1、Genetic Variation in Populations,ZHANG Xian-Ning, PhD E-mail: Tel：13105819271; 88208367Office: A713, Research Building 2008/09,Learning Objectives,The nature and amount of genetic variation in human populations, and the role of genetic variation in liability to disease.,Ethnic Groups: Caucasoid, Neg
2、roid, Mongoloid,Genes in Human Populations,Study of distribution and frequency of genes in populations reasons for different gene frequencies in different populations burden of genetic disease is related to frequency and severity of genetic disorders- to an individual and to the population as a whol
3、e.,Mendelian population,An interbreeding population of sexually reproducing individuals sharing a common gene pool.,Gene pool, genotypes, and gene frequency,Gene pool : the genetic constitution of a population of a given organism.All the genes of all the individuals in population make up the gene po
4、ol. Genotypes: the genetic constitution of a single individual. Gene frequency (allelic frequency): the frequencies of the members of a pair of allele genes in a population. Phenotype frequency,Hardy Weinberg Equilibrium,The law that relates allele frequency to genotype frequency, used in population
5、 genetics to determine allele frequency and heterozygote frequency when the incidence of a disorder is known. assumes: large populationrandom matingno new mutationsno immigration in or out.,(Hardy Weinberg equation p2 + 2pq + q2 = 1)If all alleles at a locus are either A or a, frequency of “A” in th
6、e population is p, frequency of “a”in the population is q then p + q = 1and, frequency of AA is p2 aa is q2Aa is 2pq,(p2 + 2pq + q2 = 1)Observed frequency of recessive disease in population is q2 (e.g., frequency of PKU = 1/10000)q2 = 1/10 000 therefore: q = 1/100 (*this is not the carrier frequency
7、)Since p + q = 1p = 1 q = 1 1/100 = 99/100,Carrier frequency (2pq)2pq = 2 (99/100 x 1/100)= 2(1 x 1/100)= 2/100= 1/50Probability that a couple will have a child with PKU (i.e. q2) is therefore1/50 x 1/50 x = 1/10 000,Exceptions to Hardy Weinberg Assumptions,Migration introduction / loss of alleles M
8、utations may occur at different frequency in different populations 3. Small population size- genetic isolate / founder effect 4. Non-random mating- consanguinity- assortative mating(=non-randommating),Factors that disturb Hardy-Weinberg equilibrium,Exceptions to random mating Exceptions to constant
9、allele frequency Genetic drift Gene flow,Changes in Allele Frequency,Can be caused by: mutation (source of genetic variation) selection (phenotypes differ in biological fitness)(deleterious mutations may be removed by early death / lack of reproduction) migration (movement in or out),Selection If in
10、dividuals having certain genes are better able to produce mature offspring than those without them, the frequency of those genes will increase.,F(Fitness) - the ability to contribute to the gene pool of the next generationS(selective coefficient),Selection,Zero fitness of AD mutations Early lethalit
11、y Condition occurs only because of new mutations Appear sporadic rather than as AD pedigreeeg. osteogenesis imperfecta, type pare: achondroplasia fitness of 0.20 - frequency results from balance between “loss by selection”, and “gain by new mutation”,Heterozygote Advantage,Mutant allele has a high f
12、requency despite reduced fitness in affected individuals.Heterozygote has increased fitness over both homozygous genotypeseg. Sickle cell anemia.,Sickle Cell Anemia in West Africa,More influences on allele frequencies,Effects of population size and non-random mating- contrary to Hardy Weinberg assum
13、ptions, humans have often lived in small populations, isolated from their neighbors,Genetic Drift,Fluctuation in allele frequency due to chance in a small population.,Founder Effect,If an original member of a sub-population has a rare allele, it may become common in the sub-population (high carrier
14、frequency), resulting in high frequency of rare disease.e.g. Huntingdon disease in Lake Maracaibo, Venezuela (AD)Gyrate Atrophy in FinlandTyrosinemia in eastern Quebec 1/685 vs. 1/100 000,Non-random mating,Assortative Mating Pakistani or Cypriot population in UK Ashkenazi Jewish population “Deaf” po
