Coronary oxygen extraction fraction

The coronary oxygen extraction fraction is the fraction of oxygen carried into a represented myocardial territory that is removed before its blood leaves. Under a matched steady-state inflow–outflow model,Here $C_a$ and $C_v$ are arterial oxygen content and venous oxygen content, $Q$ is volume flow, $D_{{\mathrm{O}}_2}$ is coronary oxygen delivery, and ${\dot{V}}_{{\mathrm{O}}_2}$ is coronary oxygen uptake. Extraction is dimensionless; uptake is an amount per time.

With chosen contents $C_a=0.20$ and $C_v=0.08\mathrm{mL}{\mathrm{O}}_2\mathrm{m}{\mathrm{L}}^{-1}$, $E=0.60$. At flow $250\mathrm{mL}\mathrm{mi}{\mathrm{n}}^{-1}$, uptake is $30\mathrm{mL}{\mathrm{O}}_2\mathrm{mi}{\mathrm{n}}^{-1}$. Maintaining that same uptake at flow $200$ and unchanged $C_a$ would require $E=0.75$. Higher extraction in this model can compensate for lower delivery; it is not by itself evidence of a higher total uptake.

The ideal algebraic limit $E\le 1$ gives $Q\ge {\dot{V}}_{{\mathrm{O}}_2}/C_a$. For the example this is $150\mathrm{mL}\mathrm{mi}{\mathrm{n}}^{-1}$. That is only a conservation bound: it does not establish that complete extraction is physiologically attainable or provide a clinical flow threshold.

Content and saturation should not be confused. When hemoglobin-bound oxygen dominates and hemoglobin concentration is the same in the paired samples, a simplified relation isIt is a relative extraction fraction, not merely the saturation difference $S_a-S_v$. Hemoglobin binding and dissolved oxygen are distinct contributions to blood oxygen transport.^[ref]

Sampling must represent the same flow path. A coronary sinus sample represents its drained venous mixture, not automatically every myocardial region. The sampling site, timing and territory therefore matter. An independent volumetric-flow measurement, such as the primary thermodilution method using saline infusion and temperature sensing, supplies a different component of the uptake calculation.^[ref]

Calibration errors can affect ratios differently from products. A common multiplicative content factor cancels from $E$, but a common additive offset does not: adding $0.02$ to both example contents gives $E=0.12/0.22\simeq 0.545$. Matching samples, measuring the intended quantity and specifying the common-error model are therefore part of the extraction estimate.