Manufacturer of Insulin Problem submitted by:Professor David Meredith, P. E., Pennsylvania State University–Fayette, Uniontown, PA

{problem # 6

Problem Statement

Bioengineers are responsible for developing processes such as the recombinant DNA process described here. They also develop artificial joints and human organs and medical tools to support surgeons. Chemical engineers are involved with sizing the process equipment and developing the flow path through the process. Electrical and mechanical engineers build the support systems such as the instrumentation and distilled water supply.

To meet this year’s increase in global demand, your team is to design a manufacturing facility capable of producing 1,800 kg/yr of human insulin. You will be using a recombinant DNA process by growing E. coli bacteria that contain Trp-LE’-Met-proinsulin. You will need to provide the proper growth media, air, heat and agitation. After a batch has matured, you must homogenize the material then separate the product using centrifuges, diafilters and chromatography in a series of steps with specific conditions as you extract and purify the product. Finally, the insulin crystals must be freeze-dried into crystals for final distribution.

Background

Human insulin is a polypeptide containing 51 amino acids arranged in two chains. The A chain contains 21 amino 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 plasmid circle of DNA (1) that exists in bacteria are genetically modified to include part of the human DNA (2). This recombinant plasmid (3) is inserted into Eschericia coli (E. coli) (4). A large colony of the inoculated E. coli bacteria cells are fermented to overproduce Trp-LE’-Met-proinsulin in the form of inclusion bodies (5), which are recovered and soluablized. Proinsulin is released by cleaving the methionine linker using CNBr (6). The proinsulin chain is subjected to a folding process to allow intermolecular disulfide bonds to form (7), and the C peptide is then cleaved with enzymes (8) to yield human insulin (9). (Note – the process described here has been simplified and several key steps have been omitted. Don’t try this at home.)

Figure 6-1 Production Process

The first step of the process is fermentation ( See Figure 6-2 ). The nutrients are premixed and added to deionized water in a large tank with an agitator. When conditions are acceptable, a seed colony of inoculated bacteria are added to the tank. The tank temperature is maintained at 37°C by a water jacket that can be either heated (early in the cycle when heat loss from the reactor is greater than the heat produced by the bacteria) or cooled (later in the cycle when the bacteria produce more heat than the reactor loses).

Assumptions and Givens

• The manufacturing facility operates around the clock for 330 days per year. A new batch is initiated every 48 hours resulting in 160 batches per year. This implies a required cycle production of about 11.3 kg of 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 10-6 meter.

• Each batch requires glucose (C6H12O6) to feed

the bacteria (CH1.8O0.5N0.2) in 30 m3 of broth to grow to a final concentration of 37 g of dry cell

weight per liter of broth. The basic unbalanced chemical equation is:

C6H12O6 + H2O + NH3 + (seed bacteria)

=> CH1.8O0.5N0.2 (6-1)

• Atomic masses for selected elements are given below:

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

51. The mass of glucose (kg) required to grow one batch of the e. coli bacteria to the final concentration is closest to:

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

Additional Assumptions and Givens• In addition to temperature and food (glucose), the

bacteria also need oxygen to grow. The oxygen is

Figure 6-2 Fermentation Process

NutrientPremixTank

AirFilterCompressor

Agitator

TemperatureControlJacket

To Cell Recovery

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

• As the air bubbles rise through the broth, some of the oxygen dissolves into the broth solution to be taken up by the bacteria. The bacteria require 0.35 g of O2 per g of living cells each hour.

• The minimum dissolved oxygen concentration to support growth is 0.2 mg / liter of broth.

• The maximum dissolved oxygen concentration in the medium is 6.7 mg/liter at 37°C.

• The amount of oxygen required to support the colony 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

(m3 O2 / 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 at the final cell concentration is closest to:

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

c. 70

Additional Assumptions and Givens• To maintain constant growing conditions throughout

the broth, the aerated mixture is stirred continuously by a large agitator that looks like a kitchen mixer on steroids.

