BIOLOGY WITH HAMZA · MDCAT REVISION
Homeostasis
Name: ____________________ Date: ______________
Recall first. Check second. Return to the ideas you missed.
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Learning targets
- Distinguish a set point from the normal range around it.
- Assign the correct role to receptors, control centres and effectors.
- Apply the model to temperature and blood-glucose regulation.
Core notes
- Homeostasis maintains variables within limits, not at perfect constancy.
- Receptors detect, control centres coordinate and effectors correct.
- The response must oppose the original deviation.
- Always identify the regulated variable before naming the components.
Read, cover and redraw
A variable leaves its normal range.
A receptor senses the change.
A control centre selects a response.
An effector reduces the deviation.
Written recall
1. Why does sweating decrease once body temperature returns toward normal?
2. What separates positive feedback from negative feedback?
3. 100 mL/min and 99% reabsorption: daily output?
4. Does filtered volume equal body water content?
5. What must remain constant to isolate reabsorption?
Apply the idea
1. A kidney model filters 180 L per day and returns 178.5 L per day to blood. If no further water is added, what volume leaves as urine?
- 1.5 L per day
- 0.15 L per day
- 178.5 L per day
- 358.5 L per day
My answer: _____ Reason: ____________________________________
2. A healthy person becomes dehydrated. Which response helps conserve water?
- Less ADH and increased water loss in urine
- More ADH, increased collecting-duct water permeability and a smaller urine volume
- More ADH and no change in water permeability
- Complete shutdown of all filtration
My answer: _____ Reason: ____________________________________
3. What is the role of sweating in this negative-feedback pathway?
- Body temperature rises
- Control system activates sweat glands
- Sweating
- Heat loss increases
- It is the temperature receptor itself
- It is the original rise in body temperature
- It acts as an effector response that increases heat loss
- It raises the temperature set point permanently
My answer: _____ Reason: ____________________________________
Answer key
Cover this section while attempting the questions.
Written recall
1. The temperature deviation becomes smaller, so thermoreceptor input and the coordinating signal to sweat glands are reduced.
2. Negative feedback reduces the initiating change; positive feedback amplifies it until a particular endpoint is reached.
3. 100 × 1,440 × 0.01 ÷ 1,000 = 1.44 L.
4. No. The same circulating water can be processed repeatedly.
5. Filtration rate and elapsed time.
Applied questions
1. A — Final water output is filtered volume minus reabsorbed volume: 180 − 178.5 = 1.5 L per day. Filtration volume is not urine volume.
2. B — Dehydration normally promotes ADH release. Increased collecting-duct water permeability allows more water reabsorption when the medullary gradient is present.
3. C — Sweat evaporation increases heat loss and helps counter the original temperature rise. The effector response reduces the deviation from the regulated condition.