Example:In a population of 20 human individuals, 4 people haveblue-colored eyes because they are homozygous for the recessive Ballele at a particular "blue eyes"locus, where the alternative allele isA.Molecular analyses showed that 12 individuals are of genotypeAA, 4 are of genotypeAB.eggs(p):frequencyof allele"A"=0.7Genotypepq(q):frequencyofallele“B"=0.30.70.3freguencies of the"AA"“AB"nextgeneration:0.7p49%21%sperm“AB"“BB"0.3q9%21%homozygote“AA"=p?=0.49homozygote“BB"=q=0.09heterozygote"AB"=2(pq)=0.421.00Fig.20.321-16
Copyright © The McGraw-Hill Companies, Inc. Permission required to reproduce or display 21-16 Genotype frequencies of the next generation: Example: In a population of 20 human individuals, 4 people have blue-colored eyes because they are homozygous for the recessive B allele at a particular “blue eyes” locus, where the alternative allele is A. Molecular analyses showed that 12 individuals are of genotype AA, 4 are of genotype AB. Fig. 20.3
The genotype frequencies of zygotes arising in a largepopulation of sexually reproducing diploid organisms are p? forAA,2pqforAB, andq?forBBAllelefrequencyequation:p+q=1Genotypefrequency equation:p?+2pq+q?=121-17
Copyright © The McGraw-Hill Companies, Inc. Permission required to reproduce or display 21-17 The genotype frequencies of zygotes arising in a large population of sexually reproducing diploid organisms are p2 for AA, 2pq for AB, and q2 for BB. Allele frequency equation: p+ q = 1 Genotype frequency equation: p2 + 2pq + q2 = 1
An example: Predicting the frequency of albinismAnalbinoAfricangirl21-18
Copyright © The McGraw-Hill Companies, Inc. Permission required to reproduce or display 21-18 An example: Predicting the frequency of albinism An example: Predicting the frequency of albinism An albino African girl
Populationof100,o00people98.100 AA individuals1,800Aacarriers100 aaalbinosThe frequency of allelesAallelefrequency:(98,100×2+1,800)/200,000=0.99-p=0.99a allelefrequency:(1,800+100×2)/200,000=0.01q=0.0121-19
Copyright © The McGraw-Hill Companies, Inc. Permission required to reproduce or display 21-19 Population of 100,000 people Population of 100,000 people 98,100 AA individuals individuals 1,800 Aa carriers carriers 100 aa albinos albinos The frequency of alleles The frequency of alleles A allele frequency: (98,100 allele frequency: (98,100 2 + 1,800)/200,000 = 0.99 + 1,800)/200,000 = 0.99 p = 0.99 a allele frequency: (1,800 + 100 allele frequency: (1,800 + 100 2)/200,000 = 0.01 )/200,000 = 0.01 q = 0.01
Hardy-Weinberg equation for the albino gene in the populationis:■p2+2pq+g=(0.99)2+2×(0.99×0.01)+0.01)2=0.9801+0.0198+0.0001=1Next generation of 100,000 people:100.000x0.9801AAindividuals=98.010100,000×0.0198Aaindividuals=1,98010100,000×0.0001aaindividuals=21-20
Copyright © The McGraw-Hill Companies, Inc. Permission required to reproduce or display 21-20 Hardy-Weinberg equation for the alb Weinberg equation for the albino gene in the population ino gene in the population is: p2 + 2pq + q2 = (0.99) = (0.99)2 + 2(0.99 0.01) + (0.01) 0.01) + (0.01)2 = 0.9801 + 0.0198 + 0.0001 = 1 0.9801 + 0.0198 + 0.0001 = 1 Next generation of 100,000 people: Next generation of 100,000 people: 100,000 100,000 0.9801 AA individuals = 98,010 individuals = 98,010 100,000 100,000 0.0198 Aa individuals = 1,980 individuals = 1,980 100,000 100,000 0.0001 aa individuals = 10 individuals = 10