A condensed summary of this video can be found in the Video summary page.

Transcription

1 Welcome to Week 3 Starting week three video Please watch the online video (1 minutes, 6 seconds). Chapter 6 - Blood and Drug Transport Introduction to Chapter 6 Chapter 6 contains two subsections. Blood ADME At the conclusion of this chapter, you should understand the composition of blood and the role it plays in transporting a drug to and from its site of action. You should also be able to give the meaning of the acronym ADME and understand the significance of each of the four letters. 6.1 Blood What is blood? video Please watch the online video (6 minutes, 3 seconds). A condensed summary of this video can be found in the Video summary page. Fluids in the body Background: Blood is the medium by which drugs are normally distributed throughout the body. A 70-kg (154-lbs) human has an approximate blood volume of 5 L (1.4 US gal). Of those 5 L of whole blood, approximately 46% consists of different cells. The remaining 54% is the fluid portion, or plasma, with a volume of 2.7 L. Plasma contains water, electrolytes, small signal molecules, and some proteins. Serum is the fluid left behind after whole blood clots. Serum is very similar to plasma except serum is missing the clotting proteins (fibrinogens).

2 Instructions: Read the passage below concerning the fluids within the body. Learning Goals: To understand the fluids of the body into which a drug can be transported. Drug concentrations are measured in plasma and reported as C p. Drugs, however, certainly travel in other fluids in the body. Below are descriptions of the most relevant fluids for drugs in a 70-kg patient. whole blood - 5 L As has been mentioned, a 70-kg human has 5 L of blood. That works out to L/kg, which is a figure that is widely used to determine the blood volume of a specific patient. Drugs do not necessarily enter all parts of whole blood. Drugs will freely move throughout the non-cellular part of blood, but they may or may not enter the red blood cells. To test the concentration of a drug in whole blood, the cells in a whole blood sample must be lysed (broken open), and then the resulting solution may be analyzed for the drug. Breaking open the cells releases drug that has entered the cell as well as any drug that was bound within the cell membranes themselves. Whole blood concentrations are normally reported as C b or C wb. plasma L Plasma is the fluid fraction of blood. Plasma, because of its uniformity, is the medium of choice for monitoring drug concentrations in the body. Plasma concentration is reported as C p. interstitial fluid - 10 L Interstitial fluid is the liquid that sits between the cells of the body. This fluid contains the nutrients and waste of the cells. Drugs reach the interstitial fluid from the capillaries. The walls of the capillaries have pores that allow passage of liquids and anything smaller than the pores. Everything in blood can pass except proteins and cells. All orally-delivered drugs are small enough to slip into the interstitial fluid and reach the cells in the body. intracellular fluid - 25 L The total water found within the cells of the body is 25 L. Note that this volume includes the cellular volume of whole blood. Only drugs that can cross a cell membrane will have access to the intracellular fluid. total body water - 38 L Total body water includes plasma, interstitial fluid, and intracellular fluid.

3 body volume - 70 L The volume of the body of a 70-kg human is approximately 70 L. This volume is not all water, but it provides a comparison for the total body water value of 38 L. Serum binding Background: Approximately 8 wt% of whole blood is protein dissolved in the fluid fraction. Instructions: Read the passage below concerning the influence of blood proteins upon drugs. Learning Goals: To understand the complicating factors of proteins and how they can affect the behavior of drugs in the body. Drugs are designed to target a specific protein in the body. By binding that particular target, the drug gives rise to a biological effect. Just because a drug has a high affinity for one target does not mean that it cannot bind to other proteins. Proteins in the blood have a large effect on the drugs that bind them. Because drugs are transported by the blood, the blood proteins cannot be avoided by drugs. Some of the blood proteins have a very high concentration. Even if a drug has only a low affinity for a protein, if the protein is present in a high enough concentration, the protein will bind a large fraction of the drug. The equilibrium below demonstrates this idea. By Le Chatlier's principle, a higher concentration of protein forces the reaction to shift to the left and form more protein-drug complex in order to reach equilibrium. The equilibrium expression below reinforces the concept. A high protein concentration increases the size of the numerator. The value of K D is restored by shifting the equilibrium to the left. This shift raises the concentration of the protein-drug complex (the denominator) while simultaneously decreasing both the unbound protein and drug concentrations (the numerator). The interaction of a drug with proteins in the blood affects both how a drug is cleared from the bloodstream as well as a drug's distribution. A drug's clearance and distribution together determine the half-life of a drug. Both concepts will be formally introduced in Chapter 7 - Pharmacokinetics.

4 The two blood proteins most relevant to drug action are discussed below. serum albumin Blood is up to 5 wt% serum albumin. This is a remarkably high concentration of a single protein. Serum albumin has multiple potential binding sites. Although serum albumin can affect almost any drug, serum albumin tends to bind acid drugs most strongly. Two acidic drugs that are extensively bound by serum albumin are warfarin and ibuprofen, both shown below. globulins Blood contains several different globulin proteins - the α-, β-, and γ-globulins. These proteins can comprise up to 2.5 wt% of blood. One specific protein, α 1-acid glycoprotein, tends to bind basic drugs, including disopyramide and lidocaine, which are shown below. Since the concentration of α 1- acid glycoprotein is lower than serum albumin, the degree of drug binding tends to be somewhat lower. FAQ, help, and tips What is blood? video Are hits that bind plasma proteins bad? Not necessarily. Compounds with very high serum binding may be downgraded relative to other compounds with lower serum binding. High serum binding, however, it not a major problem for a molecule. Other issues like toxicity or an inability to be absorbed into the blood stream are far bigger issues.

