Chapter 2 - Basic Aspects of Biochemistry: Organic Chemistry, Acid-Base Chemistry, Amino Acids, Protein Structure and Function, and Enzyme Kinetics Flashcards
The bonds labeled A and B in the compound shown are best described as which one of the following?
(A) C–O–P
(B) P–OCH3
(A) Anhydride / Ester (B) Ester / Anhydride (C) Ether / Ester (D) Ester / Ether (E) Phosphodiester / Anhydride (F) Anhydride / Phosphodiester
The answer is A. Bond A is an anhydride bond, which is formed when a carboxylic acid and a phosphoric acid react, releasing H2O. Bond B is a phosphate ester, formed when phosphoric acid reacts with an alcohol (methanol in this case), releasing water. An ether linkage is not found in this structure (a–C–O–C–linkage), nor is a phosphodiester (when a phosphate group contains two ester linkages, as in the structures of the nucleic acids).
Both β-hydroxybutyrate and acetoacetate are ketone bodies, which are produced by the liver under conditions of extended fasting. The two ketone bodies are easily interconverted depending upon the conditions present within the mitochondria, where they are synthesized. The conversion of β-hydroxybutyrate to aceto- acetate occurs by what type of reaction? (A) Oxidation (B) Reduction (C) Dehydration (D) Dehydroxylation (E) Decarboxylation
The answer is A. An alcohol is oxidized to a ketone when β-hydroxybutyrate is converted to acetoacetate (look at the change in the β-carbon, carbon 3). These compounds are classically described as ketone bodies, although technically only acetoacetate contains a ketone.
When the pH of a solution of a weak acid, HA, is equal to the pKa, the ratio of the concen- trations of the salt and the acid ([A–]/[HA]) is which one of the following? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4
The answer is B. The pKa is the pH at which the functional group is 50% dissociated, which in this case is the pH at which [A−] = [HA]. The Henderson–Hasselbalch equation, pH = pKa + log10 [A−]/[HA], gives the relationship between these parameters. If pH = pKa, log10 [A−]/ [HA] = 0, and [A−]/[HA] = 1.
Human serum albumin, the most abundant blood protein, has multiple roles, including acting as a buffer to help maintain blood pH. Albumin can act as a buffer because of which one of the following?
(A) The protein contains a large number of amino acids.
(B) The protein contains many amino acid residues with different pKa values.
(C) The amino and carboxyl ends of albumin can donate and accept protons in the range of physiologic pH.
(D) Albumin contains peptide bonds that readily hydrolyze, consuming hydrogen and hydroxyl ions.
(E) Albumin contains a large number of hydrogen bonds in α-helices, which can accept and donate protons.
The answer is B. The side chains of the amino acid residues in proteins contain functional groups with different pKa’s. Therefore, they can donate and accept protons at various pH values and act as buffers over a broad pH spectrum. There is only one N-terminal amino group (pKa ∼ 9) and one C-terminal carboxyl group (pKa ∼ 3) per polypeptide chain. At physiologic pH, these groups would not be accepting or donating protons, as the amino terminal group would always be protonated and the carboxy terminal carboxylic acid would always be deprotonated. Peptide bonds are not readily hydrolyzed, and such hydrolysis would not provide buffering action. Hydrogen bonds have no buffering capacity, as the hydrogen in these bonds is not donated or accepted once the bond is formed.
The following question is based on the hexapeptide with the following sequence: D-A-S-E-V-R
The C-terminal amino acid of the hexapeptide is which one of the following? (A) Ala (B) Asn (C) Asp (D) Arg (E) Glu
The answer is D. By convention, peptides are written with the N-terminal amino acid on the left and the C-terminal amino acid on the right. Therefore, this peptide contains arginine (single-letter code R, three-letter code arg) at its C-terminus. The sequence of this peptide is aspartic acid (D, asp), alanine (A, ala), serine (S, ser), valine (V, val), and arginine.
The following question is based on the hexapeptide with the following sequence: D-A-S-E-V-R
At physiologic pH (7.4), this hexapeptide will contain a net charge of which one of the following? (A) –2 (B) –1 (C) 0 (D) +1 (E) +2
The answer is B. The N-terminal aspartate contains a positive charge on its N-terminal amino group and a negative charge on the carboxyl group of its side chain. The side chains of alanine and serine have no charge at physiologic pH. Glutamate contains a negative charge on the carboxyl group of its side chain. The valine side chain is hydrophobic, and has no charge. The C-terminal arginine contains a negative charge on its C-terminal carboxyl group and a positive charge on its side chain. Thus, the overall charges are +2 and -3, which gives a net charge of -1. The amino acids within the interior of this hexapeptide (alanine, serine, glutamate, and valine) have their amino and carboxy ends involved in peptide bond formation, so there are no charges associated with those groups of the internal amino acids.
