Exam 1: End of Chapter Problems
11. Meanings of Ka and pKa (a) Does a strong acid have a greater or lesser tendency to lose its proton than a weak acid? (b) Does the strong acid have a higher or lower Ka than the weak acid? (c) Does the strong acid have a higher or lower pKa than the weak acid?
(a) STRONG acid = Greater tendency to LOSE H+ (b) STRONG acid = Higher Ka (c) STRONG acid = Lower pKa & pH
Peptide (a); it has more amino acid residues that favor an α-helical structure (see Table 4-1).
13. Predicting Secondary Structure Which of the following peptides is more likely to take up an α-helical structure, and why? (a) LKAENDEAARAMSEA (b) CRAGGFPWDQPGTSN
The activity of the enzyme that converts sugar to starch is destroyed by heat denaturation.
1. Keeping the Sweet Taste of Corn The sweet taste of freshly picked corn (maize) is due to the high level of sugar in the kernels. Store-bought corn (several days after picking) is not as sweet, because about 50% of the free sugar is converted to starch within one day of picking. To preserve the sweetness of fresh corn, the husked ears can be immersed in boiling water for a few minutes ("blanched"), then cooled in cold water. Corn processed in this way and stored in a freezer maintains its sweetness. What is the biochemical basis for this procedure?
(a) Shorter bonds have a higher bond order (are multiple rather than single) and are stronger. The peptide C—N bond is stronger than a single bond and is midway between a single and a double bond in character. (b) Rotation about the peptide bond is difficult at physiological temperatures because of its partial double-bond character.
1. Properties of the Peptide Bond In x-ray studies of crystalline peptides, Linus Pauling and Robert Corey found that the C—N bond in the peptide link is intermediate in length (1.32 Å) between a typical C—N single bond (1.49 Å) and a C=N double bond (1.27 Å). They also found that the peptide bond is planar (all four atoms attached to the C—N group are located in the same plane) and that the two α-carbon atoms attached to the C—N are always trans to each other (on opposite sides of the peptide bond). (a) What does the length of the C—N bond in the peptide linkage indicate about its strength and its bond order (i.e., whether it is single, double, or triple)? (b) What do the observations of Pauling and Corey tell us about the ease of rotation about the C—N peptide bond?
(a) 2,000 s−1 (b) Measured Vmax = 1 μM s−1; Km = 2 μM
10. Applying the Michaelis-Menten Equation II An enzyme catalyzes the reaction M ⇌ N. The enzyme is present at a concentration of 1 nM, and the Vmax is 2 μM s−1. The Km for substrate M is 4 μM. (a) Calculate kcat. (b) What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an α′ of 2.0?
Protein (a), a β barrel, is described by Ramachandran plot (c), which shows most of the allowable conformations in the upper left quadrant where the bond angles characteristic of the β conformation are concentrated. Protein (b), a series of α helices, is described by plot (d), where most of the allowable conformations are in the lower left quadrant.
10. Interpreting Ramachandran Plots Examine the two proteins labeled (a) and (b) below. Which of the two Ramachandran plots, labeled (c) and (d), is more likely to be derived from which protein? Why?
5. Calculation of Hydrogen Ion Concentration from pH What is the H + concentration of a solution with pH of (a) 3.82 (b) 6.52 (c) 11.11?
10^(-pH) (a) 1.51 × 10^−4 M (b) 3.02 × 10^−7 M (c) 7.76 × 10^−12 M
(a) 400 s−1 (b) 10 μM (c) α = 2, α′ = 3 (d) Mixed inhibitor
11. Applying the Michaelis-Menten Equation III A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY ⇌ SAD. The researchers begin to characterize the enzyme. (a) In the first experiment, with [Et] at 4 nM, they find that the Vmax is 1.6 μM s−1. Based on this experiment, what is the kcat for happyase*? (Include appropriate units.) (b) In another experiment, with [Et] at 1 nM and [HAPPY] at 30 μM, the researchers find that V0 = 300 nM s−1. What is the measured Km of happyase* for its substrate HAPPY? (Include appropriate units.) (c) Further research shows that the purified happyase* used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation and the two experiments repeated, the measured Vmax in (a) is increased to 4.8 μM s−1, and the measured Km in (b) is now 15 μM. For the inhibitor ANGER, calculate the values of α and α′. (d) Based on the information given above, what type of inhibitor is ANGER?
(a) at pH 3, +2; at pH 8, 0; at pH 11, −1 (b) pI = 7.8
11. Net Electric Charge of Peptides A peptide has the sequence Glu−His−Trp−Ser−Gly−Leu−Arg−Pro−Gly (a) What is the net charge of the molecule at pH 3, 8, and 11? (Use pKa values for side chains and terminal amino and carboxyl groups as given in Table 3-1.) (b) Estimate the pI for this peptide.
