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in a dihybrid cross involving pea plants heterozygous for two traits (r…

Question

in a dihybrid cross involving pea plants heterozygous for two traits (rryy × rryy), how many of the offspring will be phenotypically green? round (r) is dominant over wrinkled (r), and yellow (y) is dominant over green (y).
a 12/16
b 9/16
c 4/16
d 3/16

Explanation:

Step1: Analyze the cross for each trait separately

For the seed shape trait (\(Rr\times Rr\)):

  • The phenotypic ratio is \(3\) (round, \(R-\)) : \(1\) (wrinkled, \(rr\)).

For the seed color trait (\(Yy\times Yy\)):

  • The phenotypic ratio is \(3\) (yellow, \(Y-\)) : \(1\) (green, \(yy\)).

Step2: Use the multiplication rule (since the two traits are independent)

We want the phenotype of green seeds. For green seeds (\(yy\)), the probability from the \(Yy\times Yy\) cross is \(\frac{1}{4}\).
For the seed - shape, we don't care about the shape (because the question only asks about the green phenotype, and the cross for shape is \(Rr\times Rr\) which has no effect on the color). The combined probability of getting green seeds (regardless of shape) is \(\frac{1}{4}\) (from the color cross). Since \(\frac{1}{4}=\frac{4}{16}\) (by multiplying numerator and denominator by \(4\)).

Answer:

C. \(4/16\)