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Let $q_g=P(g\mid g_f,g_m)$ be the Mendelian segregation probability, and assume the child's disease status depends on the parental genotypes only through its own genotype, with penetrance proportional to $R_g$. Bayes' theorem gives the displayed conditional law. For a diallelic parent pair there are four equally likely labelled transmission outcomes. If an unordered genotype occurs in $m_g$ of them, $q_g=m_g/4$, so its probability is $m_g{R}_g/{\sum }_h{m}_h{R}_h$. Dropping multiplicities would, for example, misweight a heterozygous child of two heterozygotes.

Fertilization is fusion of gametes. Fusion of two haploid gametes restores the diploid state in the inheritance model used for Mendelian segregation.

Experimental Reconstruction of Functional Gene Transfer from the Tobacco Plastid Genome to the Nucleus by Sandra Stegemann and Ralph Bock (2006, Results, pp. 2870–2874) separated relocation of DNA from activation of a relocated gene. Engineered tobacco chloroplasts carried an nptII gene with nuclear expression signals beside an aadA gene with plastid expression signals. Selection on kanamycin recovered plants with the segment in the nuclear genome. Crosses then replaced the engineered chloroplasts with ordinary chloroplasts: the nuclear segment remained, but the plants were initially sensitive to spectinomycin and streptomycin. The transferred aadA gene was not yet functional in its new setting.

Leaf pieces were cultured on regeneration medium containing both drugs. Resistant regenerated plants supplied the functional readout. Using both drugs reduced selection of spontaneous plastid mutations conferring resistance mainly to one drug. Reciprocal crosses and selfing distinguished maternal plastid inheritance from nuclear Mendelian segregation; eight of sixteen initial candidates had the latter pattern. ^[ref]

The molecular readout came from RNA hybridization and DNA sequencing. A Northern blot detected aadA messenger RNA in activated lines, with different transcript sizes. Thermal asymmetric interlaced polymerase chain reaction recovered upstream DNA; products present in activated lines but absent in original-line and ordinary-plant controls were sequenced. The mapped rearrangements connected aadA to the upstream nuclear gene promoter. Thus resistance, inheritance, RNA detection and changed nucleotide sequences supported activation of the transferred nuclear gene, rather than resistance alone. ^[ref]

This engineered test demonstrates one route to functional nuclear gene expression after plastid DNA transfer. It does not establish a natural transfer rate, replay ancient endosymbiosis or demonstrate that the resulting protein is imported into a chloroplast. The authors explicitly identify acquisition of targeting information as an additional step not reproduced by this experiment. ^[ref]

A genetic genetic carrier has one disease-associated recessive allele and one normal allele at a genetic locus, and is unaffected under a fully penetrant recessive model. Two genetic carriers have affected-offspring probability $1/4$ under Mendelian segregation; a genetic carrier and an individual with no disease allele have affected-offspring probability zero under that same mutation-free model.

Why can two individually fair allele transmissions fail to be independent? Genetic loci on the same chromosome can travel together through meiosis. This tendency is genetic linkage; in an ordinary two-locus model, detectable genetic linkage means a recombination fraction below $1/2$. A genetic marker can therefore help track a neighboring genetic locus through a pedigree. See Clark, Douglas and Choi, Biology 2e: chromosomal theory and genetic linkage.

Take a diploid individual heterozygous at two genetic loci, with phase-known genotype $AB/ab$: one homologous chromosome carries alleles $A,B$, the other $a,b$. The parental haplotypes are $AB,ab$; the recombinant haplotypes are $Ab,aB$. Assume fair Mendelian segregation, equal probabilities within each reciprocal pair of gametes, and no allele-dependent differences in gamete recovery. If the recombination fraction is $r=0.2$, the gamete probabilities are $0.4,0.4,0.1,0.1$ in that order. Each genetic locus still transmits its two alleles equally, but $P(AB)=0.4$ exceeds $P(A)P(B)=0.25$: fairness at each genetic locus does not imply independence between genetic loci.

At $r=0$, only the parental haplotypes occur, each with probability $0.5$: this is complete genetic linkage. At $r=0.5$, all four haplotypes have probability $0.25$, giving independent assortment. A recombination fraction near $0.5$ alone cannot establish whether the genetic loci lie on different chromosomes or far apart on one chromosome.

Genetic linkage concerns joint transmission through meiosis. Linkage disequilibrium concerns haplotype frequencies in a population, while genetic association compares allele or genotype frequencies with a phenotype. These questions require different observations; pedigrees directly reveal co-transmission.

Genetics studies how genes are inherited, how their alleles vary, and how that variation contributes to phenotypes. Its central distinction is between an inherited description, the genotype, and what an organism actually expresses, the phenotype. Following alleles through a pedigree answers a family-level question; following allele frequencies through generations leads to population genetics and evolution.

The beauty of Mendelian segregation is that discrete inherited copies can generate predictable statistical patterns without making every offspring identical. A single genetic locus already connects meiosis, probability and observable variation. The Hardy-Weinberg principle then supplies a population baseline: it separates reshuffling inherited copies from processes such as natural selection that change their frequencies. These simple models are useful precisely because their assumptions can be stated and checked.

