In a population that is in Hardy–Weinberg equilibrium for two alleles, C and c, 16% of the population shows a recessive trait. Assuming C is dominant to c, state the percent of the population shows the dominant trait? Show your work.
Thank you!
c=0.16
C+c=1, so
1-c=C
C=1-0.16=0.84
(C+c)^2=C^2+2Cc+c^2=1
(0.84)^2+[2(0.84)(0.16)]+(0.16)^2=1
Cc=Heterozygous dominant
Cc=2(0.84)(0.16)=0.27
CC=Homozygous dominant
CC=(0.84)^2=0.48
Frequency of CC + the frequency of Cc multiplied by 100 =percentage dominant for the trait
(0.27 +0.48)*100=75%
But if you need to know,
cc=0.25
or 25 % for the recessive trait.
In a population that is in Hardy-Weinberg equilibrium, the frequencies of alleles C and c will remain constant over generations.
Let's assume that p represents the frequency of the dominant allele C and q represents the frequency of the recessive allele c. Since there are only two alleles, p + q = 1.
According to the information given, 16% of the population shows the recessive trait. This means that q^2 (the frequency of individuals who are homozygous recessive) is equal to 0.16.
To find q, we take the square root of 0.16:
q = √0.16 = 0.4
Since p + q = 1, we can substitute the value of q:
p + 0.4 = 1
Solving for p:
p = 1 - 0.4
p = 0.6
The percent of the population that shows the dominant trait (those with at least one dominant allele) is given by p + 2pq (the frequency of individuals who are heterozygous) since C is dominant to c.
Substituting the values:
p + 2pq = 0.6 + 2(0.6)(0.4) = 0.6 + 0.48 = 1.08
Therefore, the percent of the population that shows the dominant trait is 108%. However, percentages cannot exceed 100%, so we can conclude that the correct answer is 100%.
To determine the percentage of the population that shows the dominant trait, we first need to calculate the frequency of the recessive allele (q). According to the Hardy-Weinberg equilibrium, p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele.
Given that 16% of the population shows the recessive trait, we can assume that q^2 = 0.16. Taking the square root of both sides, we find that q = √0.16 = 0.4.
Since p + q = 1, we can deduce that p = 1 - q = 1 - 0.4 = 0.6.
Now, to find the percentage of the population that shows the dominant trait (C), we use the equation p^2. Substituting the value of p, we have 0.6^2 = 0.36.
Therefore, the percentage of the population that shows the dominant trait is 36%.
To double-check our work, we can calculate the percentage of the population that shows the recessive trait by using q^2. Substituting the value of q, we have 0.4^2 = 0.16, which confirms our initial information.
So, 36% of the population shows the dominant trait, while 16% shows the recessive trait, assuming Hardy-Weinberg equilibrium.