Intro to Structure Part 1 Amino Acids, Primary Structure, and Secondary Structure

Transcription

1 Welcome to Week 2 Starting week two video Please watch the online video (1 minutes 24 seconds). Chapter 3 Protein Structure Introduction to Chapter 3 Chapter 3 contains two subsections. Intro to Structure Part 1 Amino Acids, Primary Structure, and Secondary Structure Intro to Structure Part 2 Tertiary Structure, Quaternary Structure, and X-Ray Crystallography At the conclusion of this chapter, you should understand how the ordering of individual amino acids in a protein can affect the localized and global folding and function of the entire protein. You should further have some appreciation of how x-ray crystallographic data is used to determine the structure of proteins. 3.1 Intro to Structure Pt 1 Amino acids to secondary structure video Please watch the online video (8 minutes, 15 seconds). Clarifications and corrections The video refers to the purple-colored helices in protein structure files as left-handed helices. This is incorrect. The purple helices are a different type of alpha-helix. Refer to the FAQ, help, and tips unit for more information. The structure of leucine has a mistake in the R-group at 1:02. The R-group should be CH 2CH(CH 3) 2.

2 PDB files can be inspected visually with Jmol. PDB entries also include considerable statistical information. The FAQ, help, and tips unit contains more information on how to find this information. A condensed summary of this video can be found in the Video summary page. Technical advice for the Protein Data Bank Background: The Protein Data Bank is a repository of structural information on proteins. Instructions: Read the passage below concerning how to make sure your computer is able to run Jmol, the application that allows three-dimensional visualization of proteins in the Protein Data Bank. Learning Goals: To make sure your computer can fully access the Protein Data Bank. Jmol Tips 1. Make sure your version of Java is up-to-date: You can check if you have the latest version of Java through this page on Oracle's Java site. 2. Check the compatibility of your browser with Jmol: Follow this link to a browser checker. 3. Note for Mac users: Jmol and Safari do not seem to work well together. We suggest using Firefox or Chrome to visualize PDB files with Jmol. 4. Ask for help on the discussion board: If all else fails, ask for help. Either the DavidsonX team or perhaps some of the technically savvy members of the course will lend a hand. Working with Protein Data Bank entries Background: The structural information freely available in the Protein Data Bank is immense. It allows anyone to be able to study protein structure and function. Instructions: Use the tools in the PDB to examine the structures mentioned below and answer the assessment questions. If the questions seem difficult, be certain to check the FAQ, help, and tips unit for assistance. Learning Goals: To learn how to manipulate proteins and identify their structural elements with the tools in the PDB. Please complete the online exercise.

3 Finding Protein Data Bank entry codes Background: Searching a specific entry in the Protein Data Bank requires one to know the corresponding PDB code. Instructions: Read the passage below concerning how PDB codes can be determined. Learning Goals: To learn how to find specific, useful information within the vast PDB database. The Protein Data Bank is only useful if one can find information of interest. As we work through this chapter, some students might wonder how to find other proteins. In the videos, I start with a PDB code. What if you do not already have a code? Searching the PDB requires one to have an idea of what is being sought. For the different examples in this course, I knew that I wanted to find proteins that are rich in one type of secondary structure. I performed an online search using terms like "proteins rich in alpha-helices" and found a few names of proteins that seemed to match what I wanted. I then searched on those particular proteins within the PDB. Any single protein may have multiple entries in the PDB. I then looked through the different entries until I found a specific PDB entry that demonstrated the properties I want to highlight. Some websites on the internet include PDB codes. One example is SCOP: Structural Classification of Proteins, an extensive site that can be searched by a number of keywords that correspond to common traits of proteins. Most proteins in SCOP are linked to a PDB entry. FAQ, help, and tips Amino acids to secondary structure video Are there other drug targets aside from proteins? Proteins - enzymes and receptors - are far and away the most common drug targets, and they are the only targets covered in this course. Oligonucleotides like DNA and RNA are much less common, although many anticancer drugs target DNA. On very rare occasions, the target of a drug is not a molecule of the body at all. Chelators, which bind to toxic metals in the body, do not act upon a typical, biomolecule target. Examples of

4 chelators include dimercaptrol, 2,3-dimercapto-1-propanesulfonic acid, and dimercaptosuccinic acid. All contain two thiol (SH) groups for strong interaction with toxic metals. Are there other tools in the PDB besides the pretty pictures? The main page for each PDB entry contains many tools for learning about a protein's structure. The Jmol visualizer was discussed in the video. The main page contains lots of information. Look down the page to the second section named Molecular Description. A horizontal graphic next to the word "Secstruc" shows the different secondary structures by color - Β-sheets in pale yellow and α-helices in red. Across the top of the screen are a number of tabs. Clicking on the Sequence tab reveals a page full of protein information, including the number of residues and a breakdown of secondary structures. 3.2 Intro to Structure Pt 2 Tertiary structure to X-ray crystallography video Please watch the online video (7 minutes, 41 seconds). A condensed summary of this video can be found in the Video summary page. Validating protein structures Background: Most protein structures are determined based on x-ray crystallographic data. The primary sequence of the protein is matched with the electron density map, and the individual amino acids are placed within the structure as closely as possible. After all the amino acids are positioned, the quality of the assigned structure can be measured with several tools. One of the most common tools is the Ramachandran plot.

