Carrier Probability Calculator (Autosomal Recessive)

Background.

This carrier probability calculator combines what is known about two prospective parents into a single number: the chance that a child of theirs would be affected by an **autosomal recessive** condition. For each partner you choose the strongest piece of evidence that applies — nothing known, a confirmed carrier result, an unaffected sibling of someone affected, or a negative carrier screen — and the calculator assigns the corresponding carrier probability, multiplies the two, and applies the Mendelian one-in-four.

**Read this before you read the number.** What comes out is an idealised expectation from a textbook model, not a clinical risk assessment. It assumes one gene, two alleles, complete penetrance, two unaffected and unrelated parents, and a population sitting in Hardy-Weinberg equilibrium. Real reproductive risk assessment is done by a genetic counsellor or clinical geneticist using your actual family history, your ancestry-specific variant frequencies, and the specific panel your laboratory ran. This page ships **no incidence figure and no detection rate for any named condition** — both of those boxes contain neutral placeholders that you are expected to replace with numbers from a source that knows your condition. And it covers autosomal recessive inheritance only: for X-linked conditions, where the hemizygous sex is affected at frequency q rather than q², use the sex-linked inheritance calculator instead.

The three priors the calculator can assign each come from a different place. The population prior is Hardy-Weinberg: if the disease incidence is q², then q is its square root and the carrier frequency is 2(1 − q)q, the form Ogino and Wilson use in the genetic-counselling literature. With an incidence of 0.0004 that is q = 0.02 and a carrier frequency of 3.92 %. The obligate-carrier prior is 100 % by definition — a parent of an affected child must carry a copy. The sibling prior of two-in-three is the one that surprises people, and it is worth seeing the derivation. GeneReviews states the Mendelian split for a sib of an affected individual: at conception, a 25 % chance of being affected, a 50 % chance of being a heterozygote, and a 25 % chance of inheriting neither variant. If you already know the sibling is unaffected, the 25 % affected branch is eliminated, and the remaining 50 and 25 renormalise to 50 ÷ 75 = two-thirds carrier and one-third non-carrier.

The fourth option, a negative carrier screen, is the one to treat most carefully. A negative result reduces your carrier probability but never to zero, because no panel detects every pathogenic variant. The calculator applies the exact Bayesian update — prior × (1 − detection rate), divided by (1 − prior × detection rate) — which at a 3.92 % prior and a 90 % detection rate leaves a residual 0.41 %. The published approximation used by Nussbaum, Slotnick and Risch is the same expression with the denominator dropped, giving 0.392 %; the exact form is about 3.7 % higher, and this page uses the exact form because understating a residual risk is the unsafe direction. Their paper is titled 'Challenges in providing residual risks in carrier testing' for good reason: the honest residual risk depends on which variants your panel covers and on the variant spectrum in your ancestry group, neither of which this page can know.

Below the calculator you will find the full derivation of each prior, the offspring arithmetic including the often-overlooked carrier-child probability, the exact points where the model is refused rather than fudged, and a plain statement of what would change the answer that this page cannot see.

What is carrier probability calculator?

A carrier — a heterozygote — is someone who has one copy of a recessive variant and one working copy, and who therefore does not show the condition themselves. GeneReviews defines a carrier as an individual with a recessive pathogenic variant at a particular locus on one chromosome of a pair who is not expected to develop manifestations of the related condition. Carrier probability is the probability that a specified person is in that state, given whatever is known about them. It is a personal probability, not a population frequency, and that distinction is the whole point of this page: the population carrier frequency 2pq says what fraction of everybody carries a copy, while a carrier probability says what to believe about one named individual who may have evidence the population average does not. Three kinds of evidence move the number. Family history moves it up — a parent of an affected child is certain to be a carrier, and an unaffected sibling of an affected individual sits at two-in-three. A negative test moves it down, but only partly, in proportion to how much of the variant spectrum the test can see. And ancestry changes the starting point entirely, because incidence and carrier frequency vary substantially between populations for most recessive conditions. Once each partner has a carrier probability, the offspring arithmetic is short. Both parents must be carriers for a child to be affected, which happens with probability a × b if the two statuses are independent; and if both are carriers, the Punnett square for Aa × Aa gives one affected child in four. So the affected-child probability is a × b × ¼. The probability that a child is an unaffected carrier is a separate and larger quantity, ½ × (a + b − ab), because a child inherits one copy whenever exactly one parent passes one on. Both numbers are quoted at conception, following the GeneReviews convention, which matters for any condition with prenatal loss.

