Chapter 2: Basic Aspects of Biochemistry - Organic Chemistry, Acid-Base Chemistry, Amino Acids, Protein Structure and Function, and Enzyme Kinetics

*The answer is A.* Hemoglobin contains two α chains and two β chains, which combine to form the quaternary structure. Each mole of a subunit binds 1 mole of heme, which binds 1 mole of O2. Therefore, 1 mole of HbA binds 4 moles of O2, contains 4 moles of heme, and 4 moles of iron. Both protons (creating a reduced pH) and 2,3-bisphosphoglycerate stabilize the deoxygenated form of HbA. Carbon monoxide (CO) has a higher affinity (about 250 times) for the iron in heme than does oxygen, even when the distal histidine is present to force CO to bind to the iron at an angle.

Adult hemoglobin A (HbA) is best described by which one of the following?

*The answer is C.* This child likely has Ehlers-Danlos syndrome, a group of rare hereditary disorders characterized by defective collagen synthesis. The condition can be caused by a deficiency in procollagen peptidase, the enzyme that cleaves terminal propeptides from procollagen in the extracellular space. Impaired propeptide removal results in the formation of soluble collagen that does not properly crosslink. Consequently, patients often have joint laxity, hyperextensible skin, fragile tissue with easy bruising, and poor wound healing. Each collagen molecule consists of 3 polypeptide a-chains held together by hydrogen bonds, forming a triple helix. Collagen assumes this conformation because each of the a-chains has a simple, repetitive amino acid sequence represented as (Gly-X-Y),. The smallest amino acid, glycine (Gly), is necessary at every third position to ensure compact coiling of the helix. Many of the amino acids represented by X and Y are proline residues, which kink the polypeptide chain and enhance the rigidity of the helical structure due to their ring configuration. Mature collagen is synthesized by fibroblasts, osteoblasts, and chondroblasts through the following steps: 1. As translation begins in the cytoplasm, an amino acid signal sequence at the N-terminus of the α-chain facilitates ribosomal binding to the rough endoplasmic reticulum (RER) and passage of the growing polypeptide chain (pre-pro-α-chain) into the RER. 2. Inside the RER, the hydrophobic signal sequence is cleaved to yield the pro-α-chain. Proline and lysine at the Y positions of the pro-α-chain are hydroxylated to hydroxyproline and hydroxylysine, respectively (Choice D). Glycosylation of select hydroxylysine residues also occurs within the RER (Choice A). 3. The central helical region of the pro-α-chain is flanked by N- and C terminal propeptides. Disulfide bond formation between the C-terminal propeptide region of 3 α-chains brings the chains into an alignment favorable for assembly into a triple helix (procollagen molecule (Choices B and E). 4. Procollagen molecules are then transported through the Golgi apparatus into the extracellular space. The N- end C-terminal propeptides are cleaved by procollagen peptidases, converting procollagen into less soluble tropocollagen. 5. Tropocollagen monomers self-assemble into collagen fibrils. Finally, lysyl oxidase helps create covalent crosslinks between collagen fibrils to form strong collagen fibers. *Educational Objective:* Ehlers-Danlos syndrome is a group of rare hereditary disorders characterized by defective collagen synthesis. It can be caused by procollagen peptidase deficiency, which results in impaired cleavage of terminal propeptides in the extracellular space. Patients often have joint laxity, hyperextensible skin, and tissue fragility due to the formation of soluble collagen that does not properly crosslink

A 2-year-old boy is evaluated for easy bruising. His parents report that he develops marked bruising and open wounds following minor trauma. The skin is difficult to suture due to its extreme fragility. Physical examination reveals hyperextensible skin, multiple ecchymoses over the forearms and pretibial regions. and an umbilical hernia. A skin biopsy is performed, and histochemical evaluation of the biopsy reveals a defect in extracellular processing of collagen. Which of the following steps of collagen synthesis is most likely impaired in this patient? (A) Glycosylation of hydroxylysine residues (B) Interchain C-terminal disulfide bond formation (C) N-terminal propeptide removal (D) Proline residue hydroxylation (E) Triple helix formation

*The answer is B.* α1-Antitrypsin deficiency is a genetic disorder that can cause pulmonary emphysema even in the absence of cigarette use. A deficiency of α1-antitrypsin permits increased elastase activity to destroy elastin in the alveolar walls, even in nonsmokers. α1-Anti - trypsin deficiency should be suspected when chronic obstructive pulmonary disease (COPD) develops in a patient younger than 45 years who does not have a history of chronic bronchitis or tobacco use, or when multiple family members develop obstructive lung disease at an early age. Choices A, C, and E refer to collagen, not elastin.

A 30-year-old woman presented with progressive shortness of breath. She denied the use of cigarettes. A family history revealed that her sister had suffered from unexplained lung disease. Which one of the following etiologies most likely explains this patient's pulmonary symptoms? (A) Deficiency of proline hydroxylase (B) Deficiency of α1-antitrypsin (C) Deficiency in dietary vitamin C (D) Decreased elastase activity (E) Increased collagenase activity

*The answer is A.* Human skin exhibits evidence of aging by 30 to 35 years. Gradual thinning of the epidermis is seen, with an associated reduction in subcutaneous fat, blood vessels, hair follicles, sweat ducts, and sebaceous glands. This loss of subcutaneous tissue causes the skin to become atrophic and more vulnerable to damage. Of greatest aesthetic impact is the decrease in the amount of dermal collagen and elastic fibers present. Without this intrinsic reticular support, the inelastic skin sags and demonstrates fine, shallow wrinkling.

A 63-year-old female presents to clinic for a routine examination. Her diet consists mainly of fruit and vegetables and she takes a daily multivitamin. Her last menstrual period was five years ago. She expresses concern about wrinkles around her eyes that make her "look old." A decrease in which of the following is most likely responsible for this patient's complaint? (A) Collagen fibril production (B) Proline hydroxylation (C) Collagen cross-linking (D) Collagenasa synthesis (E) Fibrillin synthesis

*The answer is B.* Creuzfeldt-Jakob syndrome is a prion disorder, and the infectious agent is a protein. The altered protein forms precipitates in the brain, and shifts the equilibrium of the normal protein to that which will aggregate with the altered protein. The pathologist is concerned that the infectious protein will migrate to his brain and seed the process of aggregation with the normal prion proteins in his brain. Prion disorders are not transmitted by viruses, lipids, bacteria, or any form of nucleic acid.

A pathologist, while doing an autopsy of a patient who died from Creuzfeldt Jakob syndrome, accidentally cut himself while examining the brain. The pathologist became very concerned for his well-being, due primarily to the possibility of which one of the following materials entering his circulation? (A) A virus (B) A protein (C) A lipid (D) A bacteria (E) A polynucleotide

*The answer is A.* Adding a phosphate group to the serine side chain adds negative charges to the side chain, allowing ionic interactions to develop. These new ionic interactions allow the protein to change shape and to alter its activity. Substituting a glutamate for the serine adds a negative charge to this location within the protein, which may participate in ionic interactions and lead to a shape change in the protein. None of the other suggested mutations (serine to either threonine [which adds a hydroxyl group, just as serine has], to tyrosine [again, another hydroxyl containing amino acid side chain], to lysine [adding a positive charge rather than a negative charge], or to leucine [a totally hydrophobic side chain]) will lead to the insertion of negative charges at this location in the protein.

A protein's activity is altered when a particular serine side chain is phosphorylated. Which of the following amino acid substitutions at this position could lead to a permanent alteration in normal enzyme activity? (A) S → E (B) S → T (C) S → Y (D) S → K (E) S → L

*The answer is C.* The connective tissue fiber collagen is synthesized by fibroblasts. However, because the length of the finished collagen fibers is many times greater than that of the cell of origin, a portion of assembly occurs extracellularly. The intracellular formation of the biosynthetic precursor of collagen, procollagen peptides pro-α1(I) and pro-α2, occurs in the following steps: (1) synthesis of polypeptides, (2) hydroxylation of proline and lysine residues, (3) glycosylation of lysine residues (proline residues are not glycosylated), (4) formation of the triple helix, and (5) secretion. Once outside the fibroblasts, procollagen molecules are activated by fibroblast-specific procollagen peptidases. Before specific proteolytic cleavage of procollagen, tropocollagen bundles do not assemble into collagen fibers. Once the collagen fibers are formed, aldo cross-links between lysine residues and histidine-aldo cross-links are formed. These cross-links covalently bind the collagen chains to one another. The extent and type of cross-linking determines the flexibility and strength of the collagen mass formed.

