Mendelian genetics — segregation and dominance

Intuition [Beginner]

Gregor Mendel, an Augustinian friar working in Brno (now the Czech Republic), discovered the fundamental rules of inheritance by breeding pea plants in the 1850s and 1860s. His experiments revealed that inheritance is particulate — traits are passed on by discrete units (now called genes) that do not blend together. This was a revolutionary idea, because most biologists at the time believed in blending inheritance.

Mendel's first key insight was the law of segregation. Every organism carries two copies of each gene (one from each parent). These two copies (called alleles) may be the same or different. When the organism produces gametes (sperm or eggs), the two alleles separate — each gamete receives exactly one. At fertilisation, the offspring receives one allele from each parent, restoring the pair.

His second key insight was the concept of dominance. When an individual carries two different alleles (e.g., Aa), one allele (A, the dominant allele) may mask the effect of the other (a, the recessive allele). The individual's phenotype (observable trait) shows the dominant form, even though the genotype (genetic makeup) contains both alleles.

Mendel's law of independent assortment states that alleles of different genes are distributed to gametes independently of one another. This holds when the genes are on different chromosomes (or far apart on the same chromosome), so the inheritance of one trait does not influence the inheritance of another.

Visual [Beginner]

A Punnett square is a grid that predicts the genotypes of offspring from a given cross. For a monohybrid cross between two heterozygotes (Aa x Aa), the Punnett square is:

The four equally likely outcomes are: AA (1/4), Aa (1/4), aA (1/4), aa (1/4). Since Aa and aA are genetically identical, the genotype ratio is 1 AA : 2 Aa : 1 aa. If A is dominant, the phenotype ratio is 3 dominant : 1 recessive.

Worked example [Beginner]

In a dihybrid cross (AaBb x AaBb), both parents are heterozygous for two independently assorting genes. Predict the phenotype ratio and calculate expected numbers from 160 offspring.

By Mendel's law of independent assortment, each gene segregates independently. For gene A alone (Aa x Aa): the phenotype ratio is 3 dominant (A_) : 1 recessive (aa). For gene B alone (Bb x Bb): the ratio is 3 dominant (B_) : 1 recessive (bb).

Combining the two genes: $3/4\times 3/4=9/16$ show both dominant phenotypes (A_B_). $3/4\times 1/4=3/16$ show A dominant, b recessive (A_bb). $1/4\times 3/4=3/16$ show a recessive, B dominant (aaB_). $1/4\times 1/4=1/16$ show both recessive phenotypes (aabb).

The 9:3:3:1 ratio. From 160 offspring: 90, 30, 30, and 10 expected respectively.

What this tells us: when two genes assort independently and each shows complete dominance, the offspring phenotype ratio is the product of the two individual monohybrid ratios — a direct consequence of the probability product rule.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Mendel's laws

First law (segregation): A diploid individual carries two alleles at each autosomal locus. During gamete formation, the two alleles segregate so that each gamete carries exactly one. Formally, for a parent with genotype $A_1{A}_2$, the gametes are $A_1$ and $A_2$ each with probability 1/2.

Second law (independent assortment): Alleles at different loci segregate independently into gametes. For a dihybrid $A_1{A}_2{B}_1{B}_2$ with the two loci on different chromosomes, the four gamete types ($A_1{B}_1$, $A_1{B}_2$, $A_2{B}_1$, $A_2{B}_2$) are produced in equal frequency (1/4 each). Independent assortment fails for linked loci on the same chromosome; recombination between them produces a proportion of non-parental gamete types.

Extensions to Mendel

Incomplete dominance: The heterozygote shows an intermediate phenotype. Example: snapdragons — RR (red) x rr (white) produces Rr (pink) in the F1, and the F2 ratio is 1 red : 2 pink : 1 white.

Codominance: Both alleles are fully expressed in the heterozygote. Example: human ABO blood groups — $I^A$ and $I^B$ alleles are codominant ($I^A{I}^B$ genotype produces AB blood type), while both are dominant to $i$ (which produces type O).

Epistasis: One gene masks or modifies the effect of another gene. In recessive epistasis, a homozygous recessive genotype at one locus (e.g., cc) prevents the expression of alleles at a second locus (e.g., B/b), regardless of the genotype at that second locus. The dihybrid ratio is modified from 9:3:3:1 to 9:3:4 (if cc masks B/b).

Pleiotropy: One gene affects multiple phenotypic traits. Example: the gene causing Marfan syndrome affects connective tissue throughout the body, producing tall stature, long limbs, lens dislocation, and aortic aneurysms from a single mutation.

Polygenic inheritance: Multiple genes contribute additively to a single quantitative trait (e.g., height, skin colour). The distribution of phenotypes in the population approaches a normal (Gaussian) distribution as the number of contributing loci increases.