5 Almost all drugs bind serum proteins to some degree, and some drugs bind to a high degree. Plasma binding alone will not prevent a molecule from becoming a drug. What could go wrong with serum binding? A classic example is warfarin, a blood thinner. Warfarin is 99% bound to serum albumin. If a patient taking warfarin takes a second drug that binds albumin in the same position as warfarin, then some of the warfarin may become unbound and released into the blood stream. If even a small amount of warfarin is released, then the free, unbound percentage of warfarin might rise from 1% to 10%. That would be a 10-fold increase in the concentration of free, unbound warfarin. Such an increase in warfarin can lead to uncontrolled bleeding in a patient. The example of warfarin is somewhat extreme, but it does demonstrate that serum albumin binding can cause complications for a drug. Serum binding What is wt%? wt% stands for weight percent. Whole blood is approximately 8 wt% protein. In other words, a 100-gram sample of whole blood would contain 8 grams of protein. Some might notice that the term wt%, while still encountered, is perhaps better expressed as a mass percentage. Albumin is said to bind to acidic drugs, but warfarin does not contain a carboxylic acid. According to DrugBank, a Canadian website that is full of reference information on different drugs, the OH group of warfarin has a pka of That value is very close to the pka of a typical carboxylic acid. We will use the DrugBank site more later in the course. If a drug is 99% bound to serum albumin, how can the remaining 1% have an adequate biological effect? If a drug is 99% bound by serum albumin, then 99% of the drug that is in the blood is bound to serum albumin. That does not mean, however, that 99% of the drug that is in the entire body is bound to serum albumin. Drug that has left the blood stream and entered other parts of the body can represent a significant amount of the total dose and will be available to bind the desired target and have a therapeutic effect.

6 6.2 ADME ADME video Please watch the online video (7 minutes, 39 seconds). A condensed summary of this video can be found in the Video summary page. Cell membranes Background: A high percentage of drugs are intended for oral delivery. In order to be absorbed, a drug must be able to cross a membrane to move from the intestinal tract to the bloodstream. Instructions: Read the passage on the structure of biological membranes. Learning Goal: To better understand why crossing membranes can be a challenge for a drug. Membranes are a lipid bilayer that separate one volume from another. A lipid bilayer consists of phospholipids. Phospholipids are a phosphorylated diglyceride. The diglyceride, with its two fatty acid chains, forms a non-polar double tail. The phosphate forms a charged group at the head of the molecule. Phospholipids can stack together as sheets with their polar ends together on one side and the nonpolar tails together on the other. In an aqueous environment, two sheets can stack together with their non-polar ends face-to-face. This stacking forms a bilayer that separates the aqueous layer on one side of the bilayer from the other side.

7 In order for an oral drug to be absorbed, a drug must leave the aqueous medium of the digestive tract, cross a non-polar bilayer, and enter the aqueous blood of the hepatic portal system. This poses a challenge to a drug discovery group. The drug must be soluble in the polar, aqueous medium of the digestive system. If the drug cannot mix with the digestive juices, then it will never be absorbed. A drug therefore must be somewhat polar. If the drug is very polar, however, it will likely never be able to cross the non-polar bilayer. A drug therefore must also be somewhat nonpolar. Designing a drug that falls within this Goldilocks zone of being a little polar and a little non-polar is a challenge. Remember that drugs are typically organic molecules that are designed to bind hydrophobic pockets on a protein target. In general, as a lead is optimized, the molecule's lipophilicity tends to increase as its activity (target binding) increases. The discovery team must continually push against this tendency in order to preserve reasonable water solubility in the lead. This process requires a very delicate balance Lipinski's rules Background: Most major drugs are delivered orally. Establishing how well a drug can be absorbed from a pill form requires considerable experimentation. Fortunately, simple methods have been developed to visually inspect a molecule and make a crude prediction of a compound's potential to be used as an oral drug. Instructions: Read the passage below concerning the structural features of a molecule that can affect a drug's absorption into the bloodstream from the digestive tract. Learning Goal: To be introduced to molecular indices, which are often used as predictive tools in drug discovery. In 1997 Chris Lipinski of Pfizer published a set of simple rules for predicting whether a molecule is likely able to diffuse across membranes and therefore be absorbed from the digestive tract and enter the bloodstream. 1 The rules became known as the Rule of Five because each "rule" involves a multiple of 5. The rules are also commonly referred to as Lipinski's rules. The rules are shown below. molecular property molecular weight (MW) 500 lipophilicity (log P) 5 hydrogen bond acceptors (HBA) 10 hydrogen bond donors (HBD) 5 maximum value

