What does the Hardy weinberg equilibrium states?How is the hardy weinberg equation used in population genetic?

What does the Hardy weinberg equilibrium states?How is the hardy weinberg equation used in population genetic?

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  • 1 decade ago
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    In population genetics, the Hardy–Weinberg principle states that the genotype frequencies in a population remain constant or are in equilibrium from generation to generation unless specific disturbing influences are introduced.

    For a population to be in Hardy-Weinberg equilibrium then there must be:

    1. no selection

    2. no mutation

    3. no migration (* the previous answer missed this out)

    4. a large population (no genetic drift)

    5. random mating (no inbreeding etc)

    [See http://en.wikipedia.org/wiki/Hardy-Weinberg_princi... ]

    The Hardy-Weinberg principle is like a Punnett square for populations, instead of individuals. A Punnett square can predict the probability of offspring's genotype based on parents' genotype or the offsprings' genotype can be used to reveal the parents' genotype. Likewise, the Hardy-Weinberg principle can be used to calculate the frequency of particular alleles based on frequency of, say, an autosomal recessive disease.

    For example, imagine a gene for eye colour. Let's call the alleles

    B - dominant, brown eyes

    b - recessive, blue eyes

    Allele frequencies

    f(B) = p

    f(b) = q

    Since all individuals MUST have one of the two alleles

    p + q = 1

    Genotype frequencies

    f(BB) = p^2

    f(Bb) = 2pq

    f(bb) = q^2

    [These frequencies cal be worked out using a Punnett square. See http://en.wikipedia.org/wiki/Hardy-Weinberg_princi... ]

    Since all individuals MUST have one of the three genotypes

    p^2 + 2pq + q^2 = 1

    Imagine we observe a population of 1,000 rabbits, and count 40 with the blue eye phenotype. We know that all rabbits showing the recessive phenotype MUST have the homozygous recessive genotype.

    f(bb) = 40 / 1000 = 0.04

    q^2 = 0.04

    q = 0.2 (20%)

    p + q = 1

    p = 1 - q

    p = 1 - 0.2

    p = 0.8 (80%)

    Now that we know the allele frequencies we can work out what proportion of the individuals showing the dominant phenotype are homozygous or heterozygous.

    f(Bb) = 2pq = 2 x 0.2 x 0.8 = 0.32 (32% or 320 individuals)

    f(BB) = p^2 = 0.8^2 = 0.64 (64% or 640 individuals)

    [Note that p^2 + 2pq + q^2 = 1

    640 + 320 + 40 = 1,000]

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  • finey
    Lv 4
    3 years ago

    Hardy Weinberg Principle Equation

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  • 1 decade ago

    The Hardy Weinberg equilibrium is a theoretical point in population dynamics in which 5 things are true of the population in question:

    1. mutation is not occurring

    2. natural selection is not occurring

    3. the population is infinitely large

    4. all members of the population breed

    5. all mating is totally random

    I'll let you consult the site to answer any further questions

    cheers

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  • Anonymous
    6 years ago
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