Chi-square test

The chi-square (${\chi }^2$) test evaluates whether observed data are consistent with predicted Mendelian ratios:

$${\chi }^2=\sum \frac{(O_i-E_i{)}^2}{E_i},$$

where $O_i$ are observed counts and $E_i$ are expected counts for each class $i$. The degrees of freedom are the number of classes minus one. A significant ${\chi }^2$ value (exceeding the critical value for the given degrees of freedom at $p=0.05$) suggests the observed data deviate from the predicted ratio, indicating either non-Mendelian inheritance or experimental sampling effects.

Counterexamples to common slips

• Dominant does not mean common. Achondroplasia (dwarfism) is caused by a dominant FGFR3 mutation but affects only about 1 in 25,000 births worldwide. Dominance describes the phenotype of the heterozygote, not the population frequency of the allele.

• Recessive alleles do not "disappear" over time. Under Hardy-Weinberg equilibrium 19.02.01 pending, a recessive allele at frequency $q$ is carried by $2pq$ heterozygotes, and for rare alleles ($q\ll 1$) the vast majority of copies are hidden in heterozygous carriers who show no phenotype. The allele persists indefinitely in the absence of selection.

• Phenotype does not always reveal genotype. Complete dominance means the heterozygote Aa is phenotypically identical to AA. A test cross (crossing to a homozygous recessive) is needed to determine whether a dominant-phenotype individual is homozygous or heterozygous.

Key theorem with proof [Intermediate+]

Theorem (Mendel's 9:3:3:1 ratio for a dihybrid cross with independent assortment). For a cross between two double heterozygotes AaBb x AaBb, where A is dominant to a, B is dominant to b, and the two loci assort independently, the phenotype ratio among offspring is 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb.

Proof. By the law of independent assortment, the outcome at each locus is independent. For locus A alone: Aa x Aa gives AA (1/4), Aa (1/2), aa (1/4). The dominant phenotype (A_) has probability 3/4; the recessive phenotype (aa) has probability 1/4. Similarly for locus B: B_ has probability 3/4; bb has probability 1/4.

By independence: $P(A\_B\_)=(3/4)(3/4)=9/16$. $P(A\_bb)=(3/4)(1/4)=3/16$. $P(aaB\_)=(1/4)(3/4)=3/16$. $P(aabb)=(1/4)(1/4)=1/16$. $\square$

The ratio 9:3:3:1 is a diagnostic signature of two independently assorting loci with complete dominance at each. Deviations from this ratio are informative: a 9:7 ratio suggests complementary gene interaction (recessive epistasis where both genes are needed); a 9:3:4 ratio suggests recessive epistasis at one locus; a 12:3:1 ratio suggests dominant epistasis.

Bridge. The 9:3:3:1 ratio is the foundational reason that two-locus genetics has a clean combinatorial structure: each locus contributes independently, and dominance collapses genotypes into phenotypes at each locus separately. This is exactly the product rule of probability applied to heritable traits, and it builds toward 19.02.01 pending Hardy-Weinberg equilibrium, where the 1:2:1 genotype ratio at a single locus is generalised to population-level allele frequencies stable across generations under random mating. The pattern recurs in 19.05.01 pending quantitative genetics, where Fisher's infinitesimal model treats quantitative traits as the sum of many independent Mendelian loci each following these same segregation rules.

Exercises [Intermediate+]

Advanced results [Master]

Extensions of Mendelian ratios

The simple 3:1 monohybrid ratio assumes complete dominance, full penetrance, and equal viability of all genotypes. Each assumption can fail, and the specific pattern of failure reveals biological mechanism.

Lethal alleles eliminate one genotype class entirely. The classic example is the yellow mouse. The agouti locus in Mus musculus has an allele $A^Y$ (yellow) that is dominant to the wild-type $A$ (agouti) for coat colour: $A^{Y}A$ mice are yellow, $AA$ mice are agouti. But $A^Y{A}^Y$ is embryonic lethal — the combination kills the embryo before birth. A cross $A^{Y}A\times A^{Y}A$ produces viable offspring in the ratio 2 yellow ($A^{Y}A$) : 1 agonti ($AA$), not 3:1, because the $A^Y{A}^Y$ class dies. The observed ratio of 2:1 instead of 3:1 is diagnostic for a recessive lethal allele linked to a dominant visible marker. In humans, achondroplasia (FGFR3 gain-of-function mutation) follows the same pattern: heterozygotes have dwarfism, homozygotes die in infancy, and two affected parents have a 2:1 ratio of affected to unaffected surviving offspring.

