healthcare insulin problem

8/13/2019 Healthcare Insulin Problem

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Manufacturer of InsulinProblem submitted by:Professor David Meredith, P. E., Pennsylvania State UniversityFayette,Uniontown, PA{

problem # 6

Problem Statement

Bioengineers are responsible for developing processessuch as the recombinant DNA process described here.They also develop artificial joints and human organs andmedical tools to support surgeons. Chemical engineersare involved with sizing the process equipment and

developing the flow path through the process. Electricaland mechanical engineers build the support systemssuch as the instrumentation and distilled water supply.

To meet this years increase in global demand, yourteam is to design a manufacturing facility capable ofproducing 1,800 kg/yr of human insulin. You will beusing a recombinant DNA process by growing E. colibacteria that contain Trp-LE-Met-proinsulin. You willneed to provide the proper growth media, air, heatand agitation. After a batch has matured, you musthomogenize the material then separate the productusing centrifuges, diafilters and chromatography in a

series of steps with specific conditions as you extract andpurify the product. Finally, the insulin crystals must befreeze-dried into crystals for final distribution.

Background

Human insulin is a polypeptide containing 51 aminoacids arranged in two chains. The A chain contains 21amino acids and the B chain includes 30 amino acids.The two chains are linked by two disulfide bonds.Human insulin has a molecular weight of 5,734 and an

isoelectric point of 5.4.

In the proinsulin process (see Figure 6-1), the plasmidcircle of DNA (1) that exists in bacteria are geneticallymodified to include part of the human DNA (2). Thisrecombinant plasmid (3) is inserted into Eschericia coli(E. coli) (4). A large colony of the inoculated E. colibacteria cells are fermented to overproduce Trp-LE-Met-proinsulin in the form of inclusion bodies (5), whichare recovered and soluablized. Proinsulin is releasedby cleaving the methionine linker using CNBr (6). Theproinsulin chain is subjected to a folding process to allowintermolecular disulfide bonds to form (7), and the C

peptide is then cleaved with enzymes (8) to yield humaninsulin (9). (Note the process described here has beensimplified and several key steps have been omitted.Dont try this at home.)

Figure 6-1 Production Process


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The first step of the process is fermentation ( SeeFigure 6-2 ). The nutrients are premixed and added todeionized water in a large tank with an agitator. Whenconditions are acceptable, a seed colony of inoculatedbacteria are added to the tank. The tank temperatureis maintained at 37C by a water jacket that can be

either heated (early in the cycle when heat loss fromthe reactor is greater than the heat produced by thebacteria) or cooled (later in the cycle when the bacteriaproduce more heat than the reactor loses).

Assumptions and Givens

The manufacturing facility operates around the clockfor 330 days per year. A new batch is initiated every48 hours resulting in 160 batches per year. Thisimplies a required cycle production of about 11.3 kgof insulin crystals per batch. See Figure 6-2.

A cubic meter is 1,000 liters and a cubic centimeter(cm3) is equal to one ml.

A micron is 10-6meter.

Each batch requires glucose (C6H

12O

6) to feed

the bacteria (CH1.8

O0.5N

0.2) in 30 m3of broth

to grow to a final concentration of 37 g of dry cell

weight per liter of broth. The basic unbalancedchemical equation is:

C6H

12O

6 + H

2O + NH

3 + (seed bacteria)

=> CH1.8

O0.5

N0.2

(6-1)

Atomic masses for selected elements are givenbelow:

Hydrogen=1.008 Carbon=12.01Nitrogen=14.007 Oxygen=15.999Sulfur=32.064 Zinc=65.37Bromine=79.91

51. The mass of glucose (kg) required to growone batch of the e. coli bacteria to the final

concentration is closest to:

a. 1,350 d. 18,500b. 7,500 e. 225,000c. 4,500

Additional Assumptions and Givens

In addition to temperature and food (glucose), thebacteria also need oxygen to grow. The oxygen is

Figure 6-2 Fermentation Process

NutrientPremixTank

Air

FilterCompressor

Agitator

TemperatureControlJacket

To Cell Recovery


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provided by air, which is 21% oxygen by volume.The air is compressed by a 50 kW compressor to 7atmospheres of pressure and is filtered through acartridge air filter before being sparged (released astiny bubbles) into the bottom of the reactor.

