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Lesson 6 of 12

Principles of Inheritance and Variation · Lesson 6 of 12

Linkage, Recombination and Genetic Mapping

“How linked loci produce excess parental combinations and how recombination frequencies estimate genetic distances.”

Learning Objectives

• Distinguish linked and independently assorting genes. • Identify parental and recombinant gametes. • Calculate a recombination percentage. • Explain the limits of interpreting genetic map distance.

Genes on the same chromosome are physically associated

Genes located on the same chromosome are linked because that chromosome tends to travel as a unit into a gamete. If two loci are close together, a crossover between them is relatively unlikely in a single meiosis, so allele combinations inherited from each parent commonly remain together. For example, a heterozygote with chromosome arrangements AB/ab has A with B on one homologue and a with b on the other; this arrangement is called coupling or cis phase. Without an intervening crossover, its gametes carry AB or ab. This contrasts with independently assorting loci, for which four two-locus combinations can be equally frequent.

Crossing over can break linkage

During prophase I, homologous chromosomes pair and nonsister chromatids may exchange DNA segments. A crossover between the A and B loci generates recombinant chromatids Ab and aB from an AB/ab arrangement. Such recombinant allele combinations differ from the two parental combinations. Crossing over is a chromosomal event, while recombination is the genetic outcome being measured. Not every crossover produces an observable recombinant at the loci being followed: a crossover outside the interval leaves their association unchanged, and multiple crossovers within the interval can restore parental associations in some chromatids.

Morgan's fruit fly experiments

Morgan studied fruit fly traits including body colour, eye colour and wing form. Certain sex-linked gene pairs did not produce the combinations predicted by independent assortment; instead parental combinations were unusually common. The physical association of loci on the X chromosome explained the pattern. Some pairs were tightly linked, while others showed greater recombination. In the chapter's examples, the white-eye and yellow-body loci show about 1.3% recombination, whereas white-eye and miniature-wing loci show about 37.2%. The larger value suggests greater genetic separation, not that chromosomes physically stretch to match a percentage.

Linked genes: parental and recombinant combinationsParental ABParental abRecombinant AbRecombinant aBCrossing over between homologues can create new allele arrangements.More recombinants generally suggest greater separation between loci.
Linked-gene allele arrangements— Original schematic; proportions assume the specified genetic model.

Calculating recombination frequency

For an informative test cross, recombination frequency is calculated as the number of offspring with recombinant phenotypes divided by the total scored offspring, multiplied by 100. If 20 offspring have Ab and 30 have aB out of 500 total, the recombinant count is 20 + 30 = 50 and the frequency is 50/500 × 100 = 10%. The estimate is about 10 centimorgans (cM) for a short interval, because one centimorgan is defined by one percentage point of observed recombination under classical mapping. We must correctly identify parental versus recombinant classes before calculating: rarity alone is suggestive but the parental haplotypes establish which are recombinant.

Why map distance is an estimate

Recombination frequency generally rises with the genetic distance between loci over short intervals. However, it does not exceed 50% as an observed two-locus recombination fraction. Loci far apart on the same chromosome can approach 50% because double or multiple crossovers obscure some exchanges. A recombination fraction of 50% is consistent with independent assortment and does not prove that loci lie on different chromosomes. Genetic map distance is an inferred measure of crossover behaviour, not a direct physical distance in base pairs. Mapping functions and additional markers help analyse long intervals.

Linkage in inheritance predictions

An AB/ab heterozygote with recombination fraction r produces parental gametes AB and ab, each with probability (1-r)/2, and recombinant gametes Ab and aB, each with probability r/2 in a simple model. With r = 0.10, parental classes each have 45% frequency and recombinant classes each 5%. Crossing this individual with aabb allows each gamete class to be observed in the offspring. If r = 0.50, each class has 25% frequency, the same proportions as independently assorting genes.

Linkage maps as a practical application

Alfred Sturtevant used recombination frequencies to arrange genes in an inferred order along chromosomes. If A–B is 3 cM and B–C is 7 cM, the order A–B–C is one plausible arrangement with A–C map distance about 10 cM across these short intervals. A separate three-point cross is needed to establish order and detect double crossovers robustly. Genetic maps help organise inherited markers and can guide physical mapping and genome research, but the map is a statistical model of recombination, not a microscopic photograph of gene positions.

Recombination percentageLaTeX

Quick Quiz

Quiz

Quick check

Which gametes are parental in AB/ab coupling phase?

Quick check

What event can produce Ab and aB from AB/ab?

Quick check

What is the recombination frequency if 40 of 400 offspring are recombinant?

Quick check

What is the maximum observable two-locus recombination fraction?

Quick check

What does a high recombination fraction generally suggest?

Quick check

What proportion of Ab gametes is expected for AB/ab with r=0.20?

Practice Problems

Practice Problems
  1. A test cross gives 450 AB, 430 ab, 60 Ab and 60 aB offspring. Calculate recombination frequency.
  2. Predict gamete probabilities from AB/ab when the recombination fraction is 0.08.
  3. Explain why 50% recombination does not prove two genes are on separate chromosomes.
  4. Describe Morgan's and Sturtevant's contributions.

Total offspring = 450 + 430 + 60 + 60 = 1000. Recombinants are Ab and aB, total 120. Frequency = 120/1000 × 100 = 12%. For a reasonably short interval, this corresponds to approximately 12 cM genetic distance.

Key Takeaways

Key Takeaways

• Linked loci lie on the same chromosome and often pass together into gametes. • An intervening crossover can produce new allele arrangements. • Morgan used fruit fly crosses to demonstrate linkage and recombination. • Recombination frequency equals recombinant offspring divided by total offspring. • For short intervals, one percentage point of recombination corresponds roughly to 1 cM. • Observed recombination fractions do not exceed 50%. • A 50% fraction does not by itself prove that genes are on different chromosomes.