Sbar Septic Shock

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ACTIVE LEARNING TEMPLATES

Medication STUDENT NAME _____________________________________

MEDICATION __________________________________________________________________________ REVIEW MODULE CHAPTER ___________

CATEGORY CLASS ______________________________________________________________________

ACTIVE LEARNING TEMPLATE:

PURPOSE OF MEDICATION

Expected Pharmacological Action

Complications

Contraindications/Precautions

Interactions

Medication Administration

Evaluation of Medication Effectiveness

Therapeutic Use

Nursing Interventions

Client Education

 

  1. STUDENT NAME:
  2. MEDICATION:
  3. REVIEW MODULE CHAPTER:
  4. CATEGORY CLASS:
  5. Therapeutic Use:
  6. Complications:
  7. Contraindications/Precautions:
  8. Interactions:
  9. Evaluation of Medication Effectiveness:
  10. Expected Pharmacological Action:
  11. Nursing Interventions:
  12. Medication Administration:
  13. Client Education:

Common reasons pharmacists call providers to clarify orders.

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The purpose of this assignment is to interview a pharmacist to learn more about interprofessional collaboration between prescribers and pharmacists. If possible, volunteer to shadow the pharmacist. Discuss the following in your interview:

  1. The process of filling a prescription.
  2. Common reasons pharmacists call providers to clarify orders.
  3. What is on a proper prescription.
  4. Common omissions on prescriptions they receive.
  5. Medication errors pharmacists encounter.
  6. The prior authorization (par) process and common par medications.

Write a 500-1,000 word paper summarizing your experience and what you learned from the pharmacist you interviewed. Reflect on how this information will help your prescription writing. Discuss intercollaboration between pharmacists and prescribers.

You are required to cite three to five sources related to intercollaboration to complete this assignment. Sources must be published within the last 5 years and appropriate for the assignment criteria and nursing content.

Prepare this paper according to the guidelines found in the APA Style Guide, located in the Student Success Center. An abstract is not required.

This assignment uses a rubric. Review the rubric prior to beginning the assignment to become familiar with the expectations for successful completion.

Zero Order Absorption

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*

Types of chemical administration or exposure that provide (or approximate) zero-order chemical input include:

  • Intravenous (IV) infusion
  • Prolonged breathing of chemical vapors
  • Prolonged skin exposure to constant chemical concentration
  • Implantable controlled/sustained release system

 

*

 

*

Model Equations

D = k0 * T

 

k0 = D/T

 

*

During Chemical Exposure (t≤T)

  • dA/dt = (rate in) – (rate out)

= (absorption rate) – (elimination rate)

 

dA/dt = k0 – (k*A)

 

dA/dt + (k*A) = k0

 

*

A = c1 + (c2 * e-kt)

 

  • The unknown constant terms are then evaluated by employing the initial condition (A=0 at t=0) and the differential equation to get:

c1 = -c2 = k0/k

or

A = (k0/k)(1-e-kt)

 

 

 

 

 

*

    • As before, it is more useful to write the final equation in terms of plasma concentration, which can be done through the transformation C = A/V, giving:

 

C = [k0/V*k] [1-e-kt] for t≤T

 

*

After Chemical Exposure Stopped (t>T)

  • After chemical exposure has stopped, the differential mass balance can be written:

dA/dt = -k*A

 

A = AT at t = T

 

*

A = AT * e-k(t-T) = (k0/k)(1- e-kT)e-k(t-T)

 

 

C = CT * e-k(t-T) = (k0/V*k)(1- e-kT)e-k(t-T)

 

 

CT = (k0/V*k)(1- e-kT)

 

*

 

 

*

31.bin

Steady-State Concentration (CSS)

  • Css occurs when the rate of chemical elimination exactly equals the rate of chemical input
  • Chemical Elimination Rate = Chemical Input Rate

CL * Css = k0

V * k * Css = k0

Css = k0/V*k

 

Note: that the equations for C at t≤T can be rewritten in terms of Css:

C = (k0/V*k)(1-e-kt) = Css(1-e-kt) for t≤T

 

*

  • Concentration drops exponentially immediately after exposure stops, so after infusion (t>T):

C = CT * e-k(t-T)

  • taking the natural logarithm of both sides gives:

lnC = lnCT -k(t-T)

or

lnC = {lnCT + kT} – kt

 

