Chapter 25: Problem 30
The forward mutation rate for piebald spotting in guinea pigs is \(8 \times 10^{-5}\); the reverse mutation rate is \(2 \times 10^{-6} .\) If no other evolutionary forces are assumed to be acting, what is the expected frequency of the allele for piebald spotting in a population that is in mutational equilibrium?
Short Answer
Step by step solution
Understanding Mutation Equilibrium
Set Up the Mutational Equilibrium Equation
Express Allele Frequencies
Solving for Allele Frequency \( q \)
Substitute the Mutation Rates
Calculate the Frequency \( q \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Allele Frequency
In a population, if an allele is more frequent, it is said to be more common. Usually, the sum of the frequencies of all alleles for a gene is equal to one. For example, if there are two alleles in a population, like 'A' and 'a', then their frequencies, denoted as 'p' and 'q', respectively, add up to one:
- p + q = 1
Mutation Rates
The forward mutation rate is how often a change occurs from one form of an allele to another, such as changing from the non-piebald to the piebald allele in guinea pigs. The reverse mutation rate is the frequency of changes from a mutant form back to the original form.
- Forward mutation rate (u): mutation from non-piebald to piebald
- Reverse mutation rate (v): mutation from piebald to non-piebald
Genetic Equilibrium
When at mutation equilibrium, the influx of new mutations converting one allele to another (e.g., non-piebald to piebald) is exactly balanced by the reverse mutations converting it back. This results in stable allele frequencies over time.
- The equilibrium condition: \(up = vq\)
- Solving the equation helps determine stable allele frequencies