Inheritance-Path Probabilities Along a Genetic Map

Why Track Inheritance Paths?

At one marker, a child's inherited marker allele reflects which of a parent's two homologues was transmitted. Across several ordered markers, that transmitted homologue may remain the same or switch. The sequence of choices is an inheritance path. Its probability is determined by the recombination fractions in the intervals between markers, together with assumptions about recombination across intervals.

Representing paths explicitly separates two questions. First, how probable is a particular sequence of transmitted parental homologues? Second, how does that sequence, combined with genotypes and disease information, contribute to an analysis? This tutorial addresses the first question and the bookkeeping needed for the second. It does not calculate a pedigree's multipoint likelihood.

Encoding a Path Along Ordered Markers

Suppose markers \(1,\ldots,K\) lie in order along a chromosome. Consider one meiosis: a parent transmits one of its two homologues to a child. At marker \(k\), encode the transmitted homologue by \(H_k\), where \(H_k=0\) denotes the parent's first labelled homologue and \(H_k=1\) denotes its second. The labels are arbitrary, but must be used consistently along the chromosome.

Definition: For a single meiosis, the inheritance path across \(K\) ordered markers is the sequence \(H_1,\ldots,H_K\), with each \(H_k\in\{0,1\}\). A change between adjacent markers, \(H_{k+1}\ne H_k\), is a switch in the transmitted homologue. Under the usual two-homologue model, its probability is the recombination fraction \(\theta_k\) for the interval between markers \(k\) and \(k+1\).

A switch records a change in which parental homologue is represented in the transmitted chromosome; it does not count every physical crossover. For example, an even number of crossovers within an interval can leave the transmitted state the same at the two markers. The recombination fraction \(\theta_k\) is the probability of a recombinant transmission, equivalently a change in this binary state between the interval's endpoints under the model.

For the path probability below, assume that the initial transmitted homologue is equally likely to be either of the parent's two homologues, so \(P(H_1=0)=P(H_1=1)=1/2\). Also assume that recombination indicators in distinct adjacent intervals are independent. This is the no-interference interval model; Haldane's mapping function, for example, corresponds to independent crossovers along the map. If an interference model is used, adjacent recombination fractions alone need not determine the joint probability of a whole path.

Probability of a Single-Meiosis Path

Let \(\theta_k\) be the recombination fraction between markers \(k\) and \(k+1\). If the path stays on the same labelled homologue across that interval, its transition probability is \(1-\theta_k\); if it switches, the transition probability is \(\theta_k\). Define \(I_k\) to be 1 when \(H_{k+1}\ne H_k\), and 0 otherwise. The initial state contributes a factor of \(1/2\), and each interval contributes the probability of its observed transition.

Result: Under equal initial transmission and independent recombination indicators across intervals, the probability of a specified path \(h_1,\ldots,h_K\) is

P(H_1=h_1,\ldots,H_K=h_K) =\frac{1}{2}\prod_{k=1}^{K-1} \theta_k^{I_k}(1-\theta_k)^{1-I_k}, \qquad I_k=\mathbf{1}(h_{k+1}\ne h_k).

The probability depends on where the path switches, not on whether its initial label is 0 or 1.

Worked Example: A Four-Marker Path in One Meiosis

Suppose four ordered markers have adjacent recombination fractions \(\theta_1=0.10\), \(\theta_2=0.20\), and \(\theta_3=0.05\). Find the probability of the path \(H=(0,0,1,1)\).

The initial state is 0, which has probability \(1/2\). The path stays on the same homologue from marker 1 to 2, switches from marker 2 to 3, and stays on the same homologue from marker 3 to 4. Thus the interval transition probabilities are \(1-\theta_1=0.90\), \(\theta_2=0.20\), and \(1-\theta_3=0.95\).

P(H=(0,0,1,1)) =\frac{1}{2}(0.90)(0.20)(0.95) =0.0855.

