Question

The bicoid (bcd) gene in Drosophila melanogaster has a role in establishing the polarity of the insect larva early in development. When homozygous in the maternal parent, a mutation in bcd has no effect on the parent but causes failure of anterior development in all of the offspring. The developing larvae have two abdomens, lack a head and fail to develop fully to adulthood. Suppose a heterozygous (bcd + /bcd - ) female is testcrossed with a homozygous (bcd - /bcd - ) male.

Answer 1

Some of the male offspring will be affected while the female offspring will have the mutant gene but not be affected by it.

Step-by-step explanation:

Drawing a Punnett square we can see:

bcd- ║ bcd- (male, XY)

bcd+║ bcd+/bcd- ║ bcd+/bcd-

bcd- ║ bcd-/bcd- ║ bcd-/bcd-

(femaleXX)

25% of the male offsprings will be affected by the mutation because they do not have a spare non mutated gene like the female offsprings.

25% of the male offspring will not be affected and will not have the mutated gene.

25% of female offsprings will have the mutated gene but not be affected by it because they have a spare non mutated gene in the other X chromosome.

25% of the female offspring will not be affected and will not have the mutated gene.