Chapter 11 Population Genetics

Transcription

1 Chapter 11 Population Genetics

2 Population = group of organisms of the same species living in the same geographical area Population genetics is the study of the distribution of genes in populations and of the factors that maintain or change the frequency of genes and genotypes from generation to generation.

3 OUTLINE Hardy-Weinberg Equilibrium Factors affecting H-W Equilibrium

4 Hardy-Weinberg Equilibrium -important concepts Allele frequency: the proportion of a specific allele at a given locus, considering that the population may contain from one to many alleles at that locus. Genotype frequency: the proportion of a specific genotype at a given locus, considering that many different genotypes may be possible. Phenotype frequency: the proportion of individuals in a population that exhibit a given phenotype.

5 Hardy-Weinberg Equilibrium -allele frequency 1. two alleles at a gene - A and a 2. frequency of the A allele = p 3. frequency of the a allele = q 4. p + q = 1 A a f A = f a = A A + a a A + a (p) (q) p+q=1

6 More than two alleles at a gene 1. let p = the frequency of the A allele 2. let q = the frequency of the B allele, 3. let r = the frequency of the i allele. 4. p + q + r = 1 I f A I = A IA + I B + i (p) I f B I = B IA + I B + i (q) I A I B i f i = i IA + I B + i (r) p+q+r=1

7 Hardy-Weinberg Equilibrium -genotype frequency A a genotype AA Aa aa genotype frequency AA AA+Aa+aa Aa AA+Aa+aa aa AA+Aa+aa Let f AA =D, f Aa =H, f aa =R, D+H+R=1 Allele A frequency Allele a frequency p= q= 2D+H 2D+2H+2R = 2D+H 2(D+H+R) 2R+H 2D+2H+2R = R+½H = D+½H

8 In a population, 50% people were AA, 20%people were Aa, 30%people were aa, calculate the frequencies of alleles Let f A =p, f a =q p=f AA +½f Aa =0.5+½ 0.2=0.6 q=f aa +½f Aa =0.3 +½ 0.2=0.4=1-p

9 In a population of 747 individuals,there were 233 type M individuals (31.2%), 129 type N individuals (17.3%), 485 type MN individuals (51.5%), calculate the allele frequencies. Let f L M=p, f L N=q p=mm+½mn=0.312+½ 0.515=0.57 q=nn+½mn=0.173+½ 0.515=0.43=1-p When the phenotype is equal to the genotype (esp. Co-dominant inheritance), the calculation of gene frequency is based on the gene counting method.

10 Law of Genetic Equilibrium Hardy-Weinberg law Hardy GH Weinberg W

11 Law of Genetic Equilibrium Hardy-Weinberg law Explains how Mendelian segregation influences allelic and genotypic frequencies in a population.

12 Law of Genetic Equilibrium Hardy-Weinberg law Assumptions Large population Random mating No natural selection No mutation No migration If assumptions are met, population will be in genetic equilibrium.

13 Law of Genetic Equilibrium Hardy-Weinberg law Allele frequencies do not change over generations. Genotypic frequencies will remain in the following proportions: p+q=1 p 2 +2pq+q 2 =1 p 2 : frequency of AA 2pq: : frequency of Aa q 2 : frequency of aa

14 Law of Genetic Equilibrium Hardy-Weinberg law Genotypic frequencies do not change over generations. After only one generation of random mating, population will be in genetic equilibrium.

15 In a population there are 3 genotypes (AA, Aa, aa) and 6 mating types. AA Aa aa AA Aa aa

16 Frequency of Mating types (random mating) Paternal genotypes AA(p 2 ) Aa(2pq) aa(q 2 ) AA (p 2 ) AA X AA (p 4 ) AA X Aa (2p 3 q) AA X aa (p 2 q 2 ) Maternal genotypes Aa (2pq) Aa X AA (2p 3 q) Aa X Aa (4p 2 q 2 ) Aa X aa (2pq 3 ) aa (q 2 ) aa X AA (p 2 q 2 ) aa X Aa (2pq 3 ) aa X aa (q 4 )