5 Instructions: Read the passage below about the use of Ramachandran plots to validate protein structural assignments. Learning Goal: To learn how Ramachandran plots graphically represents dihedral angles of individual amino acids to predict the validity of a proposed protein structure and folding. The Ramachandran plot is one of the primary methods for validating proposed protein structures based on x-ray crystallographic data. The plot compares selected dihedral angles in each amino acid found within the proposed protein. The key dihedral angles for each amino acid are located along the backbone of the protein and are labeled φ (phi), ψ (psi), and ω (omega). To repeat, each amino acid residue contributes three rotatable bonds and three distinct dihedral angles to the backbone of a peptide chain. In theory, all dihedral angles can range in value from 180 to In practice, within a protein, the dihedral angles tend to fall in well-defined ranges. Because of interactions between the nitrogen and carbonyl, ω is either 0 or 180, typically 180. φ has a value near -50 in an α-helix and ranges between -50 and -160 in a β-sheet. ψ also has a value near -50 in an α-helix but ranges from +100 and +180 in a β-sheet. When ψ is plotted against φ for each amino acid, most amino acids fall within tightly defined regions bounded by the angle ranges above and another small area for a specific subtype of α-helix. Such a graph is called a Ramachandran plot, shown below.

6 While keeping track of dihedral angles in a protein may seem complex, the Ramachandran plot makes the process very simple. In the process of validating a protein's assigned structure, the amino acids that fall outside of the expected, shaded regions of the Ramachandran plot are of greatest concern. If a plot has too many of these outliers, then the structure of the protein may be incorrectly assigned. In general, any protein with over 5% of its amino acids as outliers should be treated with suspicion. For this reason, the Ramachandran plot is a simple, visual tool for quickly checking the validity of an assigned protein structure. Evaluating Ramachandran plots Background: Ramachandran plots are a simple, visual tool for validating proposed protein structures. A website that allows users to generate Ramachandran plots from many PDB entries is the Ramachandran Server at Uppsala University in Sweden. Instructions: Compare the Ramachandran plots below to answer the questions. Learning Goals: To learn how to read and compare data from Ramachandran plots. Please complete the online exercise. FAQ, help, and tips Tertiary structure to x-ray crystallography video Do all parts of an x-ray structure have the same resolution? No. Not all parts have the same resolution. In proteins, the parts that are most important for the function of the protein in the body tend to be contained in more rigid regions. Regions of high rigidity tend to give a higher resolution x-ray structure. Parts that are less important for the protein's function are often more flexible and give a lower resolution in the x-ray structure. Are there proteins with structures that have not been solved? Yes. Many, many proteins have unknown structure. In fact, there is a website called Foldit that invites members of the scientific community to work with protein sequences as puzzles and try to solve structures of stubborn proteins.

7 Validating protein structures Why not make a 3D plot with all three of the dihedral angles of the protein backbone? Researchers do not use all three dihedral angles because all three angles are not helpful for distinguishing the folding of the protein backbone. The angle ω, because of resonance between the nitrogen and carbonyl, tends to have values of either 0 or 180. Those two values are not helpful for determining or validating protein structure. Evaluating Ramachandran plots Is there an easy way to evaluate a Ramachandran plot? Each plot generated by the Ramachandran Server at Uppsala University lists the percentage of outliers at the bottom. Sometimes one can look at the plot and tell something is amiss. Other times it is more convenient just to look at the bottom of the plot. Are there other ways to generate a Ramachandran plot? The PDB can generate Ramachandran plots. They do not have the classic background shading to show outliers, but similar ideal regions are marked. To use the PDB function, go to a PDB entry and select the Geometry tab at the top of the screen. Within the Geometry tab is an option to download a Ramachandran plot. Selecting the download brings up a new window with four different plots. The first one is the standard Ramachandran plot. Why is glycine treated differently in Ramachandran plots? Glycine is an amino acid with an R-group of just hydrogen. The simplicity of glycine allows the residue to assume dihedral angles that other amino acids do not. Therefore, glycine residues fall outside the standard regions of a Ramachandran plot. The special nature of glycine as well as proline has been discussed in the literature.