Multiple alleles arise when a locus has more than two allelic forms in the population. The ABO blood group system has three alleles ($I^A$, $I^B$, $i$) producing six genotypes ($I^A{I}^A$, $I^A{I}^B$, $I^{A}i$, $I^B{I}^B$, $I^{B}i$, $ii$) and four phenotypes (A, AB, B, O). The dominance relationships are: $I^A$ and $I^B$ are codominant with each other, and both are dominant to $i$. The biochemical basis is that $I^A$ encodes a glycosyltransferase that adds N-acetylgalactosamine to the H antigen on red blood cells; $I^B$ encodes a different transferase adding galactose; $i$ is a loss-of-function allele producing unmodified H antigen. The specificity of the two enzymes explains the codominance: both modifications coexist on the cell surface in $I^A{I}^B$ heterozygotes.

The HLA (human leukocyte antigen) system on chromosome 6 extends this to hundreds of alleles across multiple loci (HLA-A, HLA-B, HLA-DRB1 each have over 1000 known alleles). HLA typing for transplant matching must contend with this extraordinary allelic diversity, where each individual's two haplotypes (one from each parent) specify a unique combination of immune-response proteins.

Penetrance is the fraction of individuals with a given genotype who show the associated phenotype. Expressivity is the degree to which the phenotype is manifested among those who show it. Retinoblastoma (Rb tumour suppressor, chromosome 13) has about 90% penetrance for the hereditary form — roughly 10% of individuals who inherit the mutant allele never develop the tumour. Neurofibromatosis type 1 (NF1 gene, chromosome 17) has nearly complete penetrance but highly variable expressivity: affected individuals in the same family may show only cafe-au-lait spots or may develop disfiguring neurofibromas, optic gliomas, and skeletal dysplasia from the same germ-line mutation. Reduced penetrance and variable expressivity both cause deviations from expected Mendelian ratios.

Sex-linked inheritance breaks the autosomal assumption that both sexes have the same genotype-phenotype mapping. X-linked recessive conditions (hemophilia A, Duchenne muscular dystrophy, red-green colour blindness) appear predominantly in males, who have only one X chromosome and therefore express every allele they carry. A carrier female ($X^{+}{X}^a$) is usually unaffected because the wild-type allele on her other X chromosome produces enough functional protein. The criss-cross inheritance pattern — affected males transmit the mutant allele to all daughters (who become carriers) and to no sons (who inherit the Y chromosome) — is diagnostic for X-linked recessive traits.

Genomic imprinting breaks the assumption that the two parental copies of a gene are functionally equivalent. Prader-Willi syndrome (paternal deletion of 15q11-q13) and Angelman syndrome (maternal deletion of the same region) produce distinct phenotypes depending on whether the deleted segment was inherited from the father or the mother. The molecular mechanism involves parent-of-origin-specific DNA methylation that silences one copy (the maternal copy of the SNRPN cluster for Prader-Willi, the paternal copy of UBE3A for Angelman). Uniparental disomy (inheriting both copies of chromosome 15 from one parent) produces the same syndromes as the corresponding deletions, confirming that the phenotypes arise from loss of expression of a parentally-imprinted gene rather than from a specific DNA deletion.

Theorem (Lethal-allele modified ratio). Let $A^L$ be a dominant visible allele that is recessive lethal (homozygous $A^L{A}^L$ is inviable). The cross $A^{L}A\times A^{L}A$ produces viable offspring in the ratio 2 dominant-phenotype : 1 recessive-phenotype, with 1/4 of zygotes non-viable.

Theorem (ABO genotype-phenotype mapping). For three alleles $I^A$, $I^B$, $i$ with $I^A$ and $I^B$ codominant and both dominant to $i$, a cross $I^{A}i\times I^{B}i$ produces offspring phenotypes in the ratio 1 A : 1 AB : 1 B : 1 O, each with probability 1/4.

Gene interactions and epistasis

When two or more genes contribute to a single phenotypic outcome, the dihybrid ratio 9:3:3:1 is reshaped by the specific nature of their interaction. Each modified ratio is a signature of a particular type of gene-gene relationship.

Recessive epistasis occurs when homozygosity for the recessive allele at one locus (ee) blocks the phenotypic expression of alleles at a second locus, regardless of the genotype at that locus. Labrador retriever coat colour is the standard example. The B locus encodes TYRP1, an enzyme in the eumelanin synthesis pathway: the B allele produces functional TYRP1 (black eumelanin), the b allele produces non-functional TYRP1 (brown eumelanin). The E locus encodes MC1R, a receptor on melanocytes that signals pigment deposition: the E allele produces a functional receptor (pigment deposited), the e allele produces a non-functional receptor (no eumelanin deposited, and the fur appears yellow from phaeomelanin instead). The cross BbEe x BbEe produces the ratio 9 B_E_ (black) : 3 bbE_ (brown) : 4 _ _ee (yellow, combining the 3 B_ee and 1 bbee classes). The ee genotype is epistatic to B/b because it acts upstream in the pathway: if the receptor for pigment deposition is absent, the type of pigment produced is irrelevant.

