Question

A virgin Drosophila female whose thorax bristles are very short is mated with a male having normal (long) bristles. The F1 progeny are 1/3 short-bristle females, 1/3 long bristle females, and 1/3 long bristle males. A cross of the F1 long bristle females with their brothers gives only long bristle F2 progeny. A cross of short bristle females with their brothers gave 1/3 long bristle females, 1/3 short bristle females, and 1/3 long bristle males. A) Explain what is going on genetically and B) why short bristled females cannot be homozygous.

Answer 1

When mating a normal male with long bristles with a female with short bristles, two phenotypes are obtained, those of long and short bristles, with an abnormal sex ratio, with a 2: 1 ratio for female and male.

the males kept the long sows and they are normal, the females with long sows and short sows in equal numbers.

It can be seen that there is an inheritance linked to x, which explains the phenotype, so the females will be heterozygous and the allele of the disease causes death, which means that the first cross is: A / a x a / Y.

obtaining crossbreeding of the F1 series with females (a / a) and siblings with long sows (a / Y) results in individuals with only long sows (a / a or a / Y).

Crossbreeding males with long bristles (a / Y) and females with short bristles (A / a) results in:

¼ male long bristles (a / Y)

¼ male short bristle (A / Y)

¼ female short bristles (Y / y),

¼ female long bristles (y / y),

Despite the prediction, the results were different when obtaining 1/3 female of short sows, 1/3 female of long sows and 1/3 male of long sows, it is inferred that there were no males with short sows due to lethality. of the mutation.

B) why female short sows cannot be homozygous because the mutation has x-linked inheritance and is lethal