8 Chapter 4 Enzymes Introduction to Chapter 4 Chapter 4 contains three subsections. Michaelis-Menten Kinetics Enzyme Inhibition Measuring Inhibition At the conclusion of this chapter, you should understand how enzyme kinetics data are presented graphically. You should also understand how different inhibitors affect enzymes, and how the inhibition is quantified. 4.1 Michaelis-Menten Kinetics Theory of action video Please watch the online video (7 minutes, 8 seconds). A condensed summary of this video can be found in the Video summary page. Working with concentrations Background: Data in medicinal chemistry, including enzyme kinetics data, rely upon numbers with various concentration units. Being comfortable with interconverting different concentration units is a basic skill for one to have. Instructions: Read the passage below and use the information to answer the subsequent assessment questions. Learning Goal: To become comfortable working with the different types of units commonly encountered in medicinal chemistry. The goal of a drug discovery program is generally to find a molecule that binds a target protein at very low concentrations. As has been mentioned before (in Chapter 2), the binding is normally determined in a biochemical assay, often in the form of a dissociation equilibrium constant (K D). For a drug, the values for K D are very small, indicating the drug and target bind very tightly and do not readily dissociate. Ideal K D values are in the nanomolar (nm) range, but during development observed K D values are much higher. Hits in an early screen might have K D values in the micromolar

9 (µm) range. The table below shows the concentrations regularly encountered in a drug discovery program. name description unit relation to molarity molar moles / liter M 1 millimolar millimoles / liter mm 10-3 micromolar micromoles / liter µm 10-6 nanomolar nanomoles / liter nm 10-9 picomolar picomoles / liter pm Beyond the reporting of binding data (pharmacodynamics), the units in the table above are also found throughout pharmacokinetics, especially in reports of the concentration of drugs in blood. Please complete the online exercise. Using a spreadsheet application Background: Interpreting enzyme kinetics data requires one to be able to perform a linear regression on the information. Instructions: Watch the videos below on performing linear regressions on x-y data points in Google Docs, Apache OpenOffice, and Microsoft Excel. Learning Goal: To how to use common spreadsheet applications to determine the slope and y- intercept from linear data points. Linest in Google Docs spreadsheet video Please watch the online video (3 minutes 43 seconds). Linest in OpenOffice spreadsheet video Please watch the online video (4 minutes 9 seconds). Linest in Microsoft Excel video Please watch the online video (3 minutes 45 seconds).

10 Sample calculation - Lineweaver-Burk plot Background: Lineweaver-Burk plots depict 1/V vs. 1/[S]. These are very useful for determining the nature of enzyme-substrate interactions. Instructions: Review the sample calculation demonstrating the use of V-[S] data points to calculate V max and K m to three significant figures and generate a Lineweaver-Burk plot. Learning Goal: To learn how to use V-[S] data points to determine V max and K m of an enzymesubstrate system. Task: Determine V max and K m based on the provided data points. V (mmol/min) [S] (mmol) Solution: Start by loading the V-[S] data points into a spreadsheet (Apache Open Office shown). One column is for [S] and the other for V. For the Lineweaver-Burk plot, we need the reciprocal of both V and [S].

11 We can use LINEST to perform the regression of the 1/V-1/[S] data points. The format in OpenOffice is LINEST(data_y;data_x;linear_type;stats). For data_y, highlight the cells for 1/V. For data_x, highlight the cells for 1/[S]. For linear_type, enter a 1. For stats, enter a 0. Remember to finish the function properly based on your spreadsheet application. In OpenOffice, press ctrl-shift-enter to execute the function. With execution of the function, two cells are filled. The one on the left is the slope, and the one on the right is the y-intercept. I manually added the slope and Y-intercept labels for clarity. From the y-intercept, V max can be directly determined. The slope can then be used to determine K m. The instructions call for the calculation of V max and K m to three significant figures, but the calculations above use all the digits provided by the spreadsheet application. Do not round numbers during your calculations even though the instructions might call for an answer with fewer significant figures or decimal places. Rounding early in a calculation can introduce error that

12 are then carried through the rest of the calculation. The instructions do ask for a specific number of significant figures, so the final answers are provided to comply with the instructions. V max is 12.3 mm/min, and K m is 225 mm. A graph can be more satisfying than the LINEST function. The graph below shows the data in a Lineweaver-Burk format. 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.

