Population Genetics

Chapter 7 Population Population Genetics Genetics

郑 大 基 础 医 学 院 ： 程 晓 丽

Segment

1

Frequency of gene and genotype

Segment 2

Law of genetic equilibrium

Segment

3

Factors that affect frequency of gene

郑 大 基 础 医 学 院 ： 程 晓 丽

instruction 1. Hardy — Weinberg law 2. Inbreeding coefficient F 3. How to calculate the F 4. fitness

郑 大 基 础 医 学 院 ： 程 晓 丽

population the group of same species that live in certain region and mate randomly can be called Mendelian population.

medical population genetics study in how to maintain or change the frequency of the genes and genotypes in the population. The population genetic data are used primarily in genetic counseling and in planning genetic screening program.

gene pool all the genetic information (genes) contained in all members in the population.

郑 大 基 础 医 学 院 ： 程 晓 丽

Segment 1

Gene frequency and genotype frequency gene frequency: the proportion of allele of specific gene among the population. The ratio of the number of certain allele to the total number ofexample all kinds of alleles. A

a

Gene frequency ： A = 0.7 Then: a = 0.3 The sum is 1 A(0.7) + a(0.3) = 1

郑 大 基 础 医 学 院 ： 程 晓 丽

genotype frequency: the ratio of number of individuals with certain genotype to the total number of the population.

exampl e The alleles of tasting are T and t. genotyp e

phenotype

TT

1/70000 ～ 1/3000000 taster

Tt

1/50000 ～ 1/400000 taster

tt

1/20000 ～ rime nontaster

郑 大 基 础 医 学 院 ： 程 晓 丽

Group 1 100 members TT : 49 Tt : 42 tt : 9

The sum ?

frequency TT : 0.49 Tt : 0.42 tt : 0.09

TT(0.49) + Tt(0.42) + tt(0.09) = 1

calculation Frequency T= TT + ½ Tt = 0.49 + 0.42/2 = 0.7 Frequency t= tt + ½ Tt = 0.09 + 0.42/2 = 0.3

郑 大 基 础 医 学 院 ： 程 晓 丽

Group 2 Blood type of MN the total number:1000. Blood type number genotype frequency M: N: MN:

400 200 400

LMLM : LNLN : LMLN :

Frequency of gene

0.40 0.20 0.40

郑 大 基 础 医 学 院 ： co-dominance程 晓 丽

LM frequency = 0.4 + 0.4/2 = 0.6 LN frequency = 0.2 + 0.4/2 = 0.4 Only used for calculating the trait of

Segment 2 Genetic equilibrium In a large population with random mating, gene frequencies and genotype frequencies do not change from generation to generation.which is called Hardy – Weinberg law Ideal population 1. Very large or limitless 3. No natural selection

2. Random mating 4. No new mutation

5. No migration

When there is difference between generations, whether or not the population is balanceable?

郑 大 基 础 医 学 院 ： 程 晓 丽

Which one is the population of equilibrium Known

Group 1

Group 2

genotype

TT

Tt

tt

LMLM LNLN LMLN

number

49

42

9

400

F of genotype .49

.42

.09

.40

200 .20

400 .40

Inferring

T

t

LM

LN

F of gene

0.70

0.30

0.60

0.40

郑 大 基 础 医 学 院 ： 程 晓 丽

We suppose there is no difference between the eugenesis of group1 and group 2. T(.7) .49TT

.21Tt

.21Tt

.09tt

F1genotype f of genotype

TT .49

Tt .42

LM(.6)

LM

)6.(

LN ) (.4

T )7.( (.3 t )

t(.3)

LN(.4)

.36LM LM

.24LM LN

.24LM LN

.16LN LN郑 大

tt .09

LM LM .36

T = .49 + .42/2 = .7 f of gene t = .09 + .42/2 = .3

LM = LN =

基 础 M N N N L L L L 医 .48 .16 学 院 .36 + .48/2 = .6： 程 .16 + .48/2 = .4晓 丽

Compare with the frequency of gene and genotype between two groups

P F1 ： :

Group 1 T t .7 .3 .7 .3 TT Tt .49 .42

tt .09

Group 2 LM LN .6 .4 .6 .4 LM LM LM LN LN LN .4 .4 .2

郑 大 基 1 础 .49 .42 .09 .36 .48 .16 医 学 ∴结论 甲 , 乙两群体基因频率 : P = F1; 基因型频率甲 : P = F1, 乙 :院 ： P≠F1 ： 程 甲 : 平衡群体 ; 乙 : 不是平衡 晓 群体 . 不是平衡群体只需经过一代随机婚配 , 即可达平衡 丽

P F ： :

formula Hardy – Weinberg:

T→p;

T(.7) t(.3) TT.49 Tt.21 T(.7) sperm Tt.21 tt.09

t →q LM(.6)

egg

t(.3) Hardy - Weinberg frequencies

p+q=1 LN(.4)

LM(.6) LMLM.36 LMLN.24 LMLN.24 LNLN.16

L (.4) N

TT → p2 2Tt → 2pq tt → q2

∴ p2 + 2pq + q2 = 1

(p + q) = 1

郑 大 基 础 医 学 院 ： 程 晓 丽

The application of the Hardy – Weinberg law 1. To infer the gene frequency for AR

example The incidence of PKU ： 1/10 000

AR

AA Aa

homozygotes heterozygotes

aa

homozygotes = 1/10 000 known

Frequency of the carrier

郑 大 基 础 医 学 院 PKU ： 程 晓 Phenylketonu 丽

ria

Calculation

Frequency of a = ? Frequency of carrier ?

