BCH 361 Lecture Notes - Lecture 3: Salt Metathesis Reaction, Enzyme Kinetics, Specific Activity

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30 Nov 2020
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BCH 361Advanced Biochemistry IFall 2020
1
Group Discussion Questions
Week 3 (September 21/22)
1. Kinetics PowerPoint slides 24/25. Solve the problems and compare your answers with your
groupmates.
A: A) Activity refers to the amount of enzyme transformed in a specific number of unit time. In my
answer, I used micromoles per minute.
12 millimoles -> 12000 μm.
Activity = enzyme transformed / unit time = (12000 μm) / (2 min) = 6000 μm/min
B) Specific activity refers to the units of activity per mg
Specific activity = activity / total mg of enzyme = (6000 μm/min) / (24 mg) = 250 μm/min*mg
Slide 25
Catalytic efficiency = Km/Kcat. This equation is helpful because it provides information about the most
effective substrate for acetylcholinesterase. When you do the calculation, 1-butenyl acetate is the most
efficient substrate because it has the highest catalytic efficiency number. Small km refers to fast and
tight binding while a large kcat refers to a fast reaction.
2. What is the biochemical advantage to having a KM approximately the same as the substrate
concentration that is normally available to the enzyme?
A: Km occurs at the concentration of the substrate where the enzyme works at half of the maximum
velocity. When the michaelis constant is about the same as concentrations of the substrate, the
advantage is that the enzyme shows that catalytic activity occurs and is sensitive in substrate
concentrations.
3. What is the defining characteristic of an enzyme catalyzing a sequential reaction? A double-
displacement reaction? (How can you tell these apart?)
A: Sequential reactions defining characteristic is the formation of the ternary complex that forms. The
complex consists of the enzyme and both substrates. In a double displacement reaction, it requires the
formation of a temporary formed enzyme intermediate which remains the same at the end of the
reaction. To tell these apart, in the double displacement reaction, only one substrate and enzyme bind
at a time.
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BCH 361Advanced Biochemistry IFall 2020
2
4. The affinity between a protein and its ligand is frequently reported as a dissociation constant,
Kd. Does KM measure the affinity between the enzyme and the substrate? Under what
circumstances might KM be approximately the same as Kd?
A: Km does not measure affinity and is not equal to Kd. This is because Km contains k2, which is the rate
constant that measures the conversion of the enzyme and substrate complex into the enzyme and
product complex. Circumstances where the dissociation constant can equal the Km value is when K2 is a
small number with respect to K-1. If this occurs the formula is the same and kd=km.
Kd = K-1/K1 and Km = (K-1 + K2)/K1
5. A substrate, S, was cleaved by an enzyme and the rate of disappearance measured every 30
seconds for 3 minutes. A series of six test tubes was arranged with 1.5 μg of enzyme (MW 30
000 Da) being added to the same volume of substrate solution, but with each at a different
concentration. The results are summarized in Table 1.
a. What are the initial velocities at each substrate concentration? [Hint: it’s easiest to use
Excel or a similar spreadsheet for this problem.]
A: Rate = [S t2] - [S t1] / t2-t1 = mol/min
Initial concentration (μmol/mL)
Initial Rate (mol/min)
0
0.2
2.32
1.164
3
1.376
4.55
1.645
12.66
2.38
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BCH 361Advanced Biochemistry IFall 2020
3
38.50
2.76
200.00
3.18
b. Determine the vmax and KM values for the enzyme using graphical techniques. Use more
than one type of graph. Compare and contrast the kinetic constants you obtain.
A:
- 1/Vmax occurs at y intercept of the lineweaver burk plot
- y=1.2466(0) +0.3211
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