Isoelectric - Novel Separation Processes - Lecture Notes, Study notes of Learning processes

some concept of Novel Separation Processes are Asymmetric Membrane, Centrifugal Separation Processes, Cloud Point Extraction, Colloidal Particles, Common Stationary Phase.Main points of this lecture are: Isoelectric, Positively Charged, Negatively Charged, Electrophoresis, Temperature Increase, Efficient Cooling, Prevent Cooling, Electroosmosis, Low Residual Charge, Ployacrylamide

Typology: Study notes

2012/2013

Uploaded on 04/27/2013

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Chapter 9
1. At isoelectric pH, protein is
(a) Positively charged (b) Negatively charged (c) Neutral
Ans: (c)
2. What major problems in electrophoresis arises due to joule heating effects?
Ans: (a) Density difference occurs leading to free convection.
(b) Temperature increase degrades protein.
3. How joule heating is prevented?
Ans: (a) Efficient cooling.
(b) Electrophoresis is used in a gel/membrane to prevent cooling.
4. How electroosmosis is prevented in electrophoresis?
Ans: Use of material for gel having low residual charge like ployacrylamide.
5. In a batch page electrophoresis, µosm=0. Mobilities of two proteins are µA =5X10-5
cm2/V.s and µB= 3X10-5cm2/V.s. DA=1.5X10-7 cm2/s and DB=1.2X10-7 cm2/s.
E=200 V/cm; t=4 hours. Find resolution (R).
Ans
2
5
510 .2
5
310 .
cm
Avs
cm
Bvs


exp
Z
Et
AA
2
5
5 10 200 (4 3600)
.
cm v s
vs cm

144cm
exp
Z
Et
BB
pf2

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Chapter 9

1. At isoelectric pH, protein is

(a) Positively charged (b) Negatively charged (c) Neutral

Ans: (c)

2. What major problems in electrophoresis arises due to joule heating effects?

Ans: (a) Density difference occurs leading to free convection.

(b) Temperature increase degrades protein.

3. How joule heating is prevented?

Ans: (a) Efficient cooling.

(b) Electrophoresis is used in a gel/membrane to prevent cooling.

4. How electroosmosis is prevented in electrophoresis?

Ans: Use of material for gel having low residual charge like ployacrylamide.

5. In a batch page electrophoresis, μosm=0. Mobilities of two proteins are μ A =5X

cm

2

/V.s and μB = 3X

cm

2

/V.s. D A =1.5X

cm

2

/s and D B =1.2X

cm

2

/s.

E=200 V/cm; t=4 hours. Find resolution (R).

Ans

cm

A

v s

cm

B

v s

exp

Z Et A A

2 5 5 10 200 (4 3600) .

cm v s v s cm

     

 144 cm

exp Z Et B B

cm v s v s cm

 86.4 cm

( ) 10 / sec

2

D cm eff

7 2 1.35 10 cm / sec

  

exp D t eff

7

     

 0.0039 cm

1 exp ( ) 1 2 4 2

t

R E D eff