SME
Analytical Chemistry · Problem type
Enzyme activity from an absorbance slope
Use this for “NADH (ε = 6220 L/(mol·cm)) falls by 0.120 absorbance per minute in a 1 cm cell holding 3.00 mL, 0.050 mL of it the sample. Find the activity in U/mL.”
Chart
This is an example. Type your own values to start a new problem.
activity (U/mL)
=
ΔA/min (min⁻¹)
ε (L/(mol·cm))
×
b (cm)
×
Vₜ (mL)
Vₛ (mL)
v
Reaction rate
calculated · μM/min
Formulas
v
=
10
6
×
ΔA/min
ε
×
b
15
=
10
6
×
0.0933
6,220 × 1
activity
=
v
×
Vₜ
1,000 × Vₛ
0.45
=
15
×
3
1,000 × 0.1
Assumptions
•
One unit (U) of enzyme converts 1 μmol of substrate each minute, so U/mL is μmol per minute per mL of sample.
•
The slope is read from the straight first minutes, while the substrate is still in excess.
•
Beer’s law turns the slope into a rate: ΔA/min ÷ (εb) is in M/min, and 10⁶ times that is μM/min.
•
Only NADH absorbs at 340 nm (ε = 6220 L/(mol·cm)); the slope’s size counts whether A rises or falls.
Step-by-step
WE KNOW
ΔA/min = 0.0933 min⁻¹, ε = 6,220 L/(mol·cm), b = 1 cm, Vₜ = 3 mL, Vₛ = 0.1 mL
FIND
reaction rate (v), enzyme activity (activity)
Step 1 · Find reaction rate (v)
Use v = 10⁶ × (ΔA/min) ÷ (ε × b)
Beer’s
law:
divide
the
absorbance
change
by
εb
for
M/min,
then
times
10
6
for
μM/min.
v
=
10
6
×
0.0933
6,220 × 1
v
=
1,000,000
×
0.0933
6,220
v
=
93,300
6,220
v
=
15
μM/min
Step 2 · Find enzyme activity (activity)
Use activity = v × Vₜ ÷ (1,000 × Vₛ)
μmol per minute in the cell (rate × Vₜ, with 1000 mL in a liter), shared out per mL of sample.
activity
=
15
×
3
1,000 × 0.1
activity
=
15
×
3
100
activity
=
45
100
activity = 0.45 U/mL
Check
Put the numbers back in. Both sides should match.
15
=
10
6
×
0.0933
6,220 × 1
✓
0.45
=
15
×
3
1,000 × 0.1
✓
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