A female fruit fly with singed bristles was mated with a male from a true-breeding wild-type...

A female fruit fly with singed bristles was mated with a male from a true-breeding wild-type stock with long bristles. All of the F1 females had wild-type long bristles and all of the F1 males had singed bristles. If the F1 flies are intercrossed, the expected ratio of long to singed bristles in the F2 flies is

Multiple Choice

3:1 in both sexes.
3:1 in females, while all the males will have singed bristles.
1:1 in both sexes.
1:1 in females, while all the males will have singed bristles.
1:0 in both sexes (i.e., males and females will all have long bristles).

Expert Solution

The bristle length gene is inherited in an X linked recessive pattern, where the wild type is long bristles and the recessive X^l X^l is the recessive pattern showing signed bristles.

Female fruit fly with singed bristle has the genotype X^l X^l

male fruit fly with wild type long bristles is XY

when they are crossed, X^l X^l x XY, we get the following progenies, two children with X^lX(wild type female), and two children with X^lY(singed bristle male).

now when these are crossed between themselves,

X^lX x X^lY, the progenies have the genotype X^lX, X^l X^l ,X^lY and XY.

The X^lX(heterozygous for the recessive allele so exhibits the wild type phenomenon) and XY genotype shows the long bristle phenotype while the other two X^l X^l(recessive gene present in homozygous condition) and X^lY show the singed bristle phenotype.

So the ratio is 1:1, for long :singed in each sex.

so the answer is 1:1 in both sexes(option c)

gladiator answered 1 year ago

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