As the air bubbles rise through the broth, some ofthe oxygen dissolves into the broth solution to betaken up by the bacteria. The bacteria require 0.35g of O

2per g of living cells each hour.

The minimum dissolved oxygen concentration tosupport growth is 0.2 mg / liter of broth.

The maximum dissolved oxygen concentration in themedium is 6.7 mg/liter at 37C.

The amount of oxygen required to support thecolony growth is given by:

(6-2)

where,

L = liter

X= final concentration of cells (dry cell weight) per liter

of broth (g/L),

kLA

= oxygen volume transfer rate

(m3O2/ m3

broth-h)

CDO* = maximum dissolved oxygen (DO) concentration

in the broth (gO2

/L)

CDO

= minimum DO concentration to

support growth (gO2

/L)

qO2

= respiration rate of cells (gO2

/gcell

h)

52. The flow rate of air (m3/s) to maintain the broth atthe final cell concentration is closest to:

a. 2.3 d. 81b. 17 e. 292

c. 70


To maintain constant growing conditions throughoutthe broth, the aerated mixture is stirred continuouslyby a large agitator that looks like a kitchen mixer onsteroids.

The agitator is 2.5 m in diameter and stirs the tankat 60 revolutions per minute. The motor inputpower, P

o(kW), required by this agitator is given by:

Po= 0.7 N3D

a

2V (6-3)

where,

N= speed of the agitator (revolution per second)

Da= diameter of the agitator (m)

V= the volume of the broth in the reactor (m3)

Because aerated mixtures are less dense than non-aerated mixtures, the motor power is reduced. Theinput, P

g(kW), for the agitator for the bioreactor is

empirically given by:

(6-4)

where,

k = proportionality constant (use 0.5)

Q= volumetric flow rate of gas per volume of tank (s-1)

53. When the volumetric flow rate of gas is equal to

15 (s-1), the electrical input (kW) to operate the

agitator is closest to:

a. 71 d. 813b. 97 e. 2,268

c. 140


At the end of the 18-hour fermentation time, thebroth is cooled from 37C to 10C to quench thebiological action. This cooling process occurs inabout 30 minutes. Chilled water at 5C is injectedinto the reactor jacket at a flow rate of 50 L/s, picks

up heat from the broth and exits the reactor jacketto return to the cooling system.

The density and specific heat of the broth are 1,020kg/m3and 3.8 kJ/kg-C respectively.

The density and specific heat of the water are 1,000kg/m3and 4.19 kJ/kg-C respectively.

LA

2

k ( )X= DO DO

O

C C

q

0.452 3

0.56

o ag

P NDP k

Q


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The energy balance equation is given by:

( ) ( )broth water

V cp T Q t cp T = (6-5)

where,

cp= specific heat of the fluid (kJ/kg-C)= density of the fluid (kg/m3)

V= volume of the broth (m3)

Q= volumetric flow rate of chilled water (m3/h)

t= time duration of chilled water flow (h)

T= temperature difference for the fluid before andafter the heat transfer occurs (C)

54. The average water temperature in (C)leaving thereactor jacket during this cooling period is closest to:

a. 6 d. 20b. 8 e. 35

c. 13

Additional Background

Once the broth has been cooled, it is transferred toa holding tank to allow the fermentation tank to becleaned and prepared for the next cycle. To reduce thevolume of material to be processed, the 30,000 litersof broth are passed through a centrifuge to reduce thevolume by a factor of four. This process also removesmost of the extra-cellular impurities.

