Dette emnet er nedlagt

Kollokvie 010105

 

Fasit til enkelte av sp�rsm�lene

 

3)

X^2=(760-750)^2/750+(240-250)^2/250=0.53

d.f.=1

P(0.53,1)=0,47

 

P-verdi er st�rre enn 0.05. Vi beholder v�r hypotese, 3:1 fordeling.

 

7b)

 

F1: Ww Ll x Ww Ll

 

 

WL

Wl

wL

wl

WL

WWLL

WWLl

WwLL

WwLl

Wl

WWLl

WWll

WwLl

Wwll

wL

WwLL

WwLl

wwLL

wwLl

wl

WwLl

Wwll

wwLl

wwll

 

 

8)

 

b) 1/8 uavhengig av sannsynligheten for � f� en resessiv homozygot som f�rste barn.

 

Pp Bb x ppBb

 

 

PB

Pb

pB

pb

pB

PpBB

PpBb

ppBB

ppBb

pb

PpBb

Ppbb

ppBb

ppbb

pB

PpBB

PpBb

ppBB

ppBb

pb

PpBb

Ppbb

ppBb

ppbb

 

c)

 

 

PB

Pb

pB

pb

pB

PpBB

PpBb

ppBB

ppBb

pb

PpBb

Ppbb

ppBb

ppbb

pB

PpBB

PpBb

ppBB

ppBb

pb

PpBb

Ppbb

ppBb

ppbb

 

 

Sannsynligheten for begge fenotypene er 3/8.

 

 

 

 

 

 

9)

 

Dihybrid kryssning; 9:3:3:1

 

X^2=(223-223.9)^2/223.9+(72-74.6)^2/74.6+(76-74.6)^2/74.6+(27-24.9)^2/24.9

=  0.2976162

 

d.f.=3

 

P(0.3,3)= 0,96

 

Vi godtar dermed hypotesen.

 

10)

 

b)

 

F0: BB HH  FFx Bb Hh Ff

 

 

BHF

BHF

BHF

BHF

BHF

BHF

BHF

BHF

BHF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BhF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BHf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

Bhf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

bHF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

bhF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

bHf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

bhf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

 

 

8/64=1/8 er sannsynligheten for � f� morens genotype.

 

c)

 

 

BHF

BHF

BHF

BHF

BHF

BHF

BHF

BHF

BHF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BBHHFF

BhF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BBHhFF

BHf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

BBHHFf

Bhf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

BBHhFf

bHF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

BbHHFF

bhF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

BbHhFF

bHf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

BbHHFf

bhf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

BbHhFf

 

 

 

8/64=1/8 er sannsynligheten for � f� farens genotype.

 

 

14)

 

i)  F0: Kk x Kk

 

 

K

k

K

KK

Kk

k

Kk

kk

 

 

15)

 

kkMm x Kkmm

 

 

kM

km

kM

km

Km

KkMm

Kkmm

KkMm

Kkmm

Km

KkMm

Kkmm

KkMm

Kkmm

km

kkMm

kkmm

kkMm

kkmm

km

kkMm

kkmm

kkMm

kkmm

 

 

75% d�ve.

 

16)

 

Genotype foreldre: Aarr x Aarr -> aarr for mannen

 

Kryssning med kvinne: AaRr x aarr

 

 

AR

Ar

aR

ar

ar

AaRr

Aarr

aaRr

aarr

ar

AaRr

Aarr

aaRr

aarr

ar

AaRr

Aarr

aaRr

aarr

ar

AaRr

Aarr

aaRr

aarr

 

75% har sykdommen.

 

 

 

 

 

 

 

 

17)

 

a)

 

F0: WWyy x wwYY

 

F1:  WwYy – Alle avkom er hvite

 

b)

 

F2: WwYy x WwYy

 

 

WY

Wy

wY

wy

WY

WWYY

WWYy

WwYY

WwYy

Wy

WWYy

WWyy

WwYy

Wwyy

wY

WwYY

WwYy

wwYY

wwYy

wy

WwYy

Wwyy

wwYy

wwyy

 

Fenotypisk ratio: 12:3:1

 

Genotypisk ratio: 1(WWYY):2(WWYy):1(WWyy):2(WwYY):4(WwYy):2(Wwyy):1(wwYY):2(wwYy):1(wwyy)

 

Dette er et eksempel p� en dihybrid kryssning hvor et av genene er epistatisk til  det andre.

 

c)

 

P: WwYy x wwYy

 

 

WY

Wy

wY

wy

wY

WwYY

WwYy

wwYY

wwYy

wy

WwYy

Wwyy

wwYy

wwyy

wY

WwYY

WwYy

wwYY

wwYy

wy

WwYy

Wwyy

wwYy

wwyy

 

Denne kryssningen gir en ratio: 4:3:1. dette stemmer ikke s� d�rlig med den observerte ratioen 58:39:16

 

Chi^2= 0.5575221

 

p-verdi=0,76

 

18)

 

P: wwS_  x W_ s _

 

F1: SsWw x SsWw

 

Hvit hvit kryssning fra F1.

 

 

SW

Sw

sW

sw

SW

SSWW (h)

SSWw (h)

SsWW (h)

SsWw (h)

Sw

SSWw (h)

SSww(s)

SsWw (h)

ssWw (h)

sW

SsWW (h)

SsWw (h)

ssWW (h)

ssWw (h)

sw

SsWw (h)

Ssww (s)

ssWw (h)

ssww (r)