Population genetic
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Population genetic. พันธุศาสตร์ประชากร. เป็นการศึกษายีนในระดับประชากร … ว่ามีการกระจายของยีนในภาพรวมของประชากรอย่างไร โดยทำการประเมินจากความถี่ของยีน ; f(gene) และความถี่ยีโนไทป์ ; f(genotype) ซึ่งเป็นการศึกษาวิวัฒนาการของสัตว์ (Evolution) ที่จะเป็นพื้นฐานของการคัดเลือก และการผสมพันธุ์.

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Population genetic

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Population genetic

Population genetic


Population genetic

  • ;f(gene) ;f(genotype) (Evolution)


Population genetic

  • Evolution f(gene) f(genotype)

  • Evolution

gene genotype


Population genetic

  • co-dominant incomplete dominant

    R = , r = R>r

    Rr x Rr

    genotype 3

Phenotype Genotype

RR 800

Rr 150

rr 50


F genotype

Phenotype Genotype

RR 800

Rr 150

rr 50

1000

f(genotype) =

D = f(RR) = 800/1000 = 0.8

H = f(Rr) = 150/1000 = 0.15

R = f(rr) = 50/1000 = 0.05


D h r 1 0 8 0 15 0 05

D+H+R = 1 (0.8+0.15+0.05)

D = f(RR) = 800/1000 = 0.8

H = f(Rr) = 150/1000 = 0.15

R = f(rr) = 50/1000 = 0.05

D = Homozygous dominance

H = Heterozygous

R = Homozygous recessive


Population genetic

  • f(R) = D+ H = p

  • f(r) = R + H = q

  • Gene frequency =


Hardy weinberg law

Hardy-Weinberg law ()

  • Gene equilibrium ()


Population genetic

  • (random mating)

  • (selection),

  • (mutation),

  • (migration)

  • (genetic drift)

  • genotype (equilibrium) Hardy-Weinberg


Population genetic

genotype

D = p2

H = 2pq

R = q2

p q


Population genetic

  • R(p) = 0.875 r (q) =0.125

  • genotype F1

  • HW

  • D= f(RR) = p2 = (0.875)2 = 0.766

  • H =f(Rr) = 2pq = (0.875)(0.125) = 0.218

  • R = f(rr) = q2 = (0.125)2 = 0.016


Population genetic

F1

  • f(R) = 0.766 + (0.218) = 0.875

  • f(r) = 0.016 + (0.218) = 0.125

f(gene) F1 f(gene)


Complete dominant

(complete dominant)

  • D =

    d =

    D>d 9 625 genotype gene


Population genetics 2

Population genetics 2


Population genetic

: R r (co-dominant alleles) Rr 250 ,450 200 genotype


Population genetic

genotype:

f(RR) = 250/900 = 0.278

f(Rr) = 450/900 = 0.500

f(rr) = 200/900 = 0.222

gene:

f(R) = (500+450)/1,800 = 0.528

f(r) = (400+450)/1,800 = 0.472


Population genetic

:

f(RR) + f(Rr) + f(rr) = 1.0

f(R) + f(r) = 1.0

  • genotype gene 1

    genotype f(RR) f(Rr) f(rr)

    f(rr) = 1 f(RR) f(Rr)

  • 1 f(R)

    f(r) f(r) = 1 f(R)


Population genetic

...

-(2-test)

goodness of fits -


Population genetic

(equilibrium population)

p = A q = a p + q = 1 genotype binomial expansion

(p+q)2 = (p+q)*(p+q)

= p2+2pq+q2= 1


Population genetic

Ex. f(R) = 0.528f(r) = 0.472

genotype

f(RR) = p2= (0.528)2 = 0.279

f(Rr) = 2pq = 2(0.528)(0.472) = 0.498

f(rr) = r2 = (0.472)2= 0.223


Population genetic

  • gene

  • f(R) = f(RR) + f(Rr)

  • = 0.279 + (0.498) = 0.528

  • f(r) = f(rr) + f(Rr)

  • = 0.223 + (0.498) = 0.472


Population genetic

Hardy-Weinberg genotype


Complete dominant1

(complete dominant)

  • d =

    D =

  • D>d 9 625 genotype gene


Population genetic


Population genetic

  • d =

  • D =

  • D>d 9 625 genotype gene


Population genetic

f(D) = p f(d) = q

f(dd) = 9/625 = 0.0144

Hardy-Weinberg

f(dd) = q2

f(d) = == 0.12

p+q = 1

f(D) = 1-0.12 = 0.88


Population genetic

0.12 0.88 DD Dd

f(DD) = p2

= (0.88)2 = 0.7744

f(Dd) = 2pq

= 2(0.88)(0.12) = 0.2112


Population genetic

R r (co-dominant alleles) Rr 250 , 450 200

Rr :: 1:2:1 R r

Ho : :: =1:2:1

HA : :: 1:2:1


Population genetic

-(2-test)

2value=

O = (observed number)

E = (expected number)

2value < 2 (df)


Population genetic

:

1.

= 3+1= 4

= 900/4= 225

2.

