Examination I Key MCMP 208 Biochemistry for Pharmaceutical Sciences I February 7, 2017

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1 Examination I Key MCMP 208 Biochemistry for Pharmaceutical Sciences I February 7, 2017 Correct answers in multiple choice questions are indicated in RED and underlined. Correct answers to essay questions are indicated in RED in comic book font. In some cases and explanation is provided in BLUE/BLUE MCMP 208 Exam I Key - 1 MULTIPLE CHOICE. For problems 1 to 17, select from the list immediately following each question the single most correct choice to complete the statement, solve the problem, or answer the question. Mark that answer on your answer sheet. [3 points each] 1. The processes in cells that make larger, more complex biomolecules from smaller molecules are collectively referred to as metabolism redox reactions anaplerotic metabolism anabolism This includes polymerization but also making larger more complex molecules by reactions that are not polymerization, such as a larger amino acid out of a smaller one or a larger carbohydrate from a smaller. catabolism polymerization 2. Which macromolecular class or classes always contains nitrogen? proteins nucleic acids polysaccharides lipids lipids and nucleic acids polysaccharides and proteins nucleic acids and proteins polysaccharides and lipids polysaccharides and nucleic acids 3. Why is compartmentalization important for living cells? Cells have a lot of transport functions which requires a lot of different compartments. Each compartment has a membrane, and cells have a lot of membranes. Each compartment has a membrane, and transmembrane signaling is important for regulating cells. Each compartment provides a controlled microenvironment that favors a selected set of reactions. Cells have many functions and having many compartments allows each compartment to handle a small number of functions. All cells have a plasma membrane and this is needed for the cell to control the intracellular environment.

2 4. The primary function of the endoplasmic reticulum and golgi is to MCMP 208 Exam I Key - 2 generate ATP for the cell degrade all types of unwanted molecules use oxygen to metabolize other molecules, most commonly lipids to process proteins destined for the cell surface or secretion from the cell to process material absorbed from outside the cell to provide structure and shape to the cell and to be involved in cell movement and mitosis 5. A permanent ion such as phosphate does what to water when it is in solution? It organizes the water that was not previously organized It disorganizes water that was previously organized It changes the organization of water by reorienting the water molecules It changes the ionization of water by reorienting the water molecules It hydrogen bonds to water It does not interact with water while it tries to find an oppositely charged ion so it can become a salt It acts as an amphipathic agent and forms a micelle 6. Animal cells are cultured in laboratory incubators that add carbon dioxide gas and thus control the concentration of carbon dioxide in the atmosphere of the incubator. The culture medium always contains bicarbonate. Which statement is correct about the relationship of the carbon dioxide concentration in the incubator atmosphere and the ph of the culture medium? Increasing the carbon dioxide concentration increases the ph of the culture medium. Decreasing the carbon dioxide concentration increases the ph of the culture medium. The direction of the effect of changing the carbon dioxide concentration on the ph of the medium depends on the concentration of bicarbonate in the medium. There is no effect of changing the carbon dioxide concentration and the ph of the culture medium. 7. A common reaction that all amino acids can do is form a Schiff base. The other molecule that must be involved in the formation of a Schiff base with an amino acid is a carboxylic acid a primary or secondary amine an imine an aldehyde a ketone an aldehyde or ketone an aldehyde, ketone, or carboxylic acid an imine or a primary or secondary amine a thiol

3 8. Levinthal s paradox is that MCMP 208 Exam I Key - 3 there is no computational way to predict the 3D structure of a protein from its amino acid sequence it is physically impossible for a polypeptide to try all the possible 3D structures to find the lowest energy one in the time proteins to fold it is impossible to predict the function of a protein from its amino acid sequence the structure of a protein depends on its function as much as the function of a protein depends on its structure there are a large number of lowest energy state 3D structures for each polypeptide, yet there is generally only one stable structure formed for each polypeptide 9. Which bond in ATP is broken to deliver energy to reactions in metabolism? the N-glycosidic bond between the ribose and the adenine the phosphoester bond between the ribose and the alpha phosphate the phosphoanhydride bond between the alpha phosphate and beta phosphate the phosphoanhydride bond between the beta phosphate and gamma phosphate either of the two phosphoanhydride bonds 10. The functions of hemoglobin and myoglobin are nearly identical. Yet the cooperativity of hemoglobin is important for its function but the reasons for that cooperativity are irrelevant for myoglobin because myoglobin in adults is the same as it is during fetal life myoglobin has a very different secondary and tertiary structure compared to hemoglobin myoglobin functions inside cells that are stationary and do not circulate in the blood myoglobin does not utilize heme myoglobin has a very different primary structure compared to hemoglobin hemoglobin functions in the extracellular environment while myoglobin functions only inside cells 11. An intrinsically unstructured protein or part of a protein is a polypeptide that does not have secondary structure but does have tertiary structure has secondary structure but does not have tertiary structure does not have secondary or tertiary structure until it stably associates with another molecule never has any secondary or tertiary structure can reversibly take on quaternary structure can switch between two different but stable structures so to have two very different functions. causes misfolded protein diseases such as Alzheimer s disease a protein that assists other proteins with the process of folding correctly

