Imagine that you study the esterase (\"Est\") locus of the European Starling in
ID: 219183 • Letter: I
Question
Imagine that you study the esterase ("Est") locus of the European Starling in Georgia using electrophoresis. You discover two alleles, Est-f and Est-s (i.e., "fast" and "slow); in a population of 640 birds, 384 have the Est-FEst-f genotype, 154 are Est-f/Est-s, and 102 are Est-s/Est-s. a) What are the frequencies of the two alleles? b) Is t his population of starlings in Hardy-Weinberg equilibrium? How can you tell? easons why a population might NOT be in Hardy-Weinberg equilibrium. (Show your work, continuing on the back if necessary) (8 pts)Explanation / Answer
a)
Given that,
Est-f/Est-f= AA= 384
Est-f/Est-s=Aa= 154
Est-s/Est-s= aa= 102
By using Hardy-Weinberg equation to solve the above problems
p2 + 2pq + q2 = 1 and p + q = 1 , where
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Total population= 384+154+102= 640
q2= 102/640
q2=0.16
= 0.4
Now put the above value in (p+q=1) to solve for p.
p+q =1
p = 1-0.4
=0.6
Therefore allelic frequencies dominant and recessive allele are 0.6 and 0.4 respectively
b) To find the population is in Hardy-weingberg equilibrium by using p2+2pq+q2=1
p = 0.6
p2 = 0.36
2pq = 2×0.6×0.4
= 0.48
q=0.4
q2=0.16
.36+ .48+ .16= 1
If the population is in Hardy-weinberg equation, the genotype frequencies should be 0.36 AA, 0.48 Aa, and 0.16 aa.
c) A population might not be in Hardy-Weinberg equilibrium because of following reasons.
i)Mutations: Mutation introduces a new allele into the population that disrupts the equilibrium of allele frequencies.
ii)Natural selection: Natural selection disrupt the allele frequencies by changing the gene frequencies.
iii) Genetic drift: Genetic drift results from a small population that changes the allele frequencies.