Question

Consider the following cross:

Parent 1: YyTt
Parent 2: YyTt

Using the rules of probability, determine the probability that the offspring will show YyTT genotype?

A.) 1
B.) 8
C.) 16
D.) 2

Answer 1

Answer: The correct answer is-

D.) 2/16

Explanation-

As per the information given in the question, the genotype of parents are-

Parent 1: YyTt

Parent 2: YyTt

Gametes produced by each parent would be- YT, Yt, yT, and yt

When these parents are allowed to cross, they will produce nine different types of genotypes out of which the probability of offspring showing YyTT would be genotype 2/16

Refer punnet square.

Answer 2

The probability that the offspring will show YyTT genotype is 1/16. To solve for this, the Punnet Square must be used. All possible allele combinations are then shown. Both parents show two traits making the Punnet Square have four columns and four rows.