Biochem+Eng+Hmk+5+Soln - PROBLEM SET 5 SOLUTION...

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Unformatted text preview: PROBLEM SET 5 SOLUTION PHARMACOKINETICS Question 14: Kidneys are vital organs that keep the chemical balance of blood and remove toxic waste from blood. Typically kidneys process about 1.0 liter of blood and filter out 0.08 liter of waste and extra water every hour, and the remainder 0.92 liter of purified blood goes back to the body. We have been monitoring the removal of two toxic wastes from the blood to urine. The table bellow show the volume fraction of toxic wastes in blood before (call this the feed) and after (call this the product) passing through the kidneys. Volume fraction in feed Volume fraction in product i.e. blood before i.e. blood after (%) (%) Waste A 10.4 8.7 Waste B 7.5 6.5 Water 82.1 84.8 a. Calculate the volumetric flow rates (l/hr) of waste streams A and B in the blood and the urine b. Calculate the transfer rate (l/hr) of waste A from blood to urine. Assume steady state c. (Skip) d. Due to the injection of a drug the volume fraction of waste B in the feed changes over time as X(B, feed)=9.2 + 2e‐2t, whereas in the product the volume fraction of B changes with time as X(B, product)=10+1.5e‐2t. You may assume that the volume flow rates of feed and product do not change. What is rate of waste B removal? Answer on next page… Question 15: A patient enters the emergency room (E.R.) and needs an infusion of acetaminophen to achieve a steady‐state concentration of 5.0 mg/L in his bloodstream. The volume of distribution of acetaminophen is 63 L. Acetaminophen can be considered to be eliminated entirely by metabolism in the liver, with a first‐order elimination rate constant ke = 0.35 h‐1. a) Find the infusion rate of acetaminophen in mg/min. The patient returns home, but eventually the pain returns. Therefore, the patient takes a 500 mg pill of acetaminophen. Acetaminophen is absorbed by the GI tract via a first order process with a rate constant ka = 2.2 h‐1. b) At what time is the maximum acetaminophen concentration reached and what is the maximum concentration? Assume that the initial concentration of acetaminophen is zero. c) If the patient has liver disease, causing a reduction in the elimination kinetics, sketch what will be the effect on the concentration vs. time profile and explain your answer: a. For the infusion case of part (a) b. For the pill absorption case of part (b) d) Acetaminophen freely crosses the blood‐brain barrier. How does the concentration vs. time profile in the brain compare with that in the bloodstream? Answer: a) kE =0.35h‐1 Ri V = 63L For constant infusion, we have C= Ri/(KEV)*(1‐ e‐KEt) At steady state, t ‐> 0 C = Ri/(KEV) => Ri = C KE V, Ri = 5.0mg/L* 0.35hr‐1 * 63L*1hr/60 min = 1.84 mg/min. b) Abs ka =2.2h‐1 V = 63L kE =0.35h‐1 For single compartment with absorption, maximum concentration occurs at tMAX= ln(ka/kE)/(ka‐kE) = ln(2.2/0.35)/(2.2‐0.35) = 0.994 h C = ka*CA0/ (kE‐ka) (e‐kat – e‐kEt) CA0 = 500 mg / 63 L = 7.94 mg/L CMAX = [2.2/(0.35‐0.22)]*7.94mg/L* (e‐2.2 *0.9936– e‐0.35*0.9936) = 8.06 mg/L c) At steady‐state the rate of infusion equals the rate of elimination. For same infusion rate, steady‐state concentration of drug will go up if kE decreases because C kE V = Ri. Note that V is constant. C drug Decreasing kE increases steady‐state drug concentration Time Pill absorption: since elimination is slower while absorption is unchanged, the drug will not be cleared as quickly from the body, and it will peak earlier and reach a higher level. C drug Decreasing kE causes faster accumulation of drug because it is not eliminated as quickly d) They are the same. Time Question 13: Enzyme‐catalyzed reactions usually follow Michaelis‐Menten kinetics. It is common to rewrite the basic form of the rate equation dP = VMAX [S] dt KM + [S] into a “linearized” form 1/v as a function of 1/[S] where v = dP/dt. a) Derive the linearized form described above. b) What do the slope and vertical intercept mean? Answer: a) dP/dt = VMAX[S]/(KM+[S]) = V 1/V = (KM+ [S])/(VMAX[S]) = (KM/ VMAX)*1/[S] + [S]/(VMAX[S]) = (KM/ VMAX) * 1/[S] + 1/VMAX b) Slope = KM/ VMAX , vertical intercept = = 1/ VMAX ...
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