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Ch. 20 - Population Genetics and Evolution at the Population, Species, and Molecular Levels
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
Chapter 20, Problem 26b

Assume that the flower population described in the previous problem undergoes a different pattern of predation. Flower-color determination and the starting frequencies of C₁ and C₂ are as described above, but the new insects attack yellow and red flowers, not orange flowers. As a result of the predation pattern, the relative fitness values are C₁C₁ = 0.40, C₁C₂ = 1.0, and C₂C₂ = 0.80.
What are the genotype frequencies among the progeny of predation survivors?

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1
Step 1: Understand the problem. The question involves calculating genotype frequencies after predation, given the relative fitness values for each genotype (C₁C₁ = 0.40, C₁C₂ = 1.0, C₂C₂ = 0.80). This requires applying the concept of relative fitness and Hardy-Weinberg principles.
Step 2: Recall the formula for calculating genotype frequencies after selection. The frequency of each genotype is proportional to its initial frequency multiplied by its relative fitness. Use the formula: \( f'(genotype) = \frac{f(genotype) \cdot w(genotype)}{\sum(f(genotype) \cdot w(genotype))} \), where \( f(genotype) \) is the initial frequency and \( w(genotype) \) is the relative fitness.
Step 3: Assign initial genotype frequencies. If not explicitly provided, assume Hardy-Weinberg equilibrium for the starting population. For alleles C₁ and C₂ with frequencies \( p \) and \( q \), the initial genotype frequencies are \( f(C₁C₁) = p^2 \), \( f(C₁C₂) = 2pq \), and \( f(C₂C₂) = q^2 \).
Step 4: Multiply each initial genotype frequency by its relative fitness. For example, \( f'(C₁C₁) = f(C₁C₁) \cdot w(C₁C₁) \), \( f'(C₁C₂) = f(C₁C₂) \cdot w(C₁C₂) \), and \( f'(C₂C₂) = f(C₂C₂) \cdot w(C₂C₂) \).
Step 5: Normalize the frequencies to ensure they sum to 1. Divide each adjusted frequency by the total sum of adjusted frequencies: \( f'(genotype) = \frac{f'(genotype)}{\sum f'(genotype)} \). This gives the final genotype frequencies among the progeny of predation survivors.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Genotype Frequencies

Genotype frequencies refer to the proportion of different genotypes in a population. In this context, it is essential to understand how the frequencies of genotypes C₁C₁, C₁C₂, and C₂C₂ are affected by natural selection and predation. These frequencies can be calculated using the Hardy-Weinberg principle or by considering the relative fitness of each genotype.
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Gamete Genotypes

Relative Fitness

Relative fitness is a measure of the reproductive success of a genotype compared to others in the population. In this scenario, the fitness values indicate how well each genotype survives predation. For example, a genotype with a fitness of 1.0 is considered the baseline, while values less than 1.0 indicate reduced survival and reproduction, which directly influences genotype frequencies in subsequent generations.
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Natural Selection

Natural Selection

Natural selection is the process by which certain traits become more or less common in a population based on their impact on survival and reproduction. In this case, the predation pattern affects which flower colors (and thus genotypes) are more likely to survive. Understanding how natural selection operates in this context is crucial for predicting changes in genotype frequencies among the progeny of the surviving flowers.
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Natural Selection
Related Practice
Textbook Question

In a population of flowers growing in a meadow, C1 and C2 are autosomal codominant alleles that control flower color. The alleles are polymorphic in the population, with f (C1) = 0.80 and f (C2) = 0.20. Flowers that are C1C1 are yellow, orange flowers are C1C2, and C2C2 flowers are red. A storm blows a new species of hungry insects into the meadow, and they begin to eat yellow and orange flowers but not red flowers. The predation exerts strong natural selection on the flower population, resulting in relative fitness values of C1C1 = 0.30, C1C2 = 0.60, and C2C2 = 1.0.

If predation continues, what are the allele frequencies when the second generation mates?

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Textbook Question

In a population of flowers growing in a meadow, C1 and C2 are autosomal codominant alleles that control flower color. The alleles are polymorphic in the population, with f (C1) = 0.80 and f (C2) = 0.20. Flowers that are C1C1 are yellow, orange flowers are C1C2, and C2C2 flowers are red. A storm blows a new species of hungry insects into the meadow, and they begin to eat yellow and orange flowers but not red flowers. The predation exerts strong natural selection on the flower population, resulting in relative fitness values of C1C1 = 0.30, C1C2 = 0.60, and C2C2 = 1.0.

What are the equilibrium frequencies of C1 and C2 if predation continues?

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Textbook Question

Assume that the flower population described in the previous problem undergoes a different pattern of predation. Flower-color determination and the starting frequencies of C₁ and C₂ are as described above, but the new insects attack yellow and red flowers, not orange flowers. As a result of the predation pattern, the relative fitness values are C₁C₁ = 0.40, C₁C₂ = 1.0, and C₂C₂ = 0.80.

What are the allele frequencies after one generation of natural selection?

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Textbook Question

Assume that the flower population described in the previous problem undergoes a different pattern of predation. Flower-color determination and the starting frequencies of C₁ and C₂ are as described above, but the new insects attack yellow and red flowers, not orange flowers. As a result of the predation pattern, the relative fitness values are C₁C₁ = 0.40, C₁C₂ = 1.0, and C₂C₂ = 0.80.

What are the equilibrium allele frequencies in the predation environment?

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Textbook Question

ABO blood type is examined in a Taiwanese population, and allele frequencies are determined. In the population, f (Iᴬ) = 0.30, f (Iᴮ) = 0.15, and f (i) = 0.55.f. Assuming Hardy–Weinberg conditions apply, what are the frequencies of genotypes, and what are the blood group frequencies in this population?

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Textbook Question
A total of 1000 members of a Central American population are typed for the ABO blood group. In the sample, 421 have blood type A, 168 have blood type B, 336 have blood type O, and 75 have blood type AB. Use this information to determine the frequency of ABO blood group alleles in the sample.
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