Name_________________________ Section #_______________
1. Define Km. The
substrate concentration and ˝ Vmax.
2. The presence of a competitive inhibitor will have what effect on Km and Vmax?
Vmax will stay
the same, Km will be larger
3. What kind of bond is responsible for the formation of secondary structures in a protein? Give an example of secondary structure.
Hydrogen
bonds. Beta sheets or alpha helices.
4. What are the enzyme, substrate, and product from the reaction studied last week in lab?
Enzyme:
acid phosphatase
Substrate:
p-nitrohphenylphosphate
Product:
p-nitrophenol
5. Amino acids in a protein are linked together by a covalent ______peptide___ bond between a ___C___ atom and ______N_____ atom.
6. During last week's experiment, each
reaction had an accompanying blank consisting of substrate, buffer and water.
Why would a blank consisting of enzyme, buffer and water be an unsuitable
choice for a blank?
The reaction can occur
NON-enzymatically. Therefore the blank must have substrate, and NO enzyme.
7. Calculate the specific activity of
an enzyme whose Vmax was 0.10 mMoles/sec with a protein
concentration of 13 mg/ml and the reaction was done in 1 ml.
SA =
Vmax/mg protein
13 mg/ml
x 1ml = 13 mg and 13
mg x 1mg/10^3 mg = 0.013 mg
0.10 mMoles/sec x 60 sec/1min = 6 mMoles/min
SA = 6 mMoles/min / 0.013 mg
SA = 461.5 6 mMoles/min/mg
8. The Following data was obtained from last week’s enzyme lab, where the incubation time was 15 min. Calculate the rates and put them in the table below. Make certain to use correct units! Use the attached standard curve.
|
[S], mM |
Absorbance |
Rate |
Km |
Vmax |
|
0.363 |
0.83 |
0.02500 |
|
|
|
0.175 |
0.42 |
0.01133 |
|
|
|
0.090 |
0.21 |
0.00467 |
0.0235 mM |
0.57 mM/min |
|
0.0452 |
0.13 |
0.00200 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
9. Now plot a Lineweaver-Burke graph for the data on the graph paper supplied. Remember to title it, and correctly label axes (don’t forget units!) Calculate Km and Vmax and enter in the table above. Use correct units!
Using the data above to calculate 1/V and 1/[S], I used my calculator to get the equation for the line (remember, y=mx+b). Here's what I got:
y = 0.0412 + 1.74
which is related to this eqn: 1/v = (Km/Vmax * 1/s) + 1/Vmax
So, 1/Vmax = 1.74
Vmax = 1/1.74 = 0.57 mM/min
Slope = Km/Vmax = 0.0412
So Km = 0.0412 x Vmax = 0.0412 x .57
= 0.0235 mM
10. On your Lineweaver-Burke plot, draw and label a line that could result if the experiment was run in the presence of a competitive inhibitor.