Complementary gene action produces the 9:7 ratio. Two genes encode enzymes in the same biosynthetic pathway, and both functional enzymes are required for the end product. In sweet peas (Lathyrus odoratus), flower colour requires functional copies of both gene C (chalcone synthase, the first committed step of flavonoid biosynthesis) and gene P (dihydroflavonol reductase, a later step). The cross CcPp x CcPp produces 9 C_P_ (purple, both enzymes functional) : 7 (white, combining C_pp, ccP_, and ccpp — in each case at least one enzyme is missing). The mathematical collapse is 9 : (3 + 3 + 1) = 9 : 7. This ratio is diagnostic for two genes in a linear biosynthetic pathway where the end product is the visible trait.

Duplicate gene action produces the 15:1 ratio. When either of two genes can independently produce the same phenotype, only the double homozygous recessive shows the alternative phenotype. In shepherd's purse (Capsella bursa-pastoris), seed shape is determined by two redundant genes: either T1 or T2 can produce triangular (heart-shaped) capsules. The cross T1t1T2t2 x T1t1T2t2 produces 15 triangular (at least one dominant allele at either locus) : 1 ovoid (t1t1t2t2, no functional copy of either gene). This ratio signals gene redundancy, common in polyploid species where genome duplication creates paralogous copies.

Dominant epistasis produces the 12:3:1 ratio. A dominant allele at one locus masks the expression of alleles at a second locus. In summer squash, fruit colour is controlled by two genes: the W locus (white vs coloured) and the Y locus (yellow vs green). A dominant W allele produces white fruit regardless of the Y genotype. The cross WwYy x WwYy produces 12 W_ (white, combining 9 W_Y_ and 3 W_yy) : 3 wwY_ (yellow) : 1 wwyy (green). The dominant allele W is epistatic because a single copy is sufficient to block all colour production.

Suppressor mutations are a specialised form of epistasis in which a mutation at one locus restores the wild-type phenotype that was lost by a mutation at a different locus. In Drosophila, the purple eye mutation (pr, affecting xanthommatin synthesis) can be suppressed by the su-pr mutation, which restores wild-type eye colour without correcting the original pr mutation. Suppressor screens are a powerful genetic tool for identifying interacting genes in the same pathway.

Synthetic lethality occurs when two mutations are individually viable but lethal in combination. This is the inverse of suppressor genetics: two genes are synthetic lethal when the cell can tolerate loss of either one but not both. In cancer biology, BRCA1-deficient tumours are exquisitely sensitive to PARP inhibitors because the combined loss of homologous recombination (BRCA1) and base-excision repair (PARP) kills the cell. This principle underlies targeted cancer therapies that exploit the synthetic-lethal relationships specific to tumour cells.

Theorem (Complementary gene action ratio). For a dihybrid cross $AaBb\times AaBb$ where both dominant alleles are required for a trait and homozygous recessive at either locus eliminates it, the phenotype ratio is 9 functional : 7 non-functional.

Theorem (Duplicate gene action ratio). When either of two independently assorting loci can produce the same phenotype, the dihybrid ratio is 15 (at least one dominant allele) : 1 (double recessive).

Pedigree analysis and Bayesian risk calculation

Pedigree analysis applies Mendelian principles to family trees to determine the mode of inheritance and calculate recurrence risks. The standard conventions use squares for males, circles for females, filled symbols for affected individuals, half-filled for known carriers, a horizontal line connecting mates, and vertical lines to offspring. Consanguinity is indicated by a double horizontal line.

Autosomal dominant inheritance shows a vertical pattern: the trait appears in every generation, affected individuals usually have at least one affected parent, and both sexes are affected in roughly equal numbers. Male-to-male transmission occurs (ruling out X-linkage). Each affected individual has a 50% chance of passing the trait to each offspring. Huntington disease (CAG repeat expansion in the HTT gene on chromosome 4) illustrates the challenges: onset is typically in the fourth or fifth decade of life, so individuals may have children before knowing their genetic status. The CAG repeat shows anticipation — expansion in successive generations leads to earlier onset and more severe disease. About 10% of Huntington cases represent de novo mutations from an intermediate-length allele in a parent.

Autosomal recessive inheritance shows a horizontal pattern: affected individuals often have unaffected parents who are both carriers, and the trait may cluster among siblings but not across generations. Both sexes are affected equally. The recurrence risk for carrier parents (Aa x Aa) is 25% per pregnancy. Consanguinity increases the risk because related parents are more likely to share rare recessive alleles inherited from a common ancestor. First cousins share approximately 1/8 of their genes; the offspring of a first-cousin union have approximately double the baseline risk of a major congenital anomaly (from about 2-3% in the general population to about 4-6%). Founder effects can elevate specific recessive disease frequencies in isolated populations: Tay-Sachs disease (HEXA deficiency, chromosome 15) has a carrier frequency of approximately 1/27 in Ashkenazi Jewish populations compared with approximately 1/250 in the general population, and cystic fibrosis (CFTR mutations, chromosome 7) has a carrier frequency of approximately 1/25 in Northern European populations.

