Hardy-Weinberg Equation

Question 1

Cystic Fibrosis affects about 1 in 2200 people. Using the Hardy-Weinberg equation, work out:
- q, frequency of the recessive for CF
- p, frequency of the dominant allele
- the frequency of the three phenotypes in the population (person with CF, carrier and ‘normal’ - person homozygous for the dominant allele that makes functioning CF protein)

Answer 1

A - 1/2200 = 0.00045
q2= 0.00045, therefore
q = square root of 0.00045
q = 0.02

B - p+q=1
therefore p=1-0.02=0.98

C - 2pq = 2(0.02x0.98) = 0.039 (39%)
q2 = 0.005 = 0.05%
p2 = 0.96 = 96%

Question 2

What is the Hardy-Weinberg equation?

Answer 2

p2 + 2pq + q2

Question 3

What do the values in the Hardy-Weinberg equation mean?

Answer 3

p= dominant alleles
q= recessive alleles