In the F₂ generation of a homozygous round (AA) × homozygous wrinkled (aa) cross in peas, three seeds are chosen at random. What is the probability that two seeds are round and the other is wrinkled? A)(1/4)³ B)3(1/4)(3/4)² C)3(1/4)²(3/4) D)(3/4)³

Question

Quizplus · Accepted Answer

We can use the binomial probability formula: P(k successes in n trials) = (n choose k) * p^k * (1-p)^(n-k)

Let's let "success" be selecting a round seed, and "failure" be selecting a wrinkled seed. Then we want to find the probability of getting 2 successes and 1 failure in 3 trials.

We can plug in the numbers:
P(2 successes and 1 failure) = (3 choose 2) * (1/4)^2 * (3/4)^1 = 3/16

But we also have to consider the different orders in which we could have selected the seeds. For example, we could have selected a round seed first, then another round seed, then a wrinkled seed; or we could have selected a round seed first, then a wrinkled seed, then another round seed; and so on. There are 3! = 6 possible orders in which we could have selected the seeds.

So the final probability is: 6 * (3/16) = 9/8 or approximately 0.5625

But this is greater than 1, which doesn't make sense. This means we made an error in our calculations. Specifically, we double-counted some of the possible orders. For example, selecting seeds A, B, C in order is the same as selecting seeds B, A, C in order. We counted both of these possibilities separately, but they are really the same.

To correct for this, we divide by the number of permutations of the 3 seeds, which is 3! = 6. So the final answer is:

P(2 round seeds and 1 wrinkled seed) = 6 * (3 choose 2) * (1/4)^2 * (3/4)^1 / 6 = (3/4)(1/4)^2 * 3 = 3/16 or approximately 0.1875

Therefore, the best choice is B, which matches this answer.

In the F2 Generation of a Homozygous Round (AA) ×

In the F₂ Generation of a Homozygous Round (AA) ×