Topic 1 Evolution: Genetic Equilibrium (Hardy-Weinberg Equation) Flashcards
1
Q
- If allele frequencies remain constant from generation to generation, then there is no evolution. In this situation, the Hardy-Weinberg equation can be used to determine allele frequencies for a population.
A
Note
2
Q
- p2 + 2pq + q2 = 1 (all individuals sum to 100%)
- p + q = 1 (all alleles sum to 100%)
- p = frequency of the dominant allele
- q = frequency of the recessive allele
- p2 = frequency of homozygous dominant individuals
- q2 = frequency of homozygous recessive individuals
- 2pq = frequency of heterozygous individuals
A
Hardy-Weinberg Equation
3
Q
- Both equations must sum to 1 in order for the population to be in Hardy-Weinberg Equilibrium.
A
Note
4
Q
- No mutations - no new alleles can be introduced to the population
- No natural selection - the environment is not impacting allele frequencies, and so traits are neutral
- No gene flow - also happens as a result of no migration. An isolated population will have no gene flow
- Large populations - this decreases the effects of genetic drift
- Random mating - this decreases the chance of any allele from changing in frequency
A
Requirements to be in Hardy-Weinberg Equilibrium
5
Q
- A plant population has 84% red flowers and 16% white flowers. The red allele (R) is dominant and the white allele (r) is recessive.
1. 16% white flowers can be written as: q2 = 0.16. Taking the square root to find q will equal 0.4
2. 0.4 can then be plugged into “p + q = 1” to get “p + 0.4 = 1.” Solving for p equals 0.6
3. Plugging 0.6 into p2 results in 0.36 Plugging 0.6 and 0.4 into 2pq results in 0.48 - p = 0.6 (60% of the alleles are R)
- q = 0.4 (40% of the alleles are r)
- p2 = 0.36 (36% of the individuals are homozygous dominant and are red)
- 2pq = 0.48 (48% of the individuals are heterozygous and are red)
- q2 = 0.16 (16% of the individuals are homozygous recessive and are white)
A
Example of a Hardy-Weinberg Problem
