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Parametric Linkage Analysis · Tutorial 224 of 989

Phase and Coupling-Repulsion in Linkage Calculations

Distinguish coupling from repulsion and see how the correct phase changes recombinant probabilities and the evidence measured by a linkage likelihood.

Intermediate 10 min read

What You'll Learn

  • Define coupling and repulsion using disease and marker alleles on parental haplotypes
  • Classify transmitted haplotypes as recombinant or nonrecombinant given a known phase
  • Calculate haplotype transmission probabilities for a specified recombination fraction
  • Construct phase-specific likelihoods from observed transmissions
  • Compare a linkage likelihood with the unlinked likelihood using a LOD score
  • Recognize when phase is unknown and why it cannot be ignored

Why Phase Matters in Linkage

A two-point linkage calculation depends on whether the disease-associated allele and a marker allele are on the same parental chromosome or on opposite parental chromosomes. That arrangement is called phase. The same transmitted marker allele can count as evidence of a nonrecombinant or a recombinant transmission depending on the phase, so phase must be established or properly handled in the likelihood.

Consider a parent who is heterozygous at a disease locus, with alleles \(D\) and \(d\), and at a marker locus, with alleles \(M\) and \(m\). A haplotype specifies the allele at both loci on one chromosome. The parent carries two haplotypes, one on each homologous chromosome. The notation below treats \(D\) as the disease-associated allele for the linkage model under consideration; it does not imply that every carrier is affected.

Definition: Coupling phase (also called cis phase) is the arrangement \(D\!-\!M\) and \(d\!-\!m\): the disease-associated allele \(D\) is on the same parental chromosome as marker allele \(M\). Repulsion phase (also called trans phase) is the arrangement \(D\!-\!m\) and \(d\!-\!M\): \(D\) is on the same chromosome as \(m\). A transmission is nonrecombinant if it carries one of the two parental haplotypes as arranged before meiosis, and recombinant if it carries one of the other two possible haplotypes.

The phase describes the parent’s chromosome arrangement, not the alleles that happen to be observed in a child. For example, \(D\!-\!m\) is a parental haplotype in repulsion phase, but it is a recombinant haplotype in coupling phase. This distinction is the central bookkeeping step in phase-specific linkage calculations.

Transmission Probabilities Under Each Phase

Let \(\theta\) denote the recombination fraction between the disease locus and the marker. In a meiosis from a parent heterozygous at both loci, \(\theta\) is the probability that the transmitted haplotype is recombinant, while \(1-\theta\) is the probability that it is nonrecombinant. In the conventional two-point linkage model, \(0\leq\theta\leq0.5\); \(\theta=0.5\) represents no linkage information between the loci.

Conditional on a particular phase, each of the two possible nonrecombinant haplotypes has probability \((1-\theta)/2\), and each of the two possible recombinant haplotypes has probability \(\theta/2\). The factor of two in each category accounts for its two possible haplotypes. The phase determines which haplotypes belong to each category.

PhaseNonrecombinant haplotypesRecombinant haplotypes
Coupling: \(D\!-\!M,\ d\!-\!m\)\(D\!-\!M,\ d\!-\!m\), each with probability \((1-\theta)/2\)\(D\!-\!m,\ d\!-\!M\), each with probability \(\theta/2\)
Repulsion: \(D\!-\!m,\ d\!-\!M\)\(D\!-\!m,\ d\!-\!M\), each with probability \((1-\theta)/2\)\(D\!-\!M,\ d\!-\!m\), each with probability \(\theta/2\)

At \(\theta=0.5\), all four haplotypes have probability \(0.25\), regardless of phase. Thus, when the loci are unlinked, the two phase arrangements give the same transmission probabilities. When \(\theta<0.5\), the parental haplotypes are more likely than the recombinant haplotypes, and the phase tells us which observed haplotypes receive those larger probabilities.

Worked Example: Assigning Probabilities in Coupling Phase

Suppose a hypothetical parent has known haplotypes \(D\!-\!M\) and \(d\!-\!m\), so the parent is in coupling phase. Let the recombination fraction be \(\theta=0.08\). Find the probability of each possible transmitted haplotype.

The parental haplotypes \(D\!-\!M\) and \(d\!-\!m\) are nonrecombinant. Each has probability \( (1-\theta)/2=(1-0.08)/2=0.92/2=0.46 \). The other two haplotypes, \(D\!-\!m\) and \(d\!-\!M\), are recombinant. Each has probability \( \theta/2=0.08/2=0.04 \).

$$ P(D\!-\!M)=P(d\!-\!m)=0.46,\qquad P(D\!-\!m)=P(d\!-\!M)=0.04. $$

The four probabilities sum to \(0.46+0.46+0.04+0.04=1.00\), as required. If the same parent instead had repulsion phase, the \(0.46\) probabilities would apply to \(D\!-\!m\) and \(d\!-\!M\), while \(D\!-\!M\) and \(d\!-\!m\) would each have probability \(0.04\).