(a) 3.85−3 10−3M−1; [glucose 6-phosphate] = 8.9 × 10−8M. No. The cellular [glucose 6-phosphate] is much greater than this, favoring the reverse reaction. (b) 14 M. No. The maximum solubility of glucose is less than 1 M. (c) 837 (ΔG′° = −16.7 kJ/mol); [glucose] = 1.2 × 10−7M. Yes. This reaction path can occur with a concentration of glucose that is readily soluble and does not produce a large osmotic force. (d) No. This would require such high [Pi] that the phosphate salts of divalent cations would precipitate. (e) By directly transferring the phosphoryl group from ATP to glucose, the phosphoryl group transfer potential ("tendency" or "pressure") of ATP is utilized without generating high concentrations of intermediates. The essential part of this transfer is, of course, the enzymatic catalysis.
11. Strategy for Overcoming an Unfavorable Reaction: ATP-Dependent Chemical Coupling The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by Pi is described by the equation Glucose+Pi→glucose6-phosphate+H2 OΔG'°=13.8kJ/mol (a) Calculate the equilibrium constant for the above reaction at 37 °C. In the rat hepatocyte, the physiological concentrations of glucose and Pi are maintained at approximately 4.8 mM. What is the equilibrium concentration of glucose 6-phosphate obtained by the direct phosphorylation of glucose by Pi? Does this reaction represent a reasonable metabolic step for the catabolism of glucose? Explain. (b) In principle, at least, one way to increase the concentration of glucose 6-phosphate is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and Pi. Assuming a fixed concentration of Pi at 4.8 mM, how high would the intracellular concentration of glucose have to be to give an equilibrium concentration of glucose 6-phosphate of 250 μM (the normal physiological concentration)? Would this route be physiologically reasonable, given that the maximum solubility of glucose is less than 1 M? (c) The phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP; that is, part of the free energy of ATP hydrolysis is used to phosphorylate glucose: (1)Glucose+Pi→glucose6-phosphate+H2 OΔG'°=13.8kJ/mol (2)ATP+H2O→ADP+PiΔG'°=−30.5kJ/mol --------------------------------------Sum: Glucose+ATP→glucose6-phosphate+ADP Calculate K′eq at 37 °C for the overall reaction. For the ATP-dependent phosphorylation of glucose, what concentration of glucose is needed to achieve a 250 μM intracellular concentration of glucose 6-phosphate when the concentrations of ATP and ADP are 3.38 mM and 1.32 mM, respectively? Does this coupling process provide a feasible route, at least in principle, for the phosphorylation of glucose in the cell? Explain. (d) Although coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, we have not yet specified how this coupling is to take place. Given that coupling requires a common intermediate, one conceivable route is to use ATP hydrolysis to raise the intracellular concentration of Pi and thus drive the unfavorable phosphorylation of glucose by Pi. Is this a reasonable route? (Think about the solubility product, Ksp, of metabolic intermediates.) (e) The ATP-coupled phosphorylation of glucose is catalyzed in hepatocytes by the enzyme glucokinase. This enzyme binds ATP and glucose to form a glucose-ATP-enzyme complex, and the phosphoryl group is transferred directly from ATP to glucose. Explain the advantages of this route.
(a) 24 nM (b) 4 μM (V0 is exactly half Vmax, so [A] = Km) (c) 40 μM (V0 is exactly half Vmax, so [A] = 10 times Km in the presence of inhibitor) (d) No. kcat/Km = (0.33 s−1)/ (4 × 10−6M) = 8.25 × 104M−1 s−1, well below the diffusion- controlled limit.
12. Applying the Michaelis-Menten Equation IV An enzyme is found that catalyzes the reaction X ⇌ Y. Researchers find that the Km for the substrate X is 4 μM, and the kcat is 20 min−1. (a) In an experiment, [X] = 6 mM, and V0 = 480 nM min−1. What was the [Et] used in the experiment? (b) In another experiment, [Et] = 0.5 μM, and the measured V0 = 5 μM min−1. What was the [X] used in the experiment? (c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an α of 10. In an experiment with the same [Et] as in (a), but a different [X], an amount of Z is added that reduces V0 to 240 nM min−1. What is the [X] in this experiment? (d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.
(a) −12.5 kJ/mol ( b) −14.6 kJ/mol
12. Calculations of ΔG′° for ATP-Coupled Reactions From data in Table 13-6, calculate the ΔG′° value for the following reactions: (a) Phosphocreatine + ADP → creatine + ATP (b) ATP + fructose → ADP + fructose 6-phosphate
Vmax ≈ 140 μM/min; Km ≈ 1 × 10−5M
13. Estimation of Vmax and Km by Inspection Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction (see Box 6-1), sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:
Lys, His, Arg; negatively charged phosphate groups in DNA interact with positively charged side groups in histones.
