Question

By applying the rules of probability to a Tt × Tt cross, determine the probability of the offspring being heterozygous (Tt).

25 percent
50 percent
75 percent

Answer 1

A heterozygous offspring can have genotype \(\mathrm{Tt}\) or \(\mathrm{tT}\), meaning one parent contributes either the dominant or recessive allele while the other parent contributes the opposite. The law of independent assortment says that the pairs occur independently of one another. The keyword here is "independent", as it allows you to use this property:
\[\mathbb P(\mathrm{Tt}\cup\mathrm{tT})=\mathbb P(\mathrm{Tt})+\mathbb P(\mathrm{tT})\]Each parent has \(\dfrac{1}{2}\) probability of contributing one of either potential allele, so \(\mathbb P(\mathrm{Tt})\) is really the probability that the first parent contributes the dominant allele *and* the second parent contributes the recessive allele, and vice versa for \(\mathbb P(\mathrm{tT})\). This means you have
\[\mathbb P(\mathrm{Tt}\cup\mathrm{tT})=\mathbb P(\mathrm{T})\mathbb P(\mathrm{t})+\mathbb P(\mathrm{t})\mathbb P(\mathrm{T})\]and since each probability is just \(\dfrac{1}{2}\), you're left with
\[\mathbb P(\mathrm{Tt}\cup\mathrm{tT})=\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}=\frac{1}{2}\]