X-linked loci

Two sexes, two frequencies, one recursion

An autosomal locus has one allele frequency, p, because every individual carries two copies. An X-linked locus does not. Females have two X chromosomes and males have one, so the population has two frequencies of allele A: pf among the X chromosomes of females and pm among the X chromosomes of males. Males are hemizygous, so pm is also the proportion of males who show the A phenotype.

If pf and pm start unequal, nothing forces them together in one generation. The reason is how the X is inherited. A son receives his only X from his mother and none from his father. A daughter receives one X from each parent. The two sexes therefore sample the parents differently, and the gap between them has to be worked off over several generations.

Random mating, no selection, mutation, migration or drift, and a 1:1 sex ratio give two equations.

• Daughters: pf(t + 1) = (pf(t) + pm(t))/2, the mean of the mother and father.

• Sons: pm(t + 1) = pf(t), a copy of the mother.

The X-linked module iterates exactly this pair. It is deterministic, so the same inputs give the same lines every time. Once the lines have met at p̂, genotypes follow Hardy-Weinberg proportions in females. A recessive X-linked trait, carried by the allele a with frequency 1 − p̂ = 0.4333, then shows in 43% of males and in only 0.4333² = 19% of females. That ratio of 1 to q is why such traits are more common in males. The defaults are pf = 0.7, pm = 0.3 and 20 generations. Generation 1 gives pf = (0.7 + 0.3)/2 = 0.5 and pm = 0.7. The sexes have swapped sides of the mean: females were above males, and now they are below.

The gap halves and flips

Let d = pf − pm be the gap. One step later the new gap is (pf + pm)/2 − pf = −(pf − pm)/2, so d(t + 1) = −d(t)/2. The gap halves and changes sign every generation. After t generations it is d(0) × (−1/2)^t.

With d(0) = 0.4 the gaps are 0.4, −0.2, 0.1, −0.05 and so on. The alternating signs draw the zigzag. The size of the gap first falls below 0.01 at generation 6, where it is 0.4/64 = 0.00625. At generation 10 it is 0.4/1024 = 0.0004.

An autosomal locus would be at Hardy-Weinberg after one generation of random mating. The X-linked locus needs 6 generations to bring the gap under 0.01, because the gap is shrinking by half and not by all of it.

Both lines converge on p̂ = (2pf(0) + pm(0))/3, not on the simple mean (pf + pm)/2. For the defaults p̂ = (2 × 0.7 + 0.3)/3 = 1.7/3 = 0.5667, where the simple mean would be 0.5. The chart does not draw p̂. You read it from the end of the lines: at generation 20 both values are 0.56667.