13. Isoelectric Point of Histones Histones are proteins found in eukaryotic cell nuclei, tightly bound to DNA, which has many phosphate groups. The pI of histones is very high, about 10.8. What amino acid residues must be present in relatively large numbers in histones? In what way do these residues contribute to the strong binding of histones to DNA?
(a) Aromatic residues seem to play an important role in stabilizing amyloid fibrils. Thus, molecules with aromatic substituents may inhibit amyloid formation by interfering with the stacking or association of the aromatic side chains. (b) Amyloid is formed in the pancreas in association with type 2 diabetes, and is formed in the brain in Alzheimer disease. Although the amyloid fibrils in the two diseases involve different proteins, the fundamental structure of the amyloid is similar and is similarly stabilized in both, so they are potential targets for similar drugs designed to disrupt this structure.
14. Amyloid Fibers in Disease Several small aromatic molecules, such as phenol red (used as a nontoxic drug model), have been shown to inhibit the formation of amyloid in laboratory model systems. A goal of the research on these small aromatic compounds is to find a drug that would efficiently inhibit the formation of amyloid in the brain in people with incipient Alzheimer disease. (a) Suggest why molecules with aromatic substituents would disrupt the formation of amyloid. (b) Some researchers have suggested that a drug used to treat Alzheimer disease may also be effective in treating type 2 (non-insulin-dependent) diabetes mellitus. Why might a single drug be effective in treating these two different conditions?
−13 kJ/mol
14. Calculations of ΔG at Physiological Concentrations Calculate the actual, physiological ΔG for the reaction Phosphocreatine+ADP→creatine+ATP at 37 °C, as it occurs in the cytosol of neurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6 mM.
(a)Vmax = 51.5 mM/min; Km = 0.59 mM (b) Competitive inhibition
14. Properties of an Enzyme of Prostaglandin Synthesis Prostaglandins are a class of eicosanoids, fatty acid derivatives with a variety of extremely potent actions on vertebrate tissues. They are responsible for producing fever and inflammation and its associated pain. Prostaglandins are derived from the 20-carbon fatty acid arachidonic acid in a reaction catalyzed by the enzyme prostaglandin endoperoxide synthase. This enzyme, a cyclooxygenase, uses oxygen to convert arachidonic acid to PGG2, the immediate precursor of many different prostaglandins (prostaglandin synthesis is described in Chapter 21). (a) The kinetic data given below are for the reaction catalyzed by prostaglandin endoperoxide synthase. Focusing here on the first two columns, determine the Vmax and Km of the enzyme. (b) Ibuprofen is an inhibitor of prostaglandin endoperoxide synthase. By inhibiting the synthesis of prostaglandins, ibuprofen reduces inflammation and pain. Using the data in the first and third columns of the table, determine the type of inhibition that ibuprofen exerts on prostaglandin endoperoxide synthase.
(a) (Glu)20 (b) (Lys-Ala)3 (c) (Asn-Ser-His)5 (d) (Asn-Ser-His)5
14. Solubility of Polypeptides One method for separating polypeptides makes use of their different solubilities. The solubility of large polypeptides in water depends on the relative polarity of their R groups, particularly on the number of ionized groups: the more ionized groups there are, the more soluble the polypeptide. Which of each pair of polypeptides that follow is more soluble at the indicated pH? (a) (Gly)20 or (Glu)20 at pH 7.0 (b) (Lys-Ala)3 or (Phe-Met)3 at pH 7.0 (c) (Ala-Ser-Gly)5 or (Asn-Ser-His)5 at pH 6.0 (d) (Ala-Asp-Gly)5 or (Asn-Ser-His)5 at pH 3.0
(a) pKa = 5 w/ N+ & H => more soluble in a strong acid 0.1 M HCl (b) pKa = 10 w/ OH => more soluble in a strong base 0.1 M NaOH (c) pKa = 10 w/ OH => more soluble in a strong base 0.1 M NaOH
15. Effect of pH on Solubility The strongly polar, hydrogen-bonding properties of water make it an excellent solvent for ionic (charged) species. By contrast, nonionized, nonpolar organic molecules, such as benzene, are relatively insoluble in water. In principle, the aqueous solubility of any organic acid or base can be increased by converting the molecules to charged species. For example, the solubility of benzoic acid in water is low. The addition of sodium bicarbonate to a mixture of water and benzoic acid raises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water. Are the following compounds more soluble in an aqueous solution of 0.1 M NaOH or 0.1 M HCl? (The dissociable protons are shown in red.)