= q = 2 pq

∵ q2 = 1/10 000 ∴ Frequency for afflicted gene: q = (1/10 000) ½ = 0.01 ∵ 　 P + q = 1 ∴ 　 p = 1 – q = 0.99 ≈ 1 ∴ Frequency for heterozygous carrier: 2pq = 2×1×0.01 = 0.02 = 1/50

郑 大 基 础 医 学 院 ： 程 晓 丽

2.to infer the gene frequency of AD Pattern ?

AD ∵ F of A: ∴ F of a:

genotype?

AA 极为罕见 Aa 频率为 H

p infinitesimal

∵ H = 2pq

q≈1

∴p=½H

example Incidence of syndactyly 1/2 000 ∴ H = 1/2000 ∴ p = ½ H =1/2 ×1/2000 = 1/4000 = 0.00025

郑 大 基 础 医 学 院 ： 程 晓 丽

analysi Infer the gene F for X-linkedsinheritance Gene F ∵ = ∴ Genotype F ∴ genotype = ∵ ∴ genephenotype F = genotype F example

AD

∴ F of

郑 大 基 础 daltonism: incidence of male is 7% 医 学 Xb q = 7% = 0.07 ∴F of XB p =1 – 0.07 = 0.93院 ： 2 2 b b X X = q = (0.07) = 0.0049 ≈0.005 ≈0.5% 程 晓 B b 2pq= 2 ×0.93 ×0.07=0.13=13% X X = 丽

Segment

3

Factors that alter gene frequencies 1. consanguinity 2. mutation 3. selection 4. genetic drift 5. migration

郑 大 基 础 医 学 院 ： 程 晓 丽

1. Consanguineous marriage The marriage of the couple with same ancestor within three generations Consanguinity: the individuals with same ancestor within three or four generation What is the influence caused by consanguineous marriage Increase in the chance of appearance of homozygotes! Increase in the risk of disease of AR!

郑 大 基 础 医 学 院 ： 程 晓 丽

The form of consanguineous marriage

cousin

Inbreeding F

Second Semi-cousin degree coefficient cousin

Semi-Second degree 郑 大 cousin 基

The probability for the couple’s children to form homozygotes at certain locus just because of couple’s consanguineous marriage.

础 医 学 院 ： 程 晓 丽

Calculate the F for autosomal inheritance 1.F of sibling A1A2 A3A4 B1 marry B2

P1

P2

1/ B1 A12 1/ B2 2 1/ B1 A22 1/ B2 2 1/ B1 A32 1/ B2 2 1/ B1 A42 1/ B2

1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/

P1

P2

B1

B2

S A1A1 =(1/2)4 S

A ２ A ２ =(1/2)4

S A３A３ =(1/2)4 S

A４A４

A1A1 A2A2 A3A3 A4A4

郑 大 S 基 础 医 S: the probability学 of homozygotes, 院 namely the F of ： relatives of first 程 =(1/2)4 degree. F=4×(1/2) 4 =1/4晓 丽

2.F of uncle(aunt)and niece(nephew) B1 marry C S: the probability of homozygotes, namely the F of relatives of second degree.

F=4×(1/2) 5 =1/8

A1A2

A3A4

P1

P2

B1

B2

S

C A1A1 A2A2 A3A3 A4A4

郑 大 基 础 医 学 院 ： 程 晓 丽

3.the F of cousins C 1 marry C 2 S: the probability of homozygotes, namely the F of relatives of third degree. F=4×(1/2) 6 =1/16 A1A2 A3A4 P1

P2

B1

B2

C1

C2

SC1

SC2 A1A1 A2A2 3A3 S A A4A4

4.F of cousin of second degree

A1A2 A3A4 P1

P2

B1

B2

C1

C2

S

SC1 Marry SC2 S: the probability of homozygotes, namely the F of relatives of fifth degree. F=4×(1/2) 8 =1/64

A1A1 A2A2 A3A3郑 A4A4 大 基 础 医 学 院 ： 程 晓 丽

Calculate the F for X-linked inheritance Only the daughter can be homozygotes,so the F is only for daughters Father must transmit the chromosome X to his daughter. Father can never transmit the chromosome X to his son

1.F of cousin from mother’s sister P1

P2

B1 1/2 X1 1/2 B 2 1 1/2 B1 1/2 X2 1/2 B 1/2 2 1

1/2 B1 1/2 X3 1/2 B 1/2 2

X1Y X2X3 1 C marry C C1 1 2 郑 S P1 P大 2 C2 1/2 1

F?

