This set covers five skills: writing limited inferences from food tests, naming monomers and polymers, using the lock and key model, reading rate data, and separating denaturation from slower collisions. Work through the questions in order, since they go from easy to harder.
Write a full answer for each one before you open the working. All data are invented for practice.
Questions
Q1. Iodine solution is added to a sample and turns blue-black. State the observation and the inference.
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Observation: orange-brown changed to blue-black. Inference: starch is present.
Q2. A sample gives a purple colour with biuret reagent and stays blue when heated with Benedict’s solution. Write a careful conclusion.
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Protein is present. A reducing sugar was not detected. The wording “not detected” is used because the test only finds reducing sugars.
Q3. Name the monomers of (a) starch and (b) protein.
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(a) Glucose. (b) Amino acids.
Q4. A fat molecule is made of glycerol and fatty acids. Explain why it is better not to call it a polymer of identical units.
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A fat is made from one glycerol and three fatty acids. These are two different kinds of part, not a long chain of the same repeating unit, so it does not fit the usual meaning of polymer.
Q5. Protease breaks down protein but has no effect on starch. Use the lock and key model to explain.
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Protease has an active site with a specific shape that is complementary to protein. Protein fits and is broken down. Starch has a different shape, does not fit the active site, so no enzyme-substrate complex forms.
Q6. A student times how long amylase takes to break down the starch in a mixture.
| Temperature (°C) | 20 | 30 | 40 |
|---|---|---|---|
| Time for starch to disappear (s) | 120 | 60 | 30 |
Using rate = 1000 ÷ time, calculate the rate at each temperature and state how many times faster the reaction is at 40 °C than at 20 °C.
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At 20 °C: 1000 ÷ 120 = 8.3. At 30 °C: 1000 ÷ 60 = 16.7. At 40 °C: 1000 ÷ 30 = 33.3 (rates in units per second, to 1 decimal place).
Times faster: 120 ÷ 30 = 4, or 33.3 ÷ 8.3 is about 4. The reaction is 4 times faster at 40 °C.
Q7. The rate of an enzyme at different pH values is shown.
| pH | 1 | 2 | 3 | 5 | 7 |
|---|---|---|---|---|---|
| Rate (units per minute) | 4 | 20 | 12 | 2 | 0 |
(a) State the pH with the highest rate and what you can say about the optimum. (b) Calculate the percentage decrease in rate from pH 2 to pH 5. (c) Suggest where in the body an enzyme like this might work.
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(a) The highest rate is at pH 2, so the optimum is about pH 2. It could be between pH 1 and pH 3, since those are the neighbouring readings.
(b) The decrease is 20 − 2 = 18. The percentage is 18 ÷ 20 × 100 = 90%.
(c) The optimum is acidic, so a suitable place is the stomach, where pepsin works.
Q8. Tube X is kept at 0 °C for 10 minutes and then warmed to 37 °C, where it reacts quickly. Tube Y is kept at 90 °C for 10 minutes and then cooled to 37 °C, where there is no reaction. Explain the difference.
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In tube X the cold only slowed collisions. The enzyme kept its shape, so warming restored the rate. In tube Y the high temperature changed the shape of the active site, so the enzyme is denatured and the substrate no longer fits. Cooling does not reverse this.
Q9. A student finds that a juice sample turns brick-red with heated Benedict’s solution and writes: “The juice contains glucose and nothing else.” Identify the two problems and rewrite the conclusion.
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Problem 1: Benedict’s solution shows a reducing sugar, not specifically glucose. Problem 2: “nothing else” claims everything else is absent, but only one test was done and the test cannot show that.
Rewritten: the juice contains a reducing sugar. The test does not show which sugar it is or what else is present.
Q10. The table shows the rate of a reaction catalysed by an enzyme.
| Temperature (°C) | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Rate (units per minute) | 3 | 6 | 12 | 24 | 4 |
(a) Describe the pattern from 10 °C to 40 °C. (b) Explain the rise. (c) Calculate the percentage decrease from 40 °C to 50 °C and explain the fall.
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(a) The rate doubles with each 10 °C rise: 3, 6, 12, 24 units per minute.
(b) Higher temperature gives faster particle movement, so enzyme and substrate collide more often and with more energy, giving more reactions per minute.
(c) The decrease is 24 − 4 = 20. The percentage is 20 ÷ 24 × 100 = 83.3%. At 50 °C the enzyme begins to denature, so the active site changes shape and the substrate fits less well.
If you got these wrong
| Where the error was | Questions | Go back to |
|---|---|---|
| Wording of observations and inferences | Q1, Q2, Q9 | Food-test inferences |
| Mixing up monomers and polymers | Q3, Q4 | Monomer and polymer |
| Explaining why an enzyme acts on one substrate | Q5 | Enzyme specificity |
| Reading graphs, rates and percentages | Q6, Q7, Q10 | Temperature and pH graphs |
| Choosing between denaturation and slower collisions | Q8, Q10 | Denaturation |
Keep a short log of the error type, not only the question number. The mistake log and retest queue helps you return to each slip after a few days with a fresh question, and the inheritance model board is a good place to practise stating model assumptions.
Return to the topic overview for the study order. If the same errors keep appearing after a second attempt, our teachers can look at them in online one-to-one Biology tuition.