Building a Phase-Specific Linkage Likelihood

Suppose the parental phase is known and the transmitted haplotypes can be classified. Let \(n\) be the number of observed meioses, \(r\) the number of recombinant transmissions, and \(n-r\) the number of nonrecombinant transmissions. Conditional on the parent’s phase and on \(\theta\), assume the meioses are independent and each has the same recombination fraction. The likelihood of the observed recombinant/nonrecombinant classifications is then proportional to \(\theta^r(1-\theta)^{n-r}\). If the specific transmitted haplotypes are recorded, the likelihood also includes a factor of \(1/2\) for each meiosis.

For example, under coupling phase, if the observed transmissions include \(a\) copies of \(D\!-\!M\), \(b\) copies of \(d\!-\!m\), \(c\) copies of \(D\!-\!m\), and \(e\) copies of \(d\!-\!M\), then the number of nonrecombinants is \(a+b\) and the number of recombinants is \(c+e\). Under repulsion phase, those two totals switch roles. For a specified ordered sequence of transmissions, the coupling likelihood is

$$ L_C(\theta) = \left(\frac{1-\theta}{2}\right)^{a+b} \left(\frac{\theta}{2}\right)^{c+e}. $$

The repulsion likelihood uses the same expression with the counts of \(D\!-\!m\) and \(d\!-\!M\) in the nonrecombinant category, and the counts of \(D\!-\!M\) and \(d\!-\!m\) in the recombinant category. If only category counts are used, a multinomial coefficient may also appear; it is the same under the linked and unlinked models for those data and therefore cancels in their likelihood ratio.

Result: For a specified phase and observed haplotypes, the likelihood is the product of the corresponding per-meiosis probabilities. The LOD score at a specified \(\theta\) is \(\log_{10}\{L(\theta)/L(0.5)\}\). The denominator is the likelihood under no linkage, since at \(\theta=0.5\) every haplotype has probability \(1/4\).

This formula shows why an incorrect phase can change the linkage likelihood sharply: it changes which observed haplotypes are multiplied by \((1-\theta)/2\) and which by \(\theta/2\). For \(\theta<0.5\), a haplotype assigned to the recombinant category receives a smaller probability than it would under the nonrecombinant category. The likelihood ratio compares the resulting probability of the observed transmissions with their probability at \(\theta=0.5\).

Worked Example: A Phase-Specific Likelihood and LOD Score

In a hypothetical family, four transmitted haplotypes are observed: three \(D\!-\!M\) haplotypes and one \(d\!-\!m\) haplotype. Assume coupling phase is known and evaluate the likelihood ratio at \(\theta=0.10\).

All four observed haplotypes are nonrecombinant under coupling. Each has probability \((1-0.10)/2=0.45\), so the likelihood of this specified sequence is

$$ L_C(0.10)=(0.45)^4=0.04100625. $$

Under no linkage, each haplotype has probability \(0.25\), giving \(L_C(0.5)=(0.25)^4=0.00390625\). The likelihood ratio and LOD score are therefore

$$ \frac{L_C(0.10)}{L_C(0.5)} = \frac{0.04100625}{0.00390625} =10.4976, \qquad \log_{10}(10.4976)\approx1.021. $$

For these specified transmissions and this assumed phase, the likelihood at \(\theta=0.10\) is about 10.50 times the unlinked likelihood. The LOD score is approximately \(1.021\). This calculation compares two likelihoods; it is not the probability that the loci are linked.

How the Same Observations Change Under Repulsion

The preceding example also illustrates why phase cannot be inferred from the disease allele alone. If the same observed haplotypes were evaluated under repulsion, \(D\!-\!M\) and \(d\!-\!m\) would both be recombinant. The observed transmissions would then be much less likely at a small recombination fraction. A phase assignment is therefore part of the model used to interpret transmissions, not a cosmetic label.

Worked Example: Comparing Coupling and Repulsion

Use the same hypothetical sequence of three \(D\!-\!M\) transmissions and one \(d\!-\!m\) transmission, but calculate its likelihood under both phases at \(\theta=0.10\).

Under coupling, all four transmissions are nonrecombinant, as above: \(L_C(0.10)=(0.45)^4=0.04100625\). Under repulsion, all four are recombinant. Each has probability \(\theta/2=0.10/2=0.05\), so \(L_R(0.10)=(0.05)^4=0.00000625\). The unlinked likelihood remains \((0.25)^4=0.00390625\) for either phase.

$$ \operatorname{LOD}_C = \log_{10}\left(\frac{0.04100625}{0.00390625}\right) \approx 1.021, \qquad \operatorname{LOD}_R = \log_{10}\left(\frac{0.00000625}{0.00390625}\right) = \log_{10}(0.0016) \approx -2.796. $$

For these data at this value of \(\theta\), coupling gives a much larger likelihood than repulsion. This is a comparison conditional on the stated model and phase assignments. In a real pedigree, phase may be uncertain, and the likelihood calculation must account for that uncertainty rather than treating an unverified phase as known.

