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What is the Hardy-Weinberg equilibrium?

What is the Hardy-Weinberg equilibrium?. The Hardy-Weinberg equilibrium (law or distribution) describes the relationship between alleles (genes) and genotypes

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What is the Hardy-Weinberg equilibrium?

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  1. What is the Hardy-Weinberg equilibrium? The Hardy-Weinberg equilibrium (law or distribution) describes the relationship between alleles (genes) and genotypes It states that allele and genotype frequencies in a population remain constant (are in equilibrium) from generation to generation unless specific disturbing influences are introduced For a bi-allelic system at locus A there are 2 alleles A1 and A2 with allele frequencies p and q the sum of all alleles must equal 1 therefore p + q = 1

  2. For a bi-allelic system at locus A there are three genotypes genotype A1A1 A1A2 A2A1 A2A2 frequency p x p p x q q x p q x q p2 2pq q2 the sum of all genotypes must equal 1 therefore p2 + 2pq + q2 = 1 this can be simplified down to (p +q)2 = 1

  3. What factors can distort this equilibrium? Hardy-Weinberg calculations are invalid if individuals genes/alleles are not drawn independently and at random from the gene pool This occurs when there is non-random (assortative) mating this is generally due to inbreeding/consanguinity skews the gene pool and increases the likelihood of allele sharing Other factors that can effect the equilibrium include migration large scale immigration of people from a population with a different allele frequencies would distort the equilibrium mortality individuals with a certain genotype may be more likely to be lost from a population depleting it of that genotype

  4. The Hardy-Weinberg equilibrium only applies if these assumptions are true 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 6.     Everyone produces the same number of offspring 7.     There is no migration in or out of the population If one or all of these assumptions occur in a population it will not evolve - this is not the case in naturally occurring populations Hardy-Weinberg Paradox: Gene pool frequencies are inherently stable (do not change), however evolution should be expected in all populations at all times www.biol.andrews.edu/gen/l16.htm (Biology Dept, Andrew University, Michigan)

  5. How can we use the Hardy-Weinberg equilibrium to calculate carrier frequencies of recessive diseases? Example – Cystic fibrosis CF has a frequency in the Caucasian population of 1/2500 CF is an autosomal recessive disease – 2 mutant alleles = clinical phenotype Therefore the proportion of the population homozygous for the mutant allele is q2 and this is equal to the incidence of the disease q2 = 1/2500 q = √1/2500 q = 0.02 p + q = 1 p = 1 - q p = 1-0.02 p = 0.98 the carrier frequency of an autosomal recessive disease is equal to the incidence of heterozygotes which is 2pq carrier frequency = 2pq = 2x 0.98 x 0.02 = 0.0392 = 1/25.5

  6. How do the calculations differ for multiple alleles? For a bi-allelic system (p + q)2 = 1 and p + q = 1 This can be expanded to p2 + 2pq + q2 = 1 For a tri-allelic system (p + q + r)2 = 1 and p + q + r = 1 This can be expanded to p2 + q2 + r2 + 2pq + 2pr + 2qr = 1 For a tetra-allelic system (p + q + r + s)2 = 1 and p + q + r + s = 1 This can be expanded to p2 + q2 + r2 + s2 + 2pq + 2pr + 2qs + 2qr + 2qs + 2rs = 1

  7. Example of a tri-allelic system – ABO blood group If the allele frequencies for the ABO blood group are O = 0.6 A = 0.3 B = 0.1 p = 0.6 q = 0.3 r = 0.1 Frequency of blood group O = p2 = 0.62 = 0.36 Frequency of blood group A = q2 + 2pq = 0.32 + 2 x 0.6 x 0.3 = 0.45 Frequency of blood group B = r2 + 2pr = 0.12 + 2 x 0.6 x 0.1 = 0.13 Frequency of blood group AB = 2qr = 2 x 0.3 x 0.1 = 0.06

  8. How do the calculations differ for X-linked inheritance? For X-linked alleles and loci (in a bi-allelic system) Males Females Genotypes A1 A2 A1A1 A1A2 A2A2 Frequencies p q p2 2pq q2 Therefore for males the genotype frequencies are equal to the allele frequencies

  9. Example – red-green colour blindness X-linked red-green colour blindness affects 1in 12 males so q = 1/12 therefore p = 11/12 For females the proportion of affected females is q2 q2 = (1/12) 2 = 1/144 = 0.007 or 0.7% For females the proportion of carriers is 2pq 2pq = 2 x 11/12 x 1/12 = 22/144 = 0.15 or 15% The frequency of affected females is lower than the calculation predicts because there are 2 forms of red-green colour blindness due to 2 adjacent loci

  10. References Strachan and Read 2010. Human Molecular Genetics 4th Edition. Key words Alleles Genotypes Allele frequencies Genotype frequencies Carrier frequencies Homozygotes Heterozygotes Random and non-random mating

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