Next the thickened cell sludge is diluted with an equalvolume of buffer solution consisting of 94.4% distilledwater, 0.7% EDTA (ethylenediaminetetraacetic acid),a sequestering agent to prevent further reactions and2.9% TRIS-Base (trishydroxymethylaminomethane), abuffering agent. This process facilitates the separationof the cell debris particles from the inclusion bodies thatcontain the proinsulin.


Figure 6-3b. Cell Recovery Process

The structure of EDTA is shown above in Figure (6-3a)

55. The molecular weight of EDTA is closest to:

a. 265 d. 292b. 276 e. 340

c. 286


A high-pressure homogenizer is used to breakthe cells and release the inclusion bodies. Passingthe 15,000 liters of buffered sludge through thehomogenizer three consecutive times through acircular nozzle at high velocity and under a pressurechange of 80 MPa each time, ruptures the cell wallsto expose the inclusion bodies.

The homogenizer capacity is 10,102 L/h and the

electrical input to the motor is 225 kW.

It takes about 1.5 hrs for each pass or 4.5 hrs totalfor the process.

The density of the sludge at this point is 985 kg/m3.

The pressure change, P (kPa), is given by:

(6-4)

where,

= density of the fluid (kg/m3)

V = throat velocity through the nozzle (m/s)

The flow rate of fluid, Q (m3/s) is given by:

Q = VA (6-5)

where,

V = throat velocity (m/s)

A= cross-sectional area of the nozzle throat (m2)Figure 6-3a EDTA Structure

2 P = V / 2,000

From Fermentation

Holding

TankHomogenizer

To Solubilization

Centrifuge

Centrifuge


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56. The nozzle diameter (mm) is closest to:

a. 3 d. 17b. 5 e. 22

c. 9

Additional Assumptions and Givens The 15,000 liters of homogenized sludge is returned

to the centrifuge to separate the inclusion bodiesfrom the cell debris. The inclusion bodies have adiameter of 1.0 microns and a density of 1.3 g/cm3.

Assume water (= 1.0 g/cm3) is the liquid mediawith a kinematic viscosity of 1.004x10-6m2/s.

The centrifuge has an acceleration factor of 1,500g(one g is equal to an acceleration of 9.81 m/s2), has147,718 m2of effective clarifying surface area anda motor input of 25.6 kW with a yield of 98%. The

centrifuge reduces the volume from 15,000 litersto 1,400 liters of slurry at the end of the recoverysection.

The throughput or flow rate of entering fluids to beseparated, (kg/s) is given by:

(6-6)

where,

d = particle diameter (m)

s= density of the particle (g/cm3)

1= density of the fluid media (g/cm3)

(r2)= acceleration factor (m/s2)

V / D= effective clarifying surface area (m2)

Fs = correction factor for fraction of solid present (use

0.05 for 20% solids)

= kinematic viscosity (m2/s) of the fluid

= shape factor (use 1.0 for spherical)

The process time, t (s) is given by:

(6-7)

where,

Q = entering flow volume to be separated (m3)

= density of the liquid media entering the filter (kg/m3)

= throughput (kg/s)

57. The length of time (h) required to process each batchis closest to:

a. 1.8 d. 8.3b. 2.3 e. 46

c. 5.3


The 1,400 L of inclusion body suspension is transferred toa glass-lined agitator tank and mixed with small quantitiesof urea and of 2-mercaptoethanol. Urea is a chaotropicagent that dissolves the denatured protein in the inclusionbodies and 2-mercaptoethanol is a reductant that reducesdisulfide bonds. It takes a reaction time of 8 hours tosolubilize 95% of the proinsulin. The next step is to

remove the excess urea and 2-mercaptoethanol with adiafilter and replace it with distilled water. This solutionis then filtered through a dead end filter to removefine particles that might overload filtering processesdownstream. See Figure (6-4).


The design flux capacity of the membrane usedis 4.2 liters per hour for each square meter ofmembrane.

2 2( )( )( )

18

s l sd r V D F

=

Qt =

Figure 6-4 Solubilization Process

From Cell Recovery

To CNBr CleavageDistilled

Water

Reactant

Diaflter

Agitated

TankHolding

Tank

Filter


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The diafiltration process takes 6 hours to complete.