4 MCMP 208 Exam I Key The urea group of biotin is critical for the enzymatic activity of carboxylase enzymes and is involved in the biosynthesis of fatty acids, as shown below in the reaction catalyzed by acetyl CoA carboxylase. Biotin is best described as an enzyme active site an enzyme substrate an apoenzyme a transferase a coenzyme a heterotropic enzyme modulator a zymogen a ligase Coenzymes are organic molecules that give enzymes additional chemical versatility that the 20 amino acids. These are often vitamins. In this case, biotin is vitamin B The enzyme argininosuccinate synthetase catalyzes the formation of argininosuccinate from citrulline and aspartate (shown below). This enzyme is best classified as a Oxidoreductase Transferase Hydrolase Lyase Isomerase Ligase Ligases catalyze bond formation between two substrate molecules. In this case, argininosuccinate synthetase EC ligates citrulline and aspartate together. Many ligases include the term synthetase. 14. The active site of alcohol dehydrogenase is shown below with a bound ethoxide substrate and cofactor nicotinamide adenine dinucleotide (NAD + ). In the catalyzed conversion of ethanol to acetaldehyde, the primary role of NAD + is to exert strain on the substrate. as a general base. as a general acid. as a hydrogen bond donor. as a hydrogen bond acceptor. as a hydride donor. as a hydride acceptor. to coordinate the active site zinc. Alcohol dehydrogenase is an oxidoreductase that catalyzes the oxidation of ethanol to acetaldehyde. NAD + acts as coenzyme accepting a hydride from the alcohol substrate to become the reduced NADH.

5 MCMP 208 Exam I Key The catalytic efficiency of enzyme aspartate transcarbamoylase (ATCase) is modulated through allosteric binding by the nucleotide triphosphates ATP and CTP. Because these molecules are not enzyme substrates these effects are described as. zymogen activation homotropic heterotropic genetic control compartmentation cooperativity noncompetitive inhibition uncompetitive inhibition Heterotropic effects are allosteric modulation of enzyme activity, which can be either positive or negative, by molecules that differ from the enzyme s substrates. Homotropic effects are allosteric modulation by the substrates. 16. The apparent Vmax and Km of an enzyme were determined under valid conditions in the absence or presence of an inhibitor at 5 µm concentration. This behavior of inhibition is characteristic of a/an inhibitor. The Ki value for the inhibitor is. competitive; 5 mm competitive; 5 µm competitive; 10 µm noncompetitive; 5 mm noncompetitive; 5 µm noncompetitive; 10 µm uncompetitive; 5 mm uncompetitive; 5 µm uncompetitive; 10 µm Apparent Km Apparent Vmax Uninhibited M inhibitor Since the Vmax s are equal, this is identified as a competitive inhibitor. The substrate is able to out compete the binding of the inhibitor at high concentration. The apparent Km s differ by a factor of 2. In competitive inhibition, the apparent Km equals Km, where i. Thus, the ratio of [I] to the Ki is 1. [I] was given at 5 µm. 17. In the Michaelis-Menten plot below, which point on the plot best represents the value of kcat[et], where [Et] is the total enzyme concentration? And what would be appropriate units for this value? M represents molar concentration (moles/liter). i; M -1 s -1 ii; M -1 s -1 iii; M -1 s -1 iv; M i; Ms -1 ii; Ms -1 iii; Ms -1 iv, M -1 (Explanation is on the next page) Initial Reaction Rate iii line slope Substrate Concentration iv [S] value i ii v o value v o value

