IntroductionBuilding the model of the kinetics of leadIdentification of the ModelCase StudyNon-lead environmentSmall lead environmentMassive medicationSafer MedicationConclusionHome PageTitle PageContentsJJ IIJ IPage 1 of 21Go BackFull ScreenCloseQuitThe Kinetics of Lead Transfer in the Human BodyMiyuki ShimokawaMay 20, 2000AbstractThe kinetics of lead in the human body are described by a system offirst order differential equations. Parameters are estimated from actualdata. The model is applied to specific situations.Home PageTitle PageContentsJJ IIJ IPage 2 of 21Go BackFull ScreenCloseQuit1. IntroductionLead is closely related to human life. Gasoline, paint, and foil contain lead. Leadis contained in the soil in a natural environment [Batschelet]. People ingest leadindirectly through natural foods and water. Human beings have been facingthe cumulative effects of lead poisoning for many years. In our society, thelead concentration is extremely high because sophisticated industries pollutethe environment with lead. To detect levels of lead in the body, a model of thekinetics of lead transfer is described below.Home PageTitle PageContentsJJ IIJ IPage 3 of 21Go BackFull ScreenCloseQuitBonesx3(t)Bloodx1(t)Tissuex2(t)LungsDigestiveTractαz4y41βz5y51y13y31y21y12z1z2Figure 1: The compartment model for the kinetics of lead.Home PageTitle PageContentsJJ IIJ IPage 4 of 21Go BackFull ScreenCloseQuit2. Building the model of the kinetics of leadWe divide the body into three compartments: the blood, the tissue, and thebones (see Figure 1). The lead is transferred from one compartment to anotherby several blood vessels. We will model this exchange with a system of firstorder differential equations.Lead is taken in from outside the bo dy via the lungs and the digestive system.These are the main sources of lead intake. From this point forward, we assumethat all lead transfer is measured in micrograms per day (µg/day). Let α be therate at which lead enters the lungs and let β be the rate at which lead entersthe digestive system.Lead also enters the digestive tract from the tissue compartment (saliva,gastric secretions, bile, etc.). Let Y25be the rate at which this occurs. Of course,a certain amount of lead escapes through the lungs and digestive tract. Let Z4be the rate at which lead escapes the lungs through normal breathing. Let Z5be the rate at which lead escapes the digestive tract through bowel movements.Lead from the lungs and digestive tract enters the blood. Let Y41be the rateat which lead enters the blood from the lungs. Let Y51be the rate at which leadenters the blood from the digestive tract.Next, le t’s describ e the lead exchange between the tissue and blood. Let Y12be the rate at which lead enters the tissue from the blood. Let Y21be the rateat which lead comes back to the blood from the tissue.Let’s describe the lead exchange between the bones and the blood. Let Y13be the rate at which lead enters the bones from the blood. Let Y31be the rateat which lead comes back to the blood from the bones.Lead can also escape from the body by way of the blood and the tissue. LetHome PageTitle PageContentsJJ IIJ IPage 5 of 21Go BackFull ScreenCloseQuitZ1be the rate at which lead escapes the blood as urine. Let Z2be the rate atwhich lead escapes the tissue through the hairs, nails, sweat, etc.The rate Y41at which lead flows from the lungs to the blood is equal to therate at which the lungs absorb lead.Y41= α − Z4(1)Furthermore, we assume that this rate is proportional to the rate at which leadenters the lungs.Y41= α − Z4= pα, where 0 < p < 1 (2)Next, the rate Y51at which lead flows from the digestive tract to the blood isequal to the rate at which the digestive tract absorbs lead. Remember that thedigestive tract absorbs lead from two sources, diet and tissue. If we assume thatthe rate Y51is proportional to the rate at which lead enters the digestive tract,then we can writeY51= q(β + Y25), where 0 < q < 1. (3)Let x1(t), x2(t), x3(t) represent the amount (µg) of lead in the blood, tissue,and bone compartments, respectively. Then ˙x1represents the rate at which leadenters and escapes the blood compartment. This is equal to the rate at whichlead comes into the blood compartment minus the rate at which lead escapesthe compartment.˙x1= −Y12+ Y21− Y13+ Y31+ Y41+ Y51− Z1(4)Home PageTitle PageContentsJJ IIJ IPage 6 of 21Go BackFull ScreenCloseQuitIn a similar manner, one can develop rate equations for the tissue and bonecompartments.˙x1= −Y12+ Y21− Y13+ Y31+ Y41+ Y51− Z1˙x2= Y12− Y21− Y25− Z2˙x3= Y13+ Y31(5)Assume that the lead flow from the blood compartment to the tissue com-partment is proportional to the amount of lead in the bloo d compartment. Insymb ols,Y12= a12x1. (6)In a similar manner,Y21= a21x2Y13= a13x1Y31= a31x3Y25= a25x2Z1= bx1Z2= cx2(7)Take these proportions and equations (2) and (3) and substitute them in theequations of system (5).˙x1= −a12x1+ a21x2− a13x1+ a31x3+ pα + q(β + a25x2) − bx1˙x2= a12x1− a21x2− a25x2− cx2˙x3= a13x1+ a31x3(8)The ai,k, b, c are positive constants. By combining coefficients the equationsHome PageTitle PageContentsJJ IIJ IPage 7 of 21Go BackFull ScreenCloseQuitcan be rewritten in the following form.˙x1= A11x1− A12x2+ A13x3+ D˙x2= A21x1+ A22x2˙x3= A31x1+ A33x3(9)The three liner nonhomogeneous differential equations are formed. The in-homogeneous term is D, whereD = pα + qβ. (10)Home PageTitle PageContentsJJ IIJ IPage 8 of 21Go BackFull ScreenCloseQuit3. Identification of the ModelAlthough the three differential equations are formed in a reasonable way, thecoefficients Ai,kare not determined. We next consider a real-life study to deter-mine some particular co effi cients for our differential equations.The following data was taken from the Journal of Mathematical Biologyin 1979[Batschelet]. The authors used a subject who was healthy 53 year oldmale, weighing 70 kg, smoking eight cigarettes per day. The subject lived in thesouthern California. For 104 days, the subject studied was considered to be ina stable state with respect to lead.x1= 1800 µg, Y12= 20 µg/d, Z1= 38 µg/d,x2= 700 µg, Y21= 8 µg/d, Z2= 4 µg/d,x3= 200000 µg, Y13= Y31= 7 µg/d, Z4= 32 µg/d,α = 49 µg, Y41= 17 µg/d, Z5= 38 µg/d,β = 367 µg, Y51= 33 µg/d, Y25= 8 µg/d.Use the data to calculate D.D = pα + qβD = 49p + 367qHome PageTitle PageContentsJJ IIJ IPage 9 of 21Go BackFull ScreenCloseQuitUse equations (2) and (3) to c