The sizing equation for a diafilter is given by:

Q = J A t (6-7)

where,

Q= volume of media to be processed (liters)

J= Design flux capacity (l/m2-h)

A= membrane area (m2)

t= process time (h)

58. The membrane area (m2) required for this process isclosest to:

a. 0.02 d. 55b. 18 e. 590

c. 41


The chimeric protein is cleaved in a well-mixed reactorwith CNBr (cyanogen bromide) in a 70% formic acidsolution into the signal sequence Trp-LE-Met, whichcontains 121 amino acids and the denatured proinsulin(82 amino acids). The reaction takes 12 hours at 20C.The final mass of proinsulin is 31% of the mass of the

initial Trp-LE-Met-proinsulin. See Figure (6-5).


A small amount of toxic cyanide gas is formed as abyproduct of the cleavage reaction. This toxic gas

and excess reactants are removed by applying avacuum and raising the temperature to 35C (theboiling temperature of CNBr at that pressure).

Assume the absolute pressure is 0.28 atmospheresat 35C and you have to boil off 20 kg of CNBr

vapor. One atmosphere (atm) is 101 kPa andassume the Ideal Gas law:

PV = nRT (6-8)

where,

R= universal gas constant and is 8.314 kPa-m3/[kmol-K]

P= pressure in kPa

V= volume in m3

n= number of kilomoles of gas

T= temperature in degrees Kelvin.

59. The minimum volume (m3) of the CNBr vaporproduced at evaporator conditions is closest to:

a. 1.940 d. 54.3b. 9.570 e. 1,800

c. 17.0


A sulfitolysis step is used to unfold the proinsulin, break

any disulfide bonds and add SO3moieties (moleculesegments) to all sulfur residues on the cysteines. Thisreaction occurs in a well-stirred reactor under alkalineconditions (pH 9 to 11) over a 12-hour period. Thereagents are filtered through a diafilter and replacedwith a 20% by weight solution of HCl guanidine.

Figure 6-5 Cleavage and Sulfitolysis

VacuumHCI + Guanidine

Sultolysis CNBr Cleavage

Evaporator

Reactant

Diaflter

Agitated

Tank

Holding

Tank

Agitated

Tank


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The human proinsulin(S-SO3-)

6is next chromatographically

purified in an ion-exchange column. The bed consistsof small beads coated with a cation (group of atomscarrying a positive electric charge) exchange resin. Anionic bond is temporarily formed with the properly

formed proinsulin(S-SO3-)6molecules as the solutionpasses slowly up through the resin. The molecules arereleased when the column is regenerated (back-flushed)with acid. See Figure 6-6.


The ion exchange column measures 140 cm indiameter with a bed height of 25 cm.

The 8,000 liters of solution at this point contains 28kg of proinsulin(S-SO

3-)

6.

The maximum holding capacity of the resin is 18 mgof proinsulin(S-SO

3-)

6for each ml of nominal bed

volume. The regeneration cycle begins when thebed reaches 70% of maximum holding capacity.


Folding of the proinsulin(S-SO3-)

6and disulfide bond

formation takes place in a a well-mixed reactor usingmercaptoethanol to facilitate the disulfide interchangereaction. The reaction is carried out at 8C for 12

hours and reaches a yield of 85%. The reactions areremoved and the protein solution is concentratedwith a diafiltration unit followed by purification in ahydrophobic interaction chromatography (HIC) columnto remove unfolded or incorrectly folded molecules.

The final reaction step is to enzymatically remove theC peptide from the human proinsulin using trypsin andcarboxypeptidase B. The reaction occurs in a well-mixedreactor at 10C for 12 hours. The reagents are removedby another diafiltration step followed by purification inanother ion-exchange chromatography (IEC) column.