46.7 kJ/mol
15. Free Energy Required for ATP Synthesis under Physiological Conditions In the cytosol of rat hepatocytes, the temperature is 37 °C and the mass-action ratio, Q, is ([ATP/][ADP][Pi])=5.33×102M−1 Calculate the free energy required to synthesize ATP in a rat hepatocyte.
Vmax = 0.50 μmol/min; Km = 2.2 mM
15. Graphical Analysis of Vmax and Km The following experimental data were collected during a study of the catalytic activity of an intestinal peptidase with the substrate glycylglycine: Glycylglycine+H2O→2glycine Use graphical analysis (see Box 6-1) to determine the Vmax and Km for this enzyme preparation and substrate.
(a) NFκB transcription factor, also called RelA transforming factor (b) No. You will obtain similar results, but with additional related proteins listed. (c) The protein has two subunits. There are multiple variants of the subunits, with the best characterized being 50, 52, or 65 kDa. These pair with each other to form a variety of homodimers and heterodimers. The structures of a number of different variants can be found in the PDB. (d) The NFκB transcription factor is a dimeric protein that binds specific DNA sequences, enhancing transcription of nearby genes. One such gene is the immunoglobulin κ (kappa) light chain, from which the transcription factor gets its name.
15. Protein Modeling on the Internet A group of patients with Crohn disease (an inflammatory bowel disease) underwent biopsies of their intestinal mucosa in an attempt to identify the causative agent. Researchers identified a protein that was present at higher levels in patients with Crohn disease than in patients with an unrelated inflammatory bowel disease or in unaffected controls. The protein was isolated, and the following partial amino acid sequence was obtained (reads left to right): (a) You can identify this protein using a protein database such as UniProt (www.uniprot.org). On the home page, click on the link for a BLAST search. On the BLAST page, enter about 30 residues from the protein sequence in the appropriate search field and submit it for analysis. What does this analysis tell you about the identity of the protein? (b) Try using different portions of the amino acid sequence. Do you always get the same result? (c) A variety of websites provide information about the three-dimensional structure of proteins. Find information about the protein's secondary, tertiary, and quaternary structures using database sites such as the Protein Data Bank (PDB; www.pdb.org) or Structural Classification of Proteins (SCOP2; http://scop2.mrc-lmb.cam.ac.uk). (d) In the course of your Web searches, what did you learn about the cellular function of the protein?
(a) Specific activity after step 1 is 200 units/mg; step 2, 600 units/mg; step 3, 250 units/mg; step 4, 4,000 units/mg; step 5, 15,000 units/mg; step 6, 15,000 units/mg. (b) Step 4 (c) Step 3 (d) Yes. Specific activity did not increase in step 6; SDS polyacrylamide gel electrophoresis.
15. Purification of an Enzyme A biochemist discovers and purifies a new enzyme, generating the purification table below. (a) From the information given in the table, calculate the specific activity of the enzyme after each purification procedure. (b) Which of the purification procedures used for this enzyme is most effective (i.e., gives the greatest relative increase in purity)? (c) Which of the purification procedures is least effective? (d) Is there any indication based on the results shown in the table that the enzyme after step 6 is now pure? What else could be done to estimate the purity of the enzyme preparation?
long-chain alkyl groups (long carbon w/ 1 less H) probably acidic so buffering it with a strong base such as sodium bicarbonate will relieve the itching (d) Bicarbonate, a weak base, titrates —OH to —O −, making the compound more polar and more water-soluble.
16. Treatment of Poison Ivy Rash The components of poison ivy and poison oak that produce the characteristic itchy rash are catechols substituted with long-chain alkyl groups. If you were exposed to poison ivy, which of the treatments below would you apply to the affected area? Justify your choice. (a) Wash the area with cold water. (b) Wash the area with dilute vinegar or lemon juice. (c) Wash the area with soap and water. (d) Wash the area with soap, water, and baking soda (sodium bicarbonate).
C elutes first, B second, A last.
17. Peptide Purification At pH 7.0, in what order would the following three peptides (described by their amino acid composition) be eluted from a column filled with a cation-exchange polymer? Peptide A: Ala 10%, Glu 5%, Ser 5%, Leu 10%, Arg 10%, His 5%, Ile 10%, Phe 5%, Tyr 5%, Lys 10%, Gly 10%, Pro 5%, and Trp 10%. Peptide B: Ala 5%, Val 5%, Gly 10%, Asp 5%, Leu 5%, Arg 5%, Ile 5%, Phe 5%, Tyr 5%, Lys 5%, Trp 5%, Ser 5%, Thr 5%, Glu 5%, Asn 5%, Pro 10%, Met 5%, and Cys 5%. Peptide C: Ala 10%, Glu 10%, Gly 5%, Leu 5%, Asp 10%, Arg 5%, Met 5%, Cys 5%, Tyr 5%, Phe 5%, His 5%, Val 5%, Pro 5%, Thr 5%, Ser 5%, Asn 5%, and Gln 5%.
