Autosomal locusX-linked locusBiology 311 Human Genetics Fall 2004Lecture 6 Hardy-Weinberg EquilibriumReading: Chap. 4 pp. 117-119Lecture Outline:1. Allele frequencies in populations2. Hardy Weinberg equilibrium3. Disease prediction/Genetic counselingLecture1. Allele frequenciesPopulation: Group of interbreeding individuals.Gene Pool: All the alleles for a particular gene in a population.Gene frequency=allele frequency (more precise phrase)=proportion of all alleles for gene A of one type (such as A1)Pick an allele at random from the gene poolChance= p that allele is A1q that allele is A2Pick a second allele at randomChance= p that allele is A1 q that allele is A2Chance that both alleles were A1 = p x p =p2Chance that both alleles were A2 = q x q = q2Chance that first allele was A1, second was A2 = pqChance that first allele was A2, second was A1 = qpOverall chance of one A1 and one A2 allele is 2pq.1A1A1 A2A2 A1A2A1A2 A2A2A2. Hardy Weinberg equilibriumGene frequencies and genotypes of a population can be predicted using probabilities.Autosomal locusGenotype FrequencyA1A1 p2A1A2 2pqA2A2 q2X-linked locusFemales Frequency Males FrequencyA1A1 p2A1 pA1A2 2pq A2 qA2A2 q2Assumptions:- Random matingo Inbreeding=non-random mating decreases heterozygosity; and increases homozygosity3. Disease prediction/Genetic counselingSee Box 4.6a. Calculate carrier frequencyAn autosomal recessive condition affects 1 newborn in 10,000. What is the expected frequency of carriers?unaffected unaffected affectedphenotypes AA Aa aafrequencies p22pq q2Affected individuals = q2 = 1/10,000 ____√ q2 = q = 1/1001 in 100 alleles at the A locus are a99 in 100 are AThe carrier frequency, 2pq = 2 (99/100)(1/100) = 0.022If a parent of an affected child remarries, what is the risk of producing an affected child in the new marriage?Apply the product ruleProbability that parent that had affected Risk of new spouse being probability ofchild is a carrier carrier affected child1 x 0.02 x ¼ = .005X-linked recessive red-green color blindness affects 1 in 12 British males. What proportion of females will be carriers? What proportion will be affected?Males A=normala=colorblindFemales AA or Aa = normal aa= colorblindFemales Frequency Males FrequencyAA p2A pAa 2pq a q = 1/12aa q2From males, q=1/12 therefore p=11/12.Female carriers =2pq =2 x 11/12 x 1/12 =22/144Females affected =q2 = 1/12 x 1/12 = 1/144Other forces that affect allele frequenciesa. mutation μ A a υμ= mutation rate/gene/generationυ= back mutation rate/gene/generationb. Natural selectionCan act on dominant or recessive traitCan have heterozygote superiority Aa>AA or aa3Can have heterozygote inferiority AA>Aa and aa>Aas=coefficient of selections=0, no selections=1, lethal geneEstimate mutation rate- Assume mutation and selection are in equilibriumfor autosomal recessive traitμ = sq2 mutation rate=selection coefficient x allele frequency of recessivefor rare autosomal dominant traitμ =sp mutation rate = selection coefficient x allele frequency of dominant allelefor X-linked recessive traitu=sq/3 mutation rate = selection coefficient x allele frequency of recessive trait/ only 1/3 of the sex chromosomes (those in males) are affected- For most genetic diseases, the mutation rates are actually rather low. Natural selection would actually drive one allele to extinction, other to fixation. - Most likely the trait remains in the population due to natural selection acting on both dominant and recessive phenotypes =heterozygote