Given that pK = 2; pK = 9; pK = 11, Draw the ionization states and calculate the pI of 1 2 R lysine . Write the PI on the line below 3points for correct ionization state structures, 1 point for correct pI PI=9+11/2=10 pI = 10 3 points for drawing and 1 point for PI 2. [4 points] Draw the structure of Tyr-Ser-Asp in the ionic form that predominates at pH 7. Draw all atoms (including hydrogens) as well as any charges necessary. O O O H H H N CH C N CH C N CH C O 3 CH2 CH2 CH2 OH C O O OH 2 points for correct structure (side chains and amide bond linkages), 1 point for putting correct amino acids at the n terminus and c terminus, 1 point for getting the ionization states correct including all side chains and the n and c terminus
The amino acid substitution of Val for Glu in Hemoglobin S results in aggregation of the protein because of ___________ interactions between molecules. A) covalent B) disulfide C) hydrogen bonding D) hydrophobic E) ionic Circle the correct answer. 4. [2 points] In the sequencing of peptides by tandem mass spectrometry, peptide molecules are fragmented into smaller molecules to determine the amino acid sequence of the peptide. The types of bonds broken to generate MSMS tandem mass spectra are A) carbon-carbon bonds within the amino acid side chains B) disulfide bonds C) amide bonds D) bonds between alpha carbons and amino groups in the polypeptide backbone E) bonds between alpha carbons and carbonyl groups in the polypeptide backbone Circle the correct answer. 5. [2 points] A B C D E Leu-Ala-Phe Tyr-Lys-Met Asp-Pro-Glu Gly-Pro-Arg Asp-Trp-Tyr Which one of the above tripeptides: __C__(a) is most negatively charged at pH 7? __A__(b) contains the largest number of nonpolar R groups? 6. [2 points] An α helix would be destabilized most by: A) interactions between neighboring Asp and Arg residues. B) interactions between two adjacent hydrophobic Val residues. C) the presence of an Arg residue near the carboxyl terminus of the α helix. D) the presence of two Lys residues near the amino terminus of the α helix. E) the presence of a Glu residue near the amino terminus of the α helix. Circle the correct answer. 7. [2 points] The major reason that antiparallel β-stranded protein structures are more stable than parallel β-stranded structures is that the latter: A) are in a slightly less extended configuration than antiparallel strands. B) do not have as many disulfide crosslinks between adjacent strands. C) do not stack in sheets as well as antiparallel strands. D) have fewer lateral hydrogen bonds than antiparallel strands. E) have weaker hydrogen bonds laterally between adjacent strands. Circle the correct answer.
Experiments on denaturation and renaturation in the enzyme ribonuclease (RNase) by Dr. Anfinsen have shown that: A) folding of denatured RNase into the native, active conformation, requires the input of energy in the form of heat. B) native ribonuclease does not have a unique secondary and tertiary structure. C) the completely unfolded enzyme, with all —S—S— bonds broken, is still enzymatically active. D) the enzyme, dissolved in water, is thermodynamically stable relative to the mixture of amino acids whose residues are contained in RNase. E) the primary sequence of RNase is sufficient to determine its specific secondary and tertiary structure. Circle the correct answer. 9. [2 points] Which one phrase best describes the hydrogen bonds in a β sheet? A) they occur mainly near the amino and carboxyl termini of the strands. B) they are perpendicular to the plane of the sheet. C) they occur mainly between atoms of the R groups. D) they occur only between some of the amino acids of each strand. E) they occur mainly between atoms of adjacent strands in a sheet. Circle the correct answer. 10. [6 points] Supply the missing words in each of the following 3 sentences. Choose from increase, decrease, or remain unchanged. Increasing the concentration of HCO3– in a solution of hemoglobin (without changing the pH) causes the affinity for O to _____decrease_. 2 Increasing the pH of a solution of hemoglobin causes the affinity for O to _______ increase __. 2 Removing 2,3 BPG from a solution of adult hemoglobin causes the affinity for O to _ increase _. 2 14. [4 points] Name two examples of proteins with atypical protein structures discussed in lecture 2 and explain briefly the key characteristics of these atypical protein structures . 2 points for naming them. Myosin, keratin or collagen 2 points for explaining the concept of coiled-coils. 1 point for either of the following concepts: – myosin and keratin contain two right-handed helices twisted in a left-handed superhelix – amphiphilic helices – collagen contains three left-handed helices in a right-handed superhelix – rope-like structures -3.5 residues per turn myosin keratin -collagen contains Gly at center of triple helix -collagen contains Pro hydroxyproline
