2. Consider a simple 2-compartment model of the fate of methyl mercury in the hu
ID: 1042043 • Letter: 2
Question
2. Consider a simple 2-compartment model of the fate of methyl mercury in the human body as shown below T12 in Qi Q2 21 Te Fin is a steady state intake rate, Qi is the amount of methyl mercury in the blood and Q2 is the amount in soft tissue. Te is the mean time for excretion (all of which is from the blood) and T12, T21 are the mean times for the linear donor controlled transfers between blood and soft tissue a. Write the algebraic equations linking these quantities in steady state. (5 pts.) b. Measurements reveal that the mean residence time of methyl mercury in the body is 100 days, that Qz is always about 30 times bigger than Qi in steady state, and that T12 1 day Given this information, solve for the values of T21 and Te. (5 pts.) c. Symptoms of methyl mercury poisoning become evident when the total body burden reaches about 20 milligrams. According to our model, with what constant daily intake of methyl mercury would this steady state body burden be associated? (5 pts.) d. Under the circumstances of c., what would the blood concentration be (in units of mg/liter) (5 pts.) e. If the body is free of methyl mercury at t 0, and thereafter the intake rate is 10 times what you calculated in c,. how long will it take for the body burden to become 20 milligrams? (5 pts.) f. Fish contaminated with methyl mercury have concentrations of this compound in the range of 10 micrograms per gram of fish flesh. Would a person eating 200 g of contaminated fish per day every day reach the danger level described in part c? (5 pts.) g. Consider a person who has consumed a meal containing 200 grams of contaminated fish once per week for 3 weeks. Estimate the body burden immediately after the 3rd meal. (5 pts.)Explanation / Answer
Answer..
Daily dietary intake of methyl mercury corresponds to a specific blood concentration, factors of absorption rate, elimination rate constant, total blood volume and % of total volume that is present in the circulatory blood.
Since our soft tissues accumulates a greater concentration of MeHg, therefore interchange of MeHg between blood and soft tissue will occur at different rates. A steady state is reached when the concentration of incoming (intake) and outgoing (elimination) of MeHg reaches a sort of equilibrium.
a) Under such circumstances applying first order kinetics, the relationships among terms can be shown in the form of a chemical equation.
d = C x b x V/A x f
Where, d = daily dietary intake ( micrograms of MeHg per day)
C = concentration in blood ( here Q1)
b = elimination constant.
V = volume of blood in the body.
A = absorption factor, expressed as a decimal fraction, viz 0.95, a unitless quantity.
f = fraction of daily intake, unit less quantity.
b) Transfer of MeHg..
Rate of transfer from blood to soft tissue
= Q1/T12
Rate of transfer from tissue to blood
= Q2/T21
Therefore, net rate of transfer of MeHg
= Q2/T21 - Q1/T12
(Since diffusion of MeHg would take place from higher to lower concentration)
Given, Q2 = 30Q1, and Q1 = 1 day,
So Q2 = 30x1 = 30
Net rate of transfer = 30/T21 - 1/T12
Again Te is related to amount of intake per day 'd' from the first formula.
Te can be calculated by the relation
Fin /100 - Q1/Te
(Since mean residence time = 100 days)
Further d = Fin/100
We know, Fin = steady state intake.
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