A recombination fraction is a proportion of observed recombinant products. A genetic map distance describes crossover separation on a map. They are related, but treating them as interchangeable can hide double crossovers and lead to an unjustified conclusion that two genes must lie on different chromosomes.
The AAMC content outline includes linkage, testcrosses, gene mapping and double crossovers. These original examples build from observable offspring to the limits of the inference.
Start with the parental arrangement
Write AB/ab to mean that one homolog carries A with B and the other carries a with b. This is different from Ab/aB, even though both individuals are heterozygous at both loci.
In a testcross with an ab/ab partner, the partner contributes ab to every offspring. If offspring genotypes can be identified accurately, their other allele combination reveals the gamete contributed by the heterozygous parent. This is the testcross logic described in OpenStax’s linkage chapter.
Assume ordinary diploid inheritance, equal viability of the offspring classes, accurate scoring and no segregation distortion. If phenotypes rather than genotypes are scored, additional dominance and penetrance assumptions are necessary. A class may look rare because it survives poorly or is misclassified, not only because its gamete is rarely produced.
Worked example: count recombinants relative to phase
Cross an AB/ab individual with an ab/ab tester. The following table records the heterozygous parent’s inferred gamete for each of 1000 offspring:
| Inferred gamete | Count |
|---|---|
| AB | 420 |
| ab | 420 |
| Ab | 80 |
| aB | 80 |
Because parental phase is specified as AB/ab, Ab and aB are recombinant combinations. The recombination fraction is:
r = (80 + 80)/1000 = 0.16, or 16%.
Under the stated assumptions, the large departure from the 50% expectation for independently assorting loci is evidence of linkage. An introductory map estimate would be approximately 16 centimorgans, with the qualification that using recombination percentage directly approximates map distance when undetected multiple crossovers are uncommon.
The answer 8% counts only one recombinant class. The answer 84% counts parental classes. The answer “16% of meioses had exactly one crossover” changes the quantity being measured. We observed recombinant gametes through offspring, not complete crossover histories of all chromatids in each meiosis.
A centimorgan is not a fixed number of DNA base pairs. Recombination rates vary across a genome and can differ by biological context, so equal map intervals need not have equal physical lengths.
Why a crossover can be hidden from two markers
Follow a transmitted chromatid between two marker loci. An odd number of switches between homologous origins changes the outer-marker combination; an even number can restore the parental outer-marker combination. The offspring then looks nonrecombinant at those two markers despite intervening exchange.
That is why recombinant fraction can underestimate crossover-based map distance. The primary methods discussion in Predicting recombination frequency from map distance distinguishes observable recombination from underlying crossover number. Its indexed methods text supports this distinction; the examples here are independently constructed.
Do not automatically interpret every parental-looking gamete as a crossover-free gamete. The markers sample a history at particular positions; they do not record every event between them.
A third marker makes some hidden events visible
Now testcross an ABC/abc parent. Assume the listed classes fit a simplified three-locus mapping model: the smallest reciprocal pair represents double-crossover products, and additional undetected multiple exchanges within each interval are negligible.
| Inferred gamete | Count | Interpretation under this model |
|---|---|---|
| ABC | 350 | Parental |
| abc | 350 | Parental |
| Abc | 60 | Single-crossover product in A–B |
| aBC | 60 | Single-crossover product in A–B |
| ABc | 80 | Single-crossover product in B–C |
| abC | 80 | Single-crossover product in B–C |
| AbC | 10 | Double-crossover product |
| aBc | 10 | Double-crossover product |
The total is 1000. Compare ABC with AbC, and abc with aBc: only the B state changes relative to its matching parental arrangement. Under the model, B is the middle marker. A–B–C and its reverse C–B–A describe the same order.
For A–B, count Abc, aBC, AbC and aBc:
rAB = (60 + 60 + 10 + 10)/1000 = 14%.
For B–C, count ABc, abC, AbC and aBc:
rBC = (80 + 80 + 10 + 10)/1000 = 18%.
The approximate interval map therefore spans 14 + 18 = 32 cM from A to C. Each double-crossover product contributes to both adjacent interval estimates.
But a two-marker A–C count detects only the 120 + 160 single-crossover products. The double-crossover classes retain AC or ac. Thus rAC = 28%, not 32%.
The gap is twice the double-crossover fraction: 28 + 2(2) = 32 percentage points in this simplified accounting. The arithmetic demonstrates why you add adjacent map intervals rather than automatically equating their sum with the outer-marker recombinant fraction. It does not establish a universally exact map distance when further events go undetected.
Transfer exercise: what does 50% establish?
A new AB/ab testcross produces 260 AB, 240 ab, 250 Ab and 250 aB offspring. Choose the strongest supported conclusion under the same scoring and viability assumptions.
A. A and B must be on different chromosomes.
B. The observed recombinant fraction is 50%, so this cross alone does not demonstrate linkage.
C. Every meiosis had a crossover between A and B.
D. A and B are exactly 50 cM apart.
Answer: B. Recombinant offspring total 500 of 1000. Independently assorting loci can produce this result, but sufficiently separated loci on the same chromosome can also behave as effectively unlinked in a two-point comparison.
A turns a compatible explanation into a unique one. C infers unobserved meiotic histories. D treats an observed fraction as an exact distance despite the possibility of hidden exchanges.
In the usual model without chromatid interference, expected recombination fractions do not exceed one-half. A finite sample can nevertheless produce an observed fraction above 50% through sampling variation. Viability differences, mistaken phase or scoring errors can also distort a count; “above 50% is impossible data” is therefore too strong.
Additional markers between the loci, a larger appropriately designed cross or physical chromosome information could distinguish explanations that this two-marker result cannot separate.
Make the next mapping answer explicit
On a new genetics passage, record parental phase, identify what each offspring class reveals, and state the denominator. Then separate three statements: the observed recombinant fraction, the map estimate under specified assumptions, and what remains unknown about the chromosome.
That final separation prevents a numerical answer from claiming more than the experiment actually observed.