Enzymes catalyze the rapid formation of product by: A) shifting the position of the equilibrium between substrate and product in favor of product. o B) lowering ΔG´ of the reaction. C) raising the free energy of the substrate. D) raising ΔG‡ E) none of the above. Circle the correct answer. 13. [5 points] Write out the simple chemical reaction that describes the mechanism for enzyme action used as a model by Michaelis and Menten. Indicate the rate constants for all steps in this reaction. List the assumptions used by Michaelis and Menten to derive a rate equation for this reaction. Finally write down the Michaelis Menten equation. The chemical reaction (with rate constants) is: 2 points k k k 1 2 3 E + S ES EP E + P k-1 k-2 k-3 Assumptions: One assumption is that significant product has not accumulated, so that the rate of the reaction depends exclusively on the breakdown of ES and is not influenced by the reverse reaction; that is, k-2 can be ignored and V = k [ES]. This condition is possible only if early reaction times are 0 2 measured; the velocity, therefore, is an initial velocity. A second assumption is that the rate of ES formation equals the rate of ES breakdown; in other words, the reaction is at a steady state. A third assumption is [S] >> [E ], so that total [S], which equals free substrate and enzyme-bound substrate, t is essentially equal to [S]. 2 points for assumptions (1 each) Michaelis Menten equation: V =V [S]/(K +[S]) 0 max m 1 point 11. [4 points] In lecture 3 we discussed the T and R states describing the quaternary structure of hemoglobin. Describe briefly the two principal models for the cooperative binding of ligands to proteins such as hemoglobin with multiple ligand binding sites. Be sure to name both models and provide an explanation of the differences between the two. Ans: In the concerted model, binding of a ligand to one site on one subunit results in an allosteric effect that converts all of the remaining subunits to the high-affinity conformation. As a result, all of the subunits are either in the low- or high-affinity conformation. In the sequential model, each subunit is changed individually to the high affinity conformation. As a result, there are many possible combinations of low- and high-affinity subunits.
An enzyme follows Michaelis-Menten kinetics. Indicate (with an “x”) which of the kinetic parameters at the left would be altered by the following factors. Give only one answer for each. K V Neither Both Factor m max X competitive inhibitor X noncompetitive inhibitor 16. [2 points] Which amino acids in chymotrypsin are found in the active site and are participants in substrate cleavage? A) his, ser, asp B) his, ser, glu C) asp, lys D) lys, arg E) his, ser, arg Circle the correct answer. 17. [2 points] How is specificity determined by chymotrypsin? A) interaction of the substrate with the oxyanion hole B) binding of the N-terminus amino acid at the active site C) covalent binding of a his residue to the substrate D) conformational change upon binding of substrate E) binding of the proper amino acid into a deep pocket on the enzyme Circle the correct answer. 18. [2 points] In the following oligopeptide, indicate by drawing vertical line(s) where chymotrypsin would hydrolyze peptide bond(s). (You cannot receive credit if your vertical line(s) is/are drawn through a letter) : D—A—V—T—W–|-C—L—A—F–|-T—K—S—G 19. [2 points] Catalytic antibodies are most likely to be active if they are made by using an antigen that: A) resembles the substrate. B) is composed of an equimolar mixture of substrate and product. C) resembles the enzyme-substrate complex. D) resembles a transition state intermediate. E) is an allosteric modulator. Circle the correct answer. 20. [4 points] The Michaelis-Menten constant, Km, is actually a summary of three rate constants. Write down the equation to calculate Km based on these rate constants. Describe a method to determine Km graphically indicating precisely how K is measured. m K = (k + k )/k m 2 –1 1 K can be determined graphically on a plot of V vs. [S] by finding the [S] at which V = 1/2 m 0 0 V . More conveniently, on a double-reciprocal plot, the x-axis intercept = –1/ K . max m 2 points for Km equation, 2 points for Km from V vs. S plot, or 2 points for double reciprocal plot x intercept of -1/Km
How is chymotrypsinogen converted to chymotrypsin? A) A protein kinase-catalyzed phosphorylation converts chymotrypsinogen to chymotrypsin. B) An increase in Ca2+ concentration promotes the conversion. C) Proteolysis of chymotrypsinogen forms chymotrypsin. D) Chymotrypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active chymotrypsin monomers. E) Two inactive chymotrypsinogen dimers pair to form an active chymotrypsin tetramer. Circle the correct answer. 22. [2 points] The enzyme ATCase is regulated by CTP, which binds to the T-state of ATCase. CTP is a: A) heterotropic positive allosteric regulator. B) heterotropic negative allosteric regulator. C) cofactor. D) competitive inhibitor. E) homotropic negative allosteric regulator. Circle the correct answer. 23. [2 points] Circle the fatty acid in each pair that has the lower melting temperature. A) 18:1Δ9 18:2Δ 9,12 B) 16:0 18:0 24. [2 points] Vitamin A (retinol), vitamin K (phylloquinone), testosterone, and cortisol are all members of a single family of lipids. Draw the structure of the basic building block that is used to assemble all of these molecules. 25. [4 points] The bacterium E. coli can grow at 20 °C or at 40 °C. At which growth temperature would you expect the membrane phospholipids to have a lower ratio of unsaturated to saturated fatty acids, and why? At 40 °C, the membranes of E. coli will contain a lower ratio of unsaturated to saturated fatty acids than at 20 °C. The cell regulates fatty acid composition to achieve the same fluidity in its membranes, regardless of growth temperature. Saturated fatty acids counterbalance the fluidizing effect of high temperature by providing additional hydrophobic interactions.
