Question

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

Explanation:

Step1: Analyze the probability for each trait

For the \(Rr\times Rr\) cross, the probability of getting \(rr\) (recessive for the first trait) is \(\frac{1}{4}\) (using the Punnett - square \(Rr\times Rr: RR:Rr:Rr:rr = 1:2:2:1\)).
For the \(Yy\times Yy\) cross, the probability of getting \(yy\) (recessive for the second trait) is \(\frac{1}{4}\) (using the Punnett - square \(Yy\times Yy: YY:Yy:Yy:yy=1:2:2:1\)).

Step2: Use the multiplication rule

Since the two traits are independent (Mendel's law of independent assortment), the probability of being recessive for both traits is the product of the probabilities of being recessive for each trait.
Let \(P\) be the probability of being recessive for both traits. Then \(P=\frac{1}{4}\times\frac{1}{4}\)

Answer:

\(d\). \(1/16\)