Red hair is a recessive trait in humans. If 2% of the US population has red hair (q2), then what percent of people in the US are heterozygous (2pq) for this trait assuming that the population is following Hardy-Weinberg equilibrium? Write your answer as a whole number percentage. Do not include decimals or the percent symbol (%).

If 2% of the US population has red hair (q²), then the percent of people in the US are heterozygous (2pq) for this trait at Hardy-Weinberg equilibrium is 24%.

The equation of the Hardy-Weinberg equilibrium is p² + 2pq + q² = 1, where,

In this case,

In consequence the percentage of heterozygous is

2pq = 2 x 0.141 x 0.859 = 0.242 > 24%

In conclusion, if 2% of the US population has red hair (q²), then the percent of people in the US are heterozygous (2pq) for this trait at Hardy-Weinberg equilibrium is 24%.

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Explanation:

For Hardy weinberg principle for a population in equilibrum we have,

P2+ 2pq+q2=1

P+q= 1

Given that q2=2%

q2=0.02

q=√0.02

q=0.141

Since p+q=1

Therefore, p=1-q

p=1-0.141

p=0.859

P2=0.737

Since p2+2pq+q2=1

2pq= 2×0.859×0.141

=0.24

0.24×100

2pq=24%

Therefore 2pq will be 24% of the population

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