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S5 Biology: Homeostasis (Paper 4)
S5 Biology: Homeostasis (Paper 4)
The fourth paper works through the numbers of heat and water balance: production against loss, heat removed by evaporation, change in urine volume and how far glucose sits above the limit, plus the feedback-type misconception.
Direct answer Temperature is steady when heat produced equals heat lost; each gram of evaporated sweat removes 2.4 kJ; water intake must match output; feedback type depends on whether the response opposes or amplifies the change.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S5 Biology: Homeostasis |
| Level | Four basic, three advanced, two challenge and one error-repair item |
02 Questions (10 items)
- FoundationA student produces 8 400 kJ of heat in a day and loses 8 400 kJ. What happens to body temperature?
- FoundationEvaporating 1 g of sweat removes about 2.4 kJ. How much heat does 500 g of sweat remove?
- FoundationBlood glucose rises from 5.2 to 7.8 mmol/dm³. What is the rise?
- FoundationBody temperature falls from 37 °C to 35 °C. What state is that, and by how much has it fallen?
- IntermediateWhere are insulin and glucagon made, and which way do they act?
- IntermediateAfter drinking 800 cm³ of water, urine output rises from 1 200 to 1 800 cm³. What is the percentage increase?
- IntermediateDaily intake is 1 400 cm³ of drink plus 600 cm³ from food; output is 1 500 in urine, 400 in sweat and 100 in breath. Is water balanced?
- ChallengeExercise raises heat production from 300 to 1 500 kJ/h. How many times more is that, and how much sweat at 2.4 kJ/g must evaporate to match it?
- ChallengeBlood glucose is 18 mmol/dm³ against an upper limit of 8. By what percentage is it above the limit?
- Error redoA student says sweating lowers temperature, so sweating is positive feedback. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | It stays the same, since net heat is zero | When heat produced equals heat lost there is no net gain, so body temperature stays at the set point. |
| 2 | 1 200 kJ | Five hundred grams times 2.4 gives 1 200 kJ of heat carried away as sweat evaporates. |
| 3 | A rise of 2.6 mmol/dm³ | Seven point eight minus five point two gives 2.6, still within the acceptable after-meal range. |
| 4 | Hypothermia, a fall of 2 °C | Thirty-seven minus thirty-five gives 2 °C; below the normal range enzymes slow and consciousness dulls. |
| 5 | Insulin comes from pancreatic beta cells and lowers glucose; glucagon from alpha cells and raises it | The two hormones act in opposite directions and hold glucose inside a narrow band, a classic negative feedback loop. |
| 6 | An increase of 600 cm³, that is 50 per cent | One thousand eight hundred minus 1 200 gives 600, and 600 over 1 200 is 0.5, that is 50 per cent. |
| 7 | Yes, both sides total 2 000 cm³ | Intake is 1 400 plus 600, giving 2 000, and output is 1 500 plus 400 plus 100, also 2 000. |
| 8 | Five times more, needing 625 g of evaporated sweat | One thousand five hundred over 300 gives five, and 1 500 over 2.4 gives 625 g, which is why sweating surges during exercise. |
| 9 | 10 mmol/dm³ higher, that is 125 per cent above | Eighteen minus eight gives ten, and ten over eight is 1.25, so it is 125 per cent above the limit. |
| 10 | Sweating pulls the raised temperature back to the set point, so it is negative feedback; positive feedback would amplify the change | Judge by direction: a response that opposes the change is negative feedback, and only amplification counts as positive. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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