Genotype Frequencies After Self-Fertilization

This problem asks us to determine the frequencies of the AA, Aa, and aa genotypes in a population after three generations of repeated self-fertilization, starting with a founder population where all individuals are heterozygous (Aa) with a frequency of 1. Initially, the frequencies are:

• Frequency of AA ($\text{P}_0$): 0
• Frequency of Aa ($\text{H}_0$): 1
• Frequency of aa ($\text{Q}_0$): 0

Impact of Self-Fertilization

When an individual self-fertilizes, the offspring genotypes are determined by the parent's genotype.

• An AA parent produces only AA offspring.
• An aa parent produces only aa offspring.
• An Aa parent (Aa x Aa cross) produces offspring with the following genotype probabilities: 1/4 AA, 1/2 Aa, 1/4 aa.

Let $\text{P}_n$, $\text{H}_n$, and $\text{Q}_n$ be the frequencies of AA, Aa, and aa genotypes in generation $n$, respectively. In the next generation ($n+1$), the frequencies change as follows:

The frequency of AA in generation $n+1$ ($\text{P}_{n+1}$) comes from AA individuals in generation $n$ selfing (producing all AA) and Aa individuals in generation $n$ selfing (producing 1/4 AA). $\text{P}_{n+1} = \text{P}_n \times 1 + \text{H}_n \times \frac{1}{4}$

The frequency of Aa in generation $n+1$ ($\text{H}_{n+1}$) comes only from Aa individuals in generation $n$ selfing (producing 1/2 Aa). $\text{H}_{n+1} = \text{H}_n \times \frac{1}{2}$

The frequency of aa in generation $n+1$ ($\text{Q}_{n+1}$) comes from aa individuals in generation $n$ selfing (producing all aa) and Aa individuals in generation $n$ selfing (producing 1/4 aa). $\text{Q}_{n+1} = \text{Q}_n \times 1 + \text{H}_n \times \frac{1}{4}$

Calculating Frequencies Across Generations

We start with generation 0: $\text{P}_0 = 0$, $\text{H}_0 = 1$, $\text{Q}_0 = 0$.

Generation 1 (n=1):

Using the formulas:

$\text{P}_1 = \text{P}_0 + \frac{1}{4}\text{H}_0 = 0 + \frac{1}{4} \times 1 = \frac{1}{4}$

$\text{H}_1 = \frac{1}{2}\text{H}_0 = \frac{1}{2} \times 1 = \frac{1}{2}$

$\text{Q}_1 = \text{Q}_0 + \frac{1}{4}\text{H}_0 = 0 + \frac{1}{4} \times 1 = \frac{1}{4}$

Frequencies after generation 1: AA = 1/4, Aa = 1/2, aa = 1/4.

Generation 2 (n=2):

Using the frequencies from generation 1: $\text{P}_1 = 1/4$, $\text{H}_1 = 1/2$, $\text{Q}_1 = 1/4$.

$\text{P}_2 = \text{P}_1 + \frac{1}{4}\text{H}_1 = \frac{1}{4} + \frac{1}{4} \times \frac{1}{2} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

$\text{H}_2 = \frac{1}{2}\text{H}_1 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$\text{Q}_2 = \text{Q}_1 + \frac{1}{4}\text{H}_1 = \frac{1}{4} + \frac{1}{4} \times \frac{1}{2} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

Frequencies after generation 2: AA = 3/8, Aa = 1/4, aa = 3/8.

Generation 3 (n=3):

Using the frequencies from generation 2: $\text{P}_2 = 3/8$, $\text{H}_2 = 1/4$, $\text{Q}_2 = 3/8$.

$\text{P}_3 = \text{P}_2 + \frac{1}{4}\text{H}_2 = \frac{3}{8} + \frac{1}{4} \times \frac{1}{4} = \frac{3}{8} + \frac{1}{16} = \frac{6}{16} + \frac{1}{16} = \frac{7}{16}$

$\text{H}_3 = \frac{1}{2}\text{H}_2 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

$\text{Q}_3 = \text{Q}_2 + \frac{1}{4}\text{H}_2 = \frac{3}{8} + \frac{1}{4} \times \frac{1}{4} = \frac{3}{8} + \frac{1}{16} = \frac{6}{16} + \frac{1}{16} = \frac{7}{16}$

Frequencies after three generations: AA = 7/16, Aa = 1/8, aa = 7/16.

Summary of Frequencies After Three Generations

These are the genotype frequencies after three generations of repeated self-fertilization starting from a population of only Aa heterozygotes. As expected with self-fertilization, the frequency of the heterozygote (Aa) decreases by half each generation, while the frequencies of the homozygotes (AA and aa) increase. The increase in AA and aa frequencies is equal because the initial population had equal numbers of A and a alleles (since all individuals were Aa).