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.

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.

For two individuals carrying labelled copies $(a,b)$ and $(c,d)$, condensed identity coefficients give the probabilities of nine identity by descent patterns, identifying states that differ only by exchanging the two homologues within either individual. They retain within-individual autozygosity, which three ordinary outbred sharing coefficients omit. One convention isEach block consists of copies identical by descent, and different blocks have distinct ancestral origins. For siblings whose parents are otherwise outbred first cousins, these coefficients are $(1,0,2,1,2,1,15,30,12)/64$. This follows by conditioning on whether the parents share one ancestral copy and enumerating the four independent Mendelian segregation choices to their two children. The kinship coefficient is ${\Delta }_1+({\Delta }_3+{\Delta }_5+{\Delta }_7)/2+{\Delta }_8/4$, here $9/32$. Weir's discussion of identity states uses this convention.

The kinship coefficient ${\varphi }_{XY}$ is the probability that one randomly selected allele copy from each of $X,Y$ is identical by descent. For a noninbred individual, ${\varphi }_{XX}=1/2$, and more generally ${\varphi }_{XX}=(1+F_X)/2$. For distinct individuals with parents $M_X,P_X$, Mendelian segregation gives ${\varphi }_{XY}=[{\varphi }_{M_{X}Y}+{\varphi }_{P_{X}Y}]/2$. The offspring's inbreeding coefficient equals the parents' kinship coefficient.

// Target: biology.bigb

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.

Meiosis reduces two sets of chromosomes to one. In the ordinary diploid life cycle, this makes it possible for fertilization of two haploid gametes to restore the diploid state rather than double the number of sets in every generation. Its two divisions perform different tasks: meiosis I separates homologous chromosomes, whereas meiosis II separates sister chromatids. One round of DNA replication precedes both divisions; there is no second round between them. See ^[ref].

Two homologous chromosomes carry corresponding genetic loci, potentially with different alleles; two sister chromatids originate by DNA replication of one chromosome. After chromosomal crossover between nonsister chromatids, the sister chromatids need not carry identical alleles everywhere. Separating the two relationships explains why a haploid product of meiosis I still contains replicated chromosomes.

The two divisions explain Mendelian segregation; variable orientation of different homologous chromosome pairs connects it to independent assortment. Genetic recombination adds another source of allele combinations. These mechanisms reshuffle inherited material without requiring a new mutation.

The symmetric diagram is useful for counting, but human oogenesis produces one functional ovum rather than four. The polar bodies receive discarded chromosomes and little cytoplasm. See Human Reproductive Anatomy and Gametogenesis (§43.3, Oogenesis and Figure 43.13: unequal divisions and polar bodies).

Nonparametric linkage analysis tests excess identity by descent among relatives with a shared phenotype, without specifying its complete penetrance model or disease-allele frequency. Its null still uses Mendelian segregation and the specified pedigree.

// Target: biology.bigb

If parental genotype frequencies are $(u,v,w)$ and the two parental gametes are independently sampled from the same allele pool, Mendelian segregation gives allele frequency $p=u+v/2$. Their independent union has the displayed Hardy-Weinberg proportions, irrespective of parental genotype proportions. Equal allele frequencies in the two sexes and no transmission or viability distortion are part of this conclusion.

Label each parent's two allele copies by its realized transmission: let $(A,B)$ be the transmitted and untransmitted alleles of the first parent and $(C,D^{\prime })$ those of the second. Under random mating and Hardy-Weinberg equilibrium, the four parental allele draws are independent with population frequencies $\pi$. Random Mendelian segregation merely swaps the two independent copies within each parent, so the transmitted/untransmitted labelled pair still has probability ${\pi }_a{\pi }_b$; the second pair has probability ${\pi }_c{\pi }_d$.

The child receives $A,C$. If $\mathcal{D}$ is its disease event, its penetrance is $k{\psi }_A{\psi }_C$ and its overall disease probability is $k{Z}^2$. ThereforeThis proves the transmitted and untransmitted alleles under multiplicative penetrance factorization. Under these assumptions, the transmitted pair $a/c$ has the affected-case distribution, while the complementary pair $b/d$ has the population distribution and is independent of the transmitted pair. It is the family pseudo-control genotype.

As in part (a), the formula records labelled transmissions. When observed genotypes are unordered, sum over every compatible parental-origin transmission. This includes multiplicities for overlapping parental alleles or two heterozygous parents producing a heterozygous child; it avoids interpreting one product as the total probability of all observationally identical configurations. The homogeneous random-mating model is also essential to the unconditional independence: mixing ancestry strata can correlate the case and pseudo-control through their shared family background.

