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Chemical Engineering 170 October 7, 2002 Midterm Exam Closed Book and Closed Notes One 8.5 x 11 in. page of notes allowed
Section 1. Short Answers
- (3 pts.) The serine protease subtilisin is the active enzyme in many household laundry detergents. This enzyme is inactivated readily by oxidation of Met 222, a residue that is conserved in all known subtilisin sequences and occupies a partially buried position next to the catalytic Ser 221. It should thus be possible to produce a more stable and useful enzyme by mutating this residue to one that is not susceptible to oxidation. Which amino acid do you believe would be best to use as a replacement for the Met residue, and why?
Valine and Leucine would be the best amino acids based on size and hydrophobicity. Cysteine is not a good answer because it has sulfur which is site of oxidation (if mention that cysteine would be a good substitute because of sulfur we gave partial credits)
- (3 pts.) Name three interactions involved in maintaining the tertiary structure of proteins.
Any 3 of 5:
Hydrogen bond; Hydrophobic Interactions; Ionic Interactions (electrostatic); Van der Waals;
Disulfide bonds
- a. (4 pts.) What are the definitions of the Dämkohler number (Da) and the Thiele Modulus (φ) for Michaelis-Menten kinetics?
MaximumTransportRate
MaximumReactionRate Da = ; KS S o
V max Da =
Transport rate
2 Reactionrate
Km Deff
R V max 3
b. (1 pt.) Which of the two deals with external mass transfer?
Damkohler number
c. (1 pt.) What does Da >>1 imply about the limiting rate of the process?
Da>>1 implies that the reaction rate is much greater than the transport rate, therefore the mass transfer rate is the limiting step of the process.
d. (1 pt.) What does φ << 1 imply about the limiting rate of the process?
φ>>1 implies that the transport rate is much greater than the reaction rate, therefore the reaction rate is the limiting step of the process.
- (3 pts.) Define the following terms: a. DNA ligase
Covalently link strands of nucleic acid on a template, the oligonucleotide may have a blunt end or sticky end
b. Plasmid
Closed circular (supercoiled) piece of DNA, found in bacteria, usually a few thousand base pairs long; Plasmids are normally double-stranded
c. Restriction endonucleases
Recognize a particular sequence of nucleotides in a polynucleotide or nucleic acid, and cleave at a specific site.
- (4 pts.) What are the two steps of protein synthesis? Give a brief description of each step.
- Transcription: Synthesis of mRNA from DNA. RNA polymerase is directed to start site of transcription by one of its subunits affinity to a particular DNA sequence (promoter) that appears at the beginning of genes. The promoter is a unidirectional sequence on one strand of the DNA and it tells RNA polymerase where to start and in which direction to continue synthesis
- Translation: Synthesis of polypeptide chain from mRNA. The mRNA has codons (triplets of bases) that specify the specific amino acid to be added to the protein. The amino acids are brought to the ribosome by tRNA, in the ribosome the amino acid is added to the growing protein chain. mRNA (messenger RNA); tRNA (transfer RNA)
Section 2. Short Problems
- (5 pts.) In class we discussed the design of inhibitors for the enzyme purine nucleoside phosphorylase (PNP), which has been identified as a potential target for inhibition by new drugs for the treatment of T-cell disorders. Shown below is the proposed structure of the transition- state for the reaction catalyzed by PNP, the phosphorylase of the inosine. Based on the information shown, estimate the rate enhancement of the enzyme-catalzyed reaction relative to the uncatalzyed reaction.
