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).
A patient seen in the ER has ingested antifreeze in a suicide attempt. Other than bicarbonate, which one of the following is the major buffer of acids to help maintain the pH in the blood within the range compatible with life? (A) Hemoglobin (B) Acetoacetate (C) β-Hydroxybutyrate (D) Phosphate (E) Collagen
The answer is A. Antifreeze contains ethylene glycol (HO–CH2–CH2–OH), and ingestion of ethylene glycol (which has a sweet taste) will lead to a metabolic acidosis due to the metabolism of ethylene glycol to glycolic and oxalic acids. As the acids form, protons are released, and bicarbonate and hemoglobin are the major buffers in the blood that will bind these protons to blunt the drop in blood pH. Acetoacetate and β-hydroxybutyrate are ketone bodies produced by the liver, and since they are both acids, their accumulation is often a cause of metabolic acidosis. Increasing their synthesis under acidotic conditions would only exacerbate the acidosis. Phosphate is an intracellular buffer, but its role is not as significant as that of either hemoglobin or bicarbonate.
Which one of the following is the amino acid in hemoglobin that accepts H+ and allows hemoglobin to act as a buffer to acids? (A) Alanine (B) Histidine (C) Serine (D) Threonine (E) Aspartate
The answer is B. The side chain of histidine has a pKa of 6.0, which, of all amino acid side chains, is the one closest to physiologic pH. The local environment of the protein can raise this pKa value closer to 7 such that the histidine side chains within hemoglobin will be the major groups that accept and donate protons when hemoglobin acts as a buffer. The alanine side chain (a methyl group) cannot accept or donate protons. The pKa for the side chains of serine or threonine are above 10.0, so at physiologic pH these side chains are always protonated, and cannot act as a binding site for excess protons generated during an acidotic event. The pKa for the side chain of aspartate is about 4.0, so at physiologic pH that group is always deprotonated, and will not accept protons generated during an acidotic event.
A patient is going skiing high in the Rockies, and is given acetazolamide to protect against altitude sickness. Unfortunately, the patient is also a Type 1 diabetic. He is admitted
to the hospital in a worsening ketoacidosis. In which of the following cells has acetazolamide inhibited a reaction that has led to the severity of the metabolic acidosis?
(A) White blood cells
(B) Red blood cells
(C) Lens of the eye
(D) Hepatocyte
(E) Muscle
The answer is B. Acetazolamide is a carbonic anhydrase inhibitor, which is found primarily in red blood cells. The red blood cells contain carbonic anhydrase that catalyzes the reaction that forms carbonic acid from CO2 and H2O. Under high-altitude conditions, the inhibition of carbonic anhydrase will lead to a decrease in blood pH, which stabilizes the deoxygenated form of hemoglobin. This is due to an increased loss of bicarbonate in the urine by the inhibition of carbonic anhydrase within the kidney. The change in pH increases oxygen delivery
to the tissues, and can overcome, in part, the symptoms of altitude sickness. However, in the case of the person with Type I diabetes who begins to produce ketone bodies, the body’s main compensatory mechanism to overcome the acidosis is blocked. As ketone bodies are formed and protons generated, the H+ will react with bicarbonate to form carbonic acid. Carbonic anhydrase, which catalyzes a reversible reaction, will then convert the carbonic acid to CO2 and H2O, with the CO2 being exhaled. These reactions soak up excess protons, and help to buffer against the acidosis. If, however, carbonic anhydrase has been inhibited by acetazolamide, then the bicarbonate cannot buffer the blood pH and the acidosis could become more severe. White blood cells, muscle cells, liver cells, and the lens of the eye do not contribute to the buffering of the blood, and inhibition of carbonic anhydrase in those cells would not affect the ability to overcome an acidosis.
A 23-year-old female patient presents to the ER with a feeling of being unable to catch her breath, light-headedness, and “tingling” of her fingers, toes, and around her mouth. This happens whenever she drives through a tunnel, and that is what set off this episode. Which of the following arterial blood pHs would be most consistent with her diagnosis? (A) 8.10 (B) 7.55 (C) 7.15 (D) 6.40 (E) 6.10
The answer is B. The patient is having a panic attack (due to driving in tunnels) and is hyper- ventilating, causing an acute respiratory alkalosis. The loss of CO2 pushes the carbonic anhydrase reaction in the direction of CO2 production, which reduces the proton concentration (and thereby raising the pH). The patient would lose consciousness with a more severe attack. A respiratory alkalosis is usually mild as compared to a metabolic alkalosis. For this reason, the pH increase is smaller (7.55 is more likely than a pH of 8.10, which could occur via a meta- bolic alkalosis). The other choices given, 7.15, 6.40, and 6.10, are all lower than physiologic pH (7.4), and would be considered acidosis, instead of alkalosis. An example of a metabolic alka- losis is hypokalemia, a reduction in normal potassium values. Owing to low serum potassium levels, potassium leaves the cells and is replaced by protons from the circulation. The loss of protons from the blood leads to the alkalosis.
