Hardy-weinberg Equilibrium Practice Problems With Answers

Unraveling the secrets of population genetics often feels like deciphering a complex code, but the Hardy-Weinberg equilibrium provides a foundational principle for understanding how gene frequencies behave in a population. This principle, however, truly comes to life when applied to practical problems. Tackling Hardy-Weinberg equilibrium practice problems, complete with detailed answers, not only solidifies your understanding of the core concepts but also equips you with the analytical skills needed to address real-world scenarios in genetics and evolutionary biology.

What is the Hardy-Weinberg Equilibrium?

The Hardy-Weinberg equilibrium, named after Godfrey Harold Hardy and Wilhelm Weinberg, independently derived, describes a theoretical state where the genetic variation in a population remains constant from generation to generation. This equilibrium occurs when there are no disturbing factors at play, meaning there are no mutations, random mating is occurring, there is no gene flow, the population size is infinitely large, and there is no selection.

This equilibrium is defined by two equations:

1. p + q = 1: This equation represents the allele frequencies, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. The sum of these frequencies always equals 1 (or 100%).
2. p² + 2pq + q² = 1: This equation represents the genotype frequencies, where p² is the frequency of the homozygous dominant genotype, 2pq is the frequency of the heterozygous genotype, and q² is the frequency of the homozygous recessive genotype. The sum of these frequencies also always equals 1.

Why Practice Hardy-Weinberg Problems?

Working through Hardy-Weinberg equilibrium practice problems is essential for several reasons:

• Conceptual Understanding: Applying the equations to different scenarios reinforces the theoretical understanding of the principles.
• Problem-Solving Skills: These problems help develop analytical skills needed to tackle more complex genetic scenarios.
• Application in Real-World Scenarios: Provides a foundation for understanding population changes in response to evolutionary pressures, such as natural selection or genetic drift.
• Exam Preparation: Indispensable for students studying biology, genetics, or evolutionary biology.

Hardy-Weinberg Equilibrium Practice Problems with Answers

Now, let's dive into some practice problems. Each problem will be followed by a detailed solution to guide you through the process.

Problem 1:

In a population of butterflies, the allele for brown wings (B) is dominant over the allele for white wings (b). If 40% of the butterflies are white, calculate the frequency of each allele and the frequency of each genotype.

Solution:

1. Identify the known variable: We know that the frequency of the homozygous recessive genotype (white butterflies) is 40%, or 0.40. This is q².
2. Calculate q: Take the square root of q² to find q:
   • q = √0.40 = 0.63
3. Calculate p: Use the equation p + q = 1 to find p:
   • p = 1 - q = 1 - 0.63 = 0.37
4. Calculate the genotype frequencies:
   • p² (homozygous dominant): p² = (0.37)² = 0.14
   • 2pq (heterozygous): 2pq = 2 * 0.37 * 0.63 = 0.46
   • q² (homozygous recessive): q² = 0.40 (given)

Therefore, the frequency of the B allele is 0.37, the frequency of the b allele is 0.63, the frequency of BB genotype is 0.14, the frequency of Bb genotype is 0.46, and the frequency of bb genotype is 0.40.

Problem 2:

In a certain population of beetles, the green coloration (G) is dominant over brown (g). If the frequency of the green allele is 0.7, what are the expected genotype frequencies?

Solution:

1. Identify the known variable: We know that the frequency of the green allele (p) is 0.7.
2. Calculate q: Use the equation p + q = 1 to find q:
   • q = 1 - p = 1 - 0.7 = 0.3
3. Calculate the genotype frequencies:
   • p² (homozygous dominant, GG): p² = (0.7)² = 0.49
   • 2pq (heterozygous, Gg): 2pq = 2 * 0.7 * 0.3 = 0.42
   • q² (homozygous recessive, gg): q² = (0.3)² = 0.09

Thus, the frequency of the GG genotype is 0.49, the frequency of the Gg genotype is 0.42, and the frequency of the gg genotype is 0.09.

Problem 3:

A population of 5000 people has 25 individuals with cystic fibrosis, a recessive genetic disorder. Assuming the population is in Hardy-Weinberg equilibrium, how many people in the population are carriers of the cystic fibrosis allele?

Solution:

1. Identify the known variable: We know the number of individuals with cystic fibrosis (homozygous recessive) is 25 out of 5000. Therefore, q² = 25/5000 = 0.005.
2. Calculate q: Take the square root of q² to find q:
   • q = √0.005 = 0.0707
3. Calculate p: Use the equation p + q = 1 to find p:
   • p = 1 - q = 1 - 0.0707 = 0.9293
4. Calculate the frequency of heterozygous carriers (2pq):
   • 2pq = 2 * 0.9293 * 0.0707 = 0.1314
5. Calculate the number of carriers in the population:
   • Number of carriers = 0.1314 * 5000 = 657

Therefore, there are approximately 657 carriers of the cystic fibrosis allele in the population.

