Worked Multipoint Linkage Analysis Across Three Markers

Why Work Through a Multipoint Example?

A marker can indicate whether a child inherited the same parental chromosome that carries a disease allele, but recombination can change which chromosome is inherited as we move along the map. With several ordered markers, the evidence for a disease-locus position depends on the full sequence of observed transmissions and on the recombination probabilities between successive loci. A multipoint calculation makes those dependencies explicit.

This tutorial works through a small pedigree with one informative meiosis and three ordered markers. The goal is to compare candidate disease-locus positions using the likelihood of the observed marker transmissions, and then express each comparison as a LOD score relative to an unlinked disease locus.

Definition: An inheritance state at a locus records which of a parent's two homologous chromosomes was transmitted to the child. Code the homolog carrying the disease allele at the candidate disease locus as state 1 and the other homolog as state 0. At marker loci, phased parental genotypes and the child's genotype identify the transmitted homolog. A transition between states at adjacent loci represents a recombination in that meiosis.

Model and Likelihood Setup

Consider a hypothetical trio. One parent is heterozygous for a disease allele, with the disease allele on parental homolog 1; the other parent is a noncarrier. Assume the affected child inherited the disease allele from the heterozygous parent. The parent's phased marker haplotypes are known, and the child's genotypes identify which homolog was inherited at each marker. Denote the ordered markers \(M_1,M_2,M_3\), and their observed inheritance states by \(B_1,B_2,B_3\).

For the worked data, the child inherited homolog 1 at \(M_1\) and \(M_2\), then homolog 0 at \(M_3\). Thus the observed marker-state sequence is \(1,1,0\). Suppose the recombination fractions are \(r_{12}=0.10\) between \(M_1\) and \(M_2\), and \(r_{23}=0.20\) between \(M_2\) and \(M_3\). Recombination events in disjoint intervals are assumed independent, so the probability of a state change across an interval is its recombination fraction.

For this illustration, assume a known phase, one informative meiosis, no genotyping errors, a fully penetrant disease allele, no phenocopies, and no other source of the child's affected status. These assumptions let us treat transmission of the disease allele as observed. The likelihood comparison is conditional on that transmission and the observed pedigree structure. In a larger pedigree, the corresponding likelihood sums over unobserved inheritance states and incorporates the specified penetrance and transmission probabilities.

Likelihood comparison: At a candidate position \(D\), let \(M\) denote the observed marker transmissions and let \(B_D=1\) denote transmission of the disease-bearing homolog at \(D\). The linked likelihood is \(P(M\mid B_D=1,D)\). Under the unlinked model, disease transmission is independent of marker transmission, so the likelihood is \(P(M)\). The LOD score is the base-10 logarithm of their ratio.

\operatorname{LOD}(D)=\log_{10}\!\left\{\frac{P(M\mid B_D=1,D)}{P(M)}\right\}

Before calculating candidate positions, it helps to obtain the likelihood of the marker sequence itself. With no information favoring either homolog at \(M_1\), its state has probability \(1/2\). The probability of no change from \(M_1\) to \(M_2\) is \(1-r_{12}=0.90\), and the probability of a change from \(M_2\) to \(M_3\) is \(r_{23}=0.20\).

Worked Example: Likelihood of the Three Marker States

For the observed sequence \(B_1=1,B_2=1,B_3=0\), multiply the initial state probability by the transition probabilities along the map:

P(M)=P(B_1=1)P(B_2=1\mid B_1=1)P(B_3=0\mid B_2=1)

Substituting the stated recombination fractions gives

P(M)=\frac{1}{2}(1-0.10)(0.20) =\frac{1}{2}(0.90)(0.20) =0.0900.

The value \(0.0900\) is the probability of this particular marker-transmission sequence under the stated map model. It is not yet a LOD score: a LOD requires comparing the marker likelihood conditional on disease-allele transmission at a candidate position with the unlinked likelihood.

Inferring the State Between Two Markers

At a candidate disease locus between adjacent markers, its inheritance state is not directly observed. Instead, calculate the probability of each possible state there, conditional on the observed states at the flanking markers. Let the recombination fractions from the left marker to \(D\) and from \(D\) to the right marker be \(r_L\) and \(r_R\).

If both flanking states are 1, the path with \(B_D=1\) requires no switch in either subinterval, with probability \((1-r_L)(1-r_R)\). The path with \(B_D=0\) requires a switch in each, with probability \(r_Lr_R\). Normalizing these two path probabilities gives the conditional probability of state 1. If the flanking states differ, the path probabilities are different: the path through state 1 requires no switch on the left and a switch on the right, while the path through state 0 requires the reverse.

