Mendelian Transmission Probabilities Along Pedigrees

Why Transmission Paths Matter

A pedigree records who is related to whom, but a path through that pedigree also gives a way to describe how a particular allele copy might pass from one individual to another. Each parent–offspring link on the path involves a meiosis: the parent passes one of the two copies at an autosomal locus to the offspring. If the question concerns a specified allele copy, it must be transmitted at every link for it to reach the path’s endpoint.

The calculation is simple once the path is clear: count the successive transmissions and use the probability for each one. The main care is conceptual. A probability for a tracked copy along one specified path is not necessarily the probability that someone carries the same allele value, such as A. Another allele copy, or another route through a pedigree, may also result in that allele value being present.

Definition: A specified parent–offspring path is a sequence of successive parent–offspring links in a pedigree, with the transmitting parent identified for each link. A tracked allele copy is one particular copy at a locus in the starting individual, followed along that path. A transmission event occurs when the copy being tracked is passed from a parent to the next individual on the path.

Mendelian Transmission Along One Path

For the basic calculation, consider a single autosomal locus in diploid individuals. Assume ordinary Mendelian segregation, no mutation at the locus, and no segregation distortion. In each meiosis the transmitting parent passes one of its two copies at that locus. A particular copy in that parent therefore has probability \(1/2\) of being passed to the offspring. The calculation does not require the other parent’s genotype, because the question is whether the specified copy passes along the chosen path.

If the path has \(m\) successive parent–offspring links, let \(E_j\) denote the event that the tracked copy is transmitted at link \(j\), for \(j=1,\ldots,m\). Given that the copy has reached the parent at link \(j\), its probability of passing at that meiosis is \(1/2\). Successive meioses are independent under the standard Mendelian model. Equivalently, applying the multiplication rule successively gives:

P(E_1\cap E_2\cap\cdots\cap E_m) =\prod_{j=1}^{m}P(E_j\mid E_1,\ldots,E_{j-1}) =\left(\frac{1}{2}\right)^m.

For \(j=1\), the conditional probability in the product is simply \(P(E_1)\). For each later link, conditioning on the preceding transmissions means the tracked copy has arrived at the current transmitting parent; it does not change the probability that this parent passes that copy in its own meiosis. This is why the probability for the whole specified path is the product of the probabilities for its individual links.

The count \(m\) is the number of parent–offspring links, or meioses, on the specified path. It is not the number of individuals named on the path. A path with a starting individual and one child has one link; a path with a starting individual, child, and grandchild has two.

Result: For a tracked autosomal allele copy, if each of \(m\) successive meioses follows ordinary Mendelian segregation and the transmission events are independent, the probability that the copy passes through all \(m\) links on the specified path is \(2^{-m}\). For \(m=0\), no transmission is required, so the probability is \(1\).

Worked Path Calculations

Worked Example: From a Grandparent to a Grandchild

Suppose a hypothetical pedigree has a path from individual G to G’s child P and then to P’s child C. At one autosomal locus, track a particular copy in G. What is the probability that this same copy passes along both links and reaches C by this specified path?

There are two parent–offspring links: G to P and P to C. Thus \(m=2\). Under Mendelian segregation, the tracked copy has probability \(1/2\) of passing from G to P. Conditional on reaching P, it has probability \(1/2\) of passing from P to C. The path probability is therefore:

P(\text{copy passes from G to C along this path}) =\frac{1}{2}\times\frac{1}{2} =\frac{1}{4} =0.25.

The result is a 25% probability that this particular copy passes along both links. It is not, without additional information, the probability that C has an allele with the same value as G’s copy. C could receive an allele with that value from another source.

Worked Example: Counting Four Successive Meioses

In another hypothetical pedigree, a specified path runs from individual A through three successive descendants, B, C, and D, to E. At each of the four links, follow the same allele copy initially present in A. Calculate the probability it reaches E along this path.

The links are A to B, B to C, C to D, and D to E, so \(m=4\). Each link contributes a factor of \(1/2\), provided the copy has reached the transmitting parent. Multiplying the four factors gives:

P(\text{copy passes along all four links}) =\left(\frac{1}{2}\right)^4 =\frac{1}{16} =0.0625.

Thus, under the stated assumptions, the probability is 6.25%. The names of the individuals are only labels: the calculation depends on the four transmissions, not on how many generations the pedigree diagram displays in some other direction.

Worked Example: Conditioning on Earlier Transmissions

Consider a hypothetical five-link path from F to K through four successive parent–offspring links before the final link to K. Suppose the first four transmissions of a tracked copy are known to have occurred. What is the probability that the copy reaches K, conditional on those four transmissions? What was the probability of the full five-link path before any of its transmissions were known?

Once the first four transmissions are given, the tracked copy has reached the parent transmitting to K. There is one remaining meiosis, so the conditional probability of the final transmission is:

P(E_5\mid E_1\cap E_2\cap E_3\cap E_4)=\frac{1}{2}.

