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Chromosomes, linkage and gene mapping

Genetics / Bioinformatics (B.Sc) · Classical genetics and cytogenetics · free study notes for exams, placements and interviews

In short

A chromosome is one long DNA–protein thread carrying hundreds to thousands of genes in a fixed linear order, so genes on the same chromosome tend to be inherited together — this is linkage, the first major exception to independent assortment. Crossing over at meiosis breaks that togetherness in a fraction of gametes, and because that fraction grows with the physical distance between two loci, it can be used as a ruler: 1% recombinants = 1 centimorgan. Gene mapping — two-point and three-point test crosses, interference, mapping functions, LOD scores and marker-based maps — is the core quantitative content of classical genetics and is examined heavily in CSIR-UGC NET Life Sciences, GATE BT/XL, ICAR ASRB NET, DBT-BET and ICMR JRF. In practice it underpins marker-assisted breeding at ICAR institutes, linkage and association studies at CSIR-IGIB and NIPGR, and seed-industry trait pyramiding.

Explained

Chromosomes as the physical carriers of genes

Linkage: definition, discovery and types

Crossing over: mechanism and evidence

Recombination frequency and two-point mapping

Three-point test cross, gene order and interference

Three loci in one cross give order, two distances and interference at once. Eight phenotypic classes appear in four reciprocal pairs.

  1. Identify the parental (non-crossover) pair — the two most frequent classes; they reveal the F1 phase.
  2. Identify the double-crossover (DCO) pair — the two rarest classes.
  3. Determine the middle gene: compare a parental class with a DCO class; the single gene whose allele has "switched" is the middle gene (a DCO flips only the central marker).
  4. Rewrite all classes in the correct order, then assign each pair to region I (left–middle), region II (middle–right) or DCO.
  5. RF (region I) = (SCO I + DCO)/total × 100; RF (region II) = (SCO II + DCO)/total × 100. Always add the DCOs — they are crossovers in both regions.
  6. Total map length = region I + region II; the directly observed outer-marker RF is smaller by 2 × DCO%.
  7. Interference: expected DCO = RF(I) × RF(II) × total. Coefficient of coincidence (CoC) = observed DCO/expected DCO; interference I = 1 − CoC.
  8. Positive interference (CoC < 1) is the norm — one crossover reduces the chance of a second nearby; it falls towards zero as the regions get farther apart. Negative interference (CoC > 1) is rare and usually indicates clustering or small samples. Complete interference gives CoC = 0.
  9. The coincidence/interference value is not a probability of linkage; it is a measure of crossover independence.

Mapping functions, and other routes to a map

Key points

Formulas and examples

RF (%) = (number of recombinant progeny / total progeny) × 100 — dimensionless percentage; 1% = 1 cM = 1 map unit
Map distance (cM) = RF(region I) + RF(region II); RF(outer pair, observed) = map distance − 2 × DCO%
Expected DCO = RF(I) × RF(II) × N (RFs as fractions, N = total progeny)
Coefficient of coincidence CoC = observed DCO / expected DCO; Interference I = 1 − CoC
Haldane: d = −0.5 ln(1 − 2r); Kosambi: d = 0.25 ln[(1 + 2r)/(1 − 2r)] — r = RF as a fraction, d in Morgans (× 100 for cM)
χ² = Σ (O − E)² / E, df = (classes − 1) = 3 for a two-locus test cross against 1:1:1:1
LOD(θ) = log10 [ θk (1 − θ)n−k / 0.5n ] — θ = recombination fraction, k = recombinants, n = informative meioses
Arm ratio = long arm (q) / short arm (p); Centromeric index = p/(p + q) × 100

Example 1 — two-point test cross and a test for linkage. *Drosophila* F1 females AB/ab crossed to ab/ab males; 1,000 progeny: AB 420, ab 400, Ab 90, aB 90.

Example 2 — three-point test cross: order and distances. F1 heterozygote test-crossed; 1,000 progeny scored for a, b, c:

Phenotype classCountInterpretation
a b c357parental
+ + +357parental
a b +93single crossover, region II
+ + c93single crossover, region II
a + +43single crossover, region I
+ b c43single crossover, region I
a + c7double crossover
+ b +7double crossover

Example 3 — interference, and why the outer distance looks short (same data).

Example 4 — mapping functions at high RF. Two markers show r = 0.45 (45% recombinants).

Example 5 — LOD score in a human pedigree. Pooling five phase-known families: 40 informative meioses, 8 recombinants, so the maximum-likelihood θ̂ = 8/40 = 0.20.

Independent assortmentLinkage
Gene locationDifferent chromosomes (or very far apart)Same chromosome
Test-cross ratio1:1:1:1Parentals ≫ recombinants
RF≈ 50%< 50%
χ² (1:1:1:1)Not significantSignificant
UseMendel's second lawGene mapping

Where it is used

Interview questions

Common mistakes

Related topics

Genetics · Mendelian inheritance: gene interaction and pedigree analysis · Cytogenetics and chromosomal disorders · Cell biology · Molecular markers and QTL mapping · Population genetics and Hardy-Weinberg equilibrium

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