Increasing the pH of Aspirin will deprotonate the molecule (making it charged) which will slow down the absorption. However, aspirin in its low pH is rabidly absorbed because it is uncharged and less polar which means in the stomach is where it will be absorbed more. Stomach; the neutral form of aspirin present at the lower pH is less polar and passes through the membrane more easily
17. pH and Drug Absorption Aspirin is a weak acid with a pKa of 3.5 (the ionizable H is shown in red): It is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption requires passage through the plasma membrane, the rate of which is determined by the polarity of the molecule: charged and highly polar molecules pass slowly, whereas neutral hydrophobic ones pass rapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the small intestine is about 6. Is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice
29,000. The calculation assumes that there is only one essential Cys residue per enzyme molecule.
19. Irreversible Inhibition of an Enzyme Many enzymes are inhibited irreversibly by heavy metal ions such as Hg2+, Cu2+, or Ag+, which can react with essential sulfhydryl groups to form mercaptides: Enz-SH+Ag+→Enz−S−Ag+H+ The affinity of Ag+ for sulfhydryl groups is so great that Ag+ can be used to titrate —SH groups quantitatively. To 10.0 mL of a solution containing 1.0 mg/mL of a pure enzyme, an investigator added just enough AgNO3 to completely inactivate the enzyme. A total of 0.342 μmol of AgNO3 was required. Calculate the minimum molecular weight of the enzyme. Why does the value obtained in this way give only the minimum molecular weight?
(a) −4.8 kJ/mol (b) 7.56 kJ/mol (c) −13.7 kJ/mol
2. Calculation of ΔG′° from an Equilibrium Constant Calculate the standard free-energy change for each of the following metabolically important enzyme-catalyzed reactions, using the equilibrium constants given for the reactions at 25 °C and pH 7.0.
(a) I (lowest pH, more acidic, more desire to donate) (b) II (equivalents is 0.2) (c) IV (reaching fully deprotonated NH2) (d) II (pKa of the carboxyl group is when we have more protonated carboxyl group) (e) IV (pKa of protonated amino acid is when the carboxyl is f (f) II and IV (g) III (the isoelectric point) (h) III (1 equivalent = isoelectric point OR when carboxyl is completely titrated) (i) V (2 equivalents = completely titrated) (j) III (zwiteron phase @ pI) (k) V (when carboxyl is fully deprotonated) (l) II (before all of the amino is deprotonated) (m) III (when the net charge is 0) (n) V (last equivalents) (o) I, III, and V ( any point more than 1 PH away from the pKa of carboxyl & pKa of amino)
2. Relationship between the Titration Curve and the Acid-Base Properties of Glycine A 100 mL solution of 0.1 M glycine at pH 1.72 was titrated with 2 M NaOH solution. The pH was monitored and the results were plotted as shown in the graph. The key points in the titration are designated I to V. For each of the statements (a) to (o), identify the appropriate key point in the titration and justify your choice. (a) Glycine is present predominantly as the species +H3N—CH2—COOH. (b) The average net charge of glycine is +1/2. (c) Half of the amino groups are ionized. (d) The pH is equal to the pKa of the carboxyl group. (e) The pH is equal to the pKa of the protonated amino group. (f) Glycine has its maximum buffering capacity. (g) The average net charge of glycine is zero. (h) The carboxyl group has been completely titrated (first equivalence point). (i) Glycine is completely titrated (second equivalence point). (j) The predominant species is +H3N—CH2—COO—. (k) The average net charge of glycine is −1. (l) Glycine is present predominantly as a 50:50 mixture of +H3N—CH2—COOH and +H3N—CH2—COO−. (m) This is the isoelectric point. (n) This is the end of the titration. (o) These are the worst pH regions for buffering power.
buffers are 1 more or less from the pKa (a) pH 8.6 to 10.6 within 2 points of the pH, the molecule will either be fully protonated or fully deprotonated (d) pH = pKa − 2
20. Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form or as the free base (—NH2), because of the reversible equilibrium (a) In what pH range can glycine be used as an effective buffer due to its amino group? (d) When 99% of the glycine is in its -NH3+ form, what is the numerical relation between the pH of the solution and the pKa of the amino group?
(a) Peptide 2 (b) Peptide 1 (c) Peptide 2 (d) Peptide 3
22. Chromatographic Methods Three polypeptides, the sequences of which are represented below using the one-letter code for their amino acids, are present in a mixture: 1.ATKNRASCLVPKHGALMFWRHKQLVSDPILQKRQHILVCRNAAG 2.GPYFGDEPLDVHDEPEEG 3.PHLLSAWKGMEGVGKSQSFAALIVILA Of the three, which one would migrate most slowly during chromatography through: (a) an ion-exchange resin, beads coated with positively charged groups? (b) an ion-exchange resin, beads coated with negatively charged groups? (c) a size-exclusion (gel-filtration) column designed to separate small peptides such as these? (d) Which peptide contains the ATP-binding motif shown in the following sequence logo?