Which one of the following types of bonds is covalent? (A) Hydrophobic (B) Hydrogen (C) Disulfide (D) Electrostatic (E) Van der Waals
The answer is C. Disulfide bonds are an example of covalent bonds. Hydrophobic interactions occur between hydrophobic groups as they come together in space to reduce their interactions with water, and to allow water to maximize its entropy. Hydrogen bonds are the sharing of a hydrogen atom between two electronegative atoms. While the hydrogen is covalently bound
to one of those atoms, it is also attracted to the other electronegative group (which creates the hydrogen bond) via partial charge interactions, in a noncovalent manner. Electrostatic interactions are the attraction of fully charged groups between each other (one negatively charged, such as a carboxylic acid, and one positively charged, such as a primary amine), due to the opposite charges attracting each other. Van der Waals interactions are nonspecific interactions between two atoms as they approach each other up to a certain distance; once they get too close, repulsion will occur between the two atoms.
One method to separate proteins is by charge, through a suitable gel-like substance. If normal HbA and sickle cell hemoglobin (HbS) were placed on such a gel, which molecule would migrate more rapidly to the positive pole of the gel?
(A) HbA
(B) HbS
(C) Both would have the same charge, so there
is no difference in migration.
The answer is A. The mutation in sickle cell is E6V of the β-globin chain. The sickle hemoglo- bin molecules have a valine in place of a glutamate in the two β chains within the tetramer. All other amino acids are the same, and so in comparison to normal hemoglobin, the sickle variant has two fewer negative charges. This means that the normal form of hemoglobin (HbA) will migrate more rapidly toward the positive pole of a gel because it contains more negative charges than does HbS.
The active site of an enzyme will bind to which set of molecules indicated below?
Substrate of the reaction / Allosteric inhibitors / Competitive inhibitors / Noncompetitive inhibitors
(A) Yes / Yes / Yes / Yes (B) Yes / No / Yes / Yes (C) Yes / No / Yes / No (D) No / No / No / No (E) No / Yes / No / Yes
The answer is c. The active site is formed when the enzyme folds into its three-dimensional configuration, and may involve amino acid residues that are far apart in the primary sequence. Substrate molecules bind at the active site, as will competitive inhibitors (since the inhibitor reduces enzyme activity by competing with substrate for binding at the active site). Allosteric inhibitors bind at a site other than the active site, as do noncompetitive inhibitors (which reduce the Vmax without affecting the Km).
An enzyme catalyzing the reaction E + A EA → E + P was mixed with 4 mM substrate (compound A). The initial rate of product formation was 25% of Vmax. The Km for the enzyme is which one of the following? (A) 2mM (B) 4mM (C) 9mM (D) 12 mM (E) 25 mM
The answer is d. The Michaelis–Menten equation is v = (Vmax × [S])/(Km + [S]). In this case, the velocity (v) is 1⁄4 Vmax, and [S] = 4 mM. Thus, the Michaelis–Menten equation becomes
1⁄4 Vmax = (Vmax × 4)/(Km + 4). When one solves that equation for Km, Km = 12 mM.
The liver enzyme glucokinase catalyzes the phosphorylation of glucose to glucose 6-phosphate. The value of Km for glucose is about
7 mM. Blood glucose is 5 mM under fasting conditions, and can rise in the liver to 20 mM after a high-carbohydrate meal. Therefore, if a person who is fasting eats a high-carbohydrate meal, the velocity of the glucokinase reaction will change in which one of the following ways?
(A) Remain at <50% Vmax
(B) Remain above 80% Vmax
(C) Increase from <50% Vmax to >50% Vmax
(D) Decrease from >50% Vmax to <50% Vmax
(E) Remain at Vmax
The answer is c. This problem is best solved using the Michaelis–Menton equation and com- paring the velocity (as a function of maximal velocity) under fasting and nonfasting condi- tions. During fasting, [S] = 5 mM, and the Km is 7 mM; so v = (5 × Vmax)/(7 + 5) = 42% Vmax. In the fed state, [S] = 20 mM, and the Km is 7 mM; so v = (20 × Vmax)/(7 + 20) = 74% Vmax. Glucokinase is more active in the fed than in the fasting state, and the velocity will increase from <50% Vmax to >50% Vmax.
Malonate is a competitive inhibitor of succinate dehydrogenase, a key enzyme in the Krebs tricarboxylic acid cycle. The presence of malonate will affect the kinetic parameters of succinate dehydrogenase in which one of the following ways?
(A) Increases the apparent Km but does not affect Vmax.
(B) Decreases the apparent Km but does not affect Vmax.
(c) Decreases Vmax but does not affect the apparent Km.
(d) Increases Vmax but does not affect the apparent Km.
(E) Decreases both Vmax and Km.
The answer is A. A competitive inhibitor competes with the substrate for binding to the ac- tive site of the enzyme, in effect increasing the apparent Km (in the presence of inhibitor, it will require a higher concentration of substrate to reach 1⁄2 Vmax, as the substrate is competing with the inhibitor for binding to the active site). As the substrate concentration is increased, the substrate, by competing with the inhibitor, can overcome its inhibitory effects, and eventually the normal Vmax is reached. A noncompetitive inhibitor will decrease the Vmax without affecting the binding of substrate to the active site, so the Km is not altered under those conditions. An activator of an allosteric enzyme will decrease the apparent Km without affecting Vmax (less substrate is required to reach the maximal velocity).