• The agitator is 2.5 m in diameter and stirs the tank at 60 revolutions per minute. The motor input power, Po (kW), required by this agitator is given by:

Po = 0.7 N3 Da2 V (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. The input, Pg (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. 813 b. 97 e. 2,268

c. 140

Additional Assumptions and Givens• At the end of the 18-hour fermentation time, the

broth is cooled from 37°C to 10°C to quench the biological action. This cooling process occurs in about 30 minutes. Chilled water at 5°C is injected into the reactor jacket at a flow rate of 50 L/s, picks up heat from the broth and exits the reactor jacket to return to the cooling system.

• The density and specific heat of the broth are 1,020 kg/m3 and 3.8 kJ/kg-°C respectively.

• The density and specific heat of the water are 1,000 kg/m3 and 4.19 kJ/kg-°C respectively.

LA

2

k ( )X= DO DO

O

C Cq∗− 0.452 3

0.56o a

gP NDP kQ

• The energy balance equation is given by:

( ) ( ) broth waterV cp T Q t cp Tρ ∆ = ρ ∆ ∆ (6-5)

where,

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

ρ = 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 and after the heat transfer occurs (°C)

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

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

c. 13

Additional BackgroundOnce the broth has been cooled, it is transferred to a holding tank to allow the fermentation tank to be cleaned and prepared for the next cycle. To reduce the volume of material to be processed, the 30,000 liters of broth are passed through a centrifuge to reduce the volume by a factor of four. This process also removes most of the extra-cellular impurities.

Next the thickened cell sludge is diluted with an equal volume of buffer solution consisting of 94.4% distilled water, 0.7% EDTA (ethylenediaminetetraacetic acid), a sequestering agent to prevent further reactions and 2.9% TRIS-Base (trishydroxymethylaminomethane), a buffering agent. This process facilitates the separation of the cell debris particles from the inclusion bodies that contain the proinsulin.

Additional Assumptions and Givens

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. 292 b. 276 e. 340

c. 286

Additional Assumptions and Givens• A high-pressure homogenizer is used to break

the cells and release the inclusion bodies. Passing the 15,000 liters of buffered sludge through the homogenizer three consecutive times through a circular nozzle at high velocity and under a pressure change of 80 MPa each time, ruptures the cell walls to 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 total for 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,

ρ = 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∆ ρ

From Fermentation

HoldingTank Homogenizer

To Solubilization

Centrifuge

Centrifuge

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

a. 3 d. 17 b. 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 bodies from the cell debris. The inclusion bodies have a diameter of 1.0 microns and a density of 1.3 g/cm3.

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

• The centrifuge has an acceleration factor of 1,500g (one g is equal to an acceleration of 9.81 m/s2), has 147,718 m2 of effective clarifying surface area and a motor input of 25.6 kW with a yield of 98%. The centrifuge reduces the volume from 15,000 liters to 1,400 liters of slurry at the end of the recovery section.

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

(6-6)

where,

d = particle diameter (m)

ρs = density of the particle (g/cm3)

ρ1 = density of the fluid media (g/cm3)

(rω2) = 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)

η = kinematic viscosity (m2/s) of the fluid

θ = 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)

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

= throughput (kg/s)

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

a. 1.8 d. 8.3 b. 2.3 e. 46 c. 5.3

Additional BackgroundThe 1,400 L of inclusion body suspension is transferred to a glass-lined agitator tank and mixed with small quantities of urea and of 2-mercaptoethanol. Urea is a chaotropic agent that dissolves the denatured protein in the inclusion bodies and 2-mercaptoethanol is a reductant that reduces disulfide bonds. It takes a reaction time of 8 hours to solubilize 95% of the proinsulin. The next step is to remove the excess urea and 2-mercaptoethanol with a diafilter and replace it with distilled water. This solution is then filtered through a dead end filter to remove fine particles that might overload filtering processes downstream. See Figure (6-4).

Additional Assumptions and Givens• The design flux capacity of the membrane used

is 4.2 liters per hour for each square meter of membrane.

