Home / iGCSE Biology (0610)-12.2 Aerobic respiration – iGCSE Style Questions Paper 4

iGCSE Biology (0610)-12.2 Aerobic respiration – iGCSE Style Questions Paper 4

Question

(a) Fig. 3.1 shows some of the structures present in the human thorax (chest). Identify the structures labelled P, Q, R and S. Write your answers in the boxes on Fig. 3.1.

(b) Table 3.1 shows the composition of inspired (inhaled) and expired (exhaled) air.

(i) Expired air has more carbon dioxide then inspired air. Approximately how many times greater was the percentage of carbon dioxide in the expired air than in the inspired air? Choose your answer from this list.
×1.3                 ×13                ×130                 ×1300

(ii) There is no difference in the nitrogen content of inspired and expired air. Suggest a reason for this.
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(iii) Expired air often contains more water vapour than inspired air. Explain how this water vapour gets into the expired air.
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(c) Table 3.2 shows the effects of exercise on the rate of oxygen uptake and the rate of energy use.

(i) Describe what happens to the rate of oxygen uptake as activity increases.
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(ii) Use the results in Table 3.2 to calculate how many kilojoules of energy are needed to run quickly for 12 minutes.

………………………………………………………………kJ

(iii) 1 g of sugar provides about 18.0 kJ of energy. Calculate how much sugar would be needed to run at maximum speed for 30 seconds. Show your working.

………………………………………………………………g

Answer/Explanation

Ans:

3. (a) P trachea/windpipe;
          Q bronchus / cartilage ring;
          R air sac /alveolus;
          S diaphragm; 

(b) (i) x 130; 

      (ii) nitrogen is not used up/ produced by (the cells of) the body;

      (iii) air sacs / alveoli have a moist lining;
             water evaporates (from lining into air);
             water (in lining) replaced by osmosis from cells (of alveoli)/AW;

(c) (i) (oxygen uptake) increases;
           reaches a maximum;
           specific reference to figures in table;

     (ii) (12 × 9 =) 108 (kJ);

     (iii) 30/ 60× 12 = 6;
            6/ 18 = 0.33;

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