If 4 percent of a population in equilibrium expresses a recessive trait, what is the probability that the offspring of two individuals who do not express the trait will express it?

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Hey everyone. Let's take a look at this question together. Which of the following is the probability that the offspring of two individuals who do not express a recessive trait will express it. Given that 5% of the population in equilibrium expresses the trait. So because this population is an equilibrium, that means that the frequency of this recessive trait is shown as Q equals the square root of 0.25, which comes out to 0.5. And the frequency of the dominant allele is P equals one minus Q. Which equals 0.5. Because one minus 0.5 equals 0.5. And this formula comes from P plus Q equals one. Where we subtract Q from both sides to give us P equals one minus Q. Two, sulfur P. And since the parents do not express that trait but their offspring will express it. That means that the parents are heterosexual. And since the probability of each parent passing on that recessive allele is 0.5 we go ahead and do 0.5 times 0. which gives us 0.25 or 25% Which is answer choice. A. The correct answer. Because we know that the probability of the offspring expressing that recessive trait is 0.5 and both parents do not express the trait but their offspring will we go ahead and multiply .55.5 to give us 0.25 which we then convert to that probability percentage which is 25% or answer choice. A the correct answer. I hope you found this video to be helpful, thank you and goodbye.

If 4 percent of a population in equilibrium expresses a recessive trait, what is the probability that the offspring of two individuals who do not express the trait will express it?

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Key Concepts

Hardy-Weinberg Equilibrium

Recessive Trait

Punnett Square