1、Genetic Variation in Populations ZHANG Xian-Ning, PhD: 88208367 Office: A713, Research Building2019/09Learning Objectives The nature and amount of genetic variation in human populations, and the role of genetic variation in liability to disease. Ethnic Groups: Cauc

2、asoid, Negroid, MongoloidGenes in Human Populations Study ofdistribution and frequency of genes in 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

3、a whole.Mendelian populationAn 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 gen

4、e pool. 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 frequencyHardy Weinberg Equilibrium The law that relates allele frequency to genotype frequency, used in populat

5、ion 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.1qpq2p22 (Hardy Weinberg equation p2 + 2pq + q2 = 1)If all alleles at a locus are either A or a, frequency o

6、f “A” in the population is p, frequency of “ain 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 = 1/10000) q2 = 1/1

7、0 000therefore: q = 1/100 (*this is not the carrier frequency)Since p + q = 1p = 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 + aAaaFreq

8、uencyp22pqq2PhenotypeNormal Carriers AffectedExceptions to Hardy Weinberg AssumptionsMigration introduction / loss of allelesMutations may occur at different frequency in different populations3. Small population size- genetic isolate / founder effect4. Non-random mating- consanguinity- assortative m

9、ating(=non-random mating)Factors that disturb Hardy-Weinberg equilibrium Exceptions to random mating Exceptions to constant allele frequencyGenetic driftGene flow Changes in Allele FrequencyCan be caused by:mutation (source of genetic variation)selection (phenotypes differ in biological fitness)(del

10、eterious mutations may be removed by early death / lack of reproduction)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

11、to the gene pool of the next generation S(selective coefficient)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 result

12、s from balance between “l(fā)oss by selection”, and “gain by new mutation”Heterozygote Advantage Mutant allele has a high frequency despite reduced fitness in affected individuals. Heterozygote has increased fitness over both homozygous genotypes eg. Sickle cell anemia.Sickle Cell Anemia in West AfricaA

13、/A“normal”Susceptibility to malariaA/SHeterozygous carrier-Resistant to malaria-Anemia rareS/SSickle cell diseaseSevere anemia, often fatal.More influences on allele frequencies Effects of population size and non-random mating - contrary to Hardy Weinberg assumptions, humans have often lived in smal

14、l populations, isolated from their neighborsGenetic 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 frequency), resulting in high frequency o

15、f rare disease. e.g. Huntingdon disease in Lake Maracaibo, Venezuela (AD) Gyrate Atrophy in Finland Tyrosinemia in eastern Quebec 1/685 vs. 1/100 000Non-random mating Assortative MatingPakistani or Cypriot population in UKAshkenazi Jewish population“Deaf” population or “blind” populationConsanguinit

16、y/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 2Clinical and Public Health Implications increased population-specific frequen

17、cies 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 providersHardy-Weinberg equilibrium law If two alleles at a gene - A and a frequency of t

18、he A allele = p frequency of the a allele = qFirst offspring： p2【AA】+2pq【Aa】+q2【aa】Gametal frequency of first offspring： A = p2 + 1/22pq）; a = q2 + 1/22pq）random matingFemalesA (p)a (q)MalesA (p) AA (p2) Aa (pq)a (q) aA (qp) aa (q2)Gametal frequency of second offspring：A = p2 + 1/22pq） a = q2 + 1/22

19、pq）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:large populationrandom matingno mutationsno mig

20、ration between populations no selection - all genotypes reproduce with equal success Hardy-Weinberg equilibrium law If two alleles at a gene - A and a frequency of the A allele = p frequency of the a allele = q3. The two fractions add up to totality p + q = l 4. the proportions of the three genotype

21、s: AA, Aa and aa are p2: 2pq : q25. Hardy-Weinberg formula: p2 (AA) + 2pq (Aa) + q2 (aa) = 1Applications 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?G

22、enotypeNumber 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 (Aa)+( aa)q2 = 1 So, 0.49 + 0.42 + 0.09 = 1 GenotypeNumber Frequency of genotypeAA 600.6Aa

23、200.2aa200.2Total1001ButAfter 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) = 0.42 aa (q2 ) = 0.09FemalesA (p)a (q)MalesA (p)AA (p2)Aa (pq)a (q)aA (qp)aa (q2)How to

24、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 Caucasian

25、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 frequency

26、 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

27、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 qExample: 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)2Gene flow The exchange of genes between different populations.Why are some people re