25 individuals in a population are homozygous dominant, then the individuals that are expected to be homozygous recessive are

The correct option/answer is 25.

Let us analyze this problem step-by-step using the principles of population genetics and the Hardy-Weinberg equilibrium.

The Hardy-Weinberg equation relates the frequencies of alleles and genotypes in a population in genetic equilibrium:
$$p^2+2pq+q^2=1$$
where:
- $p$ represents the frequency of the dominant allele in the gene pool.
- $q$ represents the frequency of the recessive allele in the gene pool, such that $p+q=1$.
- $p^2$ represents the frequency of the homozygous dominant genotype.
- $2pq$ represents the frequency of the heterozygous genotype.
- $q^2$ represents the frequency of the homozygous recessive genotype.

Under symmetric conditions where the allele frequencies are equal (i.e., $p=q=0.5$), we have:
- The homozygous dominant genotype frequency: $p^2={0.5}^2=0.25$ (or 25% of the population).
- The homozygous recessive genotype frequency: $q^2={0.5}^2=0.25$ (or 25% of the population).

Given that the number of homozygous dominant individuals in the population is 25, and their frequency matches that of the homozygous recessive individuals in this balanced state, the number of individuals expected to be homozygous recessive is also exactly 25.