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1、Learning ObjectivesThe nature and amount of genetic variation in human populations, and the role of genetic variation in liability to disease. Ethnic Groups: Caucasoid, Negroid, MongoloidGenes in Human Populations Study ofdistribution and frequency of genes in populationsreasons for different gene f
2、requencies 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.- Mutations may or may not result in an expressed phenotype.- Mutations can alter RNA expression, processing and/or stability.- Mu
3、tations 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 inher
4、ited (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 population.
5、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 (fou
6、nder effect, small group of individuals found a new population).PolymorphismsMendelian populationAn interbreeding population of sexually reproducing individuals sharing a common gene pool.Gene pool, genotypes, and gene frequencyGenotype: the genetic constitution of a single individual. Gene pool : t
7、he 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): the freq
8、uencies 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- pHardy Weinberg LawStat
9、es the relationships between the frequency of alleles at a locus, and the genotypes resulting from these alleles.assumes: large populationrandom matingno new mutationsno immigration in or out no natural selection (Hardy Weinberg equation: p2 + 2pq + q2 = 1)If all alleles at a locus are either A or a
10、, 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 AAAa aAaap22pqq2homozygous normalcarriersaffected(p2 + 2pq + q2 = 1)Observed frequency of recessive disease in population is q2 (e.g., frequency of PKU = 1/1
11、0 000) q2 = 1/10 000 q = 1/100 (*this is not the carrier frequency!) p + q = 1 p = 1 q = 1 1/100 = 99/100Carrier 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 = 1/10,000GenotypeAAAa + aAaaFr
12、equencyp22pqq2PhenotypeNormalCarriersAffectedExceptions to Hardy Weinberg Assumptions1.Mutation-selection equilibrium may occur at different frequency in different populations2.Heterozygote Advantage3.Founder effect and genetic drift Small population size, genetic isolate4.Non-random mating consangu
13、inity; assortative mating(=non-random mating)5.Migration and gene flow introduction / loss of allelesChanges 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 / lack of rep
14、roduction)migration (movement in or out) 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(selective coefficien
15、t)SelectionZero fitness of AD mutations Early lethalityCondition occurs only because of new mutationsAppear sporadic rather than as AD pedigreeeg. osteogenesis imperfecta, type 2. compare: achondroplasia fitness of 0.20 - frequency results from balance between “loss by selection”, and “gain by new m
16、utation”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.Heterozygote AdvantageGeneHomozygote PhenotypeHeterozygote AdvantageAlternative ExplanationsCFTRC
17、ystic fibrosisIncreased resistance to choleraCFTR mutant alleles appeared before Cholera epidemics; increased resistance to typhoid and/or asthma.HLA (MHC)Highly polymorphicEnhancing resistance to infectious disease The loci too polymorphic to be accounted for by heterozygote advantage; may have oth
18、er benefits. TDSTay-Sachs mainly in Ashkenazi JewsResistance to TBThe high frequency of disease can be explained by founder effect and genetic driftGemmell NJ and Slate J. PLoS One. 2006;1:e125.21Sickle Cell Anemia in West AfricaA/A“normal”Susceptibility to malariaA/SHeterozygous carrier-Resistant t
19、o malaria-Anemia rareS/SSickle cell diseaseSevere anemia, often fatal.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 high frequency of rare disease.e.g. Huntingdon disease in Lake Maracaibo,
20、 Venezuela (AD) Gyrate Atrophy in Finland Tyrosinemia in eastern Quebec 1/685 vs. 1/100 000Genetic DriftFluctuation in allele frequency due to chance in a small population.Non-random mating Assortative MatingPakistani or Cypriot population in UKAshkenazi Jewish population“Deaf” population or “blind”
21、 populationConsanguinity/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 2Clinical and Public Health Implicationsincreased popula
22、tion-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 providersGene flow The exchange of genes between different populatio
23、ns.Hardy-Weinberg equilibrium law If two alleles 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
24、(q)aA (qp) aa (q2)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 generation to generation. If disequilibrium occurs, equilibrium will be reestablished after one generation of random mati
25、ng. H-W law rests on several assumptions:1.large population2.random mating3.no mutations4.no migration between populations 5.no selection - all genotypes reproduce with equal success Applications of Hardy-Weinberg law:How to judge the population equilibrium?For example, consider the hypothetical pop
26、ulations: AA 60 persons, aa 20 persons, Aa 20 persons, how about the population balance?GenotypeNumber Frequency of genotypeAA 600.6Aa200.2aa200.2Total1001The 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 (
27、Aa)+( aa)q2 = 1 So, 0.49 + 0.42 + 0.09 = 1 GenotypeNumber Frequency of genotypeAA 600.6Aa200.2aa200.2Total1001ButAfter 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)
28、= 0.42 aa (q2 ) = 0.09FemalesA (p)a (q)MalesA (p)AA (p2)Aa (pq)a (q)aA (qp)aa (q2)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
29、using the Hardy Weinberg equation and its assumptions as follows: i.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 ii.The square root of the frequency of recessives is equal to the allele frequency o
30、f the CF allele q = 0.024 iii The frequency of the normal allele is equal to 1 - the 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 i
31、s p = (0.976) = 0.953 vi Thus the population is composed of three genotypes at the calculated frequencies of homozygous normal = 0.952576 heterozygous carriers = 0.046848 homozygous affected = 0.000588 2. Frequency of sex-linked genesThe distribution of the recessive phenotype (aa) is equal to qExam
32、ple: Color blindness q (XaY)= frequency of Male sufferer = 0.07 So the frequency of genotype in Female Xa Xa = q2 = ( 0.07 )2 XA XA= p2 = (1-q)2 = (1-0.07)2Summary of H-W calculationsAD(if homozygote dominants are NOT lethal)Allele f (p) of disease gene (A) = disease f /2Allele f (q) of normal gene
33、(a) = 1 - pARAllele f (q) of disease gene (a) = disease f Allele f (p) of normal gene (A) = 1 qf of heterozygote carriers = 2pq XRAllele f (q) of disease gene (a) = disease f Allele f (p) of normal gene (A) = 1 qf of heterozygote carriers = 2pqWhy are some people resistant to HIV?Duncan SR, et al. R
34、eappraisal 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-Delta32 deletion; the average frequency of this allele is 10% in European populations. A mathematical model based on the ch
35、anging demography 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 pers
36、istent 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 WeinbergChing Chun Li (19122003): A Hero of Genetics Am J Hum Genet 74:789792, 20041. Which is the balanced population in the following populations? C. Population: (1)AA 64% (2)Aa 20% (3)aa 16% q (f(aa) = 16% = 0.16 q (f(a) = 0.4 p (f(A) = 1-q = 1- 0.4 = 0.6 p (f(AA) = 0.6 = 0.36 2pq (f(Aa) = 2 0.6 0.4 = 0.48 NOT 0.2! This pop is not a balanced pop.