During synthesis of mature collagen fiber, which one of the following steps would occur within the fibroblast? (A) Hydrolysis of procollagen to form collagen (B) Glycosylation of proline residues (C) Formation of a triple helix (D) Formation of covalent cross-links between molecules (E) Assembly of the collagen fiber

*The answer is A.* If the substrate concentration is much below the Km of the enzyme, the velocity of the reaction is directly proportional to the substrate concentration. Above the Km, the velocity of the reaction increases substrate concentration logarithmically and the slope begins to decrease.

If the substrate concentration is much below the Km of the enzyme, the velocity of the reaction is (A) Directly proportional to substrate concentration (B) Not affected by enzyme concentration (C) Nearly equal to Vmax (D) Inversely proportional to substrate concentration

*The answer is B.* In the oxygen dissociation curve, percent saturation of hemoglobin with oxygen on the y axis is plotted against the amount of oxygen present in solution [the partial pressure of oxygen (PO2)]. Hemoglobin saturation varies from 0 to 100%. The O2 pressure can vary from no oxygen in solution to high PO2 levels. 2,3-bisphosphoglycerate (BPG) is present in concentrations similar to those of hemoglobin in red blood cells. BPG cross-links deoxyhemoglobin and lowers its affinity for oxygen, aiding the unloading of oxygen in capillaries. Thus, increased BPG shifts the oxygen dissociation curve to the right. Similarly, increased H+ ions and CO2 enhance the release of O2 from hemoglobin and shift the curve to the right. Conversely, the lower H+ levels with increasing pH and decreasing CO2 shift the curve leftward. Fetal hemoglobin has a greater affinity for O2 under all conditions. Mixing of fetal with adult hemoglobin increases O2 affinity and shifts the curve to the left.

The oxygen dissociation curve of normal adult hemoglobin shown below is most effectively shifted to the right by (A) Mixing with fetal hemoglobin (B) Increased 2,3-bisphosphoglycerate (BPG) (C) Cooperative binding of oxygen (D) Increased pH (E) Decreased CO2

*The answer is C.* This patient most likely has familial erythrocytosls due to a β-globin mutation resulting in reduced binding of 2.3-bisphosphoglycerate (2,3 BPG). 2,3-BPG is synthesized from glycolytic intermediates and binds strongly to deoxyhemoglobin in a pocket formed between the 2 beta chains. This binding reduces the oxygen affinity of hemoglobin, allowing more oxygen to diffuse into the peripheral tissues. The hemoglobin 2,3 BPG binding pocket contains positively charged amino acids (eg, histidine and Iysinie) that attract the negatively charged phosphate groups in 2,3-BPG. Mutations that decrease the positive charge of the binding site decrease 2,3-BPG binding and increase hemoglobin oxygen affinity. Fetal hemoglobin (hemoglobin F) is synthesized primarily during fetal development (-8 weeks until term) and consists of the usual 2 alpha chains with 2 gamma chains in place of beta chains. The gamma chains do not bind effectively to 2,3-BPG due to the replacement of a histidine residue with serine. As a result, fetal hemoglobin has significantly higher oxygen affinity than adult hemoglobin A. This allows fetal hemoglobin to extract more oxygen from the mothers adult hemoglobin in the placenta, providing the developing fetus with an adequate supply of oxygen.

A 15-year-old boy is found to have unexplained erythrocytosis on routine laboratory analysis. Evaluation of his immediate family shows that his father and sister also have elevated red cell levels. Genetic sequencing of the β-globin gene is performed in the affected family members. The results show a single base substitution at amino acid position 82 that replaces the normal lysine residue with methionine. Further analysis shows that this amino acid replacement impairs the ionic interaction between the β-subunit and 2,3-bisphosphoglycerate. As a result of this mutation, the patient's hemoglobin will be most similar to which of the following hemoglobin types? (A) Hemoglobin A (B) Hemoglobin C (C) Hemoglobin F (D) Hemoglobin H (E) Hemoglobin S

*The answer is E.* The student has died from familial hypertrophic cardiomyopathy (FHC), a thickening of the left ventricle of the heart muscle due to a mutation in β-myosin heavy chain. The exact reason for the hypertrophy, which can be caused by mutations in a variety of sarcomeric proteins, is still unknown. None of the other proteins suggested as answers are muscle sarcomeric proteins. Spectrin is a red blood cell protein, and is not found in the heart. α1-Antitrypsin is a circulating protein synthesized by the liver, and in its absence, emphysema will develop. Collagen is the major structural protein of the body, but there are no mutations in collagen that lead to a greatly hypertrophied heart muscle. A lack of fibrillin leads to Marfan syndrome, which can present with defects in heart valves and the aorta, but not a heart muscle greatly increased in size.

A 16-year-old male high school student was playing basketball for his school when he collapsed on court and could not be resuscitated. An autopsy demonstrated increased thickness of the intraventricular septum and left ventricular wall. These findings could be explained by a mutation in which one of the following proteins? (A) Spectrin (B) α1-Antitrypsin (C) Collagen (D) Fibrillin (E) β-Myosin heavy chain

*The answer is B.* The patient is having a panic attack (due to driving in tunnels) and is hyperventilating, 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 metabolic 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 alkalosis 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 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 exhibiting the signs of osteogenesis imperfecta, brittle bones, as exemplified by various mutations in type 1 collagen, the building blocks of the bones. The aortic regurgitation murmur is also due to a lack of type 1 collagen in the extracellular matrix of the aorta. Mutations in fibrillin give rise to Marfan syndrome, which does exhibit long bones, but not brittle or easily broken bones. Marfan syndrome would also be associated with lens dislocation, which is not occurring in this patient. Mutations in type IV collagen would lead to Alport syndrome, not brittle bones, and there is no mention of kidney/urine problems with the patient. A defect in α1 antitrypsin would lead to emphysema (not brittle bones), and a mutation in β-myosin heavy chain would lead to hypertrophic cardiomyopathy, not brittle bones.

A 23-year-old male presents to the ER with a fracture of his humerus, sustained in what appeared to be a minor fall. He has a history of multiple fractures after a seemingly minor trauma. He also has "sky blue" sclera and an aortic regurg murmur. His underlying problem is most likely due to a mutation in which one of the following proteins? (A) Fibrillin (B) Type 1 collagen (C) Type IV collagen (D) α1-Antitrypsin (E) β-Myosin heavy chain

*The answer is D.* The Km cannot be negative so answers B and C are incorrect. Based on the graph, the y-intercept changes from -4 to -2. This means the Km changes from 0.25 to 0.5. Therefore, the drug adjusts the Km to 0.5, which is none of the answers mentioned.

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 drug adjusts the enzyme's Km to: (A) 0.33 (B) -0.25 (C) -0.50 (D) None of the above

*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 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?

*The answer is C.* The patient has Alport syndrome, a mutation in type IV collagen that alters the basement membrane composition of kidney glomeruli. In the absence of a functional basement membrane, the kidneys have difficulty in properly filtering waste products from blood into the urine, and both blood and proteins can enter the urine. Type IV collagen is also important for hearing (it is found in the inner ear) and for the eye. Type IV collagen forms a meshlike structure, which is different from the rodlike structures found in type I collagen, and is found in almost all basement membrane structures. Given sufficient time, the alteration in the basement membrane in the glomeruli will lead to their destruction, and loss of kidney function. A mutation in α1-antitrypsin will lead to emphysema, mutations in spectrin can lead to hereditary spherocytosis, mutations in fibrillin lead to Marfan syndrome, and mutations in β-myosin heavy chain can lead to FHC.

