Bioreactors
Assignment 2 Solutions
The correct option is highlighted
1. Which of these outputs from a bioreactor can be a product of interest?
(a) Cell biomass
(b) Biopolymers produced by cell
(c) Organic acids excreted by the cell
(d) All of the above
2. A chemistry student needs to make a product P from the reaction 2A + B → P. He combines 3
moles of A and 2 moles of B. Which reactant would limit P formation under these conditions?
(a) A
(b) B
(c) P
(d) None are limiting
3. Which of these can be classified as substrates?
(a) Precursor molecules that can be converted to product by cells/enzymes
(b) Carbon sources such as starch and cellulose that allow cell growth
(c) Both (a) and (b)
(d) None of the above
4. In the presence of low substrate, it is utilized by the cell mainly for:
(a) cell multiplication
(b) cell maintenance
(c) cell autolysis
(d) cell proliferation
5. Which of the following are used for product formation in bioreactors?
(a) Microorganisms
(b) Mammalian cells
(c) Enzymes
(d) All of the above
6. Where does the inhibitor bind in ‘Uncompetitive Inhibition’ of enzymes?
(a) Substrate
(b) Enzyme
(c) Enzyme – Substrate complex
(d) Product
7. Enzymatic reaction of an Enzyme E on substrate S in the presence and absence of another
compound I generates the following data. Assume that the enzyme follows MM kinetics
Substrate S
Reaction rate v (mmolesL-1h-1)
(mM)
I=0
I = 1.07 mM
I = 1.95 mM
1.0
2.5
2.17
1.82
0.75
2.44
1.82
1.39
0.60
2.08
1.41
1.28
0.50
1.89
1.30
1.00
0.40
1.67
1.09
0.85
0.33
1.39
0.87
0.75
0.25
1.02
0.73
0.56
What type of inhibition does I exhibit?
(a) Competitive inhibition
(b) Uncompetitive inhibition
(c) Non-competitive inhibition
(d) Mixed inhibition
Solution
The nature of the inhibition can be determined from the plot of 1/v versus 1/S (LineweaverBurke plot). The data points for the plot can be determined from the question.
S
(mM)
1/S
v1 at
I=0 mM
1/v1
v2 at I =
1.07mM
1/v2
v3 at I =
1.95mM
1/v3
0.25
4
1.02
0.98
0.73
1.36
0.56
1.78
0.33
3.03
1.39
0.71
0.87
1.14
0.75
1.33
0.4
2.5
1.67
0.59
1.09
0.91
0.85
1.17
0.5
2
1.89
0.52
1.3
0.76
1
1
0.6
1.66
2.08
0.48
1.41
0.70
1.28
0.78
0.75
1.33
2.44
0.40
1.82
0.54
1.39
0.71
1
1
2.5
0.4
2.17
0.46
1.82
0.54
It is clear from the above plot that Km (that can be obtained from the slope = Km/vm) varies in
the presence of the inhibitor whereas vm (that can be obtained from the y-intercept = 1/vm)
remains a constant. Thus the inhibition is ‘competitive’.
8. From the data in question 7, determine MM parameters vm and Km
(a) vm = 0.15 mmolesL-1 h-1 and Km = 0.775 mM
(b) vm = 6.66 mmolesL-1h-1 and Km = 1.29 mM
(c) vm = 0.19 mmolesL-1 h-1 and Km = 0.15 mM
(d) vm = 8.2 mmolesL-1h-1 and Km = 4.1 mM
Solution:
vm and the different Km values can be obtained from the plot in the solution to question 7.
From the y intercept we have: 1/vm = 0.15, therefore, vm = 6.667 mmolesL-1 h-1
Similarly, determine the various Km values from the slope values
Km/vm (at I = 0 mM) = 0.194; therefore, Km = 0.194 x 6.667 = 1.29 mM.
Similarly,
Km’ (at I = 1.07 mM) = 2.064 mM
Km’’ (at I = 1.95 mM) = 2.69 mM
9. From the data in question 7, determine KI
(a) 1.79 mM
(b) 0.79 mM
(c) 2.3 mM
(d) 3.2 mM
Solution:
Since the nature of the inhibition has been identified to be competitive in question 7, the value of
KI can be determined from the expression for Km’.
Km’ that arises as a result of the presence of a competitive inhibitor can be given as:
=
(1 +
) …… (1)
As discussed in the solution to problem 8, the values of Km’ and Km are 2.064 mM and 1.29
mM respectively. Km’ corresponds to the inhibitor concentration 1.07 mM. Therefore
substituting these in (1), we have
⇒ 2.064 = 1.29(1 +
1.07
)
Therefore KI = 1.79 mM
The same KI value can be obtained using the Km’’ value of 2.69 mM that corresponds to the
inhibitor concentration of 1.95 mM.
10. Viable cell concentration can be determined from measurements of
(a) Optical density
(b) Colony forming units (CFUs)
(c) Cell dry weight
(d) All of the above