Chapter 23: Problem 14
A population of Hawaiian Drosopbila is segregating two alleles, \(P^{1}\) and \(P^{2}\), of the phosphoglucose isomerase \((P G I)\) gene. In a sample of 100 flies from this population, 30 were \(P^{1} P^{1}\) homozygotes, 60 were \(P^{1} P^{2}\) heterozygotes, and 10 were \(P^{2} P^{2}\) homozygotes. (a) What are the frequencies of the \(P^{1}\) and \(P^{2}\) alleles in this sample? (b) Perform a chisquare test to determine if the genotypes in the sample are in Hardy-Weinberg proportions. (c) Assuming that the sample is representative of the population, how many generations of random mating would be required to establish Hardy-Weinberg proportions in the population?
Short Answer
Step by step solution
Calculate Total Number of Alleles
Determine Allele Counts
Calculate Allele Frequencies
Calculate Expected Genotype Frequencies under Hardy-Weinberg
Calculate Expected Genotype Numbers
Perform Chi-square Test
Determine Chi-square Significance
Determine Generations for Hardy-Weinberg Equilibrium
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Allele Frequency
- In our example, there are two alleles: \(P^{1}\) and \(P^{2}\).
- To calculate allelic frequencies, first find the total number of alleles. Since there are 100 flies, and each fly carries two alleles, we have a total of 200 alleles in the sample.
- The frequency of an allele is calculated by dividing the number of times that allele appears by the total number of alleles.
- For allele \(P^{1}\), we have homozygous \(P^{1}P^{1}\) contributing 60 alleles and heterozygous \(P^{1}P^{2}\) contributing 60 alleles, giving a total of 120 \(P^{1}\) alleles.
- Thus, the frequency of \(P^{1}\) is \(\frac{120}{200} = 0.6\).
- Similarly, \(P^{2}\) has 80 alleles accounting to a frequency of \(\frac{80}{200} = 0.4\).
Allele frequencies remain constant across generations in an ideal population, providing a baseline to study changes over time.
Chi-Square Test
In population genetics, the chi-square test assesses whether a population's genotype frequencies align with Hardy-Weinberg proportions.
- Calculate expected frequencies using the Hardy-Weinberg equation, which assumes random mating and no other factors influencing allele frequencies.
- The main formula for chi-square is: \[\chi^2 = \sum \frac{(O_{i} - E_{i})^2}{E_{i}} \]where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency.
- For our given population of Drosophila, the calculated chi-square value is 6.25.
This example indicates the population is not in Hardy-Weinberg equilibrium, as the chi-square statistic exceeds the critical value for 1 degree of freedom.
Genotype Frequencies
- According to the principle, the frequencies of genotypes are represented as: - Homozygous dominant \(p^2\) - Heterozygous \(2pq\) - Homozygous recessive \(q^2\)
- Here, \(p\) and \(q\) are the allele frequencies of \(P^{1}\) and \(P^{2}\).
- For our example: - \(P^{1}P^{1}\) frequency \((p^2)\) is \((0.6)^2 = 0.36\) - \(P^{1}P^{2}\) frequency \((2pq)\) is \(2 \times 0.6 \times 0.4 = 0.48\) - \(P^{2}P^{2}\) frequency \((q^2)\) is \((0.4)^2 = 0.16\)
- These calculations provide a theoretical distribution if the population were in Hardy-Weinberg equilibrium.
Population Genetics
- The Hardy-Weinberg equilibrium is a cornerstone concept in this field, setting the baseline expectation for genetic distribution.
- Populations are subject to forces such as genetic drift, mutation, migration, and natural selection.
- Understanding these processes in the context of allele and genotype frequencies helps determine how a population's genetic structure shifts over time.
- For instance, only one generation of random mating is required to reach Hardy-Weinberg equilibrium if no other forces are at play.