2 2( )( )( )18

s l sd r V D Fρ −ρ ωφ =

ηθ

Qt ρ∆ =

φ

Figure 6-4 Solubilization Process

From Cell Recovery

To CNBr CleavageDistilled Water

ReactantDiafilter

Agitated Tank

HoldingTank

Filter

• 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 is closest to:

a. 0.02 d. 55

b. 18 e. 590 c. 41

Additional BackgroundThe chimeric protein is cleaved in a well-mixed reactor with CNBr (cyanogen bromide) in a 70% formic acid solution into the signal sequence Trp-LE’-Met, which contains 121 amino acids and the denatured proinsulin (82 amino acids). The reaction takes 12 hours at 20°C. The final mass of proinsulin is 31% of the mass of the initial Trp-LE’-Met-proinsulin. See Figure (6-5).

Additional Assumptions and Givens

• A small amount of toxic cyanide gas is formed as a byproduct of the cleavage reaction. This toxic gas

and excess reactants are removed by applying a vacuum and raising the temperature to 35°C (the boiling temperature of CNBr at that pressure).

• Assume the absolute pressure is 0.28 atmospheres at 35°C and you have to boil off 20 kg of CNBr vapor. One atmosphere (atm) is 101 kPa and assume 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 vapor produced at evaporator conditions is closest to:

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

c. 17.0

Additional BackgroundA sulfitolysis step is used to unfold the proinsulin, break any disulfide bonds and add SO3 moieties (molecule segments) to all sulfur residues on the cysteines. This reaction occurs in a well-stirred reactor under alkaline conditions (pH 9 to 11) over a 12-hour period. The reagents are filtered through a diafilter and replaced with a 20% by weight solution of HCl –guanidine.

Figure 6-5 Cleavage and Sulfitolysis

Vacuum HCI + Guanidine

Sulfitolysis CNBr Cleavage

Evaporator

ReactantDiafilter

Agitated Tank

HoldingTank

Agitated Tank

The human proinsulin(S-SO3-)6 is next chromatographically purified in an ion-exchange column. The bed consists of small beads coated with a cation (group of atoms carrying a positive electric charge) exchange resin. An ionic bond is temporarily formed with the properly formed proinsulin(S-SO3-)6 molecules as the solution passes slowly up through the resin. The molecules are released when the column is regenerated (back-flushed) with acid. See Figure 6-6.

Additional Assumptions and Givens

• The ion –exchange column measures 140 cm in diameter with a bed height of 25 cm.

• The 8,000 liters of solution at this point contains 28 kg of proinsulin(S-SO3-)6.

• The maximum holding capacity of the resin is 18 mg of proinsulin(S-SO3-)6 for each ml of nominal bed volume. The regeneration cycle begins when the bed reaches 70% of maximum holding capacity.

Additional Background

Folding of the proinsulin(S-SO3-)6 and disulfide bond formation takes place in a a well-mixed reactor using mercaptoethanol to facilitate the disulfide interchange reaction. The reaction is carried out at 8°C for 12

hours and reaches a yield of 85%. The reactions are removed and the protein solution is concentrated with a diafiltration unit followed by purification in a hydrophobic interaction chromatography (HIC) column to remove unfolded or incorrectly folded molecules.

The final reaction step is to enzymatically remove the C peptide from the human proinsulin using trypsin and carboxypeptidase B. The reaction occurs in a well-mixed reactor at 10°C for 12 hours. The reagents are removed by another diafiltration step followed by purification in another ion-exchange chromatography (IEC) column.

Several additional purification steps are included in the process. A reversed-phase, high-pressure liquid-chromotography (PR-HPLC) step removes structurally similar insulin-like components. A diafiltration process removes the reactants and concentrates the solution by a factor of two. The final purification step is a gel filtration chromatographic column followed by another diafiltration process to concentrate the solution by a factor of ten. The 500-liter solution at this point contains 12.8 kg of insulin. An ultrafilter captures particles larger than 100 nanometers (>10-7 m). See Figure 6-7.

The final step is to crystallize the final insulin. The insulin solution is mixed with ammonium acetate and zinc chloride in an agitated reaction tank. The crystallization into insulin6-Zn2 is carried out at 5°C for twelve hours.

