Linkage disequilibrium from gamete counts

Four gametes, two loci

Take two loci on one chromosome. Locus A has alleles A and a, locus B has alleles B and b. A gamete carries one allele at each locus, so there are four gamete types: AB, Ab, aB and ab. The module takes a count of each type and works out whether the two loci are associated.

Start with what independence predicts. If the allele at B tells you nothing about the allele at A, the frequency of a gamete is the product of its allele frequencies: f(AB) = pA × pB, f(Ab) = pA × pb, and so on. This state is linkage equilibrium. Any departure from it is linkage disequilibrium (LD).

The default counts are 230 AB, 270 Ab, 120 aB and 380 ab, so N = 1000. Allele A is in 230 + 270 = 500 gametes, so pA = 0.5. Allele B is in 230 + 120 = 350 gametes, so pB = 0.35. Independence predicts 0.5 × 0.35 = 0.175 for AB, which is 175 gametes. The sample has 230.

D and the chi-square test

The disequilibrium coefficient is D = f(AB) × f(ab) − f(Ab) × f(aB). It also equals f(AB) − pA × pB. For the defaults, D = 0.23 × 0.38 − 0.27 × 0.12 = 0.0874 − 0.0324 = 0.055. Multiply by N and you get the 55 gametes of excess in the figure. Every bar is off by exactly D: plus D for AB and ab, minus D for Ab and aB. Both loci keep their allele frequencies, so what AB gains, Ab and aB lose.

D depends on allele frequencies, so it is hard to compare across pairs of loci. The squared correlation r² = D² / (pA × pa × pB × pb) rescales it. Here r² = 0.0532. The test statistic is chi-square = N × r², with one degree of freedom. For the defaults that is 1000 × 0.0532 = 53.19. The 5% critical value is 3.84, so these loci are clearly not independent.

D′: how much of the possible D is used

D cannot take any value. Given the allele frequencies, there is a largest D the sample could have. For positive D it is the smaller of pA × pb and pa × pB. For the defaults that is the smaller of 0.5 × 0.65 = 0.325 and 0.5 × 0.35 = 0.175, so D max = 0.175.

Lewontin's D′ is D divided by D max: 0.055 / 0.175 = 0.314. D′ = 1 means at least one of the four gametes is missing. D′ = 0 means linkage equilibrium. The sign says which pair of gametes is in excess: negative D means Ab and aB are over-represented (repulsion).

D′ and r² answer different questions. D′ asks whether recombination has ever separated the alleles. r² asks how well one locus predicts the other. They disagree when allele frequencies differ.