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Genotyping determines the alleles carried at chosen genetic loci. A diploid genotype call need not identify its two haplotypes; that additionally requires a phase-known genotype or haplotype estimation.

A genotyping error is an incorrect observed allele or genotype call. A probabilistic observation model replaces the deterministic genetic marker factor by $P(X=x\mid G=g)$, allowing nonzero probability of an observed call inconsistent with the latent genotype. Such error can bias family transmission tests and linkage likelihoods.

Haplotype estimation reconstructs which alleles share a chromosome from observed genotypes and other information, such as pedigrees. A doubly heterozygous genotype can be $AB/ab$ or $Ab/aB$; a phase-known genotype distinguishes them. Uncertainty in that reconstruction must propagate to estimated haplotype frequencies and linkage disequilibrium.

The Hardy-Weinberg principle describes the genotype proportions produced by independently combining two parental gametes from the same allele pool. For alleles $A,a$ with allele frequencies $p,q$, where $p+q=1$, the Hardy-Weinberg proportions areThe two ordered heterozygote outcomes contribute $pq+qp$. These proportions sum to $(p+q{)}^2=1$, and their implied $A$ allele frequency is $p^2+(2pq)/2=p$. Reshuffling copies therefore preserves their frequencies.

For stable Hardy-Weinberg equilibrium across generations, use a large population with random mating, equal allele frequencies in the parental pools, ordinary Mendelian segregation, and no natural selection, mutation, gene flow or genetic drift. Negligible genetic drift is an idealization of large population size. The one-generation random-mating equilibrium needs independent union of gametes; persistence additionally requires processes that change allele frequencies to be absent.

At $p=0.70,q=0.30$, the predicted genotype proportions are $0.49,0.42,0.09$. For 100 sampled individuals the expected values of the counts are $49,42,9$, rather than the $25,50,25$ of a specified $Aa\times Aa$ cross. The latter fixes both parental genotypes; Hardy-Weinberg proportions average over population mating.

An observed population need not have these proportions. For example, $70AA$, no $Aa$, and $30aa$ among 100 individuals still gives $p=0.70$, but has no heterozygotes. Random mating predicts $0.42$ heterozygosity in the next generation under the stated model. A finite sample can also depart by chance, so a discrepancy is a starting point for investigating inbreeding, population stratification or genotyping error, rather than proof of one cause.

Hardy's 1908 note made the conceptual payoff explicit: a dominant phenotype does not by itself make its allele spread. Complete dominance describes expression; changes in frequency require a separate mechanism such as natural selection. See Hardy, Mendelian Proportions in a Mixed Population, Science 28 (1908), 49–50.

Heterozygosity means that the two allele states at a diploid genetic locus differ; an individual with that property is a heterozygote. Heterozygous parental genotypes provide alternative alleles for a transmission disequilibrium test.

For a biallelic locus with frequencies $p=1-q$ and $q$, and inbreeding coefficient $F$, ancestral-copy identity with probability $F$ gives genotype probabilities $P(AA)=p^2+Fpq$, $P(Aa)=2pq(1-F)$ and $P(aa)=q^2+Fpq$. In the IBD component both copies equal a single ancestral draw; in the remaining component the two ancestral alleles are independent population draws. A fully penetrant recessive disease therefore has risk $q^2+Fq(1-q)$.

// Target: biology.bigb

If one parent is heterozygous and the other parent's genotype is missing, a homozygous child reveals the observed parent's transmitted allele, whereas a heterozygous child need not. Selecting only revealed transmissions can therefore bias the transmission disequilibrium test. For allele-1 frequency $p$ under random mating and fair transmission, the probabilities of an allele-1 or allele-2 homozygous child from an observed heterozygous parent are $p/2$ and $(1-p)/2$. Among selected homozygous children, the observed allele-1 transmission probability is $p$, not generally $1/2$. A complete-trio analysis avoids this reconstruction rule; methods retaining missing-parent families need an explicit likelihood function or a valid conditional adjustment. Family-based association tests using genotype data with uncertainty discusses this bias.

Independent assortment means that transmission at different genetic loci has conditional independence given the parental genotype. With fair Mendelian segregation, a parent heterozygous at two genetic loci, $AaBb$, produces $AB,Ab,aB,ab$ gametes with probability $1/4$ each. For example,Here $AB$ labels an allele combination; it does not assert that the two genetic loci lie on the same chromosome.

