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How can we tell whether a population is evolving at a gene?

Topic 7.5 Hardy-Weinberg Equilibrium: use the Hardy-Weinberg equations to calculate allele and genotype frequencies and test whether a population is evolving.

A focused answer to AP Biology Topic 7.5, covering the Hardy-Weinberg conditions, the equations p + q = 1 and p squared plus 2pq plus q squared = 1, and how to calculate and interpret allele and genotype frequencies, with worked calculations.

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  1. What this topic is asking
  2. The equations and conditions
  3. How to use it
  4. Try this

What this topic is asking

The College Board (Topic 7.5) wants you to use the Hardy-Weinberg equations to calculate allele and genotype frequencies, state the conditions for equilibrium, and use the model to test whether a population is evolving at a gene.

The equations and conditions

How to use it

The recessive phenotype (aa) is the only one you can count directly from phenotypes, so it gives q2q^2; take the square root to get qq, then find pp, then p2p^2 and 2pq2pq. You cannot find pp directly from the dominant phenotype, because the dominant phenotype includes both homozygous dominant (p2p^2) and heterozygous (2pq2pq) individuals, which look the same. This is why every Hardy-Weinberg problem starts from the recessive phenotype.

The five conditions map neatly onto the five mechanisms of evolution from the population-genetics topic. "No mutation" rules out new alleles, "no gene flow" rules out migration, "no selection" rules out differential reproduction, "random mating" rules out non-random mating, and "very large population" rules out genetic drift. Because each condition blocks one mechanism, a deviation from the predicted frequencies points to which mechanism is acting, which is what makes the model such a useful null hypothesis for detecting evolution.

The model is also useful for estimating the frequency of carriers of a recessive condition in a population, which is why it is applied in genetic counselling and conservation. Even though the conditions are rarely fully met, the predictions are often close enough to be informative, and large differences are the signal worth investigating.

Try this

Q1. State the two Hardy-Weinberg equations. [1 point]

  • Cue. p+q=1p + q = 1 and p2+2pq+q2=1p^2 + 2pq + q^2 = 1.

Q2. Explain how the Hardy-Weinberg model can show that a population is evolving. [2 points]

  • Cue. It predicts the genotype frequencies expected if the population is not evolving; if the observed frequencies differ significantly, one of the five conditions is violated, so the population is evolving at that gene.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2020 (style)4 marksSection II (long FRQ excerpt, statistics). In a population in Hardy-Weinberg equilibrium, 9% of individuals are homozygous recessive (aa) for a trait. (a) Calculate the frequencies of the recessive and dominant alleles. (b) Calculate the percentage of the population that are heterozygous carriers (Aa).
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A 4-point Hardy-Weinberg calculation FRQ.

(a) Calculate (2 points): (1 point) q2=0.09q^2 = 0.09, so q=0.09=0.3q = \sqrt{0.09} = 0.3; (1 point) p=1q=10.3=0.7p = 1 - q = 1 - 0.3 = 0.7.
(b) Calculate (2 points): (1 point) heterozygotes are 2pq2pq; (1 point) 2pq=2×0.7×0.3=0.422pq = 2 \times 0.7 \times 0.3 = 0.42, so 42% are carriers.

Markers reward taking the square root of q2q^2 to get qq, finding pp, and using 2pq2pq for the heterozygotes.

AP 2017 (style)1 marksSection I (multiple choice). Which of the following is a condition required for a population to remain in Hardy-Weinberg equilibrium? (A) Natural selection is occurring. (B) The population is very small. (C) There is no gene flow (no migration). (D) Mating is non-random.
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The correct answer is (C).

Hardy-Weinberg equilibrium requires no gene flow, no mutation, no natural selection, random mating, and a very large population. Options (A), (B) and (D) each violate a condition, so they would cause allele frequencies to change.

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