03.03 Enzyme: Product Formation

Question 1

Describe how the initial rate of an enzyme-catalyzed reaction can be measured using catalase and hydrogen peroxide. (5 marks)

Mark Scheme:

1. Set up an experiment where catalase is added to hydrogen peroxide, and the oxygen gas produced is collected in a gas syringe. (1 mark)
2. Measure the volume of oxygen produced over a specific time period, such as the first 30 seconds. (1 mark)
3. Calculate the initial rate by dividing the volume of oxygen produced by the time taken (e.g., cm³/s). (1 mark)
4. Alternatively, draw a tangent to the curve at time 0 on a reaction graph and calculate the slope. (1 mark)
5. Ensure other factors (temperature, pH, enzyme concentration) are controlled for reliable results. (1 mark)

Question 2

Explain why the initial rate of reaction is the highest in an enzyme-catalyzed reaction. (4 marks)

Mark Scheme:

1. At the start, substrate concentration is highest, so enzyme active sites are saturated. (1 mark)
2. More frequent collisions between substrate molecules and enzyme active sites occur. (1 mark)
3. The reaction proceeds at its maximum rate, as active sites are continuously occupied. (1 mark)
4. Over time, substrate concentration decreases, reducing the rate as active sites become unoccupied. (1 mark)

Question 3

Sketch and annotate a graph showing the course of an enzyme-catalyzed reaction, highlighting the initial rate, progression, and end of the reaction. (5 marks)

Mark Scheme:

1. Correctly shape the graph: Steep initial slope, gradually leveling off. (1 mark)
2. Label axes: X-axis = Time, Y-axis = Volume of product (e.g., oxygen gas). (1 mark)
3. Annotate the steep slope at the start as the initial rate of reaction. (1 mark)
4. Show and annotate the progression phase where the slope decreases due to substrate depletion. (1 mark)
5. Annotate the flat region at the end where the reaction has stopped due to substrate exhaustion. (1 mark)

Question 4

Discuss how enzyme concentration affects the rate of product formation in an enzyme-catalyzed reaction. (6 marks)

Mark Scheme:

1. Increasing enzyme concentration increases the number of active sites available. (1 mark)
2. This leads to more frequent enzyme-substrate collisions and faster product formation. (1 mark)
3. At low enzyme concentration, the reaction rate is limited by enzyme availability. (1 mark)
4. Beyond a certain point, increasing enzyme concentration has no effect as substrate becomes the limiting factor. (1 mark)
5. Graph explanation: Rate of reaction rises initially and then plateaus as substrate is depleted. (1 mark)
6. Link to example: Catalase activity depends on the enzyme turnover rate and substrate availability. (1 mark)

Question 5

Explain why the reaction rate decreases as the enzyme-catalyzed reaction progresses. (4 marks)

Mark Scheme:

1. As the reaction proceeds, substrate molecules are converted into products, reducing substrate concentration. (1 mark)
2. Fewer substrate molecules are available to occupy enzyme active sites. (1 mark)
3. Enzyme active sites are idle for longer periods, slowing the reaction rate. (1 mark)
4. The reaction stops when all the substrate is depleted. (1 mark)

Question 6

Why is it important to measure the initial rate of an enzyme-catalyzed reaction? (3 marks)

Mark Scheme:

1. The initial rate reflects the maximum enzyme activity when substrate concentration is at its highest. (1 mark)
2. It avoids inaccuracies caused by substrate depletion over time. (1 mark)
3. Provides a consistent basis for comparing the effects of different variables on enzyme activity. (1 mark)

Question 7

Outline how the reaction rate can be calculated from a graph of product formation over time. (4 marks)

Mark Scheme:

1. Draw a tangent to the curve at the desired time point (e.g., time 0 for initial rate). (1 mark)
2. Calculate the slope of the tangent: Rate=change in Y-axischange in X-axis\text{Rate} = \frac{\text{change in Y-axis}}{\text{change in X-axis}}Rate=change in X-axischange in Y-axis​. (1 mark)
3. Ensure correct units (e.g., cm³/s for volume of oxygen over time). (1 mark)
4. Slope of the tangent represents the reaction rate at that specific time point. (1 mark)

Question 8

Explain why the reaction rate levels off during an enzyme-catalyzed reaction. (3 marks)

Mark Scheme:

1. As the reaction progresses, substrate molecules are used up, so fewer remain to bind to active sites. (1 mark)
2. The reaction becomes substrate-limited, and the enzyme cannot process substrates faster. (1 mark)
3. Once all substrate is converted into product, the reaction rate drops to zero. (1 mark)

Question 9

How does the rate of oxygen production in the catalase-hydrogen peroxide reaction relate to enzyme turnover rate? (4 marks)

Mark Scheme:

1. The enzyme turnover rate is the speed at which an enzyme processes substrates and releases products. (1 mark)
2. Catalase has a high turnover rate, breaking down millions of hydrogen peroxide molecules per second. (1 mark)
3. The rate of oxygen production depends on both enzyme concentration and turnover rate. (1 mark)
4. High turnover rate ensures rapid detoxification of hydrogen peroxide. (1 mark)

Question 10

Describe how substrate concentration affects the course of an enzyme-catalyzed reaction. (6 marks)

Mark Scheme:

• At high substrate concentrations, all enzyme active sites are saturated, leading to the maximum reaction rate. (1 mark)
• The rate is limited by the number of active sites and enzyme turnover. (1 mark)
• As substrate concentration decreases over time, the reaction rate slows because fewer substrate molecules are available. (1 mark)
• Active sites spend more time unoccupied, reducing the overall rate. (1 mark)
• Graph explanation: Initial steep slope, gradually flattening as substrate concentration drops. (1 mark)
• Reaction stops when no substrate is left. (1 mark)