BCOR 102 Homework 4, Migration and drift
Due Wednesday March 1
1. Imagine two populations that each exchange a fraction m migrants each
generation. Migration is symmetrical, so the same fraction travels into and out of
the two populations. Right now the two populations are fixed for alternative
alleles at the “G” locus. In population 1, all individuals have genotype GG and in
population 2 all individuals have genotype gg.
a. What are the new allele frequencies after one generation of migration and
after two generations of migration?
b. What will the allele frequency be after many generations of migrations?
m
Pop 1
Pop 2
m
Let p=freq(G)
Starting frequencies: p1=1.0, p2=0.0
We don’t have a specific number for migration rate so we will use “m”
P1’ = (1-m) p1 + m(p2) = 1-m
P2’ = (1-m) p2 + m(p1) = m
By similar reasoning, you can calculate the third generation frequencies:
P1’’ = (1-m) p1’ + m(p2’) = (1-m)(1-m) + m(m) = 1-2m+2m2
P2’’ = (1-m)m + m(1-m) = 2 m (1-m)
b. after many generations, the frequency will be the average of the two starting
frequencies, since migration is symmetrical both directioins. p1* = p2* = 0.5
2. A large population of prairie flowers contained two alleles for flower color, B and
b. The allele frequencies were 0.6 and 0.4, respectively. After the prairie was
plowed into farm fields, the plant was restricted to hedgerows and the original
large population was split into several dozen small, isolated, populations. Assume
that flower color is completely neutral and that there is no gene flow among the
small populations.
a. What will be the likely frequencies of alleles in the small populations after
many years of isolation?
After many many years, the subpopulations will become fixed for one allele or the other:
either all BB or all bb. Because the probablility of fixation of an allele is equal to p0,
about 60% of the populations will be fixed for B and 40% will become fixed for b.
3. Here are some genetic data showing the number of individuals of each genotype
at the “B” locus in two different populations.
Population
BB
Bb
BB
1
128
64
4
2
4
64
128
a. Calculate the expected heterozygosity in each subpopulation. How does
that compare to the observed heterozygosity? Allele frequencies for the
two populations: p1=0.8; p2=0.2 The expected heterozygosity in each
population is 2pq = 0.32.
In each population the observed heterozygosity is 64/196 = 0.32 (same as
expected)
b. Calculate the expected heterozygosity for the species as a whole (first you
need to determine the overall allele frequency). How does the expected
heterozygosity compare to the observed overall heterozygosity? For the
species as a whole, the combined allele frequency is 0.5. That makes the
expected heterozygosity = 2pq = 0.5.
The observed heterozygosity overall is 128/396 = 0.326
c. Why are they different? Because the allele frequencies are different in the
two populations.
d. What is Fst for these populations? Fst = (HT - HS)/HT = 0.36
(Remember that HT- and HS are the expected heterozygosity in the species
as a whole and the average expected heterozygosity in the
subpopulations.)
e. Assuming equilibrium between migration and drift, what is the value of
Nm, the number of migrants?
At migration drift equilibrium, Fst = 1 / (4Nm + 1
So, Nm=(1-Fst)/(4 Fst)
Nm= 0.444