How Recombination Changes Linkage Disequilibrium

Why Recombination Changes Linkage Disequilibrium

Linkage disequilibrium (LD) describes a statistical association between alleles at different loci. Recombination can weaken that association by reshuffling alleles onto new haplotypes. The amount of change depends on the recombination fraction between the loci and on the LD present before recombination. This tutorial derives that relationship and uses it to track LD across generations.

We consider two biallelic loci, with alleles \(A\) and \(a\) at the first locus and \(B\) and \(b\) at the second. The four haplotypes are \(AB\), \(Ab\), \(aB\), and \(ab\). Their frequencies in the gamete pool before recombination are denoted \(x_{AB}\), \(x_{Ab}\), \(x_{aB}\), and \(x_{ab}\), and sum to one. The allele frequencies are \(p_A=x_{AB}+x_{Ab}\) and \(p_B=x_{AB}+x_{aB}\).

Definition: For these two loci, linkage disequilibrium is \(D=x_{AB}-p_Ap_B\). It measures the difference between the observed frequency of haplotype \(AB\) and the frequency expected if alleles at the two loci were independent. Equivalently, \(D=x_{AB}x_{ab}-x_{Ab}x_{aB}\). The recombination fraction \(c\) is the probability that a gamete is recombinant between the two loci in a meiosis; for two loci, \(0\leq c\leq 0.5\).

A positive \(D\) means \(AB\) occurs more often than expected from the two allele frequencies; a negative \(D\) means it occurs less often. Recombination does not, by itself, change \(p_A\) or \(p_B\). In the simple model below, it changes how alleles at the two loci are associated.

Model and Recurrence

The recurrence describes expected frequencies in a large, diploid, randomly mating population. Individuals form by random union of gametes from the current pool. During meiosis, recombination between the two loci occurs with the same fraction \(c\). Assume no selection, mutation, migration, or other force changing the frequencies, and ignore finite-population sampling effects. Thus, the recurrence is a deterministic description of recombination alone.

Under random mating, the two haplotypes that form an individual's pair of homologous chromosomes are independent draws from the gamete pool. The two kinds of double heterozygote have different phase. A coupling-phase individual has haplotypes \(AB/ab\); a repulsion-phase individual has \(Ab/aB\). Their population frequencies are \(2x_{AB}x_{ab}\) and \(2x_{Ab}x_{aB}\), respectively. These are the individuals in which recombination can change which alleles occur together.

Theorem: Under random mating and the assumptions stated above, the next-generation gamete frequencies satisfy \(D_{\text{next}}=(1-c)D\). The allele frequencies \(p_A\) and \(p_B\) remain unchanged.

To see the effect on \(AB\), consider a coupling-phase individual \(AB/ab\). Without recombination, half its gametes carry \(AB\); with recombination fraction \(c\), the expected \(AB\) fraction is \((1-c)/2\). Its \(AB\) contribution therefore falls by \(c/2\). A repulsion-phase individual \(Ab/aB\) has no parental \(AB\) haplotype, but produces \(AB\) recombinant gametes at expected fraction \(c/2\). Weighting these changes by the frequencies of the two classes gives the net change in \(x_{AB}\):

x_{AB,\text{next}}-x_{AB} = (2x_{AB}x_{ab})\left(-\frac{c}{2}\right) + (2x_{Ab}x_{aB})\left(\frac{c}{2}\right) = -c(x_{AB}x_{ab}-x_{Ab}x_{aB}) = -cD.

The same calculation shows that \(x_{ab}\) changes by \(-cD\), while \(x_{Ab}\) and \(x_{aB}\) each change by \(+cD\). The changes sum to zero, as required for frequencies. The changes to the marginal frequencies cancel: the \(AB\) and \(Ab\) changes offset one another for allele \(A\), and the \(AB\) and \(aB\) changes offset one another for allele \(B\). Thus, \(p_A\) and \(p_B\) stay fixed. Since \(D=x_{AB}-p_Ap_B\), it follows that:

D_{\text{next}} = x_{AB,\text{next}}-p_Ap_B = (x_{AB}-cD)-p_Ap_B = (1-c)D.

This argument also gives the equivalent identity \(D=x_{AB}x_{ab}-x_{Ab}x_{aB}\): the difference in expected recombinant gains and losses is determined by the excess of coupling over repulsion products. The recurrence scales the original \(D\), so its sign is preserved for \(c<1\), while its magnitude declines whenever \(c>0\) and \(D\ne 0\).

Worked Example: One Generation of Recombination

Suppose a hypothetical population has haplotype frequencies \(x_{AB}=0.40\), \(x_{Ab}=0.10\), \(x_{aB}=0.10\), and \(x_{ab}=0.40\), with recombination fraction \(c=0.20\). Assume random mating and no other forces, as in the model.

The allele frequencies are \(p_A=0.40+0.10=0.50\) and \(p_B=0.40+0.10=0.50\). The initial LD is \(D=0.40-(0.50)(0.50)=0.15\). The recurrence gives:

D_{\text{next}}=(1-0.20)(0.15)=0.80(0.15)=0.12.

The corresponding haplotype-frequency changes are \(-cD=-0.20(0.15)=-0.03\) for \(AB\) and \(ab\), and \(+0.03\) for \(Ab\) and \(aB\). Thus the next-generation frequencies are \(0.37, 0.13, 0.13,\) and \(0.37\), respectively. They sum to one, and the allele frequencies remain \(p_A=p_B=0.50\). Checking directly, \(D_{\text{next}}=0.37-(0.50)(0.50)=0.12\). Recombination has reduced the positive LD without changing either allele frequency.

