If a normal (AA) male and a female with the disease (aa) mate, what percentage of offspring will be carriers of the disease?

In this genetic scenario, we are examining the mating between a male with a normal genotype (homozygous dominant, AA) and a female who has the disease (homozygous recessive, aa). The male has two dominant alleles, and the female has two recessive alleles.

To determine the genotypes of the offspring, we can use a Punnett square, which outlines the possible combinations of alleles from both parents. The male can only contribute the dominant allele (A) to the offspring, while the female can only contribute the recessive allele (a). Therefore, all offspring will inherit one dominant allele from the father and one recessive allele from the mother, leading to the genotype Aa for all offspring.

Being carriers is denoted by having one dominant allele and one recessive allele (Aa). Since all offspring will have this genotype, this means every single offspring will be a carrier of the disease. Thus, 100% of the offspring will be carriers, making the correct answer 100%.

In this case, the other options do not align with the genetic outcomes produced by the specific alleles contributed by each parent.