Bridge probabilities: Conditional on equal flanking states of 1, the probability of state 1 at \(D\) is \(\frac{(1-r_L)(1-r_R)}{(1-r_L)(1-r_R)+r_Lr_R}\). Conditional on left state 1 and right state 0, it is \(\frac{(1-r_L)r_R}{(1-r_L)r_R+r_L(1-r_R)}\). These expressions condition on the flanking states; the candidate locus's location determines \(r_L\) and \(r_R\).

The bridge formula follows directly from conditional probability. For example, when the observed endpoints are 1 and 0, there are exactly two possible states at \(D\). The probability of the endpoint sequence given \(B_D=1\) is \((1-r_L)r_R\); given \(B_D=0\), it is \(r_L(1-r_R)\). The probability of the endpoints is the sum of these mutually exclusive path probabilities. Dividing the first path probability by that sum yields \(P(B_D=1\mid B_L=1,B_R=0)\). The same argument gives the equal-endpoint expression.

Equal-length intervals make \(r_L=r_R\), so the two paths for differing endpoints have equal probabilities. That equality does not hold for unequal intervals. In the numerical comparison below, the unequal-interval formula is important: the conditional probability is not obtained by simply assigning one-half to each state.

Worked Example: A Candidate One-Third of the Way from \(M_2\) to \(M_3\)

The observed flanking states are \(B_2=1\) and \(B_3=0\). Place \(D\) one-third of the genetic distance from \(M_2\) toward \(M_3\). Use the Haldane mapping function, \(r(d)=(1-e^{-2d})/2\), where \(d\) is distance in Morgans. Since \(r_{23}=0.20\), the full interval has \(e^{-2d}=1-2r_{23}=0.60\). The two distance fractions therefore give \(e^{-2d_L}=0.60^{1/3}\) and \(e^{-2d_R}=0.60^{2/3}\).

r_L=\frac{1-0.60^{1/3}}{2}\approx 0.0782837, \qquad r_R=\frac{1-0.60^{2/3}}{2}\approx 0.144311.

Because the flanking states differ, use the differing-endpoint bridge probability:

P(B_D=1\mid B_2=1,B_3=0) = \frac{(1-r_L)r_R}{(1-r_L)r_R+r_L(1-r_R)}

Substitution gives

\frac{(1-0.0782837)(0.144311)} {(1-0.0782837)(0.144311)+(0.0782837)(1-0.144311)} \approx \frac{0.133014}{0.200000} \approx 0.66507.

Thus, conditional on the observed flanking states, the probability of inheriting homolog 1 at this candidate position is about \(0.6651\), not \(0.5\). The larger share comes from the unequal recombination probabilities on the two sides of \(D\).

From a Conditional State Probability to a LOD

The bridge probability becomes a likelihood ratio through Bayes' rule. Before observing marker transmissions, either parental homolog is equally likely to have been transmitted at \(D\), so \(P(B_D=1)=1/2\). Consequently,

\frac{P(M\mid B_D=1,D)}{P(M)} = \frac{P(B_D=1\mid M,D)}{P(B_D=1)} = 2P(B_D=1\mid M,D).

The first equality is Bayes' rule rearranged. It shows why the posterior probability of state 1, conditional on all observed marker states, is the relevant quantity. In this example, the marker states outside the flanking interval do not change the bridge probability: conditional on the state at a shared endpoint, transitions on the other side of that endpoint are independent. With a larger pedigree, calculate the corresponding posterior using the full inheritance-path likelihood.

Worked Example: LOD at the One-Third Position

From the previous calculation, \(P(B_D=1\mid M,D)\approx0.66507\). Therefore the likelihood ratio and LOD score are

\frac{P(M\mid B_D=1,D)}{P(M)} =2(0.66507)\approx1.3301, \qquad \operatorname{LOD}(D)=\log_{10}(1.3301)\approx0.1239.

The same result can be checked by calculating the linked likelihood directly. The joint probability of the marker sequence and \(B_D=1\) is \(P(M)P(B_D=1\mid M,D)\). Divide by the prior probability \(P(B_D=1)=0.5\) to obtain the linked likelihood conditional on disease transmission:

P(M\mid B_D=1,D) =\frac{(0.0900)(0.66507)}{0.5} \approx0.1197.

