All questions and explanatory responses
1. Calculate cardiac output at rest and during exercise, in dm³ min⁻¹.
At rest, output = 72 × 70 = 5040 cm³ min⁻¹ = 5.04 dm³ min⁻¹. During exercise it is 144 × 110 = 15840 cm³ min⁻¹ = 15.84 dm³ min⁻¹. Both heart rate and stroke volume contribute to the increase.
- Use heart rate × stroke volume.
- Obtain 5040 and 15840 cm³ min⁻¹.
- Convert to 5.04 and 15.84 dm³ min⁻¹ respectively.
Common confusion: A doubled heart rate does not imply only a doubled output when stroke volume also changes.
2. Compare the fall in saturation from 13 to 6 kPa with the fall from 13 to 4 kPa. Express the differences in percentage points.
Saturation falls by 22 percentage points between 13 and 6 kPa, but by 47 between 13 and 4 kPa. The second fall is larger by 25 percentage points under these conditions. These differences concern saturation, not a measured volume of oxygen consumed by muscle.
- Calculate 97 − 75 = 22 percentage points.
- Calculate 97 − 50 = 47 percentage points.
- State that the second fall is larger by 25 percentage points.
Common confusion: Percentage-point differences and percentage changes relative to an initial value are different calculations.
3. Explain how increased carbon dioxide in an active tissue can promote oxygen unloading at a given oxygen partial pressure.
Increased carbon dioxide and associated acidity can reduce haemoglobin’s oxygen affinity. At a given oxygen partial pressure, haemoglobin then retains less oxygen. This favours unloading in an active tissue. It changes the dissociation relationship, rather than converting carbon dioxide into oxygen.
- Connect increased carbon dioxide with altered acidity and haemoglobin affinity.
- State that lower affinity means lower saturation at the same oxygen partial pressure.
- Relate this to greater oxygen release in the active tissue.
Common confusion: Red-cell shape alone does not explain a change in haemoglobin affinity.
4. Predict how doubling gas-exchange barrier thickness would affect diffusion rate if area and partial-pressure difference remain unchanged in a simple inverse-distance model.
In the stated model, rate is inversely proportional to barrier thickness, so doubling thickness gives approximately half the original rate. The diffusion path is longer. This comparison assumes that exchange area and the driving partial-pressure difference remain unchanged.
- Use the stated inverse relationship with diffusion distance.
- Predict approximately half the original diffusion rate.
- Link this to a longer diffusion path and retain the other-factor assumptions.
Common confusion: A conditional model prediction is not a measurement of every physiological consequence of a thicker barrier.
5. Evaluate the claim that the higher exercise cardiac output proves every muscle consumes three times as much oxygen.
Cardiac output increases about 3.14-fold in the supplied model, but this is whole-heart flow. Oxygen uptake by a particular muscle also depends on how flow is distributed, blood oxygen content and extraction. The data therefore do not prove a threefold increase in every muscle’s consumption. Tissue-level measurements would be needed.
- Distinguish whole-heart blood flow from individual-muscle oxygen uptake.
- Note that oxygen content, extraction and blood-flow distribution also matter.
- Conclude that tissue-level uptake measurements or equivalent evidence are needed.
Common confusion: A related physiological measurement can support an explanation without directly measuring the claimed outcome.
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