= Lysosomal enzyme latency
<Lysosomal enzyme latency> is the difference between the <enzyme> activity accessible to an added <substrate> before disruption of the surrounding membrane and the activity revealed after disruption. Low accessible activity can reflect separation of <enzyme> from <substrate>, rather than absence of <enzyme>. This is an assay-accessibility statement, not a claim that the intact <lysosome> is inactive in the living <biological cell>.<Exploring Cells with a Centrifuge by Christian de Duve>[Christian de Duve, Exploring Cells with a Centrifuge (Nobel Lecture, 12 December 1974), printed pp. 157–159, enzyme-latency discussion and Figure 6]{ref}
\Image[/lysosome-latency-disruption-assay.png]
{title=<Lysosomal enzyme latency>: matched assay conditions and disruption controls}
{description=Original assay-logic diagram and explicitly chosen activity example. The values $10$, $100$ and $25$ are arbitrary blank-corrected rate units, not historical observations. The soluble-<enzyme> comparison is an additional control proposed for the schematic, not an asserted measurement from the <rat-liver acid-phosphatase latency experiment>.}
Let $F$ be the blank-corrected initial rate measured with a membrane-preserving preparation and $T>0$ the rate after a treatment that exposes all assayable <enzyme>. Use matched amounts of starting material, <pH>, <substrate> and measurement time. Define the apparent latent fraction by
$$
\ell=1-\frac{F}{T}.
$$
This definition requires more than a large treated signal. Disruption must preserve <enzyme> activity, and the treatment must not independently alter the readout. A soluble-<enzyme> comparison with and without treatment helps test direct activation or inhibition; an appropriate blank tests signal unrelated to <enzyme>-catalysed <hydrolysis>. These are controls for interpreting the constructed assay, rather than additional historical observations.
For a deliberately chosen example, take $F=10$ and $T=100$ arbitrary rate units. Then $\ell=0.90$: ninety percent of the assayable activity was initially inaccessible. After hypothetical partial damage, $F=25$ with unchanged $T=100$ gives $\ell=0.75$. The preparation has lost latency without gaining total assayable activity. These percentages describe activity accessibility; they do not count intact <lysosomes>. Unequal <enzyme> contents, partial leakage and mixed particle populations prevent that identification. If $F>T$, the formula is negative, flagging failure of this simple accessibility interpretation rather than a negative population of hidden <enzymes>.
To check the logic, divide the same uniformly mixed preparation into two equal portions. If dilution halves both measured rates, $(F/2)/(T/2)=F/T$, so $\ell$ is unchanged. In contrast, comparing an undiluted untreated portion with a diluted treated portion changes the ratio artificially. Normalization to the same amount of original material is therefore part of the definition, not an optional numerical correction.
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