Selfing And The Decline Of Heterozygosity

Introduction

Self-fertilization changes how alleles are paired into genotypes. When a heterozygous individual produces offspring by selfing, some offspring are homozygous even though the parent carried one copy of each allele. Repeating this process steadily reduces the fraction of heterozygotes. Yet, without forces that change the representation of alleles themselves, the allele frequencies can remain constant. Separating these two patterns is central to understanding selfing.

We will describe a single locus with two alleles, \(A\) and \(a\), in a population that undergoes complete self-fertilization. The key result is simple: each generation has half the heterozygote frequency of the preceding generation. The reasoning also shows why this decline need not imply any change in allele frequency.

The Selfing Model

Definition: Let \(f_{AA,t}\), \(f_{Aa,t}\), and \(f_{aa,t}\) be the frequencies of genotypes \(AA\), \(Aa\), and \(aa\) in generation \(t\). These frequencies sum to one. Write \(H_t=f_{Aa,t}\) for the heterozygote frequency, and let \(p_t=f_{AA,t}+\frac{1}{2}f_{Aa,t}\) be the frequency of allele \(A\). The frequency of allele \(a\) is \(q_t=1-p_t\).

Generation \(t+1\) consists of the offspring produced by selfing in generation \(t\). Assume every individual self-fertilizes, all individuals contribute equally to the next generation, and there is no selection, mutation, migration, or other process that changes which genotypes contribute. We use deterministic population frequencies: the calculations describe the frequencies implied by the model, rather than random deviations in a finite set of offspring.

The offspring genotype probabilities follow from Mendelian transmission. An \(AA\) parent produces only \(A\) gametes, so all its offspring are \(AA\); an \(aa\) parent produces only \(a\) gametes, so all its offspring are \(aa\). An \(Aa\) parent produces \(A\) and \(a\) gametes with probability one-half each. When those gametes unite in selfing, the offspring probabilities are \(1/4\) \(AA\), \(1/2\) \(Aa\), and \(1/4\) \(aa\).

Weighting these outcomes by the genotype frequencies of the parents gives the one-generation recurrences:

\begin{aligned} f_{AA,t+1} &= f_{AA,t}+\frac{1}{4}f_{Aa,t},\\ f_{Aa,t+1} &= \frac{1}{2}f_{Aa,t},\\ f_{aa,t+1} &= f_{aa,t}+\frac{1}{4}f_{Aa,t}. \end{aligned}

For example, the next generation's \(AA\) frequency includes all offspring of \(AA\) parents and one-quarter of the offspring produced by \(Aa\) parents. The corresponding reasoning gives the other two expressions. The three next-generation frequencies still sum to one: their sum is \(f_{AA,t}+f_{Aa,t}+f_{aa,t}\).

Why Heterozygosity Halves and Allele Frequency Is Preserved

Only heterozygous parents produce heterozygous offspring under this model, and each such parent produces heterozygous offspring with probability one-half. Therefore, the heterozygote frequency obeys \(H_{t+1}=H_t/2\). Repeated substitution gives an expression for any number of generations.

Result: Under complete selfing with equal contribution and no allele-changing forces, after \(t\) generations the heterozygote frequency is \(H_t=H_0(1/2)^t\). The allele frequency is constant: \(p_t=p_0\) and \(q_t=q_0\).

To check allele-frequency preservation directly, calculate the frequency of \(A\) in generation \(t+1\) from its genotype frequencies:

\begin{aligned} p_{t+1} &=f_{AA,t+1}+\frac{1}{2}f_{Aa,t+1}\\ &=f_{AA,t}+\frac{1}{4}f_{Aa,t} +\frac{1}{2}\left(\frac{1}{2}f_{Aa,t}\right)\\ &=f_{AA,t}+\frac{1}{2}f_{Aa,t}\\ &=p_t. \end{aligned}

The same conclusion has a transmission interpretation: each parent contributes gametes whose allele frequencies match the parent's own allele frequencies. When all parents contribute equally, the average frequency of \(A\) among those gametes is \(p_t\). Selfing changes how gametes are paired, not the overall proportion of \(A\) alleles in the gamete pool. Since \(q_t=1-p_t\), preserving \(p_t\) also preserves \(q_t\).