8 Lipinski's rules involve four different qualities of a molecule. Each of these qualities represents a different molecular index, which is an attempt to correlate structural or physical properties of a molecule to its behavior as a drug. The four molecular properties tracked by Lipinski's rules are all examples of molecular indices. The use of molecular indices to predict drug behavior is not perfect or infallible, but the simplicity of molecular indices makes them very attractive. The first molecular property associated with Lipinski's rules is molecular weight. Molecules with a molecular weight of over 500 have more difficulty crossing a membrane. Compounds with a MW of over 500 are therefore considered to have a lower probability of being absorbed from the digestive system. Lipophilicity is a polarity measure for a molecule. Less polar molecules are more lipophilic. The use of lipophilicity in Lipinski's rules emphasizes the fact that drugs must have reasonable solubility in the aqueous, polar environment of the digestive juices. If a drug cannot dissolve in this medium, then it will likely pass straight through the patient and not be absorbed. Although drugs are organic molecules and typically interact with non-polar binding sites on proteins, drugs must be somewhat water soluble. Formalized by the late Corwin Hansch 2, lipophilicity is measured as the base-10 logarithm of an equilibrium constant (P) for the partitioning of a drug in a biphasic system of 1-octanol (non-polar) and water (polar). A molecule that too strongly favors the octanol layer (log P > 5) is likely too non-polar to adequately dissolve in digestive fluids. Programs have been developed to predict log P values. These numbers are denoted as calculated log P or clog P (pronounced see-log). The number of hydrogen bond acceptors (HBAs) is, at most, equal to the number of oxygen and nitrogen atoms in the structure. The lone pairs on nitrogen and oxygen atoms are normally able to accept a hydrogen bond. Exceptions occur if the lone pair is involved extensively in resonance. Examples include amide nitrogens and nitrogens in aromatic rings that require the lone pair for aromaticity of the ring.

9 The number of hydrogen bond donors (HBDs) is determined by the number of O-H and N-H bonds in the structure. Keep in mind that O-H and N-H groups can be deprotonated depending on the ph of the surrounding environment. For example, a carboxylic acid contains an O-H bond, but in most parts of the body, the acid is deprotonated to a carboxylate. Therefore, the number of O-H and N-H bonds in a structure may not equal the exact number of HBDs in the same structure. Through both HBAs and HBDs a molecule can very strongly interact with water. While strong interactions with water are good for water solubility, they are not good for diffusing across membranes. A drug must shed its shell of water molecules as it crosses from an aqueous medium to the non-polar lipids of a cell membrane. The more strongly the water molecules interact with a drug, the harder (less energetically favorable) shedding water molecules is. Collectively, Lipinski's rules attempt the balance a drug's need to interact both with an aqueous environment as well as cell membranes. Lipinski's rules are frequently adjusted and criticized. Lipinski's rules have regardless become a fixture in drug discovery discussions. 1. Lipinski, C. A.; Lombardo, F.; Dominy, B. W.; Feeney, P. J. Experimental and Computational Approaches to Estimate Solubility and Permeability in Drug Discovery and Development Settings. Adv. Drug Dev. Rev. 1997, 23, Fujita, T.; Iwasa, J.; Hansch, C. A. A New Substituent Constant, π, Derived from Partition Coefficients. J. Am. Chem. Soc. 1964, 86, Please complete the online exercise. FAQ, help, and tips ADME video Are any other letters ever added to the ADME acronym? Sometimes an 'L' is added to the beginning to represent Liberation - the breaking down of the drug product (normally a pill) to free the drug substance (active pharmaceutical ingredient). More frequently, a 'T' is added to the end to represent Toxicity.

10 Is the first pass effect only associated with the liver? In addition to the hepatic first pass effect is the intestinal first pass effect. Some drugs are broken down by digestive enzymes, enzymes in the wall of the small intestine, and even enzymes of bacteria found in the digestive tract. A recent, freely available review on the topic can be found on the web. What are other common sites for excretion? Some drugs are excreted through the bile (biliary excretion). Less common but still encountered is excretion by the lungs through exhalation. Does the 'E' in ADME stand for excretion or elimination? In a search for ADME on the web, about half of the entries list excretion, and the other half are elimination. The terms are not identical, but both are functional for the purposes of our discussions. What are the different routes of administration recognized by the FDA? The routes are listed on a web page maintained by the FDA. Lipinski's rules When is a nitrogen atom not a hydrogen bond acceptor? There are three common situations in which a nitrogen atom will not act as a hydrogen bond acceptor. 1. The nitrogen is part of a nitro group. A nitrogen in a nitro group does not have a lone pair and therefore cannot participate in hydrogen bonding. 2. The nitrogen is next to a carbonyl (C=O) and therefore part of an amide. The electronwithdrawing nature of the carbonyl decreases the tendency of the lone pair to undergo hydrogen bonding. 3. The nitrogen is in an aromatic ring, and the lone pair is required for the ring to be aromatic. (If necessary, review the Khan Academy videos on aromaticity.) Nitrogens in an aromatic ring that

11 are not part of a double bond, normally have a lone pair that will not engage in hydrogen bonding. Some examples are shown below. Why are fluorine atoms not counted as HBAs? Fluorine atoms can indeed serve as strong hydrogen bond acceptors. For whatever reason, they are not included in the Lipinski count for hydrogen bond acceptors. Chapter 7 - Pharmacokinetics Introduction to Chapter 7 Chapter 7 contains eight subsections. IV Bolus - C p vs. time (begin during Week 3) Clearance I Clearance II Volume of Distribution I Volume of Distribution II Oral Delivery I Oral Delivery II (start of Week 4) Clearance and Volume of Distribution Revisited Upon completing this chapter, you should understand the mathematical relationships between drug concentration in the plasma and time for both IV and oral drugs. You will also understand the concepts of clearance and volume of distribution, which together determine the half-life of a drug. With knowledge of clearance and distribution, you should be able to propose changes to a drug's structure to shorten or lengthen the half-life as desired.