Let the initial genotype frequencies be arbitrary $u,v,w$ for $a/a,a/b,b/b$, with $u+v+w=1$. Mendelian segregation makes the allele frequency among gametesIn random mating, the two parents are selected independently. If $g(G)$ is a parent's transmission probability for allele $a$, then independence gives $\mathbb{E}[g(G_1)g(G_2)]=\mathbb{E}g(G_1)\mathbb{E}g(G_2)=p^2$. Thus the two independently united gametes have allele probabilities $p,1-p$, even though the two alleles within an original parent need not have been independent. The offspring frequencies are thereforeNo initial condition $v^2=4uw$ was used. This proves one-generation random-mating equilibrium. The usual assumptions are a common allele pool in the two sexes, random union of gametes, ordinary Mendelian segregation, and no intervening mutation or genotype-dependent survival selection. Unequal male and female allele frequencies, for example, would instead give $p_m{p}_f$ for $a/a$, not generally the square of a single common frequency.

Treat the two parents as unordered, so three possible genotypes give six mating types. A homozygous parent transmits its allele with probability one; a heterozygous parent transmits either allele with probability one half by Mendelian segregation. Independent parental transmissions give the following complete offspring distributions:Each row sums to one. In particular, heterozygote-by-heterozygote mating can produce all three offspring genotypes, whereas the two opposite homozygotes produce only heterozygotes. These conditional offspring probabilities do not require Hardy-Weinberg equilibrium of the parents.

Use the conventional additive pedigree relatedness $r_{UV}=2{\varphi }_{UV}$, where ${\varphi }_{UV}$ is the kinship coefficient, the probability that randomly selected gene copies from $U$ and $V$ are identical by descent. For noninbred individuals this convention gives ordinary full siblings relatedness $1/2$, as required. It is important that this is not the probability of sharing at least one copy: that probability is $3/4$ for ordinary full siblings.

Let the parents be $M,F$. They are otherwise outbred first cousins, so ${\varphi }_{MF}=1/16$ and ${\varphi }_{MM}={\varphi }_{FF}=1/2$. For two children $C,D$, independent Mendelian segregation givesTherefore the autosomal additive relatedness isThe additional $1/16$ beyond $1/2$ comes from maternal-paternal cross-origin matches. Each child has inbreeding coefficient $1/16$. The additive coefficient is a sum of descent probabilities rather than a literal bounded probability in all inbred pedigrees; a relationship normalized to self-relatedness would be $r_{CD}/(1+F)=9/17$. Likewise, identity-by-descent sharing of inbred full siblings gives a different matched-copy statistic. The exam's calibration to $1/2$ supports the additive convention, but the quoted probability wording needs this qualification.

For sex chromosomes the route of cousinship matters, not merely the fact that the parents have opposite sex. Let $P_M$ be the mother-cousin's parent in the shared sibling pair, and $P_F$ the father-cousin's parent in that pair. Let $F_X$ be the probability that the mother's randomly transmitted X matches the father's single X by descent. With no founder inbreeding, the possibilities areIn the first two rows the father-cousin is related through his father, whose X he does not inherit. In the third row his mother is a sister of the mother-cousin's father: the relevant brother and sister match their maternal X copies with probability $1/4$, and the mother-cousin transmits her paternal X with probability $1/2$, giving $1/8$. In the last row the two sisters' random X gametes match with probability $3/8$: their common father's X contributes $1/4$, and their common mother's two X copies contribute $1/8$. Transmission by the mother-cousin halves this to $3/16$. This derives the sex-linked cousin kinship values.

For an unweighted comparison by chromosome pair, two daughters have sex-pair additive relatedness $3/4+F_X$: their paternal X always matches, maternal X matches with probability $1/2$, and the two cross-origin terms each contribute $F_X$. Two sons have mean sex-pair relatedness $3/4$, from an X match of probability $1/2$ and a common paternal Y with probability one. A daughter-son pair has sex-pair value $1/4+F_X/2$, with the Y contribution zero. These calculations treat chromosomes as unrecombined blocks and weight X and Y equally; chromosome-length-weighted or locus-specific coefficients are different.

For two children with independent equally likely sexes, these three cases have weights $1/4,1/4,1/2$. Their mean sex-pair relatedness is $1/2+F_X/2$, so averaging the 22 autosomal pairs and the sex pair givesIf the cousins' route has $F_X=0$, this is $103/184$, slightly below $9/16$. If, as an additional sampling assumption, the four parent-sex routes in the table are equally likely, $\mathbb{E}{F}_X=5/64$ and the answer is $1653/2944$. These are different legitimate numerical corrections under different specified pedigrees. There is no unique sex-chromosome correction from opposite-sex cousinhood alone. The table and formula give it once the missing pedigree and offspring-sex information is supplied.

Genetic linkage analysis asks whether a genetic marker and a disease genetic locus cosegregate within families more often than would be expected under independent assortment. It uses pedigrees and Mendelian segregation to estimate the recombination fraction $\theta$, with evidence for linkage arising when the data favor $\theta <1/2$. Even when founders are in linkage equilibrium, linked loci can show familial cosegregation.

Genetic association analysis instead asks whether observed allele or genotype frequencies differ with disease status in a sampled population, often comparing cases with controls. It can detect a causal variant or a correlated genetic marker through linkage disequilibrium; population structure and selection can also create association. Linkage is about familial transmission; association is about population dependence. Neither an association alone nor a low family recombination estimate establishes that the marker itself causes disease.