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Chemical Engineering 170 October 16, 2003 Midterm Exam Closed Book and Closed Notes One 8.5 x 11 in. page of notes allowed
Section 1. Short Answers
- Name four interactions involved in maintaining the tertiary structure of proteins.
Hydrogen bond
Hydrophobic Interactions
Ionic Interactions (electrostatic)
Van der Waals
Disulfide bonds
- Define the Damkohler number for an immobilized enzyme reaction and explain its physical
meaning.
KS S o
V max Da = Maximum TransportRate
Maximum Reaction Rate Da = ;
- Explain why, in physical terms, ηI << 1 when φ >> 1. What is the controlling resistance in such a case?
Km D eff
R V max 3
φ =
If φ >>1 then Vmax>> KmDeff, hence the system must be transport- limited. Reactant molecules are consumed before they diffuse very far, and the reaction is limited to a thin region near the periphery of the particle. Thus, the observed reaction rate is much smaller than the rate in the
absence of an internal concentration gradient (ηI<<1)
- What are three of the major differences between RNA and DNA that we discussed in class?
- RNA has a hydroxyl group at the 2’ carbon of its sugar
- RNA uses uracil as one of its base, while DNA uses thymine as one of its base
- The extra hydroxyl group of RNA prevents it from forming a stable double stranded structure, hence RNA exists as a single stranded molecule.
- Some regions of RNA can form double helix (hairpin loops) (i.e. secondary structure)
- RNA can form many different tertiary structures, or stable three-dimensional structures
- In site-directed mutagenesis, a neutral mutation has no effect on the properties of the mutated protein. Indicate whether the following would likely be neutral or non- neutral mutations, and briefly explain why. (Ignore possible effects on protein secondary structure.)
i. glutamate changed to aspartate
Neutral – similar size and pKa’s.
ii. leucine changed to valine
Neutral – both branched-chain hydrophibics of similar size.
iii. valine changed to glycine
Not neutral – G is very small, allows significant free rotation around peptide bonds.
iv. cysteine changed to alanine
Not neutral – A is hydrophobic, C is polar and can even dissociate.
v. serine changed to threonine
Neutral – both contain –OH in side chains and similar size.
vi. tryptophan changed to phenylalanine
Neutral – both are hydrophobic aromatics of similar size.
vii. arginine changed to lysine
Neutral – both are large and basic.
viii. histidine changed to alanine
Not neutral – H is polar and ionizes with a pKa around neutrality, A is non-polar.
- What is the two-stage model for enzyme inactivation, and what is the corresponding expression for kobs, the observed rate constant for inactivation?
The two-stage model of enzyme inactivation is: N ←→ K^ D → k I
The corresponding expression for kobs is:
K
k kobs
, where
[ ]
[ D ]
N
K =
Section 2. Short Problems
- Enzyme that is immobilized on a nonporous support obeys the following reaction scheme:
E + S ES E + 2 P
You are also given the following information:
kp = 5 x 10-3^ cm/sec Km = 0.5 mM v 'max= 100 μM cm-2^ min- So = 1 mM ηE = 0.
a) Write an expression that relates the flux of product from the surface to the rate of substrate consumption at the surface. Please define any parameters in your expression that are not given above.
m
' * max o
p K S
2 v S k P P
− = OR^ (^ )^
m o
o
' E max o
p (^) K S
2? v S k P P
P* = concentration of product at surface S* = concentration of substrate at surface
b) Calculate the measured rate of appearance of product in the bulk per unit area.
0.5mM 1 mM
1 mM cm min
2 0.40 100 μM K S
2? v S k P P 2 m o
o
' E max o
p 53.3^ μM cm
-2 (^) min-
- A biochemical engineer, who was attempting to enhance the rate of a particular enzyme by manipulating its structure via site-directed mutagenesis, produced a mutant that exhibited rather unusual behavior. The enzyme-substrate complex was found to reversibly aggregate into an inactive dimer, according to the following reaction:
ES + ES D
However, the mutant enzyme still follows general enzyme kinetics:
E + S ES E + P
Use these definitions in answering the following questions:
3
3 (^3) k
k K = − ; 1
1 2 m k
k k K
a) Assuming the reaction occurs in a closed system, begin by writing mass balances for the enzyme complexes (ES and D):
[ ][ ] [ ] [ ] 2 [ ] 2 [ ]
[ ]
3
2 k 1 E S k 1 ES k 2 ES k 3 ES k D dt
d ES = − − − − + −
[ ] [ ]
[ ]
3
2 k 3 ES k D dt
d D = − −
b) Using the pseudo-steady-state-hypothesis for each enzyme complex, derive an expression for [ES] in terms of Eo (and other parameters, including constants, but no other enzyme complexes):
0 [ ] [ ]
[ ]
3
2 k 3 ES k D dt
d D = = − − => 3
2 2 3
[ ]^3 [ ] [ ]
K
ES
ES
k
k D = = − also, since k (^) 3 [ ES ]^2 = k − 3 [ D ]:
0 [ ][ ] [ ] [ ] 2 [ ] 2 [ ]
[ ]
3
2 k 1 E S k 1 ES k 2 ES k 3 ES k D dt
d ES = = − − − − + −
[ ]
[ ]
[ ]
[ ]
[ ]
1
1 2 S
ES
K
S
ES
k
k k E = m
k 1
k (^) -
k 2
k 3
k (^) -
- Consider our “oil-eating” organism discussed in class. In addition to hexadecane, the same organism can also grow on glucose according to the following biological reaction:
C 6 H 12 O 6 + a O 2 + b NH 3 → c (C4.4H7.3N0.86O1.2) + d H 2 O + e CO 2
Assume that the cells can convert 2/3 (wt/wt) of the substrate carbon to biomass.
a) Calculate each of the stoichiometric coefficients, a, b, c, d , and e.
Amount of C per mol of glucose: 72g; 2/3*(72g) = 48g converted to biomass
Biomass balance: 48 g/mol = (4.4)(c)12g/mol c = 0.
N balance: 14b = (0.86)(0.909)14 b = 0.
H balance: 12 + 3(0.782) = (0.909)(7.3) + 2d d = 3.
C balance: 6(12) = 0.909(12)(4.4) + 12e e = 2
O balance: 6(16) + 2(16)a = (0.909)(1.2)(16) + (3.86)(16) + 2.16e a = 1.
b) Calculate the yield coefficient YX/S (g dw cell/g substrate).
YX/S = (0.909)(MW biomass)/(MW glucose) = (0.909)(91.34)/
YX/S = 0.461 g biomass/ g glucose
c) How would you expect this coefficient to compare to YX/S for hexadecane (i.e., larger or smaller)? Why?
It should be lower because hexadecane is more reduced than glucose (i.e. YX/SHex^ > YX/SGlu^ ). In other words, hexadecane has a greater degree of reductance, so it has more electrons/energy to form biomass.