12 7.1 IV Bolus Cp vs. time video Please watch the online video (7 minutes, 52 seconds). A condensed summary of this video can be found in the Video summary page. Sample calculation - IV bolus Cp vs. time Background: C p vs. time relationships for most drugs follow a predictable, first-order relationship and show a linear relationship of ln C p vs. time. Instructions: Review the sample calculation demonstrating the use of C p vs. time data points to determine the half-life of a drug as well as the initial C p (C po ) value. Learning Goal: To learn how to use concentration-time data to determine a drug's properties. Task: Determine the half-life of the drug based on the provided data. C p (ng/l) time (h) Solution: Start by loading the C p-time data points into a spreadsheet (Apache Open Office shown). One column is for time and the other for C p.

13 The question explicitly asks for the half-life, so we need a linear form of the data. That means making an ln C p-time plot. The slope will give us k el. From there we can determine half-life. The next step is therefore to create and fill a column for ln C p. It is now linear regression time. LINEST is an easy option. The format in OpenOffice is LINEST(data_y;data_x;linear_type;stats). For data_y, highlight the cells for ln C p. For data_x, highlight the cells for time. For linear_type, enter a 1. For stats, enter a 0. Remember to finish the function by pressing ctrl-shift-enter to execute the function. Just pressing enter is not sufficient.

14 Upon pressing ctrl-shift-enter, two cells are filled. The one on the left is the slope, and the one on the right is the y-intercept. Formatting the cells to show more decimal places might be necessary to see the number of significant figures that you want. At this point, just run the values through the equations. So, the half-life of the drug is 13.9 h based on a k el of /h. Calculation of C p o requires use of the y-intercept of the best-fit line. The y-intercept is equal to ln C po. In order to convert ln C p o to C po, one must take the inverse natural log of the y-intercept. In most spreadsheet applications, the inverse natural log of a number is entered as EXP(x). In other words, if ln X = Y, then X = EXP(Y). For this question, ln C po = Therefore, C p o = e or EXP(4.605) in a spreadsheet. The question has been fully answered, but many people prefer to see a graph and trend line to determine the slope and y-intercept rather than use LINEST. Below is the graph. Right click on a data point, select a Format Trend Line..., select a Linear regression and Show Equation, and the

15 graph will display the best-fit equation for the data points. The slope and y-intercept should match the output of the LINEST function.

16 Graphing practice Background: Most C p vs. time relationships for drugs follow a predictable, first-order relationship and show a linear relationship of ln C p vs. time. Instructions: Answer the assessment questions below by graphing ln C p data points. Learning Goal: To gain practice graphing concentration-time data points to determine drug properties. Please complete the online exercise. FAQ, help, and tips Sample calculation - IV bolus Cp vs. time What is "ln"? Unfortunately, in the display font for this course, the mathematical operation for natural logarithm (ln) is indistinguishable from the capitalized preposition In. The term "ln" is shorthand for the base e logarithm of a number. In other words, ln x = log e x. In a spreadsheet one can take the natural logarithm of a number (X) by typing =LN(X) and pressing Enter.

17 Just as important for this chapter is the ability to take an inverse natural logarithm. If ln X = 3, then X = e 3 or In a spreadsheet, type =EXP(X) and then the enter key. Later in the course we will switch to base 10 logarithms. Those calculations will use the standard "log" notation instead of "ln". It is simpler to do the calculations by fitting the data to an exponential equation. That is true, and performing an exponential fit on Cp-time data in Excel and OpenOffice is fairly easy. Many students, however, will be using Google Docs, which does not have all the capabilities of other applications. Graphing practice If a drug has a larger ln C p o value, will the drug have a longer half-life? No, half-life is related to the slope of the line, not the value of the y-intercept. In fact, the ln C p o value of a drug is actually not a property of a drug. Instead, the ln C p o value is largely determined by the dose of a drug. We will cover this idea later in the chapter. Is one of the recommended spreadsheet applications better than the others? I would recommend that you use the one with which you are most comfortable. Personally, I like to look at the graphs of the data points, so I tend to avoid Google Docs. All of the graphs in the course are generated with OpenOffice Calc. Keep in mind that you do not need to generate any graphs. For all the graphing exercises, you only need to be able to determine the slope and y- intercept with LINEST or a similar function 7.2 Clearance I Clearance demo video Please watch the online video (7 minutes 50 seconds). A condensed summary of this video can be found in the Video summary page. Blood flow and extraction Background: Clearance is the process of removing drug from the bloodstream. As blood circulates through various tissues and organs, drug is removed.

