Unit 2 - Mathematical Fundamentals in Pharmacokinetics.pdf

Mathematical Fundamentals in Pharmacokinetics Gerard Lee L. See, RPh, PhD, SRPharmS Beatrix B. Loyao,...

Mathematical Fundamentals in Pharmacokinetics Gerard Lee L. See, RPh, PhD, SRPharmS Beatrix B. Loyao, RPh, MSc. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY Biopharmaceutics & Pharmacokinetics Outline „ Calculus „ Graphs „ Mathematical Expressions and units „ Units for Expressing Blood Concentrations „ Exponential and Logarithmic Functions „ Rates and Orders of Processes SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY Calculus „ Pharmacokinetic models consider drugs in the body to be in a dynamic state. „ Calculus is an important mathematic tool for analyzing drug movement quantitatively. „ Differential equations are used to relate to the concentrations of drugs in various body organs over time. „ Integrated equations are frequently used to model the cumulative therapeutic or toxic responses of drugs in the body Calculus is essential for studying how drugs move through the body by providing a way to analyze these processes mathematically. Differential equations help us track how drug levels in different organs change over time, while integrated equations are used to predict the overall effects, whether beneficial or harmful, as the drugs accumulate in the body. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 3 Calculus „ Differential Calculus - It is a branch of calculus that involves finding the rate at which a variable quantity is changing. - For example, a specific amount of drug X is placed in a beaker of water to dissolve. - The rate at which the drug dissolves is determined by the rate of drug diffusing away from the surface of the solid drug and is expressed by the Noyes-Whitney equation: Differential calculus is a branch of calculus that focuses on understanding how things change, specifically the rate at which a variable quantity changes. 𝒅𝑿 𝑫𝑨 Dissolution Rate = = (𝑪𝟏 − 𝑪𝟐 ) 𝒅𝒕 𝒍 Legends: d denotes a very small change; X = drug X; t = time; D = diffusion coefficient; A = effective surface area of drug; l = length of diffusion layer; C1 = surface concentration of drug in the diffusion layer; C2 = concentration of drug in the bulk solution SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 4 Calculus „ Integral Calculus - Integration is the reverse function of differentiation and is considered the summation of f (x) ⋅ dx; the integral sign ∫ implies summation - For example, given the function y = ax, plotted in Fig. 2-1, the integration is ∫ ax ⋅ dx. Compare Fig. 2-1 to a second graph (Fig. 2-2), where the function y = Ae–x is commonly observed after an intravenous bolus drug injection. Differential calculus and integral calculus are two key branches of calculus that focus on different aspects of change. Differential calculus is about finding the rate at which something changes, like how fast a drug dissolves in water. On the other hand, integral calculus is about summing up parts to find the whole, such as calculating the total amount of drug in the body over time after it’s been injected. While differential calculus looks at small, instantaneous changes, integral calculus adds up these changes to S C I E N TtheI Aoverall Ÿ Veffect. IRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 5 understand  SUMMATION - TOTALITY OF CONCENTRATION Calculus „ Integral Calculus - Definite Integral of a mathematical function is the sum of individual areas under the graph of that function. - Trapezoidal rule is a numerical method frequently used in pharmacokinetics to calculate the area under the plasma drug concentration-versus-time curve, called area under the curve (AUC) Integral calculus is about finding the total area under a curve, which represents the sum of all the small parts of that area. In pharmacokinetics, this is used to calculate the area under the curve (AUC) of a drug's concentration in the blood over where [AUC] = area under the curve, tn = time of observation of drug time. The Trapezoidal rule is a concentration Cn, and tn–1 = time of prior observation of drug method used to estimate this area by breaking it down into concentration corresponding to Cn–1. smaller sections, helping to understand how the drug concentration changes over time after it’s been administered SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 6 Calculus „ Calculation of AUC using Trapezoidal Rule Time (hours) Plasma Drug Level (µg/mL) 0.5 38.9 1.0 30.3 2.0 18.4 where [AUC] = area under the curve, tn = time of observation of drug concentration 3.0 11.1 Cn, and tn–1 = time of prior observation of 4.0 6.77 drug concentration corresponding to Cn–1. 