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 I trait mates with a male leopard that is also
heterozygous for both traits. Both the S and I traits assort independently from one another.
What is the probability that they will have an offspring that is sassy and timely?
The probability that the leopards will have an offspring that is sassy and timely is 3/16
This question involves two different genes. The allele for being shy (S) is dominant to the allele for being sassy (s) in the first gene while the allele for being timely (T) is dominant to being tardy (t).
According to this question, A female leopard that is heterozygous for both the S and T traits i.e. SsTt mates with a male leopard that is also heterozygous for both traits i.e. SsTt
The two heterozygous leopards (SsTt) will produce the following gametes: ST, St, sT, and st
Using these gametes in a punnet square (see attached image), the following proportion of offsprings will be produced:
S_T_ = shy, timely (9)
S_tt = shy, tardy (3)
ssT_ = sassy, timely (3)
sstt = sassy, tardy (1)
According to this question, the probability that the two leopards will have an offspring that is sassy and timely is 3/16.
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