18 Instructions: Read the passage below concerning how total clearance is the cumulative effect of clearance by the liver and kidneys. Learning Goal: To understand better how drugs are cleared from the body, especially with regard to the role of the liver and kidneys, and how the contributions from each can be determined. In an ln C p vs. time plot, the slope of line is equal to k el, the elimination rate constant. Back in Chapter 6.2, we mentioned elimination within the larger picture of ADME. Elimination was linked to metabolism and excretion, which are dominated by the liver and kidneys, respectively. By breaking down drugs and filtering drugs from the blood, the liver and kidneys play a key role in clearance. The elimination rate constant is directly related to the overall process of clearance, called total clearance (CL T), in the body. Total clearance, which has units of volume over time, describes the volume of blood that is scrubbed of drug per unit time. Most drugs are cleared almost exclusively by the liver and kidneys. Therefore, total clearance is the sum of the clearance caused by the liver (hepatic clearance, CL H) and the kidneys (renal clearance, CL R). Clearance at an individual organ is a function of two factors, blood flow (Q) and extraction ratio (E). Blood flow to an organ is simply the volume of blood that passes through an organ per unit time. Extraction ratio is a bit more complicated. If an organ clears a drug, the plasma concentration of a drug that enters the organ (C p in ) is higher than the plasma concentration of the drug that leaves the organ (C p out ). The difference between these two concentrations divided by C p in is the extraction ratio. Extraction ratio is a dimensionless number that falls within the range of 0 to 1. With these ideas in mind, one can re-express CL T as a combination of the blood flow and extraction ratio of both the liver and kidneys.

19 Values for Q are easy to handle. For a 70-kg human, blood flow to the liver is approximately 1,500 ml/min (Q H = 1,500 ml/min). Blood flow to the kidneys is around 1,100 ml/min, but the kidneys are only able to filter approximately 220 ml/min (Q R = 220 ml/min). The extraction ratios for the kidneys and liver are more challenging. Assuming we can determine CL T (to be covered in the next chapter section), we can estimate E H and E R. To approach these two variables, we need to put a drug into one of three categories: cleared by kidneys only, cleared by liver only, or cleared by both the liver and kidneys. Category #1 - cleared by liver only If a drug is not cleared by the kidneys (i.e., no drug is found in the urine because E R = 0), then the contribution of renal clearance to CL T can be ignored. CL T reduces to just hepatic clearance, CL H. If CL T is known and Q H = 1,500 ml/min, then we can calculate the hepatic extraction ratio, E H. In general, the hepatic extraction ratio is equivalent to the bioavailability (F) of an oral drug. For a drug that is well absorbed from the digestive system (assume 100% absorption), then the only barrier to a drug's reaching the bloodstream is liver metabolism. Bioavailability (F) will be reduced from 100% by the percent extraction of the liver (E H). So, if a drug is only cleared by the liver, then the hepatic extraction ratio can generally be determined through the drug's bioavailability. The key assumption is that the drug is well absorbed when administered orally. Category #2 - cleared by kidneys only If a drug is not metabolized by the liver (i.e., bioavailability, F, is very high because E H = 0), then the contribution of hepatic clearance to CL T can be ignored. CL T reduces to just renal clearance, CL R. If CL T is known and Q R = 220 ml/min, then we can calculate the renal extraction ratio, E R. The value we get for E R may not make sense. The underlying problem is that the kidneys filter 220 ml/min but they receive 1,100 ml/min. The two volumes of blood are not completely separate because of processes of reabsorption and secretion that can occur between the filtered and unfiltered blood in the kidneys. For this and other reasons, renal clearance is typically not broken down to Q R and E R. It is just reported as CL R.

20 Category #3 - cleared by both the liver and kidneys If a drug is cleared by both the liver and kidneys, then some outside information is required to tease apart CL R and CL H. One useful piece of information is oral bioavailability, F. Since E H = 1 F, E H can be calculated. With Q H known (1,500 ml/min), CL H may be determined as the product of E H and Q H. CL R is then just the difference between CL T and CL H. This entire discussion assumes that one can determine CL T. Calculation of CL T will be covered in the next section of Chapter 7 Illustrating clearance Background: Clearance is the process of removing drug from the bloodstream. Clearance is one of the fundamental properties of a drug that contributes to a drug's observed half-life. Instructions: Read the passage below, which outlines two approaches for visualizing the effect of hepatic clearance on a drug. Learning Goal: To appreciate exactly how drug is cleared by the liver and the relationships between the different pharmacokinetic variables. Diphenhydramine is an antihistamine that is frequently used to treat allergy symptoms such as watery eyes, nasal congestion, etc. According to Goodman and Gilman's The Pharmacological Basis of Therapeutics, diphenhydramine has clearance of 6.2 ml/min/kg. For a 70-kg patient, that equates to a clearance of approximately 430 ml/min as CL T. The urinary excretion of diphenhydramine is approximately 0, so renal clearance is negligible (CL R 0) and total clearance hepatic clearance (CL T CL H). Therefore, CL H = 430 ml/min. Hepatic clearance is 430 ml/min. We can work with number in two different ways. One is a literal interpretation, and the other is more realistic. While the literal interpretation is less accurate, both methods of interpretation give us the same result. All the discussions below focus on capturing one minute of blood flow through the liver. literal interpretation of hepatic clearance Hepatic clearance is 430 ml/min. In other words, every minute, of all the blood that passed through the liver (Q H = 1,500 ml/min), 430 ml of that blood is completely cleared of