The unlinked likelihood is \(P(M)=0.0900\), so \(0.1197/0.0900\approx1.3301\), as before. A LOD of \(0.1239\) means the observed marker transmissions are about \(1.330\) times as likely under linkage at this candidate position as under the unlinked model, conditional on the stated assumptions. It does not by itself establish linkage.

Comparing Candidate Positions

Now compare several positions on the same map. At \(M_1\), the observed state is 1, so the conditional probability of disease-homolog transmission is 1. At the midpoint of \(M_1\)–\(M_2\), both observed endpoint states are 1. Under the Haldane map, splitting an interval with total recombination fraction \(0.10\) into two equal distances gives

r_L=r_R=\frac{1-\sqrt{1-2(0.10)}}{2} =\frac{1-\sqrt{0.80}}{2} \approx0.0527864.

The probability of state 1 at that midpoint, conditional on the two observed endpoint states, is

\frac{(1-0.0527864)^2}{(1-0.0527864)^2+(0.0527864)^2} \approx0.996904.

For comparison, the midpoint of \(M_2\)–\(M_3\) has equal subintervals and different endpoint states, so the two paths have equal probability and the conditional probability of state 1 is \(0.5\). The candidate scores follow from \(2P(B_D=1\mid M,D)\).

Candidate position	\(P(B_D=1\mid M,D)\)	Likelihood ratio	LOD
At \(M_1\)	1.000	2.000	0.3010
Midpoint \(M_1\)–\(M_2\)	0.9969	1.994	0.2997
One-third from \(M_2\) to \(M_3\)	0.6651	1.330	0.1239
Midpoint \(M_2\)–\(M_3\)	0.5000	1.000	0.0000

These values illustrate how a recombination between \(M_2\) and \(M_3\) weakens support for a disease locus in that interval, while the two matching states at \(M_1\) and \(M_2\) favor positions near those markers. The LOD at the midpoint of \(M_2\)–\(M_3\) is zero because the conditional probability of state 1 is \(1/2\), making the linked and unlinked likelihoods equal in this simplified example. These are scores at specified candidate positions, not a claim that the disease locus must lie at the largest listed score.

A Practical Calculation Sequence

Resolve marker transmissions.
Use the parental phase and child genotypes to assign an inheritance state at each observed marker.

Write the ordered path.
Record the state sequence and the recombination fraction for each adjacent interval.

Insert the candidate locus.
Use the map to obtain the recombination fractions on each side of the candidate; do not assume they are equal unless the distances are equal.

Calculate the candidate likelihood.
For an unobserved state, sum the probabilities of paths compatible with the observed data, or use the bridge probability for a single meiosis between known flanking states.

Compare with the unlinked model.
Divide the likelihood conditional on disease transmission by the marker likelihood under independence, then take the base-10 logarithm.

Common Mistake

A tempting shortcut is to say that when the flanking states differ, the two possible midpoint states are equally likely. That is true for equal-length subintervals under a symmetric transition model, but not for unequal intervals. For example, with \(r_L=0.10\) and \(r_R=0.20\), and endpoint states 1 and 0, the probability of the path through state 1 is \((1-0.10)(0.20)=0.18\), while the path through state 0 has probability \((0.10)(1-0.20)=0.08\). Conditional on the endpoints, the probability of state 1 is \(0.18/(0.18+0.08)\approx0.6923\), not \(0.5\). Use the two path probabilities and normalize them; the map position determines whether the subinterval recombination fractions match.

A second useful check is to keep the likelihood conditioning consistent. The linked likelihood here is conditional on transmission of the disease allele; the unlinked likelihood is the marker likelihood under independence. Comparing a joint probability in one model with a conditional probability in the other would produce an invalid likelihood ratio.

Key takeaway: In a multipoint linkage calculation, observed marker transmissions define a path through inheritance states. At each candidate disease position, use the map-specific transition probabilities to evaluate disease-homolog transmission, compare the resulting conditional marker likelihood with the unlinked likelihood, and report the base-10 LOD with its model assumptions.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

For the observed marker sequence \(1,1,0\), why is its likelihood \(\frac{1}{2}(1-r_{12})r_{23}\) under the stated model?
When the flanking states are 1 and 0, write the conditional probability that the candidate locus has state 1 for general \(r_L\) and \(r_R\).
Why does a candidate with \(P(B_D=1\mid M,D)=0.5\) have LOD zero in this example?
For a candidate between markers with equal observed states, which two path probabilities must be compared to find the probability of state 1?
What does a LOD of \(0.1239\) say about the likelihood ratio, and what does it not establish?