Ex1: Given that the disposition kinetics of a drug is described by a one-compartment model, which one(s)of the following statements is correct? The half-life of a drug following therapeutic doses in humans is 4 hr, therefore

  1. The elimination rate constant of this drug is 0.173 hr^-l.
  2. It takes 16 hr for 87.5% of an i.v. bolus dose to be eliminated.
  3. It takes twice as long to eliminate 0.375 g following a 0.5-g bolus dose as it does to eliminate 0.5 g following a 1-g dose.
  4. Complete urine collection up to 12 hr is needed to provide a good estimate of the ultimate amount of drug excreted unchanged.
  5. The fraction of the administered dose eliminated by a given time is independent of the size of the dose.

Answer

1a: Calculate elimiation rate based on the equation \[ t_1/2 = \frac{ln2}{k}\]

then we have: \[ k = \frac{ln2}{t_{1/2}}\]

t_1/2 = 4 hr, applying to the equation \[ k = \frac{ln2}{4} \]

k = log(2)/4
k
## [1] 0.1732868

then elimination rate is 0.1732868, so the statement 1.a is correct

1b: Because, intravenous so, Bioavaibility (F) = 1, after 4 half-lives, the fraction of dose remaining in the body is \[ \frac{1}{2}^4\] the percentage of dose was eliminated is \[ 1 - (\frac{1}{2})^4\]

PercentgeOfEliminatedDose <- (1 - 1/2^4) * 100
PercentgeOfEliminatedDose
## [1] 93.75

so, we have the percentage of eliminated dose is 93.75%, this means 1.b is incorrect

1c: to eliminate from 1g to 0.5g it takes a 1 half-live

To eliminate 0.375g from 0.5, it will takes 2 t half-lives because the remaining of drug in the body is 0.5 - 0.375 = 0.125. this means 1 half life for reduce from 0.5g to 0.25g and an other t half-life was needed for reduce from 0.25g to 0.125. This means, it takes 2 half-lives to eliminate 0.375g following 0.5g. Thus 1c is correct

1d: Complete urin collection up to 12h, this mean there are 3 t half lives. The excreted drug is \[ 1 - (\frac{1}{2})^3 \]
and equal to 87.5%. This mean 1d is incorrect. ***

Ex2: Calculate the following:

  1. The fraction of an i.v. dose remaining in the body at 3 hr, when the half-life is 6 hr
  2. The half-life of a drug, when 18%of the dose remains in the body 4 hr after an i.v. bolus dose

Answer

Related equations: EliminationRateConstant: \[ k = \frac{ln2}{t_{1/2}}\]

Fraction of Dose Remaining in the body: \[ FractionOfDoseInBody = e^{-kt}\]

t_hf <- 6
t <- 3
k <- log(2)/t_hf

FractionOfDoseInBody = exp(-k*t)

2a: Applying two above equations, we have Fracntion of Dose Remaining in the body is 0.7071068. 2b: After 4hr the Fraction of Dose remaining in the body is 0.18, so the rate of elimination is \[ k = \]

From the equation \[ FractionOfDoseInBody = e^{-kt}\] we have \[ 0.18 = e^{-k*4hr} \]

thus, elimiation rate constant is \[ k = \frac{log(\frac{100}{18})}{4} \]

k <- log(100/18)/4

t_hf = log(2)/k

then t half-life is 1.6168598 hr-1

Ex3: Prepare a semilogarithmic plot of the following plasma concentration-time relationship \[ C = 0.9*e^{-0.347t}\] from above equation we have t half life equal \[ t_{1/2} = \frac{ln2}{0.347}\]

t_hf <- round(log(2)/0.347,1)

t half-life is 2 hr-1

Preparing data for graph by time

library(ggplot2)
time <- c(0:10)
concentration <- 0.9*exp(-0.347*time)

data <- data.frame(as.factor(time), concentration)

data

Plot

ggplot(aes(x = time, y = concentration), data = data) +
    geom_point() +
    scale_x_continuous(breaks = c(0:10), labels = c(0:10), expand = c(0,0.1)) +
    theme_bw() +
    geom_text(aes(time, y = concentration, label = round(concentration,2)), nudge_x = 0.3) +
    ggtitle("Time vs Concentration") +
    xlab("Time (hr)") +
    ylab("Drug concentration")