13 Graphing practice Background: Interpreting enzyme kinetics data requires one to be able to graph the information. An earlier unit contained videos with instructions on how to generate graphs from enzyme kinetics data in linear form (Lineweaver-Burk equation). Instructions: Use the data below and a spreadsheet application to determine the K m and V max of an enzyme-substrate system. The sample calculation in the previous unit may be helpful. If the question gives you trouble, consider looking back to see one way to approach this problem. Learning Goal: To learn how to graph kinetics data and use the resulting plot to understand the activity of an enzyme. Please complete the online exercise. FAQ, help, and tips Theory of action video Is Km the concentration at which half of the enzyme is bound with the substrate? Yes, when the system reaches 1/2V max (when [S] = K m), half of the enzyme is bound with substrate. The video does not provide a full derivation of the Michaelis-Menten equation, but the fractional binding of the enzyme by the substrate is directly proportional to the rate of conversion.

14 Why bother with linearized forms when modern software can fit hyperbolic functions? Many software packages can directly fit hyperbolic Michaelis-Menten kinetics data. Not everyone has access to these applications, and many of those applications are not free. How can I use LINEST in Excel with a Mac? F2-Ctrl-Shift-Enter is definitely a Window type of a key sequence. The University of Colorado Department of Chemistry hosts a file (pdf) that details using LINEST in Excel on a Mac. Using a spreadsheet application What does the zero signify at the end of the argument list of LINEST? The zero at the end signifies that the analysis of variance (ANOVA) statistics should not be shown with the slope and y-intercept. LINEST is actually a very powerful little function. It can generate information like the regression coefficient and much more. Sample calculation - Lineweaver-Burk plot Are there other functions that will give the slope and y-intercept? Yes, there are two functions. The slope can be found with the SLOPE function. The intercept can be found with the INTERCEPT function. Both functions call for the y- and x-data cells. I need help on inverting the values. An easy way to invert a value works as follows... If the value you want to invert is in cell A1. Move to an empty cell ad type "=1/A1" and press Enter. The value in the new cell will be the reciprocal of A1. How are V values determined experimentally? Each V-[S] data point is found through a separate experiment. The enzyme and substrate are mixed and the rate of the reaction is measured at the very start. This initial rate is V o. It is important to measure the initial rate because the rate will slow over time. The value of V o for a particular [S] represents a single V-[S] data point.

15 Graphing practice Help. I can't get a correct answer for these graphs. There are a few common mistakes. First, make sure all your V and [S] values are correct. Second, make sure your inverted values are correct. Third, in your LINEST (or SLOPE and INTERCEPT) function, make sure you properly select all the appropriate x- and y-values. Why can't I get a trendline to appear on my graph? Are you using Google Docs? Google Docs Spreadsheet does not currently have the option of adding a best-fit trendline to a plot. The answers do not always match the number of significant figures as the data. Why? In this course, all the answers are machine graded. In order to make sure the grader works properly, the questions often ask for a number of significant figures or decimal places that are unreasonable. Note that a student can enter a number with 50 significant figures, and the grader will not blink. All that matters is that the student's answer matches the given answer or falls within the allowed range. What is the importance of K m? K m is very important in enzymology, but it is not very interesting for medicinal chemistry. The idea of K m is as a measure of the affinity between an enzyme and a substrate. This idea of affinity - a binding affinity - is very important in medicinal chemistry. We will see many different K values in this course as we talk about drugs, but K m will not be one of them. Are international versions of Excel different from the US version? Some students have noted that changing the Excel to US settings can minimize problems. To change your settings to the US, go to "File" (upper left) -> "Spreadsheet Settings" -> Change locale to "United States" and save the settings

16 4.2 Enzyme Inhibition Reversible inhibitors video Please watch the online video (7 minutes 12 seconds). A condensed summary of this video can be found in the Video summary page. Distinguishing types of enzyme inhibitors Background: Inhibitors can be readily distinguished by visually inspecting plots of V vs. [S] at varying inhibitor concentrations ([I]) and noting the effect of the inhibitor upon K m and V max. In the previous video clip, we observed how an inhibitor changes the shape of a traditional Michaelis-Menten-type plot. Instructions: Use the Lineweaver-Burk plots below to classify the type of inhibitor present in each system. Remember that in a Lineweaver-Burk plot, K m can be observed by looking at the x-intercept (1/K m), and V max can be noted from the y-intercept (1/V max). Learning Goal: To learn how to interpret effects of an inhibitor upon Lineweaver-Burk plots. Please complete the online exercise. FAQ, help, and tips Reversible inhibitors video Where does an uncompetitive inhibitor bind the E-S complex? When a substrate binds an enzyme, the conformation of the enzyme may change somewhat. Therefore, the E-S complex has a different shape and potentially different binding sites. One of these different binding sites in the E-S complex is where an uncompetitive inhibitor binds the E-S complex. Distinguishing types of enzyme inhibitors Reading a Lineweaver-Burk plot is harder than reading a Michaelis-Menten plot. Lineweaver-Burk plots are difficult because they involve reciprocal variables. As an axis variable (like 1/[S] of the x-axis) increases in value, it moves closer to the origin rather farther away. This is not a hard concept, but it does require more concentration.

