What carrier risk is, and why it is a chain of probabilities
Carrier risk answers one question: what is the chance this couple's child inherits two pathogenic copies of a gene and is therefore affected? For an autosomal recessive condition the answer is a product of three independent probabilities — that the mother carries a variant, that the father carries one, and that both happen to transmit it to the same child.
The third factor is the only one that is fixed. Two heterozygous carriers, Aa × Aa, produce offspring in the ratio 1 AA : 2 Aa : 1 aa, so exactly one quarter are affected and one half are unaffected carriers. You can read that off a Punnett square. The first two factors are where all the uncertainty lives, and they are what a carrier screen exists to reduce.
Because the factors multiply, the risk collapses fast as either parent's carrier probability falls. Two confirmed carriers face a 25% risk per pregnancy. Two untested people at a 1 in 25 population carrier frequency face 0.04 × 0.04 × 0.25 = 0.0004, which is 1 in 2,500 — 625 times lower — and that figure is nothing other than the population incidence of the condition, which is the consistency check that tells you the chain is set up correctly.
Where the carrier probabilities come from, and how a negative screen changes them
Three sources feed the two prior probabilities, in descending order of certainty. A confirmed pathogenic variant on a laboratory report makes the probability 100%. A pedigree relationship gives a Mendelian fraction: the unaffected sibling of an affected person has a 2 in 3 chance of being a carrier, because the affected aa outcome is already excluded from the 1:2:1 ratio, leaving 2 Aa against 1 AA. An obligate carrier — a parent of an affected child — is 100%. Failing either of those, the population carrier frequency for that person's ancestral group applies, which you can obtain from a published incidence with the Hardy-Weinberg equilibrium calculator.
A negative carrier screen does not make the probability zero, because no panel detects every pathogenic variant at a locus. Bayes' theorem gives the residual. If the prior is P and the panel's detection rate is d, the only way to be a carrier and still screen negative is to carry an undetected variant, with probability P(1 − d). The only ways to screen negative at all are that, plus genuinely not being a carrier, with probability (1 − P). Divide the first by the sum and you have the posterior.
The behaviour of that formula is worth internalising. At a 4% prior and 90% detection, the residual is 0.004 ÷ 0.964 = 0.41%, roughly a tenfold reduction. Pushing detection from 90% to 99% takes it to 0.0004 ÷ 0.9604 = 0.042%, another tenfold. Residual risk falls roughly in proportion to (1 − d), so the last few percentage points of detection matter far more than the first fifty.
The other two patterns rearrange the same logic. X-linked recessive: a carrier mother transmits the variant to half her children; it causes disease in sons, who have no second X, and carrier status in daughters. Across children of both sexes that is one quarter affected and one quarter carrier daughters, and the father's status is irrelevant unless he is himself affected. Autosomal dominant: one variant copy suffices, so an affected heterozygous parent transmits it to half their children, and with complete penetrance there is no unaffected-carrier category at all.
Worked example: a couple where she screens negative and he is a known carrier
A couple ask about a recessive condition with a population carrier frequency of 1 in 25 in their ancestral group. He has a confirmed pathogenic variant. She has screened negative on a panel with a stated 90% detection rate.
- Her prior. 1 ÷ 25 = 0.04, so 4%.
- Chance she is a carrier and screens negative anyway. 0.04 × (1 − 0.90) = 0.04 × 0.10 = 0.004.
- Chance she is not a carrier. 1 − 0.04 = 0.96. Anyone in this group screens negative.
- Total chance of a negative screen. 0.004 + 0.96 = 0.964.
- Her residual carrier probability. 0.004 ÷ 0.964 = 0.004149, or 0.415% — about 1 in 241.
- His probability. Confirmed carrier, so 1.00.
- Chance a child is affected. 0.004149 × 1.00 × 0.25 = 0.001037, or 0.1037%.
- As odds. 1 ÷ 0.001037 = 1 in 964.
- Chance the child is an unaffected carrier. Half of 0.004149 + 1 − 0.004149 gives 0.5 × (0.004149 + 1 − 0.004149) = 50%, because he transmits his variant to half his children regardless.
Compare that to where they started. Before her screen, with him confirmed, the risk was 0.04 × 1.00 × 0.25 = 1%, or 1 in 100. Her negative result moved it to 1 in 964 — a tenfold improvement, exactly the (1 − d) factor. It did not move it to zero, and the difference between 1 in 964 and zero is the whole reason residual risk is quoted rather than a clean negative.
This is arithmetic, not counselling
These numbers are population-genetic probabilities computed from the figures you type in. A reproductive risk assessment requires a clinician who can see the actual variants reported, the full pedigree, the ancestry-specific detection rate for the panel that was run, and the prenatal, preimplantation and donor options available. Use this calculator to understand the structure of the arithmetic, and take any real decision to a certified genetic counsellor or clinical geneticist.
