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?

1

8

1

16

1

2

The question is one of a dihybrid cross.

The picture that I attach below is a punnett cross of the parents (Yy Tt and  Yy Tt).  

So first you have to separate the two parents into gametes by pairing the alleles ( Yt , YT, yT, yt).  Then I crossed then with the punnett square and from that we see that there are two possible off spring that will would possess the genotype Yy TT (shaded gray in the diagram).

Thus the probability that a Yy TT offspring would be 2 out of 16
                   =   
                   =  
                   =   12.5 %

If there were no memes, there would be no life.