Problem 4:

In a flock of sheep, the allele for white wool (W) is dominant over the allele for black wool (w). In a sample of 800 sheep, 722 are white and 78 are black. Estimate the allelic and genotypic frequencies within this population.

Solution:

1. Identify the known variable: The number of black sheep (homozygous recessive) is 78 out of 800. Therefore, q² = 78/800 = 0.0975.
2. Calculate q: Take the square root of q² to find q:
   • q = √0.0975 = 0.3122
3. Calculate p: Use the equation p + q = 1 to find p:
   • p = 1 - q = 1 - 0.3122 = 0.6878
4. Calculate the genotype frequencies:
   • p² (homozygous dominant, WW): p² = (0.6878)² = 0.4731
   • 2pq (heterozygous, Ww): 2pq = 2 * 0.6878 * 0.3122 = 0.4296
   • q² (homozygous recessive, ww): q² = 0.0975 (given)

Thus, the frequency of the W allele is 0.6878, the frequency of the w allele is 0.3122, the frequency of the WW genotype is 0.4731, the frequency of the Ww genotype is 0.4296, and the frequency of the ww genotype is 0.0975.

Problem 5:

Phenylketonuria (PKU) is an autosomal recessive metabolic disorder that results in decreased metabolism of the amino acid phenylalanine. In the United States, PKU affects approximately 1 in 10,000 newborns. Assuming Hardy-Weinberg equilibrium, what is the carrier frequency in the US population?

Solution:

1. Identify the known variable: The frequency of individuals with PKU (homozygous recessive) is 1 in 10,000, or 0.0001. This is q².
2. Calculate q: Take the square root of q² to find q:
   • q = √0.0001 = 0.01
3. Calculate p: Use the equation p + q = 1 to find p:
   • p = 1 - q = 1 - 0.01 = 0.99
4. Calculate the frequency of heterozygous carriers (2pq):
   • 2pq = 2 * 0.99 * 0.01 = 0.0198

Therefore, the carrier frequency in the US population is approximately 0.0198, or about 2%. This means that about 2% of the population carries one copy of the PKU allele.

Problem 6:

In a population of frogs, the allele for spotted skin (S) is dominant over the allele for non-spotted skin (s). After sampling the population, it was found that 64% of the frogs had spotted skin. If the population is in Hardy-Weinberg equilibrium, what is the frequency of the recessive allele?

Solution:

This problem is a bit trickier because we are given the percentage of frogs with spotted skin, which includes both the homozygous dominant (SS) and heterozygous (Ss) genotypes.

1. Determine the frequency of the recessive phenotype: Since 64% of the frogs have spotted skin, it means that 36% have non-spotted skin (100% - 64% = 36%). The non-spotted frogs are homozygous recessive (ss), so q² = 0.36.
2. Calculate q: Take the square root of q² to find q:
   • q = √0.36 = 0.6

Therefore, the frequency of the recessive allele (s) is 0.6.

Problem 7:

Consider a population of birds where feather color is controlled by a single gene with two alleles: red (R) and white (r). If you observe that 16% of the birds have white feathers, and assuming the population is in Hardy-Weinberg equilibrium, what percentage of the birds are heterozygous?

Solution:

1. Identify the known variable: The frequency of birds with white feathers (homozygous recessive) is 16%, or 0.16. This is q².
2. Calculate q: Take the square root of q² to find q:
   • q = √0.16 = 0.4
3. Calculate p: Use the equation p + q = 1 to find p:
   • p = 1 - q = 1 - 0.4 = 0.6
4. Calculate the frequency of heterozygous birds (2pq):
   • 2pq = 2 * 0.6 * 0.4 = 0.48

Therefore, 48% of the birds are heterozygous.

Problem 8:

In a population of wildflowers, the color of the petals is governed by two alleles: red (C^R) and white (C^W). These alleles exhibit codominance, and the heterozygotes (C^RC^W) have pink petals. In a sample of 500 plants, there are 100 red, 250 pink, and 150 white flowers. Calculate the allele frequencies for C^R and C^W.

Solution:

Since the alleles are codominant, we can directly count the alleles.

