The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in
Hardy-Weinberg equilibrium.
(a) Calculate the percentage of heterozygous individuals in the population.
According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq
term. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals
2 × 0.19 × 0.81 = 0.31 or 31%
(b) Calculate the percentage of homozygous recessives in the population.

a

step-by-step explanation:

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!