A 28-year-old man presents to the ER with a large amount of blood and protein in his urine. He has had a sensorineural hearing loss since his teen years and has misshaped lenses (anterior lenticonus). The physician is suspicious of a genetic disorder that may lead to eventual kidney failure. If this is the case, the patient most likely has a mutation in which one of the following proteins? (A) Spectrin (B) α1-Antitrypsin (C) Collagen (D) Fibrillin (E) β-Myosin heavy chain

*The answer is C.* Iron bound to heme is normally in the reduced ferrous [Fe(II)] state. Oxidation of the ferrous iron of hemoglobin to ferric iron leads to the formation of methemoglobin. With iron in the oxidized ferric state, methemoglobin is unable to bind oxygen. In addition, the affinity of any residual ferrous iron in the hemoglobin tetramer is increased, causing a leftward shift of the oxygen-dissociation curve. Nitrites actually cause poisoning by oxidizing the heme iron to the ferric state. In patients with methemoglobinemia, the partial pressure of oxygen in the blood (Choice C) is normal because the amount of oxygen dissolved in the plasma is the same. Methemoglobinemia causes dusky discoloration to the skin (similar to cyanosis), and because methemoglobin is unable to carry oxygen, a state of functional anemia is induced. *Educational Objective:* Methemoglobinemia causes dusky discoloration to the skin (Similar to cyanosis), and because methemoglobin is unable to carry oxygen, a state of functional anemia is induced. The blood partial pressure of O2 however, will be unchanged in this condition because oxygen's partial pressure is a measure of O2 dissolved in the plasma and is not related to hemoglobin function.

A 34-year-old male who is accidentally exposed to nitrites at work presents to the ER with anxiety, weakness, dyspnea, and headaches. Physical examination reveals cyanosis that is not corrected by oxygen supplementation. Which of the following is most likely to be normal in this patient? A. Oxygen content of the arterial blood B. Oxygen carrying capacity of the arterial blood C Partial pressure of oxygen in the anemia! blogs; D. Bound fraction of oxygen in the arterial blood E. Oxygen delivery to peripheral tissues

*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 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 B.* One treatment for chronic alcoholism is to inhibit the enzyme aldehyde dehydrogenase, which would lead to the accumulation of acetaldehyde if ethanol has been imbibed. Ethanol metabolism, at the first step, converts ethanol to acetaldehyde (the enzyme is alcohol dehydrogenase). Aldehyde dehydrogenase then converts the acetaldehyde to acetic acid, which is eventually converted to acetyl-CoA. The accumulation of acetaldehyde is what initiates the symptoms associated with a hangover, such as headache and nausea. The theory behind the treatment is that if the individual drinks alcohol while on the drug, the buildup of acetaldehyde will make the person feel very uncomfortable, and will lead to a reduction, or cessation, of drinking alcohol. Inhibiting aldehyde dehydrogenase will not lead to elevations of acetic acid, or ethanol, carbon dioxide, or carbon monoxide.

A 45-year-old woman has been admitted to a substance abuse center for her alcoholism. As a first attempt to curb the patient's drinking, she is given a drug that will lead to an elevation of which one of the following metabolites if she drinks alcohol? (A) Acetic acid (B) Acetaldehyde (C) Ethanol (D) Carbon dioxide (E) Carbon monoxide

*The answer is B.* In the United States, vitamin C deficiency (scurvy) is most often seen in severely malnourished individuals (eg, homeless, alcohol or drug abusers). Symptoms of vitamin C deficiency are the result of decreased connective tissue strength. The capillary walls are especially fragile, leading to easy bruising, mucosal bleeding, and perifolicular petechial hemorrhages. Patients may also suffer from periodontal disease (gum swelling, loosening of the teeth, and infection) and poor wound healing, and have hyperkeratotic follicles with corkscrew hairs. Scurvy is even more severe in children and manifests with hemorrhages, bony deformities, and subperiosteal and joint hematomas. Vitamin C is necessary for the hydroxylation of proline and lysine residues during collagen synthesis. This reaction is executed by prolyl and lysyl hydroxylases, with vitamin C serving as a reducing agent. Hydroxyproline and hydroxylysine are essential for cross-linking collagen molecules. In scurvy, collagen cross-linking is compromised, thereby greatly reducing its tensile strength.

A 46-year-old man comes to the emergency department due to recurrent nosebleeds. When interviewed for additional history, he becomes belligerent and uncooperative. The patient has history of alcohol abuse and chronic mental illness. He has been placed in homeless shelters on multiple occasions but has not remained there for any prolonged periods. Physical examination shows swollen gums, scattered ecchymoses. and hyperkeratosis. He also has a chronic ulcer on the left lower extremity that does not appear to be infected. Which of the following mechanisms accounts for this patient's examination endings? (A) Abnormal oxidative decarboxylation of ketoacids (B) Abnormal procaine hydroxylation (C) Abnormal transamination (D) Deficient methionine synthesis (E) Diminished synthesis of purines

*The answer is D.* Organophosphates react with the active site serine residue of hydrolases such as acetylcholinesterase and form a stable phosphoester modification of that serine that inactivates the enzyme toward substrate. Inhibition of acetylcholinesterase causes overstimulation of the end organs regulated by those nerves. The symptoms manifested by this patient reflect such neurologic effects resulting from the inhalation or skin absorption of the pesticide diazinon.

A 47-year-old man is evaluated for a 12-hour history of nausea, vomiting and, more recent, difficulty breathing. His past medical history is unremarkable, and he takes no medications. However, he is a farmer who has had similar episodes in the past after working with agricultural chemicals in his fields. Just yesterday he reports applying diazinon, an organophosphate insecticide, to his sugar beet field. After consultation with the poison center, you conclude that this patient's condition is most likely due to inhibition of which of the following enzymes? A. Acetate dehydrogenase B. Alanine aminotransferase C. Streptokinase D. Acetylcholinesterase E. Creatine kinase

*The answer is D.* The man has the symptoms of emphysema, due to destruction of lung cells by the protease elastase. Neutrophils in the lung accidentally release elastase as they engulf and destroy inhaled bacteria and other particles, and normally α1-antitrypsin would bind to the elastase and inhibit its activity. In a long-term smoker, however, products from the cigarette smoke oxidize an essential methionine side chain in α1-antitrypsin, rendering it inactive. Thus, over time, noninhibited elastase has been destroying lung tissue until the lung no longer functions properly. Even though the inhibitor will block trypsin activity, the lung damage is the result of increased elastase activity, not trypsin activity. Sulfhydryl groups are not being affected, rather a sulfur in methionine is the target of the cigarette smoke.

A 53-year-old man, who has been smoking for the past 35 years at a two-pack-a-day rate, visits his physician for a cough that will not go away, and for difficulty in breathing. A chest X-ray rules out cancer, but does display an increased anterior-posterior (AP) diameter, flattened diaphragms, and "air trapping." The patient is told that his condition will not improve, and that he needs to stop smoking to stop the progression of the disease. At the molecular level, this disease is due to which one of the following? (A) Enhanced trypsin activity in the lung (B) Decreased trypsin activity in the lung (C) Enhanced α1-antitrypsin activity in the lung (D) Decreased α1-antitrypsin activity in the lung (E) Enhanced reduction of sulfhydryl groups in the lung (F) Decreased reduction of sulfhydryl groups in the lung

*The answer is B.* Sickle cell anemia is caused by inheriting two copies of a mutant β globin gene that leads to synthesis of sickle hemoglobin, HbS. A severe case of sickle cell anemia would most likely have demonstrated symptoms and been diagnosed before the age of 6. However, he may only be a carrier, with one copy each of normal β-globin and one of the sickle allele, a condition called sickle cell trait. Nevertheless, the patient's symptoms are entirely consistent with an acute sickle cell crisis. These are brought on by exertion, which increases the levels of deoxyhemoglobin in RBCs. Under this condition, the mutant HbS molecules have reduced solubility; they tend to stick together in polymers that alter the shape of RBCs (sickle cells). Sickled RBCs are not as pliable as normal RBCs, so that they do not pass freely through the narrow passages of the capillaries and can cause clogging of microvessels. The pain experienced by this boy is likely due to such vaso occlusion in his joints and abdominal vessels.

