Final answer:
To calculate the percentage of heterozygous students in a population that is in Hardy-Weinberg equilibrium, we use the recessive allele frequency to find the dominant allele frequency and then apply the 2pq part of the equation. In this case, 48% of the students are heterozygous.
Step-by-step explanation:
The question deals with the concept of Hardy-Weinberg equilibrium, which is a principle that defines the genetic variation in a population that is not subject to evolutionary forces and therefore remains constant from generation to generation. According to the Hardy-Weinberg principle, the frequency of alleles can be represented by the equation p^2 + 2pq + q^2 = 1, where p represents the frequency of the dominant allele and q represents the frequency of the recessive allele.
In this scenario, the allele T (ability to taste) is dominant over allele t (inability to taste). Given that 64 out of 400 students are non-tasters (genotype tt), we can calculate q^2 as 64/400 = 0.16. Taking the square root of 0.16 gives us q = 0.4. Since p + q = 1, p must be 0.6. The frequency of heterozygous individuals (Tt) is 2pq, which equals 2(0.6)(0.4) = 0.48 or 48%.
Therefore, the percentage of heterozygous students (Tt) is 48%.