Hardy-Weinberg Equations - ANSWERS

Hardy-Weinberg Equations - ANSWERS

In the following scenarios, use the Hardy-Weinberg equation of p2 + 2pq + q2 = 1, and knowing that p + q = 1.0 (p% + q% = 100%).

1.  In a species of magical mice, green fur is dominant over orange fur. If there are 300 orange mice in a population of 130,000, find the following:

a.  The frequency of the recessive (orange) allele

b.  The frequency of the dominant (green) allele

c.  The frequency of each genotype

p2 = GG (green) = ? q2 = gg (orange) = 300 2pq = Gg (green) = ?

a.  q2 = 300/130,000 = 0.002 q=√0.002 = 0.044 = 0.04

b.  p + q = 1.0 1.0 – 0.04 = p 1.0 – 0.04 = 0.96

c.  GG = p2 = (0.96)2 = 0.92 Gg = 2pq = 2[(0.96)(0.04)] = 0.08 gg = q2 = 0.042 = 0.0016 = 0.002

2.  In a certain population of 1000 fruit flies, 640 have red eyes while the remainder has sepia (light brown) eyes. The sepia eye trait is recessive to red eyes. How many individuals would you expect to be homozygous for red eye color?

p2 = RR (red) = ? q2 = rr (sepia) 2pq = Rr (red)

1000-640 = 360 sepia = rr

q2 = 360/1000 = 0.36

q = √0.36 = 0.6

p + q = 1 1 – q = p 1 – 0.6 = 0.4 p = 0.4

p2 = (0.4)2 = 0.16 = 16% of population is RR, so 160 fruit flies

3.  Phenylketonuria (PKU) is a human metabolic disorder that results in intellectual disabilities if it is untreated in infancy. In the United States, one out of approximately 10,000 babies is born with the disorder. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?

p2 = PP (normal) q2 = pp (PKU disease) 2pq = Pp (carriers) = ?

q2 = 1/10,000 = 0.0001

q = √0.0001 = 0.01

p + q = 1 1 – q = p 1 – 0.01 = 0.99 p = 0.99

2pq = 2(0.99)(0.01) = 0.0198 = 1.98% or 2% of the population is Pp