17 If Lineweaver-Burk plots are harder to read, why are they the subject of one of our exercises? Michaelis-Menten plots are excellent for comparing the V max of one system to another. Michaelis- Menten plots are not as useful for comparing K m values, which tend to be tightly compressed at low values on the x-axis. The Lineweaver-Burk is much better for visualizing changes in K m. 4.3 Measuring Inhibition IC50 and Ki video Please watch the online video (6 minutes 41 seconds). A condensed summary of this video can be found in the Video summary page. Determining inhibition Background: Measuring the affinity of an inhibitor for an enzyme is a fundamental part of preliminary screening in a drug discovery program. Inhibitor-enzyme affinity is normally reported as either a K i or IC 50 value. Both are related to the potency of an inhibitor. A smaller K i or IC 50 value indicates that the inhibitor has a stronger affinity for the enzyme. Instructions: Use what you know concerning K i and IC 50 values as well as the Cheng-Prussoff equation to answer the questions below. Learning Goal: To practice working through enzyme inhibition data. Please complete the online exercise.

18 Cheng-Prussoff equation Background: The Cheng-Prussoff equation is a useful mathematical relationship for either converting an IC 50 value to a K i value or comparing IC 50 values determined under different experimental conditions. Instructions: Use the Cheng-Prussoff equation to answer the questions below. Learning Goal: To gain experience converting IC 50 values to K i values. Please complete the online exercise. FAQ, help, and tips IC50 and Ki video How are K m and K i values determined? Are they different? K m relates to the affinity of an enzyme for a substrate. K m is actually approximately equal to the dissociation equilibrium constant (K D) of the enzyme-substrate complex. (We will learn much more about K D values in Chapter 5.) K m can be determined by analysis of kinetic data from an enzyme that obeys Michaelis-Menten kinetics. This was the subject of Chapter 4.1. K i is instead the dissociation equilibrium constant (K D) of an enzyme-inhibitor complex, and normally apply to a reversible competitive inhibitor. K i values are often determined from IC 50 values and the Cheng-Prussoff equation. IC 50 values are determined through inhibition studies on enzymatic reactions as shown in Chapter 4.2. K m and K i values are very closely related because both give information about the complexation of an enzyme with another molecule, either the substrate or an inhibitor. They differ in that K m applies to the substrate while K D applies to an inhibitor.

19 If I find an IC 50 value in an online database, is that number worth anything? Certainly it is worth something. That values gives you an approximate idea of how potent of an inhibitor a molecule, and that information definitely has value. The only problem is that you cannot directly compare two molecules from different databases based on their IC 50 values. For direct comparison, you must have a K i value for each Determining inhibition I am trying to make the graph, but I get an error when I take the reciprocal of the first data point. This question is not a Lineweaver-Burk problem, so you do not take the inverse of any of your data points. This question only requires you to plot K m obs vs. [I]. Why calculate K m when the data points already give that value? The first data point does give K m obs when [I] = 0, which is the same as K m. This value for K m is based on a single data experimental data point. The value that comes from the best-fit line arises from a series of experiments and should be a more reliable value. Why is smaller better for K m and K i values? Both are essentially K D values. A strong inhibitor binds the enzyme well, so the enzyme-inhibitor complex concentration will be high, and the free enzyme concentration will be low. This situation will result in small values for K i. The smaller K i is, the stronger the inhibitor binds to the enzyme, and presumably the more potent the inhibitor will be. How does one determine K m obs? K m obs is determined in nearly the same way as K m. For K m, an enzymatic reaction with a known [S] is set up, and the initial rate (V) of the reaction is monitored. This process is repeated at several different [S] values. The different V-[S] data points form a Michaelis-Menten graph which can be analyzed for K m. For K m obs, the series of enzymatic reactions is set up in the same way (varying [S]) all the reactions will have a constant concentration of an inhibitor [I]. The different V-[S] data points form an inhibited Michaelis-Menten graph which can be analyzed for the apparent K m of the system. This apparent K m is called K m obs.

