Nov 20, 2013

Sample SPM Biology essay question

Diagram 6.1 shows the part of the regulatory mechanism of oxygen and carbon dioxide contents in the body.
Based on the diagram, explain how the concentration of carbon dioxide in the blood is regulated during a vigorous activity.
[8 marks]
Sample answer
During vigorous activity, the concentration / the partial pressure of carbon dioxide increases as a result of active cellular respiration
Carbon dioxide reacts with water to form carbonic acid which results in a drop in the pH level of the blood and tissue fluid that bathing the brain
The drop in pH is detected by the central chemoreceptors in the medulla oblongata
and also detected by the peripheral chemoreceptors (carotid bodies and aortic bodies)
The central chemoreceptors and pheripheral receptors send nerve impulses to the respiratory centre in the medulla oblongata
The respiratory centre sends nerve impulses to the diaphragm and the intercostal muscles, causing the respiratory muscle to contract and relax faster
As a result, the breathing rate and ventilation rate increase causes more oxygen inhaled and the oxygen concentration return to the normal level
As excess carbon dioxide is eliminated from the body, the carbon dioxide concentration and pH value of the blood return to normal level.

ii) Explain why the pulse rate takes several minutes to return to normal after a vigorous activity.
[4 marks]
Sample answer
After vigorous activity, the pulse rate takes several minutes to return to normal because during the activity the oxygen intake is not able to meet the oxygen demand of the body.
Respiration has to take place anaerobically/anaerobic respiration occurs
As a result, lactic acid accumulates in the muscle.
So more oxygen is needed to oxidise the lactic acid and to provide the energy for the recovery of the muscle