6 MCMP 208 Exam I Key - 6 The Vmax = kcat/[et]. Recalling the M/M equation: v 0 = V max[s] [S]+K m At very high substrate concentration, the equation simplifies to vo = Vmax. Reaction velocities have units of [product concentration] per unit time. ESSAY PROBLEMS. Write your answers to problems 18 to 22 in the space immediately below each problem. 18. [6 points] The RGD sequence is found in fibronectin, a protein in the extra cellular matrix which is bound by specific cell surface receptors as part of cellular attachment. Draw the 2-dimensional structure of a middle section (residues 1524 through 1529) of human fibronectin that has the amino acid sequence RGDSPA showing any ionizations as they would be at physiological ph. In your drawing, you do not need to label carbon atoms or hydrogens that are bonded to carbon atoms, but you must show all bonds (except bonds between C and H) thereby implicitly showing locations of C atoms by angles in the bonds. You also must explicitly label all other atoms (including all O, N, and S, and H atoms that are bonded to O, N, or S atoms.) Note that for full credit the peptide must be drawn with the provided sequence interpreted to be N to C (as shown here), not C to N. However, the peptide can be drawn in either direction (N to C is either left to right or right to left) in the answer space as long as the sequence is correct. The guanidino group charge may be shown using any of the three correct resonance structures (two of which are indistinguishable) or the combined structure shown above. The carboxyl on the N terminus and the amino group on the C terminus are option and do not need to be drawn. 19. A. [4 points] Chymotrypsin is a serine protease with a well characterized catalytic mechanism. Using the blanks below on the Cleland notation diagram, indicate the order of addition of the enzyme substrates, the order of release of products, and all enzyme bound forms and intermediates. The first blank is filled in for you, E = free enzyme. peptide Pep. frag. NH2 H2O Pep. frag.-cooh E E. peptide Acyl E E. H2O. Pep. C-term E B. [2 points] Multisubstrate reactions can be classified into two classes based on their mechanisms. What classification describes the mechanism of chymotrypsin? Double displacement or ping-pong mechanism

7 MCMP 208 Exam I Key [6 points] The figure below shows a stable structure of a protein that is to be used for each part of this problem. The dashed line divides the structure into three regions labeled A, B, and C in the figure. A A B C A. [2 points] Does this protein have quaternary structure and how do you reach that conclusion? No; because all three regions are connected on the same polypeptide chain. B. [1 point] Explain why it is likely that each of these three regions folds independently. The backbones of the regions consist of a contiguous region of the backbone of the protein, and there is no intertwining of backbone from different regions. Thus, each region is a globular domain that folds independently. The regions interact only via surface-surface interactions. C. [1 point] Which one of the three parts of this protein is comprised mostly of α-helical secondary structure? A D. [1 point] Which one of the three parts of this protein is comprised mostly of β-sheet secondary structure? B E. [1 point] For your answer to part D, what kind of β-sheet structure is present in this part of the protein? Anti-parallel Note that the arrow heads of the beta-strands in region B are pointed in opposite directions. While this is mostly seen with two adjacent pairs, the entire beta sheet structure is rolled into an asymmetrical barrel (called a beta-barrel) and if any two of the strands are anti-parallel, then all the strands in the sheet must be antiparallel.

8 MCMP 208 Exam I Key A. [4 points] The Michaelis-Menton equation is frequently rearranged to its reciprocal form to simplify plots to a form known as the Lineweaver-Burk plot. Draw a Lineweaver-Burk plot with a single line, and indicate the values generally plotted on the x and y axes as well as the values generally located at the x and y intercepts. The Michaelis-Menton equation is given for your information. v 0 = V max[s] [S] + K m 1 point for each label (two axes and two intercepts.) The slope label is not required. Full credit requires that the plot line be correctly placed relative to the axes. B. [2 points] A particular enzyme can be inhibited by 2 different inhibitors, A and B, which both display mixed noncompetitive inhibition, but with significantly different parameters. Inhibitor A binds the free enzyme with greater affinity than the ES complex. Inhibitor B binds the free enzyme with less affinity than the ES complex. Draw an additional Lineweaver-Burk plot with 3 labeled lines (uninhibited, +A, and +B) that demonstrates how the x-intercepts of these lines differ? The x-intercept value is normally -1/Km. In the presence of noncompetitive inhibitors, this value is αʹ/(αkm), where α = 1 + [I]/Ki (Ki for free E) and αʹ = 1+ [I]/Kiʹ (Ki for ES complex). Thus, the ratio of ʹ to determines if the apparent Km increases or decreases in the presence of the inhibitor. If it binds the free E with greater affinity than the ES complex Ki is smaller than Ki and α > αʹ. Thus, in this case, the apparent Km increases. The opposite is the case when α < αʹ. One point each if the plot line s x- intercept is on the correct side of x-intercept of the uninhibited reaction plot.