How to read the number you get back
The per-pregnancy risk is the number that matters, and it does not change. Each conception is an independent draw. A couple with a 25% risk who have had one affected child still face 25% next time, and a couple who have had three unaffected children still face 25%. The gambler's instinct that risk is somehow used up is the single most common misunderstanding in a counselling room.
Sibship risk is a different question with a different answer. The chance that at least one of k children is affected is 1 − (1 − r)k. At r = 0.25 and two children that is 1 − 0.752 = 43.75%; at four children, 68.4%. Both figures are correct and they answer different questions, so be explicit about which one you are quoting.
Calibrate against the population baseline. The most useful comparison is almost always the general-population risk for the same condition, which for a 1 in 25 carrier frequency is 1 in 2,500. A residual risk of 1 in 964 is above that baseline; a residual of 1 in 232,000, which is where two negative screens at 90% detection land a couple, is far below it. Quoting a risk without its baseline leaves people unable to interpret it.
Watch the carrier-child figure separately. It is often larger than people expect and it has different consequences: an unaffected carrier child faces no health effect from the condition, but will face the same reproductive question a generation later. With two carrier parents, half of all children are carriers — twice as many as are affected.
Risk of an affected child by parental status, at a 1 in 25 carrier frequency
| Mother | Father | Chance of an affected child | That is 1 in |
|---|---|---|---|
| Confirmed carrier | Confirmed carrier | 25% | 4 |
| Confirmed carrier | Untested, population risk | 1.00% | 100 |
| Confirmed carrier | Negative screen | 0.1037% | 964 |
| Untested, population risk | Untested, population risk | 0.0400% | 2,500 |
| Untested, population risk | Negative screen | 0.00415% | 24,100 |
| Negative screen | Negative screen | 0.000430% | 232,324 |
Every entry is the product of the two carrier probabilities and one quarter. The fourth row equals the population incidence of the condition, which is the check that the whole scheme is internally consistent.
Mistakes that produce a misleading risk figure
- Treating a negative screen as a zero. No panel detects every variant. The residual is prior × (1 − detection) ÷ [prior × (1 − detection) + (1 − prior)], and it is never zero for a non-zero prior.
- Using a detection rate from the wrong ancestral group. The same panel can have very different detection rates in different populations, because the variants it targets were characterised in specific groups.
- Applying a carrier frequency across ancestries. Carrier frequencies for recessive conditions vary widely between populations, and importing one group's figure to another gives a number with no meaning.
- Believing risk is used up by previous children. Each pregnancy is independent. Three unaffected children do not make the fourth safer, and one affected child does not make the next safe.
- Confusing per-pregnancy risk with sibship risk. 25% per child and 68.4% for at least one affected among four children are both correct; quoting one while meaning the other misleads badly.
- Using the sibling-of-an-affected prior as 1/2. An unaffected sibling of an affected person has a 2 in 3 carrier probability, not 1 in 2, because the affected outcome has already been excluded.
- Forgetting consanguinity. Related parents are far more likely to carry the same variant, so the independence assumption behind multiplying the two priors no longer holds.
Consanguinity, and why this calculator does not simply multiply
The whole scheme assumes the two parents' carrier statuses are independent events. For a related couple that is false: they may have inherited the same variant from a shared ancestor. Multiplying two population priors then understates the risk, sometimes substantially.
For a couple whose carrier status is unknown, the standard population treatment uses the coefficient of inbreeding F of their offspring, which is 1/16 for the children of first cousins, 1/32 for first cousins once removed and 1/64 for second cousins. The chance the child is homozygous is then F·q + (1 − F)·q2, where the first term covers the two alleles being identical by descent and the second covers the ordinary independent route.
Put numbers on it. For a condition with q = 0.02 — a carrier frequency near 1 in 25 and an incidence of 1 in 2,500 — first-cousin parents give (1/16 × 0.02) + (15/16 × 0.0004) = 0.00125 + 0.000375 = 0.001625, about 1 in 615. That is roughly four times the population baseline of 1 in 2,500, and the multiple grows as the condition gets rarer, because the identity-by-descent term scales with q while the independent term scales with q2. This calculator does not fold that route in automatically, because doing so requires knowing which route each entered probability came from; work it by hand from the formula above when the couple are related.
Key terms
- Carrier
- A heterozygote for a recessive pathogenic variant. Unaffected, but able to transmit the variant to half their children.
- Obligate carrier
- Someone whose carrier status is certain from the pedigree rather than from a test — for example either parent of a child affected by an autosomal recessive condition.
- Detection rate
- The proportion of pathogenic variants at a locus that a screening panel identifies in a given ancestral group. It sets the size of the residual risk after a negative result.
- Residual risk
- The probability that someone is still a carrier after a negative screen. Computed by Bayes' theorem from the prior and the detection rate.
- Penetrance
- The proportion of people with a genotype who show the phenotype. The dominant calculation here assumes it is complete; reduced penetrance lowers the affected risk and creates unaffected carriers.
- Coefficient of inbreeding (F)
- The probability that an individual's two alleles at a locus are identical by descent. 1/16 for the child of first cousins.