Question

when two heterozygous parents are crossed in a dihybrid cross, what fraction of offspring will be homozygous recessive? a 2/16 b 8/16 c 1/16 d 4/16

Explanation:

Step1: Determine the genotypes of parents

Let the two - gene loci be \(A/a\) and \(B/b\). The heterozygous parents have the genotype \(AaBb\).

Step2: Use the Punnett square or probability rules

For a single - gene cross (\(Aa\times Aa\)), the probability of getting a recessive homozygote (\(aa\)) is \(\frac{1}{4}\) (from \(AA:Aa : aa=1:2:1\)). Similarly, for \(Bb\times Bb\), the probability of getting \(bb\) is \(\frac{1}{4}\).
Since the two gene crosses are independent events (Mendel's law of independent assortment), we use the multiplication rule. The probability of getting \(aabb\) (homozygous recessive for both genes) is \(P(aabb)=P(aa)\times P(bb)\)

$$P(aabb)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$$

Answer:

C. \(1/16\)