-Ti le bo ba chua 2A, 1U

(1/6)² x (2/6)=1/108

-3A

(1/6)³ =1/216

-2U,1A

(2/6)² x (1/6)=1/54

-It nhat 1A

1-((2/6)² x(3/6)+(3/6)²x(2/6)+(2/6)³+(3/6)³)=151/216

a)2⁵=32(te bao)

b)2⁵ * 78=2496(NST)

c)78*(2⁵-1)=2418(NST)

a)Co n kieu

b)co n kieu

c)co C_n² x C²_n-2 kieu

d)Co n kieu

10 nguoi I^OI^O\(\Rightarrow\)f(I^OI^O)=10/1000=0,01\(\Rightarrow\)f(I^O)=\(\sqrt{ }\)0,01=0,1

80 nguoi I^BI^O \(\Rightarrow\)f(I^BI^O)=80/1000=0,08\(\Rightarrow\)f(I^B)=0,08/0,1=0,8

\(\Rightarrow\)f(I^A)=1-(0,1+0,8)=0,1

De cap vo chong co nhom mau A va B sinh duoc con co nhom mau O thi cap vo chong nay phai co kieu gen di hop.

Xac suat mau A co kieu gen di hop la

(2*0,1*0,1)/(0,1²+2*0,1*0,1)=2/3

Xac suat mau B co kieu gen di hop la

(2*0,8*0,1)/(0,8²+2*0,8*0,1)=1/5

Xac suat de cap vo chong nay sinh duoc con nhom mau O la

2/3*1/5*1/4=1/30

 ( 4x - 3)(3x - 5)( 7 - 8x) 

= ( 12x^2 - 20x - 9x + 15 )( 7 -8x)

= 84x^2 - 140x - 63x + 105  - 96x^3 + 160x^2 + 72x^2 - 120x

= -96x^3 + 316x^2-323x +105

8 x 2n = 8 x 4 = 32(NST)

Tong so cromaic o ki giua quan sat thay trong cac te bao la:

3 x 4n =3 x 8 =24

a)Lai hai ruoi TC than xam canh ngan va than den canh dai F1 toan ruoi than xam canh dai \(\Rightarrow\)Than xam canh dai troi hoan toan so voi than den canh cut.Vay P: Ab/Ab x aB/aB. F2 thu duoc 101 ruoi than xam canh ngan : 199 ruoi than xam canh dai : 100 ruoi than den canh dai tuong ung ti le 1:2:1\(\Rightarrow\)4 to hop giao tu nen tinh trang di truyen theo quy luat di truyen lien ket.

So do lai

P: Ab/Ab (xam ngan) x aB/aB(den dai)

Gp: Ab aB

F1: Ab/aB(Xam dai)

F1xF1: Ab/aB(xam dai) x Ab/aB(xam dai)

Gf1xf1 : Ab aB Ab aB

F2: 1/4 Ab/Ab (xam ngan) ; 1/4 Ab/aB (xam dai)

1/4 aB/Ab (xam dai) ; 1/4 aB/aB(den dai)

b) AB/Ab (Xam dai)

=> abc x 10 - abc = 1215

=> abc x 9 = 1215

=> abc = 1215 : 9

=> abc = 135

aaaa:x=a

=>    x=aaaa:a

=>    x=1111

Vậy x=1111