012 X 1800 432

The interference between these genes is 0.4.

A fly with ebony body, rough eyes, and brevis bristles is crossed with a fly that is homozygous for the wild-type traits. The resulting F1 females are test-crossed with males that have ebony body, rough eyes, and brevis bristles; 1800 progeny are produced. Give the phenotypes and expected numbers of phenotypes in the progeny of the testcross.

The crosses are:

However, some interference occurs; so the observed number of double crossovers will be less than the expected. The interference is 1 — coefficient of coincidence; so the coefficient of coincidence is:

coefficient of coincidence = 1 — interference

The interference is given as 0.4; so the coefficient of coincidence equals 1 — 0.4 = 0.6. Recall that the coefficient of coincidence is:

coefficient of coincidence =

number of observed double crossovers number of expected double crossovers

Rearranging this equation, we obtain:

number of observed double crossovers =

coefficient of coincidence X number of expected double crossovers number of observed double crossovers = 0.6 X 43.2 = 26

A total of 26 double crossovers should be observed. Because there are two classes of double crossovers ( e+ / ro / bv+ and e / ro+ / bv ), we should observe 13 of each.

Next, we determine the number of single-crossover progeny. The genetic map indicates that there are 20 m.u. between e and ro; so 360 progeny (20% of 1800) are expected to have resulted from recombination between these two loci. Some of them will be single-crossover progeny and some will be double-crossover progeny. We have already determined that the number of double-crossover progeny is 26; so the number of progeny resulting from a single crossover between e and ro is 360 — 26 = 334, which will e e e ro e ro

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