Question

in a dihybrid cross involving pea plants heterozygous for two traits (rryy × rryy), how many of the offspring will be dominant for both traits? a 9/16 b 8/16 c 3/16 d 1/16

Explanation:

Step1: Determine the gametes

For \(RrYy\) plant, the gametes are \(RY, Ry, rY, ry\) (using Mendel's law of independent assortment).

Step2: Create a Punnett square

When crossing \(RrYy\times RrYy\), we can consider each trait separately first. For the \(R - r\) trait: \(Rr\times Rr\) gives \(RR:Rr:rr = 1:2:1\), and the probability of dominant (\(RR\) or \(Rr\)) is \(\frac{3}{4}\). For the \(Y - y\) trait: \(Yy\times Yy\) gives \(YY:Yy:yy=1:2:1\), and the probability of dominant (\(YY\) or \(Yy\)) is \(\frac{3}{4}\).

Step3: Use the multiplication rule

Since the two traits are independent, the probability of being dominant for both traits is \(\frac{3}{4}\times\frac{3}{4}\) (using the rule for independent events \(P(A\cap B)=P(A)\times P(B)\)).

Answer:

A. \(9/16\)