The drawing shows the locations and map distances among three linked Drosophila

Question

The drawing shows the locations and map distances among three linked Drosophila genes on the same chromosome. Your friend Jane is a Drosophila geneticist who carried out a trihybrid test cross using P parents with the genotypes ABC/ABC and abc/abc. Jane is specifically looking for flies with the genotypes AbC/abc (because they have some very interesting phenotype). She needs at least 30 of those flies with that special phenotype for another experiment. Help Jane out: tell her how many F2 offspring she should raise from that test cross in order to have a good chance of collecting at least 30 flies with that interesting genotype? Show your work: why that many flies?

Explanation / Answer

Answer:

AbC/aBc is the double crossover progeny.

Expected double cross over progeny = (recombinaiton frequency between A & B ) * (recombinaiton frequency between B & C )

Recombinaiton frequency (%)= Map distance

Expected double cross over = .0.12 * 0.25 = 0.03 = 3%

For every 100 offsprings, 3 offsprings would be double cross overs (AbC/aBc)

We need to have 30 double cross overs (AbC/aBc) means total progeny should be 1000

100------3

1000---30

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