17 AA AA AA Aa AA aa Aa Aa Aa aa aa aa frequency p 2 p 2 =p 4 2(p 2 2pq)=4p 3 q 2(p 2 q 2 )=2p 2 q 2 2pq 2pq=4p 2 q 2 2(2pq q 2 )=4pq 3 q 2 q 2 =q 4 offspring's genotypes AA Aa aa p p 3 q 2p 3 q 0 0 2p 2 q 2 0 p 2 q 2 2p 2 q 2 p 2 q 2 0 2pq 3 2pq q 4 (p+q) 4 =1 p 2 (p 2 +2pq+q 2 ) q 2 (p 2 +2pq+q 2 ) =p 2 =q 2 2pq(p 2 +2pq+q 2 ) =2pq

18 In a population study, individuals were genotyped for A locus. There were 5000 AA individuals, 2000 Aa individuals, and 3000 aa individuals. Is this population in Hardy-Weinberg equilibrium? (1) f AA =0.5, f Aa =0.2, f aa =0.3 allele A frequency: p=0.5+½ 0.2=0.6, allele a frequency: q=0.3+½ 0.2=0.4 (2) If this population were at Hardy-Weinberg equilibrium, one would predict that f AA =0.6 2 =0.36, f Aa = =0.48, f aa =0.4 2 =0.16. It is quite different than the observation. Thus the population is not at Hardy-Weinberg equilibrium.

19 offspring s genotype Mating types AA Aa aa AA AA AA Aa 2( ) AA aa 2( ) Aa Aa Aa aa 2( ) aa aa After one generation with random mating, the population is at Hardy-Weinberg equilibrium

20 Hardy-Weinberg Equilibrium - application Application of Hardy-Weinberg Equilibrium -Determine allele frequency in a population -Determine inheritance pattern

21 Determine allele frequency - Autosomal recessive diseases Genotype Genotype frequency Phenotype phenotype frequency AA Aa aa p 2 2pq q 2 Unaffected affected F u F a Disease allele frequency(q)=(f a ) 1/2 e.g. disease frequency (F a )=1/10000 disease allele frequency=1/100 carrier frequency=2pq=2x0.01x0.99=0.0198

22 Determine allele frequency - Codominant alleles MN blood group: In a population, there were 233 type M individuals, 485 type MN individuals, and 129 type N individuals. L M allele frequency= (233X2+485)/[2X( )]=0.57 L N allele frequency= (129X2+485)/[2X( )]=0.43 =1- L M

23 Determine allele frequency - Multiple alleles ABO blood group: type A=41.72%; type B=8.56%; type O=46.68%; type AB=3.04% I A allele frequency= p; I B allele frequency= q i allele frequency= r Phenotype Type A Type B Type AB Type O Genotype I A I A, I A i I B I B, I B i I A I B ii Genotype frequency p 2 +2pr q 2 +2qr 2pq r 2 r=(o) 1/2 =(0.4668) 1/2 =0.683 A+O= p 2 +2pr+r 2 =(p+r) 2 =(1-q) 2 q=1-(a+o) 1/2 =1-( ) 1/2 =0.06 p=1-(b+o) 1/2 =1-( ) 1/2 =0.257

24 Determine allele frequency - X-linked recessive disease Phenotype Unaffected affected Genotype Genotype frequency X A X A p 2 2pq p XA Xa XA Y X a X a q 2 X a Y q Disease allele frequency (q)= disease frequency in male or Disease allele frequency (q)= (disease frequency in female) 1/2 Example - Color blindness: disease frequency is 7% in male, 0.49% in female Disease allele frequency = 7%

25 Determine allele frequency - X-linked recessive disease Allele frequency Genotype Genotype frequency X A X a X A X A p 2 /2 2p 2 /3 - X A X a 2pq/2 2pq/3 2pq/3 X A frequency =2p 2 /3 + 2pq/3 + p/3 = [2p 2 +2pq+p]/3 = p/3 [2p+2q+1] = p/3 [2(p+q)+1] = p/3 3 = p X a X a q 2 /2 X A Y p/2 X a Y q/2 - p/3-2q 2 /3 - q/3

26 Determine inheritance pattern Example - high myopia (P61) population study Disease frequency: 0.724% In the families with one affected parent, there were 104 offspring, of which 8 affected, frequency was 7.69%; In the families with unaffected parents, there were 1637 offspring, of which 10 affected, frequency was 0.61%. Based on these results, please determine inheritance pattern of high myopia in the population.

