Download Welcome to Week 3 Chapter 6 - Blood and Drug Transport 6.1

 WelcometoWeek3
Startingweekthreevideo
Please watch the online video (58 seconds). OPTIONAL‐Please participate in the online discussion forum. Chapter6‐BloodandDrugTransport
IntroductiontoChapter6
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. OPTIONAL‐Please participate in the online discussion forum. 6.1Blood
Whatisblood?video
Please watch the online video (6 minutes, 4 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. Fluidsinthebody
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). 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 Cp. 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 0.071 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 Cb or Cwb. plasma ‐ 2.7 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 Cp. 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. 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. OPTIONAL‐Please participate in the online discussion forum. Serumbinding
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 KD 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. 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, α1‐acid glycoprotein, tends to bind basic drugs, including disopyramide and lidocaine, which are shown below. Since the concentration of α1‐
acid glycoprotein is lower than serum albumin, the degree of drug binding tends to be somewhat lower. OPTIONAL‐Please participate in the online discussion forum. 6.2ADME
ADMEvideo
Please watch the online video (7 minutes, 39 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. Cellmembranes
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 non‐
polar 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. In order to enter the bloodstream from the digestive tract, a drug must cross at least two bilayers. 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 non‐polar. 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. OPTIONAL‐Please participate in the online discussion forum. Lipinski'srules
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 help it to be absorbed to the bloodstream from the digestive tract. Use the information in the passage and the DrugBank to answer the assessment questions that follow. Learning Goals: To be introduced to molecular indices, which are often used as predictive tools in drug discovery, and to be able to search for this information online. 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 maximum value molecular weight (MW) 500 lipophilicity (log P) 5 hydrogen bond acceptors (HBA) 10 hydrogen bond donors (HBD) 5 Lipinski's rules involve four different qualities of a molecule. The first is molecular weight. Molecules with a molecular weight of over 500 have more difficulty crossing a membrane. 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. 2
Formalized by the late Corwin Hansch, lipophilicity is measured as the 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. 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, counting O‐H and N‐H bonds in a structure may be somewhat inaccurate method for determining the exact number of HBDs. 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. Indeed, most online databases on drugs include values for the various Lipinski rules. The DrugBank is an example. If one searches for a specific drug in DrugBank, the Information section (top of the page) lists the Weight (molecular weight) of the drug. The other Lipinski terms are shown further down the page in the Properties section. 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, 3‐25. 2. Fujita, T.; Iwasa, J.; Hansch, C. A. A New Substituent Constant, π, Derived from Partition Coefficients. J. Am. Chem. Soc. 1964, 86, 5175‐5180. Please complete the online exercise. OPTIONAL‐Please participate in the online discussion forum. Chapter7‐Pharmacokinetics
IntroductiontoChapter7
Chapter 7 contains eight subsections. 
IV Bolus ‐ Cp 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. 7.1IVBolus
Cpvs.timevideo
Please watch the online video (7 minutes, 52 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. Samplecalculation‐IVbolusCpvs.time
Background: Cp vs. time relationships for most drugs follow a predictable, first‐order relationship and show a linear relationship of ln Cp vs. time. Instructions: Review the sample calculation demonstrating the use of Cp vs. time data points to determine the half‐life of a drug. 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. Cp (ng/L) time (h) 90 2 82 4 67 8 55 12 45 16 30 24 Solution: Start by loading the Cp‐time data points into a spreadsheet (Apache Open Office shown). One column is for time and the other for Cp. The question explicitly asks for the half‐life, so we need a linear form of the data. That means making an ln Cp‐time plot. The slope will give us kel. From there we can determine half‐life. The next step is therefore to create and fill a column for ln Cp. 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 Cp. 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. 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. This satisfies the question, but I 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 graph will display the best‐fit equation for the data points. The slope and y‐intercept should match the output of the LINEST function. OPTIONAL‐Please participate in the online discussion forum. Graphingpractice
Background: Most Cp vs. time relationships for drugs follow a predictable, first‐order relationship and show a linear relationship of ln Cp vs. time. Instructions: Answer the assessment questions below by graphing ln Cp data points. Learning Goal: To gain practice graphing concentration‐time data points to determine drug properties. Please complete the online exercise. OPTIONAL‐Please participate in the online discussion forum. 7.2ClearanceI
Clearancedemovideo
Please watch the online video (7 minutes 50 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. Bloodflowandextraction