21 diphenhydramine. If C p = 50 ng/ml, a typical concentration of diphenhydramine, then the liver receives blood with a concentration of 50 ng/ml (C p in = 50 ng/ml). In one minute, the volume received is 1,500 ml. In the literal picture of clearance, what exits the liver in one minute would be 430 ml of blood that is completely cleared of diphenhydramine (CL H = 430 ml/min). That 430 ml of blood has a concentration of 0 (C p out = 0 ng/ml). The balance of the blood exiting the liver (1,070 ml) still has the original concentration of 50 ng/ml. A picture of this literal interpretation is shown below. Note that blood that exits the liver is shown as having a drug concentration of either 50 ng/ml or 0 ng/ml. Perhaps the best way to evaluate any picture of an organ's clearance is to consider how much drug, by mass, enters and leaves the organ. For the picture above, the amount of drug entering the liver is equal to the concentration of the drug multiplied by the fluid volume. The amount of drug exiting the liver can be calculated in a similar manner. Note that we must use a correction factor of 0.54 in these calculations to convert the volume of whole blood to a volume of plasma because our concentration term references plasma. Keep these numbers - 40,500 ng into the liver and 28,890 ng out of the liver - in mind as we look at the more realistic interpretation of clearance.

22 realistic interpretation of hepatic clearance A more realistic picture for clearance is based around the extraction ratio of the liver (E H). How do we determine E H? E H can be calculated from the provided bioavailability (F). Assuming that diphenhydramine is fully absorbed from the gastrointestinal system, we know that F is the fraction of drug that manages to pass from the hepatic portal system to the general circulatory system by way of the liver. For diphenhydramine, F = 0.72, so 72% of the dose survives the first-pass effect. Therefore, 28% is broken down in the first-pass effect. This value, 28%, is the basis of E H. E H = Continuing our assumption that C p in = 50 ng/ml, we can determine C p out with the equation below. C p out = 36 ng/ml. In this picture of clearance, in the period of minute, 1,500 ml of blood enters the liver at a plasma concentration 50 ng/ml. The same blood volume exits the liver with a uniform plasma concentration of 36 ng/ml.

23 We can calculate the mass of drug that enters and leaves the liver in this more realistic picture. conclusion Under the literal interpretation of clearance, the mass of drug leaving the liver was 28,890 ng/min. Under the more realistic interpretation of clearance based on use of the hepatic extraction ratio, the mass of drug leaving the liver was 29,160 ng/min. These are essentially the same number and equivalent when significant figures are considered. Through the discussion above, hopefully you understand two different yet equal methods to consider the clearance of a drug from the bloodstream. FAQ, help, and tips Clearance demo video How do you know that the slope of the best-fit line for the demo should be ? The total volume of the beads and water was 2,000 ml. The "scooper" is a measuring cup with a volume of approximately 250 ml. Because the volume of the scooper is around 1/8th the volume of the total liquid volume, the slope (-k el) should be

24 The data in the demonstration seem a bit scattered? Yes, the demo data have plenty of scatter. When I practiced in my office, I got much cleaner data. I was rushing in front of the camera. Below are graphs of the practice run from my office.

25 Blood flow and extraction How do we determine bioavailability (F)? That is coming in Chapter 7.6. Illustrating clearance Why are there two different values for hepatic clearance for diphenhydramine in the passage? The method based on CL H of diphenhydramine of 6.2 ml/min/kg gives value of 430 ml/min (for a 70-kg patient). If CL H is determined through the bioavailability (F) of diphenhydramine, it comes out to 420 ml/min. While those numbers are not identical, they are roughly the same value 7.3 Clearance II Area under curve (AUC) video Please watch the online video (6 minutes 55 seconds). A condensed summary of this video can be found in the Video summary page. Two routes to AUC Background: The two ways for calculating area under curve (AUC) of a C p-time plot are integration of the C p-time curve and using the trapezoid approximation. Both methods provide a value for determining AUC, which can then be used to calculate total clearance. Instructions: Read the passage below concerning the advantages and disadvantages of both methods. Use this information to answer the assessment questions below. Learning Goal: To understand both the different approaches for calculating AUC and how to use both methods. Integration Integration is an easy method for determining the AUC of a particular drug dose. Pharmacokinetic data, whether from animals or humans, are found in the form of C p-time data points. As we have

26 seen in the assessment questions of Chapter 7.1, we can plot these points, especially in a linear ln C p vs. time form, and quickly determine k el and the hypothetical C po. AUC is simply C po /k el. Although integration is easy, it is not always accurate. The problem is that a series of ln C p-time data points can be forced to fit to a linear equation without the data actually being linear. If the data do not fit, then the simplicity of integration is of little gain because the AUC value will be inaccurate. Trapezoid Approximation Using the trapezoid rule or approximation sounds less than desirable. We want an exact number, not an approximation. Despite being an approximation, the trapezoid method does provide a very useful estimate of AUC. The trapezoid rule is applied to C p-time data by calculating the area between each C p-time data point and then adding all the areas for an AUC estimate. The area between two adjacent data points can be approximated as a trapezoid. The area is equal to the average of the two C p values multiplied by the time interval between the two points. The area for all the time intervals can be determined except for two special regions. One, the interval between time=0 and the first data point. Two, the area after the last data point.