The optimum activity occurs when Glu35 is protonated and Asp52 is unprotonated.
23. pH Optimum of Lysozyme The active site of lysozyme contains two amino acid residues essential for catalysis: Glu35 and Asp52. The pKa values of the carboxyl side chains of these residues are 5.9 and 4.5, respectively. What is the ionization state (protonated or deprotonated) of each residue at pH 5.2, the pH optimum of lysozyme? How can the ionization states of these residues explain the pH-activity profile of lysozyme shown below?
(a) 262 (b) 608 (c) 0.30
3. Calculation of the Equilibrium Constant from ΔG′° Calculate the equilibrium constant K′eq for each of the following reactions at pH 7.0 and 25 °C, using the ΔG′° values in Table 13-4.
a) in the zwitterionic, both groups are ionized which allows interactions whereas when they are not charged, not much interaction will occur b) the pI is 6.01
3. How Much Alanine Is Present as the Completely Uncharged Species? At a pH equal to the isoelectric point of alanine, the net charge on alanine is zero. Two structures can be drawn that have a net charge of zero, but the predominant form of alanine at its pI is zwitterionic. (a) Why is alanine predominantly zwitterionic rather than completely uncharged at its pI? (b) What fraction of alanine is in the completely uncharged form at its pI? Justify your assumptions.
pKa1 = 2.34, pKa2= 9.69, pI = 6.11 (a) fully protonated: (+) on -NH3 (alpha amino group) & fully deprotonated: (-) on -COO (alpha carboxyl group) (b) low pH => Fully protonated amino group (c) 2 below the pI =>. Zwitterion (d) 2 above the pI => Zwitterion (e) high pH => Fully deprotonated carboxyl
34. Ionic Forms of Alanine Alanine is a diprotic acid that can undergo two dissociation reactions (see Table 3-1 for pKa values). (a) Given the structure of the partially protonated form (or zwitterion; see Fig. 3-9) below, draw the chemical structures of the other two forms of alanine that predominate in aqueous solution: the fully protonated form and the fully deprotonated form. Of the three possible forms of alanine, which would be present at the highest concentration in solutions of the following pH: (b) 1.0; (c) 6.2; (d) 8.02; (e) 11.9. Explain your answers in terms of pH relative to the two pKa values.
At pH > 6, the carboxyl groups of poly(Glu) are deprotonated; repulsion among negatively charged carboxylate groups leads to unfolding. Similarly, at pH 7, the amino groups of poly(Lys) are protonated; repulsion among these positively charged groups also leads to unfolding.
4. Effect of pH on the Conformation of α-Helical Secondary Structures The unfolding of the α helix of a polypeptide to a randomly coiled conformation is accompanied by a large decrease in a property called specific rotation, a measure of a solution's capacity to rotate circularly polarized light. Polyglutamate, a polypeptide made up of only L-Glu residues, has the α-helix conformation at pH 3. When the pH is raised to 7, there is a large decrease in the specific rotation of the solution. Similarly, polylysine (L-Lys residues) is an α helix at pH 10, but when the pH is lowered to 7 the specific rotation also decreases, as shown by the following graph. What is the explanation for the effect of the pH changes on the conformations of poly(Glu) and poly(Lys)? Why does the transition occur over such a narrow range of pH?
K′eq = 21; ΔG′° = −7.6 kJ/mol
4. Experimental Determination ofK′eqKeq′and ΔG′° If a 0.1 M solution of glucose 1-phosphate at 25 °C is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Glucose1-phosphate (4.5×10−3M)⇌glucose6-phosphate (9.6×10−2M) ] Calculate K′eq and ΔG′° for this reaction.
a) imidazole charged, COO-, NH3+ at pK1 = 1.82 imidazole uncharged, COO-, NH3+ at pKR = 6.0 imidazole uncharged, COO-, NH2 at pK2 = 9.17 b) pH1: imidazole chraged, NH3+, COOH (fully protonated) pH4: imidazole charged, COO-, NH3+ half protonated) pH8: imidazole uncharged charged, COO-, NH3+ (half way deprotonated) pH12: imidazole uncharged, COO-, NH2 (deprotonated)
4. Ionization State of Histidine Each ionizable group of an amino acid can exist in one of two states, charged or neutral. The electric charge on the functional group is determined by the relationship between its pKa and the pH of the solution. This relationship is described by the Henderson-Hasselbalch equation. (a) Histidine has three ionizable functional groups. Write the equilibrium equations for its three ionizations and assign the proper pKa for each ionization. Draw the structure of histidine in each ionization state. What is the net charge on the histidine molecule in each ionization state? (b) Draw the structures of the predominant ionization state of histidine at pH 1, 4, 8, and 12. Note that the ionization state can be approximated by treating each ionizable group independently. (c) What is the net charge of histidine at pH 1, 4, 8, and 12? For each pH, will histidine migrate toward the anode (+) or cathode (−) when placed in an electric field?