Glucose transport through intestinal epithelial cells involves two types of transporters. Glucose is first transported into the intestinal epithelial cell by _________ and then it is transported out by _________. A) secondary active transport / primary active transport B) facilitated diffusion / secondary active transport C) primary active transport / secondary active transport D) secondary active transport / facilitated diffusion E) secondary active transport / secondary active transport Circle the correct answer. + + 27. [2 points] In one catalytic cycle, the Na /K ATPase transporter transports: + + A) 2 Na out, 3 K in, and converts 1 ATP to ADP + P . i + + B) 3 Na out, 2 K in, and converts 1 ATP to ADP + P . i + + C) 3 Na in, 2 K out, and converts 1 ATP to ADP + P . i + + D) 1 Na out, 1 K in, and converts 1 ATP to ADP + P . i + + E) 2 Na out, 3 K in, and converts 1 ADP + P to ATP. i Circle the correct answer. 28. [2 points] Which one of the following chemical modifications (I-IV) would result in the conversion of D-fructofuranose from a reducing carbohydrate to a non-reducing carbohydrate? I CH3 A) II H C O 3 IV B) I O a HO a OH e O f C) IV H C 2 b b H OH e OH D) III H H c d H C f E) None of the above OH H 2 OH Circle the correct answer. H III CH3 c II O d O CH3 H 29. [2 points] Label each of the following as an aldose, a ketose, or neither. H H H H H H C OH O C H C OH HO C H C O C O H C OH HO C H H C OH H C OH H C OH HO C H HO C H C O H C H H H H H H (a) (b) (c) (d) (e) ___Ketose__ __Aldose___ __neither__ ___aldose_
Indicate in the blanks provided the match between the molecules and their biological roles. (a) peptidoglycan ____d____ homopolysaccharide of glucose in animals (b) starch ____b____ homopolysaccharide of glucose in plants (c) chitin ____c____ exoskeleton of lobsters (d) glycogen ____a____ structural component of bacterial cell walls 33. [3 points] Indicate the oxidation state of each carbon in the fructose molecule shown below: Oxidation State C1 _____5______ C2 _____2______ C3 _____4______ C4 _____4______ C5 _____4______ C6 _____5______ 31. [2 points] What factor contributes most significantly to the thermodynamic stability of the monosaccharide glucose relative to other monosaccharide six carbon D-aldoses? Most thermodynamically stable form of glucose is the form with all non-hydrogen substituents in the equatorial position. No other 6 carbon D-aldose is fully equatorial 32. [4 points] (a) Describe one biological advantage of storing glucose units in branched polymers (glycogen, amylopectin) rather than in linear polymers. The enzymes that act on these polymers to mobilize glucose for metabolism act only on their nonreducing ends. With extensive branching, there are more such ends for enzymatic attack than would be present in the same quantity of glucose stored in a linear polymer. In effect, branched polymers increase the substrate concentration for these enzymes. (b) What is the biological advantage to an organism that stores its carbohydrate reserves as starch or glycogen rather than as an equivalent amount of free glucose? The polymers are essentially insoluble and contribute little to the osmolarity of the cell, thereby avoiding the influx of water that would occur with the glucose in solution. They also make the uptake of glucose energetically more feasible than it would be with free glucose in the cell. Full credit for recognition of concentration difference or full credit for concept of less reducing ends
Standard reduction potentials are tabulated as: Oxidized form → Reduced form + electrons. A) True B) False Catabolic pathways diverge, anabolic pathways converge. A) True B) False Circle the correct answer. 36. [2 points] One molecule of NADH is produced per glucose in the conversion of glucose to lactate A) True B) False Circle the correct answer. Triose phosphate isomerase decreases ΔG for the step in glycolysis catalyzed by aldolase through rapid removal of dihydroxyacetone phosphate. A) True B) False Circle the correct answer 37. [2 points] The pentose phosphate pathway can alternatively be called the pentose phosphate cycle because _______ is a net product of the pathway that can be recycled. A) phosphate B) NADP+ C) carbon dioxide D) ribose 5-phosphate E) UDP-glucose Circle the correct answer. 38. [2 points] The conversion of pyruvate to ethanol also causes the ______. A) Oxidation of NADH B) Production of ADP C) Release of CO 2 D) Generation of an ion gradient across mitochondrial membranes E) Both A and C Circle the correct answer. 39. [2 points] ATP is both a substrate and a negative effector (inhibitor) for the allosteric, glycolytic enzyme phosphofructokinase (PFK). Which of the following answers best explains this observation? A) ATP has equal affinity for the active site and the regulatory (allosteric) site of PFK B) ATP only binds the regulatory site of PFK at low concentrations C) ATP only binds the regulatory site of PFK at high concentrations D) ATP and AMP are both negative effectors of PFK Circle the correct answer.