27 Determine inheritance pattern Suppose high myopia in the population was inherited in AR Disease allele frequency (q) = (0.724%) 1/2 =0.085 For the families with one affected parent Mating type frequency Offspring AA Aa aa AA X aa 2p 2 q 2 2p 2 q 2 Aa X aa 2 2pq q 2 =4pq 3 2pq 3 2pq 3 Expected disease frequency in these families =2pq 3 /(2p 2 q 2 +4pq 3 )=q/(p+2q)=q/(1+q)=0.085/( )=7.83% Observed disease frequency in these families was 7.69%, which is close to 7.83%.

28 Determine inheritance pattern For the families with unaffected parent Mating type frequency Offspring AA Aa aa AA X Aa 4p 3 q 2p 3 q 2p 3 q AA X AA p 4 p 4 Aa X Aa 4p 2 q 2 Expected disease frequency in these families =p 2 q 2 /(p 4 +4p 3 q+4p 2 q 2 ) =q 2 /(p 2 +4pq+4q 2 ) =q 2 /(p+2q) 2 =q 2 /(1+q) 2 = /( ) 2 =0.647% p 2 q 2 2p 2 q 2 p 2 q % (expected) Vs. 0.61% (observed)

29 AD diseases Derivation (Affected of Heterozygote) 2pq p 2 + 2pq 1 (Affected) Almost all of the affected are heterozygote

30 AR diseases Derivation 2pq = 2q(1-q) = 2q - 2q 2 2q 2pq 2q The heterozygote carrier frequency is twice the frequency of the mutant allele.

31 AR diseases Derivation 2pq q 2 2q q 2 = 2 q The number of heterozygote carriers in the population is much larger than the affected. The ratio increases as the disease frequency decreases.

32 XD diseases Derivation p p 2 +2pq = 1 p + 2q = 1 p + 2(1-p) = 1 2-p 1 2 The ratio of affected females to the males is approximately 2 to 1.

33 XR diseases Derivation q q = 1 2 q More affected males than affected females The ratio increases as the disease frequency decreases.

34 OUTLINE Hardy-Weinberg Equilibrium Factors affect H-W Equilibrium

35 Factors that Alter Genetic Equilibrium Mutation Selection Genetic Drift Isolation Migration Consanguineous Marriage

36 Factors affect H-W Equilibrium -Nonrandom mating In human population, mating is seldom random Defined by racial, ethnic, religious, or other criteria Consanguinity- mating among close relatives, a special form of nonrandom mating in human population Consanguinity does increase the proportion of homozygotes in the next generation, thereby exposing disadvantageous recessive phenotypes to selection. Such selection may in turn alter allele frequencies in subsequent generations.

37 堂兄妹姑表亲姨表亲舅表亲 First cousins 隔山表亲 Half first cousins Second cousins First cousins once removed

38 Pedigree of royal hemophilia

39 Royal Consanguinity Czarina Alexandra I was a carrier of hemophilia Her son, Grand Duke Alexis was affected with hemophilia A well-documented case of consanguinity to retain power occurred in European Royal Families of many countries.

40 Factors affect H-W Equilibrium -Nonrandom mating Coefficient of relationship (r) : the proportion of all genes in two individuals which are identical by descent. expresses the fraction of genes shared by two individuals 1 st degree r=(½) 1 2 nd degree r=(½) 2 3 rd degree r=(½) 3

41 Factors affect H-W Equilibrium -Nonrandom mating 1/2 1/2 1/4 1/2 1/4 1/8

42 Factors affect H-W Equilibrium -Nonrandom mating 2. Inbreeding Coefficient (F) : The probability of identical homozygosity due to common ancestor Coefficient of Inbreeding (F) can be calculated by the formula F= Σ(1/2) n Where n = the number of steps between the proband and the common ancestor and Σ says sum it for the number of alleles that can become homozygous by descent

43 P1 P2 A1: F1 A1A2 A3A4 F2 P1 ½ F1 ½ G1 ½ F2 ½ G2 ½ ½ S G1 A3A3 A4A4 S A1A1 A2A2 G2 A1A1: A2A2: A3A3: A4A4: (½) 6 =1/64 (½) 6 =1/64 (½) 6 =1/64 (½) 6 =1/64 F=4 X 1/64=1/16

44 Second cousin A 1 A 1 :(½) 8 =1/128 A 2 A 2 :(½) 8 =1/128 A 3 A 3 :(½) 8 =1/128 A 4 A 4 :(½) 8 =1/128 F=4 (½) 8 =1/64 For autosome loci:f=r/2 A 1 A 2 A 3 A 4 P 1 P 2 1/2 1/2 B 1 B 2 1/2 C 1 C 2 1/2 1/2 S 1/2 1/2 D 1 D 2 1/2 A 1 A 1 A 2 A 2 A 3 A 3 A 4 A 4