(a) Disulfide bonds are covalent bonds, which are much stronger than the noncovalent interactions that stabilize most proteins. They cross-link protein chains, increasing their stiffness, mechanical strength, and hardness. (b) Cystine residues (disulfide bonds) prevent the complete unfolding of the protein.
5. Disulfide Bonds Determine the Properties of Many Proteins Some natural proteins are rich in disulfide bonds, and their mechanical properties (tensile strength, viscosity, hardness, etc.) are correlated with the degree of disulfide bonding. (b) Most globular proteins are denatured and lose their activity when briefly heated to 65 °C. However, globular proteins that contain multiple disulfide bonds often must be heated longer at higher temperatures to denature them. One such protein is bovine pancreatic trypsin inhibitor (BPTI), which has 58 amino acid residues in a single chain and contains three disulfide bonds. On cooling a solution of denatured BPTI, the activity of the protein is restored. What is the molecular basis for this property?
(a) Asp (b) Met (c) Glu (d) Gly (e) Ser
5. Separation of Amino Acids by Ion-Exchange Chromatography Mixtures of amino acids can be analyzed by first separating the mixture into its components through ion-exchange chromatography. Amino acids placed on a cation-exchange resin (see Fig. 3-17a) containing sulfonate ( —SO−3—SO3−) groups flow down the column at different rates because of two factors that influence their movement: (1) ionic attraction between the sulfonate residues on the column and positively charged functional groups on the amino acids, and (2) aggregation of nonpolar amino acid side chains with the hydrophobic backbone of the polystyrene resin. For each pair of amino acids listed, determine which will be eluted first from the cation-exchange column by a pH 7.0 buffer. (a) Aspartate and lysine (b) Arginine and methionine (c) Glutamate and valine (d) Glycine and leucine (e) Serine and alanine
(a) −1.68 kJ/mol (b) −4.4 kJ/mol (c) At a given temperature, the value of ΔG′° for any reaction is fixed and is defined for standard conditions (here, both fructose 6-phosphate and glucose 6-phosphate at 1 M). In contrast, ΔG is a variable that can be calculated for any set of reactant and product concentrations.
6. Difference between ΔG′° and ΔG Consider the following interconversion, which occurs in glycolysis (Chapter 14): Fructose6-phosphate⇌glucose6-phosphate K′eq=1.97 (a) What is ΔG′° for the reaction ( K′eq measured at 25 °C)? (b) If the concentration of fructose 6-phosphate is adjusted to 1.5 M and that of glucose 6-phosphate is adjusted to 0.50 M, what is ΔG? (c) Why are ΔG′° and ΔG different?
(a) Bends are most likely at residues 7 and 19; Pro residues in the cis configuration accommodate turns well. (b) The Cys residues at positions 13 and 24 can form disulfide bonds. (c) External surface: polar and charged residues (Asp, Gln, Lys); interior: nonpolar and aliphatic residues (Ala, Ile); Thr, though polar, has a hydropathy index near zero and thus can be found either on the external surface or in the interior of the protein.
7. Amino Acid Sequence and Protein Structure Our growing understanding of how proteins fold allows researchers to make predictions about protein structure based on primary amino acid sequence data. Consider the following amino acid sequence. 1Ile−2Ala−3His−4Thr−5Tyr−6Gly−7Pro−8Phe9-Glu−10Ala−Ile1−Ala2−His3−Thr4−Tyr5−Gly6−Pro7−Phe8-Glu9−Ala10− 11Ala−12Mat−13Cys−14Lys−15Trp−16Glu−17Ala−18Gln19-Pro−20Asp−Ala11−Mat12−Cys13−Lys14−Trp15−Glu16−Ala17−Gln18-Pro19−Asp20− 21Gly−22Met−23Glu−24Cys−25Ala−26Phe−27His−28ArgGly21−Met22−Glu23−Cys24−Ala25−Phe26−His27−Arg28 (a) Where might bends or β turns occur? (b) Where might intrachain disulfide cross-linkages be formed? (c) Assuming that this sequence is part of a larger globular protein, indicate the probable location (external surface or interior of the protein) of the following amino acid residues: Asp, Ile, Thr, Ala, Gln, Lys. Explain your reasoning. (Hint: See the hydropathy index in Table 3-1.)
a) Structure at pH 7: pK2 = 8.03 (NH3+), pK1 = 3.39 (COO-) (b) Electrostatic interaction between the carboxylate anion and the protonated amino group of the alanine zwitterion favorably affects ionization of the carboxyl group. This favorable electrostatic interaction decreases as the length of the poly(Ala) increases, resulting in an increase in pK1. (c) Ionization of the protonated amino group destroys the favorable electrostatic interaction noted in (b). With increasing distance between the charged groups, removal of the proton from the amino group in poly(Ala) becomes easier and thus pK2 is lower. The intramolecular effects of the amide (peptide bond) linkages keep pKa values lower than they would be for an alkyl-substituted amine.