*

 

*

Effect of Model Parameters on Plasma Concentrations

Dosing Rate (k0): Concentration proportional to k0 at all times

 

*

Effect of Model Parameters on Plasma Concentrations

Distribution Volume (V): Concentration proportional to 1/V at all times

 

*

Effect of Model Parameters on Plasma Concentrations

Elimination Rate Constant (k) and Half-life (t1/2): Affect approach to plateau and rate of drop after stopping input, k and t1/2 have opposite effects

 

*

    • The approach to steady state (plateau) conditions is dependent on half-life in a manner similar to elimination:

 

at t1/2 → 50% of way to equilibrium → C=0.5Css

at 2t1/2 → 75% of way to equilibrium → C=0.75Css

at 5t1/2 → ~97% of way to equilibrium → C~0.97Css

at 7t1/2 → ~99% of way to equilibrium → C~0.99Css

at 10t1/2 → ~99.9% of way to equilibrium → C~0.999Css

 

*

 

*

Determination of Chemical Input Rate (k0)

    • IV Infusion
    • Known dose (D) infused over known time period (T) → k0 = D/T
    • k0 = D/T → 100mg/2hr = 50 mg/hr

 

 

  • Skin Exposure Absorption
  • k0 = Absorption Rate = Pskin*Sskin [Cskin – Cblood]

 

*

Determination of Chemical Input Rate (k0)

    • Lung Exposure Absorption

 

    • Low Solubility Gas (Kblood/air <<1, perfusion limited)
    • K0 = Absorption Rate = Qlungblood [Cblood out – Cblood in] ~ Qlung blood [Kblood/air*Cair in-Cvein]

 

  • High Solubility Gas (Kblood/air >>1, ventilation limited)
  • K0 = Absorption Rate = Qlung air [Cair in – Cair out] ~ Qlungair *Cair in

 

*

Example: Estimate the toluene absorption rate for rats exposed to 100ppm toluene via lung absorption

    • kblood/air = 18; MW = 92.15; Qlung air = 3.75 L/hr; Qlung blood = 3.75 L/hr

 

    • remember: C(mg/L) = C(ppm)*MW/24450
    • Cair in = 100ppm * [92.15/24450] = 0.377 mg/L

 

  • kblood/air = 18, therefore kblood/air >> 1 → High solubility gas
  • k0 = Qlung air * Cair in ~ 3.75 L/hr * 0.377mg/L = 1.4mg/hr

 

*

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

  • Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)
  • Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
  • Estimate theoretical Css: Calculate Css if exposure lasted for a very long time (e.g., T>10 t1/2,elim)
  • Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), same as C at 2hr after exposure ends (t-T = 6hr – 4hr = 2hr)
  • Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L
  • Estimate t-T for given C if t>T: calculate t-T when C falls back down to 0.5mg/L
    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)

 

CT = [k0/Vk][1-e-kt]

 

CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]

 

CT = 1.31mg/L * 1-0.0138

 

CT = 1.29 mg/L

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)

 

C = [k0/Vk][1-e-kt]

 

C = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*2hr]

 

C = 1.31mg/L * 1-0.1177

 

C = 1.16mg/L

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)

 

CT = [k0/Vk][1-e-kt]

 

CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]

 

CT = 1.31mg/L * 1-0.0138

 

CT = 1.29 mg/L

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate theoretical Css: Calculate Css if exposure lasted for a very long time (e.g., T>10 t1/2,elim)

 

Css = [k0/Vk]

 

Css = [1.4mg/hr/(1.0L * 1.07hr-1)]

 

Css = 1.31mg/L

 

 

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), [same as C at 2hr after exposure ends (t-T = 6hr – 4hr = 2hr)]

 

C = CT * e-k(t-T)

 

C = 1.29mg/L * e-1.07hr-1(2hr)

 

C = 1.29mg/L * 0.1177

C = 0.152 mg/L

 

 

 

*

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L

 

C = [k0/Vk][1-e-kt]

 

0.5mg/L = 1.31mg/L * e-1.07hr-1(t)

 

(0.5mg/L)/(1.31.mg/L) = 0.3817 = 1 – e-1.07hr-1(t)

e-1.07hr-1(t) = 1 – 0.3817 = 0.6183

 

ln e-1.07hr-1(t) = ln0.6813

 

