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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 law of linkage

Chapter4 Lawof Linkage


Chapter4 law of linkage

()

W. Bateson(1906)(linkage)

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


Chapter4 law of linkage


Chapter4 law of linkage


Chapter4 law of linkage

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


Chapter4 law of linkage

3:1BatesonPunnett PLplPlpL


Chapter4 law of linkage

n ()


Chapter4 law of linkage

2.

  • ()

    • =(4831+390)(393+1338)=31

    • =(4831+393)(1338+390)=31

    • :=(226+95)(97+1)=31

    • =(226+97)(95+1)=31


Chapter4 law of linkage

  • ()

    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 +


Chapter4 law of linkage

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


Chapter4 law of linkage

3.

  • F150%50%

  • F2


Chapter4 law of linkage

    • F1

    • F150%50%


Chapter4 law of linkage


Chapter4 law of linkage


Chapter4 law of linkage

C-Sh


Chapter4 law of linkage

  • I ()

    1.

    2.

    3.

    4. (/)

    5.


Chapter4 law of linkage

1

P

F1

()

()

()


Chapter4 law of linkage

250%


Chapter4 law of linkage

3

B

A

4 0

A

B

a

b

a

b


Chapter4 law of linkage

B

A

2 2

A

B

a

b

a

b


Chapter4 law of linkage

B

A

0 4

A

B

a

b

a

b


Chapter4 law of linkage

  • 1=(/)100%

    005050

    2

    = 1/2


Chapter4 law of linkage

3


Chapter4 law of linkage

    • (Ft)(F1)

      =/100%

  • C-Sh


Chapter4 law of linkage

  • 1

    =2 /F2

    =12 /F2

    2

    =2 /F2


Chapter4 law of linkage

  • [0~50%]

  • (genetic distance)1%(1cM)


Chapter4 law of linkage

  • 1.

  • 2.

  • 3.

  • 4.


Chapter4 law of linkage


Chapter4 law of linkage


Chapter4 law of linkage

1.

  • 1.()

  • 2.()

  • 3.(linkage analysis)


Gene location

(gene location)

  • C/cSh/shC-Sh3.6%


Chapter4 law of linkage

2.


Chapter4 law of linkage

3.

  • (two-point testcross)

  • (three-point testcross)


3 1 1 1 3

3.1.11/3

1. F1

9

(C)(c)

(Sh)(sh)

(Wx)(wx).

(1).(CCShShccshsh)F1ccshsh

(2).(wxwxShShWxWxshsh)F1wxwxshsh

(3).(WxWxCCwxwxcc)F1wxwxcc


Chapter4 law of linkage

2/3

2.


Chapter4 law of linkage

3/3

3.

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


3 1 2

3.1.2

1.

2.5


3 2 1 1 7 2 7

3.2.1(1/7-2/7)

1.

P:

shsh ++ ++ ++ wxwx cc F1:

+sh +wx +c shsh wxwx cc

?

2.


Chapter4 law of linkage

  • A B C

    a b c

    (1)AaBbABCabcAbcaBC

    (2)Bb CcABCabcABcabC

    (3)AaBbCcABCabcAbCaBc

    (4)ABCabc


Chapter4 law of linkage

  • 11:1:1:1:1:1:1:1

    2

    (1)1:1:1:1

    (2)

    3


3 7 4 7

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.


Chapter4 law of linkage

(5/7)

5.

()()

+wxcshwxcshshwxcsh+++++shshwxc


Chapter4 law of linkage


Chapter4 law of linkage

(6/7)

6.


Chapter4 law of linkage

(7/7)

7.

Shwxc

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


3 2 2

3.2.2

  • 1.

    • ()()

    • wxshcwxc0.1840.035=0.64%


Chapter4 law of linkage

  • 2. (interference)

    • wxc0.09wx-shsh-c


Chapter4 law of linkage

  • 3.(coefficient of coincidence)

  • =0.09/0.64=0.14

  • [0, 1]00


Chapter4 law of linkage

  • AbD/aBdab

    10%bd20%

    (1)0?

    (2)1?


Chapter4 law of linkage

4.

  • 1.(n)

  • 2.

  • 3.

    • %


Chapter4 law of linkage

4.1


Chapter4 law of linkage

4.2


4 2 1

4.2.1

  • ()

    • ()


4 2 2

4.2.2

  • (tetrad analysis)()


Chapter4 law of linkage

  • (centromere mapping)


4 2 21

4.2.2

  • (lys)

    • lys+()

    • lys

  • lys+lys-lys+lys-4:4

  • lys+lys-


Chapter4 law of linkage


Chapter4 law of linkage


4 2 3

4.2.3

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

    • lys

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


Chapter4 law of linkage


4 2 4

4.2.4

  • ()


Chapter4 law of linkage

  • 9lys-5lys-

  • =[5(1/2)/9+5]100%

  • =18%

  • lys+/lys-18


Chapter4 law of linkage

  • 1.

  • 2.

  • 3.

  • 4.


Chapter4 law of linkage


Chapter4 law of linkage

1.

  • (XY,ZW,XO,ZO)


Chapter4 law of linkage

1.1

sex-chromosome

autosome


Chapter4 law of linkage

  • XY

    AA+XXA+XAA+XYA+XA+Y

  • XYX0

    AA+XXA+XAA+X0A+XA


Chapter4 law of linkage

  • ZW

    AA+ZZA+ZAA+ZWA+ZA+W

  • ZWZ0

    AA+ZZA+ZAA+Z0A+ZA


Chapter4 law of linkage

1.2

(n=16)(2n=32)


Chapter4 law of linkage

1.3

1.8

1/500

()


Chapter4 law of linkage

1.4

Ba

Ts

Ba-Ts-

Ba-tsts

babaTs-

babatsts


Chapter4 law of linkage

2.

1

2011

30

2


Chapter4 law of linkage

3.


Chapter4 law of linkage

3.1

P (X+X+)(XwY)

F1(X+Xw)(X+Y)

F2X+X+ X+Xw X+Y XwY

:=3:1

:=11


Chapter4 law of linkage

P (XwXw)(X+Y )

F1 (X+Xw)(XwY)

F2 X+Xw XwXw X+Y XwY

:=1:1

:=1:1

:=1:1


Chapter4 law of linkage

3.2

  • 1.

  • 2.

  • 3.


Chapter4 law of linkage

3.3

P (XBXB)(XbY )

F1 (XBY)

P (XBXb)(XbY )

F1 (XBY)(XbY )11

P (XBXb)(XBY )

(XBXBXBXb)

(XBY ):(XbY )=1:1


Chapter4 law of linkage

3.4

ZBW() ZbZb()

F1 ZbW () ZBZb ()

ZBZb ZbZb ZBW ZbW


Chapter4 law of linkage

3.5


Chapter4 law of linkage

3.6

P (X+X+)(XcYc)

F1 (X+Xc)(X+Yc)

F2X+X+ X+Xc X+Yc XcY

:=3:1

:=11


Chapter4 law of linkage

3.7

HH

Hh

hh


Chapter4 law of linkage

  • AbD/aBdab

    10%bd20%

    (1)0?

    (2)1?


Chapter4 law of linkage

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


Chapter4 law of linkage

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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