A Short Likelihood Derivation

The phase-specific likelihood follows directly from multiplying the transmission probabilities. Under coupling, let \(N\) be the number of observed nonrecombinant transmissions and \(R\) the number of recombinants, so \(N+R=n\). Each nonrecombinant haplotype has probability \((1-\theta)/2\), and each recombinant haplotype has probability \(\theta/2\). Conditional independence gives

$$ L_C(\theta) = \left(\frac{1-\theta}{2}\right)^N \left(\frac{\theta}{2}\right)^R = 2^{-n}(1-\theta)^N\theta^R. $$

At \(\theta=0.5\), each of the \(n\) specified haplotypes has probability \(1/4\), so \(L_C(0.5)=4^{-n}=2^{-2n}\). Dividing and simplifying yields

$$ \frac{L_C(\theta)}{L_C(0.5)} = \frac{2^{-n}(1-\theta)^N\theta^R}{2^{-2n}} = 2^n(1-\theta)^N\theta^R. $$

Equivalently, each observed nonrecombinant contributes a factor \(2(1-\theta)\) to the likelihood ratio, and each observed recombinant contributes \(2\theta\). Under repulsion, the phase-specific labels determine which observed haplotypes contribute each factor. This derivation also confirms that at \(\theta=0.5\), the ratio is \(1\) and the LOD score is \(0\), regardless of phase.

When Phase Is Uncertain

In some pedigrees, the available genotypes do not establish whether a parent is in coupling or repulsion. There is then no justified basis for assigning every transmission to one phase’s recombinant category as though the phase were known. Depending on the specified linkage model, the likelihood may need to sum over possible phase and inheritance configurations with appropriate probabilities, or compare likelihoods while treating phase as an unknown nuisance parameter. The chosen treatment must be stated and applied consistently.

A simple identity helps explain the relationship between phase and recombination fraction. At a fixed value of \(\theta\), a coupling nonrecombinant probability \((1-\theta)/2\) equals a repulsion recombinant probability when the recombination fraction is \(1-\theta\). Thus \(L_C(\theta)=L_R(1-\theta)\) for the same fully specified transmissions. In the conventional range \(0\leq\theta\leq0.5\), phase is used to describe which arrangement makes the observed transmissions more compatible with a low recombination fraction. One should not switch phases after inspecting the data and report the resulting score as if that phase had been fixed in advance without accounting for the phase choice.

Common Mistakes and Practical Checks

  • Calling a haplotype recombinant without stating phase. The label depends on the parent’s chromosome arrangement. First write both parental haplotypes, then classify each possible transmitted haplotype.
  • Using the disease allele to define phase by itself. \(D\) does not say whether the marker allele on that chromosome is \(M\) or \(m\). That association is precisely what phase specifies.
  • Giving a recombinant haplotype probability \(\theta\) rather than \(\theta/2\). The total probability of either recombinant outcome is \(\theta\); each of the two recombinant haplotypes has probability \(\theta/2\).
  • Changing phase but keeping the same recombinant count. Switching from coupling to repulsion switches which haplotypes are classified as recombinant. Recount before multiplying probabilities.
  • Interpreting a LOD score as a probability. A LOD score is a base-10 logarithm of a likelihood ratio. A score of \(1\) means the specified linked likelihood is ten times the unlinked likelihood for the data and model used; it does not mean a 10% probability of linkage.
Key takeaway: Write the two parental haplotypes before calculating. In coupling, \(D\!-\!M\) and \(d\!-\!m\) are nonrecombinant; in repulsion, \(D\!-\!m\) and \(d\!-\!M\) are nonrecombinant. Apply \((1-\theta)/2\) to each parental haplotype and \(\theta/2\) to each recombinant haplotype, then form the likelihood under the stated phase.

Check Your Understanding

Answer each question by writing the relevant phase or probability explicitly.

  1. A parent carries \(D\!-\!m\) and \(d\!-\!M\). Is this coupling or repulsion, and which two haplotypes are recombinant?
  2. Under coupling and \(\theta=0.12\), what is the probability of transmitting \(d\!-\!m\)? What is the probability of transmitting \(d\!-\!M\)?
  3. Why do all four haplotypes have probability \(0.25\) at \(\theta=0.5\), regardless of phase?
  4. For a known phase, explain why the likelihood of a specified sequence of \(n\) transmissions is a product of per-meiosis probabilities under the stated independence assumption.
  5. In a linkage calculation, what does a LOD score of \(1\) say about the linked and unlinked likelihoods, and what does it not say?
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