In leopards, being shy (S) is dominant to being sassy (s), and being timely (T) is dominant to being tardy (t).

A female leopard that is heterozygous for both the S and T trait mates with a male leopard that is also heterozygous for both traits. Both the S and T traits assort independently from one another.

What is the probability that they will have an offspring that is sassy and timely?

Choose 1

(Choice A)
A
6.25%

(Choice B)
B
18.75%

(Choice C)
C
0%

(Choice D)
D
56.25%

B

Explanation:

Start by making a punnet square with the female’s genotype on top and the male’s along the left side. Because the traits are independent of one another, you have to put all possibilities for what the mother and the father can give to the offspring:

ST St sT st
ST SSTT SsTt SsTT SsTt
St SSTt SStt SsTt Sstt
sT SsTT SsTt ssTT ssTt
st SsTt Sstt ssTt sstt

Now for a leopard offspring to be sassy, their genotype would need to be ss because the S gene is dominant over s. Following that same logic, a timely offspring can have the genotype TT or Tt. Now we look back to the punnet square to see which squares match a sassy and timely offspring. These will be marked by a * next to it:

ST St sT st
ST SSTT SsTt SsTT SsTt
St SSTt SStt SsTt Sstt
sT SsTT SsTt ssTT* ssTt*
st SsTt Sstt ssTt* sstt

So the probability of having a sassy and timely offspring would be 3 out of a total of 16 squares. Convert that to a percentage and voila, you have your answer.

3/16 = 0.1875 x 100 = 18.75%