X-linked recessive inheritance shows the criss-cross pattern: affected males inherit the mutant allele from carrier mothers, and affected males cannot transmit the trait to their sons (who inherit the Y chromosome). All daughters of an affected male are carriers. A carrier female ($X^{+}{X}^a$) has a 50% chance of passing the mutant allele to each son (who would be affected) and a 50% chance of passing it to each daughter (who would be a carrier). Hemophilia A (Factor VIII deficiency, F8 gene at Xq28) and Duchenne muscular dystrophy (dystrophin deficiency, DMD gene at Xp21) are the classic examples. About 1/3 of DMD cases arise from de novo mutations, reflecting the large gene size (2.2 Mb, the largest known human gene).

Bayesian risk calculation updates carrier probabilities using conditional information from pedigree data and genetic testing. The framework is Bayes' theorem applied to genotype probabilities:

$$P(\text{carrier}\mid \text{data})=\frac{P(\text{data}\mid \text{carrier})\times P(\text{carrier})}{P(\text{data})}$$

Consider a woman whose brother has cystic fibrosis (autosomal recessive). Her parents must both be carriers (Aa x Aa). Given that she is phenotypically normal, her prior genotype probabilities are: P(AA) = 1/3, P(Aa) = 2/3. The 1/3 : 2/3 ratio (not 1/4 : 1/2 : 1/4) reflects the conditioning on her being unaffected (eliminating the aa class and renormalising).

If she then has a negative result on a CF carrier screen that detects 90% of pathogenic CFTR mutations, Bayes' theorem updates her carrier probability:

$$P(\text{carrier}\mid \text{negative test})=\frac{0.10\times 2/3}{0.10\times 2/3+1.0\times 1/3}=\frac{0.067}{0.40}\approx 1/6$$

The negative test reduced her carrier probability from 2/3 to 1/6 — a substantial reduction but not elimination, because the test misses 10% of mutations. If she also has two unaffected children by a partner known not to be a carrier, each unaffected child provides additional evidence against her being a carrier (because a carrier mother with a non-carrier father produces each child with a 50% chance of being a carrier, and each unaffected non-carrier child is twice as likely if the mother is AA than if she is Aa). The posterior after $n$ unaffected children is:

$$P(\text{carrier}\mid n\text{ unaffected children})=\frac{(1/2{)}^n\times 2/3}{(1/2{)}^n\times 2/3+1\times 1/3}=\frac{2\times (1/2{)}^n}{2\times (1/2{)}^n+1}=\frac{1}{1+2^{n-1}}$$

For $n=2$ unaffected children, posterior = 1/5. Combined with the negative test: $(0.10)(1/5)/((0.10)(1/5)+(1.0)(4/5))=0.02/0.82\approx 1/41$.

Theorem (Bayesian posterior carrier probability with test and pedigree). Let the prior carrier probability be $\pi$. Let a genetic test have sensitivity $\alpha$ (probability of detecting a carrier) and specificity 1 (non-carriers always test negative). Let $n$ unaffected children exist from a union with a known non-carrier partner. Then the posterior carrier probability is

$$P(\text{carrier}\mid \text{data})=\frac{(1-\alpha )(1/2{)}^n\pi }{(1-\alpha )(1/2{)}^n\pi +(1-\pi )}.$$

Statistical testing and probabilistic prediction

Mendelian ratios are statistical expectations. The chi-square (${\chi }^2$) goodness-of-fit test quantifies whether observed data are consistent with a hypothesised ratio, and the binomial distribution predicts the probability of specific outcomes in families of known size.

Chi-square goodness-of-fit. Under the null hypothesis that the data follow a specified Mendelian ratio, the test statistic

$${\chi }^2=\sum _{i=1}^k\frac{(O_i-E_i{)}^2}{E_i}$$

follows approximately a chi-square distribution with $k-1$ degrees of freedom (where $k$ is the number of phenotype classes and $E_i=N\times p_i$ for total sample size $N$ and expected proportion $p_i$). The approximation is adequate when all expected counts $E_i\ge 5$.

Mendel's own data illustrate the test. For the F2 generation of round (R) vs wrinkled (r) seeds, Mendel reported 5474 round : 1850 wrinkled (total 7324). Under the null hypothesis of a 3:1 ratio, expected counts are $E_R=5493$ and $E_r=1831$:

$${\chi }^2=\frac{(5474-5493{)}^2}{5493}+\frac{(1850-1831{)}^2}{1831}=\frac{361}{5493}+\frac{361}{1831}\approx 0.066+0.197=0.263$$

With 1 degree of freedom, the critical value at $\alpha =0.05$ is 3.84. The test statistic 0.263 is far below this threshold ($p\approx 0.61$), so the data are consistent with 3:1. Fisher ^{[Fisher 1936]} noted that Mendel's data across all experiments fit the expected ratios too well — the aggregate ${\chi }^2$ is improbably small, with an overall $p$ value near 0.93 suggesting the data cluster too tightly around expectation. This "too good to be true" observation fuelled the long-running controversy over whether Mendel (or an assistant) stopped counting when ratios looked favourable, or selectively reported the best experiments.