45 P1 X1 P2 X2X3 X1: P1 1 F1 ½ G1 1 F2 ½ G2 1 ½ S F1 G1 F2 G2 X2: P2 ½ F1 ½ G1 ½ F2 ½ G2 1 ½ S X3X3 S X1X1 X2X2 X1X1: X2X2: (½) 3 =1/8 (½) 5 =1/32 F=3/16 X3X3: (½) 5 =1/32

46 P1 B1 1 1/2 X 1 Y X 2 X 3 0 1/2 1/2 P2 B2 C1 C2 1 1/2 1 S X 2 X 2 (½) 4 =1/16 X 3 X 3 (½) 4 =1/16 F=2 1/16=1/8

47 P1 P2 X1 X2X3 F1 G1 F2 G2 F=0 X3X3 S X1X1 X2X2

48 P1 B1 C1 0 0 X 1 Y X 2 X 3 0 1/2 1/ /2 P2 B2 C2 S F=0

49 Generation Factors affect H-W Equilibrium -Mutation I A A A a a p=0.6; q=0.4 II A A a a a p=0.4; q=0.6

50 A u v a f A =p f a =q pu=(1-q)u > qv pu=(1-q)u < qv pu =qv (1-q)u = qv u-qu=qv u=qv+qu=q(v+u) q=u/(v+u)

51 Factors affect H-W Equilibrium -Selection Selection represents the action of environmental factors on a particular phenotype, and hence its genotype Selection may be positive or negative Selection is the consequence of differences of biological fitness (f). Therefore, selection coefficient (s) =1-f Biological fitness (f) is a measure of fertility

52 Selection for/against dominant alleles is efficient Mutant allele encoding dominant traits are expressed in heterozygotes and thus exposed to direct selection Aa aa A very weak selective change can rapidly alter the allele frequency

53 Selection for/against recessive alleles is inefficient AA Aa aa Why? - because then most recessive alleles are in heterozygotes Thus, rare disease-causing recessive alleles persist in the population in heterozygote carriers, even if they are lethal when homozygous

54 Selection for/against recessive alleles is inefficient # alleles in carriers 2pq = # alleles in affected 2q 2 ~ = 1 q q 2 = 1/10,000; q = 1/100,

55 Carrier/Affected Ratio for an autosomal recessive disease Disease Incidence Carrier Frequency Ratio Carrier/Affected q2 2pq

56 Selection against /for X-linked recessive alleles X A X A X A X a X a X a X A Y X a Y Efficiency: AD>XR>AR

57 Factors affect H-W Equilibrium -Genetic drift Change in a populations allele frequency due to chance characteristic of small populations

58 Before: 8 RR 8 rr 0.50 R 0.50 r After: 2 RR 6 rr 0.25 R 0.75 r

59 I Generation Factors affect H-W Equilibrium -Genetic drift A A A a a p=0.6; q=0.4 II A A a a p=0.5; q=0.5 III a a a a a p=0; q=1 IV a a a a a q=1 N a a a a a q=1

60 Random Genetic Drift Frequency of allele A Generation

61 Random Genetic Drift ABO frequency of American indians North America:I A 0.018; I B 0.009; i Blackfeet: I A 0.5

62 Factors affect H-W Equilibrium -Founder effect Founder effect A few individuals colonize a new habitat Probably accounts for high frequency of inherited disorders in some human population

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66 Factors affect H-W Equilibrium -migration Generation I II A A A a a A A A a a A a a p=0.6; q=0.4 p=0.5; q=0.5

67 例 : 已知 Ⅱ 1 为 AR 患者, 该病的 PF=1/10000, 求 Ⅱ 2 与其姨表兄弟结婚后代的再发风险? Ⅱ 2 与群体中一个无关个体婚配, 后代的再发风险又是多少? Aa Aa Aa 1/2 aa Aa 2/3? Aa 1/4

68 1) 如果 Ⅱ 2 与其姨表兄弟婚配, 子代发病风险为 2 3 2) 如果 Ⅱ 2 与群体中无亲缘关系的人婚配, 子代发病风险为 : = = 由于 PF=1/10000 则 :q 2 =1/10000, 则 q=1/100,2pq 2q=1/

69 The End