医学遗传学种群的遗传与变异精.ppt

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1、医学遗传学种群的遗传与变异第1页，本讲稿共42页Learning ObjectivesThe nature and amount of genetic variation in human populations,and the role of genetic variation in liability to disease.第2页，本讲稿共42页Ethnic Groups:Caucasoid,Negroid,Mongoloid第3页，本讲稿共42页Genes in Human Populations Study ofdistribution and frequency of genes i

2、n populationsreasons for different gene frequencies in different populationsburden of genetic disease is related to frequency and severity of genetic disorders-to an individual and to the population as a whole.第4页，本讲稿共42页第5页，本讲稿共42页Mendelian populationAn interbreeding population of sexually reproduc

3、ing individuals sharing a common gene pool.第6页，本讲稿共42页Gene pool,genotypes,and gene frequencyGene pool:the genetic constitution of a population of a given organism.All the genes of all the individuals in population make up the gene pool.Genotypes:the genetic constitution of a single individual.Gene f

4、requency(allelic frequency):the frequencies of the members of a pair of allele genes in a population.Phenotype frequency第7页，本讲稿共42页Hardy Weinberg EquilibriumThe law that relates allele frequency to genotype frequency,used in population genetics to determine allele frequency and heterozygote frequenc

5、y when the incidence of a disorder is known.assumes:large populationrandom matingno new mutationsno immigration in or out.第8页，本讲稿共42页 (Hardy Weinberg equation p2+2pq+q2 =1)If all alleles at a locus are either A or a,frequency of“A”in the population is p,frequency of“a”in the population is q then p+q

6、=1 and,frequency of AA is p2 aa is q2 Aa is 2pq AAAa aAaap22pqq2homozygous normalcarriersaffected第9页，本讲稿共42页(p2+2pq +q2=1)Observed frequency of recessive disease in population is q2 (e.g.,frequency of PKU=1/10000)q2=1/10 000therefore:q =1/100(*this is not the carrier frequency)Since p+q=1p=1 q=1 1/1

7、00=99/100第10页，本讲稿共42页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 therefore 1/50 x 1/50 x =1/10 000第11页，本讲稿共42页GenotypeAAAa+aAaaFrequencyp22pqq2PhenotypeNormalCarriersAffected第12页，本讲稿共42页Exceptions to Hardy Weinberg

8、 Assumptions1.Migration introduction/loss of alleles2.Mutations may occur at different frequency in different populations3.Small population size-genetic isolate/founder effect4.Non-random mating-consanguinity-assortative mating(=non-random mating)第13页，本讲稿共42页Factors that disturb Hardy-Weinberg equil

9、ibrium 1.Exceptions to random mating 2.Exceptions to constant allele frequency3.Genetic drift4.Gene flow 第14页，本讲稿共42页Changes in Allele FrequencyCan be caused by:mutation(source of genetic variation)selection(phenotypes differ in biological fitness)(deleterious mutations may be removed by early death

10、/lack of reproduction)migration(movement in or out)第15页，本讲稿共42页SelectionIf individuals 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 generation S(se

11、lective coefficient)第16页，本讲稿共42页SelectionZero fitness of AD mutations Early lethalityCondition occurs only because of new mutationsAppear sporadic rather than as AD pedigreeeg.osteogenesis imperfecta,type pare:achondroplasia fitness of 0.20 -frequency results from balance between“loss by selection”,

12、and“gain by new mutation”第17页，本讲稿共42页第18页，本讲稿共42页Heterozygote AdvantageMutant allele has a high frequency despite reduced fitness in affected individuals.Heterozygote has increased fitness over both homozygous genotypes eg.Sickle cell anemia.第19页，本讲稿共42页Sickle Cell Anemia in West AfricaA/A“normal”Su

13、sceptibility to malariaA/SHeterozygous carrier-Resistant to malaria-Anemia rareS/SSickle cell diseaseSevere anemia,often fatal.第20页，本讲稿共42页More influences on allele frequenciesEffects of population size and non-random mating -contrary to Hardy Weinberg assumptions,humans have often lived in small po

