The ability to taste the bitter compound phenylthiocarbamide (PTC) is an autosomal dominant trait. The inability to taste PTC is a recessive condition. In a sample of 500 people, 360 have the ability to taste PTC and 140 do not. Calculate the frequency of the recessive allele.
Table of contents
- 1. Introduction to Genetics51m
- 2. Mendel's Laws of Inheritance3h 37m
- 3. Extensions to Mendelian Inheritance2h 41m
- 4. Genetic Mapping and Linkage2h 28m
- 5. Genetics of Bacteria and Viruses1h 21m
- 6. Chromosomal Variation1h 48m
- 7. DNA and Chromosome Structure56m
- 8. DNA Replication1h 10m
- 9. Mitosis and Meiosis1h 34m
- 10. Transcription1h 0m
- 11. Translation58m
- 12. Gene Regulation in Prokaryotes1h 19m
- 13. Gene Regulation in Eukaryotes44m
- 14. Genetic Control of Development44m
- 15. Genomes and Genomics1h 50m
- 16. Transposable Elements47m
- 17. Mutation, Repair, and Recombination1h 6m
- 18. Molecular Genetic Tools19m
- 19. Cancer Genetics29m
- 20. Quantitative Genetics1h 26m
- 21. Population Genetics50m
- 22. Evolutionary Genetics29m
21. Population Genetics
Hardy Weinberg
Problem 10c
Textbook Question
Consider a population in which the frequency of allele A is p = 0.7 and the frequency of allele a is q = 0.3 and where the alleles are codominant. What will be the allele frequencies after one generation if the following occurs?
wAA = 1, wAa = 0.99, waa = 0.98
Verified step by step guidance1
Identify the initial allele frequencies: \(p = 0.7\) for allele \(A\) and \(q = 0.3\) for allele \(a\). Confirm that \(p + q = 1\).
Calculate the initial genotype frequencies assuming Hardy-Weinberg equilibrium: \(f(AA) = p^2\), \(f(Aa) = 2pq\), and \(f(aa) = q^2\).
Apply the given fitness values to each genotype to find the weighted genotype frequencies after selection: multiply each genotype frequency by its respective fitness, i.e., \(f'(AA) = f(AA) \times w_{AA}\), \(f'(Aa) = f(Aa) \times w_{Aa}\), and \(f'(aa) = f(aa) \times w_{aa}\).
Calculate the mean fitness of the population, \(\bar{w}\), by summing the weighted genotype frequencies: \(\bar{w} = f'(AA) + f'(Aa) + f'(aa)\).
Normalize the weighted genotype frequencies by dividing each by \(\bar{w}\) to get the genotype frequencies after selection. Then, calculate the new allele frequencies: \(p' = f'(AA) + \frac{1}{2} f'(Aa)\) and \(q' = f'(aa) + \frac{1}{2} f'(Aa)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Allele Frequency
Allele frequency refers to how common an allele is in a population, expressed as a proportion or percentage. In this question, p = 0.7 for allele A and q = 0.3 for allele a, meaning 70% and 30% of the alleles in the gene pool are A and a, respectively. These frequencies are the basis for predicting genotype frequencies and evolutionary changes.
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Codominance
Codominance occurs when both alleles in a heterozygote are fully expressed, resulting in a phenotype that shows traits of both alleles simultaneously. Unlike dominance/recessiveness, neither allele masks the other. This affects how genotype fitness values influence allele frequency changes in the population.
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Fitness and Selection Coefficients
Fitness (w) measures the reproductive success of a genotype relative to others. Here, w_AA = 1, w_Aa = 0.99, and w_aa = 0.98 indicate slight differences in survival or reproduction. These differences cause natural selection, altering allele frequencies over generations by favoring genotypes with higher fitness.
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