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BME501_Practice_Exam_Solutions_100808

BME501_Practice_Exam_Solutions_100808 - BME 501 Respiratory...

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Unformatted text preview: BME 501 Respiratory Section Exam Fall 2007 Oct 15, 2007 Closed book Respiratory Section (100 points) 1. Mark True(T) or False(F) 40 points toal, 2 point each 1.1 ( T) A low surface tension in the alveoli increases the compliance of the lung. 1.2 ( F) Increased compliance of the lung increases the work of breathing. 1.3 ( T) The base of the lungs in the upright posture has a higher(more positive) pleural pressure. 1.4 ( F) The vapor pressure of water increases as a fiinction of barometric pressure. 1.5 ( F) Oxygen in blood is 20 times more soluble than carbon dioxide. 1.6 ( T) Dexoxygenation of the blood increases its ability to carry carbon dioxide. 1.7 ( F) About 10 percent of the oxygen carried by venous blood into the lungs is in the dissolved form. 1.8 ( T) Total compliance (analogous to electrical capacitances in series) of the lung and chest wall is the sum of the reciprocals of the lung and chest wall compliances measured separately or: 1/CT =1/CL+1/ch __i l l ’ C? a C‘ g k.) C C 1.9 ( T-pg133) A reduction in the partial pressure of oxygen of arterial blongf—romfi'm) to 60 mm Hg causes almost no change in ventilation if the partial pressure of carbon dioxide is 35.8 mm Hg. 1.10 ( T) Hypoxia in the lungs leads to constriction of pulmonary blood vessels. 1.11 ( T) Mismatching ventilation and blood flow in the lungs cause both hypoxemia and hypercapnia, other things being equal. 1.12 ( T) Contraction of the diaphragm muscle flattens it. f i ‘ 1.13 ( T) According to Laplace’s equation a soap bubble with a smaller radius has a 4 higher pressure than one with a larger radius. 0 1.14 ( T) If flow in a tube is laminar the corresponding pressure drop is a linear fiinction of flow rate. 1.15 ( F-pg21) In the upright normal lungs ventilation is lower at the base (bottom) compared to the apex(top). 1.16 ( T-pg113) Reducing lung volume below the normal resting level is expected to lead to an increase in airway resistance. 1.17 ( T) Jumping in a pool in an upright position and being immersed up to the neck will remove the pleural pressure gradient in the lungs and could improve ventilation- perfusion inequality. 1.18 ( F) If a subject dives to 66 feet (3X sea level pressure) and remains there long enough to saturate the tissues the number of molecules of nitrogen in the tissue will approximately double. 1.19 ( F) Anaerobic metabolism during exercise refers to the production of ketone acids during fat metabolism. 1.20 ( T) Diffusing capacity of the lungs depends on the volume of blood in the pulmonary capillaries. 2. Fill in the blanks. 10 points, 2 points each 2.1 An increase in red blood cell concentration of the blood is called polycythemia 2.2 The alveoli secrete a material called surfactant which lowers the surface tension of the alveolar lining and stabilizes the lungs. 2.3 Tiny cilia lines the airways and propel mucus mouthward by rhythmic motion. 2.4 The Hering-Breuer reflex involves sensors whose fibers travel in the vagus nerve. 2.5 The_Haldane effect causes a shifl in the CO2 dissociation curve due to the higher ability of reduced hemoglobin to take up hydrogen ions. 3. (15) Assume a 2 compartment ventilation/perfusion model with total ventilation = 10 and total perfusion=10 (units unimportant). If the net alveolar PAO2=101 mm Hg, net arterial blood PaO2=97 mm Hg, for compartment 1: PalO2=PA102=132 mmHg, for compartment 2: Pa202=PA202==89 mm Hg. Calculate the individual compartment ventilations and perfusions(ie. V1,Q1,V2,Q2). Ca02=20.1*((1-exp(-.046*Pa02))"2) [out/(U , 0744/33 71 00-01,);ggf7 794; W +— 8‘7 0 am w : @757? ) Lu: 72/ ' ' ” ’0 «,9646 if) ; .777/ D 54020) 1,9525% 56(oé(2)\ / z a .. O H 5,77/ a: an Aer/(AV +—(xz)~<m)% "W / / r cw: 3257?? 47;; 6.49” 4. (15) An arterial blood sample with pH=7.5 , PaC02=20 mm Hg, Pa02=100 mm Hg. a) Calculate [HCO3-] b) Calculate [HCO3-]7,4 or standard bicarbonate c) What are the respiratory and metabolic acid/base characteristics of this sample. d) Discuss how this condition could be reached via a primary disturbance and secondary compensation . pH=6.1 + log([HCO3-]/(.O3*PaC02)) u/ normal buffer slope=-3 0.6 mEq/l per pH unit. W , 6710353 4/) 75; Mr“ M; m -341, 75”“ ZHCOZJ /' m i W mafia ' may: 575713 f 7 j l I ' k‘ - 7% r5“ 5/ fbéwg’] : @le ., 301‘, , 7' (f :1 570 7/ 3 —%— '3 ‘ (m m3 CRUCLLlOS’l > ‘_ (i)?“‘ “j _ . , h I, IA C x’e'lc‘) )l V971;- M , 'jyufilml p22“: . ) k/L (Lj l’zg (/6 i“ Cj CQ C. \ ¢\CDS( ,3 l C CK'VK VL kaOL ()é .13.“ /JO‘/§/3 l‘%’/< ‘3 at Ma \08’ I x 'K’WCVN 5.(10) Oxygen consumption rate measured using a spirometer (wet at 20 deg C(vapor pressure of water = 18 mm Hg), barometric pressure = 730 mm Hg) was .310 l/min what is the value corrected for STPD (standard temp and pressure dry)conditions? (3(0) 6,20 ~13)" \/J“T‘PD 740 (173 +20) 9:73 , Q7flé VQwawa «m : 6.(10) Based on a two compartment model of lung compliance (electrically analogous to two capacitances in parallel) Cbase=0.1 l/cm H20 and Capex=0.05 l/cm H20 for a constant flow 0f 1 liter.sec filling the parallel compliances calculate the flow to the base compliance. Starting with V=C*P where V=volume, C=compliance, and P =pressure, take the derivative to get flow dV/dt and compare flows to each compliance and combined. (H/ ) " q (56. ‘53 :1 C‘Vdofi L .. T76: fl ‘ ’ —-— 6 K3? Wbc V A t: Cytgc ’VCHP‘ /\ ...
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