= 1x225= 225

= 2x225= 450


Population genetic

2value=

=

2(2) 0.05 df 2 5.99

2value< 2 (2) Ho (null hypothesis)


Population genetic

Expected p2(N), 2pq(N), q2(N) RR, Rr rr

H-W

= p2(N) = (0.278)(900) = 250.2

= 2pq(N)=(0.5)(900) = 450

= q2(N) = (0.222)(900) = 199.8

N =


Population genetic

2(1) 0.05 df1 3.84

2value< 2(1) Ho (null hypothesis)

H-W

: R r df = 1


Population genetic


Population genetic

Hardy-Weinberg

gene force


Population genetic

2

Systematic process

Dispersive process

Positive assortive mating

Non-random mating

Negative assortive mating

Migration

Genetic drift

Non-recurrent mutation

Mutation

Recurrent mutation

Selection

Culling

Favouring


Population genetic

(non-random mating)

  • ..... positive assortive mating

  • negative assortivemating genotype


Population genetic

(non-random mating)

positive assortive mating


Population genetic

(non-random mating)

negative assortive mating


Population genetic

f(a) = q1

N= n1

f(a) = q1

N = nm

f(a)= q0

N= n0

(migration)


Population genetic

(migration)


Population genetic

  • /

  • /

  • (, )


Population genetic

q0= q0 + m(q1- q0)

q0=

q0 =

q1 =

m = = migration rate

n0=

nm=


Population genetic

Ex. 8,000

(q0) 0.2

(q1) 0.6

2,000

q0= q0 + m(q1-q0)

=

= 0.2 + 0.08

= 0.28

q= q0 - q0

= 0.28 - 0.2

= 0.08


Population genetic

  • /


Population genetic

(mutation)

  • DNA (A, T, C G)

  • 10-4 10-5


Mutation

(mutation)


Population genetic

(mutation)

  • 2

    1. (non-recurrent mutation)

    2. (recurrent mutation)


Population genetic

(mutation)

(non-recurrent mutation)

(one- way mutation) 0 1

:R r

:p q

f(R) , f(r) p0 , q0 u 1

f(R) = p1= p0 - up0

= p0(1-u)

f(r)= q1= q0 + up0 1 p1

p >> q f(R) up0 f(r) up0


Population genetic

(mutation)

q

q= q1 - q0

= (q0 + up0) - q0

= up0

(q0) (qt) (t) (q)

q= up ; t =

= u(1-q)


Population genetic

(mutation)

  • (non-recurrent mutation)

    ( one way mutation)

    0

    1


Population genetic

(mutation)

Ex. B (p) 0.8 (u) B b 10-4 B 0.1

t =

t =

= 1,335.31


Population genetic

(mutation)

(recurrent mutation)

(two-way mutation)

:B b

:pq

u v 1

f(B) = p1= p0 - up0 + vq0

= p0(1-u) + vq0

f(b)= q1= q0 + up0 - vq0

U

V


Mutation1

(mutation)

  • up0 > vq0 f(B)

    up0 < vq0 f(B)

    up0 = vq0 f(B)


Population genetic

(mutation)

vqE= upE

vqE= u(1-qE)

vqE= u - uqE

vqE+ uqE= u

(v+u)qE= u

qE=


Population genetic

(mutation)


Population genetic

Ex. q0 = 0.2 (B->b) u 4.2 x 10-5 (b->B) v 2.1 x 10-5 1

q1 = q0 + up0 vq0

= 0.2+(4.2x10-5)(0.8) - (2.1x10-5)(0.2)

= 0.2+ (0.294x10-4)

qE =

== 0.6667


Population genetic

Ex. b 10%


Population genetic

2


Selection

(selection)


Natural selection

Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant

Natural selection


Population genetic

Natural selection

Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant


Population genetic

Natural selection

Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant

Generation 2: 0.96 not resistant

0.04 resistant

mutation!


Population genetic

Natural selection

Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant

Generation 2: 0.96 not resistant

0.04 resistant

Generation 3: 0.76 not resistant

0.24 resistant


Population genetic

Natural selection

Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant

Generation 2: 0.96 not resistant

0.04 resistant

Generation 3: 0.76 not resistant

0.24 resistant

Generation 4: 0.12 not resistant

0.88 resistant


Selection1

(selection)

fitness adaptive value


Selection2

(selection)


Selection3

(selection)

1. recessive

2. dominant

3. recessive dominant


Selection4

(selection)

1. recessive

recessive (aa)

dominant (AA, Aa)


Selection5

(selection)

1. recessive

genotype

genotype


Selection6

(selection)

1. recessive

genotype


Selection7

(selection)

1. recessive


Selection8

(selection)

Ex. aa

0.5 f(AA), f(Aa), f(aa)

0.36, 0.48, 0.16

1


Selection9

(selection)

recessive (q0) (qt) (t)


Selection10

(selection)

Ex. homozygous recessive

20%


Population genetic

2. dominant

dominant (AA, Aa)

recessive (aa)


Population genetic

2. dominant

genotype


Population genetic

2. dominant


Population genetic

Ex. A 0.5f(AA), f(Aa), f(aa) 0.36, 0.48, 0.16 1


Population genetic

dominant (q0) (qt) (t)


Population genetic

Ex. (feather shank)

dominant 10% 1%


Population genetic

3. dominant recessive

homozygous recessive dominant

(aa, AA) heterozygous genotype (Aa)


Population genetic

3. dominant recessive

genotype


Population genetic

3. dominant recessive


Population genetic

Ex. AA 0.5 aa 0.2 f(AA), f(Aa), f(aa) 0.36, 0.48, 0.16 1


Population genetic


Population genetic

Ex. (chestnut) genotype AA, (creamello) genotype aa, (palomino) genotype Aa palomino


Population genetic

(chance)

0 1


Population genetic


Genetic drift

0.50 R

0.50 r

0.25 R

0.75 r

Genetic drift

Before:

8 RR

8 rr

After:

2 RR

6 rr


Population genetic

/ (genetic variation)

Why is genetic variation important?

potential for change in genetic structure

  • adaptation to environmental change

  • - conservation

  • divergence of populations

  • - biodiversity


Population genetic

global

warming

survival

Why is genetic variation important?

variation

EXTINCTION!!

no variation


Population genetic

Why is genetic variation important?

variation

no variation


Population genetic

Why is genetic variation important?

divergence

variation

NO DIVERGENCE!!

no variation


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