Assortative Mating And Genotype Frequencies

Introduction

Random pairing is a useful reference for genotype frequencies, but mates need not pair at random with respect to traits related to genotype. When similar types pair more often, positive assortative mating can produce more homozygotes and fewer heterozygotes than the random-pairing baseline. When unlike types pair more often, negative assortative mating can shift frequencies in the opposite direction.

The key is to describe how the alleles in the two gametes that form a zygote are paired. This gives a direct way to calculate genotype frequencies and to distinguish a change in genotype proportions from a change in allele frequency. The calculations below concern one generation of offspring under a specified mating pattern; they do not assume that the same pattern continues indefinitely.

This definition describes allele pairing at the locus. Mating based on a trait does not automatically imply a particular value of \(C\): the association depends on how that trait is related to the alleles and on which individuals mate. For the calculations here, assume both gamete pools have the same allele frequency \(p\), alleles are transmitted according to ordinary Mendelian segregation, and there is no allele-specific transmission, mutation, or selection between gamete production and the formation of the zygotes.

Genotype Frequencies From Gamete Pairing

Let \(X\) equal 1 if the maternal gamete carries \(A\), and 0 otherwise. Define \(Y\) in the same way for the paternal gamete. Then \(E(X)=E(Y)=p\), and their covariance is \(C=E(XY)-p^2\). A zygote is \(AA\) when both indicators are 1, \(Aa\) when exactly one is 1, and \(aa\) when both are 0.

These equations are valid only when the specified joint gamete probabilities are possible. Each genotype frequency must be nonnegative, which requires \(p^2+C\geq 0\), \(2pq-2C\geq 0\), and \(q^2+C\geq 0\). Equivalently, \(-\min(p^2,q^2)\leq C\leq pq\). These bounds are probability constraints, not additional assumptions about how mates were selected.

The last statement follows directly: \(p^2+C+\tfrac12(2pq-2C)=p^2+pq=p\). The same calculation shows why assortment can change genotype frequencies without changing the allele frequency: it changes the joint pairing of gametes, not either gamete pool’s marginal frequency of \(A\).

What Positive and Negative Assortment Change

When \(C>0\), both homozygotes are more frequent than under independent pairing. Their changes from the baseline are exactly equal: \(P(AA)-p^2=C\) and \(P(aa)-q^2=C\). The resulting frequencies need not be equal, because their baselines \(p^2\) and \(q^2\) may differ. At the same time, heterozygosity falls by \(2C\). When \(C<0\), both homozygote frequencies fall by the same absolute amount \(|C|\), while the heterozygote frequency rises by \(2|C|\), provided the probabilities remain valid.

Calculating From a Specified Mating Pattern

Sometimes a mating pattern is given as a set of parental genotype crosses rather than as a value of \(C\). In that case, calculate the offspring probabilities for each cross using Mendelian transmission, then weight each cross’s outcomes by its proportion of matings. The resulting genotype frequencies can also be checked against the covariance formulas.

A Short Derivation and Practical Checks

The formulas can also be understood by listing the four possible ordered pairs of gamete alleles. Let \(u=P(A\text{ from maternal gamete},A\text{ from paternal gamete})\). Since each gamete has \(A\)-frequency \(p\), the two mixed pairs must each have probability \(p-u\), and the \(aa\) pair has probability \(1-2p+u\). By the definition of covariance, \(u=p^2+C\). Therefore the mixed-pair probability in each direction is \(p-(p^2+C)=pq-C\); adding the two directions gives \(P(Aa)=2pq-2C\). The remaining pair has probability \(1-2p+p^2+C=q^2+C\), giving the stated \(aa\) frequency.

For any specified mating scheme, three checks make calculation errors easier to detect:

• Probability check: the three offspring genotype frequencies must be nonnegative and sum to 1.
• Allele-frequency check: \(P(AA)+\tfrac12P(Aa)\) must equal the \(A\)-allele frequency in the gamete pools under the assumptions stated here.
• Direction check: a positive \(C\) adds \(C\) to each homozygote baseline and subtracts \(2C\) from heterozygosity; a negative \(C\) reverses those changes.

Common Mistake and Why the Distinction Matters

A common mistake is to say that positive assortment makes both homozygotes increase, but not by the same amount because their baseline frequencies differ. That reasoning confuses the resulting frequencies with the changes from baseline. The changes are exactly equal: \(P(AA)-p^2=C\) and \(P(aa)-q^2=C\). If \(p=0.60\) and \(C=0.02\), for example, the frequencies change from \(0.36\) to \(0.38\) for \(AA\), and from \(0.16\) to \(0.18\) for \(aa\). Each increase is \(0.02\), even though the final frequencies differ.

Another useful distinction is between allele frequency and genotype frequency. A change in how gametes are paired can alter the fraction of heterozygotes even when the fraction of \(A\) alleles remains fixed. In applications, this matters because interpreting observed genotype proportions requires attention to the mating pattern that produced them; a genotype-frequency shift alone does not show that the allele frequency changed.