5.0 4.10 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 7 Calculus „ Calculation of AUC using Trapezoidal Rule Calculate the AUC from 1 to 4 hours. Step 1: The AUC between 1 and 2 hours is calculated by proper substitution of the where [AUC] = area under the equation curve, tn = time of observation of drug concentration Cn, and tn–1 = time of prior observation of drug !" 30.3 + 18.4 concentration corresponding to Cn–1. 𝐴𝑈𝐶 !! = 2−1 2 Time Plasma Drug Level = 24.35 µ𝑔 : ℎ/𝑚𝐿 (hours) (µg/mL) 0.5 38.9 1.0 30.3 2.0 18.4 3.0 11.1 4.0 6.77 5.0 4.10 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 8 Calculus „ Calculation of AUC using Trapezoidal Rule Calculate the AUC from 1 to 4 hours. Step 2: The AUC between 2 and 3 hours is calculated by proper substitution of the where [AUC] = area under the equation curve, tn = time of observation of drug concentration Cn, and tn–1 = time of prior observation of drug !# 18.4 + 11.1 concentration corresponding to Cn–1. 𝐴𝑈𝐶 !" = 3−2 2 Time Plasma Drug Level = 14.75 µ𝑔 : ℎ/𝑚𝐿 (hours) (µg/mL) 0.5 38.9 1.0 30.3 2.0 18.4 3.0 11.1 4.0 6.77 5.0 4.10 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 9 Calculus „ Calculation of AUC using Trapezoidal Rule Calculate the AUC from 1 to 4 hours. Step 3: The AUC between 3 and 4 hours is calculated by proper substitution of the where [AUC] = area under the equation curve, tn = time of observation of drug concentration Cn, and tn–1 = time of prior observation of drug !$ 11.1 + 6.77 concentration corresponding to Cn–1. 𝐴𝑈𝐶 !# = 4−3 2 Time Plasma Drug Level = 8.94 µ𝑔 : ℎ/𝑚𝐿 (hours) (µg/mL) 0.5 38.9 1.0 30.3 2.0 18.4 3.0 11.1 4.0 6.77 5.0 4.10 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 10 Calculus „ Calculation of AUC using Trapezoidal Rule Calculate the AUC from 1 to 4 hours. Step 4: The total AUC between 1 and 4 hours is obtained by adding the three where [AUC] = area under the smaller AUC values together curve, tn = time of observation of drug concentration Cn, and tn–1 = !$ !" !# !$ time of prior observation of drug 𝐴𝑈𝐶 !! = 𝐴𝑈𝐶 !! + 𝐴𝑈𝐶 !" + 𝐴𝑈𝐶 !# concentration corresponding to Cn–1. !$ 𝐴𝑈𝐶 !! = 24. 3 + 14.3 + 8.94 Time Plasma Drug Level (hours) (µg/mL) !$ 0.5 38.9 𝐴𝑈𝐶 !! = 48.04 µ𝑔 : ℎ/𝑚𝐿 1.0 30.3 2.0 18.4 3.0 11.1 4.0 6.77 5.0 4.10 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 11  curve fitting method - prepared data to follow a trend -fitting in the plot at every time point Graphs „ The construction of a curve or straight line by plotting the observed or experimental data on a graph is an important method of visualizing relationships between variables „ In general, the values of the independent variable (x) are placed on the horizontal line in a plane, or abscissa ( x axis) „ The values of the dependent variable are placed on the vertical line in the place, or on the ordinate (y axis) „ The values are usually arranged so that they increase linearly or logarithmically from left to right and from bottom to top. Plotting observed or experimental data on a graph is a key way to visualize the relationship between different variables. Typically, the independent variable (x) is placed on the horizontal axis (x-axis), while the dependent variable is placed on the vertical axis (y-axis). The data points are arranged so that the values increase either linearly or logarithmically as you move from left to right and from bottom to top on the graph. This method helps to clearly see how one variable affects another. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 12  Graphs „ In pharmacokinetics, time is the independent variable and Is plotted on the abscissa (x axis) „ Drug concentration is the dependent variable and is plotted on the ordinate ( y axis) „ Two types of graphs are usually used in pharmacokinetics In pharmacokinetics, graphs are often used to show the relationship between time and drug concentration. Time, being the independent variable, is plotted on the x- axis (abscissa), while drug concentration, the dependent variable, is plotted on the y- axis (ordinate). There are typically two types of graphs used in pharmacokinetics to analyze how drug concentration changes over Cartesian or Rectangular coordinates Semilog coordinates time. