A fish of genotype a/a;B/b is crossed to a fish whose genotype is A/a;B/b.What proportion of the progeny will be heterozygous for at least one of the genes? (Assume independent assortment.)

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The progeny can be heterozygous for at least one gene in several combinations. For the first gene, the possible genotypes are A/a (from A/a x a/a), which is always heterozygous. For the second gene, the possible genotypes from B/b x B/b are BB, Bb, bB, and bb, where Bb and bB are heterozygous. Therefore, the combinations that result in progeny being heterozygous for at least one gene are A/a;BB, A/a;Bb, A/a;bB, and A/a;bb. This accounts for 4 out of the 4 possible combinations for the first gene (since it's always heterozygous) and 3 out of the 4 possible combinations for the second gene (since 3 out of 4 are heterozygous or homozygous recessive, which still meets the condition when considering the first gene's guaranteed heterozygosity). Thus, 6 out of 8 possible combinations result in progeny that are heterozygous for at least one gene.

A Fish of Genotype A/a;B/b Is Crossed to a Fish