-1.07hr-1 * t = -0.4807 → -0.4807/-1.07hr-1 = 0.45 hr

or 0.45 hr after start of exposure

 

 

 

 

*

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)

 

CT = [k0/Vk][1-e-kt]

 

CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]

 

CT = 1.31mg/L * 1-0.0138

 

CT = 1.29 mg/L

    • Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1

 

    • Estimate t-T for given C if t>T: calculate t-T when C falls back down to 0.5mg/L

 

C = CT * e-k(t-T)

 

0.5 mg/L = (1.29 mg/L) * e-(1.07/hr)(t-T)

 

(0.5mg/L)/((1.29 mg/L) = 0.3876 = e-(1.07/hr)(t-T)

 

ln0.3876 = lne-(1.07/hr)(t-T)

 

-1.07hr-1 * (t-T) = -0.9478

 

t-T = (-0.9478/-1.07hr-1) = 0.86hr or 0.86hr after exposure ends

 

*

Analysis of plasma concentration data (C vs t)

  • Parameters from Post-Exposure Measurements (t≥T):
  • for t>T: C = CT * e–k(t-T)
  • taking the natural logarithm of both sides gives:

lnC = ln CT – k(t-T)

 

  • Plot post-exposure data in the form:

y = lnC vs x = t-T

  • To yield: slope = m = -k and y = intercept = lnCT

 

*

 

*

    • CT = [k0/Vk][1-e-kT]

 

    • V = [k0/CT*k ][1-e-kT]

 

  • CL = V * k

 

*

 

*

Example: Estimate toluene k, t1/2, CT, V and CL from rat concentrations following a 4hr exposure to 100ppm toluene (from earlier graph, data provided below). Previous example gave k0 ~ 1.4mg/hr

Time, t (hr) Concentration, C (mg/L)
0.00 0.000
0.25 0.278
0.50 0.537
1.00 0.830
2.00 1.114
3.00 1.243
4.00 1.277
4.08 1.201
4.50 0.722
5.00 0.392
5.50 0.261
6.00 0.153

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

*

How do we go about solving this problem?

  • First, calculate (t-T) and lnC for time points after exposure stopped (t≥4hrs)
Time, t (hr) t – T (hr) C (mg/L) lnC
4.00 0 1.277 0.245
4.08 0.08 1.201 0.183
4.50 0.50 0.722 -0.326
5.00 1.00 0.392 -0.936
5.50 1.50 0.261 -1.343
6.00 2.00 0.153 -1.877

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  • Next, plot lnC vs (t-T)
  • Determine useable points (points falling on linear portion of line)

 

*

  • Perform linear regression analysis

m = -1.07hr-1 → k = -m = -(-1.07hr-1) → k = 1.07 hr-1

 

t1/2 = 0.693/k → 0.693/1.07hr-1 → t1/2 = 0.65hr

 

b = 0.228 → CT = eb = e0.228 → CT = 1.26mg/L

 

V = [k0/(CT*k)][1-e-kT] = [(1.4mg/hr)/(1.26mg/L * 1.07hr-1)][1 – e-1.07hr-1 * 4hr] → V = 1.02L

 

CL = V*k = 1.02L * 1.07hr-1 = 1.10 L/hr (intercept method)

 

*

Parameters from Two-Post Exposure Plasma Samples (monitoring after accidental exposure)

Ct1 at t1, Ct2 at t2 with t1 and t2 ≥ T

 

Slope = (ln Ct1 – ln Ct2) / (t1 – t2)

= [ln(Ct1 / Ct2)]/(t1 – t2)

 

k = -slope = [ln(Ct1 / Ct2)]/(t2 – t1)

 

Ct1 = [k0/V*k] [1-e-kT] e-k(t1-T)

 

V = [k0/(Ct1*k)][1-e-kT]e-k(t1-T)

Example: Use 4ht and 6hr measurements from toluene example.