7. Comparing the pKa Values of Alanine and Polyalanine The titration curve of alanine shows the ionization of two functional groups with pKa values of 2.34 and 9.69, corresponding to the ionization of the carboxyl and the protonated amino groups, respectively. The titration of di-, tri-, and larger oligopeptides of alanine also shows the ionization of only two functional groups, although the experimental pKa values are different. The trend in pKa values is summarized in the table. (a) Draw the structure of Ala-Ala-Ala. Identify the functional groups associated with pK1 and pK2. (b) Why does the value of pK1increase with each additional Ala residue in the oligopeptide? (c) Why does the value of pK2decrease with each additional Ala residue in the oligopeptide?
(b), (e), (g)
7. Effect of Enzymes on Reactions Which of the listed effects would be brought about by any enzyme catalyzing the following simple reaction? (a) Decreased K′eq (b) increased k1 (c) increased K′eq (d) increased ΔG‡ (e) decreased ΔG‡ (f) more negative ΔG′° (g) increased k2.
a. amino, hydroxyl b. hydroxyls c. phosphoryl, carboxyl d. amino, carboxyl, hydroxyl, methyl e. methyl, carboxyl, amido, hydroxyl, methyl, hydroxyl f. hydroxyl, aldehyde, amino, hydroxyls
7. Identification of Functional Groups Figures 1-17 and 1-18 show some common functional groups of biomolecules. Because the properties and biological activities of biomolecules are largely determined by their functional groups, it is important to be able to identify them. In each of the compounds below, circle and identify by name each functional group.
(c) The upper curve corresponds to enzyme B ([X] > Km for this enzyme); the lower curve, enzyme A.
8. Relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation (c) An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species. The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme?
#residues = Molecular Weight/110 (682)(110)=75,000
8. The Size of Proteins What is the approximate molecular weight of a protein with 682 amino acid residues in a single polypeptide chain?
(a) 0.2 μM s−1 (b) 0.6 μM s−1 (c) 0.9 μM s−1
9. Applying the Michaelis-Menten Equation I An enzyme catalyzes the reaction A ⇌ B. The enzyme is present at a concentration of 2 nM, and the Vmax is 1.2 μM s−1. The Km for substrate A is 10 μM. Calculate the initial velocity of the reaction, V0, when the substrate concentration is (a) 2 μM, (b) 10 μM, (c) 30 μM.
10
9. The ΔG′° for Coupled Reactions Glucose 1-phosphate is converted into fructose 6-phosphate in two successive reactions: Glucose1-phosphate→glucose6-phosphate Glucose6-phosphate→fructose6-phosphate Using the ΔG′° values in Table 13-4, calculate the equilibrium constant, K′eq, for the sum of the two reactions: Glucose1-phosphate→fructose6-phosphate
3. Solubility of Ethanol in Water Explain why ethanol (CH3CH2OH) is more soluble in water than is ethane (CH3CH3 ).
Ethanol has OH making it polar, whereas ethane does not & so ethanol can H-bond with water
2. Biological Advantage of Weak Interactions The interactions between biomolecules are often stabilized by weak interactions such as hydrogen bonds. How might this be an advantage to the organism?
For organisms to easily break and reform bonds without the need for too much energy... reversible interactions
10. Physical Meaning of pKa Which of the following aqueous solutions has the lowest pH: 0.1 M HCl; 0.1 M acetic acid (pKa = 4.86); 0.1 M formic acid (pKa = 3.75)?
low pH = low pKa => STRONG ACID 0.1M HCl = 1, acetic acid= 4.86, formic acid= 3.75... lowest given pKa is formic acid & between fomic acid & HCl, HCl is a stronger acid 0.1 M HCl
4. Calculation of pH from Hydrogen Ion Concentration What is the pH of a solution that has an H + concentration of (a) 1.75 × 10^(- 5) mol/L (b) 6.50 × 10^(-10) mol/L (c) 1.0 × 10^(-4) mol/L (d) 1.50 × 10^(-5) mol/L?
pH = -log[H+] (a) -log91.75 x 10^(-5)= 4.76 (b) 9.19 (c) 4.0 (d) 4.82