Before learning whether any of the five transmissions occurred, all five must succeed. The unconditional path probability is:

P(E_1\cap E_2\cap E_3\cap E_4\cap E_5) =\left(\frac{1}{2}\right)^5 =\frac{1}{32} =0.03125.

The conditional probability is 50%, while the probability of the entire five-link path before observing any transmissions is 3.125%. These answer different questions: conditioning on four successful transmissions removes those four events from the remaining calculation.

Derivation and Scope of the Product Rule

The product calculation can be justified directly from the chain rule for probabilities. For events \(E_1,\ldots,E_m\), the probability that all occur is the probability of the first event multiplied by the conditional probability of each later event given all earlier events. In the pedigree setting, given that the copy has reached the parent at each successive link, Mendelian segregation assigns probability \(1/2\) to its transmission. Therefore every factor in the chain-rule product is \(1/2\):

\begin{aligned} P(E_1\cap E_2\cap\cdots\cap E_m) &=P(E_1)P(E_2\mid E_1)\cdots P(E_m\mid E_1\cap\cdots\cap E_{m-1})\\ &=\underbrace{\frac{1}{2}\times\frac{1}{2}\times\cdots\times\frac{1}{2}}_{m\ \text{transmissions}}\\ &=2^{-m}. \end{aligned}

This argument also shows exactly which assumptions support the result: each meiosis must give the tracked copy probability \(1/2\), and earlier transmissions must not alter the probability at a later meiosis once the copy has reached that later parent. The standard model of independent Mendelian transmissions supplies these conditions. If a problem specifies segregation distortion, mutation, or a non-autosomal inheritance pattern, the relevant per-link probabilities and assumptions may differ; the simple \(2^{-m}\) result should not be applied automatically.

For a path with different per-link probabilities \(q_1,\ldots,q_m\), the same chain-rule reasoning gives a product of the appropriate conditional probabilities. The familiar \(2^{-m}\) formula is the special case in which every \(q_j=1/2\). In this tutorial’s ordinary autosomal setting, following one particular copy gives that same one-half probability at every meiosis.

Allele Copies and Allele Values

A distinction between a copy and its value prevents a common misinterpretation. At a locus, an individual may carry alleles with values A and a. A tracked copy means one particular inherited copy, such as the A copy received from a specified parent. An allele value means the label A, regardless of which physical copy carries it. A person with genotype AA has two copies with value A, but transmission of the value A is certain from that person; transmission of one designated copy is still a one-half event.

Worked Example: A Homozygous Parent and a Tracked Copy

Suppose a hypothetical parent has genotype AA at an autosomal locus and has one child. What is the probability the child receives the allele value A from this parent? What is the probability the child receives one particular, designated A copy from this parent?

Because both copies in the parent have value A, whichever copy is passed has value A. Therefore:

P(\text{child receives allele value A from the parent})=1.

If instead one of the parent’s two A copies is designated in advance and tracked, the parent passes that particular copy with probability \(1/2\):

P(\text{child receives the designated A copy})=\frac{1}{2}.

These probabilities differ because they describe different events. The first asks whether the transmitted allele has value A; the second asks whether a particular copy is the one transmitted. Along a multistep path, the \(2^{-m}\) formula concerns the latter kind of event unless the problem explicitly defines a different allele-value event.

Common Pitfall and Practical Use

A frequent error is to count the generations shown in a pedigree rather than the transmission links on the specified route. Another is to treat the probability that a tracked copy follows one route as the probability that the endpoint carries the same allele value by any means. Both errors change the event being calculated. First trace the exact parent–offspring path, then state whether the question follows one particular copy or asks about an allele value.

Transmission-path probabilities are useful when describing how inheritance can connect individuals across several meioses. They give a clear probability for one specified route under a stated Mendelian model. The calculation does not, by itself, combine multiple routes or determine whether two relatives carry copies inherited from a shared ancestor; those are separate questions requiring additional definitions.

Before calculating, check that the locus and inheritance assumptions fit the one-half rule, mark each link on the chosen path, and count the links. Then multiply the per-link transmission probabilities. If the events are all ordinary Mendelian transmissions of one tracked copy, this reduces to \(2^{-m}\).

Check Your Understanding

Use the specified path and event in each question. Unless stated otherwise, assume independent Mendelian transmission at an autosomal locus.

A tracked copy passes along a path with three parent–offspring links. What is the probability it reaches the endpoint along that path?
A path has two meioses. Explain why its transmission probability is not counted as three merely because three individuals appear on the path.
For a parent with genotype AA, compare the probability of transmitting allele value A with the probability of transmitting one designated A copy.
On a four-link path, the first three transmissions of a tracked copy are known to have occurred. What is the conditional probability of the fourth transmission?
Explain why the probability that a particular copy follows one specified path need not equal the probability that the endpoint carries the same allele value.