Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the F1 plants had round seeds with yellow cotyledons. Based on this cross, what is the probability that an organism in the F2 generation will have round seeds and green cotyledons and be true breeding? Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the F1 plants had round seeds with yellow cotyledons. Based on this cross, what is the probability that an organism in the F2 generation will have round seeds and green cotyledons and be true breeding? 1/16 1/32 1/8 1/4

1/16

Step-by-step explanation:

This question involves two distinct genes; one coding for seed shape and the other for cotyledon color. The alleles for round seeds (R) and yellow cotyledons (Y) are dominant over the alleles for wrinkled seed (r) and green cotyledon (y) respectively.

In a cross between a truebreeding (i.e. same alleles for both genes) pea having round seeds and yellow cotyledon (RRYY) and a truebreeding pea having wrinkled seeds and green cotyledon (rryy), the F1 offsprings will all possess a heterozygous round seed and yellow cotyledon (RrYy).

The F1 offsprings (RrYy) will produce the following gametes: RY, Ry, rY, and ry. Using these gametes in a punnet square (see attached image), 16 possible offsprings will be produced in a ratio 9:3:3:1.

According to the question, 3/16 of the F2 offsprings will possess round seeds and green cotyledons, however, only 1 of them will be truebreeding i.e. RRyy. Hence, 1/16 of the F2 offsprings will be truebreeding for round seeds and green cotyledons.

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer