The correct answer for this2(4m+7)-(m-5)
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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 %
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 %
