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.
<h3>
<u>Beginning of the term</u></h3>
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
<u>The initial number of students with each genotype was:</u>
- aa = 90
- AA = p² × N = 0.7 × 1000= 490
- Aa = 2 × p × q × N = 2 × 0.7 × 0.3 × 1000 = 420
<h3>
<u>End of the term</u></h3>
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
