Catalase Practical: One Peak or Two?
A Year 12 class investigates catalase activity in potato tissue. Each group blends 2.0 g of fresh potato for exactly 30 s in 20 cm³ of pH 7.0 buffer, filters the slurry, and keeps the extract on ice. For each trial, 5.0 cm³ of 3% hydrogen peroxide is held at the test temperature for 5 min; then 1.0 cm³ of extract is added and the volume of oxygen collected in 60 s is recorded. Group A varies temperature and obtains: 20 °C, 12 cm³; 30 °C, 31 cm³; 40 °C, 28 cm³; 50 °C, 15 cm³; 60 °C, 3 cm³; 70 °C, 9 cm³. Group B varies pH at 30 °C and obtains: pH 3, 8 cm³; pH 5, 19 cm³; pH 7, 30 cm³; pH 9, 6 cm³. The 70 °C tube was run last, and because the group had run out of extract it received the final 2.0 cm³, while every other tube received 1.0 cm³. The class concludes that the enzyme has two optimum temperatures and that pH 7 is optimal for all plant enzymes.
1. Assess the class's conclusion that potato catalase has two optimum temperatures. Use the data and identify the most likely flaw in the 70 °C trial.
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The conclusion is not supported. The data show a single clear peak at 30 °C (31 cm³), with a decline to 28, 15 and 3 cm³ at 40, 50 and 60 °C. That pattern is what denaturation predicts: beyond an optimum, heat breaks hydrogen bonds and ionic interactions, distorting the active site. The 70 °C value (9 cm³) is inconsistent with this trend and is almost certainly an artefact. The trial used 2.0 cm³ of extract instead of 1.0 cm³, so it contained about twice as much enzyme; under substrate-rich conditions the initial rate is roughly proportional to enzyme concentration. Halving 9 cm³ gives about 4.5 cm³, close to the 60 °C value and consistent with progressive denaturation. The 70 °C result should be discarded and repeated with 1.0 cm³ of extract, ideally with the enzyme kept on ice until just before use. The class should not report a second optimum.
2. Group B added 1.0 cm³ of pH 7.0 extract to 5.0 cm³ of buffer at each stated pH. Explain why the actual pH in the tube may differ from the stated value, and how this affects the conclusion that pH 7 is optimal.
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The extract itself contains buffer and plant sap, so the final pH in the tube is a mixture, not the label on the buffer bottle. The effect is largest at the extremes: a pH 3 tube may end up nearer pH 4, and a pH 9 tube nearer pH 8, so the activity attributed to pH 3 or 9 can be overestimated and the curve flattened or shifted. Before drawing the conclusion that pH 7 is optimal, the class should measure the actual pH of each reaction mixture with a calibrated pH meter, or use a pH stat, and plot activity against measured pH. They should also use buffers of the same ionic strength and molarity, because buffer ions can inhibit the enzyme. The nominal pH values alone are not a reliable independent variable.
3. The class now wants to compare potato extract with liver extract. State two controlled variables and explain why 'time to collect 20 cm³ of oxygen' is a better dependent variable than 'volume after 60 s'.
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Two controlled variables are the mass of tissue per cm³ of extract (2.0 g in 20 cm³) and the volume of extract added (1.0 cm³); others include substrate concentration and volume (3% H₂O₂, 5.0 cm³), temperature, and storage time on ice. 'Time to collect 20 cm³' is better because the reaction slows as substrate is used up. Here, 31 cm³ of oxygen in 60 s means that well over half of the hydrogen peroxide in 5.0 cm³ of 3% solution has already been consumed in that minute. The 60 s volume is therefore an average over a changing rate and underestimates the initial rate. Timing a fixed smaller volume measures the early, near-linear phase. The trade-off is that very fast reactions are hard to time accurately, so a smaller volume or diluted extract may be needed.