Several additional purification steps are included inthe process. A reversed-phase, high-pressure liquid-chromotography (PR-HPLC) step removes structurallysimilar insulin-like components. A diafiltration processremoves the reactants and concentrates the solutionby a factor of two. The final purification step is a gelfiltration chromatographic column followed by anotherdiafiltration process to concentrate the solution bya factor of ten. The 500-liter solution at this pointcontains 12.8 kg of insulin. An ultrafilter capturesparticles larger than 100 nanometers (>10-7m). SeeFigure 6-7.

The final step is to crystallize the final insulin. The insulinsolution is mixed with ammonium acetate and zincchloride in an agitated reaction tank. The crystallizationinto insulin

6-Zn

2is carried out at 5C for twelve hours.

Filtered

solution

Acid Re eneration

Proinsulin solution

To Folding

From

Sulfitolysis

Figure 6-6 Ion Exchange Chromatography

Agitated

Tank

Holding

Tank

Reactant

Diaflter

Distilled

Water

Refolding

Figure 6-7 Refolding and Enzymatic Conversion

Distilled

WaterAgitated

TankHolding

Tank

Reactant

Diaflter

Enzymatic Conversion

Final Purication

IEC RP-HPLC Gel Filter


8/16

The 11.3 kg of insulin crystals are separated in a basketcentrifuge and freeze-dried at 20C under a vacuum(0.0015 atmospheres).


An average diabetic individuals daily insulin usage is 60

units. (This quantity varies with body weight, activitylevel and diet.)

A unit is 1/22

mg of crystallized insulin.

60. The number of diabetics supported by the insulinproduced at this facility is closest to:

a. 2,100 d. 1.8x106b. 4,200 e. 4x106c. 11,500


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Competition{ }Part I

and

Part II

Solution2007


10/16

Problem #6

Manufacture of Insulin

51. Answer a

The molecular weight of glucose is:

(C 180(6x16)(12x1)(6x12))OH 6126 =++=

The molecular weight of the bacteria is:

(CH = (1x12) + (1.8x1) + (0.5x16) + (0.2x14) = 24.6)NO 0.20.51.8

Six moles of bacteria require one mole of glucose, or

6x24.6 = 147.6g of bacteria require 180g of glucose.

Each batch requires enough glucose to feed the bacteria in 30 m of broth to a

concentration of 37g of dry cell weight per liter of broth.

3

(37g/lt) (30x10 lt) = 1,110 kg of bacteria.3

This requires:

bacteriaof147.6g

bacteriaof1,110kgx 180g of glucose = 1,353 kg of glucose

52. Answer d

Using equation (6-2), and solving in terms of K yields:LA

KDODO

*

02LA

CC

)(X)(q

=

where,

K =LA Oxygen Volume Transfer (in m h)-/m broth3

023

X = final concentration of cells = 37g /Lcells

q =02 Respiration rate of cells = 0.35g h/g cell02

TEAMS 2007

Part I and Part II Solutions 37


11/16

C DO = Minimum DO concentration to support growth = 0.2mg L/02

C DO* = Maximum DO concentration in the broth = 6.7mg L/02

K /L)0.2mg/L(6.7mg

)/g)(1,000mg/g/L)(0.35g(37g

0202

0202cell02cellLA

=

h

= 1,992 m h/m broth3

023

Each batch is 30m 3 and the O2represents 21% of the air:

Also there are 3,600 s/h.

Therefore the Volumetric rate of air required is:

/s79.04m)/m(0.21m(3,600s/h)

h)/m992m/batch)(1,(30mair

3

air3

023

broth30233

=

53. Answer a

Using equation (6-3) with:

N = Speed of the Agitator = 60 rev/min = 1 rev/s

D a = Diameter of the Agitator = 2.5m, and

V = Broth volume in the reactor = 30m 3 , yields:

kW131.25)(30m(2.5m)/s)(0.7)(1rev 3230 ==

Then using equation (6-4) with:

k = 0.5

kW131.250 =

N = 1 rev/s

D a = 2.5m, and

Q = 15/s, yields:

TEAMS 2007



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0.45

0.56

32

g(15/s)

.5m)(1rev/s)(2kW)(131.25(0.5)