Ex4: Summary: Co = 10mg/L; t = 60 min (1 hr), Ct = 2.5 mg/L
a. Complete the picture the fall of drug concentration in plasma with time
b. Complete the picture the accumulation of drug in unrine with time

Answer

Preparing data plasma concentration

# Calculate k and t half-live
t <- 1
k <- (log(1/0.25))/1

t_hf <- log(2)/k

Time <- seq(from = 0, to = 5, by = 0.25)

PlasmaConcentration <- round(exp(-k*Time),2)

data <- data.frame(Time = Time, PlasmaConcentration)

point_data_time <- c(0:5)
point_data_plasma <- round(exp(-k*point_data_time),2)

# Plot plasam concentraion vs time

ggplot(data = data, aes( x = Time, y = PlasmaConcentration)) +
    geom_line() +
    geom_point(aes(x = Time[5], y = PlasmaConcentration[5]))

Preparing data for the cummlative amount excreted

# Ae_inf unchanged is 60 mg
Unchange_Ae <- 60

# t half life = 0.5 hr-1
# after 5h, we will have 10 t half-lives
t_hf_times <- c(1:10)
CumulativeUrine <- round((1 - (1/2)^t_hf_times) * Unchange_Ae, 1)

data = data.frame(t_hf_times, CumulativeUrine)

ggplot(data = data, aes(x = t_hf_times, y = CumulativeUrine)) +
    geom_point() +
    geom_line()

Ex5 Dose = 50mg
AUC(0-3hr) = 5.1 mg-hr/L
AUCtotal = 22.4 mg-hr/L Aeinf = 11mg.

  1. What the percentage of the administered dose remain in the body as drug at 3 hr?
  2. Calculate total clearance
  3. renal clearance
  4. What is the fraction of the dose that is eliminated by renal excretion?

Answer

# A. Percentage of the administered dose remain in the body as drug at 3 hr.
AUC_3h <- 5.1
AUC_total <- 22.4
PercentageRemaining = (1 - AUC_3h/AUC_total)*100

5a: after 3 hr the AUC = 5.1, this means the total of drug eliniminated in 3 hours is 5.1, and the total of AUC is 22.4. so the percentage of drug remain in the body was calculated using this equation \[ PercentageRemain = (1 - \frac{AUC_{(0-3h)}}{AUC_{total}}) * 100 \] then percentage of drug remaining in the body is 77.23%

5b: Calculating total clearance follows this equation \[ Dose = CL. AUC \], then we have \[ CL_{total} = \frac{Dose}{AUC_{total}} \]

Dose <- 50
CL <- round(Dose / AUC_total,2)

Thus, Total clearance is 2.23L/hr

5c: Renal Clearance

Equation \[Renal Clearance = \frac{Ae_{inf}}{AUC_{total}}\]

Ae_inf <- 11
RenalClearance <- round(Ae_inf / AUC_total, 2)

then, Renal Clearance is 0.49 L/hr

5d. Fraction of the dose that is eliminated (fe) is calcualted using the equation \[ fe = \frac{CL_R}{CL} \]

fe = round(RenalClearance / CL, 2)

then fraction of the eliminiated dose is 0.22.

Ex6: Summary

Dose <- 100
C0 <- 7.14
k <- 0.173

\[ C = 714*e^{-0.173t} \] a. Volume of distribution?
b. Elimination half-life?
c. Total AUC
d. Total Clearance
e. The plasma concentration 20min after a 250-mg i.v bolus dose

Answer
6a: Caculate Vd \[ Volume of Distribution = \frac{Dose}{C_0}\]

Vd <- round(Dose/C0, 2)

then, Volume of distribution is 14.01 L.

