Spike-recovery estimate (source code)

= Spike-recovery estimate

A <spike-recovery estimate> compares the measured increment after adding a known <analyte> amount to a sample with the amount added. It tests the response to the added material under the stated preparation and <measurement calibration>; it does not automatically measure recovery of <analyte> originally bound or embedded in the <sample matrix>.

If matched aliquots give inferred amounts $\widehat q_s$ and $\widehat q_0$ after accounting for dilution, and $q_a$ was added, the incremental <apparent recovery> is
$$
R_s=\frac{\widehat q_s-\widehat q_0}{q_a}.
$$
An original dilution example starts with a $2.00\,\mathrm{mL}$ aliquot whose unspiked result is $3.00\,\mathrm{nmol\,mL^{-1}}$. Add $0.100\,\mathrm{mL}$ containing $20.0\,\mathrm{nmol\,mL^{-1}}$ of <analyte>; the added amount is $2.00\,\mathrm{nmol}$. If the spiked $2.10\,\mathrm{mL}$ solution measures $3.60\,\mathrm{nmol\,mL^{-1}}$, then
$$
R_s=\frac{(3.60)(2.10)-(3.00)(2.00)}{(20.0)(0.100)}=0.78.
$$
Subtracting the two <concentrations> without accounting for the changed <volume> would give a different answer. Alternatively, bring the spiked and unspiked aliquots to the same final <volume> before comparing them.

The spike's point of addition defines what is tested. Added before extraction, it experiences preparation loss and matrix-dependent response; added afterward, it cannot establish the earlier <extraction recovery>. <Spike equilibration> also matters: freshly added dissolved <analyte> may not access the same binding sites or compartments as native material. IUPAC's recovery guidance distinguishes recovery of native and added surrogate material and their possible differences.<Harmonised guidelines for the use of recovery information in analytical measurement by Michael Thompson and others>[Eurachem-hosted text, §§3.2–3.2.2, native/surrogate recovery and spiking.]{ref}

Use multiple spike levels and matched unspiked aliquots to look for concentration-dependent response. An incremental result near 100% can coexist with a constant interference in both aliquots, because subtraction removes that shared offset. The result therefore supports a particular recovery comparison, rather than proving every source of analytical bias absent.