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第四章 连锁遗传规律 Chapter4 Law of Linkage PowerPoint PPT Presentation


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第四章 连锁遗传规律 Chapter4 Law of Linkage. 连锁遗传理论的由来. 根据遗传的染色体学说与独立分配规律:   位于非同源染色体上的非等位基因遗传时独立分配;   如果有一些基因位于同一染色体上,必然会出现非独立分配的现象,否则各种性状的数目(基因对数)就超过细胞内染色体对数。   W. Bateson(1906) 在香豌豆 两对相对性状杂交试验中发现连锁遗传( linkage) 现象。   T. H. Morgan et al.(1910) 提出连锁遗传规律以及连锁与交换的遗传机理,并创立基因论( theory of the gene)。. 目 录.

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第四章 连锁遗传规律 Chapter4 Law of Linkage

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Chapter4 Lawof Linkage


()

W. Bateson(1906)(linkage)

T. H. Morgan et al.(1910)(theory of the gene)




1.

P PPLL ppll

F1 PpLl

F2 P-L- P-ll ppL- ppll

O 284 21 21 74

E 225 75 75 25

X23=189.27 X2 0.01(3)=11.34


3:1BatesonPunnett PLplPlpL


n ()


2.

  • ()

    • =(4831+390)(393+1338)=31

    • =(4831+393)(1338+390)=31

    • :=(226+95)(97+1)=31

    • =(226+97)(95+1)=31


  • ()

    P pr+pr+vg+vg+prprvgvg

    F1 pr+prvg+vgprprvgvg

    F2

    pr+prvg+vg + + 1339 + +

    prprvgvg prvg 1195 prvg

    pr+prvgvg + vg 151 + vg

    prprvg+vg pr + 154 pr +


P ++vgvgprpr++

F1 +pr+vgprprvgvg

F2

+prvgvg + vg 965 + vg

prpr+vg pr + 1067 pr +

+ + + + + + 157 + +

prprvgvg pr vg 146 prvg

F11:1:1:1


3.

  • F150%50%

  • F2


    • F1

    • F150%50%




C-Sh


  • I ()

    1.

    2.

    3.

    4. (/)

    5.


1

P

F1

()

()

()


250%


3

B

A

4 0

A

B

a

b

a

b


B

A

2 2

A

B

a

b

a

b


B

A

0 4

A

B

a

b

a

b


  • 1=(/)100%

    005050

    2

    = 1/2


3


    • (Ft)(F1)

      =/100%

  • C-Sh


  • 1

    =2 /F2

    =12 /F2

    2

    =2 /F2


  • [0~50%]

  • (genetic distance)1%(1cM)


  • 1.

  • 2.

  • 3.

  • 4.




1.

  • 1.()

  • 2.()

  • 3.(linkage analysis)


(gene location)

  • C/cSh/shC-Sh3.6%


2.


3.

  • (two-point testcross)

  • (three-point testcross)


3.1.11/3

1. F1

9

(C)(c)

(Sh)(sh)

(Wx)(wx).

(1).(CCShShccshsh)F1ccshsh

(2).(wxwxShShWxWxshsh)F1wxwxshsh

(3).(WxWxCCwxwxcc)F1wxwxcc


2/3

2.


3/3

3.

Rf C-Sh= 3.6 Rf Wx-Sh =20 Rf Wx-C=22


3.1.2

1.

2.5


3.2.1(1/7-2/7)

1.

P:

shsh ++ ++ ++ wxwx cc F1:

+sh +wx +c shsh wxwx cc

?

2.


  • A B C

    a b c

    (1)AaBbABCabcAbcaBC

    (2)Bb CcABCabcABcabC

    (3)AaBbCcABCabcAbCaBc

    (4)ABCabc


  • 11:1:1:1:1:1:1:1

    2

    (1)1:1:1:1

    (2)

    3


F1

+ wx c

2708

sh + +

2538

+ + c

626

sh wx +

601

sh + c

113

+ wx +

116

+ + +

4

sh wx c

2

6708

(3/7-4/7)

3.

