Question

A cross in Drosophila involved the recessive, X-linked genes yollow (y), white (w), and cut (ch). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut • The F, females were wild type for all three traits. The F, males expressed the yellow-body and white-eye traits. The cross was carried to an F2 progeny, and only male offspring were talied. On the basis of the data shown here, a genetic map was constructed Phenotype Male Offspring + + + + + + + + + + + + + * 2

im not sure how to calcualte the double crossover offspring

Answer 1

Answer:

• Parent’s genotype:
• Female: y w ct / y w +
• Male    : + + ct / - - -
• Gametes : y w + × + + ct, and y w ct × + + ct
• Genotype of F1 offspring:
• Female: y w + × + + ct, and y w ct × + + ct
• Male   : y w + / - - -, and y w ct / - - -
• Gametes of F1 generation:

$\rightarrow$Male : y w ct

Female:

• Non-recombinant: y w ct and + + ct
• SCO with respect to w and ct: y w ct, and + + +
• SCO with respect to y and w: y + ct and + w +
• Double recombinants: + w ct and y + +

$\rightarrow$F2 progeny:

	Genotype	No. of offspring	Frequency of recombination (% recombination)
Parental	+ + ct / y w ct y w + / y w ct	424 376	0
Single crossover between w and ct	y w ct / y w ct + + + / y w ct	90 95
Single crossover between y and w	y + ct / y w ct + w + / y w ct	9 6
Double crossover	+ w ct / y w ct y + ct / y w ct	0 0	0

The distance map would be as follows:

y__(1.5)______w____________(18.5)________________Ct

• The number of expected double crossovers =

  18.5 2x - 1.5 xtotal number of offspring 100 100 18.5 1.5 x1000 100 100 = 2.775 or 3

$\rightarrow$Therefore, the expected number of double crossovers is 3.

Please Rate My Answer..... Thank....u...