Recover interval recombination without losing double crossovers

1. Frame the inheritance question

Genes on one chromosome can travel together more often than independently assorting markers. Homologous recombination exchanges corresponding segments between nonsister chromatids during meiosis. A testcross can reveal the gamete combinations made by a heterozygote when the partner’s recessive alleles make marker states distinguishable and scoring and viability are comparable.

2. Explain mechanisms and assumptions

The original packet fixes gene order A–B–C and parental phase ABC/abc. Single exchange classes in the first interval are Abc and aBC; second-interval classes are ABc and abC; double classes are AbC and aBc. Listing the actual marker strings makes the classification checkable rather than relying on a memorized class order.

To estimate recombination in A–B, count its single classes plus both double classes; do the same for B–C. Double exchanges restore the parental association of the outer A and C markers, so the observed outer recombinant fraction excludes them. Consequently, summing adjacent interval map estimates can exceed the observed outer recombination percentage.

For short intervals, recombinant percentage approximates map distance in centimorgans. Multiple unobserved exchanges and crossover interference complicate longer intervals. Under ordinary segregation assumptions the population recombinant fraction approaches a ceiling of one-half, whereas map distances can exceed 50 cM; finite samples can fluctuate beyond one-half. The product of interval fractions predicts double frequency only under an additional independence approximation.

3. Work a quantitative example

Among 1000 offspring, let A–B single classes total 100, B–C singles 160 and doubles 20. The remaining 720 are parental. Interval fractions are 0.12 and 0.18; summed map estimate is 30 cM. Outer recombinant fraction is (100+160)/1000=0.26 because the double classes restore the outer association.

4. Predict and investigate

Predict first: Why must a double class enter both adjacent intervals but not the outer recombinant count?

Compare double-class totals zero and 20 while keeping both single-class totals fixed. Write all eight marker classes, check the total and recompute interval and outer fractions. Explain why this is a count-comparison model rather than a mechanistic claim that changing only double exchanges leaves every real process unchanged.

Retain the first prediction, settings, raw values and interpretation in the lesson notebook. All records are original supplied or calculated examples; no breeding, biological intervention or clinical inference is requested.

5. Apply the reasoning to a changed case

Apply the mechanism to a changed case

A 1000-offspring packet with the same order has first-interval singles 80, second-interval singles 140 and doubles 20. Find both interval fractions, summed short-interval map estimate and observed outer recombinant fraction.

Retain your calculation, units and assumptions in the lesson notebook before opening this public practice response.

Compare a worked response

The fractions are 0.10 and 0.16, giving a summed estimate of 26 cM. The outer fraction is 0.22. The four-percentage-point difference equals twice the 0.02 double fraction.

6. Reflect and retain evidence

Name the conditioning event, count boundary and biological replication unit. Which independent observation could distinguish the inheritance mechanism from scoring, viability or environmental alternatives? A public worked response or saved prose does not establish independent assessment or supervised practical participation.

Sources and course continuation

Scientific supporting reference, checked 30 October 2026. Original prose and numerical cases; no source exercise or figure reproduced. Continue through General Biology I for mixed written evidence and the retained multistage project.