SME
Bioinstrumentation · Problem type
An enzyme sensor saturates
Use this for “Why does a glucose sensor read low at 20 mM?”
Chart
Horizontal asymptote i = 100
i(20) = 80
Half bound at C = K = 5: i(K) = 50, half of 100
This is an example. Type your own values to start a new problem.
i (nA)
=
iₘₐₓ (nA)
×
C (mM)
Kₘ (mM)
+
C (mM)
p
Share of the straight-line reading
worked out (not typed) · %
Formulas
i
=
iₘₐₓ
×
C
Kₘ +
C
80
=
100
×
20
5 + 20
p
=
100
×
Kₘ
Kₘ +
C
20
=
100
×
5
5 + 20
Assumptions
•
One enzyme site turns over the analyte, so the current follows i = iₘₐₓ C ÷ (Kₘ + C).
•
Kₘ is the concentration at half the largest current.
•
The reading is close to a straight line only well under Kₘ.
Step-by-step
WE KNOW
C = 20 mM, iₘₐₓ = 100 nA, Kₘ = 5 mM
FIND
current (i), share of the straight-line reading (p)
Step 1 · Find current (i)
Use i = iₘₐₓ × C ÷ (Kₘ + C)
The
current
is
iₘₐₓ
times
the
share
of
enzyme
busy,
C
Kₘ +
C
.
i
=
100
×
20
5 + 20
i
=
100
×
20
25
i
=
2,000
25
i
=
80
nA
Step 2 · Find share of the straight-line reading (p)
Use p = 100 × Kₘ ÷ (Kₘ + C)
The
straight
line
from
the
start
reads
iₘₐₓ
C
Kₘ
;
the
real
current
is
this
share
of
it.
p
=
100
×
5
5 + 20
p
=
100
×
5
25
p
=
500
25
p
=
20%
Check
Put the numbers back in. Both sides should match.
80
=
100
×
20
5 + 20
✓
20
=
100
×
5
5 + 20
✓
Topic check
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