4.


(5/7)

5.

()()

+wxcshwxcshshwxcsh+++++shshwxc



(6/7)

6.


(7/7)

7.

Shwxc

Rf C-Sh= 3.5% Rf Wx-Sh =18.4% Rf Wx-C=21.9%


3.2.2

  • 1.

    • ()()

    • wxshcwxc0.1840.035=0.64%


  • 2. (interference)

    • wxc0.09wx-shsh-c


  • 3.(coefficient of coincidence)

  • =0.09/0.64=0.14

  • [0, 1]00


  • AbD/aBdab

    10%bd20%

    (1)0?

    (2)1?


4.

  • 1.(n)

  • 2.

  • 3.

    • %


4.1


4.2


4.2.1

  • ()

    • ()


4.2.2

  • (tetrad analysis)()


  • (centromere mapping)


4.2.2

  • (lys)

    • lys+()

    • lys

  • lys+lys-lys+lys-4:4

  • lys+lys-




4.2.3

  • (1-2)lys+lys- AI (first division segregation)

    • lys

  • (3-6) AII (second division segregation)



4.2.4

  • ()


  • 9lys-5lys-

  • =[5(1/2)/9+5]100%

  • =18%

  • lys+/lys-18


  • 1.

  • 2.

  • 3.

  • 4.



1.

  • (XY,ZW,XO,ZO)


1.1

sex-chromosome

autosome


  • XY

    AA+XXA+XAA+XYA+XA+Y

  • XYX0

    AA+XXA+XAA+X0A+XA


  • ZW

    AA+ZZA+ZAA+ZWA+ZA+W

  • ZWZ0

    AA+ZZA+ZAA+Z0A+ZA


1.2

(n=16)(2n=32)


1.3

1.8

1/500

()


1.4

Ba

Ts

Ba-Ts-

Ba-tsts

babaTs-

babatsts


2.

1

2011

30

2


3.


3.1

P (X+X+)(XwY)

F1(X+Xw)(X+Y)

F2X+X+ X+Xw X+Y XwY

:=3:1

:=11


P (XwXw)(X+Y )

F1 (X+Xw)(XwY)

F2 X+Xw XwXw X+Y XwY

:=1:1

:=1:1

:=1:1


3.2

  • 1.

  • 2.

  • 3.


3.3

P (XBXB)(XbY )

F1 (XBY)

P (XBXb)(XbY )

F1 (XBY)(XbY )11

P (XBXb)(XBY )

(XBXBXBXb)

(XBY ):(XbY )=1:1


3.4

ZBW() ZbZb()

F1 ZbW () ZBZb ()

ZBZb ZbZb ZBW ZbW


3.5


3.6

P (X+X+)(XcYc)

F1 (X+Xc)(X+Yc)

F2X+X+ X+Xc X+Yc XcY

:=3:1

:=11


3.7

HH

Hh

hh


  • AbD/aBdab

    10%bd20%

    (1)0?

    (2)1?


1

/0

AbD/aBd

A-b

ABd 10 /2=5%

abD 10 /2=5%

b-D

Abd 20 /2=10%

aBD 20 /2=10%

AbD 11020/2=35%

aBd 11020/2=35%

6

AbD: aBd: ABd: abD : Abd: aBD=7 : 7 : 2 : 2 : 1 : 1


11

== 10% 20%2%

AbD/aBd

ABD 2%/2= 1%

abd 2%/2= 1%

A-b

ABd (10 - 2% )/2=4%

abD (10 - 2% )/2=4%

b-D

Abd (20 - 2% ) / 2=9%

aBD (20 - 2% ) / 2=9%

AbD 1- 2%-8-18/2=36%

aBd 1- 2%-8-18/2=36%

8

AbD: aBd: ABd: abD : Abd: aBD: ABD : abd

=36 : 36 : 9 : 9 : 4 : 4 : 1 : 1


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