The physical model is stochastic orientation of different homologous chromosome pairs toward the two poles of the spindle apparatus in meiosis I. One pair's orientation does not determine another's. For two pairs, one division can send $A,B$ together and $a,b$ together; another can send $A,b$ together and $a,B$ together. See The Process of Meiosis (§11.1, Metaphase I and Figure 11.5: random orientation of chromosome pairs).

With $k$ distinguishable homologous chromosome pairs, there are $2^k$ choices of parental homologous chromosome origin for a product at a specified pole, each with probability $2^{-k}$ under this model. For $k=3$, enumerate $MMM,MMP,MPM,MPP,PMM,PMP,PPM,PPP$, where M and P mark parental origins. Whole-pole reversal pairs complementary choices, leaving $2^{k-1}$ unordered partitions for $k\ge 1$. For $k=2$ there are four choices for a specified pole but only the two relative orientations drawn. This counts possible products across meioses, not eight different products of one division when $k=3$. Identical alleles at scored genetic loci can also make different parental-origin choices observationally indistinguishable.

Independent assortment of different chromosome pairs does not imply independence of every pair of genes. Nearby genetic loci on the same chromosome can show genetic linkage. Chromosomal crossover can create new haplotypes, but does not automatically erase genetic linkage. See Laws of Inheritance (§12.3, Law of Independent Assortment and Linked Genes Violate the Law of Independent Assortment). In the ordinary fair two-locus model with parental haplotypes $AB/ab$ and recombination fraction $r$,Thus $r=0$ gives only the two parental haplotypes, while $r=1/2$ gives four equal probabilities. A recombination fraction of $1/2$ alone does not establish that the genetic loci lie on different chromosomes.

A motivating experiment appears in Experiments in Plant Hybridization by Gregor Mendel (§9, “The Reproductive Cells of the Hybrids”). In peas, hybrids for round versus wrinkled seeds and yellow versus green seeds were crossed reciprocally with constant wrinkled-green plants. The readout was visible seed shape and color: hybrid seed parents gave 31 round-yellow, 26 round-green, 27 wrinkled-yellow and 26 wrinkled-green; the reciprocal cross gave 24, 25, 22 and 27.

The apparatus was controlled pollination: opening immature flowers, removing stamens with forceps, and applying selected pollen to the stigma. Two-year trials established constant parental characters; greenhouse pots checked interference from insects in outdoor beds. Reciprocal crosses tested both transmission directions. These procedures are described in Experiments in Plant Hybridization by Gregor Mendel (§§2–3).

With complete dominance at both genetic loci, this test cross makes each phenotype report one hybrid gamete combination. The ideal expected values are $110/4=27.5$ and $98/4=24.5$ per class. The observations support four transmitted combinations, approximately equally frequent. They do not directly image chromosome orientation or prove independence for all genes; failed development and seed germination also limit the inference, as §9 notes.

Linkage disequilibrium means that knowing an allele at one genetic locus changes the probability of an allele at another genetic locus in population haplotypes. It compares their joint probability distribution with the product of their marginal distributions. This is useful because a genetic marker can carry information about another genetic locus: tag SNPs exploit this redundancy, without establishing a causal genetic association.

Consider two genetic loci with alleles $A,a$ and $B,b$. Sample one haplotype, rather than an unphased diploid genotype. Write its four haplotype frequencies as $f_{AB},f_{Ab},f_{aB},f_{ab}$, summing to one, and set the marginal allele frequencies to $p=f_{AB}+f_{Ab}$ and $q=f_{AB}+f_{aB}$. The linkage disequilibrium coefficient isThe second expression follows by substituting $f_{ab}=1-f_{AB}-f_{Ab}-f_{aB}$. The entire two-locus joint probability distribution is thenEvery row and column sum is unchanged when $D$ varies. Linkage equilibrium means $D=0$: all four haplotype frequencies equal products of allele frequencies, so the two allele indicators are independent random variables. For $p>0$, the conditional probability $P(B\mid A)=q+D/p$ shows directly how positive $D$ enriches $B$ among copies carrying $A$, while negative $D$ depletes it. Positive $D$ also enriches $ab$; negative $D$ enriches $Ab$ and $aB$. Swapping the labels $A$ and $a$ reverses the sign of $D$, so neither sign is intrinsically favorable.

For example, fix the allele frequencies at $p=0.60,q=0.40$. At linkage equilibrium, the haplotype frequencies in the order $AB,Ab,aB,ab$ are $(0.24,0.36,0.16,0.24)$. Positive linkage disequilibrium $D=0.12$ instead gives $(0.36,0.24,0.04,0.36)$, with $P(B\mid A)=0.60$; negative linkage disequilibrium $D=-0.12$ gives $(0.12,0.48,0.28,0.12)$, with $P(B\mid A)=0.20$. Counting each allele separately would miss the difference.

