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CHE2165 Bio-nano Engineering (2013)

Assignment 2 (Open-book test)

Solution Guide

Question 1 (20 marks)

The chromatogram below represents the separation of proteins A, B, C from a solution via

anion-exchange chromatography at a flow rate of 2 ml/min. The sample was dissolved in

Buffer 1 which has an ionic strength of 0.8 M [NaCl] and eluted with Buffer 2 with 1.5 M

[NaCl] ionic strength. Assume 1 OD280 = 0.75 mg protein / mL.

a) What charge does the functional groups of stationary phase carry? Which protein

has the strongest ionic interaction with the stationary phase? [2 marks]

Positive charge. Protein C.

b) Calculate the retention factors for proteins A, B and C. [3 marks]

; = 6.6 min ; ;

0

10

20

30

40

50

60

70

80

90

100

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Buffer 2, % OD 280

Time, min

Peak: 1. Mobile phase 2. Protein A 3. Protein B 4. Protein C

1

4

3 2

2

c) Calculate the selectivity factor for proteins A and C. [2 marks]

d) At what [NaCl] did proteins A and B elute? [4 marks]

NaCl]1 = 0.8 M [NaCl]2 = 1.5 M

From the graph:

Buffer 2 Buffer 1

Protein A 10% 90%

Protein B 31% 69%

[NaCl]A = 10%(1.5 M) + 90% (0.8 M) = 0.15 + 0.72 = 0.87 M

[NaCl]B = 31%(1.5 M) + 69% (0.8 M) = 0.465 + 0.552 = 1.017 M

e) Calculate the resolution between peak of protein B and peak of protein C. [2 marks]

Flow rate = 2 ml/min

Vr1 = 2 ml/min × 13.5 min = 27 ml

Vr2 = 2 ml/min × 16.5 min = 33 ml

Wb1 = 15 – 12 = 3 × 2 = 6 mL

Wb2 = 18.5 – 15.0 = 3.5 × 2 = 7 mL

f) Calculate the elution gradient. [2 marks]

g) Calculate the weight of protein A eluted. [5 marks]

Convert OD280 against time to C (mg/ml) against volume

,where 1 OD280 unit = 0.75mg/ml ; volume = flow rate × time

Use trapezoidal rule: Area in the peak = mass of eluted protein

3

∫

of the mass per unit volume (concentration) of eluted solute eluted with respect to time. At

constant flow rate and concentration sensitive detector, the integration will also be with

respect to volume flow of mobile phase through the column. Thus, peak area is proportional

to the mass of solute. Measurement of peak area accommodates peak asymmetry and even

peak tailing without compromising the simple relationship between peak area and mass. Thus,

peak area measurements give more accurate results under non-ideal chromatographic

conditions.

[ ]

OD280 C (mg/ml)

y1 0 0

y2 0.08 0.060

y3 0.34 0.255

y4 0.79 0.593

y5 0.54 0.405

y6 0.10 0.075

y7 0 0

[ ]

Mass of the eluted protein B = 1.39 mg

OR

[ ]

Question 2 (15 marks)

The following data have been obtained for an enzyme catalyzed reaction. The initial enzyme

concentration is 3.2×10-2

M.

4

ν, (Ms-1) 2.52 2.37 2.25 2.12 2.04 1.94 1.87 1.8

[S], (mM) 20 10 6.7 5 4 3.3 2.9 2.5

a) Plot the data using the Lineweaver-Burk method (

[ ]) and determine the

following constants graphically:

i. The maximum forward velocity of the reaction, [4 marks]

ii. The equilibrium constants, [4 marks]

ν = rp 1/v S 1/S

2.52 0.397 20 0.050

2.37 0.422 10 0.100

2.25 0.444 6.7 0.149

2.12 0.472 5 0.200

2.04 0.490 4 0.250

1.94 0.515 3.3 0.303

1.87 0.535 2.9 0.345

1.8 0.556 2.5 0.400

b) Calculate the rate constant for the decomposition of enzyme-substrate complex,

[3 marks]

[ ]

y = 0.4555x + 0.3765 R² = 0.9984

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.0 0.1 0.2 0.3 0.4 0.5

1/V

1/[S]

5

c) Calculate the percentage of enzyme concentration used up at a point where the

substrate concentration is 42 mM. [4 marks]

Concentration of the complex at this point.

[ ] [ ][ ]

[ ]

OR

[ ]

[ ]

[ ]

[ ]

[ ] [ ]

[ ]

[ ] [ ] – [ ]

[ ]

[ ]

Question 3 (5 marks)

Show that the specific growth rate (µ) of cell at steady state in the continuous operation of

fermentation is equal to the dilution rate (D).

Mass balance on biomass (cell): Accumulation = Incoming + Cell Growth – Removal – Death

X : cell concentration in fermenter (g dry cells / l) t : time (min) F : flow rate (l/min) V : liquid phase volume of bioreactor (l) X0 : cell concentration in incoming medium (g dry cells / l) μ : specific growth rate (hr-1) a : specific death rate (hr-1)

aXXV

FXX

V

F

dt

dX 0

6

Mass Balance on biomass at steady state in Continuous Stirred Tank (Bio)Reactor:

No cell mass entering in the incoming medium :

No death term :

No accumulation :

Hence →

– Where F/ V = D or “dilution rate”.

00 XV

F

0aX

0dt

dX

000 XV

FX D

V

F