coursework report biochemical engg

8/7/2019 coursework report biochemical engg

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ABHISEKH UMRAO SINGH

MSc BIOTECHNOLOGY

7810595

FUNDAMENTALS OF BIOCHEMICAL ENGINEERING

= 2RT

() +

4 ()

1

This represents the equation of state which is mc millan mayer free energy.

Where,

T = temperature

V= volume of the solution

2= protein molar density

R= ideal gas constant

Nav= avogadros number

d2= protein diameter

= fraction of the solution taken up the protein molecules

r= centre-to-centre separations between a pair of molecules.

1) Derive an analytical expression for the osmotic pressure and the proteinchemical potential.

Chemical potential is given by,

=id

+ex

We have the equation,


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=

=

() + () +2

()

2

Now solve the partial differentiation part

()

4 3 (1 )

We know that=

,

Using the above formula we get,

() =

()()()()()

Now simplifying this equation we get,

() =

()

Substituting this in eqn 2,

2RT

()

3

Now solve ,

We know =

=

4

Now substitute eqn 4 in eqn 3 we get,

2RT

()

5

Now solve the second term in eqn 2 i.e ,

4 () 6

We know that w(r)= -(

)7


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Now substitute eqn 7 in 6,

= 4 (-(

))dr

Simplifying the above eqn,

=

8

Substitute eqn 8 and 5 in eqn 2 we get,

= RT

() +RT

()

+

Since we know =

so substituting this in the second term of the above eqn,

= RT () +RT

()+

9

Now simplifying eqn 9 we get,

ex=RT(

() )+

10

total chemical potential is given by,

=i.d+e.x 11

=e.x12

i.d=RT ln213

Now substitute 10 & 13 in eqn 11 we get,

= RT ln2+ RT(

() )+

14

eqn 14 depicts the protein chemical potential.

i) Osmotic pressure

The relation between osmotic pressure and macmillan mayer free energy is given by,


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F=-V+2N2 15

Considering excess conditions,

Fe.x

=-e.x

V+2N216

Divide eqn 16 by v,

. =-

e.x+

17

Since =218

Substitute 18 in 17 we get,

. = -

e.x+

e.x2

e.x= e.x2-. 19

Substitute 10 & 1 in 19 we get,

e.x=2RT(

() )+

20

But we know that =i.d+

e.x21

i.d=2RT 22

After Substituting 22 and 20 in 21 and simplifying we get \,

=2RT(

() )+

23.

The equation 23 depicts the osmotic pressure derivation.

2) Determine the spinodial(T vs ) and the critical point for a solution withsalt concentration equal to 1.0 M and the range of the potential n equal to6.

Spinodal curve is used to find the stability of the solution. Basically its a limit; within this limit

any small fluctuation (infinitesimal) with respect to temperature, pressure and other parameters

will lead to spinodal decomposition. spinodal decomposition is the decomposition of the solution

When there is a change in the thermodynamic parameters such as temperature into multiple


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changes. Above this curve the solution is exposed to fluctuations but it is metastable. Now we

have the lysozyme solution on which the salting out process is done and now ready for the phase

separations. We have to calculate all the thermodynamic parameters for the liquids which are

later used in phase separations. Now we have to find the spinodial and the critical temperature

for the lysozyme solution. Now in a phase separation, there are 2 phases one phase which is light

and is the top layer whereas the dense phase forms the bottom layer. So we are splitting it into

two phases 2L which represents the light phase (top layer) and 2D represents the dense phase

(bottom layer) which are the protein molar densities of the lysozyme solution. In mathcad, we

first derived the osmotic pressure and the chemical potential and we split it into dense and light

components. Since we have two phases which are light and dense there are two chemical

potentials (2L and 2D). Now to check the equation we calculated the chemical potentials of

both light and dense as a function of molar density and temperature.

The figure above depicts the spinodal curve for light molar density lysozyme solution in

the protein phase diagram where chemical potential (as function of molar density and

temperature) is taken versus its molar density.


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The figure above depicts the spinodal curve for dense molar density lysozyme solution in

the protein phase diagram where chemical potential (as function of molar density and

temperature) is taken versus its molar density.

From both graph the curve looks similar because they both are the same components but with

dense and light phases.any point above the curve will result in a single phase and below that

will result into two phases (spinodal decomposition).then we took the graph of osmoticpressure versus the molar density (both light and dense) to get the spinodal curve.

2D


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The above graph depicts the spinodal where the osmotic pressure is plotted versus molar

density.

Now we have to find the critical point which is the extreme point of spinodal curve. In order

to find this we equate the osmotic pressure (both first derivative and second deriavative) to

zero.

02

2

2 T,( )d

d

2

Using mathcad we get the following solution,

2L=9.372

T=328.264

Now plotting the graph temperature versus molar density of the protein we get the spinodal

curve of the lysozyme protein density.

