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8. Phenolphthalein is an indicator - Edexcel - GCSE Biology - Question 9 - 2016 - Paper 1

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8. Phenolphthalein is an indicator. It is pink in alkaline solutions and turns colourless as the pH decreases. It can be used to measure the activity of the enzyme ... show full transcript

Worked Solution & Example Answer:8. Phenolphthalein is an indicator - Edexcel - GCSE Biology - Question 9 - 2016 - Paper 1

Step 1

Explain why phenolphthalein turns colourless when lipase breaks down the lipids.

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Answer

Phenolphthalein is a pH indicator that exhibits a colour change based on the acidity of the solution. When lipase breaks down lipids, fatty acids are released, lowering the pH of the solution. As the pH decreases below its threshold, phenolphthalein transitions from pink to colourless, indicating the breakdown of lipids.

Step 2

Describe the effect of temperature on the activity of lipase, as shown in Figure 10.

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Answer

From Figure 10, it is evident that the rate of lipase activity increases with temperature up to an optimal point, around 40°C. Beyond this temperature, the activity declines. This suggests that the enzyme operates most effectively at this optimal temperature, but higher temperatures may denature the enzyme, reducing its activity.

Step 3

Explain why the activity of lipase changes above a temperature of 40°C.

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Answer

Above 40°C, the activity of lipase decreases due to denaturation. The heat alters the enzyme's structure, disrupting the active site which is crucial for substrate binding. This conformational change prevents the enzyme from catalyzing the reaction effectively.

Step 4

Calculate the rate of amylase enzyme activity for the 10% starch solution.

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Answer

To calculate the rate of amylase activity, first find the average time taken from the provided results. The times are 120, 125, 110, 115, and 118 seconds. The average time is:

extAveragetime=120+125+110+115+1185=5885=117.6extseconds ext{Average time} = \frac{120 + 125 + 110 + 115 + 118}{5} = \frac{588}{5} = 117.6 ext{ seconds}

The rate of enzyme activity is then calculated as the inverse of the average time:

Rate of activity=1Average time=1117.60.0085extunits/second\text{Rate of activity} = \frac{1}{\text{Average time}} = \frac{1}{117.6} \approx 0.0085 ext{ units/second}

Step 5

State one other variable that the student should have controlled during this investigation.

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Answer

One important variable that should be controlled is the pH of the starch solution, as different pH levels can significantly affect enzyme activity.

Step 6

Explain why enzymes can only catalyse specific reactions.

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Answer

Enzymes are highly specific due to their unique active sites, which are shaped to fit particular substrate molecules. This specificity is often described by the 'lock and key' model, where only the correct 'key' (substrate) can fit into the 'lock' (enzyme). Consequently, each enzyme is tailored to catalyse its specific reaction without affecting other processes.

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