A young black man was brought to the emergency room (ER) owing to severe pain throughout his body. He had been exercis-
ing vigorously when the pain started. He has had such episodes about twice a year for the past 10 years. An analysis of the blood shows a reduced blood cell count (anemia), and odd- looking red blood cells that were no longer concave and looked like an elongated sausage. An underlying cause in the change of shape of these cells is which one of the following?
(A) Increased ionic interactions between hemo- globin molecules in the oxygenated state.
(B) Increased ionic interactions between hemoglobin molecules in the deoxygenated state.
(C) Increased hydrophobic interactions be- tween hemoglobin molecules in the oxygenated state.
(D) Increased hydrophobic interactions between hemoglobin molecules in the de- oxygenated state.
(E) Increased phosphorylation of hemoglobin molecules in the oxygenated state.
(F) Increased phosphorylation of hemoglobin molecules in the deoxygenated state.
The answer is d. The man has sickle cell disease, and his hemoglobin consists of mutated β chains, along with normal α chains. The glutamate at position 6 in the β chains of HbA
is replaced by valine in HbS. Valine contains a hydrophobic side chain, whereas glutamate contains an acidic side chain. Under low oxygenation conditions (such as vigorous exercise), the HbS molecules will polymerize owing to hydrophobic interactions between the valine on the β chain and a hydrophobic patch on another HbS molecule. Under well-oxygenated con- ditions, the valine in the β chain is not exposed on the surface of the molecule, and it cannot form an interaction with the hydrophobic patch on another hemoglobin molecule. Once the HbS polymerizes, it forms a rigid rod within the red blood cells, which deforms the cell and gives it the “sickle” appearance. Once sickled, the red blood cells cannot easily deform and pass through narrow capillaries, leading to loss of oxygen to certain areas of the body, which is what leads to the pain experienced by the patient. The sickling is not due to increased or decreased ionic interactions between HbS molecules, or to phosphorylation of the HbS monomers.
An individual is visiting Mexico City, which is at an altitude of 7,350 feet. The person is having trouble breathing due to difficulty in getting sufficient oxygen to the tissues. Which one of the following treatments might the per- son try to get hemoglobin to release oxygen more readily?
(A) Take a drug that initiates a metabolic alkalosis.
(B) Take a drug that increases the production of BPG.
(C) Hyperventilate, which will lead to de- creased levels of carbon dioxide in the blood.
(D) Take a drug that induces the synthesis of the γ subunits of hemoglobin.
(E) Take a drug that induces the synthesis of the β subunits of hemoglobin.
The answer is B. In order for hemoglobin to release oxygen more readily, the deoxygenated state of hemoglobin needs to be stabilized. This can occur by decreasing the pH (the Bohr ef- fect), increasing the CO2 concentration, or increasing the concentration of BPG. Fetal hemoglo- bin (HbF = α2γ2) has a greater affinity for O2 than does HbA (α2β2), so inducing the synthesis of the γ genes would have the opposite of the intended effect. Inducing the concentration of the β chains would not decrease oxygen binding to hemoglobin (in fact, if there is not a concurrent increase in α-gene synthesis, this may be quite detrimental to the individual, as an imbalance in the synthesis of the hemoglobin chains leads to a disorder known as thalassemia, and overall oxygen transport to the tissues would be decreased). Increased BPG would cause O2 to be more readily released. A metabolic alkalosis would raise the pH of the blood, which would stabilize the oxygenated form of hemoglobin. Reducing carbon dioxide levels in the blood through hyperventilation will also stabilize the oxygenated form of hemoglobin, and make it more difficult to deliver oxygen to the tissues.
An environmentalist attempted to live in a desolate forest for 6 months, but had to cut his experiment short when he began to suffer from bleeding gums, some teeth falling out, and red spots on the thighs and legs. This individual is suffering from an inability to properly synthe- size which one of the following proteins? (A) Myoglobin (B) Hemoglobin (C) Collagen (D) Insulin (E) Fibrillin
The answer is c. The environmentalist is suffering from scurvy, a deficiency of vitamin C. The hydroxylation of proline and lysine residues in collagen requires vitamin C and oxygen. In the absence of vitamin C, the collagen formed cannot be appropriately stabilized (owing, in part, to reduced hydrogen bonding between subunits due to the lack of hydroxyproline) and is easily degraded, leading to the bleeding gums and loss of teeth. Globin synthesis might be indirectly affected because absorption of iron from the intestine is stimulated by vitamin C, but globin is not modified through a hydroxylation reaction. Iron is involved in heme synthesis, which regu- lates globin synthesis. Insulin and fibrillin synthesis are not dependent on vitamin C (lack of insulin will lead to diabetes, and mutations in fibrillin lead to Marfan syndrome).