Following LD Across Generations

If the same recombination fraction applies each generation and the other assumptions continue to hold, the one-generation recurrence can be applied repeatedly. Writing \(D_0\) for the initial LD and \(D_t\) for LD after \(t\) generations:

D_{t+1}=(1-c)D_t \qquad\Longrightarrow\qquad D_t=(1-c)^tD_0.

The second expression follows by repeated substitution: \(D_1=(1-c)D_0\), \(D_2=(1-c)^2D_0\), and each additional generation contributes one more factor of \(1-c\). When \(c=0\), there is no recombination between the loci in this model and \(D\) is unchanged. When \(c=0.5\), the magnitude of \(D\) is halved each generation. If \(D_0=0\), the loci are in linkage equilibrium for these alleles, and recombination alone does not create LD.

Worked Example: Four Generations of LD Decay

In a hypothetical population, let \(p_A=0.30\), \(p_B=0.60\), and \(D_0=-0.080\). These values correspond to \(x_{AB}=p_Ap_B+D_0=(0.30)(0.60)-0.080=0.10\). The other haplotype frequencies are \(x_{Ab}=0.20\), \(x_{aB}=0.50\), and \(x_{ab}=0.20\), so all frequencies are nonnegative and sum to one. Suppose \(c=0.25\), random mating holds, and there are no other forces.

After four generations, the recurrence gives:

$$ D_4=(1-0.25)^4(-0.080) =(0.75)^4(-0.080) =0.31640625(-0.080) =-0.0253125. $$

The marginal allele frequencies remain \(0.30\) and \(0.60\). Therefore \(x_{AB,4}=p_Ap_B+D_4=0.18-0.0253125=0.1546875\), or about \(0.1547\). The remaining frequencies follow from the marginals: \(x_{Ab,4}=0.30-0.1546875=0.1453125\), \(x_{aB,4}=0.60-0.1546875=0.4453125\), and \(x_{ab,4}=1-0.1546875-0.1453125-0.4453125=0.2546875\). They are nonnegative and sum to one. The negative sign of \(D_4\) indicates that \(AB\) remains less frequent than its independence expectation, but the magnitude has fallen from \(0.080\) to about \(0.0253\).

Worked Example: Comparing Recombination Fractions

Consider a hypothetical starting population with \(x_{AB}=0.30\), \(x_{Ab}=0.20\), \(x_{aB}=0.10\), and \(x_{ab}=0.40\). Its allele frequencies are \(p_A=0.50\) and \(p_B=0.40\), so \(D_0=0.30-(0.50)(0.40)=0.10\). Under the model assumptions, compare the LD after three generations for \(c=0.10\) and \(c=0.50\).

For \(c=0.10\), the multiplier per generation is \(0.90\):

$$ D_3=(0.90)^3(0.10)=0.729(0.10)=0.0729. $$

For \(c=0.50\), the multiplier is \(0.50\):

$$ D_3=(0.50)^3(0.10)=0.125(0.10)=0.0125. $$

Both values remain positive, but the larger recombination fraction produces a smaller remaining LD after the same number of generations. These are dimensionless measures of association; the comparison describes expected LD under the stated model, not a causal effect of one allele on another.

Interpreting the Recurrence

The factor \(1-c\) is the fraction of LD retained per generation in this model. The closed form makes clear that decay is geometric, not a fixed subtraction of \(c\) units from \(D\). For example, \(c=0.20\) means that 80% of the current \(D\) remains after one generation; it does not mean that \(D\) decreases by 0.20 regardless of its starting value.

The result concerns LD, not allele frequency. Recombination redistributes alleles among haplotypes while preserving the marginal allele frequencies under the model. In real populations, selection, mutation, migration, drift, or nonrandom mating can also affect haplotype frequencies, so observed changes need not follow this recurrence by themselves. The equation is useful as a baseline for understanding the contribution of recombination under its assumptions.

Common Mistake

Wrong reasoning: “A recombination fraction of 0.10 reduces \(D\) by 0.10 each generation.” This confuses the recombination fraction with an absolute change in LD.
Correction: The recurrence is \(D_{\text{next}}=(1-c)D\). With \(c=0.10\), the next generation retains 90% of the current \(D\). If \(D=0.20\), the next value is \(0.90(0.20)=0.18\), a decrease of 0.02, not 0.10.

A related caution is to check that the model is appropriate before interpreting an observed temporal change as recombination alone. Random mating, a stable recombination fraction, and the absence of other frequency-changing forces are assumptions, not conclusions guaranteed by the recurrence.

Key takeaway: With random mating and no other forces changing frequencies, recombination leaves the two allele frequencies unchanged and multiplies LD by \(1-c\) each generation. After \(t\) generations, \(D_t=(1-c)^tD_0\).

Check Your Understanding

Use the assumptions and recurrence in this tutorial to answer the following questions.

If \(D_0=0.12\) and \(c=0.25\), what is \(D_2\)?
Why does recombination alone preserve the marginal allele frequencies in this model?
A population has \(D=-0.06\) and \(c=0.10\). What is the next-generation \(D\), and does its sign change?
What does \(c=0.50\) imply about the fraction of the current LD retained after one generation?
Name two assumptions that would need reconsideration before attributing an observed change in LD entirely to recombination.