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


  • ;f(gene) ;f(genotype) (Evolution)


  • Evolution f(gene) f(genotype)

  • Evolution

gene genotype


  • co-dominant incomplete dominant

    R = , r = R>r

    Rr x Rr

    genotype 3

Phenotype Genotype

RR 800

Rr 150

rr 50


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 = 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


  • f(R) = D+ H = p

  • f(r) = R + H = q

  • Gene frequency =


Hardy-Weinberg law ()

  • Gene equilibrium ()


  • (random mating)

  • (selection),

  • (mutation),

  • (migration)

  • (genetic drift)

  • genotype (equilibrium) Hardy-Weinberg


genotype

D = p2

H = 2pq

R = q2

p q


  • 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


F1

  • f(R) = 0.766 + (0.218) = 0.875

  • f(r) = 0.016 + (0.218) = 0.125

f(gene) F1 f(gene)


(complete dominant)

  • D =

    d =

    D>d 9 625 genotype gene


Population genetics 2


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


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


:

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)


...

-(2-test)

goodness of fits -


(equilibrium population)

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

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

= p2+2pq+q2= 1


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


  • 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


Hardy-Weinberg genotype


(complete dominant)

  • d =

    D =

  • D>d 9 625 genotype gene



  • d =

  • D =

  • D>d 9 625 genotype gene


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


0.12 0.88 DD Dd

f(DD) = p2

= (0.88)2 = 0.7744

f(Dd) = 2pq

= 2(0.88)(0.12) = 0.2112


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

Rr :: 1:2:1 R r

Ho : :: =1:2:1

HA : :: 1:2:1


-(2-test)

2value=

O = (observed number)

E = (expected number)

2value < 2 (df)


:

1.

= 3+1= 4

= 900/4= 225

2.

= 1x225= 225

= 2x225= 450


2value=

=

2(2) 0.05 df 2 5.99

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


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 =


2(1) 0.05 df1 3.84

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

H-W

: R r df = 1



Hardy-Weinberg

gene force


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


(non-random mating)

  • ..... positive assortive mating

  • negative assortivemating genotype


(non-random mating)

positive assortive mating


(non-random mating)

negative assortive mating


f(a) = q1

N= n1

f(a) = q1

N = nm

f(a)= q0

N= n0

(migration)


(migration)


  • /

  • /

  • (, )


q0= q0 + m(q1- q0)

q0=

q0 =

q1 =

m = = migration rate

n0=

nm=


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


  • /


(mutation)

  • DNA (A, T, C G)

  • 10-4 10-5


(mutation)


(mutation)

  • 2

    1. (non-recurrent mutation)

    2. (recurrent mutation)


(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


(mutation)

q

q= q1 - q0

= (q0 + up0) - q0

= up0

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

q= up ; t =

= u(1-q)


(mutation)

  • (non-recurrent mutation)

    ( one way mutation)

    0

    1


(mutation)

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

t =

t =

= 1,335.31


(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


(mutation)

  • up0 > vq0 f(B)

    up0 < vq0 f(B)

    up0 = vq0 f(B)


(mutation)

vqE= upE

vqE= u(1-qE)

vqE= u - uqE

vqE+ uqE= u

(v+u)qE= u

qE=


(mutation)


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


Ex. b 10%


2


(selection)


Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant

Natural selection


Natural selection

Resistance to antibacterial soap

Generation 1: 1.00 not resistant

0.00 resistant


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!


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


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


(selection)

fitness adaptive value


(selection)


(selection)

1. recessive

2. dominant

3. recessive dominant


(selection)

1. recessive

recessive (aa)

dominant (AA, Aa)


(selection)

1. recessive

genotype

genotype


(selection)

1. recessive

genotype


(selection)

1. recessive


(selection)

Ex. aa

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

0.36, 0.48, 0.16

1


(selection)

recessive (q0) (qt) (t)


(selection)

Ex. homozygous recessive

20%


2. dominant

dominant (AA, Aa)

recessive (aa)


2. dominant

genotype


2. dominant


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


dominant (q0) (qt) (t)


Ex. (feather shank)

dominant 10% 1%


3. dominant recessive

homozygous recessive dominant

(aa, AA) heterozygous genotype (Aa)


3. dominant recessive

genotype


3. dominant recessive


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



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


(chance)

0 1



0.50 R

0.50 r

0.25 R

0.75 r

Genetic drift

Before:

8 RR

8 rr

After:

2 RR

6 rr


/ (genetic variation)

Why is genetic variation important?

potential for change in genetic structure

  • adaptation to environmental change

  • - conservation

  • divergence of populations

  • - biodiversity


global

warming

survival

Why is genetic variation important?

variation

EXTINCTION!!

no variation


Why is genetic variation important?

variation

no variation


Why is genetic variation important?

divergence

variation

NO DIVERGENCE!!

no variation


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