Final answer:
In a cross between a bicoid-null male Drosophila and a wild-type female, resulting embryos will have a wild-type phenotype due to the maternal contribution of at least one functional bicoid allele. The genotype at the bicoid locus will be dependent on the alleles provided by the female parent.
Step-by-step explanation:
When a bicoid-null male Drosophila (fruit fly) is crossed with a wild-type (wt) female, it will affect the phenotype and genotype of the resulting embryos at the bicoid gene locus. The bicoid gene is crucial for early embryonic development in Drosophila, dictating the anterior-posterior axis formation. The wild-type female (XX) will contribute a functional bicoid allele, since females carry two copies of the X chromosome and the bicoid gene is maternally provided. Consequently, all embryos from this cross are expected to have at least one functional bicoid allele from the wild-type female, leading to a wild-type phenotype with normal anterior-posterior patterning. However, the bicoid-null male (XY) will not contribute a functional allele of this gene.
Regarding the genotype at the bicoid locus for these embryos, it will depend on the particular allele inherited from the female. If she is homozygous for the functional allele, all embryos will have this genotype at the bicoid locus; if she is heterozygous, the embryos could have either the functional or the null phenotype from her, but since the mutation in bicoid is recessive, the phenotype will still be wild-type as long as one functional allele is present. It's important to note that the bicoid gene is not sex-linked, so the X-linked genetics concerning eye color in Drosophila discussed in the reference materials do not directly pertain to the bicoid gene.