Carrier Probability Calculator (Autosomal Recessive)

Background.

Two people with no sign of a recessive condition can still have a child who has it, if both quietly carry one altered copy of the same gene. How likely that is depends on what each partner knows about themselves, and that is the question worked through here.

Each partner gets a carrier probability from the evidence you choose — nothing known, a known carrier, the unaffected sibling of an affected person, or a negative screen — and the two multiply. Both must be carriers, and if both are, each child has a one-in-four chance of inheriting two altered copies. Multiplying small numbers is what keeps recessive conditions rare: with a carrier frequency of 3.92% for each partner, the chance that both carry is 0.1537% and the chance of an affected child is 1 in 2,603.

Everything starts from the disease incidence you enter. The 0.0004 in the box, one in 2,500, is a neutral placeholder rather than the figure for any named condition; incidence differs between conditions and between ancestry groups, so replace it with a published figure for your condition and population.

This is a textbook model — one gene, complete penetrance, two unaffected partners who are not related to each other, a population in Hardy-Weinberg equilibrium — meant for understanding how the numbers combine. It is not a clinical risk assessment. Real reproductive risk depends on your family history, your ancestry and the exact test performed, and should be assessed and interpreted with a genetic counsellor or clinical geneticist.

What is carrier probability calculator?

A carrier of an autosomal recessive condition has one working copy of the gene and one disease-causing variant, and is usually unaffected. Carrier probability is the chance that a particular person is in that state given what is known about them, which can differ sharply from the population carrier frequency: a stranger and the sibling of an affected child start from very different places. The offspring risks are Mendelian. A carrier passes the variant to each child with probability one half, independently in every pregnancy, so two carrier parents face, at conception, a 25% chance of an affected child, 50% of a carrier child and 25% of a child with neither variant — the split GeneReviews gives for the brothers and sisters of an affected person.

How to use this calculator.

1. 01Replace the disease incidence, written as a decimal (0.0004 is one in 2,500), with the published figure for your condition and the ancestry group that matches you.
2. 02For each partner, choose the single strongest piece of evidence: nothing known, known carrier, unaffected sibling of an affected person, or negative carrier screen.
3. 03If either partner had a negative screen, enter the detection rate printed on that laboratory's report for that panel and ancestry; the 90% already filled in is a placeholder. A rate of 100% is not accepted.
4. 04Compare the population carrier frequency output with a published carrier frequency for your condition. A large mismatch means the incidence you entered describes a different population.
5. 05If the partners are blood relatives, the condition is X-linked, or one partner has both a family history and a test result, this model does not fit; take the question to a genetic counsellor.

The formula.

Three steps. First, Hardy-Weinberg: affected people are the homozygotes, so incidence = q², the variant frequency is q = √incidence, and the population carrier frequency is 2pq with p = 1 − q. Second, each partner's carrier probability: 2pq when nothing is known; 1 for a known carrier; 2/3 for an unaffected sibling, because knowing the sibling is unaffected removes the 25% affected branch and leaves 50 parts carrier out of 75; and after a negative screen with detection rate DR, the Bayesian posterior c(1 − DR) ÷ (1 − c·DR), where c is the population figure. For small carrier frequencies that posterior is close to c(1 − DR), the simpler residual-risk approximation discussed by Nussbaum, Slotnick and Risch; the exact form is slightly larger. Third, with partner probabilities a and b, P(both carriers) = ab, P(affected child) = ab ÷ 4, and P(unaffected carrier child) = ½(a + b − ab), the chance that exactly one parent passes the variant on. The 1-in-N figure is 1 ÷ P(affected child), rounded to a whole number. An incidence of 0 or 1 is refused because it leaves no carriers to count, and a 100% detection rate because it would claim a residual risk of exactly zero, which no real panel can support.

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.