Distinguish internal variables from external changes
Identify what is regulated in an unfamiliar homeostasis record.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Recall hormone production and blood transport, respiration, liver storage and electrical impulses along neurones.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity is an internal composition variable?
What you will learn
- Name the controlled internal condition
- Interpret a supplied internal and external comparison
Distinguish internal variables from external changes
Identify the location of each measurement before interpreting stability. Outside air and internal blood temperature are distinct variables. Original teaching model.
| Feature | Meaning |
|---|---|
| Internal | Blood glucose concentration; internal temperature |
| External | Surrounding air temperature |
| Model comparison | Outside change 12°C; internal change 0.1°C |
| Inference | Internal stability does not imply control of outside air |
Name the controlled internal condition
A variable is a quantity that can change. In a homeostasis question, first identify the internal variable being maintained, such as blood glucose concentration or internal body temperature. Then identify any external disturbance, such as a change in surrounding air temperature. The disturbance and regulated variable need not use the same unit or be the same substance.
For example, blood glucose concentration is an internal composition variable, whereas the amount of food offered outside the body is not itself a blood concentration. Air temperature outside an animal is external, while the temperature of its circulating blood is internal. Naming both correctly helps prevent a description of weather or behaviour from being mistaken for the homeostasis definition.
Interpret a supplied internal and external comparison
An invented temperature comparison may show outside air changing from 20 to 8°C while an internal reading changes from 37.0 to 36.9°C. The external change is 12°C and the internal change is only 0.1°C. That contrast is consistent with regulation of internal temperature. It does not show that the animal made the outside air warmer.
Read the labels before calculating. A table headed blood glucose is not automatically a temperature record, and an external thermometer does not measure internal body conditions. After identifying the variable, describe how its stability relates to homeostasis. Detailed negative-feedback mechanisms are added in the Supplement lessons rather than required by this Core identification task.
Sources for this lesson
Pause and explain
What does outside 20→8°C with internal 37.0→36.9°C support?
Worked example
Classify surrounding air temperature and blood glucose concentration.
Show the worked solution
- Surrounding air temperature is external.
- Blood glucose concentration is an internal variable.
Answer Air temperature is external; blood glucose concentration is internal.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Classify blood glucose concentration and outside air temperature as internal or external.
Give me a hint
Use where the measurement applies.
Compare my reasoning
- Blood glucose concentration is internal.
- Outside air temperature is external.
Blood glucose concentration: internal. Outside air temperature: external.
Look for these in your work
- Classify both quantities
- Use the measurement location
Calculate the magnitudes of changes for outside 20→8°C and internal 37.0→36.9°C.
Give me a hint
Subtract each pair separately.
Compare my reasoning
- Outside magnitude is 12°C.
- Internal magnitude is 0.1°C.
Outside changes by 12°C; internal changes by 0.1°C.
Look for these in your work
- Calculate both magnitudes
- Retain °C
An answer says an animal maintained constant outside air because its blood temperature changed little. Correct the variable being maintained.
Give me a hint
The blood reading is an internal condition.
Compare my reasoning
- Outside temperature may change.
- The comparatively stable blood temperature is internal.
The comparatively stable condition is internal blood temperature, not the outside air. That pattern is consistent with homeostasis.
Look for these in your work
- Identify the internal variable
- Correct the external-control claim
Common mistakes
- Calling food outside the body blood glucose concentration.
- Treating an outside thermometer as an internal reading.
What can you explain now?
What does outside 20→8°C with internal 37.0→36.9°C support?
Compare with the explanation
Internal temperature remains comparatively stable
It is consistent with homeostasis.
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Make it stick
Classify the regulated variable and compare internal stability with an external disturbance.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.