Transcript Lab 7: Mutation, Selection and Drift

Lab 10: Mutation, Selection
and Drift
Goals
1. Effect of mutation on allele frequency.
2. Effect of mutation and selection on allele frequency.
3. Effect of mutation and selection and drift on allele
frequency.
4. Compare mutation rates in different groups.
Change in allele frequency due to mutation
p
μ
A1
n
q
A2
q  p nq
where:
 (mu) = mutation
n (nu) = reverse mutation
Change in allele frequency due to mutation
q  p nq
where:
 (mu) = mutation
n (nu) = reverse mutation
at equilibrium:
μ
qeq  μ  ν
q = 0
Assumptions:
1. Large population size.
2. No selection.
Problem 1. Eye color in humans was once believed to be controlled by a single gene,
with the brown eye allele being the dominant wild-type. Recent studies, however,
revealed that eye color is actually a polygenic trait. Although 74% of the variation for
eye color is determined by the Eye Color 3 (EYCL3) locus located on chromosome 15
(with most variation explained by only 3 single nucleotide polymorphisms in the OCA2
gene!), there are at least five other loci with weaker but significant control over this
trait.
The frequency of the recessive allele associated with blue eyes (let this be allele A2)
is 0.83 in Europeans.
Assuming that an equilibrium has been reached between forward and backward
mutation, and the rate of forward mutation (i.e., from wild-type A1 alleles to A2 alleles)
is μ = 10-5, calculate the rate of backward mutation (i.e., from A2 to A1).
What are the assumptions of this calculation?
Change in allele frequency due to mutation and selection
p
A1
q
μ
A2
n
Selection (h,s)
p’
A1
μ
n
q’
A2
Change in allele frequency due to mutation
and selection
Case 1: When ν ≈ 0; population size is large and h=0 (A1
completely dominant to A2):
q eq 

s
Case 2: When ν ≈ 0; population size is large and h>>0 :
q eq 

hs
Problem 2. Sequence analysis of EYCL3 reveals that this locus is likely to have been
under very strong selection, and the allele associated with blue eyes (as well as with
light brown hair and pale skin color) is likely to have been favored in Europe, but not
in Africa and East Asia.
Let us assume that melanoma (skin cancer, which is more likely to develop in people
with light skin color) reverses the direction of selection and the blue eye/light skin
allele now becomes selected against with s = 0.12.
Calculate the equilibrium value of q (the frequency of the blue eye allele A2) in an
infinitely large population if the rate of forward mutations is μ = 10-6, the rate of
backward mutations is ν = 0, and if:
a) A1 is completely dominant to A2.
b) The allele effects are additive (i.e., h=0.5). Provide output from Populus to support
your calculations.
c) If the equilibrium frequencies of A2 in a) and b) are different, explain why. If not,
explain why not.
d) Include graphs from Populus showing the approach to equilibrium for each case
based on 6 different initial allele frequencies. Under which type of dominance is
equilibrium approached faster and why?
Mutation, selection and genetic drift
p
A1
q
μ
A2
n
Drift
xij  P(Yt 1  i | Yt  j ) 
(2 N )!
 p'i q'2 N i
(2 N  i)!i!
p’
A1
Selection (h,s)
μ
n
q’
A2
Mutation, selection and genetic drift
Mean frequency of A2 (harmful mutant) allele,
when h=0 and Ne and n are very small :
2N e
q
s
Mean frequency ofA2 allele, when h >> √s and n
is very small:
q 

hs
Problem 3. Starting with the conditions in Problem 2, calculate
the expected value of q in a finite population if:
a) A1 is completely dominant to A2 and Ne = 50.
b) Same as a) but with Ne = 10,000.
c) There is additivity, and Ne = 50.
d) Same as c) but with Ne = 10,000.
Comparing mutation rates in different groups
Mutation rates can be estimated directly by comparing the
genotypes of parents to those of their offsprings.
Sometimes it is of great interest to compare rates of
mutation between different groups (e.g., between a group
exposed to radiation and a control group).
This can be done using contingency table tests.
Two methods to analyze contingency tables are the
Chi-square test and Fisher’s exact test.
Example:
Group 1(Control)
Group 2 (Exposed group)
No. of
No. of alleles No. of
mutations tested
mutations
No. of alleles
tested
3
60
50
5
(Oi  Ei )
 
Ei
i 1
4
2
2
df = (number of rows − 1)(number of columns − 1) = 1
2x2 contingency table and χ2 test
Om Ou
gr1
3
47
gr2
5
55
Em
Eu
(50*8)/110= (50*102)/110
3.63
= 46.36
(60*8)/110= (60*102)/110
4.36
= 55.63
((Om-Em)^2)/Em
((Ou-Eu)^2)/Eu
0.111
0.008
0.092
0.007
0.204
0.016
chi square
0.2202
Here, the calculated χ2 value (0.2202) is less than the critical value
(3.8415) for α=0.05. Hence we fail to reject null hypothesis i.e.
there is no significant difference in mutation rates of the two groups.
Problem 4. The following numbers of mutations have been observed in minisatellite
loci of humans exposed to radiation in Chernobyl compared to a control group:
Control
# mutations
# alleles tested
Alleles inherited 9
701
from mother
Alleles inherited 26
706
from father
Exposed
# mutations
# alleles tested
28
1727
108
1748
Use the Chi-square test to determine whether mutation rates differ significantly
between:
a) Alleles inherited from the mother and alleles inherited from the father in the control
group.
b) Alleles inherited from the mother and alleles inherited from the father in the
exposed group.
c) The control and the exposed groups for alleles inherited from the mother.
d) The control and the exposed groups for alleles inherited from the father.
e) GRADUATE STUDENTS ONLY: Repeat all tests using Fisher’s exact test and
compare the P-values to those obtained by Chi-square tests.
Provide both a statistical and a biological interpretation for each test.