SME
Bioinstrumentation
Biosensors
Use this for “A glucose sensor with slope 2.5 nA/mM and blank 0.4 nA reads 15.4 nA. What is the concentration?”
Chart
Line of fit: S = 2.5C + 0.4
2 points above the line, 2 below, 1 on it.
Prediction for C of 6: S = 2.5 × 6 + 0.4 = 15.4
Least-squares line (dashed): S = 2.5C + 0.4, sum of squared residuals = 0.1, the least any line gives
This is an example. Type your own values to start a new problem.
C (mM)
=
S (nA)
−
b (nA)
m (nA/mM)
LOD (mM)
=
3
×
σ (nA)
m (nA/mM)
Formulas
C
=
S
−
b
m
6
=
15.4 − 0.4
2.5
LOD
=
3
×
σ
m
0.06
=
3
×
0.05
2.5
Assumptions
•
The reading lies inside the linear range, where the current is S = mC + b.
•
The blank b is the current with no analyte; its noise σ is one standard deviation.
•
The limit of detection is where the signal stands three noise widths above the blank.
Step-by-step
WE KNOW
m = 2.5 nA/mM, b = 0.4 nA, σ = 0.05 nA, S = 15.4 nA
FIND
concentration (C), limit of detection (LOD)
Step 1 · Find concentration (C)
Use C = (S − b) ÷ m
Take
away
the
blank,
then
divide
by
the
sensitivity:
each
mM
adds
m
nA.
C
=
15.4 − 0.4
2.5
C
=
15
2.5
C
=
6
mM
Step 2 · Find limit of detection (LOD)
Use LOD = 3 × σ ÷ m
Three noise widths of current, turned into a concentration by the sensitivity.
LOD
=
3
×
0.05
2.5
LOD
=
0.15
2.5
LOD = 0.06 mM
Check
Put the numbers back in. Both sides should match.
6
=
15.4 − 0.4
2.5
✓
0.06
=
3
×
0.05
2.5
✓
Topic check
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