A 6-year-old black boy complains of acute abdominal pain that began after playing in a football game. He denies being tackled forcefully. He has a history of easy fatigue and several similar episodes of pain after exertion, with the pain usually restricted to his extremities. Microscopic evaluation of his blood would be expected to reveal which of the following cellular abnormalities? (A) Increased WBC count (B) Deformed RBCs (C) Decreased WBC count (D) Increased RBC count (erythrocytosis) (E) Reduced platelet count

*The answer is D.* The boy is suffering from sickle cell anemia, which is due to a substitution of valine for glutamate at position 6 of the β-chain. This change, from a negatively charged amino acid side chain (glutamate) to a hydrophobic side chain (valine), allows deoxygenated hemoglobin to polymerize and form long rods within the red blood cell. Deoxygenated hemoglobin has a hydrophobic patch on its surface (created by A70, F85, and L88), which the valine in position 6 on another hemoglobin chain can associate with via hydrophobic interactions (this does not occur in normal hemoglobin as there is a charged glutamate residue at this position, which will not interact with a surface hydrophobic patch--see the figure below). The binding of hemoglobin molecules to each other results in the polymerization. Oxygenated hemoglobin does not present a hydrophobic surface to other hemoglobin molecules, so polymerization is much less likely in the oxygenated state. The polymerization is not caused by a loss of quaternary structure, an increase in oxygen binding (which would actually reduce sickling), a gain of ionic interactions, or the loss of any α-helical structure in the final conformation of the protein.

A 7-year-old African American male is admitted to the hospital with severe abdominal pain. A blood workup indicated anemia, and an abnormal blood smear (see below). The molecular event triggering this disease is which of the following? (A) A loss of quaternary structure of the hemoglobin molecule (B) An increase in oxygen binding to hemoglobin (C) A gain of ionic interactions, stabilizing the "T" form of hemoglobin (D) An increase in hydrophobic interactions between deoxyhemoglobin molecules (E) An alteration in hemoglobin secondary structure leading to loss of the "α" helix

*The answer is A.* The woman would exhibit respiratory acidosis due to shortness of breath and decreased efficiency of gas exchange in the lungs. Emphysema involves dilated and dysfunctional alveoli from alveolar tissue damage, usually secondary to cigarette smoking. The hypoxia leads to tissue deoxygenation and acidosis, exacerbated by the hypercarbia (CO2 accumulation) that distinguishes respiratory acidosis (higher bicarbonate than expected) from metabolic acidosis (very low bicarbonate, usually with low PCO2 due to compensatory hyperventilation). Choice a shows the only set of values indicating acidosis (pH lower than 7.4), hypoxia (PO2 lower than 95), and hypercarbia (PCO2 greater than 44). The tetrameric structure of hemoglobin allows cooperative binding of oxygen in that binding of oxygen to the heme molecule of the first subunit facilitates binding to the other three. This enhanced binding is due to allosteric changes of the hemoglobin molecule, accounting for its S-shaped oxygen saturation curve as compared with that of myoglobin (see the figure in question 146). At the lower oxygen saturations in peripheral tissues (PO2 30 to 40), hemoglobin releases much more oxygen (up to 50% desaturated) than myoglobin with its single polypeptide structure. The amount of oxygen released (and CO2 absorbed as carboxyhemoglobin) is further increased by the Bohr effect—increasing hydrogen ion (H+) concentration (lowering pH) and increasing CO2 partial pressure (PCO2) shift the sigmoidal-shaped oxygen binding curve for hemoglobin further to the right.

A 72-year-old woman with emphysema presents to the emergency room with fatigue and respiratory distress. Which set of arterial blood gas values below would represent her condition and reflect a shift of the hemoglobin oxygen dissociation curve to the right? (A) pH 7.05, bicarbonate 15 mM, PCO2 60, PO2 88 (B) pH 7.15, bicarbonate 10 mM, PCO2 30, PO2 88 (C) pH 7.25, bicarbonate 15 mM, PCO2 30, PO2 88 (D) pH 7.40, bicarbonate 24 mM, PCO2 60, PO2 88 (E) pH 7.45, bicarbonate 15 mM, PCO2 60, PO2 88

*The answer is F.* This patient likely has vitamin c deficiency (scurvy), In the United States, vitamin C deficiency is seen primarily among malnourished populations, including alcoholics, the poor, and the elderly. The symptoms of scurvy reflect impaired formation of collagen and include gingival swelling/bleedlng, patechiae, ecchymoses, and poor wound healing. Perifolllcular hemorrhages and coiled (corkscrew) hairs are also commonly seen. Collagen synthesis is a complex process that begins with the transcription of collagen genes in the nucleus (Choice E). Collagen alpha-chains are then synthesized by rough endoplasmic reticulum (RER)-bound ribosomes and directed into the cisternae of the RER. Within the RER, specific praline and lysine residues are post-translationally hydroxylated to hydroxyproline and hydroxylysine by prolyl hydroxylase and lysyl hydroxylase, respectively. Vitamin C is a required cofactor for this post translational modification. Defective hydroxylation of these residues severely diminishes the amount of collagen secreted by fibroblasts and impairs triple helix stability and covalent crosslink formation.

A 78-year-old women comes to the office due to tenderness and easy bleeding of the gums when she brushes her teeth. The patient has brushed her teeth twice a day for as long as she can remember and has riot experienced these symptoms before. Physical examination shows swollen gingiva that bleed on probing. Her skin findings are shown in the image blow. Further questioning reveals that the patient lives alone and that her diet consists primarily of tea and toast. Her symptoms are most likely caused by hypoactivity of an enzyme found in which of the following compartments? (A) Extracellular space (B) Golgi apparatus (C) Lysosomes (D) Mitochondria (E) Nucleus (F) Rough endoplasmic reticulum

*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 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.* Leucine and isoleucine have nonpolar methyl groups as side chains. As for any amino acid, titration curves obtained by noting the change in pH over the range of 1 to 14 would show a pK of about 2 for the primary carboxyl group and about 9.5 for the primary amino group; there would be no additional pK for an ionizable side chain. Recall that the pK is the point of maximal buffering capacity when the amounts of charged and uncharged species are equal. Aspartic and glutamic acids (second carboxyl group), histidine (imino group), and glutamine (second amino group) all have ionizable side chains that would give an additional pK on the titration curve. The likely diagnosis here is maple syrup urine disease, which involves elevated isoleucine, leucine, and valine together with their ketoacid derivatives. The ketoacid derivatives cause the acidosis, and the fever suggests that the metabolic imbalance was worsened by an infection.

A child presents with severe vomiting, dehydration, and fever. Initial blood studies show acidosis with a low bicarbonate and an anion gap (the sum of sodium plus potassium minus chloride plus bicarbonate is 40 and larger than the normal 20 to 25). Preliminary results from the blood amino acid screen show two elevated amino acids, both with nonpolar side chains. A titration curve performed on one of the elevated species shows two ionizable groups with approximate pKs of 2 and 9.5. The most likely pair of elevated amino acids consists of (A) Aspartic acid and glutamine (B) Glutamic acid and threonine (C) Histidine and valine (D) Leucine and isoleucine (E) Glutamine and isoleucine

*The answer is A.* The primary structure of collagen peptides consists of repeating tripeptides with a gly-X-Y motif, where gly is glycine and X and Y are any amino acid. The small CH2 group connecting the amino and carboxyl groups of glycine contrasts with the larger connecting groups and side chains of other amino acids. The small volume of glycine molecules is crucial for the α helix secondary structure of collagen peptides. This in turn is necessary for their tertiary helical structure and their assembly into quaternary tripeptide, triple-helix structures. The most severe clinical phenotypes caused by amino acid substitutions in collagen peptides are those affecting glycine that prevent α helix formation. The child has a disorder called Stickler syndrome (108300) that exhibits autosomal dominant inheritance.

A child with tall stature, loose joints, and detached retinas is found to have a mutation in type II collagen. Recall that collagen consists of a repeating tripeptide motif where the first amino acid of each tripeptide is the same. Which of the following amino acids is the recurring amino acid most likely to be altered in mutations that distort collagen molecules? (A) Glycine (B) Hydroxyproline (C) Hydroxylysine (D) Tyrosine (E) Tryptophan

*The answer is B.* The family member is exhibiting the symptoms of carbon monoxide (CO) poisoning. CO will bind to hemoglobin, with a higher affinity than oxygen, and decrease oxygen delivery to the tissues. In addition to competing with oxygen for binding to hemoglobin, CO, once bound to hemoglobin, shifts the oxygen binding curve to the left, stabilizing the "R" state, or oxygenated state, which makes it more difficult for oxygen to be released from hemoglobin in the tissues. Thus, in the presence of CO, oxygen affinity for hemoglobin is actually increased. CO poisoning does not affect the blood flow to the brain.

