## Enzyme Immobilization in Gel Beads & Strings

### Biochemical Engineering

Problem Statement: Consider a substrate-inhibited enzyme that is immobilized via physical entrapment in spherical gel beads. Unless noted otherwise, use the following parameters.

```  v(s)=vm*s/(Km+s+Ki*s2)
where  vm=0.01g/L-sec
Km=1g/L
Ki=10L/g
Diffusion coefficient: De=10-5cm2/sec
Substrate concentration in the solution: sb=1g/L
```
1. The diffusion coefficient given above is for which species (substrate, product, inhibitor, or enzyme)? In what medium (air, water, substrate solution, substrate-product mixture, polymer gel, across a membrane, across the reactor wall)?

Solution:

For substrate diffusing through the polymer gel.

2. List the differential equation that governs the substrate profile within a spherical bead. Provide applicable boundary conditions. Briefly describe how to find a solution to the equation you set up.

Solution:

```  differential diffusional mass transfer rate = differential reaction rate
De*d[4*p*r2*(ds/dr)] = 4*p*r2*dr*v(s)
which is equivalent to:
De*[(d2s/dr2)+2/r*(ds/dr)]=v(s)
B.C. s(R)=sb
ds(0)/dr=0
Dimensionless form (normalize s wrt sb, and r wrt R):
(d2s/dr2)+2/r*(ds/dr)=f2*v(s)
B.C. s(1)=1
ds(0)/dr=0
f=R*sqrt(vm/De/Km)
Transform into two coupled 1st-order odes
ds/dr = z
dz/dr = -2/r*z + f2*v(s)
B.C. s(1)=1
ds(0)/dr=0
Solution Approach:
1. Guess s(0)
2. Start from s(0) and z(0)=0, integrate dimensionless odes from r=0 to r=1
3. See if s(1)=1.  If yes, stop; otherwise, make another guess of s(0) and
iterate until s(1)=1 is satisfied.
```

The above equations were solved numerically in a series of Mathcad files posted as the class handouts. The following is one such file.

Solution:

Numerical Solution of the Two-Point Boundary Value Problem.

3. When we solve the nondimensional differential equation from part b), we obtain s(0)=0.00606 to reach s(1)=1 (in nondimensional form). With this value at r=0, we get: ds/dr=2.98 at r=1, where s is the substrate concentration normalized with respect to sb and r is the distance from the center of the bead normalized with respect to R. Find the overall rate of reaction in a spherical gel bead.

Solution:

```  In dimensional units, ds/dr=2.98*sb/R=2.98*(1g/L)/(0.3cm)=9.93g/L-cm
Rate of reaction = mass transfer across the surface
= area*De*(ds/dr)
= 4*p*(0.3cm)2*(10-5cm2/sec)*(9.93g/L-cm)
= 1.124*10-7g/sec
```

4. Find the initial rate of reaction in a spherical drop of the gelling solution (before gel formation, i.e., enzyme in solution) that contains substrate at a concentration of sb=1g/L.

Solution:

```  Rate of reaction /wo mass transfer = volume*v(sb)
= volume*vm*sb/(Km+sb+Ki*sb2)
= 4/3*p*(0.3cm)3*(0.01g/L-sec)*(1g/L)/(1g/L+1g/L+10L/g*(1g/L)2)
= 9.4*10-8g/sec
```

5. Briefly describe the physical meaning of the effectiveness factor. Find it for the given set of parameters.

Solution:

```                         rate /w mass transfer resistance
Effectiveness factor = ---------------------------------
rate /wo mass transfer resistance
= (1.12*10-7g/sec)/(9.4*10-8g/sec)
= 1.192
```

6. An optimal bead radius of 0.22cm was found for the above given set of parameters. What makes this radius an optimal one?

Solution:
A small radius fails to shield the enzyme from the effect of substrate inhibition. On the other hand, a large bead offers too much mass transfer resistance and the enzyme sees very low substrate level. Either way, the rate of reaction is diminished.

7. An optimal bead radius was given to us in the last problem, here do find the optimum bead size that maximizes the rate of substrate conversion. Plot the effectiveness factor (i.e., dimensionless rate of reaction integrated over the bead volume) as a function of the Thiele modulus (like Fig 3.20 of Shuler and Kargi) and locate the maximum in the plot.

Solution:

• Plot of Effectiveness Factor Versus Thiele Modulus with Substrate Inhibition

8. You entrapped twice the amount of enzyme per liter of gel. Which of the given parameters are affected by this extra enzyme? Will more enzyme give you a higher rate of conversion? What is the optimal size of the beads, given that 0.22cm was the optimum size for the original recipes?

Solution:
vm is indicative of the amount of enzyme in the preparation; other parameters are not affected. The more enzyme, the higher the value of vm, and the higher the conversion rate. Since the substrate profile depends on f, identical f gives rise to an identical dimensionless substrate profile. Thus, keep f at the optimal level, i.e., maintain the same R2*vm. If we double vm, we reduce R by a factor of sqrt(2), i.e., R=0.22cm/sqrt(2)=0.16cm.

9. During the class demo, some of you made long strings of gel rather than spherical droplets. Analogous to part b), give the differential equation that governs the substrate concentration profile within these strings. Provide appropriate boundary conditions.

Solution:

```  differential diffusional mass transfer rate = differential reaction rate
De*d[2*p*r*(ds/dr)] = 2*p*r*dr*v(s)
which is equivalent to:
De*[(d2s/dr2)+1/r*(ds/dr)]=v(s)
B.C. s(R)=sb
ds(0)/dr=0
Dimensionless form (normalize s wrt sb, and r wrt R):
(d2s/dr2)+1/r*(ds/dr)=f2*v(s)
B.C. s(1)=1
ds(0)/dr=0
f=R*sqrt(vm/De/Km)
```

10. You fluidize (i.e., suspend) 1000 of these spherical beads in a well-stirred continuous reactor. Given the feed flow rate F, the substrate concentration in the feed sf, and reactor volume, find the rate of conversion at steady-state. Set up all applicable equations, but do not solve them as you will not have time to do so.

Solution:

```  De*[(d2s/dr2)+2/r*(ds/dr)]=v(s)
B.C. ds(0)/dr=0
Replace s(R)=sb in part b) with:
rate of conversion = F*(sf-s(R)) = A*De*(ds(R)/dr),
where A is the surface area of the beads: A = 1000*4*p*R2 = 1131cm2
Incidentally, rate of conversion is also equal to h*Vb*v(s(R))
where Vb is the volume of the beads: Vb = 1000*4/3*p*R3 = 113cm3
```