9 E activity MCMP 208 Exam I Key [5 points] A biochemist studying a new enzyme activity E in tissue extracts found a small molecule inhibitor I that could inhibit all the E activity in the tissue extracts at 1.0 mm. A dose-response study of I showed the following result: 100% 80% 60% 40% 20% 0% [I] mm After covalently linking I to uncharged hydrophilic chromatography beads, these I-coupled-beads were used for affinity chromatography of E from tissue extracts. After binding the extract to the beads and washing the beads, the bound protein was eluted with 30 column volumes (the volume of the beads in the column) of a solution containing just 0.02 mm of I. During the elution, 60 fractions of equal volume were collected (numbered 1 to 60, with 1 eluted first and 60 eluted last). Each fraction was dialyzed and assayed for enzyme activity. Two peaks of activity were found, one in fractions 5 to 8 and a second peak in fractions 28 to 33. Dialyzed fractions 5 to 8 were pooled and called Peak 1 while dialyzed fractions 28 to 33 were pooled and called Peak 2. The amount of E activity recovered in each of the two peaks was approximately the same and each was completely pure as determined by SDS-PAGE. Full kinetic analysis for inhibition of E by I was done for Peak 1 and Peak 2 and each was found to be inhibited by I via the same mechanism but with a quantitative difference. Peak 1 showed a KI for I of 0.35 mm and the Peak 2 showed a KI for I of 0.09 mm. From your understanding of affinity chromatography, explain: (a) why there are two peaks of E in this affinity chromatography separation and (b) why the KI value for I was higher for E in Peak 1 than for E in Peak 2. (a) Note that the unusual shape of the dose-response curve suggests, but does not prove, that there two forms of E with different inhibition by inhibitor I. This is the reason for doing the affinity chromatography elution with a solution containing a low concentration of I rather than a step-gradient to a high concentration of I. There are two well-separated peaks of activity because each one has a very different affinity for the molecules of I that are covalently coupled to the chromatography beads. [This first sentence, however, is simply restating information given in the problem without a full explanation of how the differing affinity causes different elution times. The remainder of this answer for part a provides that explanation.] The rate of elution in the moving (or liquid) phase is directly related to the faction of the time the enzyme stays in the liquid phase. The two forms of E spend different fractions of the time bound to the I molecules that are linked to the beads (i.e., associated with the stationary or solid phase) because the two forms differ in their affinity for I (different KI values). [The KI is the dissociation equilibrium constant for the IE complex, so it is inversely proportional to affinity of E for I.] Different fractions of the time spent on the stationary phase means the two forms will spend different amounts of time in the moving phase, and hence elute at different times.

10 MCMP 208 Exam I Key - 10 (b) Note that there are two parts to this answer with a large amount of explanation between the two parts: From the different KIs (0.35 mm versus 0.09 mm) for the enzyme in the two peaks eluted from the affinity resin, peak 1 enzyme has a lower affinity for I than the peak 2 enzyme. [The KI is the dissociation equilibrium constant for the IE complex, so it is inversely proportional to affinity of E for I.] In the absence of added I, the E in the tissue extract binds to the I bound to the beads in the column. This is a reversible binding, but one that is fairly strong, so it would take a lot buffer without any I in it to elute the E from the beads. To speed up the elution, some I is included in the elution solution. Normally, affinity chromatography involves elution with a high concentration of the molecule that is bound to the beads (I in this case). But from the unusual shape of the inhibition dose-response curve, it may be that there are two forms of E, one with a higher affinity for I than the other. By eluting with a low concentration of I, the E will be eluted off the beads faster than if buffer without I is used for elution, but the elution still will be slow enough to see a difference in affinity between the two forms. Using a higher concentration of I, such as 1mM (which inhibits all the activity) will cause both forms to spend so little time associated with the beads that they will elute in the first column volume of elution buffer used. In any given concentration of [I] (even zero) the time spent by E in association with the beads (stationary phase) is longer for the form of E with the higher affinity for I than for the form of E with the lower affinity for I. Thus, the first peak eluted should have the lower affinity (higher KI value) than the second peak. Using a higher concentration of I, such as 1.0 mm (which inhibits all the activity) will cause both forms to spend so little time associated with the beads that they will elute in the first column volume of elution buffer used.