The full sequence follows by induction. The recurrence gives \(H_1=H_0/2\). If \(H_t=H_0(1/2)^t\), then \(H_{t+1}=H_t/2=H_0(1/2)^{t+1}\); thus the expression holds for every nonnegative integer \(t\). The allele-frequency equality follows by applying \(p_{t+1}=p_t\) at each generation.

Genotype Frequencies Across Generations

The recurrences also give genotype frequencies directly when the initial frequencies are known. Because the allele frequency \(p\) is fixed and \(H_t\) is known, the definition \(p=f_{AA,t}+H_t/2\) gives \(f_{AA,t}=p-H_t/2\). The frequencies must sum to one, so \(f_{aa,t}=1-p-H_t/2\). Hence:

f_{AA,t}=p_0-\frac{H_0}{2^{t+1}},\qquad f_{Aa,t}=\frac{H_0}{2^t},\qquad f_{aa,t}=q_0-\frac{H_0}{2^{t+1}}.

These formulas do not require the initial population to be in Hardy–Weinberg equilibrium. They require only valid initial genotype frequencies and the stated selfing model. They show that the lost heterozygote frequency is redistributed equally between the two homozygote classes: at each generation, each homozygote class gains one-quarter of the preceding heterozygote frequency.

Worked Example: Following an Initially Non-Equilibrium Population

Suppose a hypothetical population begins with \(f_{AA,0}=0.30\), \(f_{Aa,0}=0.50\), and \(f_{aa,0}=0.20\). Assume complete selfing under the model above. Find its allele frequency and genotype frequencies after three generations.

First, the initial frequency of \(A\) is \(p_0=f_{AA,0}+H_0/2=0.30+0.50/2=0.55\), so \(q_0=0.45\). The initial genotype frequencies sum to \(0.30+0.50+0.20=1\), as required.

After three generations, heterozygosity is \(H_3=H_0(1/2)^3=0.50/8=0.0625\). Thus:

\begin{aligned} f_{AA,3}&=p_0-\frac{H_3}{2}=0.55-\frac{0.0625}{2}=0.51875,\\ f_{Aa,3}&=H_3=0.0625,\\ f_{aa,3}&=q_0-\frac{H_3}{2}=0.45-\frac{0.0625}{2}=0.41875. \end{aligned}

The values sum to \(0.51875+0.0625+0.41875=1.0000\). The \(A\)-allele frequency is \(0.51875+0.0625/2=0.5500\), equal to its initial value. The initial genotype frequencies were not Hardy–Weinberg proportions, but the selfing recurrences still apply.

Worked Example: Starting in Hardy–Weinberg Equilibrium

Consider a different hypothetical population that begins in Hardy–Weinberg equilibrium with \(p_0=0.70\) and \(q_0=0.30\). Calculate its genotype frequencies after four generations of complete selfing.

Initially, \(H_0=2p_0q_0=2(0.70)(0.30)=0.42\). After four generations, \(H_4=0.42(1/2)^4=0.42/16=0.02625\). Since \(p\) remains \(0.70\) and \(q\) remains \(0.30\), the genotype frequencies are:

\begin{aligned} f_{AA,4}&=0.70-\frac{0.02625}{2}=0.686875,\\ f_{Aa,4}&=0.02625,\\ f_{aa,4}&=0.30-\frac{0.02625}{2}=0.286875. \end{aligned}

They sum to \(1.000000\), and the \(A\)-allele frequency is \(0.686875+0.02625/2=0.700000\). The initial Hardy–Weinberg frequencies were \(0.49\), \(0.42\), and \(0.09\); after selfing, they are no longer those proportions. Hardy–Weinberg equilibrium is not maintained by repeated selfing, even though allele frequencies are preserved.