27 For the initial trapezoid one needs a value for C po, which is a hypothetical value and must be estimated. One method is to extrapolate the ln C p data points of #1 and #2 backwards to C po. Once a value is chosen for C po, calculation of the area of the first trapezoid is the same as the others. With all the areas in hand, they can be added together to provide an AUC estimate. The area after the last data point also requires an estimation. In this case one needs to estimate k el for the last data points. Using the last two C p-time points, one can generate a line which has a slope of -k el. The AUC of the unplotted area can be estimated as C p last /k el. The advantage of the trapezoid rule is that the areas are based solely on the experimental data. The data points are not forced into a model that may not fit the data at hand. Practicing AUC Background: The two ways for calculating area under curve (AUC) of a C p-time plot are integration of the C p-time curve and using the trapezoid approximation. Instructions: Calculate AUC using both integration and the trapezoid approximation. Learning Goal: To practice the methods involved in both methods of AUC determination. Please complete the online exercise. FAQ, help, and tips Area under curve (AUC) video Please explain again how to determine the area of each trapezoid. Each trapezoid is defined by two Cp values, which we call Cp and Cpnext, and a time interval along the x-axis. To determine the area of each trapezoid, use the formula below. Note that the units on AUC will be concentration time. In discussions of drugs, concentration normally has units of mass/volume instead of mole/volume (molarity).

28 Two routes to AUC What is the relevance of AUC? AUC is a measure of a patient's overall exposure to a drug. A greater AUC from a drug dose equates to more drug exposure for a patient. While AUC is important for drug exposure, C p is also very important. AUC describes overall or total exposure, and Cp gives instantaneous exposure information. Instantaneous exposure through C p allows one to know whether the patient is receiving a toxic, therapautic, or ineffective amount of drug (based on the therapeutic window of the drug). The units on AUC (mass*time/volume) are not very intuitive. The units on AUC are related to the axes of a C p-time plot. The area of such a graph will always have units equal to the product of the units of the axes. Practicing AUC Please walk through determining AUC last. Determining AUC last is a challenge because the last region cannot be reduced to a simple trapezoid. In order to calculate AUC last, one needs a value for k el at the end of the data as well as the final C p point (C p last ). The problem provides C p last. Calculating the terminal k el value requires determining the slope between the final C p-time points (in ln C p form). Once k el and C p last have been found, AUC last can be calculated with the formula AUC last = C p last /k el. 7.4 Volume of Distribution I One-compartment model video Please watch the online video (5 minutes 33 seconds). A condensed summary of this video can be found in the Video summary page.

29 Detailing the one-compartment model Background: The one-compartment model is a pharmacokinetic model that describes the fluid that contains a drug in the body as a single volume of plasma. Although simple, the one-compartment model can be interpreted usefully to understand a drug's behavior. Instructions: Read the passage below concerning the V d of ranitidine. Learning Goal: To understand how to make sense of the numerical value of a drug's V d based on the one-compartment model. Ranitidine is a drug that can be used to treat heartburn, acid reflux, and occasionally ulcers. The V d of ranitidine is 1.4 L/kg, or 98 L for a 70-kg patient. To be clear, that is 98 L hypothetical volume of plasma. The volume is hypothetical because a 70-kg patient only has an actual plasma volume of 2.7 L (54% of the volume of whole blood). While ranitidine is normally administered orally, IV dosages of 25 mg are available. When administered by IV, D o = 25 mg. We know V d is 98 L. From this information we can calculate C p o to be mg/l. While the hypothetical V d may be 98 L, we know for certain that the actual volume of plasma in a 70-kg patient is only 2.7 L. We can therefore calculate how much of the original 25-mg dose is contained in the plasma at time=0. Wow! Of the 25-mg IV bolus dose, only 0.69 mg resides in the plasma. That means over 24 mg of the drug is NOT in the plasma. Where is it? It is elsewhere in the body - in the interstitial fluid, perhaps in muscles, even in fatty tissues. The drug is mostly outside the central compartment. We can model the drug as if it all resides in the plasma, but the fact is the drug, based on its properties, distributes into many parts of the body. Although V d is only a hypothetical number, it certainly provides a sense of how extensively a drug distributes out of the plasma and into other parts of the body. The more a drug leaves the plasma,

30 the less it is subject to hepatic and renal clearance because those organs only act on a drug that is contained in the plasma. Drugs with a higher V d take longer to be cleared from the body and have a longer half-life. Calculating Vd from Cp-time data Background: Volume of distribution is one of the key pharmacokinetic parameters for a drug. It describes the theoretical volume of plasma that is required to contain a drug in the body. Instructions: Use the C p-time data points below with the initial dose mass to determine the V d of a drug. Learning Goal: To practice how to calculate the V d of a drug from C p-time data points. Please complete the online exercise. Parameters by the kilo Background: Unlike other pharmacokinetic parameters, both CL and V d are reported on a per mass basis. Instructions: Read the passage below on the units of the most common pharmacokinetic parameters. Learning Goal: To understand why clearance and volume of distribution are listed as per unit mass. When drugs are reported in the literature and on the web, values for both CL and V d are listed per kilogram mass of the patient. CL is reported as ml/min/kg, and V d is listed as L/kg. These two variables are reported in this fashion because they both scale with a patient's body mass. A larger patient has a larger liver and kidneys and a higher rate of clearance. A larger patient also has a larger volume of distribution. What is the net effect of body mass on parameters like k el and t 1/2? Nothing. Because k el is defined as the ratio of CL and V d. The kg term cancels out, and both k el and t 1/2 are unaffected by patient mass. Rarely one will encounter a drug with a V d that is reported not based on patient mass but based on body surface area (BSA). The use of BSA in place of body mass is most commonly seen in certain types of cancer drugs. Almost all major oral medications list V d values based on body mass.

