Page 22 - Applied Probability

1. Basic Principles of Population Genetics
(g) equal initial genotype frequencies in the two sexes. Suppose for the sake
of simplicity that there are two alleles A 1 and A 2 at some autosomal locus
in this infinite population and that all genotypes are unordered. Consider
the result of crossing the genotype A 1 /A 1 with the genotype A 1 /A 2 . The
first genotype produces only A 1 gametes, and the second genotype yields
gametes A 1 and A 2 in equal proportion. For the cross under consideration, 5
gametes produced by the genotype A 1 /A 1 are equally likely to combine
with either gamete type issuing from the genotype A 1 /A 2 . Thus, for the
1
cross A 1 /A 1 × A 1 /A 2 , the frequency of offspring obviously is A 1 /A 1 and
2
1
2 A 1 /A 2 . Similarly, the cross A 1 /A 1 × A 2 /A 2 yields only A 1 /A 2 offspring.
1
1
The cross A 1 /A 2 ×A 1 /A 2 produces offspring in the ratio A 1 /A 1 , A 1 /A 2 ,
4
2
1
and A 2 /A 2 . These proportions of outcomes for the various possible crosses
4
are known as segregation ratios.
TABLE 1.2. Mating Outcomes for Hardy-Weinberg Equilibrium
Mating Type Nature of Offspring Frequency
u 2
A 1 /A 1 × A 1 /A 1 A 1 /A 1
1 1
A 1 /A 1 × A 1 /A 2 A 1 /A 1 + A 1 /A 2 2uv
2 2
2uw
A 1 /A 1 × A 2 /A 2 A 1 /A 2
1 1 1 2
A 1 /A 2 × A 1 /A 2 A 1 /A 1 + A 1 /A 2 + A 2 /A 2 v
4 2 4
1 1
A 1 /A 2 × A 2 /A 2 A 1 /A 2 + A 2 /A 2 2vw
2 2 2
w
A 2 /A 2 × A 2 /A 2 A 2 /A 2
Suppose the initial proportions of the genotypes are u for A 1 /A 1 , v for
A 1 /A 2 , and w for A 2 /A 2 . Under the stated assumptions, the next genera-
tion will be composed as shown in Table 1.2. The entries in Table 1.2 yield
for the three genotypes A 1 /A 1 , A 1 /A 2 , and A 2 /A 2 the new frequencies
1 2 1 2
2
u + uv + v = u + v
4 2
1 2 1 1
uv +2uw + v + vw =2 u + v v + w
2 2 2
1 2 2 1 2
v + vw + w = v + w ,
4 2
respectively. If we define the frequencies of the two alleles A 1 and A 2 as
2
p 1 = u + v and p 2 = v + w, then A 1 /A 1 occurs with frequency p , A 1 /A 2
2 2 1
2
with frequency 2p 1p 2 , and A 2 /A 2 with frequency p . After a second round
2
of random mating, the frequencies of the genotypes A 1 /A 1 , A 1 /A 2 , and
A 2 /A 2 are
1 2
2
2
p + 2p 1p 2 = p 1 (p 1 + p 2 )
1
2
= p 2 1