Ct1 = 1.277 mg/L at t1 = 4.0hr

Ct2 = 0.153 mg/L at t2 = 6.0hr

 

k = [ln(1.277mg/L / 0.153mg/L)]/(6.0 – 4.0 hr) = 1.06 hr-1

 

t1/2 = 0.693/1.06hr-1 = 0.65hr

 

V = [(1.4mg/hr)/(1.277mg/L * 1.07hr-1)][1-e-1.07hr-1*4hr)][e -1.06hr-1(4hr – 4hr)]

= 1.01 L

 

CL = V*k = 1.01 L * 1.06hr-1 = 1.08 L/hr

 

*

Parameters from measurements during exposure period (t≤T)

  • In order to perform this analysis, it is necessary to first estimate Css
  • Analysis only works for data from long chemical exposure period that clearly reaches a steady-state plateau
  • Css estimate is more accurate when 2 – 3 plateau points can be averaged
  • For a chemical fitting a one-compartment model, concentrations during the exposure period are given by:

C = Css * (1-e-kt)

 

  • This equation can be rearranged as follows:

C = Css – Css*e–kt → Css – C = Css * e–kt

 

  • Then taking the natural logarithm of each side gives:

ln(Css – C) = lnCss – kt

 

  • Which is now in a linear form with

Slope = m = -k

Intercept = b = lnCss

 

 

*

 

ln(CSS – C)

 

 

 

t

 

lnCSS

 

 

slope = -k

 

 

  • then if one performs a linear regression with

y = ln(Css – C) and x = t

 

  • the values of k and V can be determined by

k = -m

Css = k0/(V*k) → V = k0/(Css * k)

 

  • one should always plot the data to determine if the one-compartment model is appropriate and to identify unusable points

 

 

*

 

ln(CSS – C)

 

 

 

t

 

Late points that

 

jump around are

 

too close to C

 

ss

 

for

 

use in the analysis.

 

If early points do not

 

fall on line, may need

 

different model.

 

 

Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)

Estimate the Css

Calculate ln(Css-C) for each measurement

Plot ln(Css-C) vs t

Determine useable points

Perform linear regression to calculate necessary variables

Time, t (hr) Concentration, C (mg/L)
0.00 0.000
0.25 0.278
0.50 0.537
1.00 0.830
2.00 1.114
3.00 1.243
4.00 1.277

 

 

 

 

 

 

 

 

 

 

 

 

Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)

Estimate the Css

 

Since concentration is still rising between last two timepoints can only use the last timepoint to estimate Css

 

Css ~ 1.277 mg/L

Time, t (hr) Concentration, C (mg/L)
0.00 0.000
0.25 0.278
0.50 0.537
1.00 0.830
2.00 1.114
3.00 1.243
4.00 1.277

 

 

 

 

 

 

 

 

 

 

 

 

 

*

Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)

Estimate the Css: Css~ 1.277 mg/L

Calculate ln(Css-C) for each measurement

Time, t (hr) Concentration, C (mg/L) ln(Css – C)
0.00 0.000 0.245
0.25 0.278 -0.001
0.50 0.537 -0.301
1.00 0.830 -0.805
2.00 1.114 -1.814
3.00 1.243 -3.381
4.00 1.277

 

 

 

 

 

 

 

 

 

 

 

 

 

Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)

Estimate the Css

Calculate ln(Css-C) for each measurement

Plot ln(Css-C) vs t

Determine useable points

*all points are useable

*can use linear regression to describe data

 

*

Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)

Perform linear regression to calculate necessary variables

 

m = -1.18hr-1 → k = -m = -(-1.18hr-1) → k = 1.18hr-1

 

t1/2 = 0.693/k → 0.693/1.18hr-1 →t1/2 = 0.59hr

 

b = 0.315 → Css,intercept = eb = e0.315 → Css = 1.37 mg/L

 

V = k0/(Css * k) = (1.4 mg/hr)/[1.28mg/L * 1.18hr-1) → V = 0.93L

 

CL = V*k = 0.93L * 1.18hr-1 → CL = 1.10L/hr

 

AUC Calculations from C vs t Data

    • Calculate AUC(t1 →tf) using the trapezoidal rule for consecutive points
    • AUC(t1 →tf) = ½[Ct1 + Ct2][t1-t2]
    • Use C0 = 0 at t=0, even if C0 is not actually measured
    • If CT is not measured, estimate it from post-exposure regression

 

  • Evaluate the AUC of the tail region just as for bolus IV injection
  • AUCtail = AUC(tf →∞) = Ctf/k

 