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 13  exponential notation Graphs „ Curve Fitting - Fitting a curve to the points on a graph implies that there is some sort of relationship between the variables x and y, such as dosage of drug versus pharmacologic effect (e.g., lowering of blood pressure) - When using curve fitting, the relationship is not confined to isolated points but is a continuous function of x and y. - A hypothesis is made concerning the relationship between the variables log is basedx onand10` y. Then, an empirical equation must satisfactorily fir the experimental or observed data. If the relationship between x and y is linearly related, then the relationship between the two can be expressed as a straight line. Curve fitting involves drawing a curve that best represents the relationship between two variables plotted on a graph, such as drug dosage and its effect on blood pressure. - The general equation of a straight line is Instead of just connecting isolated points, curve fitting creates a continuous function that shows how one variable changes in relation to another. To do this, a hypothesis is made 𝑌 = 𝑚𝑥 + 𝑏 about how the variables are related, and an empirical equation is used to match the observed data. If the relationship is linear, it can be expressed as a straight line with the general equation = Ł Ł + Y=mx+b, where m is the slope and b is the y-intercept. where m = slope and b = y intercept. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 14  exponential notation Graphs „ Curve Fitting log is based on 10` Graphic demonstration of variations in slope (m). SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 15  exponential notation Graphs „ Linear Regression/Least Squares Method - This method is often used in clinical pharmacy studies to construct a linear relationship between an independent variable (also known as the input factor or the x factor) and a dependent variable (commonly known as an output variable, an outcome, or the y factor) - In pharmacokinetics, the relationship between the plasma drug concentrations versus time can be expressed as a linear function. log is based on 10` - The strength of the linear relationship is assessed by the correlation coefficient ( r ). The value of r is positive when the slope is positive and it is negative when the slope is negative. When r takes the value of either +1 or =1, a perfect relationship exists between the variables. - A zero value for the slope (or for r) indicates that there is no linear relationship existing between y and x. Linear regression, or the least squares method, is commonly used in clinical pharmacy studies to establish a straight-line relationship between an independent variable (x) and a dependent variable (y). In pharmacokinetics, this can show how plasma drug concentrations change over time. The strength of this linear relationship is measured by the correlation coefficient (r). A positive r indicates a positive slope, while a negative r indicates a negative slope. If r is +1 or -1, it means there is a perfect linear relationship between the variables. If r is zero, it suggests there is no linear relationship between x and y. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 16 Mathematical Expressions and Units Common Units Used in Pharmacokinetics „ Mathematics is a basic science that helps to explain relationships among variables. „ For an equation to be valid, the units or dimensions must be constant on both sides of the equation. F - Bioavailability symbol Mathematics is fundamental for understanding how different variables are related. For an equation to be accurate, the units or dimensions on both sides of the equation must match. This ensures that the relationships being described are consistent and meaningful. F - Bioavailability symbol SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 17 Units for Expressing Blood Concentrations „ Drug concentrations or drug levels should be expressed as mass/volume. „ The expressions mcg/mL, µg/mL, and mg/L are equivalent and are commonly reported in the literature. „ Drug concentration may also be reported as mg% or mg/dL, both of which indicate milligrams of drug per 100 mL (1 deciliter) Drug concentrations are typically expressed as mass per volume. Common units include mcg/mL (micrograms per milliliter), μg/mL (micrograms per milliliter), and mg/L (milligrams per liter), which are equivalent and frequently used in research. Additionally, drug concentrations might be reported as mg% (milligrams per percent) or mg/ dL (milligrams per deciliter), both of which represent milligrams of the drug per 100 milliliters of solution. