Genetics and Molecular Biology · Unit 2: Mapping and gene function

Linkage and Recombination Mapping

Genes that sit close together on one chromosome break Mendel's law of independent assortment: in a test cross the parental combinations of alleles come back more often than the new ones. This lesson shows how crossing over between chromatids makes the new combinations, how Sturtevant turned their frequency into a map distance, and why that frequency can never pass 50%. Afterwards you can read the phase of a heterozygote from test-cross data, calculate a two-point map distance in centimorgans, and say when a large distance is being underestimated.

Watch first Distance Is How Often Crossovers Separate · 3:07

Test-cross 1,000 fruit flies that carry two mutations, and Mendel’s law of independent assortment predicts that 500 will show one grandparent’s combination of traits or the other’s. You count 830.

Something is holding those alleles together. The independent assortment of Mendel’s peas, and the free multiplying of probability and crosses, work for genes on different chromosomes. For genes on one chromosome they fail, and the way they fail lets you measure where the genes are.

When 1:1:1:1 fails: coupling in sweet peas and flies

A locus is a gene’s place on a chromosome; its versions are alleles. Take two loci, A and B, with recessive alleles a and b. A test cross mates a double heterozygote to an a b / a b tester. The tester’s gametes carry only recessive alleles, so each offspring’s phenotype shows exactly which alleles came in the heterozygote’s gamete. Counting the offspring counts the heterozygote’s gametes.

If the loci assort independently, the heterozygote makes A B, A b, a B and a b gametes in equal numbers, and the test cross gives 1:1:1:1. When it does not, and the combinations the heterozygote inherited from its own parents come back too often, the loci are linked: they lie near each other on one chromosome.

Those inherited combinations are the parental classes. The other two are recombinant. Which is which depends on phase, the way the alleles were arranged in the heterozygote:

  • Coupling (cis), A B / a b: the dominant alleles came in on one chromosome. Parental classes are A B and a b.
  • Repulsion (trans), A b / a B: each chromosome carries one dominant allele. Parental classes are A b and a B.

So read the parents first. The parental classes are usually the larger ones, but you define them by what the heterozygote received, not by counting.

An early clear case came from sweet peas. In 1905 William Bateson, Edith Saunders and Reginald Punnett reported that purple flowers and long pollen, entering a cross together, stayed together in the next generation far more often than independent assortment allows. They called it coupling. The explanation they developed in later papers was a quirk of gamete formation, not genes sharing a chromosome. In 1911 Thomas Hunt Morgan, whose Columbia laboratory was finding X-linked mutants in Drosophila (see sex linkage), proposed the chromosome explanation.

Crossing over makes recombinants

Morgan’s argument rested on meiosis. In prophase I each chromosome has already copied itself into two sister chromatids, and the two partner chromosomes (homologues) pair, so four chromatids lie side by side (see chromosomes and meiosis). Crossing over is the reciprocal exchange of matching segments between non-sister chromatids, one from each homologue. Where it has happened the chromatids stay crossed, and the cross-shaped contact is a chiasma. Morgan drew on F. A. Janssens, who in 1909 had proposed from such figures that partner chromosomes exchange parts.

Picture two ribbons laid together, one red and one white, cut at the same place and swapped. A spot near the cut ends up on a new background; a spot far from it keeps its old one. Two loci are separated only when a crossover falls between them, so the farther apart they lie, the more often they are separated.

Follow one crossover between A and B in an A B / a b cell. Of the four chromatids, the two that did not take part stay A B and a b. The two that did become A b and a B. One crossover gives two parental and two recombinant products. Because the exchange is reciprocal, the two recombinant classes should be about equal in a test cross, as should the two parental classes. That is a check on your data: a large difference within a pair suggests that one genotype survives less well.

Was crossing over a real swap of chromosome parts? In 1931 Harriet Creighton and Barbara McClintock answered with maize. They used a plant whose two copies of chromosome 9 looked different: one had a visible knob at one end and an extra segment from another chromosome at the other. Offspring with new combinations of the genes on that chromosome carried new combinations of knob and segment. The same year Curt Stern reported the same correlation in Drosophila, using X chromosomes that carried their own visible structural changes. How the DNA strands are broken and rejoined is the subject of DNA repair and recombination.

Sturtevant’s map: frequency as distance

Morgan’s idea implied a ruler. Alfred Sturtevant, an undergraduate in Morgan’s laboratory, proposed using the proportion of crossover progeny as an index of distance and built a map of six X-linked factors, published in 1913. Distances measured over short intervals added up along a single line.

recombination frequency (RF) = recombinant progeny / total progeny

map distance (cM) = 100 x RF, for short intervals

One map unit, or centimorgan (cM), named for Morgan, is 1% recombinant progeny. It is a frequency, not a length. Sturtevant warned that his distances need not match physical spacing, and the ratio of DNA to centimorgans varies along chromosomes and between species. It also varies between the sexes. Drosophila males make no crossover gametes at all, and in humans and many other species the two sexes give different maps.

Worked problem. Assume complete penetrance, equal survival of all classes, and accurate scoring. A Drosophila female heterozygous for recessive mutations a and b on one autosome received A B from one parent and a b from the other, so she is in coupling. She is test-crossed to an a b / a b male. (She, not a male, is the heterozygote, because only females cross over.) Of 1,000 progeny:

ClassCount
A B420
a b410
A b85
a B85
  1. Name the classes from the phase. She received A B and a b, so those are parental: 420 + 410 = 830. A b and a B are recombinant: 85 + 85 = 170. Here they are also the smallest classes, as expected for linked loci.
  2. Calculate RF. RF = 170 / 1,000 = 0.17, or 17%. The map distance is 17 cM.
  3. Test for linkage. Independent assortment predicts 250 in each class. Chi-square = (420 - 250)^2/250 + (410 - 250)^2/250 + 2 x (85 - 250)^2/250 = 115.6 + 102.4 + 108.9 + 108.9 = 435.8, with 3 degrees of freedom. The 5% critical value is 7.81. The excess of parentals is far beyond chance.
  4. Predict the other phase. A female carrying the same loci in repulsion, A b / a B, would still give 17% recombinants, but now A B and a b are the recombinant classes: 8.5% each, with 41.5% each of A b and a B.

