = p0e-nμ 1 7 21 35 35 21 7 1
heterozygote frequency = (2pq)(½)n SCO > DCO k = fraction of segregating sites
p = average frequency of nucleotide differences in pairwise sequence comparisons
