Chapter 15.

Multiple Choice Questions

I          Answers A, B, C, and D are true statements, whereas E is false.

II        Answers A, B, D, and E are true statements, whereas C is false.

III         Answers A, B, C, and E are true statements, whereas D is false.

Case History A

 1.        FACO2 = PACO2/(101.3 - 6.2).  PACO2  = 5.32 kPa.

 2.        C = (a × P)

a is the Bunsen solubility coefficient. Its size must decrease with increasing temperature, because of the gas escapes with thermal movements.

 3.        22.4 ml = 1 mmol. FACO2  = 0.056 and (101.3 - 6.2) = 95.1; Physically dis­solved CaCO2 = (0.51 × 0.056 × 95.1 × 1000)/101.3 = 26.8 ml STPD l-1 or 26.8/22.4 = 1.2 mM.

Physically dissolved CaO2 = (0.022 * 13.3 * 1000)/101.3 = 2.89 ml STPD l-1 or 2.89/22.4 = 0.13 mM.

 4.        The mixed venous blood PCO2 is 6.1 kPa and her PO2 is 6.0 kPa. 

Physically dissolved carbon dioxide in the mixed venous blood: (0.51 × 6.1 × 1000)/101.3 = 30.7 ml STPD l-1 or 30.7/22.4 = 1.37 mM.

Physically dissolved oxygen in the mixed venous blood: (0.022 × 6.0 × 1000)/101.3 = 1.3 ml STPD l-1 or 1.3/22.4 = 0.06 mM.

Case History B

 1.            Calculate the carbon dioxide and the cardiac output. R = V°CO2 / V° O2max; V°CO2 = 0.9; V°O2max  = 4.05 L STPD/min.

Since R is 0.9 the ratio: (mixed venous CCO2 - CaCO2)/(CaO2 - mixed venousCO2) must be the same. Thus, (650 - 500)/(200 - mixed venous CO2) is 0.9. Accordingly, mixed venous CO2 is 33.3 ml STPD per l.  

Cardiac output = 4500/(200 - 33.3) = 27 l of blood per min.

 2.        Calculate the coronary bloodflow. 

Coronary bloodflow = 420/(200 - 30) = 2.47 l of blood per min. 

 3.        During exercise the myocardial energy resources switch from mainly b-oxidation of fatty acids and only 30 % carbohydrate combustion, to mainly carbohydrate utilisation including lactate combustion and only 30 % b-oxidation of fatty acids.  

 4.        Cardiac output = V°O2 / (CaO2 - mixed venous CO2) or 5450  = 273/ (200 -mixed venous CO2).

Thus mixed venous CO2 equals 150 ml STPD per l, and the O2 difference equals 50. These results are typical for the resting condition.

Case History C

 1.        RQ = (542 - 500)/(200 - 150) = 0.84.

 2.        The calculated RQ value corresponds to mixed diet.

 3.        V°O2 = (10,450/20.6) = 507.3 l STPD daily = 0.352 l per min.

 4.        V°CO2 = (V°O2 × RQ) = (0.352 × RQ) = 0.29 6 l per min.

FACO2 = 5.3/(101.3 - 6.25) = 0.056;    [FACO2= (V°CO2/ V°A].

V°A  = V°CO2 /0.056 = 5.28 l per min.    V° E  = V°CO2 /FECO2 = 0.296/[4.4/(101.3-6.2)] = 6.43 l per min.     V°D = (V°E - V°A) = 1.15 l per min. VD = 1.15/14 = 0.082 litre.

 5.        Cardiac output = V°O2 /(CaO2  - mixed venous CO2) = 352/(200 -150) = 7.04 l per min.

6.         The cause of unconsciousness is neither respiratory nor cardiovascular. By exclusion the most likely cause is brain damage by the traffic accident.

Case History D

1. Fick’s principle for cardiac output is used: 

Q°  = V° O2/(CaO2 --m ixe d  venou s  CO2 ).

Accordingly, the cardiac output is 250/50  =  5 l min-1.

2. The dissolved oxygen in his arterial plasma must be:  

 [0.022 * 13 * 1000/101.3] =  2.82 ml STPD l-1.

  3.   75% of the arterial oxygen remains in the venous blood, when only 25% are utilised by the tissues. The O2 uptake is still 250 ml STPD min-1 .  

Thus, the hypothetical  cardiac output is 250/(2.82 * 0.25) =  355 l of blood each min! This is unrealistic. 

  4.   Life without haemoglobin is not possible for humans at one atmosphere of air pressure (PaO2 = 13 kPa). However, life without haemoglobin is possible for 10-20 min by hyperbaric oxygenation until oxygen toxicity sets in (see Chapter 19).

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