The probability of this specified path is 0.0855. The path \(H=(1,1,0,0)\) has the same probability: it has the same switch pattern and the other equally likely initial homologue.

Derivation and Checks

The path probability follows by conditioning successively on the state at the preceding marker. Given \(H_k\), the next state equals \(H_k\) with probability \(1-\theta_k\) and differs from it with probability \(\theta_k\). Under the assumed independent interval transitions, the probability of a specified sequence is the initial-state probability multiplied by these successive conditional probabilities:

\begin{aligned} P(H_1=h_1,\ldots,H_K=h_K) &=P(H_1=h_1)\prod_{k=1}^{K-1} P(H_{k+1}=h_{k+1}\mid H_k=h_k)\\ &=\frac{1}{2}\prod_{k=1}^{K-1} \begin{cases} 1-\theta_k, & h_{k+1}=h_k,\\ \theta_k, & h_{k+1}\ne h_k. \end{cases} \end{aligned}

This factorization also provides a useful normalization check. For either value of \(H_k\), the two possible next states have probabilities \(1-\theta_k\) and \(\theta_k\), which sum to 1. Summing over all paths therefore sums over the two initial states, whose probabilities sum to 1, and then over each set of possible transitions, each of which also sums to 1. Hence the probabilities assigned to all \(2^K\) paths sum to 1.

Equivalently, the transition matrix for interval \(k\), with states ordered as 0 and 1, is

T_k= \begin{pmatrix} 1-\theta_k & \theta_k\\ \theta_k & 1-\theta_k \end{pmatrix}.

Each row describes the probabilities of the next state given the current state. This matrix representation makes the same path calculation possible without listing cases individually.

Worked Example: Probability of Exactly One Switch

For one meiosis across four ordered markers, let the three adjacent recombination fractions be \(0.10\), \(0.20\), and \(0.30\). Find the probability that the inheritance path has exactly one switch, without specifying which interval switches.

There are three mutually exclusive possibilities: the switch occurs only in interval 1, only in interval 2, or only in interval 3. The initial state can be either 0 or 1; summing its two probabilities gives 1, so it does not add a factor of \(1/2\) to the probability of the event. For a switch in interval 1, the probability is \((0.10)(0.80)(0.70)=0.056\). For a switch in interval 2, it is \((0.90)(0.20)(0.70)=0.126\). For a switch in interval 3, it is \((0.90)(0.80)(0.30)=0.216\).

P(\text{exactly one switch}) =(0.10)(0.80)(0.70) +(0.90)(0.20)(0.70) +(0.90)(0.80)(0.30) =0.398.

Thus the probability of exactly one switch across the three intervals is 0.398. This is an event probability that sums over all paths with one switch, not the probability of any single fully specified path.

Representing Paths Across a Pedigree

A pedigree contains several meioses, each with its own transmitted-homologue path. At a given marker, collect the binary states for the meioses into an inheritance vector. For example, with two meioses, the vector \((0,1)\) says that the first transmission uses its parent's first labelled homologue and the second transmission uses its parent's second labelled homologue at that marker. The component labels refer to each transmitting parent's own homologues; 0 in one component does not identify the same physical homologue as 0 in another.

If there are \(M\) meioses and each has an independent, equally likely initial state, then each of the \(2^M\) inheritance vectors at the first marker has probability \(2^{-M}\). From one marker to the next, count how many components switch. If the meioses have independent recombination processes, the probability of the vector transition is the product of the appropriate switch and no-switch probabilities for its components.

Result: Let \(H_k=(H_{1k},\ldots,H_{Mk})\) be the inheritance vector at marker \(k\). If meioses are independent, the initial states are fair binary choices, and interval transitions are independent, then a specified sequence of vectors has probability

P(H_1=h_1,\ldots,H_K=h_K) =2^{-M}\prod_{k=1}^{K-1}\prod_{r=1}^{M} \theta_{rk}^{I_{rk}}(1-\theta_{rk})^{1-I_{rk}}, \qquad I_{rk}=\mathbf{1}(h_{r,k+1}\ne h_{r,k}).

