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Biology · Meiosis & Patterns of Inheritance

Linkage, recombination frequencies and genetic maps

AaBb leaves one detail unstated: are A and B on the same homologue, or is A travelling with b? When the loci are linked, that arrangement changes which gametes are parental combinations. Put the alleles on chromosome strips before counting recombinants.

Begin with the cell-division lesson and Heredity, Variation & Evolution: recognise homologous chromosomes, sister chromatids, a locus and an allele. DNA, Replication & Repair explains how copying makes two sister DNA molecules. Gene Expression helps with the later distinction between transmitting a sequence and expressing a trait. Here, fractions and conditional probabilities are explained alongside the cases. All chromosome-count examples use a specified diploid model; the final lessons introduce extensions rather than requiring advanced genetics beforehand.

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By the end, you can

Identify parental phase, calculate a two-point recombinant fraction and distinguish an approximate genetic map from physical distance.

01

Specify the arrangement on the homologues

Linked loci can be transmitted together more often than expected under independent assortment. In AB/ab, A and B share one homologue and a and b the other; the slash separates homologues. In Ab/aB, the same diploid alleles have a different phase. Both can be abbreviated AaBb, so that abbreviation alone does not give phase.

Parental combinations are defined relative to the heterozygous parent's chromosome arrangement. For AB/ab they are AB and ab, with Ab and aB recombinant. For Ab/aB, the labels reverse. 'Recombinant' does not mean rare by definition, newly mutated, or biologically better.

In a test cross, pair AB/ab with ab/ab. The tester contributes ab at the two loci, so a reliably identified offspring genotype reveals the heterozygous parent's gamete: AaBb records AB, aabb records ab, Aabb records Ab and aaBb records aB. If phenotypes rather than genotypes are used to identify the classes, their expression and scoring must distinguish these possibilities reliably.

Parental phase determines which combinations are recombinant

Parental phase determines which combinations are recombinantSide-by-side chromosome-strip pairs contrast AB/ab and Ab/aB. A test-cross arrow from the assigned AB/ab pair joins an ab/ab tester, followed by a four-row table and horizontal bars starting at zero. Bar lengths correspond to the constructed counts; parental and recombinant classes have different patterns.AB/ab phaseABabAb/aB phaseAbaBAB/ab × ab/abTester contributes ab onlyAB → AaBbParentalcombination84ab → aabbParentalcombination84Ab → AabbRecombinantcombination16aB → aaBbRecombinantcombination16020406080100Constructed offspring countSlash separates homologuesTester contributes ab onlyr = (16 + 16) / 200 = 0.16First short-interval approximation: about 16 cMNo base-pair distance assigned
Constructed count table for a two-point test cross. Gamete allele combinations can be identified reliably in offspring; tester contributes ab only; no segregation distortion, classification error or differential survival. The 16 cM conversion is a short-interval approximation, not a physical measurement or exact long-interval map. Strip coordinates are illustrative and the bar axis counts offspring, not cM.
Read the diagram in words

AB and ab are parental combinations for AB/ab, while Ab and aB are recombinant. The test-cross counts are 84, 84, 16 and 16, totalling 200. Recombinants total 32, giving 16%. For Ab/aB phase, the class labels would reverse.

AB/ab phase

AB/ab phase
Displayed homologueParental combination
1AB
2ab
Parental combination
AB, ab
Recombinant combination
Ab, aB

Ab/aB phase

Ab/aB phase
Displayed homologueParental combination
1Ab
2aB
Parental combination
Ab, aB
Recombinant combination
AB, ab

AB/ab × ab/ab

Constructed offspring count
Gamete contributionOffspring genotypeConstructed offspring countCombination class
ABAaBb84Parental combination
abaabb84Parental combination
AbAabb16Recombinant combination
aBaaBb16Recombinant combination

Constructed offspring count: 0–100

84 + 84 + 16 + 16 = 200; 16 + 16 = 32

  • Slash separates homologues
  • Tester contributes ab only
  • r = (16 + 16) / 200 = 0.16
  • First short-interval approximation: about 16 cM
  • No base-pair distance assigned
02

Count recombinant products, not crossover events

Our constructed table has 84 AaBb, 84 aabb, 16 Aabb and 16 aaBb offspring. The recombinant classes total 16 + 16 = 32 among 200, so the recombinant fraction is r = 32/200 = 0.16, or 16%. The table assumes that gamete types are scored correctly and that no class is preferentially lost.

r describes how often the two marker alleles have a recombinant combination in the sampled products. It is not the fraction of all chromosomes that moved, the percentage of DNA exchanged, or a direct count of meiotic cells with crossovers. One crossover can involve only two of a bivalent's four chromatids.

For a short interval, 100r gives a useful first approximation to map distance in centimorgans: about 16 cM here. Additional exchanges can make the actual genetic distance greater than this two-point approximation. A finite real sample would also have uncertainty; our assigned counts do not provide a significance test or a measured confidence interval.

03

A genetic map measures recombination, not a ruler length

A centimorgan is a genetic-map unit. It does not have one fixed conversion to base pairs across every organism and chromosomal region, because recombination is not uniform along DNA. Equal cM intervals can contain different physical lengths.