Background: Clearance is the process of removing drug from the bloodstream. As blood circulates through various tissues and organs, drug is removed. 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 Cp vs. time plot, the slope of line is equal to −kel, 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 (CLT), 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, CLH) and the kidneys (renal clearance, CLR). 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 (Cpin) is higher than the plasma concentration of the drug that leaves the organ (Cpout). The difference between these two concentrations divided by Cpin 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 CLT as a combination of the blood flow and extraction ratio of both the liver and kidneys. Values for Q are easy to handle. For a 70‐kg human, blood flow to the liver is approximately 1,500 mL/min (QH = 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 (QR = 220 mL/min). The extraction ratios for the kidneys and liver are more challenging. Assuming we can determine CLT (to be covered in the next chapter section), we can estimate EH and ER. 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 ER = 0), then the contribution of renal clearance to CLT can be ignored. CLT reduces to just hepatic clearance, CLH. If CLT is known and QH = 1,500 mL/min, then we can calculate the hepatic extraction ratio, EH. 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 (EH). 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 EH = 0), then the contribution of hepatic clearance to CLT can be ignored. CLT reduces to just renal clearance, CLR. If CLT is known and QR = 220 mL/min, then we can calculate the renal extraction ratio, ER. The value we get for ER 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 QR and ER. It is just reported as CLR. 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 CLR and CLH. One useful piece of information is oral bioavailability, F. Since EH = 1 − F, EH can be calculated. With QH known (1,500 mL/min), CLR may be determined as the product of EH and QH. CLR is then just the difference between CLT and CLH. This entire discussion assumes that one can determine CLT. Calculation of CLT will be covered in the next section of Chapter 7. OPTIONAL‐Please participate in the online discussion forum. Illustratingclearance
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 CLT. The urinary excretion of diphenhydramine is approximately 0, so renal clearance is negligible (CLR ≈ 0) and total clearance ≈ hepatic clearance (CLT≈ CLH). Therefore, CLH = 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. 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 (QH = 1,500 mL/min), 430 mL of that blood is completely cleared of diphenhydramine. If Cp = 50 ng/mL, a typical concentration of diphenhydramine, then the liver receives blood with a concentration of 50 ng/mL (Cpin = 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 (CLH = 430 mL/min). That 430 mL of blood has a concentration of 0 (Cpout = 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. realistic interpretation of hepatic clearance A more realistic picture for clearance is based around the extraction ratio of the liver (EH). How do we determine EH? EH 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 EH. EH = 0.28. Continuing our assumption that Cpin = 50 ng/mL, we can determine Cpout with the equation below. Cpout = 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. 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. OPTIONAL‐Please participate in the online discussion forum. 7.3ClearanceII
Areaundercurve(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. OPTIONAL‐Please participate in the online discussion forum. PracticingAUCestimation
Background: The two ways for calculating area under curve (AUC) of a Cp‐time plot are integration of the Cp‐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 Cp‐time data points. As we have seen in the assessment questions of Chapter 7.1, we can plot these points, especially in a linear ln Cp vs. time form, and quickly determine kel and the hypothetical Cpo. AUC is simply Cpo/kel. Although integration is easy, it is not always accurate. The problem is that a series of ln Cp‐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 Cp‐time data by calculating the area between each Cp‐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 Cp 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. For the initial trapezoid one needs a value for Cpo, which is a hypothetical value and must be estimated. One method is to extrapolate the ln Cp data points of #1 and #2 backwards to Cpo. Once a value is chosen for Cpo, 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 kel for the last data points. Using the last two Cp‐time points, one can generate a line which has a slope of ‐kel. The AUC of the unplotted area can be estimated as Cplast/kel. 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. Please complete the online exercise. OPTIONAL‐Please participate in the online discussion forum. 7.4VolumeofDistributionI
One‐compartmentmodelvideo
Please watch the online video (5 minutes 33 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. Detailingtheone‐compartmentmodel
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 Vd of ranitidine. Learning Goal: To understand how to make sense of the numerical value of a drug's Vd based on the one‐compartment model. Ranitidine is a drug that can be used to treat heartburn, acid reflux, and occasionally ulcers. The Vd 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, Do = 25 mg. We know Vd is 98 L. From this information we can calculate Cpo to be 0.255 mg/L. While the hypothetical Vd 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 Vd 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, 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 Vd take longer to be cleared from the body and have a longer half‐life. CalculatingVdfromCp‐timedata
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 Cp‐time data points below with the initial dose mass to determine the Vd of a drug. Learning Goal: To practice how to calculate the Vd of a drug from Cp‐time data points. Please complete the online exercise. OPTIONAL‐Please participate in the online discussion forum. Parametersbythekilo?
