I: Answers A, B, and D are true statements, whereas C and E are false.
II: Answers A and E are true statements, whereas B, C and D are false.
1.The profuse bleeding reduces the total blood volume, whereby the venous return and the feeding pressure to the ventricles decrease. This is why the stroke volume falls, which is illustrated by the low blood pressure and pressure amplitude: (65 - 40) = 25 mmHg. The low arterial and dynamic blood pressure reduce the tension of the arterial baroreceptors, so the signal frequency to the cardiovascular control centres falls. This increases the sympathetic tone and reduces the parasympathetic tone towards the heart and the peripheral vessels (resistance and capacitance vessels). Thus the heart rate increases, and all the signs of shock develop.
2.The effect of the blood transfusions is to increase the blood volume and blood pressure. Hereby, the signal frequency from the arterial baroreceptors is normalised leading to a normal autonomic tone towards the heart and the peripheral vessels.
3.Let us assume that the soldier had a normal PCV of 0.45 and a normal blood volume of 8% of body weight before he was shot. This is (70 kg * 0.08) = 5.6 kg of whole blood, with (5.6 * 0.45) = 2.52 kg of red cells. The blood transfusions supplied him with (2 * 0.45) = 0.9 litres or kg of red cells.
After 24 hours the soldier contains (5.6 * 0.35) = 1.96 kg of red cells out of which 0.9 kg are derived from the transfusions. Accordingly, the soldier lost (2.52 - 1.06) = 1.46 kg of red cells. The size of the blood loss was 1.46/0.45 = 3.24 kg of whole blood. This is more than 50% of the original volume and above the survival threshold for most persons.
4. The blood transfusions were insufficient namely only 2 out of the 3.24 kg (litres) of whole blood.
1. Since the circulating total blood volume is 5400 ml and the cardiac output is 5400 ml per min, then the bloodflow per second must be 5400/60 or 90 ml per s. Thus the mean transit time is 5400/90 = 60 s.
2. The method is shown in 1.: transit time is 650/90 or 7.2 s.
3. Flow units (FU) are ml of blood per 100 g tissue and per minute. The calculation goes like this: (3*35 000/100) = 1050 ml per min. Bloodflow per second: 1050/60 = 17.5 ml per s.
4. The volume is equal to (5 s*17.5 ml per s) = 88 ml.
1. The most likely explanation is a bleeding peptic ulcer caused by gastric hypersecretion of HCl. - Actually, a large bleeding lesion was found at the cardia, but the ulcer was located in the bulbar part of duodenum.
2. The hypotension, arteriolar constriction and falling venous pressure during the severe haemorrhage lower the hydrostatic pressure in the capillaries. This promotes a net reabsorption of interstitial fluid into the blood. Hereby, the blood [haemoglobin] is diluted to half its original value within an hour. Thus the blood loss must approach half the total blood volume (5 l) or 2.5 l.
3. Endotoxins from the intestinal flora are not inactivated by the depressed macrophage system of the shock patient. He has lost about 50% of his circulating macrophages. As a consequence the patient develops fever and his haemorrhagic shock is aggravated.
4. The massive oedema of 8 l is clearly an overdose of physiologic saline - so-called iatrogenous oedema or overhydration. This is usually avoided by daily use of a bed weight for body weight control and daily serum electrolyte measurements.
1. Steno-Fallots tetralogy.
2. The oxygen capacity of haemoglobin is 1.34 ml per g: 1.34 * 164.2 = 220 ml oxygen per l. The blood leaving the lung capillaries is fully oxygenated, so the saturation degree is 100%.
3. The Fick principle states: Q°lung = 164/(220 - 138); Q°lung = 2 l per min.
4. Q°lung + Q°shunt = Q°total ; Q°shunt = 0.3 - so Q°lun g= 0.7
The 0.7 is equal to 2 l per min. Accordingly, Q°total = 2 + 2*0.3/0.7 = 2.857 l per min.
5. The venous return equals Q°total (2.857 l per min) in circulatory steady state.
6. A patient with polycythaemia has a high total concentration of haemoglobin and thus easily pass the threshold of 50 g reduced haemoglobin (Hb4) per l of average capillary blood - in contrast to anaemia.
The arterial blood of the little girl already contains (220-195)/1.34 = 18.66 g Hb4. As the blood passes the systemic capillaries, the oxygen extraction is: (195 - 138) = 57 ml per l, or 28.5 ml per l (28.5/1.34= 21.27 g Hb4 per l) in the middle of the capillary. Thus the total concentration of Hb4 is 39.93 g per l, and cyanosis is not visible at rest. Exertion produces cyanosis.
1.The cardiac output from the right ventricle (Q°right) is 310/(195- 130) = 4.77 l min-1.
2. . The cardiac output from the left ventricle (Q°left) is 310/(195- 160) = 8.86 l min-1.
3. Since the oxygen concentration of the right ventricle is larger than that of the right atrium, oxygenated blood must have passed from the left to the right ventricle. Thus, a ventricular septal defect must be present.
4. Exercise increases the oxygen demand of the body. Let us assume a 3-fold rise in Q°right for this patient. Such a rise implies a similar rise in Q°left , which is approximately 27 l min-1 that is definitely a maximum. A two-fold rise in the arterio-venous oxygen content difference (65 ml per l of blood at rest) is 130 ml per l of blood and close to maximal extraction of oxygen from muscle blood. The total estimate is a 6-fold rise in oxygen uptake or (0.310 * 6) = 1.86 l of oxygen per min. This is nothing for a male of 18, where many healthy persons can increase the oxygen uptake at rest 20-fold during maximal exercise. This is why the patient has been sedentary all his childhood.
5. The reason for the large appetite of this patient is obvious. This is the large cardiac work accomplished at every beat for as long as he has lived.
Compared to a normal 75 kg male with an oxygen uptake at rest of 250 ml min-1, the 60 kg patient is expected to have an uptake of 200 ml min-1. The patient was found to have an oxygen uptake at rest of 310 ml min-1. This is 155% of normal, reflecting an enormous cardiac work