Two exchanges can hide each other at the outer markers. In our selected three-marker ancestry model, ABC becomes AbC after exchanges on both sides of B with a chromatid from abc. The outer A-C combination is again parental, although the middle marker records changed ancestry. Two-point endpoint scoring therefore misses that exchange history.

In the usual two-locus model without transmission distortion, the underlying recombinant fraction ranges up to 1/2. A value near 1/2 is compatible with loci on separate chromosomes or sufficiently separated loci on the same chromosome. It does not prove a particular physical location. Observed finite-sample fractions can exceed 1/2 through sampling and other departures from the model.

A long chromosome map can exceed 50 cM when shorter mapped intervals are combined. Three-point crosses can help identify order and some otherwise hidden exchanges. Sturtevant's 1913 fruit-fly work used inheritance associations to infer a linear arrangement and already distinguished that map from uniform physical spacing.

Two exchanges can restore the outer-marker combination

Two exchanges can restore the outer-marker combinationThree successive snapshots trace one chromatid from ABC, through Abc after one exchange, to AbC after a second exchange. Origin patterns alternate in the final three segments. A bracket compares outer markers A and C, while the B marker remains clearly changed. The reciprocal final product is recorded separately.A/aB/bC/cBefore either exchangeABCAfter exchange between A and BAbcAfter second exchange between B and CAbC01530507085100Relative drawing positionA-C combination looks parentalMiddle B/b marker records changed ancestryOne selected history; no event frequency assignedaBc
An ancestry-segment model for two exchanges involving the same nonsister chromatid pair, one between A and B and one between B and C. Only one resulting chromatid is displayed; the reciprocal product is aBc. Endpoint A-C ancestry returns to its parental combination despite a changed middle marker. These relative coordinates are neither physical base-pair positions nor genetic-map distances.
Read the diagram in words

Two exchanges between the same nonsister chromatid pair turn the displayed ABC chromatid into AbC and its partner into aBc. The outer A-C marker combination matches the original, while the middle allele differs. An endpoint-only test misses this particular double-exchange history.

Relative drawing position
LocusRelative position
A/a15
B/b50
C/c85
Relative position range
0–100
Exchange positions
30, 70

Before either exchange

ABC

Relative drawing position
Relative position rangeSegment ancestry
0–100ABC

After exchange between A and B

Abc

Relative drawing position
Relative position rangeSegment ancestry
0–30ABC
30–100abc

After second exchange between B and C

AbC

Relative drawing position
Relative position rangeSegment ancestry
0–30ABC
30–70abc
70–100ABC
Reciprocal chromatid product
aBc
  • A-C combination looks parental
  • Middle B/b marker records changed ancestry
  • One selected history; no event frequency assigned

Worked example

Read the denominator and preserve the approximation

Add all four counts: 84 + 84 + 16 + 16 = 200. Use both recombinant classes, not just one: r = (16 + 16)/200 = 0.16. The complementary parental fraction is 168/200 = 0.84. Within a symmetric prediction model fitted to these proportions, each parental class would have probability 0.42 and each recombinant class 0.08; the counts themselves remain constructed.

If all other test-cross conditions hold, the displayed first map approximation is 16 cM between A and B. Reversing the strip gives B–A with the same separation; two loci alone do not specify which end points toward the centromere. Neither the strip's drawing width nor 16 cM supplies a number of DNA base pairs.

Check your understanding

  1. For the same AaBb genotype but phase Ab/aB, which gametes count as recombinant between the two loci?

    Show reasoning

    AB and ab. The parental combinations are Ab and aB in that phase. Phase determines the classification, rather than uppercase letters alone.

  2. Why is the table's recombinant fraction 16%, rather than 8% or 32%?

    Show reasoning

    Both recombinant classes contribute: 16 + 16 = 32 products, and the denominator is all 200 products. Thus 32/200 = 16%. The numerator alone and one class alone answer different questions.

  3. A real test cross gives approximately 50% recombinants. Must the loci be on different chromosomes?

    Show reasoning

    No. Loci sufficiently separated on the same chromosome can behave as unlinked in this two-point measure. Location needs further mapping or physical evidence, and observed fractions also have sampling uncertainty.

What to carry forward

Preserve parental phase, the counted products and the distinction between genetic and physical distance.

References

Resources supporting this lesson’s explanations and examples.

  1. Introduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)Natasha Ramroop Singh · Thompson Rivers UniversityOpen university textbookIntroduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)DNA amount and chromosome numberAssumptions of a dihybrid ratioConditional pedigree probabilitiesAllele phaseLinkage and recombinant fractionsGenetic maps and their limitsTwo-point and three-point crossesTypes of dominanceBiochemical basis of dominanceGene interactions and phenotype groupingPseudoautosomal regionsopengenetics.pressbooks.tru.ca
  2. The linear arrangement of six sex-linked factors in Drosophila, as shown by their mode of association (Journal of Experimental Zoology 14:43–59, 1913; Electronic Scholarly Publishing edition, 1998)Alfred H. SturtevantOriginal research — electronic reprint · PDFgarcialab.berkeley.eduRead source
  3. Talking Glossary — Chromosome sets, maps and polygenic traits (Live glossary)National Human Genome Research InstituteScientific glossaryTalking Glossary — Chromosome sets, maps and polygenic traits (Live glossary)Genetic mapPolygenic traitgenome.gov