A, B and C are three linked genes, A and B are separated by 40cM and B and C are

Question

A, B and C are three linked genes, A and B are separated by 40cM and B and C are separated by 20 cM. A women with the genotype AbClaBc marries a man with the genotype abc/abc.What is the probability of producing a child with the aBC phenotype assuming the interference is zero? a. 1.5% b. 3% d. 6% d. 8% e. 12% The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome Il in Drosophila. Assume that a mahogany-eyed female was mated to an ebony bodied male and the resulting F1 phenotypically wild type females were mated to mahogany ebony males. Of 1000 offspring, what would be the expected number of flies that are both mahogany and ebony? a. 125 b. 250 c. 375 d. 750 e. none

Explanation / Answer

Answer:

1). d. 6%

Explanation:

Expected double cross over frequency = single cross over frquency between a & b * single cross over frquency between a & c

Distance between the genes = (%) recombination frequency

= 40% * 20%

= 0.4 * 0.2 = 0.08 = 8%

The geneotype, aBC will be obtained from AbC/aBc by single cross over between b & c. we need to substract the double crossover frequency from single crossover frequency between b&c

= 20 - 8 = 12%

When single crossover is happened between b & c, two types of genotypes are produced Abc & aBC in 12% (each with 6%)

So the frequency of aBC is 6%.

2). a. 125

Explanation:

The data shows the trans configuration.

mahogany = m; normal eyes = M

Ebony body = e ; normal body = E

m E / m E(mahogany) x M e / M e (ebony body)----Parents

mE Me ----------------Gametes

mE / Me ----------------------------Wild (F1)

mE / M e (F1) x me/me (tester)------Test cross

distance = % recombinants

Recombinants are ME & me in 25% (each with 12.5%)

Non-recombinants are mE & Me in 75% (each with 37.5%)

So the phenotype mahogany eyes and ebony body (eb/eb) is 12.5%.

12.5% * 1000 = 125

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