Background: Unlike other pharmacokinetic parameters, both CL and Vd 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 Vd are listed per kilogram mass of the patient. CL is reported as mL/min/kg, and Vd 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 kel and t1/2? Nothing. Because kel is defined as the ratio of CL and Vd. The kg term cancels out, and both kel and t1/2 are unaffected by patient mass. Rarely one will encounter a drug with a Vd 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 Vd values based on body mass. OPTIONAL‐Please participate in the online discussion forum. 7.5VolumeofDistributionII
Two‐compartmentmodelvideo
Please watch the online video (6 minutes 48 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. Calculatinghalf‐life
Background: Together Vd and CL determine kel and t1/2 of a drug according to the relationship below. Instructions: Use the reported Vd and CL values to calculate the t1/2 of a drug. Learning Goal: To become comfortable working with pharmacokinetic parameters and carefully watching the units on different values. Please complete the online exercise. OPTIONAL‐Please participate in the online discussion forum. Volumesoffluidsinthebody
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 Vd 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 everyone has a mass of 70 kg, the volume of plasma is very frequently reported as 0.039 L/kg. If a drug has a Vd of only 0.039 L/kg (which would be uncommon), then the drug would most likely be confined to the plasma. interstitial fluid Interstitial fluid is the liquid that sits between the cells of the body. The 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. A 70‐kg patient has 10 L of interstitial fluid. That is 0.14 L/kg. Oral drugs have full access to both the plasma and interstitial fluid and should therefore have a Vd of at least 0.18 L/kg (sum of volume of plasma and interstitial fluid). intracellular fluid The total water found within the cells of the body is 25 L for a 70‐kg person. That is about 0.36 L/kg. For a drug to enter cells, it must be able to cross cell membranes. Therefore, a drug that has access to cell water will also have access to the volume of all the cell membranes in the body. The Vd of such a drug is hard to predict, but it could certainly be much higher than 0.54 L/kg ‐ the sum of the plasma, interstitial fluid, and intracellular fluid. body volume The volume of the body of a 70‐kg human is approximately 70 L, or 1 L/kg. It may be surprising, but drugs regularly have Vd values in excess of 1 L/kg. Remember that Vd is a hypothetical value and not limited to a real physical space like the volume of the body. Drugs with very high Vd values are likely very hydrophobic and bury themselves into cell membranes and fatty tissues. OPTIONAL‐Please participate in the online discussion forum. PlasmaproteinsandVd
Background: Back in the video for section 2 of Chapter 6, a statement was made that plasma proteins can affect the pharmacokinetics of a drug. This statement has been left without justification until now. Instructions: Read the passage below on drug‐plasma protein binding. Learning Goal: To understand how plasma proteins can affect both the Vd and CL of a drug. Plasma can be a challenging medium for drugs. Drugs are designed to bind a protein ‐ typically a receptor or enzyme ‐ but drugs often bind the proteins in plasma. Because the concentration of protein in the blood is high, even a low affinity between a plasma protein and a drug can dramatically decrease the amount of unbound drug in the plasma. The concentration of a drug in plasma (Cp) includes both bound and unbound drug. The fraction of drug that is not bound to a protein (fu) is simply the ratio of the free drug to Cp. The binding of proteins in the plasma has a direct impact on both clearance and volume of distribution. impact on clearance Drug clearance is primarily attributed to the action of the liver and kidneys. Enzymes in the liver metabolize drugs. Liver enzymes only interact with drugs that are not bound to plasma proteins. Protein binding therefore protects a drug from clearance in the liver. Similarly, the kidneys do not filter proteins from the blood because the proteins are too large to be removed. Drugs that are bound to the blood proteins are also not filtered by the kidneys. Protein binding therefore protects a drug from clearance by both the liver and kidneys. Drugs with a higher plasma protein binding, or lower fu, show a lower rate of clearance. impact on volume of distribution Plasma proteins are only found in the circulatory system. If a drug is highly bound to plasma proteins, then that drug will be largely confined to the volume of the plasma. This binding can greatly reduce a drug's Vd. overall impact The overall impact of plasma protein binding must be interpreted on a case‐by‐case basis. Both Vd and CL are decreased by protein binding, so the overall effect on kel and t1/2 cannot be immediately predicted. Regardless, protein binding does affect the pharmacokinetics of drugs and is monitored by drug discovery groups within a pharmaceutical company. 7.6OralDeliveryI
Oralbioavailabilityvideo
Please watch the online video (7 minutes 9 seconds). A condensed summary of this video can be found in the Video summary page. OPTIONAL‐Please participate in the online discussion forum. EstimatingAUCforanoraldrug
Background: The calculation of area under curve (AUC) for a given dose of a drug is the first step for determining CL for a drug. Instructions: Use the Cp‐time data points for an oral drug to determine the AUC of the dose. Learning Goal: To understand how to use the trapezoid rule to approximate AUC for an oral drug. Please complete the online exercise. OPTIONAL‐Please participate in the online discussion forum.