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A chromatid is one of the two copies of a replicated chromosome. Each homologous chromosome contributes two sister chromatids before separation in meiosis; chromosomal crossovers can exchange segments between nonsister chromatids.

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A chromosomal crossover is a reciprocal exchange of segments between nonsister chromatids of homologous chromosomes during meiosis. It can change the haplotypes inherited at selected genetic loci; whether it does so depends on the transmitted chromatid and interval.

Experimental tractability is the practical ability to cultivate a system, manipulate the variables of interest, and obtain an interpretable readout with appropriate scientific controls. In Saccharomyces cerevisiae, inexpensive cultivation, accessible haploid and diploid states, and tools for introducing DNA make controlled changes of genotype feasible. A haploid biological strain exposes a changed allele without a second copy at that genetic locus, while crosses and meiosis permit tests of inheritance. ^[ref]

An accessible readout is not necessarily a direct measurement of the mechanism. Growth of a biological colony integrates many cellular functions. Failure to grow could reflect the intended gene deletion, unsuitable conditions, failed genetic transformation, or another mutation. Matching the genetic background and using scientific controls narrows these alternatives. Ease of doing an experiment and strength of its inference are distinct advantages.

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.

Genetic recombination rearranges inherited genetic material into new combinations. In the meiosis example considered here, chromosomal crossover exchanges segments between nonsister chromatids of homologous chromosomes. A transmitted chromatid can consequently contain DNA from both parental homologues. This produces new haplotypes without requiring a new mutation at either scored genetic locus. See Meiosis.

For phase-known genotype $AB/ab$, a single chromosomal crossover between the selected genetic loci can yield $Ab$ and $aB$ on the participating chromatids; the other two chromatids can remain $AB$ and $ab$. Thus an exchange in the four-chromatid bundle does not mean every resulting gamete is recombinant. Only a chromosomal crossover encountered along the transmitted chromatid within the selected interval affects its two-locus score.

Follow parental origin along the transmitted chromatid from the first genetic locus to the second. Each exchange switches that origin. An odd number of such chromosomal crossovers changes the endpoint combination; an even number restores parental phase. Two exchanges can therefore leave the scored haplotype as $AB$ even though an internal segment came from the other parental homologous chromosome. Recombinant status is relative to the chosen genetic loci, not a complete record of exchanges along the chromosome.

This integer parity distinction connects the physical process of genetic recombination to the observed recombination fraction. It also explains why a long genetic map interval can accumulate many exchanges while its endpoint recombination fraction approaches $1/2$ in the Haldane mapping function model.

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.

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]

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).

The first division of meiosis separates replicated homologous chromosomes; their sister chromatids remain associated.

The second division of meiosis separates sister chromatids, without an intervening round of DNA replication.

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.

Oogenesis is development of female gametes; the unequal divisions of human meiosis produce polar bodies alongside the developing ovum.