6b. Calculate t half life \[ t_{1/2} = \frac{ln2}{k}\]

t_hf <- round(log(2)/k, 2)

then, t1/2 = 4.01 hr-1.

6c. Total AUC following the equation \[Total AUC = \frac{C_{0}}{-k}[0 -1] = \frac{C_0}{k} \]

AUC_total <- round(C0 /k, 2)

Then, The total AUC = 41.27 mg-hr/L.

6d. Total Clearance \[ CL = \frac{Dose}{AUC_{total}} \]

CL <- round(Dose / AUC_total, 2)

then, Total Clearance is 2.42 L/hr.

6e: The plasma concentration at 20min and dose is 250mg \[ C = C_0 * e^{-kt}\] and we have \[ C_0 = \frac{Dose}{Vd}\] Then \[C = \frac{Dose}{Vd}*e^{-kt} \]

time_in_hour <- 20/60
C <- round(250/Vd * exp(-k*time_in_hour),2)

Finally, C = 16.84.

Ex7 

# Input data
ex7_Dose <- 184
Time <- c(1, 6, 12, 24, 48, 72, 96, 144)
Concentration <- c(137, 120, 103, 76, 42, 23, 12, 3.7)

ex7 <- data.frame(Time, Concentration)
ex7
# plot graph
ggplot(data = ex7) +
    geom_point(aes(x = Time, y = Concentration)) +
    scale_y_continuous(trans = "log2") +
    scale_x_continuous(labels = as.character(Time), breaks = Time) +
    theme_classic() +
    ggtitle("Plasma concentration by time") +
    xlab("Time (hr)") +
    ylab("Plasma concentration (mg)") 

# Because it it semilogarit so I try to esitmate elimiation rate constant using linear model.
model <- lm(log(ex7$Concentration[1:3]) ~ ex7$Time[1:3])
k <- abs(model$coefficients[2])

# Then t half-life is
ex7_t_hf <- log(2)/k
  1. t half-life is 26.7454532 hr-1.
  2. Total AUC \[ Total AUC = \frac{C_0}{k}\] and \[C_0 = \frac{C}{e^{-kt}}\]
# calculate C0 at time = 1hr
ex7_C0 <- 137/exp(-k)
ex7_AUC <- round(ex7_C0/k,0)

So AUC is 5425 mg-hr/L.

  1. Calculate Clearance \[CL = \frac{Dose}{AUC}\]
ex7_CL <- ex7_Dose/ ex7_AUC

Clearance is 0.0339171 L/hr.

  1. Calculate Volume of Distribution \[ Vd = \frac{Dose}{C_0}\]
ex7_Vd <- ex7_Dose / ex7_C0

Vd is 1.3087054 L.

Ex8

ex8_Time <- c(0.16, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0)
ex8_Conc <- c(170, 122, 74, 45, 28, 17, 10)

ex8 <- data.frame(Time = ex8_Time, Conc = ex8_Conc)

ex8
  1. Plot semilogarithmic
ggplot(data = ex8, aes(x = Time, y = Conc)) +
    geom_point(pch = 10, size =2, color = "red") +
    geom_line(lty = 2) +
    geom_text(aes(label = Conc), nudge_x = 0.1) +
    theme_classic() +
    ylab("Plasma Concentration (ug/L)") +
    xlab("Time (hr)")

  1. Estimate t half life and CL

Estimate elinimation rate constant using linear model by log of plasma concentration by time.

ex8_Dose <- 33
ex8_model <- lm(data = ex8, formula = log(Conc) ~ Time)

# k is slope of this model
ex8_k <- abs(ex8_model$coefficients[2])

# calculate t/12
ex8_t_hf <- log(2)/k

# calculate C0 at time = 1 (hr)
ex8_C0 <- 74/exp(-k)

# Calculate AUC
ex8_AUC <- ex8_C0/k

# Calculate CL
ex8_CL <- ex8_Dose / ex8_AUC