20 Chapter 5 Receptors Introduction to Chapter 5 Chapter 5 contains three subsections. Types of Receptors Ligands Occupancy Theory At the conclusion of this chapter, you should know the different types of receptors and what structural features distinguish one from another. You should also know the types of receptor ligands and be able to identify each based on a response vs. log [L] graph. Finally, you should understand the fundamental assumptions of occupancy theory and situations in which the assumptions fail. 5.1 Types of Receptors Receptor superfamilies video Please watch the online video (6 minutes 26 seconds). A condensed summary of this video can be found in the Video summary page. PDB example of an ion channel Background: Ligand-gated ion channels are membrane-bound receptors. All membrane-bound receptors are characterized by a large number of parallel α-helices. The α-helices allow the protein to criss-cross the cell membrane and anchor itself in place. Instructions: In a separate window, bring up the Protein Data Bank website and look up entry 4HFH. Learning Goals: To appreciate the overall structure of ligand-gated ion channels and gain more experience examining proteins in the Protein Data Bank. Please access the online Protein Data Bank website. Please complete the online exercise.

21 PDB example of a G-protein-coupled receptor Background: G-Protein-coupled receptors are membrane-bound receptors. All membrane-bound receptors are characterized by a large number of parallel α-helices. The α-helices allow the protein to criss-cross the cell membrane and anchor itself in place. Instructions: In a separate window, bring up the Protein Data Bank website and look up entry 2RH1. Learning Goals: To appreciate the overall structure of G-protein-coupled receptors and gain more experience examining proteins in the Protein Data Bank. Please access the online Protein Data Bank website. Please complete the online exercise. 5.2 Ligands Ligand types video Please watch the online video (7 minutes 40 seconds). A condensed summary of this video can be found in the Video summary page. Digging deeper into antagonists Background: Antagonists are ligands that block the action of an agonist without causing a response from a receptor. Different subtypes of antagonists are known to exist. Instructions: Read the passage below on competitive and noncompetitive antagonists. Learning Goals: To learn about the two types of antagonists and understand how they are distinguished in response curves. The two types of antagonist, competitive and noncompetitive, are distinguished by how the antagonist binds the receptor. A competitive antagonist binds at the same site as an agonist. A

22 competitive antagonist increases the EC 50of any agonist present without decreasing the maximum response. By increasing the EC 50 of the agonist, the antagonist decreases the agonist's potency. A noncompetitive antagonist binds at an allosteric site. Such binding does not affect the EC 50 value (potency) agonist, but it does diminish the response caused by the agonist. Depending on the needs of the discovery program, one type of antagonist might be more effective than the other. Competitive antagonists, however, tend to be more common. If the drug discovery group knows the structure of a receptor's endogenous ligand, then the team often designs antagonists to resemble the natural ligand. The designed antagonist will likely the bind the same site as the endogenous ligand.

23 Interpreting dose-response graphs Background: Dose-response relationships are commonly encountered in drug development programs. The characteristic sigmoidal plots convey a large amount of information concisely and can be readily interpreted. Instructions: Read the passage below and interpret the dose-response graphs that follow. Each graph shows how variation of a ligand concentration (log [L] on x-axis) affects the response from a receptor. Learning Goals: To gain experience analyzing dose-response relationships and determining how different types of ligands affect dose-response graphs. Please complete the online exercise. FAQ, help, and tips Ligand types video How can a receptor spontaneously generate a response with binding to a ligand? Receptors that show constituent activity are perhaps best explained through something called the two-state theory. In the two-state theory, a receptor exists in an equilibrium. The equilibrium is between two conformations. One conformation is the relaxed form (R), and the other is the tensed form (T). The relaxed form of the receptor does not cause a response, and the tensed form does. For most receptors, the equilibrium strongly favors the relaxed form (R). The concentration of the tensed form is negligible, so the response from the tensed form is negligible (E 0). For some receptors - constituently active receptors - the equilibrium allows a relatively high amount of the tensed state and the response is greater than 0. A "relatively high amount" normally means that the response level from constituent activity is low, typically 20% or less of E max. Where does the ligand fit into this picture? In the two-state model, binding of an agonist to the receptors shifts the position of the equilibrium to favor the tensed form. A higher concentration of tensed form gives a higher response. An antagonist binds both the R and T forms in a way that does not change the equilibrium, so an antagonist has no effect on the response. An inverse agonist binds and causes the equilibrium to favor the R form, so the constituent activity is reduced close to 0.

