AGB 121: Principles of Animal Genetics and Population Genetics (2+1)

APPLICATIONS OF HARDY-WEINBERG LAW

Hardy-Weinberg law has the following applications:

• Calculation of frequencies of recessive and dominant genes in a population
• Calculation of frequency of carriers or heterozygotes in a population
• To test for agreement with a population in Hardy -Weinberg equilibrium

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).

	Genotype				Gene frequency
	BB	Bb	bb	Total	p	q
Number Observed	500	400	100	1000	{500+(400x0.5)}/1000 =0.7	1-0.7 = 0.3
Number Expected	(0.7)2x1000 = 490	(2x0.7x0.3)x1000 = 420	(0.3)2x1000 = 90	1000
Difference (O –E)	10	-20	10
(O-E)2 / E	0.204	0.952	1.111

• 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

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