17.5 Genetic Linkage & Mapping in Eukaryotes

Genetic Linkage & Mapping in Eukaryotes

Quick Answer: Genes on the same chromosome are linked and do not assort independently. The fraction of recombinant offspring in a testcross defines the recombination frequency (RF), which equals the map distance in centimorgans (cM). In a three-point cross, the two rarest classes are double crossovers and reveal the middle gene.

Linkage and Recombination Frequency

Mendel's law of independent assortment holds only for genes on different chromosomes. Genes on the same chromosome are linked and tend to be inherited together. However, crossing over during prophase I can reshuffle alleles between homologs, producing recombinant combinations.

If two genes are linked, a heterozygote written in coupling phase (AB/ab) produces mostly parental gametes (AB, ab) and fewer recombinant gametes (Ab, aB). In repulsion phase (Ab/aB), the parentals are Ab and aB, recombinants AB and ab. Recombinants arise from a crossover between the two loci in prophase I.

Recombination frequency = (recombinant offspring ÷ total offspring) × 100%.

• RF = 0% → complete linkage (loci very close; crossing over between them essentially never occurs).
• RF = 50% → independent assortment (loci far apart on the same chromosome, or on different chromosomes). Recombination frequency never exceeds 50% for a two-point cross.
• 1% recombinant = 1 cM on a genetic map. 1 cM ≈ ~1 megabase in humans, but this varies with recombination hotspots and sex (females have higher recombination rates).

Worked two-point example

In a testcross of GgWw × ggww, 800 offspring are scored:

Recombinants = 18 + 22 = 40; RF = 40/800 = 5.0%; map distance G–W = 5.0 cM. Note the two parentals are roughly equal and the two recombinants are roughly equal — a hallmark of a clean testcross.

Three-Point Crosses and Gene Order

A three-point testcross (abc/+++ × abc/abc) samples more recombinant classes and resolves gene order. Procedure:

1. Identify the two most common classes = parental (non-recombinant).
2. Identify the two rarest classes = double crossovers (DCO). A DCO requires two simultaneous crossovers, one in each interval, so it is the least frequent event.
3. Compare parental and DCO genotypes; the gene that changes position in the DCO relative to parental is the middle gene — because only the middle gene's allele pairing is swapped by a double crossover.
4. Single crossovers (SCO) in each interval give distances for each adjacent pair.

Worked three-point example

Suppose 1000 progeny from a b c / + + + × a b c / a b c:

DCOs (rarest): a + c and + b +. Comparing parentals (+ + +, a b c) with DCOs: the b locus has switched — so b is in the middle. Gene order: a – b – c.

Map distances:

• Interval a–b = (SCO in a–b + DCO) ÷ total = (22 + 18 + 3 + 1)/1000 = 4.4 cM.
• Interval b–c = (SCO in b–c + DCO) ÷ total = (70 + 66 + 3 + 1)/1000 = 14.0 cM.
• Total a–c = 18.4 cM.

Interference and Coefficient of Coincidence

A double crossover requires two crossovers in the same meiosis. If crossovers occur independently, expected DCO frequency = (RF interval 1) × (RF interval 2). Here expected DCO = 0.044 × 0.140 × 1000 = 6.16; observed DCO = 4.

Coefficient of coincidence (c.o.c.) = observed DCO ÷ expected DCO = 4/6.16 = 0.65. Interference = 1 − c.o.c. = 0.35 — meaning ~35% of expected double crossovers are suppressed by the first crossover's presence. Interference is highest near the first crossover and decreases with distance.

Sex Linkage

• X-linked recessive: males are hemizygous (one X), so a single mutant allele is expressed. Examples: hemophilia A (F8), Duchenne muscular dystrophy (DMD), red-green color blindness (OPN1LW/MW), X-linked agammaglobulinemia (Bruton). Pedigree pattern: affected males in each generation connected through carrier females; no male-to-male transmission. Carrier mothers (X^A X^a) pass to 50% of sons (affected) and 50% of daughters (carriers); affected fathers (X^a Y) pass to all daughters (carriers), no sons.
• X-linked dominant: hypophosphatemic rickets (XLH), Rett syndrome (MECP2, lethal in males). Affected fathers pass to all daughters, no sons; affected heterozygous mothers pass to 50% of children of either sex.
• Y-linked (holandric): father-to-son only. SRY (sex determination), some spermatogenesis factors (AZF regions). Very few traits.
• X-inactivation (dosage compensation): in female mammals, one X is inactivated per cell early in embryogenesis (Lyon hypothesis), forming the Barr body. Choice is random and clonally inherited, producing mosaicism in heterozygous females (e.g., calico cat coat). Skewed X-inactivation can cause female carriers of X-linked recessive diseases to show symptoms.

PA-CAT Application

Per the PA-CAT Bulletin of Information, rev. 20240815, Patterns of Inheritance includes linkage analysis. Expect testcross ratio interpretation, RF-to-cM conversion, three-point gene-order determination from rarest classes, and sex-linkage pedigree reasoning (e.g., identifying an X-linked recessive pattern from a pedigree where affected males appear in each generation through carrier females, with no male-to-male transmission). Remember: parental classes are the most frequent; DCO classes are the least frequent; the gene that flips between parental and DCO is the middle gene.