The ability to taste PTC is due to a single dominate allele "T". This means that individuals who can taste PTC are either homozygous dominant (TT) or heterozygous (Tt).
Out of the 215 individuals sampled, 150 could detect the bitter taste of PTC. These 150 individuals could be either TT or Tt genotypes. Let's represent the number of TT individuals as x and the number of Tt individuals as y.
Therefore, we have:
x + y = 150
Similarly, the individuals who could not detect the bitter taste of PTC could be either tt or Tt genotypes. Out of the 215 individuals sampled, 65 could not detect the bitter taste of PTC. Let's represent the number of tt individuals as z.
Therefore, we have:
y + z = 65
Since the frequency of the heterozygous genotype (Tt) is the same as the frequency of the ability to taste PTC, we can substitute y = 150 - x into the second equation:
(150 - x) + z = 65
z = x - 85
Now, we need to find the frequency of the heterozygous genotype, which is given by the ratio of Tt individuals to the total number of individuals sampled:
Frequency of Tt genotype = y / 215 = y / (x + y + z) = y / (x + y + (x-85)) = y / (2x + 65)
Since we want to find the frequency of the heterozygous genotype when 150 can taste PTC and 65 cannot, let's substitute y = 150 - x and z = x - 85 into the equation:
Frequency of Tt genotype = (150 - x) / (2x + 65)
To find the frequency of the heterozygous genotype, we need to know the value of x (representing the number of TT individuals). However, this information is not given in the question provided. Without the value of x, we cannot accurately calculate the frequency of the heterozygous genotype.
Therefore, given the information provided in the question, we cannot determine the exact frequency of the heterozygous genotype.
The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and
determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate the frequency of the
heterozygous genotype.
a
34%
b
72%
c
50%
d
17%
1 answer