Final answer:
When a homozygous loss-of-function bicoid female fruit fly is mated with a wild-type male, all offspring will inherit one nonfunctional allele from the mother and one normal functional allele from the father. It is expected that the offspring will exhibit wild-type phenotypes due to the dominance of the wild-type allele over the nonfunctional allele.
Step-by-step explanation:
The question is asking about the phenotypes of the progeny when a female fruit fly that is homozygous for a loss-of-function bicoid allele is mated with a wild-type male. In fruit fly genetics, bicoid is a critical gene for early development. If the female is homozygous for a loss-of-function allele of this gene, all of her offspring will inherit one copy of this nonfunctional allele. Since the male is wild-type, he would contribute a normal functioning bicoid allele. Consequently, all offspring will be heterozygous for the bicoid allele, carrying one functional and one nonfunctional allele. In Drosophila, typically, if a gene is essential for life and an individual is heterozygous, having one wild-type and one nonfunctional allele, the wild-type allele is usually dominant and sufficient for a normal or wild-type phenotype. Therefore, it is expected that the offspring will exhibit wild-type phenotypes, assuming that the bicoid gene behaves in a similar dominant-recessive manner as many other genes do in fruit flies.