Sex-Linked Inheritance Calculator

Background.

Sex-linked inheritance is where a Punnett square stops giving one answer and starts giving two, because sons and daughters do not face the same odds. This calculator takes the mother's genotype, the father's genotype and whether the condition is X-linked recessive or X-linked dominant, and returns the affected and heterozygous probabilities separately for sons and for daughters.

The asymmetry comes from the chromosomes themselves. A mother has two X chromosomes and passes one at random, exactly as with any autosomal gene. A father has one X and one Y, and which one he passes determines the child's sex: his X goes to every daughter, his Y to every son. Two consequences follow immediately. First, no X-linked condition ever passes from father to son, because a son never receives his father's X — a pattern so distinctive that its absence in a pedigree is one of the strongest clues that a condition is X-linked. Second, males are hemizygous: they carry one copy of every X-linked gene with no second copy to mask it, which is why X-linked recessive conditions such as red-green colour vision deficiency and haemophilia are so much more common in males.

Working through the default case makes the shape clear. A mother who is a heterozygous carrier (X^A X^a) with an unaffected father (X^A Y), for an X-linked recessive condition: half her sons inherit the affected X and are affected, half do not; none of her daughters are affected, because each receives a working allele from her father, but half of them are carriers like their mother. Averaged over both sexes that is 25% affected, but the per-sex figures — 50% of sons, 0% of daughters — are the numbers that actually mean something, which is why the son figure is the headline output.

The inheritance-pattern selector changes which allele causes the condition, not what the genotypes are. Under X-linked recessive inheritance the recessive allele is the disease allele and a daughter needs two copies to be affected. Under X-linked dominant inheritance the dominant allele is the disease allele and one copy is enough, so a heterozygous daughter who would be an unaffected carrier in the first case is affected in the second. That is why the output is labelled 'heterozygous daughters' rather than 'carriers' — the genotype is the same in both modes, the clinical meaning is not.

The limits matter here more than on most genetics pages, because real families search for these conditions. These are exact ratios for one X-linked locus with full penetrance, not a personal risk figure. A heterozygous female is not guaranteed to be unaffected: X-inactivation silences one X in each cell at random, and when that process is skewed — defined as more than 75% of cells silencing the same parental X, seen in an estimated 1.5–23% of females — a carrier can show symptoms. In Duchenne and Becker muscular dystrophy, 2.5–18% of carriers present with some degree of muscle weakness. New mutations arise that no cross prediction can anticipate, and germline mosaicism in a parent can produce an affected child from a genotype this model calls unaffected. Sex-chromosome aneuploidies, pseudoautosomal genes that recombine between X and Y, and Y-linked genes are all outside the model entirely. Carrier testing and real risk counselling belong with a clinical geneticist or a certified genetic counsellor.

What is sex-linked inheritance calculator?

A sex-linked gene is one carried on a sex chromosome rather than on an autosome. In humans that almost always means the X chromosome, which carries around 800 protein-coding genes, against the Y chromosome's few dozen. This calculator models X-linked genes outside the pseudoautosomal regions — the small segments at each end of X and Y that pair and recombine, and whose genes therefore behave like autosomal ones.

Females are XX, so they carry two copies of every X-linked gene and can be homozygous or heterozygous like any autosomal locus. Males are XY, so they carry exactly one copy and are described as hemizygous rather than homozygous or heterozygous: there is no partner allele, so whatever they inherit is expressed. That single fact drives the entire pattern. For a recessive condition, a female needs two copies to be affected while a male needs only one, which is why X-linked recessive conditions appear predominantly in males, and why an affected male's mother is almost always a carrier.

Transmission follows the chromosomes. A father passes his X to all of his daughters and his Y to all of his sons, so an affected father transmits the allele to every daughter and to no son. A mother passes one of her two X chromosomes at random to each child regardless of sex, so a carrier mother transmits the allele to half her sons and half her daughters. The resulting pedigree signature — affected males in several generations, connected through unaffected carrier females, with no male-to-male transmission — is the classic X-linked recessive pattern.

X-linked dominant inheritance inverts part of that. One copy is enough in either sex, so affected females outnumber affected males in most pedigrees simply because females have twice the chance of receiving an affected X. An affected father has all affected daughters and no affected sons; an affected heterozygous mother transmits to half her children of either sex. Affected females often, though not always, show milder features than affected males, which the NLM's MedlinePlus Genetics attributes to their second, unaffected X chromosome.

How to use this calculator.

1. 01Choose the inheritance pattern first. X-linked recessive means the recessive allele causes the condition and a daughter needs two copies; X-linked dominant means the dominant allele causes it and one copy is enough in either sex.
2. 02Set the mother's genotype. She has two X chromosomes: X^A X^A, X^A X^a or X^a X^a. If you do not know it but you do know the family history, work it out with the pedigree probability calculator first — that is a different calculation, and doing it by eye is the most common way these numbers go wrong.
3. 03Set the father's genotype. He has one X, so only two options exist: X^A Y or X^a Y. There is no heterozygous male at an X-linked locus.
4. 04Read the per-sex figures first. 'Affected sons' and 'affected daughters' are conditional on the child's sex and assume nothing about the sex ratio. They are the numbers a genetics problem is normally asking for.
5. 05Treat the overall figure as a model marginal. It averages the two per-sex results assuming sons and daughters are exactly equally likely, which is close but not exactly true in real populations.
6. 06Read 'heterozygous daughters' carefully. Under recessive inheritance these are usually unaffected carriers; under dominant inheritance they are affected. The same genotype, opposite clinical meaning — which is why the row is not labelled 'carriers'.
7. 07Check the genotype ratios to see the actual chromosomes involved, and use them to confirm the no-father-to-son rule: son genotypes never change when you change the father's X.