24 The two-state theory is just one model for explaining receptor activity. The most common model is Clark's occupancy theory. In its original form occupancy theory has difficulty explaining constituent activity. Some sources show a negative response with inverse agonists. Is this valid? Receptor theory has changed dramatically over the years. Before constituent activity was understood, all receptors were assumed to have zero response in the absence of a ligand (E = 0). When inverse agonists were first observed, a receptor causing a response of 0 suddenly could cause a response of less than 0. At that point, it became normal to report a negative response (E < 0) in discussions of inverse agonists. This presentation can be found widely on the internet, and Wikipedia's entry on inverse agonists follows this line of thinking. Over time, however, more presentations (an example) on constituent activity represent inverse agonists as reducing response to 0 instead of a negative value. Presenting constituent activity as a response of 0 or a value greater than 0 only depends on how a response of 0 is defined. For the purposes of this course, E = 0 is defined as the observed level of response when the receptor is not causing any response. How did people learn about inverse agonists? Response is difficult to measure with high precision and accuracy. The experimental data often show quite a bit of scatter. In other words, error bars on response data tend to be large. Constituent activity is often very low, perhaps a response of 5%. For such a receptor, distinguishing a ligand that leaves the response at 5% (an antagonist) from a ligand that might decrease the response to 2-3% (an inverse agonist) is very challenging. Experimentally, those levels of response are identical and almost indistinguishable. A breakthrough occurred when genetically engineered cells could be designed to overexpress receptors. As the concentration of receptors increased in a cell, the baseline response from those receptors also increased. Constituent activity levels rose to easily measured values, perhaps from 5% to 20%. Immediately, some ligands were identified as being able to decrease response. Those ligands no longer fit the label of antagonists. They were something different. Eventually, the term inverse agonists became the accepted label. Much of this pioneering research occurred in the 1980s. One of the early articles on GPCRs and inverse agonists appeared in 1989

25 5.3 Occupancy Theory The pros and cons of the Clark model video Please watch the online video (8 minutes 42 seconds). Clarifications and corrections At approximately 2:15 the transcript twice shows "Emacs" instead of "Emax". Around 7:05 the video discusses ligands for EGFR, which is incorrectly identified twice! The correct meaning of the acronym EGFR is epidermal growth factor receptor. A condensed summary of this video can be found in the Video summary page. EC50 and Kd revisited Background: The inflection point of a log [L]-response plot corresponds to the log EC 50 value of the ligand. At theec 50 value, half of the total receptor concentration is bound in the receptor-ligand complex (R-L) and the other half remains as free receptor, R. At EC 50, the ligand concentration is equivalent to K D. Instructions: Read the passage below that establishes the relationship between EC 50 and K D. Learning Goal: To more fully understand the equivalence of EC 50 and K D in Clark's Occupancy Theory. The graph below is a typical response vs. log [L] curve for a full agonist. The point of inflection of this curve occurs at 50% maximum response. The ligand concentration required to achieve 50% of E max is called EC 50, or the concentration required to achieve an effect of 50%.

26 The rest of the discussion assumes that [L] is equal to EC 50. If we are at EC 50, then the system is also at 50% of E max. According to Clark's theory, the only way to achieve 50% of E max is for half of the receptors to be bound by the ligand (a full agonist) as the receptor-ligand complex (R-L). If half of the receptors are bound, then an equal number (half) of the receptors are not bound and exist in the form of free receptor (R). Remember that the binding of a ligand to a receptor is a reversible process. This equilibrium binding is quantified by the dissociation equilibrium constant (K D). The equation for K D is shown below. As has been stated, at 50% E max, [R-L] = [R]. In the equation for the equilibrium dissociation constant, two terms cancel and leave the relationship K D = [L]. At 50% E 50, [L] also equals EC 50. Through a simple substitution, EC 50 = K D. This relationship is important. Experimentally, K D is not easy to measure directly, but through a log [L]-response plot, EC 50 is easy fairly easy to determine. Because EC 50 and K D are equal, a log [L]- response plot is a relatively simple method for indirectly determining the equilibrium dissociation constant (K D) of a receptor and its agonist or partial agonist.

27 Determining an EC50 value Background: The EC 50 value for an agonist is equal to the K D value of the receptor-ligand complex. The number occurs at the inflection point of a sigmoidal curve. Instructions: Read the passage below on the easiest method for determining an EC 50 from a set of experimental data. Then, use the data points to estimate the EC 50 value of the ligand. Learning Goals: To understand how to manipulate ligand-response data and gain quantitative information on the ligand. Data points fitting a sigmoidal relationship are much more difficult to manipulate than linear data. Unfortunately, software packages that best handle sigmoidal data are not freely available. In the absence of a commercial data processing software package, linearizing ligand-response data is a viable option. Because Clark's occupancy theory can be modeled with what is essentially the Michaelis-Menten equation, the ligand-response data for many receptors can be linearized with what amounts to the Lineweaver-Burk equation. The linearized, double-reciprocal version of Clark's equation is shown below. With this equation, any set of response (E) vs. [L] data points can be converted to their inverse forms as 1/E vs. 1/[L], plotted, and matched to a best-fit line with a spreadsheet function like LINEST. The y-intercept can be used to determine E max and then K D can be derived from the slope of the line. Remember that K D is equal to EC 50, at least for agonists and partial agonists that follow the Clark model. Please complete the online exercise.

