Catalase, a Mystery Solution, and an Unmeasured pH
A student grinds 20 g of potato in cold pH 7.0 buffer, filters the extract, and measures catalase activity as mL of O2 collected in 2 minutes from 10 mL of 1% H2O2 at 25 °C. Tube 1 (fresh extract) gives 48 mL. Tube 2 (extract boiled 5 minutes, then cooled to 25 °C) gives 3 mL. Tube 3 (fresh extract plus 1 mL of an unknown clear solution) gives 20 mL. When the whole series is repeated with 10× H2O2, tube 3 rises to 45 mL while tube 2 stays at 4 mL. All tubes contained the same final extract concentration. The unknown solution is unbuffered, and its pH was never measured.
1. Using the substrate-saturation data, classify the unknown solution as a competitive inhibitor, a noncompetitive inhibitor, or neither. Justify each step.
Read the worked answer
Classification: competitive inhibitor (provisionally).
- At 1× substrate, activity falls to 20/48 ≈ 42% of control - inhibition is real.
- At 10× substrate, activity returns to 45/48 ≈ 94% of control. Activity is reversible by excess substrate.
- That is the signature of competitive inhibition: inhibitor and substrate compete for the same active site, raising apparent Km while leaving Vmax unchanged. Raising [S] outcompetes the inhibitor.
- A noncompetitive inhibitor would not be rescued: at 10× substrate tube 3 would have stayed near ~20 mL, because an allosteric inhibitor lowers Vmax regardless of [S].
- Tube 2 is a different phenomenon. 3 → 4 mL is insensitive to substrate, consistent with irreversible destruction of enzyme protein (denaturation), not reversible inhibition.
Caveat to carry forward: "reversible by substrate" is only diagnostic if pH and ionic strength are held constant across the 1× and 10× runs - see the next task.
2. Explain why the boiled tube still produces 3–4 mL of gas rather than zero, and state what this implies for how the data should be reported.
Read the worked answer
A crude boiled extract is not a perfect zero. Three contributions:
- Incomplete inactivation - boiling 20 g of homogenate for 5 minutes may leave some catalase active, especially in flocs or at the tube center.
- Non-enzymatic decomposition of 1% H2O2, which evolves O2 slowly at 25 °C and faster in crude extracts containing transition metals.
- Displaced air - injecting extract into the collection system pushes gas into the measuring volume; that is air, not O2.
Implication: 3–4 mL is a blank floor, not a result. The biologically meaningful values are the excess over blank:
- Tube 1: 48 − 3 = 45 mL enzymatic O2
- Tube 3: 20 − 3 = 17 mL enzymatic O2 (38% of control, not 42%)
- Tube 3 at 10×: 45 − 3 = 42 mL (93% of control)
A proper run includes a substrate-only blank (buffer instead of extract) alongside the boiled-extract blank, and reports corrected volumes with the blank stated.
3. The unknown solution turns out to have a pH of 3.0. How does this change your interpretation, and what controls must be run before any inhibitor claim is published?
Read the worked answer
It introduces a serious alternative explanation. Potato catalase has a pH optimum near 7 and loses activity sharply in acid; adding 1 mL of an unbuffered pH 3.0 solution to 10 mL of buffer may drop the working pH enough (depending on buffer capacity) to protonate and partly unfold the enzyme. A 42% residual is entirely compatible with pH inactivation, so the observation no longer isolates a specific inhibitor molecule.
Note the tension in the data: acid denaturation is usually not reversed by adding substrate, yet tube 3 recovered to 94% at 10× H2O2. Either (a) the effect is genuinely competitive and pH is irrelevant, or (b) the 10× H2O2 stock itself shifted the pH or diluted the acid enough to mask the effect. The existing experiment cannot distinguish these.
Required controls:
- pH-matched control - add 1 mL of HCl diluted to pH 3.0 to the same final volume and buffer; if activity also drops to ~20 mL, the effect is pH, not the molecule.
- Neutralized-inhibitor control - titrate the unknown solution to pH 7.0 before adding, or add it into concentrated buffer so final pH stays 7.0; inhibition that disappears was a pH artifact.
- Measure and report final pH in every tube, including the 10× substrate tubes.
- Volume control - 1 mL of an inert solute at pH 7.0, to rule out dilution or osmotic effects.
Only if inhibition survives at pH 7.0 with full buffer and is still substrate-reversible should a competitive inhibitor be claimed.