A new antibiotic has been developed that shows a strong affinity for attacking amino acids with a specific orientation in space. In order for it to work well in humans, the antibiotic must be effective against amino acids in which one of the following configurations? (A) R-configuration (B) L-configuration (C) Aromatic ring configuration (D) Polypeptide chain configuration (E) D-configuration
The answer is E. Amino acids in humans are in the L-configuration (except glycine which is neither L nor D), whereas bacterial amino acids can be in either the L- or D-configuration. An antibiotic would need to be effective against bacterial proteins and not human proteins, so developing an antibiotic that recognizes proteins or polypeptides that contain D amino acids would only be effective against bacterial products. All amino acids are in polypeptide chains, and phenylalanine, tyrosine, and tryptophan are amino acids that contain aromatic rings, and are present in both bacteria and humans. The R and S nomenclature is not commonly used in biochemistry to describe the configuration of amino acids.
A 40-year-old tobacco farmer is seen in the ER with bradycardia, profuse sweating, vomiting, increased salivation, and blurred vision. He was spraying his field with malathion when the hose ruptured and he was covered with the malathion. Which of the following types of inhibition of enzymes does this poisoning represent? (A) Competitive (B) Noncompetitive (C) Irreversible (D) Reversible (E) The drug does not work by inhibiting enzyme activity.
The answer is c. Malathion is an organophosphate that inhibits the action of acetylcholinesterase in an irreversible manner. It is one of the most common causes of poisoning worldwide. Malathion forms an irreversible covalent bond between the inhibitor and the active site serine side chain of the enzyme. Without acetylcholinesterase, acetylcholine accumulates in the neuromuscular junction and causes the symptoms described in the case. Both competitive and noncompetitive inhibition are reversible forms of inhibition, and their mechanism of action does not apply to malathion.
A baby did well with no discernable problems until 3 months of age when he began having cyanotic spells and later had a myocardial infarction. An autopsy revealed a congenital defect, where the left main coronary artery arose from the pulmonary artery instead of the aorta. Of the following, which is the most likely reason that he showed no symptoms until
3 months of age?
(A) He had abnormal α chains of hemoglobin.
(B) He had abnormal β chains of hemoglobin.
(C) He had abnormal γ chains of hemoglobin.
(D) HbF has a lower affinity for BPG.
(E) HbF has a higher affinity for BPG.
The answer is d. Fetal hemoglobin (HbF) is composed of two α subunits and two γ subunits. It has a lower affinity for BPG and therefore a higher affinity for oxygen. The congenital defect in the child has led to a reduced oxygenation of the blood supplying the heart (the end organ of the coronary arteries). The coronary arteries normally receive oxygenated blood from the aorta, but in this case, the left main coronary artery is supplying the left ventricle with deoxygenated blood from the pulmonary artery. However, as the child matures and begins producing HbA instead of HbF, there is insufficient oxygen being delivered to the aorta, leading to a myocardial infarction due to the lack of oxygen reaching the heart. As the HbF was replaced with HbA (two α subunits and two β subunits), the lower affinity for oxygen became manifest since the pulmonary artery blood is lower in oxygen than the left ventricle, leading to a myocardial infarction from the lack of O2 delivered to the myocardium. If the child had been born with abnormal γ chains, the deficiency would have been manifest at birth, and not first appeared at 3 months of age.
A 28-year-old female presents with fluctuating fatigue, drooping of her eyelids, difficulty swallowing, and slurred speech. The patient is given a drug that affects an enzyme’s activity, and kinetic analysis of the enzyme-catalyzed reaction, in the presence and absence of the drug, is shown below. The effect of this medication can best be described by which set of terms below? (see page 42 question 23)
Type of inhibition / Effect on Km (as compared to no drug) / Effect on Vmax (as compared to no drug)
(A) Competitive / Increased / No change
(B) Competitive / Decreased / Decreased
(C) Noncompetitive / Increased / No change
(D) Noncompetitive / Decreased / Increased
(E) Irreversible / Increased / No change
(F) Irreversible / Decreased / Decreased
The answer is A. The patient has myasthenia gravis, and the treatment is pyridostigmine,
a competitive, reversible inhibitor of acetylcholinesterase. Myasthenia gravis is caused by autoantibodies to the acetylcholine receptor, reducing the effectiveness of acetylcholine at the neuromuscular junction. By reversibly inhibiting acetylcholinesterase, the effective levels of acetylcholine are increased, thereby providing sufficient acetylcholine to bind to the few functional receptors that remain. The graph is classic for a competitive inhibitor. Competitive inhibitors display an increased apparent Km, and a constant Vmax.
A patient who wanted to go skiing in the Rockies took a medication to combat altitude sickness. A graph of this medication’s mechanism of inhibition is shown below. What type of inhibitor is this medication? (see page 42 question 24)
(A) Competitive
(B) Noncompetitive
(C) Irreversible
(D) Allosteric
(E) The drug is not an inhibitor, rather it is an
activator.
The answer is B. In noncompetitive inhibition, the Km is not altered, whereas the Vmax is de- creased. The drug used to treat altitude sickness is acetazolamide, which is a noncompetitive inhibitor of carbonic anhydrase. The graph is one of a pure noncompetitive inhibitor.