Nonnegative haplotype frequencies constrain $D$. Define the nonnegative bounds $L=\min (pq,(1-p)(1-q))$ and $U=\min (p(1-q),(1-p)q)$. The four entries above giveThus $|D|\le 1/4$ globally, but a particular pair of allele frequencies usually permits a smaller, asymmetric interval. The standardized linkage disequilibrium convention used here retains the sign:This convention requires $0<p,q<1$; if either genetic locus has only one allele in the population, $D=0$ but normalized measures are undefined. Some presentations report only $|D^{\prime }|$. At $D^{\prime }=1$ or $-1$, at least one of the four haplotypes has zero frequency; this alone does not prove that genetic recombination has never occurred.

For the indicator random variables $X=1_A$ and $Y=1_B$, their covariance is $D$, and their variances are $p(1-p)$ and $q(1-q)$. The allelic correlation measure is their squared correlation coefficient,The subscript distinguishes it from the recombination fraction $r$ in linkage-disequilibrium decay under recombination. Squaring loses the association's sign. In the illustrated positive and negative examples, $r_{\mathrm{LD}}^2=0.25$ in both, whereas $D^{\prime }=0.75$ and $-0.50$, respectively. For another original example, take $p=0.10,q=0.50,f_{AB}=0.10$: then $D=0.05,D^{\prime }=1$, yet $r_{\mathrm{LD}}^2=1/9$. Every $A$ carries $B$, but most copies carrying $B$ lack $A$. The frequency constraints and the distinction between these normalizations are analyzed in Mathematical properties of the r² measure of linkage disequilibrium by Jenna M. VanLiere and others (sections 2.1 and 2.5, pp. 131–132 and 135–136).

Genetic linkage describes transmission through meiosis; linkage disequilibrium describes a population's haplotype frequencies. Nearby genetic loci on a chromosome can be in linkage equilibrium. Conversely, mixing populations can create linkage disequilibrium even between genetic loci on different chromosomes. For example, mix equal numbers of independently sampled haplotypes from two populations, each internally at linkage equilibrium, with $p=q=0.80$ in one and $p=q=0.20$ in the other. The pooled allele frequencies are $p=q=0.50$, but $f_{AB}=(0.64+0.04)/2=0.34$, giving $D=0.09$. This is a consequence of population stratification, not a measurement of the recombination fraction.

Measured haplotype frequencies also depend on the sample. Finite sample size, related individuals, genetic-study ascertainment, genotyping error and uncertain haplotype estimation can change an estimate. A missing rare haplotype can give empirical $|D^{\prime }|=1$ without demonstrating its absence from the population. Report the sample definition and allele frequencies alongside the association measure; a nonzero sample estimate alone does not establish nonzero population linkage disequilibrium.

A concrete dataset is the 2005 A haplotype map of the human genome by The International HapMap Consortium. The International HapMap Consortium used genotyping platforms including Illumina BeadArray to obtain SNP genotype calls from 269 DNA samples across four collections: 30 parent–offspring trios from Ibadan, 30 trios from Utah, 45 unrelated individuals from Beijing and 44 from Tokyo. Parent–offspring trios and the PHASE algorithm supplied haplotype estimation, rather than direct observation of every chromosome. In Figure 7, 36 SNPs resolve seven haplotypes in a chromosome 2 region in the Utah panel; seven tag SNPs capture the 36 SNPs at $r_{\mathrm{LD}}^2\ge 0.8$. The readout demonstrates reusable allelic correlation, not causal effects.

Controls required over 80% complete calls, at most one discordance among five duplicate genotypes, at most one Mendelian segregation error per trio panel, and a Hardy-Weinberg equilibrium test with $P>0.001$; blind reassessment checked performance. Minor allele frequencies of at least 0.05 were targeted, the collections were not population-representative, and inferred haplotypes retain phase uncertainty. These limits restrict extrapolation of linkage disequilibrium.^[ref]

Genetic recombination erodes linkage disequilibrium by reshuffling alleles between haplotypes, while preserving marginal allele frequencies in the idealized model. Let $f_{AB,t}$ denote the haplotype frequency in the gamete pool of generation $t$, let $p,q$ denote its marginal allele frequencies, and let $r\in [0,1/2]$ be the constant recombination fraction between two autosomal genetic loci. Assume random mating within one population, nonoverlapping generations, fair Mendelian segregation, equal reciprocal gamete production, the same $r$ in both sexes, and a deterministic population large enough to neglect genetic drift. Exclude natural selection, mutation and gene flow. ThenHere $p$ and $q$ remain constant: recombining changes which alleles occur together, not their total numbers. The other haplotype frequencies change by $+r{D}_t$ for $Ab,aB$ and $-r{D}_t$ for $ab$. Consequently the joint probability distribution stays within its allowed bounds and approaches the product of its marginal distributions when $r>0$.

