Evolution__3rd_Edition

..
A simpler proof of the Hardy–
Weinberg equilibrium
CHAPTER 5 / The Theory of Natural Selection 101
Diploid organisms produce haploid gametes. We could imagine that the haploid
gametes are all released into the sea, where they combine at random to form the next
generation. This is called random union of gametes. In the “gamete pool” A gametes
will have frequency p and a gametes frequency q. Because they are combining at
random, an a gamete will meet an A gamete with chance p and an a gamete with chance
q. From the a gametes, Aa zygotes will therefore be produced with frequency pq and
aa gametes with frequency q 2 . A similar argument applies for the A gametes (which
have frequency p): they combine with a gametes with chance q, to produce Aa zygotes
(frequency pq) and A gametes with chance p to form AA zygotes (frequency p 2 ). If we
now add up the frequencies of the genotypes from the two types of gamete, the Hardy–
Weinberg genotype frequencies emerge. We have now derived the Hardy–Weinberg
theorem for the case of two alleles; the same argument easily extends to three or more
alleles (Box 5.1).
(Some people may be puzzled by the 2 in the frequency of the heterozygotes. It is
a simple combinatorial probability. Imagine flipping two coins and asking what the
chances are of flipping two heads, or two tails, or one head and one tail. The chance of
two heads is (1/2) 2 and of two tails (1/2) 2 ; the chance of a head and a tail is 2 × (1/2) 2 ,
because a tail then a head, and a head then a tail, both give one head and one tail. The
head is analogous to allele A, the tail to a; two heads to producing an AA genotype, and
one head and one tail to a heterozygote Aa. The coin produces heads with probability
1/2, and is analogous to a gene frequency of p = 1/2. The frequency 2pq for heterozygotes
is analogous to the chance of one head and one tail, 2 × (1/2) 2 . The 2 arises
because there are two ways of obtaining one head and one tail. Likewise there are two
ways of producing an Aa heterozygote: either the A gene can come from the father and
the a from the mother, or the a gene from the father and the A from the mother. The
offspring is Aa either way.)
Box 5.1
The Hardy–Weinberg Theorem for Three Alleles
We can call the three alleles A 1 , A 2 , and A 3 , and define their gene
frequencies as p, q, and r, respectively. We form new zygotes by
sampling two successive gametes from a large pool of gametes.
The first gamete we pick could be A 1 , A 2 , or A 3 . If we first pick (with
chance p) an A 1 allele from the gamete pool, the chance that the
second allele is another A 1 allele is p, the chance that it is an A 2
allele is q, and the chance that it is an A 3 allele is r: from these three,
the frequencies of A 1 A 1 , A 1 A 2 , and A 1 A 3 zygotes are p 2 , pq, and pr.
Now suppose that the first allele we picked out had been an A 2
(which would happen with chance q). The chances that the second
allele would again be A 1 , A 2 , or A 3 would be p, q, and r, respectively,
giving A 1 A 2 , A 2 A 2 , and A 2 A 3 zygotes in frequency pq, q 2 , and qr.
Finally, if we had picked (with chance r) an A 3 allele, we produce
A 1 A 3 , A 2 A 3 , and A 3 A 3 zygotes in frequency pr, qr, and r 2 .
The only way to form the homozygotes A 1 A 1 , A 2 A 2 , and A 3 A 3 is by
picking two of the same kind of gamete and the frequencies are p 2 ,
q 2 , and r 2 . The heterozygotes can be formed from more than one
kind of first gamete and their frequencies are obtained by addition.
The total chance of forming an A 1 A 3 zygote is pr + rp = 2pr; of
forming an A 1 A 2 zygote is pq + qp = 2pq; and of an A 2 A 3 zygote is
2qr. The complete Hardy–Weinberg proportions are:
A 1 A 1 : A 1 A 2 : A 1 A 3 :A 2 A 2 : A 2 A 3 : A 3 A 3
p 2 2pq 2pr q 2 2qr r 2