64% of a population is able to taste a chemical called PTC. Tasters are either homozygous dominant or?
64% of a population is able to taste a chemical called PTC. Tasters are either homozygous dominant or heterozygous for the trait. For the parts that follow, show all formulas, calculations and statements:
a) Calculate the percentage of the population that is Unable to taste PTC ?
b) Calculate the percentage of the population that is heterozygous for the trait?
c) Calculate the percentage of the population that has the recessive allele ? (Hint: heterozygous + homozygous recessive.)
- Anonymous8 years agoFavorite Answer
Using the Hardy-Weinberg equations:
f(non tasters) = q^2 = 0.36 a), so q = 0.6 and p=0.4
b) f(heterozygotes) = 2pq = 0.6*0.4*2 = 0.48
c) 0.48 + 0.36 = 0.84 = 84%
- 8 years ago
64% = DD (must carry a dominant allele)
so D = square root of 64% (0.64) is 0.8
d + D = 1 d = recessive
1 - 0.8 = 0.2 carry recessive allele (d)
So dd (people that cant taste it as they must carry a dominant allele, and so no dominant allele = cant taste it) is 0.2 squared = 0.04 (4%) homozygous recessive. That is the answer to A.
You must use the equation:
q^2 + p^2 + 2pq = 1
2pq is for the heterozygous.
You know that the d (recessive) is 0.2 and D (dominant) is 0.8
So heterozygous is 2pq so, (0.2 x 0.8) x 2 = 0.32 = 32% (heterozygous). This is the answer to B.
You know this is correct because it must add up to 1 (or 100%). 32% is heterozygous, 4% is homozygous recessive and 64% is homozygous dominant = 100%
Wow this took LONG! Points please!