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 18  exponential notation Exponential and Logarithmic Functions „ The two mathematical functions are related to each other. - Logarithm of a number is the power or exponent when that number is converted into exponential form. The power or exponent will clearly depend on the base. If the base 10 is used, the logarithm is know as the log. Exponential and logarithmic functions Examples of Base 10 Logarithms (Logs) are closely related in mathematics, where an exponential function expresses numbers as powers of a log is based on 10` given base, while a logarithmic function finds the exponent to which the base must be raised to get that number. For example, if we have an exponential expression like 103=100010 3 =1000, the logarithm (log) with base 10 tells us that 3 is the power needed to reach 1000. Essentially, logarithms are the inverse operations of exponentiation, making them useful for solving equations where the unknown is an exponent. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 19 Exponential and Logarithmic Functions „ When the base e is used, the logarithm is known as the natural logarithm or ln. When working with logarithms, the base \(e\) (approximately 2.718) is used in natural logarithms, denoted Examples of Natural Logarithms (Lns) as \(\ln\). Sometimes, it's necessary to convert a common logarithm (base 10, denoted as \(\log\)) to a natural logarithm. This conversion uses the factor 2.303, so \(\ln X = \log X \times 2.303\). For example, if \(\log 5 = 0.6989\), then \(\ln 5\) is calculated by multiplying 0.6989 by 2.303, resulting in \(\ln 5 = 1.609\). This relationship helps in switching between logarithmic scales in mathematical calculations. When the base 10 logarithm scale is used for expressions containing e, it may be necessary to convert a log to an ln. This is done using the factor 2.303: ln X = log X × 2.303 Thus, if log 5 = 0.6989, then ln 5 = 0.6989 × 2.303 = 1.609 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 20 Exponential and Logarithmic Functions „ Calculations using Exponential Expressions and Logarithms - Example: Solve for x in the following expressions: (a) ex = 3, (b) ex = 23, (c) ex = 106, and (d) ex = 563 - Solutions: (a) ex =3 x = ln 3 x = 1.10 (b) ex =23 x = ln 23 x = 3.14 (c) ex = 106 x = ln 106 x = 4.66 (d) ex = 563 x = ln 563 x = 6.33 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 21 Exponential and Logarithmic Functions „ Calculations using Exponential Expressions and Logarithms - Example: Solve for x in the following expressions: (a) ln x = 4.3, (b) ln x = −1.4, and (c) ln x = −0.3. - Solutions: (a) ln x = 4.3 x = e4.3 x =73.7 (b) ln x = -1.4 x = e-1.4 x = 0.247 (c) ln x = -0.3 x = e-0.3 x = 0.741 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 22 Exponential and Logarithmic Functions „ Calculations using Exponential Expressions and Logarithms - Example: Evaluate e-1.3 - Solutions: e-1.3 = 0.273 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 23 Exponential and Logarithmic Functions „ Calculations using Exponential Expressions and Logarithms - Example: Find the value of k in the following expression: e−1.3k = 2. - Solutions: Take the logarithms k ! ln e−1.3 = ln 2 k ! (-1.3) = 0.693 -1.3 k = 0.693 !.#$% k=- &.% k = -0.533 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 24 Exponential and Logarithmic Functions „ Calculations using Exponential Expressions and Logarithms - Example: A common expression in pharmacokinetics is Cp = Cp0 ) e−kt. Evaluate Cp when Cp0 = 35, k = 1.5, and t = 2. - Solutions: Cp = Cp0 ) e−kt Cp = 35 ) e−1.5x2 Cp = 35 e−3 Cp = 35 x 0.0498 Cp = 1.74 SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 25  Rates and Orders of Processes „ A process such as drug absorption or drug elimination may be described by the rate by which the process proceeds. The rate of a process, in turn, may be defined in terms of specifying its order. „ In pharmacokinetics, two orders are of importance, the zero order and the first order In pharmacokinetics, the rate at which a process like drug absorption or elimination occurs can be described Zero-Order Example: Suppose a medication is administered via a transdermal patch, and it releases 10 mg of and measured. This rate can be defined by its "order," drug per hour. Regardless of the drug concentration in the blood, the patch consistently releases 10 mg of the which indicates how the rate depends on the drug every hour. If the drug concentration in the blood starts at 100 mg/L, after 1 hour it would be 90 mg/L, after concentration of the drug. 2 hours it would be 80 mg/L, and so on, but the rate of drug release remains constant at 10 mg per hour.First- Zero Order: In a zero-order process, the rate is constant and does not change with the concentration of the drug. This means the drug is absorbed or eliminated at a Order