The figure above depicts the spinodal curve and the critical point (the peak of the curve) of the

lysozyme solution where above the curve it will give a stable solution and if it is below it leads to

decomposition into two phases (slow crystallization). Also if its even below the co existence curve

spontaneously decompose.

02

2 T,( )d

d=


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3) calculate the liquid-liquid co-existence curve (T vs ) for solutions at salt

concentration equal to 0.5, 1.0 and 1.5M with n equal to 6.

Liquid liquid co-existence curve is defined as the condition where the 2 distinct liquid phases

may co-exist. In this condition the chemical potential of a particular solution component is equal

in that distinct phase of the solution in which it exists. In order to calculate the Liquid liquid co-

existence we defined the chemical potential and osmotic pressure of both light and dense

components in mathcad. Then we adjusted the salt concentration(3) to 0.5,1.0 and 1.5. after

adjusting we input values of T(200),2L(0.5) and 2D(20). The critical point of 2 was found to

be 9.372 so we adjusted 2L9.4 so that in lies in the range of critical point.then

we equated the chemical potential and osmotic pressure of both light and dense fluids to each

other.

L 2L T,( ) D 2D T,( )

From the above equation we found various values CO as a function of T with find command in

mathcad to give us various values of temperature and the corresponding 2L and 2D which was

written in a excel sheet. Plotting the graph give us Liquid liquid co-existence curve for 3=0.5,

1.0 and 1.5 with n=6 as constant.

2L 2L T,( ) 2D 2D T,( )=

0

50

100

150

200

250

300

350

0 10 20 30

Light fluid

Dense fluid

3=0.5

temperature

2


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The above three graphs shows di

constant n=6.the critical tempera

solution has less components of

concentration increases and after

temperature at more and more sa

dense fluid starts at the critical te

0

50

100

150

200

250

300

350

0 5 10

temperature

0

50100

150

200

250

300

350

400

0 1

tem

perature

ABHISEK

MSc BIO

7810595

fferent Liquid liquid co-existence curve for di

ture rises as the concentration of the salt is incr

rotein and more concentration of the salt as the

the critical point the curve slides down becaus

lt concentration. The light fluid may end at criti

mp and slide down with changing molar densit

15 20 25 30

Light fluid

Dense flui

2

0 20 30

Light fluid

dense fluid

3=1.5

2

UMRAO SINGH

TECHNOLOGY

ferent 3 but

ased. The

salt

of less

cal temp and

(2).


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4) Determine the effect of changing n on the liquid-liquid co-existencecurve for a solution with salt concentration equal to 1.0 M. What are the

implications of this result for the shape of the protein phase diagram?

For this problem, in mathcad we have previously defined chemical potential, osmotic pressure

and also the critical point for both light and dense. Now we change n from values 1 to 9 so

totally 8 cases keeping everything constant and salt concentration to 1.0 M. we feeded the values

into an excel and plotted the graphs.

-1000

-800

-600

-400

-200

0

200

400

600

800

1000

0 5 10 15 20 25 30

n=1(light)

n=1(dense)

n=2(light)

n=2(dense)

n=4(light)

n=4(dense)

n=5(light)

n=5(dense)

tem

erature

2

0

50

100

150

200

250

300

350

0 5 10 15 20 25 30

n=6(light)

n=6(dense)

n=7(light)

n=7(dense)

n=8(light)

n=8(dense)

n=9(light)

n=9(dense)

temperature

2


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From the temperature vs. 2 gra

molecules that surrounds the pro

temperature (becomes positive)

that harbour the protein. at n=3 t

potential equation (n-3).liquid li

disordered phases. One dilute wi

concentrated with a protein (whi

process, first we take little amou

concentration increases the attrabelow the spinodal curve, there i

buries into the spinodal giving a

liquids. From the critical temper

decreases. This results in less are

increased are above the curve w

critical temperature decreases an

5) What happens if theWhy does this happ

We know that w(r) is the

center-to-center separatio

attractive forces between

-1500

-1000

-500

0

500

1000

1500

0 5

critical

n

ABHISEK

MSc BIO

7810595

h, it implies that as the n which is the variable

tein increases the temperature also decreases. A

ecreases as n increases. This is due to more pro

e whole function becomes infinite due the ter

uid equilibrium exists where there is a equilibri

th the protein (which forms liquid after salting

h forms crystals) which is disordered phases.

nt of the salt to be added to the protein solution.

tive forces between the proteins also increasess a slow crystallization but as the salt is increas

liquid- liquid phase which the solution decomp

ture vs. n graph as the n increases the critical t

a which is below the curve where the protein cr

ich represents stable protein. So as more the n i

d the liquid-liquid separation will be ineffective

range ofw(r) becomes very large (i.e.

n?

attractive forces between the protein molecules

ns between a pair of proteins. As the w(r) range

a pair of protein increases which results in prec

10 15

emperature vs n

critical

temperature

UMRAO SINGH

TECHNOLOGY

f proteins

t n=4, the

tein molecules

of the chemical

um for two

ut) and the other

uring salting out

As the salt

nd it goesd further it

ses into 2

mperature

ystallizes and

ncreases the

.

mall n)?

where r is the

increases the

ipitation. As


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previously explained we know that as n reduces the critical temp increases which results

in liquid- liquid separations. Increasing salt solution moves the protein into crystal-liquidphase where the concentrated protein becomes crystal and the dilute ones becomes liquid.