Filtered solution

Acid Regeneration

Proinsulin solution

To Folding

From Sulfitolysis

Figure 6-6 Ion Exchange Chromatography

Agitated Tank

HoldingTank

ReactantDiafilter

Distilled Water

Refolding

Figure 6-7 Refolding and Enzymatic Conversion

Distilled WaterAgitated

TankHoldingTank

ReactantDiafilter

Enzymatic Conversion

Final Purification

IEC RP-HPLC Gel Filter

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

Additional Assumptions and GivensAn average diabetic individual’s daily insulin usage is 60 units. (This quantity varies with body weight, activity level and diet.)

A unit is 1/22 mg of crystallized insulin.

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

a. 2,100 d. 1.8x106 b. 4,200 e. 4x106 c. 11,500

Competition{ }Part I and Part II Solutions2007

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:

bacteriaof147.6gbacteriaof1,110kg x 180g of glucose = 1,353 kg of glucose

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

KDODO

*02

LA CC)(X)(q

−=

where, K =LA Oxygen Volume Transfer (in m h)-/m broth

302

3

X = final concentration of cells = 37g /Lcells

q =02 Respiration rate of cells = 0.35g h/gcell02 −

TEAMS 2007 Part I and Part II Solutions 37

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 broth

302

3 − Each batch is 30m 3 and the O2 represents 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

broth3

0233

=−

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)(1revΡ 323

0 == Then using equation (6-4) with: k = 0.5 kW131.25Ρ0 = N = 1 rev/s D a = 2.5m, and Q = 15/s, yields:

TEAMS 2007 Part I and Part II Solutions 38

0.45

0.56

32

g (15/s).5m)(1rev/s)(2kW)(131.25(0.5)Ρ ⎥

⎦

⎤⎢⎣

⎡=

= 70.2 kW 54. Answer c Density of broth = 1,020 kg/m 3 ρ =broth

C =pbroth specific heat of broth = 3.8 kJ/kg- C 0

T ∆ C270

broth = ρ 3

water kg/m1,000= C CkJ/kg4.19 0

pwater −= 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/h30min∆t ==

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

t))()(C(Q)(ρ

)T)()(C)(30m(ρ∆T

pwaterwater

brothpbroth3

brothwater ∆

∆=

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

C)C)(27)(3.8kJ/kg)(30m(1,020kg/m033

0033

−−

= 8.32 C0

Water Temperature = 5 = 13.32 0 C ∆TC0 +

TEAMS 2007 Part I and Part II Solutions 39

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 = ρ

P)(2,000)(∆

where, = Pressure Change = 80 MPa = 80x10 3 kPa ∆Ρ = density of fluid = 985 kg/m ρ 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/m15,000L/h

TimeVolume 3

=

= 2.7x10 /sm33−

A = 2633

m6.892x10m/s403.03

/sm2.7x10VQ −

−

==

A = 6.892 mm 2

TEAMS 2007 Part I and Part II Solutions 40

A = ,4

πD2

or

D = π

)m(4)(6.892mπ

4A 2

=

D = 2.96 mm 57. Answer b d = particle parameter = 1.0 microns = 1x10 m 6−

ρ = density of the particle = 1.3g/cm s

3

ρ = 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/s 2 2

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

= shape factor = 1.0 θ 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)(1x10φ 26

223326

−

− −=

= 1.8 kg/s Using equation (6-9) yields: Q = (15,000 lt)(1 m 33 15mlt) /1,000 = ρ= 1.0g/cm 333 kg/m1.0x10=

==1.8kg/s

)`kg/m)(1.0x10(15m∆t333

8,333s = 2.3h

TEAMS 2007 Part I and Part II Solutions 41

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 -h 2

t = process time = 6h ∆

A = h)(6h)l/m(4.2l1,400

2 −

= 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 Part I and Part II Solutions 42

= )101kPa/atm(0.28atm)(

K)273 K)(35/kmolm)(8.314kPa(0.189kmol 3 oo +−−

= 17.11 m 3 60. Answer d Facility production = (1,800kg/yr)(1yr/365 days)(10 )mg/kg6

= 4,931,506 mg/day

Average daily dose in mg = (60 units/day) ⎟⎠⎞

⎜⎝⎛

unitmg

221

= 2.73 mg/daily dose

Number of supported diabetics = 2.73mg/day

g/day4,931,506m

= 1,806,412 diabetics = 1.8x10 diabetics 6

TEAMS 2007 Part I and Part II Solutions 43