For a dihybrid cross testing the 9:3:3:1 ratio, there are 3 degrees of freedom. The critical value at $\alpha =0.05$ is 7.81. Mendel reported 315 round yellow : 108 round green : 101 wrinkled yellow : 32 wrinkled green (total 556) for seed shape and seed colour. Expected under 9:3:3:1: 312.75, 104.25, 104.25, 34.75. ${\chi }^2=(2.25{)}^2/312.75+(3.75{)}^2/104.25+(3.25{)}^2/104.25+(2.75{)}^2/34.75\approx 0.016+0.135+0.101+0.218=0.470$. Again, consistent with the hypothesised ratio ($p\approx 0.92$) and again suspiciously close.

Binomial distribution. For a cross Aa x Aa, each offspring independently has probability 1/4 of being aa. For $n$ total offspring, the number $K$ of affected individuals follows a binomial distribution:

$$P(K=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(\frac{1}{4}\right)}^k{\left(\frac{3}{4}\right)}^{n-k}.$$

For two carrier parents planning a family of 4 children, the probability of exactly 2 affected children is $P(K=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)(1/4{)}^2(3/4{)}^2=6\cdot 1/16\cdot 9/16=54/256\approx 0.211$. The probability of at least one affected child is $1-P(K=0)=1-(3/4{)}^4=1-81/256\approx 0.684$. For genetic counselling of carrier couples, these probabilities are directly relevant to reproductive decision-making.

Bayes' theorem for carrier status was treated in the preceding section. The general principle is that each piece of evidence (negative test, unaffected offspring, negative family history) updates the carrier probability multiplicatively through the likelihood ratio. Sequential updating gives the same result as simultaneous updating — the order does not matter, only the total likelihood ratio.

Linkage and recombination. Independent assortment fails when two loci are on the same chromosome. The recombination frequency $r$ (the proportion of gametes that are recombinant rather than parental) measures the genetic distance between loci. For a dihybrid in coupling phase AB/ab, the gamete frequencies are: AB and ab at $(1-r)/2$ each, Ab and aB at $r/2$ each. When $r=0.5$, the four gametes are equally frequent and the loci appear to assort independently (they may be on the same chromosome but far apart). When $r=0$, the loci are completely linked and only parental gametes are produced. Map distance in centiMorgans (cM) approximates $r\times 100$ for small $r$; for larger distances, mapping functions (Haldane, Kosambi) correct for the nonlinearity arising from double crossovers.

Theorem (Chi-square under Mendelian null). Under the null hypothesis that observed counts $(O_1,\dots ,O_k)$ are drawn from a multinomial distribution with probabilities $(p_1,\dots ,p_k)$ and total $N$, the statistic ${\chi }^2={\sum }_i(O_i-N{p}_i{)}^2/(N{p}_i)$ converges in distribution to ${\chi }_{k-1}^2$ as $N\to \infty$.

Theorem (Binomial probability for Mendelian crosses). For a monohybrid cross Aa x Aa with n offspring, the probability of exactly k recessive-phenotype (aa) offspring is $P(K=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)(1/4{)}^k(3/4{)}^{n-k}$.

Theorem (Hardy-Weinberg equilibrium from Mendelian segregation). Consider a large, randomly mating population with a biallelic locus (alleles A and a at frequencies p and q = 1-p). After one generation of random mating, the genotype frequencies reach $P(AA)=p^2$, $P(Aa)=2pq$, $P(aa)=q^2$ and remain at these values in all subsequent generations in the absence of evolutionary forces.

Synthesis. Putting these together — the extensions of Mendelian ratios through lethal alleles, penetrance, and sex-linkage; the epistatic interactions that reshape the 9:3:3:1 signature into 9:7, 15:1, 9:3:4, and 12:3:1; the Bayesian framework for inferring hidden genotypes from pedigree data; and the statistical machinery for testing whether observed data match Mendelian expectations — the central insight is that Mendel's particulate inheritance provides a complete probabilistic calculus for predicting trait transmission. The foundational reason this calculus works is the combinatorial structure of meiosis: segregation and independent assortment generate gamete frequencies that are simple products of allele frequencies, and dominance merely collapses the resulting genotype classes into phenotype classes. This is exactly the structure that 19.02.01 pending Hardy-Weinberg equilibrium generalises from individual crosses to population-level allele frequencies, and the bridge is from the single-family predictions of Mendelian genetics to the population-genetic theory of allele-frequency change under selection 19.03.01 pending, drift 19.04.01, migration, and mutation. The framework appears again in 19.05.01 pending quantitative genetics, where Fisher's 1918 variance decomposition identifies the additive genetic variance with the summed effects of individual Mendelian loci obeying these same segregation rules, and the pattern generalises to the modern genomic prediction models used in plant and animal breeding.