For the original positive example $p=0.60,q=0.40,D_0=0.12$, a recombination fraction $r=0.10$ gives $D_1=0.108$ and $D_{10}\approx 0.04184$. The negative example $D_0=-0.12$ has exactly the opposite sequence: genetic recombination reduces the magnitude without reversing the sign. With $r=0$, the initial linkage disequilibrium persists in this model; even independent assortment at $r=1/2$ only halves $D$ each generation. A population association does not vanish after one generation merely because the genetic loci are unlinked.

The geometric sequence gives a generation scale: for $r>0$, the continuous interpolation reaches $|D_0|/2$ at $t_{1/2}=\log (1/2)/\log (1-r)$. Actual generations are integers; the first at or below half is $\lceil t_{1/2}\rceil$. For small $r$, $t_{1/2}\approx (\log 2)/r$. Because the denominator of allelic correlation stays fixed, $r_{\mathrm{LD},t}^2=(1-r{)}^{2t}{r}_{\mathrm{LD},0}^2$; its decay is faster than the magnitude of $D_t$.

A short derivation uses random mating to assign frequency $2f_{AB}{f}_{ab}$ to the double heterozygote $AB/ab$ and $2f_{Ab}{f}_{aB}$ to $Ab/aB$. Relative to $r=0$, each loses or gains an $AB$ gamete fraction $r/2$, respectively. Their net contribution is $r(f_{Ab}{f}_{aB}-f_{AB}{f}_{ab})=-rD$. The other genotypes produce no net change in $f_{AB}$, proving the recurrence and, by iteration, its solution.

This temporal model does not identify elapsed generations from physical separation alone. The recombination fraction depends on genetic recombination, including recombination hotspots, while population stratification, natural selection, gene flow or genetic drift can create or maintain linkage disequilibrium. Repeated sampling across time would also require accounting for sample size and haplotype estimation. A cross-sectional plot of association against distance between genetic loci is therefore a different observation from the idealized trajectory plotted here.

For two genetic loci with two alleles each, linkage equilibrium is independence of their states on sampled haplotypes. It concerns association between genetic loci; Hardy-Weinberg equilibrium concerns pairing of alleles into diploid genotypes.

In ordinary diploid meiosis, the two copies at a genetic locus separate: each gamete receives one copy. With fair transmission, a heterozygote $Aa$ produces gametes carrying $A$ or $a$ with probability $1/2$ each. This is a transmission mechanism conditional on a parental genotype, rather than a statement about how common either allele is in a population.

For the one-locus cross $Aa\times Aa$, assume independent parental transmissions, no new mutation, and no differences in fertilization or survival between the resulting genotypes. Multiplying the two gamete probabilities gives four equally likely ordered outcomes: $AA,Aa,aA,aa$, each with probability $1/4$. Combining the two orders of the heterozygote givesOnly under the additional ideal model of complete dominance of $A$, complete penetrance, and no environmental changes to the scored characteristic do $AA$ and $Aa$ share the dominant phenotype. Its probability is then $0.25+0.50=0.75$. The inheritance calculation alone does not determine a phenotype or predict a generic disease risk.

These are probabilities, not a requirement that every four offspring contain one $AA$, two $Aa$ and one $aa$. For 100 independent offspring under the ideal model, the expected values of the three counts are $25,50,25$; the $aa$ count has a binomial distribution with standard deviation $\sqrt{100(0.25)(0.75)}\approx 4.33$. Each new transmission keeps the same conditional probabilities given the parental genotypes.

In his paper published in 1866, Mendel reported 5,474 round and 1,850 wrinkled pea seeds: $5474/7324\approx 0.7474$ round, close to the ideal $0.75$. Following descendants also distinguished two inherited classes within the round phenotype, illustrating why genotype cannot always be read from appearance.^[ref] At several genetic loci, transmission patterns additionally depend on genetic linkage and genetic recombination.