Rate enhancement can be estimated ratio of catalyzed reaction to uncatalyzed reaction:
[ ] [ ]
[ ][ ] [ ][ ]
[ ] [ ][ ]
[ ][ ] ≠ [ ] ≠
≠ ≠
≠ ≠
≠
≠
≠
= ⋅
×
× =
d
d n
e
n
b
e
b
n
e K
K
k
k
k h
kT
k h
kT
k
k PNP Inosine
PNP Inosine PNP Inosine
PNP Inosine PNP Inosine
PNP Inosine PNP Inosine
PNP Inosine
12 18
6
- 67 10 3 10
17 10 = × ×
× = = −
− ≠ d
d n
e K
K k
k , the rate enhancement is 5.67x10^12
An alternative solution is:
12 12 6 10 1
6 10 = × × = (^) ≠ =
≠
n
e n
e k
k k
k
- Consider the two-stage enzyme inactivation model we discussed in class:
I
k
D
k
k
N ↔ →
3
2
1
where N represents the native enzyme, D the reversibly denatured enzyme, and I the irreversibly inactivated enzyme.
a) (6 pts.) Assuming the process occurs in a closed system, write unsteady-state mass balances for N, D, and I.
[ ] k [ N ] k [ D ] dt
d N =− 1 + 2
[ ] k [ D ] k [ D ] k [ N ] dt
d D =− 3 − 2 + 1
[ ] k [ D ] dt
d I = 3
b) (3 pts.) Under what conditions is the loss of native enzyme, -dN/dt, equivalent to the appearance of irreversibly inactivated enzyme, dI/dt? Answer this question with both a mathematical expression and a brief statement of what it means.
[ ] [ ] [ ]
dI dt
dD dt
dN
If [D] is constant with time then:
[ ] = 0 dt
dD
Therefore [ ] [ ] dt
dI dt
d N − =
The loss of native enzyme is equivalent to the appearance of irreversibly inactivated enzyme, if the reversibly denatured enzyme concentration does not change with time.
- (6 pts.) In class we considered the effect of internal diffusion on the apparent stability of
immobilized enzymes, which required that we relate dlnφ/dt to dlnEa/dt, where Ea is the
concentration of active immobilized enzyme, and dEa/dt = -kd Ea. What would be the
corresponding relationship between dlnΦ/dt and dlnEa/dt, where Φ is the observable modulus?
m o
cat a o obs I K S
k E S v
=η
dt
d E dt
d dt
d ln vobs ln I ln a = +
(assuming kcat , K (^) m and So are constant with time) (1)
dt
d d
d dt
d (^) I I Φ Φ
ln ln
lnη ln η
o eff
obs S D
R v 2
3
( ) dt
d v dt
d S D dt
d R dt
d ln 3 ln o eff ln obs 2
ln = − +
(assuming Deff and So are constant with time)
dt
d v dt
d ln ln obs
dt
d E dt
d v d
d dt
d E dt
d d
d dt
d vobs (^) I a I ln obs ln a ln
ln ln ln ln
ln (^) ln × + Φ
×
ln
ln 1
ln ln ln
d
d
dt
d E
dt
d dt
d v I
a obs
An alternative way to get to (1) is:
2 I
( ) dt
d dt
d dt
d dt
d lnφ ln 1 β ln η I
ln
(assuming β is constant with time)
dt
d dt
d dt
d lnφ ln η I
ln = +
From class: dt
d E dt
d ln a 2
ln 1
dt
d dt
d E dt
d ln ln a ln η I
- (23 pts.) In class we considered the carbon and energy requirements for growth of an organism on hexadecane, C16H34. In this problem we shall consider the growth of a yeast, Candida utilis ,
on ethanol, which may be simply described by the following equation: αC2H6O + βO2 + γNH3 → CH1.82N0.19O0.46 (cells) + κCO2 + 1.60H2O
a) (4 pts.) Assuming the yield coefficient, Yx/s, is equal to 0.664, calculate the value of α.
( 2 12 6 16 )
masssubstrateconsumed
masscellproduced / (^) × + +
+ + × + ×
X S α
Y
×
b) (3 pts.) What is the value of κ?
