What is the minimum number of people shown in the family pedigree who must be heterozygous for the two situations described in the table in the absence of any new mutations?

Answer:

25) E

Let the dominant allele be: T
Let the recessive allele be : t

If only U has a recessive condition, their genotype must be tt. This means that S and T must be heterozygous.
Since we want the minimum number of heterozygous people, we must assume that R has to be TT.
In order for S to be Tt and R to be TT, only one of P and Q needs to heterozygous.
This means that if only U has the recessive people, 3 people must be heterozygous.

If R and U both have the recessive condition, P and Q must both be heterozygous.
This means that if R and U have the recessive condition, 4 people (P, Q, S and T) must be heterozygous. Hence, E is the answer.