B1

C1 S Mother’s sisters C2 1/2 C1 1

C1 S C2 1/2

F=(1/2)3+2×( 1/2)5=3/16

S

基 础 B2医 学 C2院 ： X1X程 1 X2X晓 2 X3X丽 3

2. F of cousin from mother’s brother C1 marry C2 X1 never can

Analysis: be homozygotes. Conclusion: F of cousin from mother’s brother F=2×(1/2) 4 =1/8 X1 Y

X2X3

P1

P2

B1

B2

C1

C2

S

X1Y

X2X3

P1

P2

B1

B2

C1

C2 2 X2 SX X3X3

3. F of cousin from father’s sister C1 marry C2 Analysis: X1, X2, X3 can never be homozygotes. Conclusion: F is 0 for cousin from father’s sister

郑 大 基 础 医 学 院 ： 程 晓 丽

3. F of cousin from father’s brother C1 marry C2

analysis X1 、 X2 、 X3 can’t be homozygotes.

conclusion

P1

P2

B1

B2

C1

F of cousin from father’s brother F=0

X2X3

X1 Y

C2

S

郑 大 基 础 医 学 院 ： 程 晓 丽

The average F The probability of producing homozygotes because of the consanguineous marriage among the population example A village ， 107 couples ， 6 of which are cousins ， 8 are second degree cousins, what is the average inbreeding coefficient The average F？ F of second degree F of cousin cousin F=1/16 F=1/64 a = (6 ×1/16 +8 ×1/64 ) / 107 = 0.467%

郑 大 基 础 医 学 院 ： 程 晓 丽

The average F of various population

nation/region era number of couples incidence a America 1958 133228 0.11% 0.00008 Germany 1946-1951 119899 0.59% 0.00019 France 1956-1958 530000 0.67% 0.00023 Italy 1956-1958 1646612 1.90% 0.00070 Japan 1950 213148 8.16% 0.004 India 1950 26042 39.37% 0.02835 郑 大 北京、湖北 1980-1981 7729 1.40% 基 0.000665 甘肃 ( 回 ) 1980-1981 1376 9.70% 0.005础 医 四川 ( 彝 ) 1980-1981 2054 14.16% 0.00913 学 贵州赤水 ( 苗 ) 1980-1981 234 16.24% 院 0.007696 ： 程 黑龙江 ( 鄂伦春 ) :: 183 1.6% 晓 0.000256 丽 : ( 鄂温克 ) :: 626 3.4% 0.000116

The harmful effect caused by consanguineous marriage

gene frequency homozygotes homozygotes increase ratio of G by random by cousin 0.2

0.04

0.05

0.01

1.25

0.10

0.01

0.015626

0.005625

1.56

0.04

0.0016

0.004

0.0024

2.5

郑 0.02 0.0004 0.001625 0.001225 4.06大 基 0.01 0.0001 0.000719 0.000619 7.19础 医 0.001 0.000001 0.0000635 0.0000625 63.5学 院 The higher of the recessive disease is, the more ： 程 harmful the effect caused by consanguineous 晓 丽 marriage will be.

Mutation

(u) A

nature

If A = p spontaneous mutation rate

a (v) a=q

∴incidence of A a / generation = pu = (1- q)u

n×10-6 / cell / generation ∴ incidence of a A /

generation = qv Affect the gene frequency Affect the genetic equilibrium

If :(1- q)u ＞ qv, a rise If :(1- q)u ＜ qv,A rise If :(1- q)u = qv, the gene frequency will be in balance

郑 大 基 础 医 学 院 ： 程 晓 丽

The effect on the equilibrium caused by selection natural selection: under the natural condition, the individuals with different genotypes have the different capability to survive and to procreate. The more adaptive it is, the more chance it will survive. Selection can affect the genetic equilibrium by means of increasing or decreasing the fitness of 郑 大 基 individuals. Fitness is the capability for certain individual to survive and to transmit its genes to its offspring. large Fitness high Next generation small Fitness low

础 医 学 院 ： 程 晓 丽

Achondroplasis object:

Danmark

example

108 patients

27 children

control: 457 their sibling

582 children

If the eugenesis of normal individual is 1, the eugenesis of Achondroplasis will be: f = (27/108) / (582/457) ≈ 0.2 suggest

The eugenesis of patients reduce The fitness is reduced by selection

郑 大 基 础 医 学 院 ： 程 晓 丽

The magnitude of effect caused by selection can be denoted by selection coefficient which is abbreviated S. S denote the portion of the eugenesis which is deduced by selection. Achondroplasis S = 1- f Relative eugenesis f = 0.2 ∴selection coefficient S = 1- f = 1 - 0.2 = 0.8 If the normal individual can give birth to 10 children, the patients can only give rise to 2 children. question ： whether or not all the disease gene can be eliminated by natural selection?

郑 大 基 础 医 学 院 ： 程 晓 丽

Key point Hardy — Weinberg law/ condition Inbreeding of coefficient, F How to calculate the F Fitness

郑 大 基 础 医 学 院 ： 程 晓 丽