1、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, Negroid, Mongoloid,Genes in Human Populations,Study of distribution and frequency of genes in populations reasons for different ge

2、ne 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 whole.,Mutations may or may not result in an expressed phenotype.,- Mutations can alter RNA expression, processing and/or stability

3、.,Mutations that have no phenotype are called neutral mutations.,A mutation is a structural change in genomic DNA sequence due to errors in DNA replication or repair.,- Mutations can also affect protein expression, processing, stability.,Mutations can be inherited (genetic/germline mutations) not in

4、herited (somatic mutations),Mutation,Mutations that are propagated and maintained in the population at relatively high frequencies are called polymorphisms. Polymorphism is defined as the existence of two or more alleles, where the rare allele appears with a frequency greater than 1% in the populati

5、on. Most mutations are quickly lost from population due to deleterious effects (natural selection) or genetic drift (random fluctuations). Mutations may become polymorphisms due to selective advantage (heterozygotes for hemoglobin sickle cell mutation are more resistant to malaria) or genetic drift

6、(founder effect, small group of individuals found a new population).,Polymorphisms,Mendelian population,An interbreeding population of sexually reproducing individuals sharing a common gene pool.,Gene pool, genotypes, and gene frequency,Genotype: the genetic constitution of a single individual. Gene

7、 pool : the genetic constitution of a population of a given organism. OR: All the genes of all the individuals in population make up the gene pool.,Genotype frequency: f(AA) = Number of individuals with AA / Total number of individuals in the pop f(Aa) = ? f(aa) = ? Gene frequency (allelic frequency

8、): the frequencies of the members of a pair of allele genes in a population. f(Allele) = Number of copies of a given allele / Sum of counts of all alleles in the pop p= f(A) = (2 count of AA) + (count of Aa) / 2 total number of individuals OR: p= f(A) = f(AA) + (1/2 f(Aa) ) q= f(a) = 1- p,Hardy Wein

9、berg Law,States the relationships between the frequency of alleles at a locus, and the genotypes resulting from these alleles. assumes: large population random mating no new mutations no immigration in or out no natural selection,(Hardy Weinberg equation: p2 + 2pq + q2 = 1) If all alleles at a locus

10、 are either A or a, frequency of “A” in the population is p, frequency of “a”in the population is q then p + q = 1 and, frequency of AA is p2 aa is q2 Aa is 2pq,(p2 + 2pq +q2 = 1) Observed frequency of recessive disease in population is q2 (e.g., frequency of PKU = 1/10 000) q2 = 1/10 000 q = 1/100

11、(*this is not the carrier frequency!) p + q = 1 p = 1 q = 1 1/100 = 99/100,Carrier frequency (2pq) 2pq = 2 (99/100 x 1/100) = 2(1 x 1/100) = 2/100 = 1/50 Probability that a couple will have a child with PKU (i.e. q2) is therefore 1/50 x 1/50 x = 1/10,000,Exceptions to Hardy Weinberg Assumptions,Muta

12、tion-selection equilibrium may occur at different frequency in different populations Heterozygote Advantage Founder effect and genetic drift Small population size, genetic isolate Non-random mating consanguinity; assortative mating(=non-random mating) Migration and gene flow introduction / loss of a

13、lleles,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 individuals having certain genes ar

14、e 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 generation S(selective coefficient),Selection,Zero fitness of AD mutations Early lethality Condition occurs only because

15、of new mutations Appear sporadic rather than as AD pedigree eg. osteogenesis imperfecta, type 2. compare: achondroplasia fitness of 0.20 - frequency results from balance between “l(fā)oss by selection”, and “gain by new mutation”,Heterozygote Advantage,Mutant allele has a high frequency despite reduced

16、fitness in affected individuals. Heterozygote has increased fitness over both homozygous genotypes eg. Sickle cell anemia.,Heterozygote Advantage,Gemmell NJ and Slate J. PLoS One. 2006;1:e125.,21,Sickle Cell Anemia in West Africa,Founder Effect,If an original member of a sub-population has a rare al

17、lele, it may become common in the sub-population (high carrier frequency), resulting in high frequency of rare disease. e.g. Huntingdon disease in Lake Maracaibo, Venezuela (AD) Gyrate Atrophy in Finland Tyrosinemia in eastern Quebec 1/685 vs. 1/100 000,Genetic Drift,Fluctuation in allele frequency

18、due to chance in a small population.,Non-random mating,Assortative Mating Pakistani or Cypriot population in UK Ashkenazi Jewish population “Deaf” population or “blind” population,Consanguinity/Inbreeding,when an individuals parents have one or more common ancestors, identifiable from a pedigree (or

19、 archival records) - because of genetic isolate, cultural practice, assortative mating -Increased likelihood of q 2,Clinical and Public Health Implications,increased population-specific frequencies of genetic disorders e.g. Saguenay region of Quebec - increased frequency of tyrosinemia, hypercholest

20、erolemia, myotonic dystrophy - dedicated treatment, screening, education for the public and health care providers,Gene flow,The exchange of genes between different populations.,Hardy-Weinberg equilibrium law,If two alleles at a gene - A and a frequency of the A allele = p frequency of the a allele =

21、 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-Weinberg equilibrium implies that gene and genotype frequencies are constant from generat

22、ion 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 between populations no selection - all genotypes reproduce with equal success,Applicati

23、ons of Hardy-Weinberg law: How to judge the population equilibrium?,For example, consider the hypothetical populations: AA 60 persons, aa 20 persons, Aa 20 persons, how about the population balance?,The gene frequency in the populations is: A = p = 0.6 + 0.2/2 = 0.7 a = q = 0.2 + 0.2/2 = 0.3 If popu

24、lation 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 = 0.6 + 0.2/2 = 0.7 a = q = 0.2 + 0.2/2 = 0.3 AA (p2) = 0.49 Aa (2pq) = 0.42 aa (q2 ) = 0.0

25、9,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 frequencies using the Hardy Weinberg equation and its assumptions as follows: 1 in 1700 US C

26、aucasian 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 of the CF allele q = 0.024,iii The frequency of the normal allele is equal to 1 - the

27、frequency of the Cf allele p = 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 is p = (0.976) = 0.953 vi Thus the population is composed of three genotypes at the ca

28、lculated frequencies of homozygous 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 Example: Color blindness q (XaY)= frequency of Male sufferer = 0.07 So the frequency of

29、 genotype in Female Xa Xa = q2 = ( 0.07 )2 XA XA= p2 = (1-q)2 = (1-0.07)2,Summary of H-W calculations,Why are some people resistant to HIV?,Duncan SR, et al. Reappraisal of the historical 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