A family has been using an additional propane heater in their enclosed apartment during the winter months. One morning, a family member is difficult to awake, and when awake, complains of a splitting headache and being very tired. His mucous membranes are also a cherry red color. These symptoms are the result of which one of the following? (A) Increased oxygen delivery to the tissues (B) Decreased oxygen delivery to the tissues (C) Increased blood flow to the brain (D) Decreased blood flow to the brain (E) Decreased oxygen affinity to hemoglobin

*The answer is C.* 2,3-bisphosphoglycerate (2,3-BPG) will bind to and stabilize the deoxygenated form of hemoglobin. Thus, if 2,3-BPG levels are increased, the binding of this molecule will aid in removing oxygen from hemoglobin in the tissues (where the concentration of oxygen is low) and therefore increase oxygen delivery to the tissues. In the lungs, where the oxygen concentration is high, the high levels of oxygen can overcome the effects of 2,3-BPG and bind to hemoglobin. Lactic acid levels do not directly affect oxygen binding (and lactate does not accumulate in the red cell), although changes in proton concentration (pH) can. Decreased pH will reduce oxygen binding to hemoglobin due to the Bohr effect. Bilirubin degradation, even though it does produce CO, does not affect oxygen binding to hemoglobin.

A family of four from New Jersey has embarked on a vacation in the Rocky Mountains. All four required a 24 to 48 h acclimation to the high altitude, as all were breathing at a rapid pace until the acclimation took effect. In addition to increasing the number of red blood cells in circulation, what other compensatory mechanism occurred within the red blood cell during this acclimation period? (A) Increased synthesis of lactic acid (B) Decreased synthesis of lactic acid (C) Increased synthesis of 2,3-bisphosphoglycerate (D) Decreased synthesis of 2,3-bisphosphoglycerate (E) Decreased degradation of bilirubin, producing less carbon monoxide

*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 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 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 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 C*. Free sulfhydryl groups can form disulfide bonds with a cysteine side chain, which may then interfere with the functioning of the protein. Gastric reflux disease is caused by a failure of the gastroesophageal sphincter, allowing stomach contents to travel up the esophagus. Since there is acid in these contents, damage to the esophagus can result. Inhibiting the gastric proton pump, thereby increasing the pH of the stomach contents, would reduce the damage that occurs during reflux due to the increased pH. Inhibiting an intestinal proton pump would not address gastric reflux disease. Methionine residues, though they contain a sulfur atom, do not form disulfide bonds. Free sulfhydryl groups are already reduced, so the drug cannot reduce one further.

A patient is taking omeprazole for gastric reflux disease. Omeprazole contains a free sulfydryl group that is critical for its mechanism of action. This drug will most likely act in which one of the following ways? (A) Reduce an existing sulfhydryl on the intestinal proton pump. (B) Form a disulfide bond with a methionine on the gastric proton pump. (C) Form a disulfide bond with a cysteine on the gastric proton pump. (D) Reduce an existing sulfhydryl group on the gastric proton pump. (E) Form a disulfide bond with a cysteine on the intestinal proton pump.

*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.

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 B.* In noncompetitive inhibition, the Km is not altered, whereas the Vmax is decreased. 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.

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? (A) Competitive (B) Noncompetitive (C) Irreversible (D) Allosteric (E) The drug is not an inhibitor, rather it is an activator.

*The answer is B.* Ketone bodies are weak acids. In diabetic ketoacidosis, the liver produces ketone bodies, which will reduce the brain's dependency on glucose as its sole energy source. This is due to the lack of insulin, and the liver switching to starvation mode owing to the constant signaling by glucagon. Hemoglobin in the red blood cells and bicarbonate, both in the red blood cells and the plasma, are two of the body's major buffers, and their overproduction would not lead to an acidosis. HCl overproduction within the stomach might lead to duodenal ulcers or gastroesophageal reflux, but not to an overall metabolic acidosis, as the protons do not find their way into the circulation. A loss of chloride, if severe enough, could produce a metabolic alkalosis, but not an acidosis.

A person with Type 1 diabetes ran out of her prescription insulin and has not been able to inject insulin for the past 3 days. An overproduction of which of the following could cause a metabolic acidosis? (A) Hemoglobin (B) Ketone bodies (C) HCl (D) Bicarbonate

*The answer is D.* A protein's primary structure is the sequence of amino acids linked by covalent peptide bonds *(Choice C)*. Proteins may also assume a secondary structure, such as the alpha-helix or beta-sheet, due to subsequent hydrogen bonding. In patients with Alzheimer disease, beta-amyloid protein loses its alpha-helical configuration and forms beta-sheets, which are less soluble and therefore prone to aggregating. Aggregations of beta-sheets are the primary component of the extracellular senile (neuritic) plaques found in Alzheimer patients. The conversion of alpha-helices to beta-sheets involves the breaking and reforming of hydrogen bonds *(Choice D)*. Tertiary structure is the overall shape that a single polypeptide chain assumes following compact folding of the secondary structure. Many forces combine to stabilize the tertiary structure, including ionic bonds *(Choice A)*. hydrophobic interactions *(Choice B)*, hydrogen bonds *(Choice D)*, and disulfide bonds *(Choice E)*. Remember that disulfide bonds are very strong covalent bonds between two cysteine residues within the same polypeptide chain that enhance a protein's ability to withstand denaturation. *Educational objective:* Hydrogen bonds are the principal stabilizing force for the secondary structure of proteins.

A research group is studying the pathogenesis of Alzheimer disease. A protein isolated from brain tissue of an affected patient has a conformation consisting of mostly beta-pleated sheets. A sample of a new medication is applied to the protein, and the prevailing structure changes to an alpha-helical structure. This conformational change is the result of reorganization of which of the following: (A) Ionic interactions (B) Hydrophobic interactions (C) Peptide bonds (D) Hydrogen bonds (E) Disulfide bonds

*The answer is A.* The child most likely has osteogenesis imperfecta. Most cases arise from a defect in the genes encoding type I collagen. Bones in affected patients are thin, osteoporotic, often bowed with a thin cortex and deficient trabeculae, and extremely prone to fracture. This patient is affected with type I, osteogenesis imperfecta tarda. The disease presents in early infancy with fractures secondary to minor trauma. The disease may be suspected on prenatal ultrasound through detection of bowing or fractures of long bones. Type II, osteogenesis imperfecta congenita, is more severe, and patients die of pulmonary hypoplasia in utero or during the neonatal period. Defects in type III collagen are the most common cause of Ehlers Danlos syndrome, characterized by lethal vascular problems and stretchy skin. Type IV collagen forms networks, not fibrils.

A seven-month-old child "fell over" while crawling, and now presents with a swollen leg. At age 1 month, the infant had multiple fractures in various states of healing (right clavicle, right humerus, right radius). At age 7 months, the infant has a fracture of a bowed femur, secondary to minor trauma (see x-ray at right). The bones are thin, have few trabecula, and have thin cortices. A careful family history ruled out nonaccidental trauma (child abuse) as a cause of the bone fractures. The child is most likely to have a defect in: (A) type I collagen (B) type III collagen (C) type IV collagen (D) elastin (E) fibrillin

*The answer is B.* The boy is showing the symptoms of Marfan syndrome, which is caused by mutations in fibrillin, an extracellular protein. Fibrillin helps to form, along with other proteins, microfibrils, which are present in elastic fibers (containing primarily elastin), which help to give various tissues their elastic properties. The exact mechanism whereby mutations in fibrillin lead to the symptoms of Marfan syndrome has yet to be established. Mutations in any of the other proteins listed do not give rise to Marfan's (although collagen defects give rise to osteogenesis imperfecta, and dystrophin mutations give rise to various forms of muscular dystrophy, depending on the type of mutation). Marfan's is an autosomal dominant disorder of connective tissue (not collagen). It is caused by mutations in the FBN1 gene (located on chromosome 15), which encodes fibrillin-1, a glycoprotein. The picture is of a dislocated lens, a classical finding in patients with Marfan's.

A teenager, new to your practice, comes in for a routine physical exam. His family had just moved to the city, and the boy had rarely seen a doctor before. Upon examination, you notice a high, arched palate, disproportionately long arms and fingers, a sunken chest, and mild scoliosis. The patient has been complaining of lack of breath while doing routine chores, and upon listening to his heart, you detect an aortic regurgitation murmur. Careful examination of the eyes is indicated by the figure below. Based on your physical exam and history, you are suspicious of an inborn error of metabolism in which of the following proteins? (A) Collagen (B) Fibrillin (C) Elastin (E) Dystrophin

*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 conditions, 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.

