Applications of Hardy – Weinberg Law
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Hardy-Weinberg law has the following applications:
Calculation of frequencies of recessive and dominant genes in a population
- Gene frequencies can be determined from their genotype frequencies, but for this it is necessary to know the frequencies of all three genotypes.
- When there is complete dominance at a locus the expression q = Q + ½ H cannot be used as the classification of genotypes is not possible.
- The heterozygote genotype frequency cannot be estimated, as it cannot be phenotypically distinguished from dominant homozygote.
- e.g.: Stem length in garden pea plant
- Tall - dominant ( TT, Tt )
- Dwarf - recessive ( tt )
- However, if all the genotypes are in Hardy – Weinberg equilibrium, we need not know the frequencies of all genotypes.
- Let T - be a dominant gene with a frequency of p and t - be a recessive gene with a frequency of q ; then the frequency of tt homozygote is q2 and therefore the frequency of recessive allele will be square root of the homozygote frequency.
- For this estimation to be valid there should not be selective elimination of recessive homozygotes.
- Since p + q = 1
- p (frequency of dominant allele) = 1 - q
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Calculation of frequency of carriers or heterozygotes in a population
- It is often of interest to know the frequency of heterozygotes or carriers of recessive genes or recessive abnormalities. This can be calculated if the gene frequency is known.
- If Hardy-Weinberg equilibrium is assumed the frequency of heterozygotes among all individuals in the population can be estimated from the formula 2pq = 2q (1-q).
- Example: if q2 is 0.04 then the gene frequency is q = (0.04)1/2 = 0.2
- The frequency of heterozygote is 2q (1-q) = 2 × 0.2 × (1-0.2) = 2× 0.2× 0.8 = 0.32
- 0.32 or 32% is of carriers or hererozygotes.
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To test for agreement with a population in Hardy -Weinberg equilibrium
- If data are available for a locus where all the genotypes are recognizable then the observed genotype frequencies are used to test for Hardy-Weinberg equilibrium.
- According to the Hardy-Weinberg law, the genotype frequencies of progenies are determined from the gene frequencies of their parents.
- If the population is in H-W equilibrium, frequency is same in progenies as in parents.
- From the gene frequency expected genotype frequencies are calculated. From them the expected numbers are arrived. The agreement between the expected and observed numbers is tested using Chi-square test.
- Example:
- In a given population of randomly mating gerbils, 500 homozygote brown (BB), 400 heterozygote brown (Bb) and 100 homozygote black (bb) gerbils were observed.
- Test whether this population is in Hardy-Weinberg equilibrium.
- NOTE: the black allele (b) is recessive to the wild type brown (agouti) allele (B).
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Genotype
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BB
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Bb
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bb
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Total
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p
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q
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Number Observed
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500
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400
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100
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1000
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{500+(400x0.5)}/1000 =0.7
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1-0.7 = 0.3
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Number Expected
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(0.7)2x1000 = 490
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(2x0.7x0.3)x1000 = 420
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(0.3)2x1000 = 90
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1000
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Difference (O –E)
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10
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-20
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10
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(O-E)2 / E
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0.204
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0.952
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1.111
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- Chi-square value = 2.267
- P value = 0.32
- The discrepancy is not significant and could easily have arisen by chance in the sampling.
- We conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population is in Hardy-Weinberg equilibrium
(TOP)
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Last modified: Saturday, 17 December 2011, 11:12 AM