Question

1. a pea plant purebred to produce round yellow peas is crossed with a plant purebred to produce wrinkled green peas. round pea shape is dominant to wrinkled, and yellow pea color is dominant to green. members of the resulting f1 generation are then crossed to produce an f2 generation, as shown in the figure. on average, what fraction of the f2 generation will display both dominant traits?

a. 12 out of 16 plants will display both dominant traits.
b. 9 out of 16 plants will display both dominant traits.
c. 16 out of 16 plants will display both dominant traits.
d. 4 out of 16 plants will display both dominant traits.

Explanation:

Step1: Recall Mendelian Dihybrid Cross

In a dihybrid cross (two traits: shape - round (R) dominant to wrinkled (r); color - yellow (Y) dominant to green (y)), the F1 generation from crossing purebred round - yellow (RRYY) and purebred wrinkled - green (rryy) will all be RrYy. When F1 (RrYy) self - pollinates, we use the Punnett square. The possible gametes from RrYy are RY, Ry, rY, ry.

Step2: Analyze the Punnett Square for F2

The Punnett square for a dihybrid cross has 16 possible combinations. The genotypes that show both dominant traits (round and yellow) are those with at least one R and at least one Y. The genotypes are: RRYY, RRYy, RrYY, RrYy, RRYy, RRyy (no, yy is recessive for color), wait no, let's list the correct ones. The phenotypic ratio for a dihybrid cross is 9 (both dominant) : 3 (dominant for shape, recessive for color) : 3 (recessive for shape, dominant for color) : 1 (both recessive). So the number of plants with both dominant traits (round and yellow) is 9 out of 16.

Answer:

B. 9 out of 16 plants will display both dominant traits.