Eye color in Jabberwocks is controlled by a single gene with two alleles that ex
ID: 213511 • Letter: E
Question
Eye color in Jabberwocks is controlled by a single gene with two alleles that exhibit simple dominance. Flame eyes (F) are dominant to brown eyes (f). In a population of 500 Jabberwocks, 300 have brown eyes (ff) and the rest have flame eyes (either FF or Ff). The population is in Hardy-Weinberg equilibrium. Let p represent the frequency of the dominant allele (F) and q represent the frequency of the recessive allele (f). The Hardy-Weinburg equation is p2 + 2pq + q2 = 1. (0.5 pts each)
a. How many total alleles for the trait are in the population?
b. What is the frequency of the recessive allele (q)? Hint: you can figure out the frequency of ff, which is equal to q2, but then you need to solve for q.
c. What is the frequency of the dominant allele (p)? Hint: recall that p + q = 1, so if you figured out q correctly above, then you can solve for p.
d. What is the frequency of the homozygous dominant individuals (FF)? Hint: this would be equal to p2.
e. What is the frequency of the heterozygous individuals (Ff)? Hint: this would be equal to 2pq.
f. Check your math: p2 + 2pq + q2 = 1.
Explanation / Answer
Answer:
a). Each individual has two alleles. 500 individuals will have 500*2=1000 alleles
b).
The frequency of recessive phenotype = ff = q^2=300/500 = 0.6
The frequency of recessive allele = f = q=0.78
c).
As p+q=1
p= 1-0.78 = 0.22
The frequency of dominant allele = F = p=0.22
d). The frequency of homozygous dominant phenotype= FF = p^2 = 0.22 * 0.22 = 0.0484
The frequency of homozygous dominant individuals = 0.0484 * 500 = 24
e).
The frequency of heterozygous dominant phenotype= Ff = 2pq = 2 * 0.22 * 0.78 = 0.3432
The frequency of heterozygous dominant individuals = 0.3432* 500 = 172
f). p^2 + 2pq+q^2 =1
0.0484+0.3432+0.6 = 0.99 = 1
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.