1. 1
   Students place ten discs of fresh potato tissue, a source of the enzyme catalase, in 10 cm³ of hydrogen peroxide solution at different temperatures. Catalase breaks down hydrogen peroxide into water and oxygen. They collect the oxygen released in 2 minutes in an inverted measuring cylinder. The volumes of oxygen collected were: 10 °C, 4.0 cm³; 20 °C, 8.2 cm³; 30 °C, 15.6 cm³; 40 °C, 20.4 cm³; 50 °C, 9.0 cm³; 60 °C, 1.2 cm³.

   (a)

   What was the mean rate of oxygen production at 30 °C?

   [1 mark]

   • A0.065 cm³ s⁻¹
   • B31.2 cm³ s⁻¹
   • C7.8 cm³ s⁻¹
   • D0.13 cm³ s⁻¹

   (b)

   Which explains the decrease in the volume of oxygen between 40 °C and 60 °C?

   [1 mark]

   • AThe substrate is used up faster at higher temperatures
   • BMolecules move more slowly at high temperature, so there are fewer collisions
   • CBonds holding the tertiary structure of catalase break, so the active site changes shape and substrate can no longer bind
   • DThe activation energy of the reaction increases at high temperature

   (c)

   Explain the increase in rate between 10 °C and 30 °C.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   In liver cells, the enzyme hexokinase adds a phosphate group to glucose as the first step in its use. X-ray crystallography shows that when glucose enters the active site, the two lobes of hexokinase close around it, and the glucose molecule is slightly distorted. Hexokinase does not act on galactose, which has the same formula as glucose and differs only in the orientation of the hydroxyl (–OH) group on one carbon atom. After a meal, other enzymes in liver cells link glucose molecules into glycogen, while in the small intestine amylase breaks starch into maltose.

   (a)

   Which statement describes the binding of glucose to hexokinase?

   [1 mark]

   • AThe active site is rigid and exactly matches the shape of glucose before binding
   • BBoth the enzyme and the substrate change shape when binding occurs
   • CGlucose binds to a site away from the active site and changes its shape
   • DGlucose changes shape but the enzyme does not

   (b)

   Which reaction described is anabolic?

   [1 mark]

   • AAddition of glucose molecules to glycogen
   • BBreakdown of starch into maltose by amylase
   • COxidation of glucose in respiration
   • DHydrolysis of proteins in the small intestine

   (c)

   Explain why hexokinase does not catalyse a reaction with galactose.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Hydrogen peroxide is a toxic by-product of metabolism. Without a catalyst, its breakdown into water and oxygen has an activation energy of about 75 kJ mol⁻¹, and at 37 °C it proceeds extremely slowly. When catalase is present, the activation energy is about 8 kJ mol⁻¹ and the reaction is millions of times faster. In both cases the reaction releases about 98 kJ of energy per mole of hydrogen peroxide.

   (a)

   Calculate the percentage reduction in activation energy caused by catalase, and explain why the energy released by the reaction is the same with or without the enzyme.

   [3 marks]

   (b)

   Using the data, explain the benefit to cells of catalysing reactions with enzymes.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A textbook sketch graph models the effect of pH on enzyme activity as a symmetrical curve with a single peak at the optimum pH. To test this model, students measure how long trypsin (a protease from the small intestine) takes to digest the gelatine coating on strips of exposed photographic film, which turns clear when the gelatine has been removed. They use buffer solutions at whole-number pH values and judge the endpoint by eye. Mean times to clear the film were: pH 5, 600 s; pH 6, 300 s; pH 7, 150 s; pH 8, 100 s; pH 9, 120 s; pH 10, 250 s. At pH 11 the film had not cleared after 900 s. The students calculate rate as 1/time.

   (a)

   Using the data, explain the effect of pH on the rate of the reaction catalysed by trypsin.

   [6 marks]

   (b)

   Evaluate the sketch graph model of the effect of pH on enzyme activity using these results, and suggest improvements to the method.

   [6 marks]

   Total for question 4: 12 marks

End of questions