How to use this calculator.

1. 01Replace the disease incidence with the real figure for your condition and ancestry group. The default of 0.0004 is a placeholder chosen to make the arithmetic easy to follow, not a value for any particular disorder — incidence for recessive conditions varies by more than an order of magnitude between populations.
2. 02Choose the strongest evidence that applies to each partner. If a partner has both a family history and a test result, that combination is beyond this page's model and needs a genetic counsellor, because the two pieces of evidence must be combined in a single Bayesian table rather than by picking one.
3. 03If you select 'negative carrier screen', set the detection rate from your laboratory report for that exact panel and ancestry. A generic number will give a generic answer.
4. 04Read the headline as a 1-in-N figure, then sanity-check the population carrier frequency output against a published carrier frequency for your condition. If they disagree badly, the incidence you entered is for a different population than the carrier frequency you are comparing against.
5. 05Treat the result as an order of magnitude, not a precise value. An incidence quoted as 'about 1 in 2,500' supports two significant figures, so the honest reading of the worked example is 'roughly 1 in 2,600', not '1 in 2,603'.
6. 06Do not use this page if the partners are blood relatives. Consanguinity makes the two carrier statuses correlated rather than independent, so multiplying them understates the risk — sometimes by a large factor.

The formula.

The calculation runs in three stages: a Hardy-Weinberg step, a per-partner evidence step, and a Mendelian offspring step.

STAGE 1 — Hardy-Weinberg. For a fully recessive condition the affected individuals are exactly the aa homozygotes, so the incidence is q² and q is its square root. With an incidence of 0.0004, q = 0.02 and p = 0.98, giving a population carrier frequency of 2pq = 2 × 0.98 × 0.02 = 0.0392, or 3.92 %. Ogino and Wilson state this step directly: assuming Hardy-Weinberg equilibrium and a disease-allele frequency q, the carrier frequency 2(1 − q)q — or approximately 2q if q is small — can be derived from the disease frequency, which equals q². One refinement is worth recording rather than hiding: a person presenting for counselling is by construction unaffected, so the strictly correct prior is 2pq ÷ (1 − q²) = 2q ÷ (1 + q) = 0.0392156863 rather than 0.0392. The difference is 0.04 % of the value, and this page uses the unconditional 2pq to match the cited source and standard clinical convention. Both numbers are stated here so the choice is visible.

STAGE 2 — per-partner evidence. A known carrier is 100 %. The population prior is 2pq. The unaffected-sibling prior of two-thirds comes from conditioning the GeneReviews 25 / 50 / 25 split on the sibling being unaffected: eliminating the 25 % affected branch leaves 50 and 25, which renormalise to 50 ÷ 75 = 2/3. The negative-screen prior is Bayes' rule in the joint-divided-by-sum-of-joints form Ogino and Wilson set out: the carrier hypothesis has joint probability c(1 − DR), the non-carrier hypothesis has joint probability (1 − c) since a non-carrier cannot test positive, and the posterior is c(1 − DR) ÷ [c(1 − DR) + (1 − c)], which simplifies to c(1 − DR) ÷ (1 − c·DR). At c = 0.0392 and DR = 0.9 that is 0.00392 ÷ 0.96472 = 0.0040634, or 0.41 %.

STAGE 3 — offspring. If a and b are the two carrier probabilities and both parents are unaffected, then P(both carriers) = ab, and given both carriers the Aa × Aa Punnett square gives one aa child in four. So P(affected child) = ab ÷ 4. The carrier-child probability is derived separately: a child is a heterozygote when both parents are carriers and only one passes the variant (probability ab × ½), or when exactly one parent is a carrier and passes it (probability [a(1 − b) + b(1 − a)] × ½). Adding and simplifying gives ½(a + b − ab). A useful internal check: set both partners to 'known carrier' and the model must collapse to the textbook Punnett square — 25 % affected, 50 % carrier, 25 % neither — and conditioning that on being unaffected regenerates exactly the 2/3 sibling prior the page uses elsewhere. It does, and the test suite asserts it.

