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Linkage, Recombination & Gene Mapping

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Linked genes break the 9:3:3:1 rule

Independent assortment only applies to genes on different chromosomes. Genes on the same chromosome are linked: they travel together into a gamete unless crossing over separates them, so a linked dihybrid testcross does not give the 1:1:1:1 ratio you would expect from unlinked genes. Instead most offspring are parental (matching the arrangements the heterozygous parent carried) and a minority are recombinant (new allele combinations made by a crossover). Which classes count as recombinant depends on how the parent’s alleles are arranged: in coupling (cis), AB / ab, the dominant alleles ride together, so AB and ab offspring are parental; in repulsion (trans), Ab / aB, the parentals are Ab and aB, and now AB and ab are the recombinants. You must read the parent’s genotype before labeling any offspring class.

Recombination frequency measures distance

The farther apart two genes sit on a chromosome, the more physical room there is for a crossover to fall between them, so the recombination frequency (RF) — the percentage of offspring that are recombinant — rises with distance. Geneticists define 1% recombination = 1 map unit = 1 centimorgan (cM), turning a breeding result into a physical distance. RF has a ceiling: it cannot exceed 50%. Genes so far apart that a crossover almost always occurs between them produce recombinants and parentals in equal numbers — exactly what unlinked genes do — so a 50% RF is indistinguishable from independent assortment. This is why very distant genes on one chromosome can appear "unlinked" in a single cross.

Building a linear map from pairwise distances

Map distances are (approximately) additive along a chromosome, so three genes can be ordered from their three pairwise recombination frequencies. The trick: the largest of the three RFs separates the two outer genes, and the remaining gene lies between them. You then check that the two shorter distances add up to the largest. In practice the directly measured outer distance is often a hair smaller than that sum, because double crossovers between the outer genes swap the middle marker but restore the outer alleles to their parental arrangement — those events go uncounted, so long spans slightly underestimate the true distance. That is exactly why maps are built by stitching together many short intervals rather than measuring one long one.

Recombination frequency
RF = (recombinant offspring ÷ total offspring) × 100%
Count both recombinant classes. 1% RF ≈ 1 map unit (centimorgan). Maximum RF is 50%, at which point linked genes assort as if independent.
Additivity of map distances
For gene order A — B — C: distance(A,C) ≈ distance(A,B) + distance(B,C)
The largest pairwise distance spans the two outer genes. Double crossovers make the measured outer distance slightly less than the exact sum, so short intervals give the most accurate maps.
Worked example

A dihybrid whose alleles are in repulsion (trans), genotype A b / a B, is testcrossed to an aabb individual. Among 500 offspring: 220 are Aabb, 210 are aaBb, 35 are AaBb, and 35 are aabb. What is the recombination frequency and the map distance between the two genes?

  1. 1.Identify the parent’s arrangement: A b / a B (repulsion), so its parental gametes are Ab and aB.
  2. 2.Match offspring to gametes (the aabb tester always contributes ab): Aabb came from an Ab gamete and aaBb from an aB gamete — these are the PARENTAL classes (220 + 210 = 430).
  3. 3.The remaining classes AaBb (from an AB gamete) and aabb (from an ab gamete) are the new combinations — the RECOMBINANTS (35 + 35 = 70).
  4. 4.Apply the formula: RF = (70 ÷ 500) × 100% = 14%.
Answer: The recombination frequency is 14%, so the two genes are about 14 map units (14 cM) apart. Note that AB and ab are the recombinants here — the opposite of a coupling (cis) cross — because the parent carried the alleles in repulsion.
Checkpoint

Two genes lie on the same chromosome, 60 map units apart. What recombination frequency will a cross reveal, and why?

Worked example

Three genes — A, B, and C — are mapped by separate testcrosses, giving pairwise recombination frequencies of A–B = 5%, B–C = 13%, and A–C = 18%. Determine the gene order and draw the linear map.

  1. 1.Convert each RF to map units: A–B = 5 cM, B–C = 13 cM, A–C = 18 cM.
  2. 2.Find the largest distance — it spans the two OUTER genes. Here A–C = 18 cM is largest, so A and C are the ends and B lies between them.
  3. 3.Confirm additivity: A–B + B–C = 5 + 13 = 18 cM, which matches the measured A–C distance, so the order A — B — C is consistent.
  4. 4.Place the genes on a line using the two short intervals: A, then 5 cM to B, then 13 cM to C.
Answer: The gene order is A — B — C. Map: A —5 cM— B —13 cM— C, with A and C 18 cM apart. (When the outer distance comes out slightly less than the sum of the inner two, the difference reflects undetected double crossovers.)
Tip

To order three genes fast: the pair with the largest recombination frequency is the outer pair, and the third gene goes in the middle. Then check that the two smaller distances sum to (or just above) the largest — if they do, your order is right.

Checkpoint

Three genes give pairwise recombination frequencies of D–E = 6%, E–F = 9%, and D–F = 15%. Which gene lies in the middle of the linear map?

On the exam

Expect a question where the directly measured distance between the two outer genes is smaller than the sum of the inner intervals. The cause is double crossovers: a second crossover flips the middle marker back but leaves the outer alleles in their parental arrangement, so those offspring are never scored as outer-gene recombinants — long spans underestimate true distance.

Checkpoint

When three genes are mapped, the recombination frequency measured directly between the two outer genes is usually slightly LESS than the sum of the two inner map distances. What best explains this?

Answer the 3 checkpoints as you read.

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