The frequency of an allele ‘a’ in a natural
population is 0.49.
What percentage of the population is homozygous
for this allele? Calculate the exact
value; do not round up or give an estimate.
Answer in units of %.
To calculate the percentage of the population that is homozygous for allele 'a', we need to know the frequency of the homozygous genotype.
In a natural population, there are three possible genotypes for a single gene locus: homozygous dominant (AA), heterozygous (Aa), and homozygous recessive (aa). Since we already know the frequency of allele 'a', we assume that it represents the frequency of the recessive allele 'a'.
The frequency of the recessive genotype (aa) can be calculated using the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
In this equation, p represents the frequency of the dominant allele ('A') and q represents the frequency of the recessive allele ('a'). Since we know the frequency of 'a' is 0.49, we can substitute q = 0.49 into the equation.
0.49^2 + 2(0.49)p + p^2 = 1
Simplifying the equation:
0.2401 + 0.98p + p^2 = 1
Rearranging the terms:
p^2 + 0.98p - 0.7599 = 0
Now we solve this quadratic equation to find the frequency of the dominant allele ('A').
Using the quadratic formula:
p = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = 1, b = 0.98, and c = -0.7599. Substituting these values into the quadratic formula:
p = (-0.98 ± √((0.98)^2 - 4(1)(-0.7599))) / 2(1)
p = (-0.98 ± √(0.9604 + 3.0396)) / 2
p = (-0.98 ± √4) / 2
Since we are looking for the frequency of the homozygous dominant genotype (AA), we only consider the positive root:
p = (-0.98 + 2) / 2
p = 0.51 / 2
p = 0.255
Now we can find the percentage of the population that is homozygous for allele 'a' (aa) by squaring the recessive allele frequency (q) and converting it to a percentage:
Percentage = q^2 * 100
Percentage = (0.49)^2 * 100
Percentage = 0.2401 * 100
Percentage = 24.01%
So, exactly 24.01% of the population will be homozygous for the recessive allele 'a'.