Question

In a certain population of genetics students at Hardy-Weinberg equilibrium, the gene for attention span has two alleles, A and a, which act in a codominant manner Individuals of the AA genotype (blabbermouths) talk too much during lectures, whereas individuals of the aa genotype (sleepyheads) fall asleep during lectures; heterozygotes are well-behaved. At the beginning of the term, in a class of 1000 students, there are 90 sleepyheads.a. What is the frequency of the A allele at the beginning of the term? b. What is the frequency of the a allele at the beginning of the term?During the course of the term, the professor kicks out all 90 sleepyheads. He also kicks out 280 blabbermouths, so that by the end of term the population has are only 630 students left.a. What is the frequency of the A allele in the population at the end of the term? b. What is the frequency of the a allele in the population at the end of the term? c. Is the population at Hardy-Weinberg equilibrium at the end of the term

Answer 1

Answer:

Beginning of term: a) 0.7; b) 0.3;

End of term: a) 0.67; b) 0.33; c) no

Explanation:

The population is in Hardy-Weinberg equilibrium, so the genotypic frequencies are:

AA = p²

Aa = 2pq

aa = q²

Where p is the frequency of the A allele and q is the frequency of the a allele.

Beginning of the term

Total population (N): 1000

aa individuals: 90

b)

q² = 90/1000

q= √0.09

q=0.3

The frequency of the a allele is 0.3

a) p + q = 1

p = 1 - 0.3

p= 0.7

The frequency of the A allele is 0.7

The initial number of students with each genotype was:

• aa = 90
• AA = p² × N = 0.7 × 1000= 490
• Aa = 2 × p × q × N = 2 × 0.7 × 0.3 × 1000 = 420

End of the term

Number of students with each genotype:

• AA = 490 - 280 = 210
• Aa = 420
• aa = 0

N = 630

a) The frequecny of the A allele can be calculated as:

p= (2 × AA + Aa)/ 2 × N

Because there are 2 × N number of alleles in the population and the A allele appears twice in the homozygous individuals and once in the heterozyogus.

p= (2 × 210 + 420)/ 2 × 630

p= 0.67

The frequency of the A allele is 0.67

q= 1-p =0.33

The frequency of the a allele is 0.33

c) The population is not at H-W equilibrium at the end of the term, because the observed genotypic frequencies are not those expected by the equilibrium.

Expected AA = p² × N = 0.67² × 630 = 283

Observed AA = 210

Expected Aa = 2pq × N = 2 × 0.67 × 0.33 × 630 = 279

Observed Aa= 420

Expected aa = q² × N = 0.33² × 630 = 69

Observed aa = 0