ROUNDING STAGE — FINAL ONLY. The square root, every product, the Bayesian ratio and the conversion to percentages are all carried in arbitrary-precision decimal arithmetic; each returned number is rounded once, to ten decimal places, on return. The 1-in-N string rounds the reciprocal to a whole number after the probability has been computed at full precision, and that integer never feeds anything else.

THREE THINGS ARE REFUSED RATHER THAN ANSWERED. An incidence of exactly 0 or exactly 1 is rejected, because a fixed locus has no carriers and the 1-in-N output would diverge. A detection rate of 100 % or more is rejected, because it would report a residual risk of exactly zero and no laboratory can support that claim — this is the substantive point of the Nussbaum, Slotnick and Risch paper, whose approximation RCR ≅ 2(q_PATH − q_KWN) is nonzero precisely because the frequency of all pathogenic variants always exceeds the frequency of the ones a panel detects. An unrecognised evidence option is rejected rather than silently defaulted.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Ogino, S. & Wilson, R. B. (2004). Bayesian analysis and risk assessment in genetic counseling and testing. Journal of Molecular Diagnostics 6(1):1–9. PMC1867463. Source for three things this page uses: the 2/3 prior for an unaffected sib of an affected individual (Figure 2B); the joint-divided-by-sum-of-joints structure of the Bayesian posterior used for a negative screen (Figure 2C); and the derivation of the carrier frequency 2(1 − q)q from the disease frequency q² under Hardy-Weinberg equilibrium. Independent, free full text; retrieved 2026-07-29.
2. [2]GeneReviews® — Gaucher Disease, section 'Genetic Counseling · Risk to Family Members · Sibs of a proband'. NCBI Bookshelf NBK1269, last update 7 December 2023. Source for the Mendelian split quoted on this page: if both parents are known heterozygotes, each sib of an affected individual has at conception a 25 % chance of being affected, a 50 % chance of being a heterozygote, and a 25 % chance of inheriting neither familial pathogenic variant. Cited for the inheritance arithmetic only — no figure specific to Gaucher disease is used anywhere on this page. Independent, free, NIH-hosted and expert-authored; retrieved 2026-07-29.
3. [3]Nussbaum, R. L., Slotnick, R. N. & Risch, N. J. (2021). Challenges in providing residual risks in carrier testing. Prenatal Diagnosis 41(9):1049–1056, doi:10.1002/pd.5975. PMC8453722. Consulted as an independent second authority on the residual-risk calculation. Its Appendix A gives RCR ≅ 2(q_PATH − q_KWN), where q_PATH is the summed frequency of all pathogenic variants and q_KWN that of the variants a panel detects — which is exactly the numerator of the exact Bayesian posterior this page implements, with the denominator (1 − prior × detection rate) dropped. The exact form used here is therefore slightly larger, and the paper's central argument is that a reliable residual risk depends on panel content and ancestry-specific variant frequencies. Independent, free via PubMed Central; retrieved 2026-07-29.
4. [4]Gregg, A. R. et al. (2021). Screening for autosomal recessive and X-linked conditions during pregnancy and preconception: a practice resource of the American College of Medical Genetics and Genomics (ACMG). Genetics in Medicine 23(10):1793. The governing US practice resource for how carrier screening should be offered and interpreted, cited to point readers at the professional standard rather than at this educational model. Citation details verified from the freely available correction notice, PMC8776567, doi:10.1038/s41436-021-01300-z. Independent; the practice resource itself is PAYWALLED at the publisher — no numeric figure on this page is drawn from it. Checked 2026-07-29.
5. [5]National Research Council (US) Committee on DNA Forensic Science (1996). The Evaluation of Forensic DNA Evidence, Chapter 4, section 'Random Mating and Hardy-Weinberg Proportions'. NCBI Bookshelf NBK232608. Source for the Hardy-Weinberg step that converts the disease incidence into a population carrier frequency: the proportion of persons with two different alleles is twice the product of the two frequencies. Independent, free, US National Academies; retrieved 2026-07-29.