Question

Enzyme is immobilised in 8 mm diameter agarose beads at a concentration of 0.018 kg protein m^-3 gel. Ten beads are immersed in a well-mixed solution containing 3.2 ×10^-3 kg m^-3 substrate. The effective diffusivity of substrate in agarose gel is 2.1 × 10^-9 m^2 s^-1. Kinetics of the enzyme can be approximated as first order with specific rate constant 3.11 × 10^5 s^– 1 per kg protein. Mass transfer effects outside the particles are negligible. Estimate the volume per bead.

(a) 2.65×10^-7 m^3

(b) 2.66×10^-7 m^3

(c) 2.67×10^-7 m^3

(d) 2.68×10^-7 m^3

I have been asked this question by my school teacher while I was bunking the class.

The origin of the question is Internal Mass Transfer and Reaction topic in section Heterogenous Reactions of Bioprocess Engineering

Answer 1

To estimate the volume per bead, we need to consider the size of the agarose beads and their relationship to the total volume they occupy.

Given Data:

• Diameter of beads = 8 mm = 8×10−38 \times 10^{-3}8×10−3 m.
• Protein concentration in the gel = 0.018 kg/m³.
• Substrate concentration = 3.2×10−33.2 \times 10^{-3}3.2×10−3 kg/m³.
• Effective diffusivity of substrate in agarose gel = 2.1×10−92.1 \times 10^{-9}2.1×10−9 m²/s.
• Specific rate constant = 3.11×1053.11 \times 10^53.11×105 s⁻¹ per kg protein.
• Number of beads = 10 beads.

Step-by-Step Calculation:

1. Volume of a single bead: The volume of a spherical bead is calculated using the formula for the volume of a sphere:

   V=43πr3V = \frac{4}{3} \pi r^3V=34​πr3

   where $r$ is the radius of the bead. Given the diameter is 8 mm, the radius is:

   r=8×10−32=4×10−3 mr = \frac{8 \times 10^{-3}}{2} = 4 \times 10^{-3} \, \text{m}r=28×10−3​=4×10−3m

   Now, calculate the volume of one bead:

   V=43π(4×10−3)3V = \frac{4}{3} \pi (4 \times 10^{-3})^3V=34​π(4×10−3)3 V=43π(64×10−9) m3V = \frac{4}{3} \pi (64 \times 10^{-9}) \, \text{m}^3V=34​π(64×10−9)m3 V≈2.68×10−7 m3V \approx 2.68 \times 10^{-7} \, \text{m}^3V≈2.68×10−7m3

Answer:

The volume of each bead is approximately 2.68×10−7 m32.68 \times 10^{-7} \, \text{m}^32.68×10−7m3, which corresponds to option (d).

Thus, the correct answer is:

(d) 2.68×10⁻⁷ m³