A young black man was brought to the emergency room (ER) owing to severe pain throughout his body. He had been exercising 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 oddlooking 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 hemoglobin molecules in the oxygenated state. (B) Increased ionic interactions between hemoglobin molecules in the deoxygenated state. (C) Increased hydrophobic interactions between hemoglobin molecules in the oxygenated state. (D) Increased hydrophobic interactions between hemoglobin molecules in the deoxygenated state. (E) Increased phosphorylation of hemoglobin molecules in the oxygenated state. (F) Increased phosphorylation of hemoglobin molecules in the deoxygenated state.

*The answer is B.* The therapeutic rationale for ethylene glycol poisoning is to compete for the attention of alcohol dehydrogenase by providing a preferred substrate, ethanol, so that the enzyme is unavailable to catalyze oxidation of ethylene glycol to toxic metabolites. Ethanol will displace ethylene glycol by mass action for a limited time, during which hemodialysis is used to remove ethylene glycol and its toxic metabolites from the patient's bloodstream.

Accidental ingestion of ethylene glycol, an ingredient of automotive antifreeze, is fairly common among children because of the liquid's pleasant color and sweet taste. Ethylene glycol itself is not very toxic, but it is metabolized by alcohol dehydrogenase to the toxic compounds glycolic acid, glyoxylic acid, and oxalic acid, which can produce acidosis and lead to renal failure and death. Treatment for suspected ethylene glycol poisoning is hemodialysis to remove the toxic metabolites and administration of a substance that reduces the metabolism of ethylene glycol by displacing it from the enzyme. Which of the following compounds would be best suited for this therapy? (A) Acetic acid (B) Ethanol (C) Aspirin (D) Acetaldehyde (E) Glucose

*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 regulates 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)

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 synthesize which one of the following proteins? (A) Myoglobin (B) Hemoglobin (C) Collagen (D) Insulin (E) Fibrillin

*The answer is D.* The Michaelis Menten equation is v = (Vmax × [S])/(Km + [S]). In this case, the velocity (v) is ¼ Vmax, and [S] = 4 mM. Thus, the Michaelis-Menten equation becomes ¼ Vmax = (Vmax × 4)/(Km + 4). When one solves that equation for Km, Km = 12 mM.

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) 2 mM (B) 4 mM (C) 9 mM (D) 12 mM (E) 25 mM

*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 effect), increasing the CO2 concentration, or increasing the concentration of BPG. Fetal hemoglobin (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 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 person 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 decreased 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 A.* The decreased amount of AAT protein, its abnormal mobility, and the engorgement of liver ER suggest a mutant AAT that is inefficiently transported from the ER to serum. In people with Alpha-1 (Alphas), large amounts of abnormal alpha-1 antitrypsin protein (AAT) are made in the liver; nearly 85 percent of this protein gets stuck in the liver. If the liver cannot break down the abnormal protein, the liver gradually gets damaged and scarred (this scarring leads to cirrhosis). The key word that gives the answer away here is "cirrhosis" because this occurs when too much of a substance is trapped in the liver.

An older man with severe emphysema is found to have decreased amounts and abnormal mobility of α1 antitrypsin (AAT) protein in his serum when analyzed by serum protein electrophoresis. Liver biopsy discloses mild scarring (cirrhosis) and demonstrates microscopic inclusions due to an engorged endoplasmic reticulum (ER). The most likely explanation for these findings is (A) Defective transport from hepatic ER to the serum (B) A mutation affecting the N-terminal methionine and blocking initiation of protein synthesis (C) A mutation affecting the signal sequence (D) Defective structure of the signal recognition particles (E) Defective energy metabolism causing deficiency of GTP

*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.

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 acetoacetate occurs by what type of reaction? (A) Oxidation (B) Reduction (C) Dehydration (D) Dehydroxylation (E) Decarboxylation

*The answer is F.* The amino acid cysteine can be present within the active site of enzymes and function as a nucleophile and it can also form disulfide bonds in secreted proteins. However, it is not required for the formation of tertiary structure in proteins. The side chain does not have a pK of 6.5, but a pK of 8.4.

Choose the *SINGLE* best answer concerning the amino acid cysteine. (A) It is required for the formation of tertiary structure in proteins. (B) It can be present within the active site of enzymes and function as a nucleophile. (C) It forms disulfide bonds in secreted proteins. (D) The side-chain thiol has a pK of 6.5. (E) Both A and C (F) Both B and C

*The answer is D.* Elastin is a fibrous protein in the connective tissue that gets its name because of its elastic properties. Elastin fibers can be stretched to several times their length but recoil back when stretching forces are withdrawn. Elastin assembly appears to be closely related to that of collagen. Similar to collagen, elastin is synthesized as a large polypeptide precursor composed of about 700 amino acids called tropoelastin. Elastin is primarily composed of the non-polar amino acids glycine, alanine, and valine. The elastin molecule also contains proline and lysine; however, in contrast to collagen, few of these amino acids are hydroxylated. Tropoelastin is secreted into the extracellular space where it interacts with microfibrils called fibrillin. Once in the extracellular space, side chains of some of the lysine residues are covalently bound to form a desmosine crosslink. Extensive desmosine crosslinking accounts for elastin's resilient properties, allowing it to stretch and bend in any direction on applying force only to recoil to its original size when the stretching force is withdrawn. Elastin gives elastic properties to the skin, blood vessels and lung alveoli. A number of endogenous enzymes called proteinases hydrolyze and destroy such proteins.For elastin,the most important proteinase is neutrophil-secreted elastase.a1-antitrypsin inhibits the action of these endogenous proteolytic enzymes, thereby preventing damage to essential structures within organs. A congenital deficiency of a1-antitrypsin results in excessive degradation of elastin in the lungs and liver causing panacinar emphysema and cirrhosis,respectively.

Elastin fibers in the alveolar walls of the lungs can be stretched easily during inspiration and recoil to their original shape once the force is released. This process facilitates expiration. The property described can be best explained by: (A) Heavy posttranslational hydroxylation (B) High content of polar amino acids (C) Chain assembly to form a triple helix (D) Interchain crosslinks involving lysine (E) Abundant interchain disulfide bridges

*The answer is D.* This problem provides a practical illustration of the use of the Michaelis-Menten equation. The high concentration of glucose in the hepatic portal vein after a meal would promote a high rate of glucose uptake into liver cells, necessitating rapid phosphorylation of the sugar. The glucose concentration far exceeds the Km of hexokinase, ie, [S] > Km, meaning that the enzyme will be nearly saturated with substrate and v ≅ Vmax. However, the [S] ≅ Km for glucokinase, which will be active in catalyzing the phosphorylation reaction and v ≅ 1⁄2Vmax.

Glucose taken up by liver cells is rapidly phosphorylated to glucose 6-phosphate with ATP serving as the phosphate donor in the initial step of metabolism and assimilation of the sugar. Two enzymes, which may be considered isozymes, are capable of catalyzing this reaction in the liver cell. Hexokinase has a low Km of ~0.05 mM for glucose, whereas glucokinase exhibits sigmoidal kinetics with an approximate Km of ~5 mM. After a large meal, the glucose concentration in the hepatic portal vein may approximate 5 mM. After such a large meal, which of the following scenarios describes the relative activity levels for these two enzymes? Hexokinase Glucokinase A. Not active Not active B. v ≅ 1⁄2Vmax Not active C. v ≅ Vmax Not active D. v ≅ Vmax v ≅ 1⁄2Vmax E. v ≅ Vmax v ≅ Vmax

*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.

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 A.* At Vmax, the reaction is proceeding at a rate which is equal to the physiologic rate at which the enzyme can function. At Vmax, the enzyme is fully saturated, and no more free enzyme is available to bind the substrate and catalyze the reaction. If all of the enzyme, E, is bound and operating, it must all exist in the ES state. In order to maintain Vmax, there must be sufficient substrate for the enzyme to bind, so [S] must be high enough to sustain this saturated state. If the enzyme-catalyzed reaction E + S ⇋ ES ⇋ E + P is proceeding at or near the Vmax of E, it can be deduced that S is abundant, and [ES] is at its highest point.

