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, selection and drift
on allele frequency.
4. Compare mutation rates in different
groups.
Change in allele frequency due to mutation
p
μ
A1
v
q  p q
q
A2
Change in allele frequency due to mutation
q  p q
Where,
µ = rate of forward mutation (From A1 to A2)
ν = rate of backward mutation (From A2 to A1)
At equilibrium, ∆q=0,
μ
qeq  μ  ν
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!), there are at least two 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 = 10-6, calculate the rate of
backward mutation (i.e., from A2 to A1). (5 minutes)
Given, A1- Brown eye ; A2- Blue Eye; A1>A2,
qeq = 0.83
μ (A1 to A2) = 10-5
ν(A2 to A1) = ?
Change in allele frequency due to mutation and selection
p
A1
q
μ
A2
v
Selection (h,s)
p’
A1
μ
v
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. There is additivity.
c. If the equilibrium frequencies of A2 in a) and b) are different, explain
why. If not, explain why not.
d. Under which type of dominance is equilibrium approached faster and
why?
Provide output from Populus to support your calculations.
In Populus, use value of s, h, and μ and for enough
# of generations with six initial allele frequency.
Mutation, selection and genetic drift
p
A1
q
μ
A2
v
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
v
q’
A2
Mutation, selection and genetic drift
Mean frequency of A2 (harmful mutant) allele,
when h=0 and Ne and ν are very small :
2N e
q
s
Mean frequency ofA2 allele, when h >> √µs and ν
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.
How does Ne affect the mean frequency of A2 with each
type of dominance? Under which level of dominance does Ne
have a stronger effect? Why? Provide output from PopG to
support your interpretation.(15 minutes).
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:
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.
Control
No. of
No. of alleles
mutations tested
Exposed
No. of
No. of alleles
mutations
tested
Alleles inherited
from mother
9
721
26
1650
Alleles inherited
from father
26
827
141
1701