Understanding Hardy-Weinberg Equilibrium in AP Biology
The Hardy-Weinberg Equilibrium is a fundamental concept in AP Biology, providing a framework to understand how allele frequencies in a population remain constant in the absence of evolutionary forces. This principle serves as a baseline to detect if a population is evolving. In this comprehensive guide, we’ll break down the Hardy-Weinberg conditions, equations, and applications with detailed explanations. Perfect for AP Bio exam prep and grasping population genetics with confidence!
1. What is Hardy-Weinberg Equilibrium?
The Hardy-Weinberg Equilibrium describes a theoretical state in which a population’s allele and genotype frequencies remain constant from one generation to the next, provided that certain conditions are met.
Key Points:
- No Evolution: Indicates no change in allele frequencies.
- Genetic Equilibrium: Serves as a null hypothesis for detecting evolution.
2. Conditions for Hardy-Weinberg Equilibrium
For a population to remain in Hardy-Weinberg Equilibrium, it must meet these five conditions:
- Large Population Size: Minimizes genetic drift.
- No Mutations: Prevents changes in allele frequencies.
- Random Mating: Ensures equal chances of reproduction.
- No Natural Selection: Prevents differential survival.
- No Gene Flow: Isolated population with no migration.
If any of these conditions are violated, evolution may occur.
3. The Hardy-Weinberg Equation Explained
The Hardy-Weinberg Equation is used to calculate allele and genotype frequencies:
Equations:
- Allele Frequency: p+q=1p + q = 1p+q=1
- p: Frequency of the dominant allele.
- q: Frequency of the recessive allele.
- Genotype Frequency: p2+2pq+q2=1p^2 + 2pq + q^2 = 1p2+2pq+q2=1
- p2p^2p2: Homozygous dominant frequency.
- 2pq2pq2pq: Heterozygous frequency.
- q2q^2q2: Homozygous recessive frequency.
4. Applying Hardy-Weinberg in AP Biology
Example Problem:
In a population of 1,000 individuals, 160 display a recessive phenotype (aa). Calculate the allele frequencies.
Solution:
- q2=1601000=0.16  ⟹  q=0.16=0.4q^2 = \frac{160}{1000} = 0.16 \implies q = \sqrt{0.16} = 0.4q2=1000160​=0.16⟹q=0.16​=0.4
- p=1−q=0.6p = 1 – q = 0.6p=1−q=0.6
- Allele Frequencies:
- Dominant allele (p): 0.6
- Recessive allele (q): 0.4
5. Factors Disrupting Hardy-Weinberg Equilibrium
- Genetic Drift: More impactful in small populations.
- Mutation: Introduces new alleles.
- Gene Flow: Migration alters allele frequencies.
- Non-Random Mating: Affects genotype frequencies.
- Natural Selection: Favors certain alleles.
6. Practice Questions for AP Biology
- Describe the five conditions necessary for Hardy-Weinberg Equilibrium.
- Calculate allele frequencies if 36% of a population shows the recessive phenotype.
- Explain how genetic drift can disrupt equilibrium.
Conclusion: Mastering Hardy-Weinberg Equilibrium for the AP Bio Exam
Understanding the Hardy-Weinberg Equilibrium equips you to identify if a population is evolving and to calculate allele frequencies confidently. Mastering this concept is crucial for the AP Biology exam and for a deeper grasp of population genetics. With this guide, you’re well-prepared to tackle any related questions.
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