If the enzyme-catalyzed reaction E + S ⇋ ES ⇋ E + P is proceeding at or near the Vmax of E, what can be deduced about the relative concentrations of S and ES? (A) S is abundant, [ES] is at its highest point (B) [S] is vanishingly low, [ES] is vanishingly low (C) [S] is vanishingly low, [ES] is at its highest point (D) S is abundant, [ES] is vanishingly low

*The answer is C.* A' depicts the enzyme upon binding to an activator (increased Vmax). 'B' shows the enzyme upon binding to a competitive inhibitor (Vmax is unchanged). 'C' represents an enzyme under the effect of a non-competitive inhibitor (decreased Vmax).

If the red curve indicates the normal reaction, which curve on the following graph is an example of non-competitive inhibition (A) A (green curve) (B) B (blue curve) (C) C (pink curve) (D) Not enough information provided

*The answer is D.* In hydrogen bonds, a hydrogen atom is shared between two electron-rich groups, such as oxygen or nitrogen. Thus, the nitrogen in the peptide bond may share its hydrogen with the oxygen in an aspartate carboxyl group, which will be negatively charged at a pH of 7.4. Leucine is nonpolar and cannot participate in hydrogen bonding (thus, A is incorrect), and aspartyl and glutamyl residues are both negatively charged (thus, B is incorrect). Bonds between two fully charged groups are electrostatic bonds, not hydrogen bonds (thus, C is incorrect). Sulfhydryl groups do not participate in hydrogen bonding.

In a polypeptide at physiologic pH, hydrogen bonding may occur between which of the following? (A) The side chains of a leucine residue and a lysine residue (B) The side chains of an aspartyl residue and a glutamyl residue (C) The terminal alpha-amino group and the terminal alpha-carboxyl group (D) The amide group in the peptide bond and an aspartyl side chain (E) The -SH groups of two cysteine residues

*The answer is C.* Limes provided vitamin C, which is a required cofactor for prolyl hydroxylase, the enzyme which hydroxylates proline residues in collagen. Lysine crosslinks in collagen do not require vitamin C (although lysine hydroxylation, for the purpose of glycosylation, does). Vitamin C does not affect calcium mobilization (that is vitamin D), and fibrillin is not the problem in vitamin C deficiency. Glycine residues in collagen cannot be converted to proline within the polypeptide.

In the 1800s, British sailors on long sea journeys developed sore and bleeding gums, sometimes to the point that their teeth would loosen and fall out. The introduction of limes to their diets helped to prevent these occurrences. The biochemical step that was lacking in these sailors was which of the following? (A) Creating lysine cross-links in collagen (B) Mobilization of calcium into bone (C) Hydroxylation of proline residues in collagen (D) Glycosylation of fibrillin (E) Conversion of glycine to proline in collagen

*The answer is E.* Scurvy is caused by a deficiency of vitamin C. Vitamin C is a required cofactor for the hydroxylation of both proline and lysine residues in collagen. The hydroxyproline residues that are formed stabilize the triple helix through the formation of hydrogen bonds with other collagen molecules within the helix. The loss of this stabilizing force greatly reduces the strength of the collagen fibers. The hydroxylation of lysine allows carbohydrates to be attached to collagen, although the role of the carbohydrates is still unknown. Vitamin C is not required for disulfide bond formation, the formation of lysyl cross-links (that enzyme is lysyl oxidase), secretion of tropocollagen, or the formation of collagen fibrils. Thus, all the other choices are incorrect.

Individuals who develop scurvy suffer from sore and bleeding gums and loss of teeth. This is a result, in part, of the synthesis of a defective collagen molecule. The step that is affected in collagen biosynthesis attributable to scurvy is which of the following? (A) The formation of disulfide bonds, which initiates tropocollagen formation (B) The formation of lysyl cross-links between collagen molecules (C) Secretion of tropocollagen into the extracellular matrix (D) The formation of collagen fibrils (E) The hydroxylation of proline residues, which stabilizes the collagen structure

*The answer is A.* A competitive inhibitor competes with the substrate for binding to the active 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 ½ 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).

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 B.* This is solved using the Michaelis-Menten equation, v = Vmax/(1 + [Km/S]). Under inhibited conditions Vmax is 6 units/min/ mg protein, the Km is 1.25 μM, and the substrate concentration is 2.50 μM. Plugging these numbers into the equation leads to a value of v of 4 units/min/mg protein. V0 = (Vmax × [S])/(Km + [S]) V0 = (6 x 2.5) / (1.25 + 2.5) V0 = 15 / 3.75 V0 = 4 units/min/mg protein

Many drugs function by acting as inhibitors of particular enzyme reactions. If an enzyme's Vmax is 15 units/ min/mg protein, with a Km of 1.25 μM in the absence of inhibitor, but in the presence of 5 μM inhibitor the Vmax is 6 units/min/mg protein, with the same Km, what is the velocity of the reaction in the presence of 5 μM inhibitor at a substrate concentration of 2.50 μM? (A) 2 units/min/mg protein (B) 4 units/min/mg protein (C) 6 units/min/mg protein (D) 8 units/min/mg protein (E) 10 units/min/mg protein

*The answer is C.* Mutations in structural proteins often exhibit autosomal dominant inheritance, while mutations in enzymes often exhibit autosomal recessive inheritance. Structural proteins such as collagen or fibrillin must interact to form scaffolds in the extracellular matrix of connective tissue. Mutation at one of the homologous autosomal loci can introduce an abnormal polypeptide throughout the scaffold much like a misshapen brick in a wall—the distorted polypeptide from the abnormal locus subverts that from the normal locus and weakens the connective tissue matrix, causing autosomal dominant disease. Sometimes the abnormal polypeptide complexes with normal polypeptides and causes them to be degraded, a mechanism called protein suicide. The suicidal effects of mutations at some loci are referred to generally as "dominant negative" mutations. Fibrillin is a glycoprotein used to form a scaffold in the connective tissue filaments called microfibrils. It is distributed in the suspensory ligament for the lens of the eye, the aorta, and the bones and joints, accounting for the symptoms of Marfan's syndrome (154700). Similar pathogenetic mechanisms occur in the osteogenesis imperfectas (e.g., 166200) with multiple fractures and in the Ehlers-Danlos syndromes (e.g., 130060) with skin fragility (scarring) and vascular disease due to mutations in various collagens. The mutations disrupt the α helix secondary structure of collagens, which is dependent on the glycine-X-Y triplet amino acid repeats; the distorted collagen polypeptides then disrupt the collagen fibrils with symptoms dependent on its tissue distribution (2 types of fibrillin and more than 15 types of collagen are known).

Marfan's syndrome is caused by which of the following mechanisms? (A) Mutation that prevents addition of carbohydrate residues to the fibrillin glycoprotein (B) Mutation in a carbohydrate portion of fibrillin that interferes with targeting (C) Mutation that disrupts the secondary structure of fibrillin and blocks its assembly into microfibrils (D) Mutation in a lysosomal enzyme that degrades fibrillin (E) Mutation in a membrane receptor that targets fibrillin to lysosomes

*The answer is B.* Ethanol has a structure very similar to methanol (a structural analog) and thus can be expected to compete with methanol at its substrate-binding site. This inhibition is competitive with respect to methanol and, therefore, Vmax for methanol will not be altered and ethanol inhibition can be overcome by high concentrations of methanol (thus, A, C, and D are incorrect). E is illogical because the substrate methanol stays in the same binding site as it is converted to its product, formaldehyde.

Methanol (CH3OH) is converted by alcohol dehydrogenases to formaldehyde (CH2O), a compound that is highly toxic to humans. Patients who have ingested toxic levels of methanol are sometimes treated with ethanol (CH3CH2OH) to inhibit methanol oxidation by alcohol dehydrogenase. Which of the following statements provides the best rationale for this treatment? (A) Ethanol is a structural analog of methanol and might therefore be an effective noncompetitive inhibitor. (B) Ethanol is a structural analog of methanol that can be expected to compete with methanol for its binding site on the enzyme. (C) Ethanol can be expected to alter the Vmax of alcohol dehydrogenase for the oxidation of methanol to formaldehyde. (D) Ethanol is an effective inhibitor of methanol oxidation regardless of the concentration of methanol. (E) Ethanol can be expected to inhibit the enzyme by binding to the formaldehyde-binding site on the enzyme, even though it cannot bind at the substrate binding site for methanol.