Full proof set [Master]

Proposition (Bayesian posterior carrier probability). Let a woman have prior carrier probability $\pi$ for an autosomal recessive condition. She undergoes a genetic test with sensitivity $\alpha$ (probability of a positive result given carrier status) and has n unaffected children with a known non-carrier partner. Then her posterior carrier probability is

$$P(\text{carrier}\mid \text{negative test},n\text{ unaffected})=\frac{(1-\alpha )\cdot (1/2{)}^n\cdot \pi }{(1-\alpha )\cdot (1/2{)}^n\cdot \pi +1\cdot (1-\pi )}.$$

Proof. By Bayes' theorem:

$$P(\text{carrier}\mid \text{data})=\frac{P(\text{data}\mid \text{carrier})\cdot P(\text{carrier})}{P(\text{data})}$$

The likelihood $P(\text{data}\mid \text{carrier})$ factorises into the test component and the offspring component. The test result is negative, which occurs with probability $(1-\alpha )$ for a carrier (the false-negative rate). Each unaffected child from a carrier mother (genotype Aa) and a non-carrier father (genotype AA) has genotype AA or Aa, each with probability 1/2. Since the child is unaffected regardless (both AA and Aa show the dominant phenotype), being unaffected provides no information in a single child. However, being a non-carrier (AA specifically) is informative: the probability that a given child is a non-carrier (AA) given the mother is Aa and father is AA is 1/2. The probability that the child is a non-carrier given the mother is AA and father is AA is 1.

So $P(\text{non-carrier child}\mid \text{carrier mother})=1/2$ and $P(\text{non-carrier child}\mid \text{non-carrier mother})=1$. For $n$ independent children:

$$P(n\text{ unaffected non-carrier children}\mid \text{carrier})=(1/2{)}^n$$ $$P(n\text{ unaffected non-carrier children}\mid \text{non-carrier})=1^n=1$$

Combining with the test:

$$P(\text{data}\mid \text{carrier})=(1-\alpha )\cdot (1/2{)}^n$$ $$P(\text{data}\mid \text{non-carrier})=1\cdot 1=1$$

Applying Bayes' theorem:

$$P(\text{carrier}\mid \text{data})=\frac{(1-\alpha )(1/2{)}^n\pi }{(1-\alpha )(1/2{)}^n\pi +1\cdot (1-\pi )}\square$$

Proposition (Complementary gene action ratio 9:7). For a dihybrid cross $AaBb\times AaBb$ with independent assortment, where the dominant phenotype requires at least one dominant allele at both loci, the phenotype ratio is 9 dominant : 7 recessive.

Proof. Each locus independently segregates with 3/4 dominant phenotype, 1/4 recessive. By the product rule, the probability of dominant phenotype at both loci is $(3/4)(3/4)=9/16$. The probability of recessive phenotype at one or both loci is $1-9/16=7/16$, which decomposes as $P(A\_bb)+P(aaB\_)+P(aabb)=3/16+3/16+1/16=7/16$. $\square$

Connections [Master]

• Hardy-Weinberg equilibrium 19.02.01 pending. The 1:2:1 genotype ratio from a monohybrid cross is the foundational building block that Hardy-Weinberg equilibrium generalises to population-level allele frequencies: under random mating, the genotype frequencies $p^2$, $2pq$, $q^2$ are reached after a single generation and remain stable indefinitely. Deviations from HWE indicate that the Mendelian null model is violated by selection, drift, non-random mating, migration, or mutation.

• Natural selection 19.03.01 pending. Selection acts on the phenotypic variation produced by Mendelian inheritance, changing allele frequencies from the values predicted by the Hardy-Weinberg null model. The selection coefficient $s$ quantifies the fitness difference between genotypes; directional selection increases the frequency of the favoured allele, while stabilising and disruptive selection reshape the phenotype distribution.

• Quantitative genetics 19.05.01 pending. Fisher's 1918 variance decomposition ^{[Fisher 1918]} showed that the correlation between relatives for quantitative traits can be explained by many independent Mendelian loci each with small additive effects. The additive genetic variance, dominance variance, and epistatic variance partition the total phenotypic variance into components traceable to specific Mendelian phenomena described in this unit.

• DNA replication and the molecular gene 17.05.01 pending. Mendel's "factors" are segments of chromosomal DNA. Alleles are variant forms of those segments arising by mutation 17.06.01 pending. Dominance arises when one allele produces a functional protein and the other does not (haploinsufficiency) or produces a product that actively disrupts function (dominant-negative). The molecular mechanisms of transcription 17.05.02 pending and translation 17.05.03 pending explain how genotype produces phenotype.

