Question

Using a Punnett square workout the distribution of an autosomal phenotypic feature in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Hint:

In a Punnett square for a single locus, a homozygous parent (AA or aa) paired with a heterozygous parent (Aa) results in either all dominant or a 1:1 phenotypic ratio, depending on the homozygous parent’s genotype.

Answer 1

Step 1: For an autosomal trait at a single locus, let’s assume the trait is controlled by alleles A (dominant) and a (recessive). A homozygous female could be AA or aa; a heterozygous male is Aa. Since the question doesn’t specify dominance, let’s assume the female is homozygous dominant (AA) for clarity.
Step 2: The cross is AA (female) × Aa (male). Using a Punnett square: Female gametes are all A; male gametes are A or a. Offspring genotypes are AA (A from female, A from male) or AA (A from female, a from male). Phenotypically, all offspring are AA (100% dominant phenotype).
Step 3: If the female were homozygous recessive (aa), the cross would be aa × Aa. Offspring genotypes would be Aa (a from female, A from male) or aa (a from female, a from male), giving a 1:1 phenotypic ratio (50% dominant, 50% recessive). Since the question asks for a typical autosomal cross, the first scenario (all dominant) is more standard unless specified otherwise.
Thus, all offspring in the first filial generation show the dominant phenotype (100% AA).