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 AbC/aBc marries a man with the genotype abclabc.What is the probability of producing a child with the aBO phenotype assuming the interference is zero? a. 1.5% b. 3% d. 6% d. 8% e. 12% 3 enes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome IlI in Drosophila. Assume that a mahogany-eyed female was mat 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 ed to an ebony-

Explanation / Answer

1) Distance between A and B is 40 cM

Distance between B and C is 20 cM

A (20 cM) C (20 cM) B

If genes are present close to each other, they are linked and hence, the chance for recombination will be less.

If genes are far apart, then recombination will be higher but less than 50 %

If genes are present in different chromosomes, recombination will be equal to 50 % of the total progeny.

Genes on different chromosomes = 50% recombinant gametes after meiosis

Genes on the same chromosome 50% recombinant gametes after meiosis.

1 Map Unit (cM) = 1 % recombination

40 map unit (cM) = 40 % recombination

Therefore, 60 % of the offspring are with parental genotypes and 40 % are recombinant.

The chance of a parental chromosome is 1/2 (100 – 40) = 30 %

Phenotype aBC is a result of recombination between B and C. And they are 20 cM apart

And the chance of a recombinant chromosome is 1/2 (20) = 10 %

The chance of both events = 30 % x 10 % = 3 %

2)

1 Map Unit (cM) = 1 % recombination

25 map unit (cM) = 25 % recombination

75 % of the offsrping are with parental chromosomes

25 % of the offspring are with recombinant chromosomes

Ebony = e, Mahogany = m

Wild type = +

Cross 1 : Wildtype body and mahogany eyed female (+ m) x Ebony body and wild type eyes male ( e +)

F1 : + m and e + : Wild type body and eyes

Cross 2 : + m . e + x m e . m e

F2 : + m e +, m e m e, m e e + , + m m e

Parental chromsomes are + m m e and m e e +

Recombinant chromosomes are m e m e and + m e +

25 % are recombinant chromsomes

750 flies with parental chromosome

250  flies with recombinant chromosome ( m e m e and + m e + )

m e m e = 250 / 2 = 125

Therefore, number of flies with m e m e chromosome is 125

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