*

AUC Calculations from C vs t Data

  • As a short-cut method, a reasonable estimate of AUC during post-exposure period is given by
  • AUC(T →∞) ~ CT/k
  • There is unfortunately no general short-cut method for the exposure period (t<T)
  • If Css can be estimated, then total AUC ~ Css * T

 

*

Evaluation of V and CL from AUC (AUC Method)

    • AUC = D/(V*k) = D/CL (always true for linear kinetics)

 

    • Remember: D = k0 * T yields:

 

V = [k0 * T]/[AUC * k]

 

CL = [k0 * T]/AUC

Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr-1 and k0 = 1.4mg/hr as determined previously)

  • Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
  • Calculate AUC for each interval and for tail region (after tf)
  • Sum AUC values to get running sum of AUC and total AUC
  • Calculate V and CL by the AUC Method (compare to intercept method used previously)

Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr-1 and k0 = 1.4mg/hr as determined previously)

  • Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
t1 (hr) t2 (hr) Ct1 (mg/L) Ct2 (mg/L)
0.00 0.25 0.000 0.278
0.25 0.50 0.278 0.537
0.50 1.00 0.537 0.830
1.00 2.00 0.830 1.114
2.00 3.00 1.114 1.243
3.00 4.00 1.243 1.277
4.00 4.08 1.277 1.201
4.08 4.50 1.201 0.722
4.50 5.00 0.722 0.392
5.00 5.50 0.392 0.261
5.50 6.00 0.261 0.153
6.00 ∞ 0.153

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr-1 and k0 = 1.4mg/hr as determined previously)

    • Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
    • Calculate AUC for each interval and for tail region (after tf)

 

AUC(0→0.25hr) = ½[0 + 0.278mg/L][0.25 – 0 hr] = 0.035 mg hr/L

 

AUCtail = (0.153mg/L)/(1.07hr-1) = 0.143 mg hr/L

3. Sum AUC values to get running sum of AUC and Total AUC

t1 (hr) t2 (hr) Ct1 (mg/L) Ct2 (mg/L) AUC(t1→t2) (mg hr/L) AUC run SUM (mg hr/L)
0.00 0.25 0.000 0.278 0.035 0.035
0.25 0.50 0.278 0.537 0.102 0.137
0.50 1.00 0.537 0.830 0.342 0.478
1.00 2.00 0.830 1.114 0.972 1.450
2.00 3.00 1.114 1.243 1.178 2.629
3.00 4.00 1.243 1.277 1.260 3.889
4.00 4.08 1.277 1.201 0.099 3.988
4.08 4.50 1.201 0.722 0.404 4.392
4.50 5.00 0.722 0.392 0.278 4.670
5.00 5.50 0.392 0.261 0.163 4.834
5.50 6.00 0.261 0.153 0.104 4.937
6.00 ∞ 0.153 0.143 5.080

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

As a comparison, using the short-cut method for the post-exposure period gives:

 

AUC(T→∞) ~ CT/k = (1.277mg/L)/(1.07hr-1) = 1.193 mg hr/L

 

AUC = AUC(0→T) + AUC(T→∞)

= 3.889 mg hr/L + 1.193 mg hr/L

= 5.082 mg hr/L

4. Calculate V and CL by the AUC Method (compare with intercept method)

V = [k0 * T] / [AUC * k]

= [1.4mg/hr * 4hr] / [5.080mg hr/L * 1.07hr-1]

= 1.03 L (compared to 1.02 L)

 

CL = [k0 * T] / AUC

= [1.4 mg/hr * 4hr] / 5.080 mg hr/L

= 1.10 L (compared to 1.10 L)

 

*

Chemical

Absorption

Rate

Time

0

Instantaneous

Absorption

Zero-Order

Absorption

T

C

k

V

k

0

Short Exposure

T

C

Time

Prolonged Exposure

T

C

Time

C

T

C

T

Short Infusion

T

lnC

Time

Prolonged Infusion

T

lnC

Time

lnC

T

lnC

T

slope = -k

slope = -k

T

C

Time

Large

k

0

Small

k

0

T

lnC

Time

Large

k

0

Small

k

0

Slope not

affected

by k

0

T

C

Time

Small

V

Large

V

T

lnC

Time

Small

V

Large

V

Slope not

affected

by V

T

C

Time

Small t

1/2

Large k

Large t

1/2

Small k

T

lnC

Time

Slope, C

�

,

and approach

to plateau all

affected

Large t

1/2

Small k

Small t

1/2

Large k

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

Toluene Concentration in Plasma (mg/L)