31 FAQ, help, and tips Detailing the one-compartment model How was a V d of 98 L determined? In this discussion, the V d of ranitidine was provided as 1.4 L/kg. It is very common for a drug to be listed with a V d on a per kilogram basis. The kg term does not refer to the drug but the mass of the patient. The "typical" patient has a mass of 70 kg. The observed V d of ranitidine in a 70-kg patient is 98 L (1.4 L/kg 70 kg). The implication is that V d for a drug changes from one patient to another. In a bigger patient a drug has a bigger V d, and in a smaller patient a drug has a smaller V d. This idea and its implications are handled later in this chapter in a section called Parameters by the kilo. Is the 0.69 mg of ranitidine in the plasma the amount that reaches the target? No. The target of ranitidine is the histamine-2 receptor found in the lining of the stomach. Therefore, the ranitidine found in the plasma is not available to the target. The other mg may be available to the target. The remaining drug might be found in the stomach lining (with the target), muscle tissue, fatty tissues, or other parts of the body. It seems unreasonable that the V d of the drug is instantaneously 98 L. It is unreasonable, and the V d of ranitidine is certainly not 98 L as soon as the IV bolus is injected into a 70-kg patient. The 98 L value is what the one-compartment model tells us, but the onecompartment model is not accurate at the time of administration. We will bring more reality into the discussion in Chapter 7.5. Does a drug with a higher V d have more side effects? It might. Side effects can be broken into two categories - on-target and off-target. An on-target side effect is an effect that is linked to a drug's intended target. Assume a drug binds a receptor to relieve pain symptoms. If that same receptor also affects appetite, the side effect of the drug will be appetite changes. Off-target effects arise from a drug that binds more than one target. The primary target gives the therapeutic effect, and the unintended secondary target causes the side effect (an off-target effect). If the secondary target is found in a different tissue from the primary target, then changes to a drug's V d might impact the side effects of the drug. For example, a drug might have an intended target (an enzyme?) in the plasma while also binding a target found in fatty tissues. If the drug can be designed so that it has a very small V d (restricted to the plasma) and only enters fatty tissues to a

32 very small degree, then its side effects might be minimized. This idea will somewhat reappear in Chapter 7.8. Is plasma the central compartment? Under the one-compartment model, the central compartment consists of plasma and its equivalents. It is because of the and its equivalents qualifier that the apparent V d of a drug can be larger than the volume of plasma found in a patient. In the next section (Chapter 7.5) we will raise the idea of multi-compartment models. In higher models, the central compartment is defined as the only the plasma. I think I am beginning to see the relationship between V d, CL, and kel. When I first learned about pharmacokinetics, I did not make it past the mathematical relationships between all the variables. To me, it was just a big math problem, and my job was to make sure the equations balanced. This was certainly true for the first couple times I taught medicinal chemistry at Davidson. Even though I was comfortable with the math, I had not fully connected the topics to medicinal chemistry. A turning point was in I attended a talk at the American Chemical Society meeting in Washington DC. The speaker was Don Middleton of Pfizer. The topic was not specifically V d and CL, but that's what I took from it. A large part of the talk focused on plots of CL vs. V d. I had never seen these plots before. (They will be introduced in Chapter 7.8.) Because I hadn't seen them, I figured they weren't very important. I later asked a couple folks I know in pharma, and they both said that CL vs. V d plots are used all the time. Whoops. I then started to think about these graphs and how they inform medicinal chemists in the design of new compounds. This is important because medicinal chemistry is not about modeling pharmacokinetics. Instead, medicinal chemistry is about designing molecules with desirable properties. Parameters by the kilo Why are some drugs dosed by a patient's surface area instead of mass? The therapeutic effects of some drugs are better managed for some drugs based on patient surface area, not mass. These drugs, which are very often for the treatment of cancer, do not fit the model that other drugs follow. This does not answer why some drugs are dosed by surface area, but it's maybe more surprising that dosing by mass works for so many drugs. We will cover some of the variability of patient populations and its impact on drugs in Chapter 8.

33 7.5 Volume of Distribution II Two-compartment model video Please watch the online video (6 minutes 48 seconds). A condensed summary of this video can be found in the Video summary page. Calculating half-life Background: Together V d and CL determine k el and t 1/2 of a drug according to the relationship below. Instructions: Use the reported V d and CL values to calculate the t 1/2 of the drugs listed below. Learning Goal: To become comfortable working with pharmacokinetic parameters and carefully watching the units on different values. Please complete the online exercise. Volumes of fluids in the body Background: Drugs are not confined to just the plasma. They can potentially distribute anywhere in the body. Instructions: Read the short section below for examples of different fluids in the body and their approximate volumes. Learning Goal: To learn how to interpret V d values of drugs and make reasonable predictions on how specific drugs distribute. Information very similar to this was presented back in section 1 of Chapter 6. It is worth repeating here since we have a better understanding of volume of distribution. plasma A 70-kg patient has 5 L of whole blood. Plasma is the non-cellular fraction of blood. At 54% of the volume of whole blood, the volume of plasma in a 70-kg patient is approximately 2.7 L. Because not