Here \(\theta_{rk}\) is the recombination fraction for meiosis \(r\) in interval \(k\). If all meioses share the same interval-specific recombination fraction, write \(\theta_{rk}=\theta_k\).

Worked Example: A Two-Meiosis Pedigree Path

Consider two independent meioses tracked across three markers. The recombination fractions are \(\theta_1=0.10\) between markers 1 and 2 and \(\theta_2=0.25\) between markers 2 and 3 for both meioses. Suppose the inheritance vectors are \(H_1=(0,0)\), \(H_2=(1,0)\), and \(H_3=(1,1)\). Calculate the probability of this complete vector path.

At the first marker, the specified vector has probability \(1/4\), since there are two independent fair initial states. From marker 1 to marker 2, the first component switches and the second stays the same, giving transition probability \((0.10)(0.90)=0.09\). From marker 2 to marker 3, the first component stays the same and the second switches, giving \((0.75)(0.25)=0.1875\).

P(H_1=(0,0),H_2=(1,0),H_3=(1,1)) =\frac{1}{4}(0.09)(0.1875) =0.00421875.

The specified pedigree inheritance-vector path has probability approximately 0.004219. This calculation concerns transmission states only; it does not yet incorporate marker genotypes, disease phenotypes, or penetrance.

What the Calculation Does—and Does Not—Assume

A recombination fraction is bounded by 0 and \(1/2\) in the usual two-locus linkage model. When \(\theta_k=0\), the model assigns probability zero to a switch in that interval. When \(\theta_k=1/2\), the two endpoint states are equally likely regardless of the current state, so the interval gives no preference for staying or switching. More generally, a small \(\theta_k\) makes a switch less probable than no switch; it does not make any particular allele or disease status more probable by itself.

The path formula relies on the interval transition probabilities being independent in the specified model. A list of adjacent recombination fractions gives the probabilities of individual interval changes, but does not by itself specify dependence between those changes under crossover interference. If interval changes are dependent, the product of marginal transition probabilities is not justified; a joint model for the sequence of transitions is needed. Likewise, meiosis-specific recombination fractions should be used when the model assigns different maps to different transmissions, rather than silently assuming a common \(\theta_k\).

Common Mistake: Confusing Switches With Allele Changes

Wrong reasoning: “The marker allele changed from one marker to the next, so the transmitted homologue must have switched.”
Correction: The inheritance state records which parental homologue was transmitted, not the allele label observed at the marker. Allele values depend on the parent's marker genotypes and phase; an allele can differ across markers even when the same labelled homologue is transmitted, or look unchanged despite a switch.
Wrong reasoning: “Multiply the recombination fractions for every interval, because each interval is part of the path.”
Correction: Use \(\theta_k\) for a switch and \(1-\theta_k\) for no switch in interval \(k\), then include the initial-state probability. Multiplying only recombination fractions assigns probability to a switch in every interval, not to an arbitrary specified path.

Key takeaway: Encode each meiosis by a binary state at each ordered marker. A specified path starts with probability \(1/2\) per meiosis and accumulates one transition factor per interval: \(\theta\) for a switch and \(1-\theta\) for no switch, subject to the stated independence assumptions.

Check Your Understanding

Use the definitions and assumptions in this tutorial to answer the following.

For a single meiosis with two markers and recombination fraction \(0.12\), what is the probability of the specified path \((1,1)\)? Show the initial-state and transition factors.
For a single meiosis across three markers, what switch pattern is represented by the path \((0,1,0)\)? Express its probability using the two adjacent recombination fractions.
Why does the probability of a specified path include an initial-state factor, while the probability of a specified switch pattern summed over both initial states does not retain a factor of \(1/2\)?
With three meioses, how many possible inheritance vectors are there at one marker, and what is the probability of each if their initial states are independent and equally likely?
Why can adjacent recombination fractions be insufficient to determine a whole-path probability under an interference model?