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.1/8
b.1/16
c.1/2"

Answer 1

The answer is a.1/8

This is an example of Mendelian dihybrid cross (a cross between two different traits).
If:
Y - (dominant) allele for the first trait
r - (recessive) allele for the first trait
S - (dominant) allele for the second trait
s - (recessive) allele for for the second trait

The cross will be like:
Parents: YyTt x YyTtAs shown in image, there will be in total 16 different genotypes. Two of them are YyTT genotypes. So the probability that the offspring will show YyTT genotype is 2 of 16:= 2/16 = 1/8