Question at position 10 Suppose the map for a particular human chromosome interval is: a——1 m.u.——b——1 m.u.——c——1 m.u.——d——1 m.u.——e——1 m.u.——f In a man heterozygous for all six genes, what fraction of his sperm would be recombinant in the a–f interval? cannot be determined0%1%5%2.5%

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BlackTom AI · Accepted Answer

Question restatement: In a man heterozygous for six genes arranged in the order a—1 m.u.—b—1 m.u.—c—1 m.u.—d—1 m.u.—e—1 m.u.—f, what fraction of his sperm would be recombinant in the a–f interval?

Option-by-option analysis:

Option 1: cannot be determined
- This is not correct because, for a defined linear interval with known genetic distances, classical genetics provides a method to estimate recombination between endpoints. Although interference and multiple crossovers can affect local patterns, the overall recombination fraction between the outermost markers (a and f) is determined by the total distance between them (5 map units in this setup) and the standard interpretation of map units. Therefore, the fraction can be estimated rather than left undetermined.

Option 2: 0%
- 0% would imply that no gametes show recombination between a and f, which contradicts the existence of any measurable distance between the markers. With gaps totaling 5 cM, there is a nonzero probability of at least one crossover between a and f in meiosis, so 0% is incorrect.

Option 3: 1%
- A 1% recombination rate would correspond to a very small distance (roughly 1 cM). Here the span from a to f covers 5 map units, not 1. Therefore, assigning a 1% recombination fraction underestimates the distance and is not consistent with the provided map.

Option 4: 5%
- This option aligns with the standard interpretation that the recombination fraction between two loci separated by 5 map units (i.e., 5 cM) is approximately 5%. For small distances and typical recombination behavior, the fraction of gametes showing recombinant chromatids between the ends reflects the total map distance. While some complexities can arise with multiple crossovers, the expected recombination fraction between a and f in a 5 cM interval is about 0.05 (5%).

Option 5: 2.5%
- A 2.5% recombination fraction would correspond to half of the interval distance or to scenarios with certain symmetry or averaging effects, but the straightforward interpretation of a 5 cM interval yields about 5% recombination. Therefore 2.5% is not the correct expectation for the end-to-end interval here.

Additional notes on reasoning pattern:
- In genetic mapping, a map unit (centimorgan) corresponds to a 1% recombination frequency between the two loci. For an interval spanning 5 cM, the expected recombination fraction between the outer markers is about 5% under standard assumptions, since the probability of an odd number of crossovers across the entire interval translates roughly to the distance in centimorgans when distances are small and interference is not overwhelming. Double crossovers can occur, but they tend to preserve the parental configuration for some chromatids and still lead to an overall 5% approximation for the endpoints in many teaching contexts.

Summary alignment:
- The correct choice is 5% because it matches the 5 cM span between a and f. Other options assert implausible or inconsistent fractions given a 5 cM distance, or claim information is undetermined, which contradicts the standard mapping interpretation.

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