The answer is
a.1/8This 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
YyTt
As 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