• Cell division and meiosis 17.08.01. Segregation is the cellular consequence of homologous chromosomes separating at anaphase I of meiosis. Independent assortment reflects the random orientation of different chromosome pairs on the meiotic spindle. Recombination (crossing over at prophase I) produces the non-parental gamete types that make linkage detectable. Mendel's laws are the macroscopic consequences of meiotic chromosome mechanics.

Historical & philosophical context [Master]

Mendel's 1866 paper "Versuche uber Pflanzenhybriden" ^{[Mendel 1866]}, published in the proceedings of the Natural History Society of Brunn (Verh. Naturforsch. Ver. Brunn 4, 3-47), reported the results of eight years of hybridisation experiments on Pisum sativum (garden pea). Mendel chose seven characters with clear alternative forms (round vs wrinkled seeds, yellow vs green cotyledons, purple vs white flowers, inflated vs constricted pods, green vs yellow unripe pods, axial vs terminal flowers, tall vs short stems). His use of quantitative counting — recording the exact numbers of each phenotype in each generation — was unprecedented in biology and allowed him to recognise the 3:1 ratio in the F2, from which he inferred the 1:2:1 genotype ratio and the law of segregation.

The paper was cited only a handful of times before 1900. The independent rediscovery by de Vries ^{[de Vries 1900]}, Correns, and Tschermak in 1900 triggered one of the most dramatic episodes in the history of biology. Within three years, Sutton and Boveri had proposed the chromosome theory of inheritance (connecting Mendel's factors to cytologically visible chromosomes), and by 1910 Morgan had identified genetic linkage in Drosophila, demonstrating that genes are arranged linearly on chromosomes.

Hardy ^{[Hardy 1908]} and Weinberg ^{[Weinberg 1908]} independently derived the equilibrium genotype frequencies in 1908, providing the null model for population genetics. Fisher's 1918 paper "The correlation between relatives on the supposition of Mendelian inheritance" ^{[Fisher 1918]} reconciled the Mendelian and biometrical schools by showing that continuous variation (championed by Galton and Pearson) is fully compatible with particulate Mendelian inheritance when many loci contribute small additive effects to a quantitative trait. Fisher 1936 ^{[Fisher 1936]} analysed Mendel's original data and noted the suspiciously close fit to expected ratios, concluding that "the data of most, if not all, of the experiments have been falsified so as to agree closely with Mendel's expectations" — a controversy that persists.

Bibliography [Master]

@article{Mendel1866,
  author = {Mendel, Gregor},
  title = {Versuche \"{u}ber Pflanzenhybriden},
  journal = {Verh. Naturforsch. Ver. Br\"{u}nn},
  volume = {4},
  pages = {3--47},
  year = {1866},
}

@article{deVries1900,
  author = {de Vries, Hugo},
  title = {Sur la loi de disjonction des hybrides},
  journal = {C. R. Acad. Sci. Paris},
  volume = {130},
  pages = {845--847},
  year = {1900},
}

@article{Correns1900,
  author = {Correns, Carl},
  title = {Gregor Mendels Regel \"{u}ber das Verhalten der Nachkommenschaft der Rassenbastarde},
  journal = {Ber. Dtsch. Bot. Ges.},
  volume = {18},
  pages = {158--168},
  year = {1900},
}

@article{Hardy1908,
  author = {Hardy, G. H.},
  title = {Mendelian proportions in a mixed population},
  journal = {Science},
  volume = {28},
  pages = {49--50},
  year = {1908},
}

@article{Weinberg1908,
  author = {Weinberg, Wilhelm},
  title = {{\"U}ber den Nachweis der Vererbung beim Menschen},
  journal = {Jh. Ver. vaterl. Naturk. W\"{u}rttemb.},
  volume = {64},
  pages = {368--382},
  year = {1908},
}

@article{Fisher1918,
  author = {Fisher, R. A.},
  title = {The correlation between relatives on the supposition of {M}endelian inheritance},
  journal = {Trans. R. Soc. Edinb.},
  volume = {52},
  pages = {399--433},
  year = {1918},
}

@article{Fisher1936,
  author = {Fisher, R. A.},
  title = {Has {M}endel's work been rediscovered?},
  journal = {Ann. Sci.},
  volume = {1},
  pages = {115--137},
  year = {1936},
}

@book{HartlClark2007,
  author = {Hartl, Daniel L. and Clark, Andrew G.},
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  edition = {4th},
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  year = {2007},
}

@book{Futuyma2017,
  author = {Futuyma, Douglas J.},
  title = {Evolution},
  edition = {4th},
  publisher = {Sinauer Associates},
  year = {2017},
}

@book{Griffiths2020,
  author = {Griffiths, Anthony J. F. and Doebley, John and Peichel, Catherine and Watson, David A.},
  title = {Introduction to Genetic Analysis},
  edition = {12th},
  publisher = {Macmillan},
  year = {2020},
}