Matthew B. Hamilton

Population Genetics


Скачать книгу

Why does Hardy–Weinberg work?

       A proof of Hardy–Weinberg.

       Hardy–Weinberg with more than two alleles.

      The Hardy–Weinberg equation is one of the most basic expectations we have in population genetics. It is very likely that you were already familiar with the Hardy–Weinberg equation before you picked up this book. But where does Hardy–Weinberg actually come from? What is the logic behind it? Let's develop a simple proof that Hardy–Weinberg is actually true. This will also be our first real foray into the type of the algebraic argument that much of population genetics in built on. Given that you start out knowing the conclusion of the Hardy–Weinberg tale, this gives you the opportunity to focus on the style in which it is told. Algebraic or quantitative arguments are a central part of the language and vocabulary of population genetics, so part of the task of learning population genetics is becoming accustomed to this mode of discourse.

      1 mating is random (parents meet and mate according to their frequencies);

      2 all parents have the same number of offspring (equivalent to no natural selection on fecundity);

      3 all progeny are equally fit (equivalent to no natural selection on viability);

      4 there is no mutation that could act to change an A to a or an a to A;

      5 it is a single population that is very large;

      6 there are two and only two mating types.

      Now, let's define the variables we will need for a case with one locus that has two alleles (A and a).

      N = Population size of individuals (N diploid individuals have 2N alleles)

Allele frequencies
p = frequency(A allele) = (total number of A alleles)/2N
q = frequency(a allele) = (total number of a alleles)/2N
p + q = 1
Genotype frequencies
X = frequency(AA genotype) = (total number of AA genotypes)/N
Y = frequency(Aa genotype) = (total number of Aa genotypes)/N
Z = frequency(aa genotype) = (total number of aa genotypes)/N
X + Y + Z = 1

      We do not distinguish between the heterozygotes Aa and aA and treat them as being equivalent genotypes. Therefore, we can express allele frequencies in terms of genotype frequencies by adding together the frequencies of A‐containing and a‐containing genotypes:

      (2.2)equation

      (2.3)equation

      Each homozygote contains two alleles of the same type, while each heterozygote contains one allele of each type so the heterozygote genotypes are each weighted by half.

Schematic illustration of random mating as a cloud of gas where the frequency of A's is 14/24 and the frequency of a's 10/24.

      A parental mating frequency table (generation t) is shown below.

Moms Frequency Dads
AA Aa aa
X Y Z
AA X X 2 XY XZ
Aa Y XY Y 2 YZ
Aa Z ZX ZY Z 2

      Next, we need to determine the frequency of each genotype in the offspring of any given parental mating pair. This will require that we predict the offspring genotypes resulting from each possible parental mating. We can do this easily with a Punnett square. We will use the frequencies of each parental mating (above) together with the frequencies of the offspring genotypes. Summed for all possible parental matings, this gives the frequency of offspring genotypes one generation later, or in generation t + 1. A table will help organize all the frequencies, like the offspring frequency table (generation t