The following seven sets of 3mL-volumes are added to seven test tubes to produce
ID: 871032 • Letter: T
Question
The following seven sets of 3mL-volumes are added to seven test tubes to produce enzyme-inhibition reactions, which vary in volume of inhibitor (1mM inorganic phosphate, Pi). Tube #1 (containing 0.0mL inhibitor) is used as a blank when the UV absorbance values (at wavelength 410nm) for Tubes #2-7 are measured by a spectrophotometer. Due to the varying volume of inhibitor, each reaction has a different absorbance and, thus, a different reaction velocity (v).
Using each reaction's measured absorbance value and volume of inhibitor, find the reaction velocity (v) and the concentration of the inhibitor [Pi] for each enzyme-inhibition reaction in terms of nmol/min and mM, respectively. The time interval for each reaction is 3 min. The molar exctinction coefficient of the reaction product is 18.5 mM-1cm-1 and the path length of the sample cuvette is 1.0 cm.
Inhibitor (I) Substrate (S) Enzyme (E) Absorbance Vol. 1mM Pi (mL) Vol. Buffer (mL) Vol. PNPP (mL) Vol. Alk phos (mL) NaOH (mL) Tube #1 -- 0.0 2.4 0.3 0.2 0.1 Tube #2 0.4149 0.1 2.3 0.3 0.2 0.1 Tube #3 0.3214 0.2 2.2 0.3 0.2 0.1 Tube #4 0.2486 0.3 2.1 0.3 0.2 0.1 Tube #5 0.2120 0.4 2.0 0.3 0.2 0.1 Tube #6 0.1795 0.5 1.9 0.3 0.2 0.1 Tube #7 0.1633 0.6 1.8 0.3 0.2 0.1Explanation / Answer
I have to correct the following statements of your question:
- Tube#1 has no enzyme (delete 0.2 from the table): see the table no enzyme activity
- for constant sample volume, add 2.6 mL buffer in tube#1
- The absorbance is in fact variation of absorbance dA/dt and I will assume that the unit is absorbance unit/min (min-1).
- the extinction coefficient e is not for the enzyme, but for for the reaction product (p-nitrophenol)
Find the method here:
http://www.worthington-biochem.com/BAP/assay.html
e = 18.5 mM-1cm-1
You works in a 1 cm cell, then use this value 18.5 mM-1cm-1 x 1cm= 18.5 mM-1 =
= 18.5 mmol/L =18 500 mmol/mL
The sample volume is V = 3 mL (all reactants, see table)
The velocity of the enzyme reaction is expressed as amount of product formed per time (usually micromol/min).The reaction rate is the number of mmol of p-nitrophenol produced per minute.
Reaction rate = Velocity v = [(dA/dt)/e] x V =
= (V/e) x (dA/dt) = 0.162x10-3 mmol x (dA/dt) = 162 micromol x(dA/dt)
dA/dt
Velocity, micromol/min
0.4149
67.21
0.3214
52.07
0.2486
40.27
0.212
34.34
0.1795
29.08
0.1633
26.45
To find the enzyme activity (I.U.), divide v by yhe enzyme sample volume (here 0.2 mL).see table:
velocity
Activity, I.U.
67.21
336
52.07
260
40.27
201
34.34
172
29.08
145
26.45
132
The concentration of the inhibitor is
1mM x added volume/ 3 mL
vol Pi added, mL
Pi conc., mmol/mL
0.1
0.033
0.2
0.067
0.3
0.100
0.4
0.133
0.5
0.167
0.6
0.200
End your work with a graph Activity vs. Inhibitor conc.
Activity, I.U.
Pi conc., mmol/mL
336
0.033
260
0.067
201
0.100
172
0.133
145
0.167
132
0.200
I
dA/dt
Velocity, micromol/min
0.4149
67.21
0.3214
52.07
0.2486
40.27
0.212
34.34
0.1795
29.08
0.1633
26.45