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Let’s consider two linked loci $A$ and $B$ that are both bi-allelic. In consequence, we have four different possible haplotypes $A_1B_1$, $A_1B_2$, $A_2B_1$, $A_2B_2$, which frequencies are $X_1$, $X_2$, $X_3$, $X_4$ respectively. Let the linkage disequilibrium $D$ be defined as:

$$D=X_1X_4 - X_2X_3$$

The change in haplotype frequencies through time is given by:

$$\delta X_1 = \frac{X_1(w_1 - \bar w) - Drw_{14}}{\bar w}$$

$$\delta X_2 = \frac{X_2(w_2 - \bar w) - Drw_{14}}{\bar w}$$

$$\delta X_3 = \frac{X_3(w_3 - \bar w) - Drw_{14}}{\bar w}$$

$$\delta X_4 = \frac{X_4(w_4 - \bar w) - Drw_{14}}{\bar w}$$

, where

  • $r$ is the recombination rate between the loci $A$ and $B$
  • $\bar w$ is the mean fitness (sum of the fitnesses of each haplotype weighted by their relative frequencies)
  • $w_{14}$ is the fitness of the (diploid) individual that has the 1st and the 4th haplotypes
  • $w_i = X_1w_{i1} + X_2w_{i2} + X_3w_{i3} + X_4w_{i4}$

The change in allele frequency of locus $A$ is $x$ and $1-x$. The change of allele frequency of locus $B$ is $y$ and $1-y$. Those changes are given by the addition of haplotypes frequencies:

$$\delta x = \delta X_1 + \delta X_2$$


$$\delta y = \delta X_1 + \delta X_3$$


I don’t fully understand these formulae! Could you please help me making sense of them. Especially of the presence of the expression $-Drw_{14}$.


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up vote 5 down vote accepted

These equations describe how the haplotype frequencies will change over time due to a combination of recombination and natural selection.

Before I proceed, I need to change your four $\delta X_i$ formulas above. Lewontin and Kojima (1960) writes the equations as:

$$\Delta X_i = \frac{X_i(w_i - \bar w) \pm Drw_{14}}{\bar w}$$

where the minus sign is used for $i = 1,4$ and the plus sign is used for $i=2,3$. (See also this chapter by Sergey Gavrilets.) Decreasing the frequency of a haplotype through recombination will necessarily increase the frequency of a different haplotype (adjusted by selection).

$D$ is the measure of linkage disequilibrium. It ranges from $-0.25$ to $0.25$ and indicates how much the haplotype frequencies deviate from an expectation of no linkage ($D=0$, the loci are independent).

First, assume no natural selection. That is, $w_i = 1$ and therefore ${\bar w} = 1$. When fitness is equal, the equations reduce down to

$$\Delta X_i = \pm Dr$$

Some recombination will occur between the linked loci proportional to the distance between them. In a large randomly mating population with no selection, $D$ will eventually decay to zero. The larger the value of $r$, the quicker that decay will occur. Eventually, the haplotype frequencies will appear as if they are not linked. (See this figure.).

For example, assume that $A$ and $B$ are linked, with these four haplotype frequencies:

$$A_1B_1 = 0.5$$ $$A_1B_2 = 0.38$$ $$A_2B_1 = 0.05$$ $$A_2B_2 = 0.07$$

In the absence of natural selection and assuming that $r=0.1$, then $\Delta X_i = \pm 0.0016$, again with plus or minus depending on the value of $i$. This is the rate the frequencies will change each generation. As $r$ decreases, the rate of decay decreases.

Differential fitness among the haplotypes will also affect the rate of decay. The fitness interactions among loci on the same chromosomes and between homologs can get rather complex depending heterozygote advantage, epistasis, etc. (see Lewontin 1964 for several fitness models).

The equation $w_i = X_1w_{i1} + X_2w_{i2} + X_3w_{i3} + X_4w_{i4}$ is the average fitness of the each haplotype in combination with all the other possible haplotypes. For example, for $w_1$, you're calculating the average fitness of four genotypes: $A_1B_1/A_1B_1$, $A_1B_1/A_1B_2$, $A_1B_1/A_2B_1$, and $A_1B_1/A_12B_2$.

I am uncertain why $w_{14}$ is used in the second term of the numerator. I thought it might be because that combination is heterozygous at both loci so recombination generates all four possible haplotypes. However, both Lewontin and Kojima (1960) and Lewontin (1964) use $w_{11}$. It may also be that a relative fitness component is necessary for comparison so the fitness of any one of the genotypes could be used there. I don't know.

Regrdless, as a simple example to illustrate the effects of selection, assume $r=0.1$, $\bar w = 0.5$, and $w_{14} = 1$ (heterozygote advantage, given maximum relative fitness). For $w_1 = 0.9$ then $\Delta X_1 = 0.397$. Selection strongly favors this haplotype in all genotypic combinations so it will increase in frequency. For $w_1 = 0.3$ then $\Delta X_1 = -0.203$. The haplotype is selected against and decreases much more rapidly, relative to no selection at all. The same idea would apply for the other haplotypes.


Lewontin, R.C. 1964. The interaction of selection and linkage. I. General considerations; heterotic models. Genetics 49: 49-67.

Lewontin, R.C. and K.-I. Kojima. 1960. The evolutionary dynamics of complex polymorphisms. Evolution 14: 458-472.

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