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Title: How do Mutations and recombinations further violate the HW equilibrium? Post by: datasian on Nov 11, 2010 I understand mutations and recombination violate the hardy weinberg equilibrium, but what I don't understand is how they are in violation of HW equilbrium.
I'm thinking it's because mutations and recombination can introduce new alleles in the population and this results in creating variation in the gene pool. More favorable mutations will accumulate in the population, resulting in evolutionary change while deleterious mutations are removed from the population by natural selection. The result of evolutionary change is what violates the HW equilibrium. Title: Re: How do Mutations and recombinations further violate the HW equilibrium? Post by: bio_man on Nov 11, 2010 Let's take allele B and b for instance.
The frequency of gene B and its allele b will not remain in Hardy-Weinberg equilibrium if the rate of mutation of B -> b (or vice versa) changes. By itself, this type of mutation probably plays only a minor role in evolution; the rates are simply too low. However, gene duplication probably has played a major role in evolution. In any case, evolution absolutely depends on mutations because this is the only way that new alleles are created. After being shuffled in various combinations with the rest of the gene pool, these provide the raw material on which natural selection can act. In terms of recombination, do you mean gene flow, genetic drift, nonrandom mating, or natural selection? Read this response: https://biology-forums.com/index.php?topic=402.0 (https://biology-forums.com/index.php?topic=402.0) View this video: Hardy-Weinberg Principle (http://www.youtube.com/watch?v=4Kbruik_LOo&feature=player_embedded#ws) |