The state of health and functioning of the liver is often assessed with dye-tracer techniques. The dye used most frequently is bromosulfophthalein (abbreviated BSP). With normal function liver, the dye will be removed from circulation after it is conjugated with cysteine or glutathione (which happens in the liver) following a first order kinetics k=0.5 hr-1 at 37°C. The usual clinical procedure is to have a single injection to the blood and then to measure the amount remaining in the liver after a certain period of time. Liver function is assessed by comparing the experimental value with the range of normal values. Assuming the flow of the dye from the blood stream to liver is donor control and of linear relationship (k12= 1 hr-1).

1)Identify the engineering principles needed to solve this problem
2)Develop the differential equations to describe the amount of drug in each compartment ( the blood stream and the liver )
3)Clearly state your assumptions
4)Find the amount of the dye in the liver 1 hours after an injection of 10 mg of the dye into the blood stream ( 5 pts)

5)If the amount of the dye in the liver is determined to be 4.0 mg after 1 hour of injection. What can potentially go wrong with this patient?

Answer

1.Radioactive or chemical decay

2. X(t) = Xoe^-kt

lnY(t) = -t + lnYo

4. 6.07mg

Explanation:

Let the liver and and blood compartment be represented by the symbol L and B respectively

For the liver

Suppose a first Order removal process started with an amount X, in which amount b disappeared in time t, the process is decay process which can be represented as follows,

∫dX/dt = -K1X

By rearrangement and integration;

∫dX/X = -K1t

ln X = -K1t + C

Since At t= 0, X = Xo then

C = lnXo

The equation becomes:

lnX = -K1t + lnXo

lnX - lnXo = -K1t

ln(X/Xo) = -K1t

X/Xo = e^-kt

X(t) = Xoe^-kt

X(t) = Xoe^-0.5t(2)

Similarly for B (blood), suppose a first order flow flow of the dye move from the blood to the liver, let Y be the initial concentration, and amount b that has flown to the liver in time t

B> B(t)

t=0 Yo 0

t=t. Y-b. b

dY/dt = -K12(Y-b)(3)

Let Y-b = Y(t)

∫dY/dt = -∫K12t

By rearrangement and integration;

∫dy/Y(t) = ∫-K12dt

lnY(t) = -K12t + C1

at t= 0, C1 = ln(Yo)

Therefore ln Y(t) = -K12t + lnYo

But K12 =1

ln Y(t) = -t + lnYo(4)

(3) The assumptions used here

is that of a decay for the liver . The amount remaining taking as the amount of a substance

(4) using the equation 2,

X(t)= Xo e^-K1t(2)

For time t = 1hour, and an initial amount X = 10mg, K = 0.5

X(t) = 10× e^-0.5

X(t) = 10 × 0.607

X(t) = 6.07mg

(5) within the scope of information presented, I have no data to make this judgment.

it has one origin of replication, but it proceeds in two directions - d)