Interpret high and low glucose feedback records
Measure departure from a supplied model reference without inventing a clinical dose.
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
What are the departure magnitudes at 10 minutes?
What you will learn
- Compare departures at matched times
- Calculate changes using separate baselines
Interpret high and low glucose feedback records
Invented reference 5 model units: high 7→6→5; low 3→4→5 at 0, 10 and 20 minutes. Both departures shrink 2→1→0. Original teaching model.
| Time / min | High start / model units | Low start / model units |
|---|---|---|
| 0 min | 7 | 3 |
| 10 min | 6 | 4 |
| 20 min | 5 | 5 |
| Feature | Meaning |
|---|---|
| High change | 2-unit decrease; 28.6% |
| Low change | 2-unit increase; 66.7% |
| Matched departure | 2, 1, 0 model units |
| Limits | Not a clinical threshold, dosing rule or sole-cause proof |
Compare departures at matched times
The invented teaching record uses reference 5 model units. At 0, 10 and 20 minutes the high-start concentration is 7, 6 and 5, while the low-start concentration is 3, 4 and 5. Both start two units away from the reference, move to one unit away and finish at the reference in this simplified model. One concentration decreases and the other increases, yet both patterns show correction.
The high-start absolute decrease is 7 − 5 = 2 model units. Its percentage decrease is 2/7 × 100, approximately 28.6% to one decimal place. The low-start absolute increase is 5 − 3 = 2 model units, but its percentage increase is 2/3 × 100, approximately 66.7%. Equal absolute changes do not establish equal percentages because the baselines differ.
Calculate changes using separate baselines
The high branch is consistent with insulin promoting liver glucose storage, and the low branch with glucagon promoting liver glycogen breakdown and glucose release. The record illustrates those taught directions. Without direct hormone measurements and controls, a concentration table alone would not prove a unique hormonal cause for each change in an actual organism.
Use model units exactly as supplied. The reference is not a diagnosis boundary and the table contains no dose-response relationship, so it cannot prescribe an insulin dose. At a later reading still one unit from the reference, describe partial correction rather than claiming full return. A data answer should combine accurate calculation with a properly limited biological interpretation.
Sources for this lesson
Pause and explain
Which percentage pair is correct?
Worked example
Compare the two complete changes.
Show the worked solution
- High: decrease 2 units; 2/7 × 100 ≈ 28.6%.
- Low: increase 2 units; 2/3 × 100 ≈ 66.7%.
Answer Equal 2-unit changes towards reference 5; high decreases 28.6% and low increases 66.7% because the baselines differ.
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.
Find departure magnitudes at 0, 10 and 20 minutes for both displayed records.
Give me a hint
Measure distance from reference 5.
Compare my reasoning
- At 0: both are two units away.
- At 10: one unit; at 20: zero units.
Both records have departure magnitudes 2, 1 and 0 model units at 0, 10 and 20 minutes.
Look for these in your work
- Use distance from reference
- Compare matched times
Calculate percentage changes from 7→5 and 3→5, to one decimal place.
Give me a hint
Use each starting value as denominator.
Compare my reasoning
- High decrease: 2/7 × 100 ≈ 28.6%.
- Low increase: 2/3 × 100 ≈ 66.7%.
High: 28.6% decrease. Low: 66.7% increase.
Look for these in your work
- Use separate baselines
- Retain directions and rounding
What additional claims are not justified by this concentration-only model?
Give me a hint
Consider direct causal evidence and personal treatment.
Compare my reasoning
- No measured hormone concentration or sole-cause control is supplied.
- No clinical dosing relationship or personal care plan is supplied.
The model does not prove a sole hormonal cause and does not supply a personal insulin dose. It illustrates corrective concentration directions.
Look for these in your work
- Limit causal inference
- Keep educational data separate from treatment
Common mistakes
- Using the reference instead of the baseline for percentage change.
- Claiming the table proves sole hormonal causation.
What can you explain now?
Which percentage pair is correct?
Compare with the explanation
High decrease 28.6%; low increase 66.7%
Each uses its own starting value.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Calculate both changes and evaluate what the invented feedback record supports.
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.