Time (hr)

100 ppm Toluene

200 ppm Toluene

0.1

1.0

10.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

Toluene Concentration in Plasma (mg/L)

Time (hr)

100 ppm

Toluene

200 ppm

Toluene

lnC

Time after Chemical Input Stopped (t – T)

b = lnC

T

m = -k

Late points may jump

around near limit of

drug assay, not reliable

=> do not use

lnC

t – T

Early high points suggest need

for more than one compartment

=> we will consider later

Useable

Points

-2.0

-1.5

-1.0

-0.5

0.0

0.5

0.0

0.5

1.0

1.5

2.0

2.5

y = 0.22844 + -1.0674x R= -0.99808

lnC

Time after Exposure Ended, t – T (hr)

C

t

If possible, average multiple

plateau points to estimate C

ss

in order to improve accuracy.

l

n

(

C

SS

–

C)

t

l

nC

SS

s

lope

 

=

–

k

l

n

(

C

SS

–

C)

t

L

ate

 

poi

n

t

s t

ha

t

j

u

m

p

a

r

ou

n

d a

r

e

t

oo

c

lo

s

e to

 

C

ss

 

fo

r

u

s

e

i

n

t

he

 

ana

l

y

s

i

s

.

I

f ea

r

l

y

poi

n

t

s d

o no

t

f

a

ll o

n

li

n

e,

m

ay

n

eed

d

i

ffe

r

e

n

t

m

ode

l

.

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

y = 0.31522 + -1.1777x R= -0.99534

ln(C

ss

-C)

Time (hr)

Pharmacology

/in /by

HMWK 1 Pharmacology BIO 107

1) Paracelsus is known for what statement

2) Define Pharmacology

3) What are the advantages of prescription drugs

What are the Disadvantages

4) What are the advantages of Over-the-Counter Drugs?

What are the disadvantages?

5) What did the The Biologics Control Act of 1902 mandate?

6) What did the Pure Food and Drug Act of 1906 mandate?

7) What did the Sherley Amendment of 1912 mandate?

8) What did the Food, Drug, and Cosmetic Act (1938) and amendments mandate?

9) Why is thalidomide significant in the history of the FDA? Name the Doctor involved.

10) What is the Center for Drug Evaluation and Research

11) Define the four phases of clinical drug research

Phase I: Pharmacokinetic Study

Phase II: Dose Ranging in Patients

Phase III: Efficacy Study in Patients

Review of new drug application (NDA)

Phase IV: Postmarketing surveillance

12) What is Preclinical Investigation?

13) How are oral Generic Drugs tested?

14) Define Addiction

15) Define Dependence

16) Define Physical dependence

17) Define Psychologically dependent

18) Why are drugs placed on schedules?

19) Define each schedule

a. I

b. II

c. III

d. IV

e. V

20) What does the Center for Drug Evaluation and Research

21) What are the Five Rights used as the basis of safe delivery of medications

22) List common abbreviations used to convey drug dosing information

23) Which dosage forms cannot be crushed

24) What are the disadvantages/advantages of oral administration

25) What are the disadvantages/advantages of the sublingual route of administration

26) What are the disadvantages/advantages of using a locally acting drugs

27) What are the disadvantages/advantages of Parenteral route of administration

28) Define ADME

29) Define diffusion

30) What determines whether or not drugs should be taken on an empty stomach? Provide a detailed answer.

31) What determines whether or not drugs can pass through a membrane?

32) Describe enzyme induction in drug metabolism

33) What is first pass metabolism?

34) How does binding to plasma proteins affect the pharmacokinetics of a drug?

35) What is the therapeutic range of a drug, what is the therapeutic index?

36) What is the half life of a drug?

37) What is a Loading Dose?

38) What does a drug do to a receptor that functions as a enzyme

39) What does a drug do to a receptor that is coupled to a G-protein

40) What does a drug do to a receptor that is coupled to an ion channel

41) What is cAMP?

42) What is different about steroid receptors

43) How does receptor occupancy relate to efficacy

44) Compare and contrast potency and efficacy

45) Define Agonist and Antagonist

46) What are the ED50 & LD50

47) Define Therapeutic Index