Question

In: Biology

In fruit flies, bristle shape is controlled by an X-linked gene. The dominant wild-type allele (+)...

  1. In fruit flies, bristle shape is controlled by an X-linked gene. The dominant wild-type allele (+) results in normal bristles, while the recessive allele (sn) results in short 'singed' bristles. A normal-bristled female offspring of a male with singed bristles is crossed with a normal-bristled male. If we consider only the progeny that have normal bristles, what is the ratio of females : males among these?

    2:3

    1:3

    3:1

    1:1

    2:1

    3:2

    1:2

Solutions

Expert Solution

Answer:

The female progeny of an affected father has normal phenotype (is normal bristled) indicating she is a carrier as the X-linked allele from father codes for short singed bristles and X-linked allele from normal mother codes for normal bristles.

This female when crossed with normal bristled male will produce following cross:

Hence we conclude that the probability of
Normal bristled male = 1/4 or 25%
Normal bristled female = 1/4 or 25%

Therefore, ratio of Normal bristled male : Normal bristled female = 1 : 1


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