But at this point we have a large activation barrier in this region which prevents the

protein crystallization and as more salt concentration is increased the protein decomposes

into liquids and the separation is ineffective. And above the co-existence curve theprotein becomes stable. So having small n will make large liquid- liquid phase and higher

critical temperature so that the protein crystallizes in the crystal-liquid phase andovercomes the activation barrier.

MATHCAD

T 200 350..:=

R 8.314:=

Nav 6.2031023

:=

3.14:=

3 1.0:=

Kb RNav

:=

Kb 750 1203+( ):=

d2

3.5 109

:=

d2

3.5 109

:=

n 6:=

d2( )

3 Nav

6:=

*

2 T,( ) 2R T2( )

22( )

3 2 ( )+ 1+

1 2 ( )3

2 Nav( )2

d2( )

32( )

2

3 n( )

+:=


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L

d2( )

3 Nav

6:=

2L 2L T,( ) T R ln 2L( ) T R8 L 2L( ) 9 L 2L( )

2 3 L 2L( )

3+

1 L 2L( )[ ]3

+

4 Nav22L d

2( )3

3 n( )

+:=

D

d2( )

3 Nav

6:=

2D 2D T,( ) T R( ) ln 2D( ) T R8 D 2D( ) 9 D 2D( )

2 3 D 2D( )

3+

1 D 2D( )[ ]3

+

4 Nav22D d

2( )3

3 n( )

+:=

2L 1 200,( ) 615.298=

2L 2 200,( ) 67.689=

2L 0.1 0.2, 60..:=

2L 2L 250,( )

3-4.84310

3-3.4610

3-2.67410

3-2.13310

3-1.72610

3-1.40410

3-1.1410

-919.031

-730.557

-567.776

-425.755

-300.852

-190.304

-91.959

-4.112

...

=


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2D 2D 250,( )

3-4.84310

3-3.4610

3-2.67410

3-2.13310

3-1.72610

3-1.40410

3-1.1410

-919.031

-730.557

-567.776

-425.755

-300.852

-190.304

-91.959

-4.112

...

=

2D 1 200,( ) 615.298=

2D 2 200,( ) 67.689=

2D 0.1 0.2, 60..:=

1 200,( ) 1.356 103

=

2 200,( ) 2.112 103

=

2 0.1 0.2, 60..:=


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2 250,( )

204.984

404.253

597.832

785.744

968.016

31.14510

31.31610

31.48110

31.64110

31.79610

31.94510

32.08810

32.22610

32.35910

32.48610

...

=

2 1:=

T 250:=

Given

02

2 T,( )d

d

02

2

2 T,( )d

d

2

Find 2 T,( )9.372

328.264

=

Given

02

2 T,( )d

d

T 2( ) Find T( ):=

2 0.1 0.2, 40..:=


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L

d2( )

3 Nav

6:=

*

L 2L T,( ) 2L( )R T2L( ) L( )[ ]

22L( ) L( )[ ]

3 2L( ) L( )[ ]+ 1+

1 2L( ) L( )[ ][ ]3

2 Nav( )2

d2( )

32L( )

2

3 n( )

+:=

2L 2L T,( ) T R ln 2L( ) T R8 L 2L( ) 9 L 2L( )

2 3 L 2L( )

3+

1 L 2L( )[ ]3

+

4 Nav22L d

2( )3

3 n( )

+:=

D

d2( )

3 Nav

6:=

*

D 2D T,( ) 2D( )R T2D( ) D( )[ ]

22D( ) D( )[ ]

3 2D( ) D( )[ ]+ 1+

1 2D( ) D( )[ ][ ]3

2 Nav( )2

d2( )

32D( )

2

3 n( )

+:=

2D 2D T,( ) T R( ) ln 2D( ) T R8 D 2D( ) 9 D 2D( )2 3 D 2D( )3+

1 D 2D( )[ ]3

+4 Nav

2

2D d 2( )3

3 n( )

+:=

T 200:=

2L 0.5:=

2D 20:=

Given

2L 2L T,( ) 2D 2D T,( )

L 2L T,( ) D 2D T,( )

2L 9.


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2D 9.>

CO T( ) Find 2L 2D,( ):=

CO 300( )3.615

17.135

=

coursework report biochemical engg

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