14、pulations,isolated from their neighbors第21页，本讲稿共42页Genetic DriftFluctuation in allele frequency due to chance in a small population.第22页，本讲稿共42页Founder EffectIf an original member of a sub-population has a rare allele,it may become common in the sub-population(high carrier frequency),resulting in hi

15、gh 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第23页，本讲稿共42页Non-random mating Assortative MatingPakistani or Cypriot population in UKAshkenazi Jewish population“Deaf”population or“blind”popul

16、ation第24页，本讲稿共42页Consanguinity/Inbreedingwhen 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第25页，本讲稿共42页Clinical and Public Health Implicationsincr

17、eased population-specific frequencies of genetic disorders e.g.Saguenay region of Quebec-increased frequency of tyrosinemia,hypercholesterolemia,myotonic dystrophy-dedicated treatment,screening,education for the public and health care providers第26页，本讲稿共42页Hardy-Weinberg equilibrium law If two allele

18、s at a gene-A and a frequency of the A allele=p 1.frequency of the a allele=qFirst offspring：p2【AA】+2pq【Aa】+q2【aa】Gametal frequency of first offspring：A=p2+1/2（2pq）;a=q2+1/2（2pq）random matingFemalesA(p)a(q)MalesA(p)AA(p2)Aa(pq)a(q)aA(qp)aa(q2)第27页，本讲稿共42页Gametal frequency of second offspring：A=p2+1/

19、2（2pq）a=q2+1/2（2pq）第28页，本讲稿共42页Hardy-Weinberg 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:1.large population2.ra

20、ndom mating3.no mutations4.no migration between populations 5.no selection-all genotypes reproduce with equal success 第29页，本讲稿共42页Hardy-Weinberg equilibrium law If two alleles at a gene-A and a 1.frequency of the A allele=p 2.frequency of the a allele=q3.The two fractions add up to totality p+q=l 4.

21、the proportions of the three genotypes:AA,Aa and aa are p2:2pq:q25.Hardy-Weinberg formula:p2(AA)+2pq(Aa)+q2(aa)=1第30页，本讲稿共42页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 ab

22、out the population balance?GenotypeNumber Frequency of genotypeAA 600.6Aa200.2aa200.2Total1001第31页，本讲稿共42页The gene frequency in the populations is:A=p=0.6+0.2/2=0.7 a=q=0.2+0.2/2=0.3If population balance,they will be:(AA)p2+2pq(Aa)+(aa)q2 =1 So,0.49+0.42+0.09=1 GenotypeNumber Frequency of genotypeAA

23、 600.6Aa200.2aa200.2Total1001But第32页，本讲稿共42页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)=0.09FemalesA(p)a(q)MalesA(p)AA(p2)Aa(pq)a(q)aA(qp)aa(q2)第33页，本讲稿共42页How to calc

24、ulate 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:i.1 in 1700 US Caucasian newborn

25、s have cystic fibrosis which means that the frequency of homozygotes for this recessive trait is q=1/1700=0.00059 ii.The square root of the frequency of recessives is equal to the allele frequency of the CF allele q=0.024 第34页，本讲稿共42页 iii The frequency of the normal allele is equal to 1-the frequenc

26、y 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 calculated frequencies of homo

27、zygous normal=0.952576 heterozygous carriers=0.046848 homozygous affected=0.000588 第35页，本讲稿共42页2.Frequency of sex-linked genesThe distribution of the recessive phenotype(aa)is equal to qExample:Color blindness q(XaY)=frequency of Male sufferer=0.07 So the frequency of genotype in Female Xa Xa=q2 =(0

28、.07)2 XA XA=p2=(1-q)2=(1-0.07)2第36页，本讲稿共42页第37页，本讲稿共42页Gene flow The exchange of genes between different populations.第38页，本讲稿共42页Why are some people resistant to HIV?第39页，本讲稿共42页Duncan SR,et al.Reappraisal of the historical selective pressures for the CCR5-Delta32 mutation.J Med Genet.2005;42(3):205

29、-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 of Europe from 1000 to 1800 AD demonstrates how plague epidemics,1347 to 1670,could hav

30、e 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.第40页，本讲稿共42页Fisher Hardy Weinberg第41页，本讲稿共42页Ching Chun Li(19122003):A Hero of Genetics Am J Hum Genet 74:789792,2004第42页，本讲稿共42页