15、pulation or “blind” population,Consanguinity/Inbreeding,when an individuals parents have one or more common ancestors, identifiable from a pedigree (or archival records)- because of genetic isolate, cultural practice, assortative mating-Increased likelihood of q 2,Clinical and Public Health Implicat
16、ions,increased population-specific frequencies of genetic disorderse.g. Saguenay region of Quebec- increased frequency of tyrosinemia, hypercholesterolemia, myotonic dystrophy- dedicated treatment, screening, education for the public and health care providers,Hardy-Weinberg equilibrium law,If two al
17、leles at a gene - A and a frequency of the A allele = p frequency of the a allele = q,First offspring： p2【AA】+2pq【Aa】+q2【aa】 Gametal frequency of first offspring：A = p2 + 1/2（2pq）; a = q2 + 1/2（2pq）,random mating,Gametal frequency of second offspring：A = p2 + 1/2（2pq） a = q2 + 1/2（2pq）,Hardy-Weinber
18、g equilibrium implies that gene and genotype frequencies are constant from generation to generation. If disequilibrium occurs, equilibrium will be reestablished after one generation of random mating.,H-W law rests on several assumptions: large population random mating no mutations no migration betwe
19、en populations no selection - all genotypes reproduce with equal success,Hardy-Weinberg equilibrium lawIf two alleles at a gene - A and a frequency of the A allele = p frequency of the a allele = q 3. The two fractions add up to totality p + q = l 4. the proportions of the three genotypes:AA, Aa and
20、 aa are p2: 2pq : q2 5. Hardy-Weinberg formula: p2 (AA) + 2pq (Aa) + q2 (aa) = 1,Applications of Hardy-Weinberg law: How to judge the population equilibrium?,For example, consider three hypothetical populations: AA 60 persons, aa 20 persons, Aa 20 persons, how about the population balance?,The gene
21、frequency in the populations is:A = p = 0.6 + 0.2/2 = 0.7a = q = 0.2 + 0.2/2 = 0.3 If population balance, they will be :(AA) p2+ 2pq (Aa)+( aa)q2 = 1 So, 0.49 + 0.42 + 0.09 = 1,But,After one generation of random mating, each of the three populations will have the same genotypic frequencies: A = p =
22、0.6 + 0.2/2 = 0.7a = q = 0.2 + 0.2/2 = 0.3AA (p2) = 0.49 Aa (2pq) = 0.42 aa (q2 ) = 0.09,How to calculate the allele frequency and the genotype frequency of the population?,1. AR traits: If the frequency of a recessive trait is known, it is possible to calculate allele frequencies and genotype frequ
23、encies using the Hardy Weinberg equation and its assumptions as follows: 1 in 1700 US Caucasian newborns have cystic fibrosis which means that the frequency of homozygotes for this recessive trait is q = 1/1700 = 0.00059 The square root of the frequency of recessives is equal to the allele frequency
24、 of the CF allele q = 0.024,iii The frequency of the normal allele is equal to 1 - the frequency of the Cf allelep = 1- q = 1 - 0.024 = 0.976 iv The frequency of carriers (heterozygotes) for the CF allele is 2pq = 2 (0.976)(0.024) = 0.047 or 1/21 v The frequency of homozygotes for the normal allele
25、isp = (0.976) = 0.953 vi Thus the population is composed of three genotypes at the calculated frequencies ofhomozygous normal = 0.952576 heterozygous carriers = 0.046848 homozygous affected = 0.000588,2. Frequency of sex-linked genes The distribution of the recessive phenotype (aa) is equal to q Exa
26、mple: Color blindnessq (XaY)= frequency of Male sufferer = 0.07So the frequency of genotype in Female Xa Xa = q2 = ( 0.07 )2XA XA= p2 = (1-q)2 = (1-0.07)2,Gene flow,The exchange of genes between different populations.,Why are some people resistant to HIV?,Duncan SR, et al. Reappraisal of the histori
27、cal selective pressures for the CCR5-Delta32 mutation. J Med Genet. 2005;42(3):205-208.,HIV strains are unable to enter macrophages that carry the CCR5-Delta32 deletion( CCR5); the average frequency of this allele is 10% in European populations. A mathematical model based on the changing demography
28、of Europe from 1000 to 1800 AD demonstrates how plague epidemics, 1347 to 1670, could have provided the selection pressure that raised the frequency of the mutation to the level seen today. It is suggested that the original single mutation appeared over 2500 years ago and that persistent epidemics of a haemorrhagic fever that struck at the early classical civilizations served to force up the frequency to about 5x10-5 at the time of the Black Death in 1347.,Fisher Hardy Weinberg,Ching Chun Li (19122003): A Hero of GeneticsAm J Hum Genet 74:789792, 2004,