Worked Example: How Many Generations Until Heterozygosity Falls Below a Threshold?

A hypothetical population has an initial heterozygote frequency of \(H_0=0.40\). Under complete selfing, find the first generation in which its heterozygote frequency is below \(0.05\).

Use \(H_t=0.40/2^t\). At generation three, \(H_3=0.40/8=0.0500\), which equals the threshold and is not below it. At generation four, \(H_4=0.40/16=0.0250\), which is below \(0.05\). Therefore, the first such generation is generation four. Each step halves the heterozygote frequency; no allele-frequency value is needed to answer this question.

Interpreting the Decline

Selfing progressively converts heterozygotes into homozygotes, while leaving the overall allele frequencies unchanged under the model assumptions. The value \(H_0/2^t\) approaches zero as \(t\) increases, but for a positive \(H_0\) it is not exactly zero after any finite number of generations. The increase in homozygosity is not evidence, by itself, that an allele has become more common: both homozygote classes receive part of the heterozygote decline.

If the initial population is in Hardy–Weinberg equilibrium, so that \(H_0=2p_0q_0\), the genotype frequencies can also be written:

\begin{aligned} f_{Aa,t}&=2p_0q_0\left(\frac{1}{2}\right)^t,\\ f_{AA,t}&=p_0^2+p_0q_0\left(1-\left(\frac{1}{2}\right)^t\right),\\ f_{aa,t}&=q_0^2+p_0q_0\left(1-\left(\frac{1}{2}\right)^t\right). \end{aligned}

This is a special case of the general recurrence, not an assumption that each later generation is in Hardy–Weinberg equilibrium. In the notation of the inbreeding coefficient, it corresponds to \(F_t=1-(1/2)^t\) when the starting generation is in Hardy–Weinberg equilibrium. The expression summarizes the growing heterozygote deficit relative to that starting equilibrium.

Common Mistake and Model Limits

Mistake: “Heterozygotes decline, so the frequency of allele \(A\) must decline.” A change in genotype proportions does not itself imply a change in allele proportions. The calculation \(p_t=f_{AA,t}+H_t/2\) shows that the increase in \(AA\) frequency offsets the loss of \(A\) copies from the declining heterozygote class.
Mistake: “The population is in Hardy–Weinberg equilibrium because \(p\) and \(q\) stay fixed.” Hardy–Weinberg proportions require particular genotype frequencies, not merely fixed allele frequencies. The example starting at \(p=0.70\) shows that selfing changes those proportions.
Mistake: “The halving rule applies under any mating system.” The recurrence \(H_{t+1}=H_t/2\) was derived from complete selfing of every individual. A different mating pattern or partial selfing requires a different recurrence.

The model also assumes equal contribution and no selection or other process that changes genotype contributions. If genotypes differ in survival or reproduction, or if alleles enter or leave the population, the simple preservation of \(p\) need not hold. In a finite population, realized offspring counts may also vary around the deterministic frequencies used here.

Key takeaway: Under complete selfing with equal contribution and no allele-changing forces, \(H_t=H_0/2^t\), while \(p_t=p_0\) and \(q_t=q_0\). Selfing changes genotype frequencies by reducing heterozygosity; it does not, by itself, change allele frequencies.

Check Your Understanding

Use the recurrence and its assumptions to answer the following questions.

A population has \(H_0=0.32\). What is its heterozygote frequency after two generations of complete selfing?
For initial frequencies \(f_{AA,0}=0.25\), \(f_{Aa,0}=0.50\), and \(f_{aa,0}=0.25\), calculate \(p_0\). What is \(p_2\) under the selfing model?
Explain why a rise in \(f_{AA,t}\) does not necessarily mean the \(A\)-allele frequency has risen.
Does starting in Hardy–Weinberg equilibrium imply remaining in Hardy–Weinberg equilibrium through repeated selfing? Explain.
Name one assumption whose failure could make the allele-frequency preservation result inapplicable.