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

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

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

gladiator answered 2 years ago

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