=

= 70.2 kW

54. Answer c

Density of broth = 1,020 kg/m 3 =broth

C =pbroth specific heat of broth = 3.8 kJ/kg- C0

T C270broth=

3water kg/m1,000=

C CkJ/kg4.19 0pwater =

Q = volumetric flow rate of chilled water (in m /h)3

= 50 L/s =)(1,000L/m

600s/h)(50L/s)(3,3

= 180 m /h3

0.5h60min/h

30mint ==

Solving equation (6-5) in terms of T yields:water

t))()(C(Q)(

)T)()(C)(30m(T

pwaterwater

brothpbroth

3

broth

water

=

=

c)(0.5h)g)(4.19kJ/kg/m/h)(1,000k(180m

C)C)(27)(3.8kJ/kg)(30m(1,020kg/m

033

0033

= 8.32 C0

Water Temperature = 5 = 13.32 0 CTC0 +

TEAMS 2007



13/16

55. Answer d

The molecular formula for EDTA is:

C and the molecular weight of the molecule is:281610 NOH

(12g x10) + (1g x16) + (8x16g) + (2x14g) = 292 g/mol

56. Answer a

Using equation (6-6) and solving it in terms of V yields:

V =

P)(2,000)(

where,

= Pressure Change = 80 MPa = 80x10 3 kPa

= density of fluid = 985 kg/m 3

therefore,

V =3

3

985kg/m

)kPa)(2,000(80x10= 403.03 m/s

Using equation (6-7) and solving it in terms of A yields:

A = Cross-Sectional Area of Nozzle Throat

Q =s/min)(90min)(60

1,000L/m

15,000L/h

Time

Volume 3=

= 2.7x10 /sm33

A = 2633

m6.892x10m/s403.03

/sm2.7x10

V

Q

==

A = 6.892 mm 2

TEAMS 2007



14/16

A = ,4

D2or

D =

)m(4)(6.892m

4A 2=

D = 2.96 mm

57. Answer b

d = particle parameter = 1.0 microns = 1x10 m6

= density of the particle = 1.3g/cms3

= density of fluid media = 1g/cm 3 1

n = kinematic viscosity of fluid = 1.004x10 /sm26

rw = acceleration factor = 1,500g = 1,500 x 9.81m/s2 2

V/D = effective clarifying surface area = 147,718m 2

= shape factor = 1.0

F s = correction factor for fraction of solid present = 0.05

Using equation (6-8) yields:

/s)(1.0)mx10(18)(1.004

)(0.05))(147,718m81m/s)(1,500x9.1.0g/cm(1.3g/cmm)(1x10

26

223326

=

= 1.8 kg/s

Using equation (6-9) yields:

Q = (15,000 lt)(1 m 33 15mlt)/1,000 =

= 1.0g/cm 333 kg/m1.0x10=

==1.8kg/s

)`kg/m)(1.0x10(15mt

333

8,333s = 2.3h

TEAMS 2007



15/16

Note:

In reality the density of the entering sludge is 13,600 lt at a density of 1.0g/cm plus

1,400 lt at a density of 1.3g/cm 3 for a total of 15,420 kg in 15,000 lt for a density of

1.03g/cm 3 .

3

58. Answer d

Solving equation (6-10) in terms of A yields:

A =t)(J)(

Q

, where

Q = volume of media to be processed = 1,4000 l

J = Design flux capacity = 4.2 l/m -h2

t = process time = 6h

A =h)(6h)l/m(4.2

l1,4002

= 55.55 m 2

59. Answer c

In order to use equation (6-11) we need to calculate (n) the number of kilomoles

of gas. The molecular weight of the CNBr vapor is:

12g + 14g + 80g = 106g/mol = 106 kg/kmol

The removal of 20 kg represents

n =106kg/kmol

20kg= 0.189 kmol of gas.

Solving equation (6-11) in terms of V yields:

V =P

(n)(R)(T)

TEAMS 2007



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healthcare insulin problem

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