The formula.

Two small Punnett squares, one per offspring sex.

A mother passes one of her two X chromosomes, each with probability ½. A father passes either his X or his Y, and that choice IS the child's sex — his X makes a daughter, his Y makes a son. So each sex has exactly two equally likely genotypes:

daughters : mother's X (2 options) × father's X (fixed) sons : mother's X (2 options) + father's Y

P(genotype | sex) = matching outcomes ⁄ 2

Affected status depends on the mode. In X-linked recessive mode the disease allele is the recessive one, so a son is affected if his single X carries it and a daughter is affected only if both of her X chromosomes carry it. In X-linked dominant mode the disease allele is the dominant one, so a son is affected if his single X carries it and a daughter is affected if either of hers does. Sons behave identically in the two modes — hemizygosity means one copy is always enough — and the entire difference between the patterns lives in the daughters.

Working the default case: mother X^A X^a, father X^A Y, X-linked recessive. Daughters get the father's X^A plus X^A or X^a from the mother, giving X^A X^A and X^A X^a in equal numbers — 0% affected, 50% heterozygous. Sons get X^A or X^a from the mother plus the Y, giving X^A Y and X^a Y in equal numbers — 50% affected. Averaging over sexes: ½ × 50% + ½ × 0% = 25% of all offspring.

ROUNDING STAGE. Rounding happens only at the final return, to ten decimal places. Each per-sex probability is a count out of two, and the overall figure is their mean, so the only values reachable anywhere are 0, 25, 50, 75 and 100. Nothing is rounded part-way through and there is no threshold in the calculation.

WHY THERE IS NO FATHER-TO-SON TRANSMISSION. It is not a probabilistic statement but a structural one: a son receives his father's Y chromosome by definition, so the father's X allele cannot reach him. The test suite pins this — changing the father's genotype leaves every son output identical in both inheritance patterns. The same fact means an affected father transmits his X allele to every one of his daughters, with no randomness at all.

WHERE THE MODEL BREAKS. X-inactivation silences one X per cell at random in females; when that is skewed toward silencing the working allele, a heterozygous female can manifest a recessive condition. Reduced penetrance and variable expressivity break the assumption that a genotype determines a phenotype. Sex-chromosome aneuploidies such as 45,X and 47,XXY cannot be represented by a model with two gametes and two sexes. Genes in the pseudoautosomal regions recombine between X and Y and are inherited like autosomal genes, so this page does not apply to them. Y-linked genes are not modelled at all. And de novo mutation and germline mosaicism can produce an affected child from parents whose genotypes predict none — no forward cross calculation, on this page or anywhere, can anticipate that.

INVALID DOMAIN. There is no singularity — denominators are the constants 2 and 4. The calculator rejects any inheritance mode, mother genotype or father genotype outside its declared options, including a two-allele genotype offered for the father, which is not a possible state at an X-linked locus. A zero probability — affected daughters when the father is unaffected under recessive inheritance, for example — is reported as 0%, which is the correct answer rather than an error.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Bin Shlhoob R., Tanaka J. & Pandya A. M. Genetics, X-Linked Inheritance. StatPearls, NCBI Bookshelf ID NBK557383, last update 22 March 2026. Primary source for every probability on this page: X-linked recessive carrier mother × unaffected father — 'each male offspring has a 50% chance of being affected, and each female offspring has a 50% chance of being a carrier'; affected father — 'none of his male offspring are affected, but all of his female offspring are carriers'; X-linked dominant — 'Affected males can transmit the mutant allele to all their daughters but not to their sons' and 'Affected females have a 50% chance of transmitting the mutant allele to each of their sons and daughters'. Free full text. Retrieved 2026-07-29.
2. [2]MedlinePlus Genetics, US National Library of Medicine. Inheritance Patterns, page last updated 19 April 2021. Independent confirmation of the affected-genotype rules: X-linked recessive — 'In males (who have only one X chromosome), one altered copy of the gene in each cell is sufficient to cause the condition. In females, a variant would have to occur in both copies of the gene to cause the disorder'; X-linked dominant — 'a variant in one of the two copies of the gene in each cell is sufficient to cause the disorder. Females may experience less severe symptoms than males.' Free. Retrieved 2026-07-29.
3. [3]Politano L. (2025). Females with X-Linked Muscle Disorders: an underestimated patient population. Acta Myologica 44(1):33–36. doi:10.36185/2532-1900-1096, PMID 40183439. Peer-reviewed source for the manifesting-carrier caveat carried beside the heterozygous-daughters output: skewed X-chromosome inactivation is defined as 'more than 75% of an individual's cells choose one parent's X chromosome as the inactive X'; 'it is estimated that 1.5-23% of females have skewed X inactivation'; and 'about 2.5-18% of DMD/BMD carriers may present with varying degrees of muscle weakness'. Open access via PMC. Retrieved 2026-07-29.
4. [4]Gulani A. & Weiler T. Genetics, Autosomal Recessive. StatPearls, NCBI Bookshelf ID NBK546620, last update 1 May 2023. Used for the contrast with autosomal inheritance — the 25% / 50% / 25% carrier × carrier outcome that X-linked inheritance departs from because males are hemizygous. Free full text. Retrieved 2026-07-29.
5. [5]Genetic Alliance & New York–Mid-Atlantic Consortium for Genetic and Newborn Screening Services. Understanding Genetics, 'Inheritance Patterns'. NCBI Bookshelf ID NBK115561, 2009. Source for the pedigree signatures quoted on this page — X-linked recessive: 'Males are more frequently affected; affected males often present in each generation'; X-linked dominant: 'Females are more frequently affected because all daughters and no sons of an affected man will be affected'. Free full text. Retrieved 2026-07-29.