1. Count the number of each allele:
   • Each red flower (C^RC^R) has two C^R alleles, so there are 100 * 2 = 200 C^R alleles.
   • Each pink flower (C^RC^W) has one C^R allele and one C^W allele, so there are 250 C^R alleles and 250 C^W alleles.
   • Each white flower (C^WC^W) has two C^W alleles, so there are 150 * 2 = 300 C^W alleles.
2. Total number of alleles in the population:
   • Total alleles = 500 plants * 2 alleles/plant = 1000 alleles.
3. Calculate the frequency of each allele:
   • Frequency of C^R (p) = (200 + 250) / 1000 = 450 / 1000 = 0.45
   • Frequency of C^W (q) = (250 + 300) / 1000 = 550 / 1000 = 0.55

Therefore, the frequency of the C^R allele is 0.45 and the frequency of the C^W allele is 0.55.

Problem 9:

A large population of laboratory mice is allowed to breed randomly for several generations. Coat color is determined by two alleles: black (B) and white (b). You sample 800 mice and find that 578 are black and 222 are white. Assuming Hardy-Weinberg equilibrium, determine if the population is in equilibrium by performing a chi-square test.

Solution:

1. Calculate the observed genotype frequencies:
   • Total mice = 800
   • Black mice (BB or Bb) = 578
   • White mice (bb) = 222
   • Observed frequency of bb (q²) = 222/800 = 0.2775
2. Calculate the allele frequencies:
   • q = √0.2775 = 0.5268
   • p = 1 - q = 1 - 0.5268 = 0.4732
3. Calculate the expected genotype frequencies based on Hardy-Weinberg equilibrium:
   • p² (BB) = (0.4732)² = 0.2240
   • 2pq (Bb) = 2 * 0.4732 * 0.5268 = 0.4984
   • q² (bb) = (0.5268)² = 0.2775
4. Calculate the expected number of each genotype:
   • Expected BB = 0.2240 * 800 = 179.2
   • Expected Bb = 0.4984 * 800 = 398.72
   • Expected bb = 0.2775 * 800 = 222
5. Perform a chi-square test:
   • We need to combine the BB and Bb categories because we cannot distinguish between them phenotypically.
     • Observed black (BB + Bb) = 578
     • Expected black (BB + Bb) = 179.2 + 398.72 = 577.92
     • Observed white (bb) = 222
     • Expected white (bb) = 222

• Chi-square (χ²) = Σ [(O-E)² / E] = 0.00001 + 0 = 0.00001
• Degrees of freedom (df) = number of categories - number of alleles estimated - 1 = 2 - 1 - 1 = 0 (since we estimated p and q). However, because we combined the BB and Bb categories, df = 1 - 0 = 1.

6. Interpret the chi-square value:
   • Using a chi-square distribution table with df = 1 and α = 0.05, the critical value is 3.841.
   • Since our calculated χ² (0.00001) is much less than the critical value (3.841), we fail to reject the null hypothesis.

Conclusion: The population is likely in Hardy-Weinberg equilibrium. The observed genotype frequencies are not significantly different from what we would expect if the population were in equilibrium.

Problem 10:

Sickle-cell anemia is an autosomal recessive disorder. In a population in West Africa, it was found that 4% of newborns have sickle-cell anemia. Assuming the population is in Hardy-Weinberg equilibrium, what proportion of the population is resistant to malaria because they are heterozygous for the sickle-cell allele?

Solution:

The resistance to malaria is conferred by being heterozygous (carriers) for the sickle-cell allele.

1. Identify the known variable: We know the frequency of individuals with sickle-cell anemia (homozygous recessive) is 4%, or 0.04. This is q².
2. Calculate q: Take the square root of q² to find q:
   • q = √0.04 = 0.2
3. Calculate p: Use the equation p + q = 1 to find p:
   • p = 1 - q = 1 - 0.2 = 0.8
4. Calculate the frequency of heterozygous carriers (2pq):
   • 2pq = 2 * 0.8 * 0.2 = 0.32

Therefore, 32% of the population is resistant to malaria because they are heterozygous for the sickle-cell allele.

Key Considerations and Assumptions

When solving Hardy-Weinberg equilibrium problems, it's crucial to keep in mind the underlying assumptions:

• No Mutation: The mutation rate is negligible.
• Random Mating: Individuals mate randomly without preference for genotype.
• No Gene Flow: There is no migration of individuals into or out of the population.
• Infinite Population Size: The population is large enough to avoid genetic drift.
• No Selection: All genotypes have equal survival and reproductive rates.

Deviations from these assumptions can lead to changes in allele and genotype frequencies over time, indicating that the population is evolving.

Conclusion

Mastering Hardy-Weinberg equilibrium practice problems is more than just an academic exercise; it’s about developing a fundamental understanding of population genetics and evolutionary processes. By understanding how to apply these principles, you can gain insights into the genetic structure of populations and how they change over time. These insights are invaluable for fields ranging from conservation biology to human genetics. Remember to carefully analyze each problem, identify the known variables, and apply the Hardy-Weinberg equations systematically.