28 Upregulation and downregulation Background: Clark's occupancy theory, while useful for modeling the behavior of many receptors, fails to accommodate many observed qualities of ligand-receptor interactions. Examples already covered include spare receptors and constituently active receptors. Instructions: Read the passage below concerning the concepts of upregulation and downregulation, two ideas that Clark's occupancy theory in its simple form cannot explain. Learning Goals: To learn about the ideas of upregulation and downregulation, which are frequently encountered in drug discovery and clinical medicine. While Clark's occupancy has a very appealing (and useful) simplicity, the fact is that receptors are not simple. The different receptor superfamilies have completely different structures, and each superfamily has its own unique behaviors and traits. These behaviors make the task of developing a single, unified ligand-receptor theory very challenging. One trait of some receptors is downregulation. For some types of cells, if a receptor is continuously stimulated to a high level by a ligand, the cell responds by decreasing its population of that particular receptor. In other words, the concentration of that receptor is decreased. Because the receptor concentration is lower, fewer receptors are available for stimulation and generation of a response. If the ligand concentration is left unchanged, that same ligand will affect a smaller response because fewer receptors are available. If the ligand is a drug, then the patient will receive a smaller therapeutic effect from the same drug dosage. The patient is said to be desensitized to the drug. Desensitization can often be observed in instances of downregulation. In order to experience the same therapeutic effect from a drug that has caused desensitization, a patient must increase his dosing levels. If a desensitized patient quits taking a medication, then it is generally true that the cells will slowly return to their original receptor concentration in a process called upregulation. Once the cells are restored to their original state and original levels of receptor concentration, the patient will once again be sensitized to the drug. If the patient begins taking the medication again, he must return to the original dosing levels. If the patient takes elevated levels of medication, an overdose may be a risk. Desensitization is very common in patients who use opiates to manage pain. Advanced patients become desensitized and require a higher dose to achieve their original level of pain relief.

29 Cheng-Prussoff for receptors Background: The Cheng-Prussoff equation was introduced in Chapter 4 as a convenient method for interconverting K i and IC 50 values of enzymes. Instructions: Read the passage below on the use of the Cheng-Prussoff equation in the area of receptors. Learning Goals: To appreciate the generality of the Cheng-Prussoff equation and more fully understand the different variables used to describe the activity of antagonists. Most drugs that bind to a receptor are antagonists, generally competitive antagonists. The competitive antagonist prevents the activation of the receptor by its endogenous ligand by binding the receptor in the same position as the endogenous ligand. A competitive antagonist is therefore much like the competitive inhibitors that we discussed in Chapter 4. Both the antagonist and inhibitor prevent the action of a protein by blocking the protein's primary binding pocket. The activity of competitive inhibitors can be quantified in two different ways: K i, which is the dissociation equilibrium constant of the enzyme-inhibitor complex, or IC 50, which is the concentration of the inhibitor required to reduce the rate of an enzyme-catalyzed reaction by 50%. K i is a constant and an inherent property of the enzyme-inhibitor complex. IC 50, however, varies depending on the concentration of the substrate used in the assay of the inhibitor's activity. If the substrate concentration in the assay is known along with the K m of the enzyme and substrate, then the Cheng-Prussoff equation can be used to interconvert K i and IC 50. Likewise, a competitive antagonist can be quantified in two different ways: K i, which is the dissociation equilibrium constant of the receptor-antagonist complex, or IC 50, which is the concentration of the antagonist required to reduce the response caused by the action of an agonist on a receptor by 50%. K i is a constant and an inherent property of the receptor-antagonist complex. IC 50 varies depending on the concentration of the agonist used in the assay of the antagonist's activity. Not surprisingly, the Cheng-Prussoff can also be used to interconvert these two values. When applied to receptors, the Cheng-Prussoff equation requires the concentration of ligand, [L], used in the assay. Also required is the K D of the receptor-ligand complex. Specifically, that is

30 the K D of the receptor-ligand complex without any antagonist present. The receptor version of the Cheng-Prussoff equation is shown below. As with enzyme assays, studies of antagonists often report IC 50 values for the antagonist. IC 50 values generated by different research groups cannot be directly compared unless the agonist concentration in the various assays are identical. The Cheng-Prussoff equation allows conversion of IC 50 values to the directly comparable K i values. Examination 1 Instructions for Examination 1 The exam is open book and open note. There are ten questions. Each is its own component within the Examination 1 subsection. After you submit your answer, you can view the solution. For multiple choice and checkbox questions, students are allowed two attempts. All other types of questions allow only one attempt. Remember that you are bound by the honor code to work independently on this exam. Postings to the forum concerning the exam are allowed but do not disclose answers in your questions. Any inappropriate posts will be promptly deleted by the course staff. Exam Questions Please complete the online questions in Examination 1.