This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Michaelis‐Menten Quiz BMB 401 P a g e  1 1. You have isolated a new enzyme that breaks down cellulose, and are beginning to characterize
it’s kinetic parameters for the substrate cellulose at 37 degrees Celsius and pH 7. At the
following concentrations of substrate, with 10 µM of your purified enzyme, you found the
following initial velocities (Vo) for the reaction. They are listed in this format:{[Cellulose](µM),
Vo (µM * sec1)} {2.50,10.00};{6.67,20.00};{20.00,40.00} Please determine the Km for this
enzyme, substrate pair. A) 1.0 µM
B) 1.3 µM
C) 13.22
D) 3.7 µM
E) 71.4 µM 2. Given these data, please determine (to two decimal places) the Vmax for this enzyme with
cellulose as it’s substrate. A) 0.05 µM /sec
B) 35.00 µM /sec
C) 7.14 µM /sec
D) 3.50 µM /sec
E) 62.5 µM /sec 3. You notice that this enzyme is markedly inhibited by something found in the crude plant
extract that is not present in the purified enzyme sample. You have isolated the inhibiting
molecule from the crude extract, and want to test the effect of this inhibitor on your purified
enzyme sample. How would you construct this experiment? What would stay the same? What
would change? Please answer these questions before going further. Michaelis‐Menten Quiz BMB 401 P a g e  2 4. You have obtained the following results with 10 µM of your enzyme, all other parameters the
same, and 100 µM of your purified small molecule added to the reaction before addition of the
substrate to begin the reaction. You found the following results: They are listed in this
format:{[Cellulose](µM), Vo (µM * sec1)} {2.50,5.00};{6.67,10.00};{20.00,20.00} Please
determine the Kmapp for this enzyme, substrate pair in the presence of your inhibitor. A) 13.21 µM
B) 0.03 µM
C) 35.8 µM
D) 1.54 µM
E) 9.7 µM 5. What is the Vmaxapp for the inhibited enzyme under these conditions? A) 0.35 µM/sec
B) 31.15 µM/sec
C) 71.43 µM/sec
D) 10.34 µM/sec
E) 0.071 µM/sec 6. What type of inhibition is seen with this small molecule? 7.Please note, this is not a question, but additional advice. You should be able to graph these
data and determine what type of competition is seen. If given this type of material on an exam,
you will be given very easy numbers to work with, a list of reciprocals, and simple linear
equations. The math will be very very simple. Michaelis‐Menten Quiz BMB 401 P a g e  3 1. You have isolated a new enzyme that breaks down cellulose, and are beginning to characterize
its kinetic parameters for the substrate cellulose at 37 degrees Celsius and pH 7. At the following
concentrations of substrate, with 10 µM of your purified enzyme, you found the following initial
velocities (Vo) for the reaction. They are listed in this format:{[Cellulose](µM), Vo (µM * sec1)} {2.50,10.00};{6.67,20.00};{20.00,40.00} Please determine the Km for this enzyme,
substrate pair. Michaelis ‐ Menten Quiz 0.25
y = 0.2115x + 0.016
inhibited
y = 0.4231x + 0.0321 0.15 B) 1.3 µM 0.1 C) 13.22 0.05 Linear (uninhibited) 0 D) 3.7 µM
E) 71.4 µM uninhibited 0.2 A) 1.0 µM ‐0.6 ‐0.4 ‐0.05 0
‐0.2
‐0.1
‐0.15
‐0.2 0.2 0.4 0.6 Linear (uninhibited)
Linear (inhibited) To Figure this out, solve for then y = 0, at that point, you will have your xintercept
For noninhibited, 0 = 0.2115x = 0.016;  0.016/ 0.2115 = X; X =  0.0757
KM = 1(1/0.0757) which equals 13.22
2. Given these data, please determine (to two decimal places) the Vmax for this enzyme with
cellulose as it’s substrate. A) 0.05 µM /sec
B) 35.00 µM /sec
C) 7.14 µM /sec
D) 3.50 µM /sec
E) 62.5 µM /sec
To do this, solve for when x = 0 to get the yintercept, y = 0.016, Vmax = 1/ 0.016
Vmax = 62.5
3. You notice that this enzyme is markedly inhibited by something found in the crude plant
extract that is not present in the purified enzyme sample. You have isolated the inhibiting
molecule from the crude extract, and want to test the effect of this inhibitor on your purified
enzyme sample. How would you construct this experiment? What would stay the same? What Michaelis‐Menten Quiz BMB 401 P a g e  4 would change? Please answer these questions before going further.
You would keep everything the same except for the addition of the inhibitor. 4. You have obtained the following results with 10 µM of your enzyme, all other parameters the
same, and 100 µM of your purified small molecule added to the reaction before addition of the
substrate to begin the reaction. You found the following results: They are listed in this
format:{[Cellulose](µM), Vo (µM * sec1)} {2.50,5.00};{6.67,10.00};{20.00,20.00} Please
determine the Kmapp for this enzyme, substrate pair in the presence of your inhibitor. Michaelis ‐ Menten Quiz 0.25 A) 13.18 µM uninhibited 0.2 B) 0.03 µM 0.15 y = 0.2115x + 0.016 C) 35.8 µM 0.1 D) 1.54 µM 0.05 inhibited
y = 0.4231x + 0.0321 E) 9.7 µM 0
‐0.6 ‐0.4 ‐0.05 0
‐0.2 0.2 0.4 0.6 Linear (uninhibited) ‐0.1
‐0.15 Linear (inhibited) ‐0.2 Same as before when y = zero, X = 0.0321/0.4231; and so KMapp = 13.18 close enough to be
equal to KM
5. What is the Vmaxapp for the inhibited enzyme under these conditions?
A) 0.35 µM/sec
B) 31.15 µM/sec
C) 71.43 µM/sec
D) 10.34 µM/sec
E) 0.071 µM/sec
Setting X equal to zero, y= 0.0321, therefore Vmaxapp = 1/0.0321 = 31.15
noncompetitive
6. What type of inhibition is seen with this small molecule?
7. Please note, this is not a question, but additional advice. You should be able to graph these data and
determine what type of competition is seen. If given this type of material on an exam, you will be given
very easy numbers to work with, a list of reciprocals, and simple linear equations. The math will be very
very simple Michaelis‐Menten Quiz BMB 401 P a g e  5 ...
View
Full
Document
This note was uploaded on 10/18/2011 for the course BMB 401 taught by Professor Kaguni during the Fall '08 term at Michigan State University.
 Fall '08
 KAGUNI

Click to edit the document details