Example: Consider a drug with a half-life of 4 hours, meaning that the concentration of the drug in the steady rate, regardless of how much drug is present. For blood decreases by half every 4 hours. If the initial concentration is 100 mg/L, after 4 hours it would drop to 50 example, if a medication is released from a patch at a mg/L. After another 4 hours (8 hours total), it would decrease further to 25 mg/L. The rate of decrease in drug constant rate, it is following zero-order kinetics. concentration is proportional to its current concentration, which is characteristic of first-order kinetics. First Order: In a first-order process, the rate changes in proportion to the concentration of the drug. As the drug concentration decreases, the rate of absorption or elimination also decreases. This is typical for many drugs where the rate at which they are metabolized or cleared from the body is directly related to their concentration.So, the main difference is that zero-order processes have a constant rate regardless of drug concentration, while first-order processes have a rate that depends on the SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO current drug concentration. D E PA RT M E N T O F P H A R M ACY 26 Rates and Orders of Processes „ Zero-Order Process - The rate of a zero-order process is one that proceeds over time (t) independent from the concentration of the drug (c). The negative sign for the rate indicates that the concentration of the drug decreases over time -dc/dt = k0 Dc = -k0dt c = c0-kot where c0 is the initial concentration of the drug at t = 0 and k0 is the zero-order rate constant. The units for k0 are concentration per unit time (eg, [mg/mL]/h) or amount per unit time (eg, mg/h) SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 27 Rates and Orders of Processes „ Zero-Order Process - Example: calculate the zero-order rate „ Zero-Order Process constant ([ng/mL]/min) if the initial - Additional Question: When concentration of the drug is 200 ng/mL and does the concentration of that at t = 30 minutes is 35 ng/mL. drug equal to 100 ng/m:? - Solution: - Solution: c = c0 −k0 t c = c0 −k0 t 35 ng/mL = 200 ng/mL −k0 (30 mins) 100 = 200 – 5.5 t −k0 = (35 ng/mL − 200 ng/mL)/30 mins (100 – 200)/ 5.5 = - t −k0 = (-165 ng/mL)/30 mins -18.2 = - t k0 = 5.5 (ng/mL)/min t = 18.2 min In pharmacokinetics, the time required for one-half of the drug concentration to disappear is known as t1⁄2. Thus, for this drug the t1⁄2 is 18.2 minutes. SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 28 Rates and Orders of Processes „ Zero– order Process - In general, t1/2 may be calculated as follows „ Zero-Order Process for a zero-order process: - Solution: c = c0 −k0 t t1/2 = (0.5 co)/k0 (0.5 c0) = c0 –k0t1/2 t1/2 = (0.5)(200)/5.5 (0.5 c0) – c0 = -k0t1/2 t1/2 = 18.2 min -0.5 c0 = -k0t1/2 t1/2 = (0.5 co)/k0 In a zero-order process the t1/2 is not constant and depends upon the initial amount or concentration of drug. need to note the dose SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 29 Rates and Orders of Processes „ First-Order Process always constant ang half-life - The rate of a first-order process is dependent upon the concentration of the drug: −dc/dt = k1c −dc/c = k1dt lnc = lnc0 − k1t - t1⁄2 = 0.693/k1 time is independent - Unlike a zero-order rate process, the t1/2 for a first-order rate process is always a constant, independent of the initial drug concentration or amount no need to note the dose SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 30 Rates and Orders of Processes „ Comparison of Zero- and First-Order Reactions Zero-Order Reaction First-Order Reaction Equation –dC/dt = k0 –dC/dt = kC C = –k0t + C0 C=C0 e–kt Rate constant - units (mg/L)/h 1/h Half-life, t1/2 (units = time) t1/2 = 0.5C/k0 (not constant) t1/2 = 0.693/k (constant) Effect of time on rate Zero-order rate is constant with respect to time First-order rate will change with respect to time as concentration changes Effect of time on rate constant Rate constant with respect to time changes as the Rate constant remains constant with respect to time concentration changes Drug concentrations versus time—plotted on Drug concentrations decline linearly for a zero-order Drug concentrations decline nonlinearly for rectangular coordinates rate process a first-order rate process Drug concentrations versus time—plotted on a Drug concentrations decline nonlinearly for a zero- Drug concentrations decline linearly for a single semilogarithmic graph order rate process first-order rate process SCIENTIA Ÿ VIRTUS Ÿ DEVOTIO D E PA RT M E N T O F P H A R M ACY 31