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Problem Set 7, Bio 4181: Due Nov. 24, 2008
1. The relative fitnesses for the genotypes at an autosomal locus with 3 alleles, A, B and C, in a
population are given in the table below:
Genotypes:
Fitness:
AA
1
AB
BB
AC
BC
CC
0.95
1.1
1.1
0.9
1.05
a. Assuming random mating and initial allele frequencies of 0.4, 0.3 and 0.3 for A, B and C
respectively, calculate the change in all allele frequencies due to natural selection in the next
generation.
1 pt genotype freq., 1 point genotypic deviations, 3 points average excesses, 1 point delta p.
Genotypes:
Geno. freq.
Fitness
Genotypic D.
Average Ex.
delta p
AA
0.160
1
-0.008
A: 0.007
0.0030
AB
0.240
0.95
-0.058
B: -0.028
-0.0082
BB
0.090
1.1
0.093
C: 0.018
AC
0.240
1.1
0.093
BC
0.180
0.9
-0.108
CC
0.090
1.05
0.043
AVERAGE/SUM
1
1.0075
AC
0.144
1.1
0.078
BC
0.108
0.9
-0.123
CC
0.174
1.05
0.028
AVERAGE/SUM
1
1.0225
0.0052
b. Assuming inbreeding with f=0.4, redo problem a.
Genotypes:
Geno. freq.
Fitness
Genotypic D.
Average Ex.
delta p
AA
0.256
1
-0.023
-0.014
-0.0053
AB
0.144
0.95
-0.073
0.006
0.0016
BB
0.174
1.1
0.078
0.013
0.0037
1 pt genotype freq., 1 point genotypic deviations, 3 points average excesses, 1 point delta p.
c. Is there any evidence for an interaction between selection and system of mating in the
contrast of part a) to part b)?
2 points. Yes, the direction of change in allele frequencies is reversed for A and for B.
2. A lethal genetic disease is due to homozygosity for a loss of function allele, a, at an autosomal
locus. The collective mutation rate to the loss of function allele a from the functional allele, A, is
10-6.
a. Assuming random mating, and complete recessiveness of a for the disease phenotype, what is
the equilibrium frequency of the a allele?
a. Note that this can be put into the fitness format 1:1-hs:1-s with s=1 and h=0 (2 pts). Use
the equation qeq =
µ
(1 pt) to get 0.001 (1 pt).
s
b. Assuming random mating and a relative fitness of AA:Aa of 1:0.99, what is the equilibrium
frequency of the a allele?
b. Note that this can be put into the fitness format 1:1-hs:1-s with s=1 and h=0.01 (2 pts).
µ
Use the equation qeq =
(1 pt) to get the qeq = 10-6/0.01 = 0.0001 (1 pt).
hs
c. Assuming random mating and a relative fitness of AA:Aa of 1:0.99 but now assuming that aa is
not lethal but rather has a relative fitness of 0.5, what is the equilibrium frequency of the a allele?
c. Note that this can be put into the fitness format 1:1-hs:1-s with s=0.5 and h=0.02 (2 pts).
µ
Use the equation qeq =
(1 pt) to get the qeq = 10-6/0.01 = 0.0001 (1 pt).
hs
d. Assuming positive assortative mating with f=0.1 and now with the fitness of Aa being 1 and the
fitness of aa 0, what is the equilibrium frequency of the a allele?
b. Use the equation qeq =
µ
(1 pt) to get qeq = 10-6/0.1 = 0.00001 (1 pt).
fs
3. Deme 1 has a frequency of the A allele of 0.2, and deme 2 has a frequency of the A allele of
0.8. Each deme exchanges 10% of its gametes per generation, and mating is inbreeding within
demes with f=0.2. The genotypes and fitnesses in deme 1 are:
AA
0.5
Aa
0.7
aa
1
What is the frequency of A in the next generation in deme 1?
The change due to selection in deme 1 is obtained by first calculating the genotype freq.
with p1=.2 and f=.2 as:
Inbreeding genotype freq.
AA
0.072
Aa
0.256
aa
0.672
(2 pts). Then, the average fitness is 0.8872 (1 pt), and the average excess is -0.2592 (2 pts),
and delta p(selection) = -0.058431 (1 pt).
The change due to gene flow is Δp1 = -m(p1-p2) = -0.1(0.2-0.8)=0.06 (2 pts).
p
The total change in allele freq. is ! p1 = 1 aA m( p1 p2 ) = 0.00156898 (2 pts). Hence, the
w
frequency of A in deme 1 in the next generation is 0.2016 (1 pt).
4. Two closely related species, X and Y, are examined for a protein coding region, with the
following estimated haplotype tree, where haplotypes are indicated by the species letter followed
by a number, and the two numbers written by each branch in the tree indicate the number of
synonymous nucleotide substitutions that occurred on that branch (non-bold) and the number of
nonsynonymous nucleotide substitutions (bold):
Species X
X1
Species Y
X2
X3
4, 2
2, 0
X4
Y1
3, 1
Y2
2, 1
3, 1
Y3
Y4
2, 2
1, 1
Y5
2, 3
X5
Y6
6, 4
1, 3
4, 9
1, 3
Y7
X6
4, 5
Y8
8, 11
4, 7
a. Test the null hypothesis of no selection on this gene using the McDonald Kreitman test. If
significant selection is detected, identify whether or not it is conservative (negative) or
directional (positive).
a. Set up the contingency table and calculate the contingency chi-square with 1 degree of
freedom (1 pt for df):
Synonymous
NonSynonymous total
expected 1
expected 2
Polymorphic
35
35
70
32.90
37.10
Fixed
12
18
30
14.10
15.90
total
47
53
100
2 pt s
chi-square
p value
0.843034926
0.358530224
2 pt s
1 pt
1 pt
The chi-square is not significant, so we cannot reject the null hypothesis of neutrality (1 pt).
b. Test the null hypothesis of no selection on this gene using the Old/Young classification test.
If significant selection is detected, identify whether or not it is conservative (negative) or
directional (positive).
b. Set up the contingency table and calculate the contingency chi-square with 1 degree of
freedom (1 pt for df):
Synonymous
NonSynonymous total
expected 1
expected 2
young
25
15
40
18.8
21.2
old
22
38
60
28.2
31.8
total
47
53
100
2 pt s
chi-square
p value
6.429813997
0.011222034
2 pt s
1 pt
1 pt
The chi-square is significant at the 5% level, so we reject neutrality (1 pt). There is an
excess of nonsynonymous substitutions over synonymous in the old (evolutionarily
successful) part of the tree, implying that selection is positive and favors amino acid
changes in this protein (2 pts).
c. Test the null hypothesis of no selection on this gene using the fixed/intraspecific interior/tip
test. If significant selection is detected, identify whether or not it is conservative (negative) or
directional (positive).
c. Set up the contingency table and calculate the contingency chi-square with 2 degrees of
freedom (1 pt for df):
Tip
Int
Fixed
total
chi-square
p value
NonSynonymous Synonymous total
25
15
40
10
20
30
12
18
30
47
53
100
2 pt s
6.697444132
1 pt
0.035129218
1 pt
expected 1 expected 2
18.80
21.20
14.10
15.90
14.10
15.90
2 pt s
The chi-square is significant at the 5% level, so we reject neutrality (1 pt). There is an
excess of nonsynonymous substitutions over synonymous in the interior and fixed
(evolutionarily successful) parts of the tree, implying that selection is positive and favors
amino acid changes in this protein (2 pts).