Question

given the following cross between parents of the genotypes aabbcc × aabbcc, assume complete dominance for each trait and independent assortment. what proportion of the progeny will be expected to phenotypically resemble the first parent?
○ 1/8
○ 1/3
○ 1/4
○ 3/4

Explanation:

Step1: Analyze each gene separately

For the \(A\) gene: \(AA\times Aa\). The offspring genotypes are \(AA:Aa = 1:1\). The proportion of offspring with the \(A -\) (dominant) phenotype (which is the same as the first parent for the \(A\) - trait) is \(1\) (since \(AA\) and \(Aa\) both show the dominant \(A\) phenotype).
For the \(B\) gene: \(BB\times bb\). The offspring genotype is \(Bb\). The proportion of offspring with the \(B -\) (dominant) phenotype (same as the first parent for the \(B\) - trait) is \(1\).
For the \(C\) gene: \(Cc\times Cc\). Using the Punnett - square or the formula for a monohybrid cross (\(Aa\times Aa\)), the offspring genotypes are \(CC:Cc:cc=1:2:1\). The proportion of offspring with the \(C -\) (dominant) phenotype (same as the first parent for the \(C\) - trait) is \(\frac{3}{4}\) (because \(CC\) and \(Cc\) show the dominant phenotype).

Step2: Use the multiplication rule

Since the genes assort independently (Mendel's law of independent assortment), we multiply the proportions of each trait together. The probability \(P\) that an offspring has the same phenotype as the first parent (\(A - B - C -\)) is \(P = 1\times1\times\frac{3}{4}\).

Answer:

\(\frac{3}{4}\)