A balance on carbon: 2 α = 1 + κ κ= 2 α − 1 = 2 × 0. 781 − 1 = 0. 561
c) (6 pts.) What is the respiratory quotient (RQ) for the above process?
rateofO consumption
rateofCO production 2
RQ^2
In order to determine RQ, we need to determine β, this can be done through an oxygen balance:
α+ 2 β= 0. 46 + 2 κ+ 1. 6 ( ) 2
κ α β
β = 1.2, thus:
RQ
d) (3 pts.) Write an overall electron balance in terms of degrees of reductance, where, as usual,
the degrees of reductance are defined per g-atom of carbon?
The degree of reductance of NH 3 , CO 2 and H 2 O is zero thus the electron balance is:
2 αγ S − 4 β= γ b
The reaction rate, v , can be written as: [ ] k [ ES ] dt
dP v = = 2 (2)
Write a balance for the enzyme complexes ( ES , EI and ESI ): [ ] k [ E ][ ] S k [ ES ] k [ ES ] dt
d ES = 1 − − 1 − 2 (3)
[ ] k [ E ][ ] I k [ EI ] dt
d EI = 3 − − 3 (4)
[ ] k [ ES ][ ] I k [ ESI ] dt
d ESI = 4 − − 4 (5)
Assume that the concentration of enzyme complexes is constant with time, hence:
[ ] = 0 dt
d ES => k (^) 1 [ E ][ ] S = ( k − 1 + k 2 )[ ES ]=> [ E ][ ] S = Km [ ES ]
[ ] [ ]
[ ES ] S
K
E = m (6)
[ ] = 0 dt
d EI => k (^) 3 [ E ][ ] I = k − 3 [ EI ]=> [ E ][ ] I = K 3 [ EI ]
[ ]
[ ][ ] [ ]
[ ] [ ES ] K
I
S
K
K
E I
EI m 3 3
[ ] = 0 dt
d ESI => k (^) 4 [ ES ][ ] I = k − 4 [ ESI ]=> [ ES ][ ] I = K 4 [ ESI ]
[ ]
[ ][ ] K 4
ES I
ESI = (8)
Substitute (6), (7) and (8) into (1):
[ ] [ ]
[ ] [ ]
[ ] [ ]
[ ] [ ] [ ES ] K
I
ES
K S
K I
ES ES
S
K
E (^) o m m 3 4
[ ] [ ]
[ ] [ ]
[ ] [ ES ] K
I
K S
K I
S
K
E (^) o m m
3 4
[ ] [ ]
[ ] [ ] [ ES ] K
I
K
I
S
K
E (^) o m
3 4
[ ]
[ ]
[ ]
[ ] [ ]
3 4
K
I
K
I
S
K
E
ES
m
o (^) (9)
Substitute (9) into (2): [ ]
[ ]
[ ] [ ]
3 4
2
1 1 K
I
K
I
S
K
k E v m
o
[ ][ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ]
3 4
max
3 4
2
1 1 1 1 K
I
S
K
I
K
V S
K
I
S
K
I
K
k E S v
m m
o (^) (10)
If you can’t derive an expression for the rate equation you can use the following expression for parts b and c (note that this is not the expression you would find in part a)
3 4
max
1 1 K
P
S
K
P
K
V S
v
m
b) (5 pts.) Based on the equation derived in part a. What it physically means if K (^) 4 =∞? What
type of inhibition do you have? How does your reaction rate look like?
Based on the reaction rate derived in part a:
K (^) 4 = ∞ means that enzyme-substrate complex has no affinity for inhibitor I, so no enzyme-
substrate-inhibitor complex is formed. Therefore you have competitive inhibition and the reaction rate is:
[ ] [ ] [ ] S K
I
K
V S
v
m (^) +
3
max
1
(1 point)
Based on the reaction rate given above:
K (^) 4 = ∞ means that enzyme-substrate complex has no affinity for inhibitor I, so no enzyme-
substrate-inhibitor complex is formed. The product is the inhibitor in this case. Therefore you have competitive inhibition and the reaction rate is:
[ ] [ ] [ ] S K
P
K
V S
v
m (^) +
3
max
1
(1 point)
