A boy holds his breath, and his pCO₂ rises from 40 mm Hg to 46 mm Hg…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A boy holds his breath, and his pCO rises from mm Hg to mm Hg. Use the model of the lesson: at mm Hg the rate is breaths a minute and each breath is mL. For every mm Hg above , the rate rises by breaths a minute and each breath grows by mL.
- What is the rate at mm Hg?
- What is the volume of each breath?
- How much air does he move in a minute?
- Is a lack of oxygen the main reason why he must breathe again? Explain.
Need a reminder? The key ideaShow
The respiratory rhythm centre is in the medulla. The pneumotaxic centre in the pons can shorten inspiration. The chemosensitive area next to the rhythm centre, and receptors in the aortic arch and the carotid artery, sense carbon dioxide and hydrogen ions. The role of oxygen is insignificant.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. Excess pCO: mm Hg.
Plan. Add the effect of the excess to the normal rate and volume, then multiply.
Step 1. Rate. breaths a minute.
Step 2. Volume. mL.
Step 3. Air in a minute. mL.
Step 4. Part (d). No. The main signal is the rise of carbon dioxide and hydrogen ions, sensed by the chemosensitive area and the receptors. The role of oxygen in the control of breathing is insignificant.
Answer: breaths a minute, mL, mL. Carbon dioxide, and not a lack of oxygen, is the main signal.
Check. At rest he would move mL, so the new value is almost three times as large.
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