Step 4 is where students go wrong. “Recombinants are the two smallest classes” is a habit, not a rule. It fails when you must predict the classes before any flies hatch, and it fails when a mutant class survives poorly enough to shrink below a recombinant class. The phase of the heterozygote, which you know from its parents, is what defines the classes.

The 50% ceiling, and what a mapping function corrects

However far apart two loci are, RF does not climb past 50%. The reason is the four chromatids. A crossover between the loci involves two of them, so a meiosis with one crossover in the interval gives half recombinant products. A second crossover between the same loci either recombines them again or puts them back, depending on which chromatids it joins; on average the recombinant share stays at one half. Loci far apart on one chromosome therefore give about 50%, the same as loci on different chromosomes. A two-point cross cannot tell those apart.

Long intervals are underestimated, because double crossovers between distant markers restore the parental combination. Sturtevant saw this in 1913. Adding his short intervals put factors B and M 57.6 units apart, but a direct B by M cross gave 37.6% recombinants. The map distances add; the raw frequencies over long intervals do not.

In 1919 J. B. S. Haldane gave a correction. If crossovers form independently of one another along the chromosome, then

RF = (1 - e^(-2d)) / 2, with d in morgans (1 morgan = 100 cM)

d = -1/2 x ln(1 - 2 RF)

Apply it to the worked problem: d = -1/2 x ln(1 - 0.34) = -1/2 x ln(0.66) = 0.208 morgans, about 20.8 cM, a little more than the 17 cM read directly. The gap grows with distance. Under the same model a 50 cM interval gives only about 31.6% recombinants. Two adjacent intervals of 20 and 40 cM add to 60 cM, yet their own frequencies, 16.5% and 27.5%, would add to 44%, while the whole 60 cM span gives 34.9%.

Recombination sampler: distance versus what you count⚙ original · interactive
Recombinants (A b + a B)
0
Observed RF
0
Expected RF at this distance
0
Distance read as 100 x RF (short-interval rule)
0

Recombination frequency tracks map distance only over short intervals; it climbs toward, and never passes, 50%, so long distances are underestimated. This is a sketch of the logic: the curve assumes crossovers form independently (Haldane), and real crossovers interfere, so it can overcorrect. Assumes equal survival of all classes and accurate scoring; a Drosophila female heterozygote.

Haldane’s model has a known weakness. Real crossovers interfere: one makes another nearby less likely, which Sturtevant suspected and which single-sperm sequencing of 20 human donors detected in every donor in 2020. With interference, fewer double crossovers hide in an interval than Haldane assumes, so his function can overcorrect. Other functions, such as Kosambi’s, build interference in. The practical rule has not changed since Sturtevant: map with short intervals and add them, and use a third marker to catch double crossovers, as in three-point mapping.

Design problem

You isolate a new recessive mutation in Drosophila, dumpy-like (dl), and suspect it lies on chromosome 2 near a known recessive marker, m. Design crosses to test whether dl and m are linked and to estimate the distance between them. Say what result would leave you unable to decide.

One way to do it

Build a heterozygote in known phase. Cross dl/dl flies to m/m flies. Every F1 fly is dl + / + m, in repulsion, and you know the phase because you made it.

Use females. Drosophila males make no crossover gametes, so an F1 male would give only parental classes whatever the distance. Test-cross F1 females to dl m / dl m males. If no double-mutant stock exists yet, you first recover a dl m recombinant chromosome from the progeny of F1 females.

Score enough flies. Several hundred progeny at least, more if the loci may be close, since a short interval yields few recombinants. Score all four classes.

Test linkage without being fooled by survival. Check each locus alone for 1:1 (dl versus +, m versus +). A deficit of dumpy-like flies, for instance, would distort the four classes even if the loci were unlinked. Then test the association between the loci in the 2 x 2 table of counts, a chi-square with one degree of freedom, which is not inflated by unequal survival at one locus.

Calculate. Here the recombinants are wild type and dl m. RF = (wild type + dl m) / total, and the distance in cM is 100 x RF, corrected with a mapping function if RF is large.

Know the limit. An RF near 50% cannot tell a distant locus on chromosome 2 from one on another chromosome. To place dl, add a third marker on chromosome 2 and do a three-point cross, which also orders the three genes.

How we measure it

Two-point test cross

An organism heterozygous at two loci is crossed to one homozygous recessive at both. Each offspring's phenotype then reads directly which combination of alleles came in the heterozygote's gamete, so counting the classes counts parental and recombinant gametes.

Cytological chromosome markers

Partner chromosomes are made to look different, with a knob, a translocated piece or a broken arm, and offspring are classified both genetically and under the microscope. If gene recombination always comes with a visible exchange of landmarks, crossing over is physical.

Chi-square test against 1:1:1:1

Observed test-cross counts are compared with the counts expected if the loci assort independently. A large chi-square says the departure is unlikely to be chance; a separate test of each locus for 1:1 checks that unequal survival is not the cause.

Finish line

Test yourself

8 questions

Practice and MCAT-style, with explained answers.

Next · Lesson 9

Three-Point Crosses: Gene Order and Double Crossovers

Two crossovers between the same pair of genes cancel out. How do you count what you cannot see?

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