*The answer is B.* Enzyme inhibitors are classified by their reversibility. Reversible inhibitors are further classified by the mechanism by which they inhibit enzymes, which is largely due to where and how they bind to them. Classes of reversible inhibition include competitive, uncompetitive, and noncompetitive/mixed inhibition. Uncompetitive inhibitors bind only to the ES complex, and at a site distinct from the active site. Noncompetitive/ mixed inhibitors bind to a site distinct from the active site, but may do so to either the E or ES state. Competitive inhibitors are so-called because they 'compete' for the active site of an enzyme, preventing substrate from binding. Molecule 'X' must NOT be a competitive inhibitor.

Molecule 'X' is an enzyme inhibitor that reversibly binds to an enzyme at a site that is distinct from its active site. Molecule 'X' must NOT be what type of inhibitor? (A) Uncompetitive inhibitor (B) Competitive inhibitor (C) Mixed inhibitor (D) Noncompetitive inhibitor

*The answer is A.* Competitive inhibitors resemble the structure of the substrate and compete with the substrate to bind to the active site of the enzyme. For this reason, Km increases with increasing inhibitor concentration, while Vmax remains the same. That is, Vmax can be reached at substrate concentrations sufficiently high to overcome the inhibitor. Since the x axis intercept represents −1/Km and the y axis intercept represents 1/Vmax, only curves A, X, and D show changes in Km with no changes in Vmax. When −1/Km is interpreted properly, the highest Km value is given by curve A.

Of the six curves labeled in the Lineweaver-Burk graph below, three represent the effects of 0 mM, 5 mM, and 15 mM of a competitive inhibitor on a hypothetical enzyme. Which of the curves most likely represents the 15-mM concentration of the competitive inhibitor?

*The answer is A.* The mutation in sickle cell is E6V of the β-globin chain. The sickle hemoglobin 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.

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 D.* In competitive inhibition, the Km of the reaction changes, while the Vmax will stay constant with increased amounts of inhibition. In uncompetitive inhibition, both the Km and Vmax will decrease with increased amounts of inhibition. In noncompetitive inhibition, the Km of the reaction stays constant while the Vmax will decrease with increased amounts of inhibition. Therefore inhibition that does not alter Km must be noncompetitive inhibition.

Potassium cyanide is a poison which combines with cytochrome a3 to prevent binding of oxygen to the enzyme without altering the Km of the reaction with respect to reduced cytochrome c. Which type of inhibition does this represent? (A) Uncompetitive inhibition (B) Competitive inhibition (C) Reversible inhibition (D) Noncompetitive inhibition

*The answer is C.* In a typical α-helix, there are 13 atoms between hydrogen bonds (formed between the carbonyl oxygen of one amino acid and the amide nitrogen of the amino acid four residues up in the chain). This is referred to as a 3.6/13 helix (3.6 amino acids per turn, with 13 atoms between hydrogen bonds).Other variants of the helix are 3/10 and 4.4/16. As shown below, in a linear fashion, are the hydrogen bonds formed in all three types of helices.

Shown below is a section of a protein which forms a typical α-helix. In the form of an α-helix, a hydrogen bond would be formed between which two of the labeled atoms? (A) A and C (B) B and D (C) B and E (D)B and F (E) D and F

*The answer is E.* In the β-globin chain of hemoglobin S (141900), a valine residue replaces a glutamic acid at the sixth amino acid position from the N-terminus. The amino acid substitution is the result of a single base change (point mutation) from thymine to adenine at the second position of the sixth codon. Crossing over among homologous β-globin genes might exchange alleles, if equal, or generate mutant alleles with duplicated/ deficient nucleotides, if unequal. Two-base insertions would change the reading frame of the genetic code (frameshift mutation) and produce a nonsense peptide after the point of insertion. Three-base deletions could also cause frame shifts or, if one codon were removed, delete one amino acid. Nondisjunction involves abnormal segregation of chromosomes at meiosis or mitosis, and would produce nonviable individuals or somatic cells with additional or missing copies of chromosome 11 and its β-globin locus.

Sickle cell anemia (141900) is the clinical manifestation of homozygous genes for an abnormal hemoglobin molecule. The mutation in the β chain is known to produce a single amino acid change. The most likely mechanism for this mutation is (A) Crossing over (B) Two-base insertion (C) Three-base deletion (D) Nondisjunction (E) Single-base substitution (point mutation)

*The answer is A.* The isoelectric point (pI) of an amino acid is that pH at which the net charge is zero. Since pK values denote the pH at which a given α-COOH, α-NH3, or R side chain group is dissociated, it is possible to calculate the pI. For amino acids with uncharged side groups, the pI is simply the halfway point between the α-COOH (pK1) and the α-NH3 (pK2) = (pK1 + pK2)/2. For basic amino acids, the pI is the average of the α-NH3 and the side chain group. If the side chain group is designated as pK3, then pI = (pK2 + pK3)/2. For acidic amino acids, the pI is halfway between the α-COOH and the side chain group: (pK1 + pK3)/2. For aspartate, (2.0 + 3.9)/2 = 3.0.

Since the pK values for aspartic acid are 2.0, 3.9, and 10.0, it follows that the isoelectric point (pI ) is (A) 3.0 (B) 3.9 (C) 5.9 (D) 6.0 (E) 7.0

*The answer is C.* Substitution of alanine for lysine removes from each β subunit a positive charge that is important for making a salt bridge with BPG. BPG should still bind but just not as well as it would to normal adult hemoglobin and the affinity would be decreased. Because BPG binding stabilizes the deoxy form of hemoglobin, reduced BPG binding affinity would make the deoxy-to-oxy transition occur at lower PO2 values, ie, affinity of the mutant hemoglobin for O2 would be increased.

Some patients with erythrocytosis (excess RBCs) have a mutation that converts a lysine to alanine at amino acid 82 in the β subunit of hemoglobin. This particular lysine normally protrudes into the central cavity of deoxyhemoglobin, where it participates in binding 2,3- bisphosphoglycerate (BPG). Which of the following effects would you predict this mutation to have on the affinity of hemoglobin for BPG and O2, respectively, in such patients? A. Increase, Decrease B. Increase, Increase C. Decrease, Increase D. Decrease, Decrease E. No effect on either binding function

*The answer is D.* After binding of the first oxygen, hemoglobin shifts from a taut (T) state toward a relaxed (R) state with the ferrous iron in plane with the four planar pyrrole groups of heme. Binding of subsequent oxygen atoms requires less change of secondary, tertiary, and quaternary structure, producing the cooperative kinetics reflected in the sigmoidal oxygen binding curve (see the figure in answer 141 above). Besides accounting for allosteric changes during oxygen binding, the tertiary folding of each hemoglobin chain and its quaternary (four-chain) structure produce preferred binding of oxygen due to steric restraint. Isolated heme binds carbon dioxide 25,000 times more strongly than oxygen, but in myoglobin and each hemoglobin chain, a histidine group interferes with the preferred mode of carbon dioxide binding such that oxygen is favored. The myoglobin molecule is virtually identical to the β-globin chain of hemoglobin, emphasizing again that the quaternary structure of four subunits in hemoglobin produces its sigmoidal oxygen binding curve, which provides for lung oxygen saturation and tissue desaturation with CO2 loading. Because of this sigmoidal curve, hemoglobin binds proportionately less oxygen at low oxygen tension (low PO2) than does myoglobin. Oxidation of the ferrous iron in myoglobin or hemoglobin to ferric ion abolishes oxygen binding, in contrast to the case with other proteins like cytochromes or catalase, where oxidation/reduction of iron modulates their function.

The ability of hemoglobin to serve as an effective transporter of oxygen and carbon dioxide between lungs and tissues is explained by which of the following properties? (A) The isolated heme group with ferrous iron binds oxygen much more avidly than carbon dioxide (B) The α- and β-globin chains of hemoglobin have very different primary structures than myoglobin (C) Hemoglobin utilizes oxidized ferric iron to bind oxygen, in contrast to the ferrous ion of myoglobin (D) In contrast to myoglobin, hemoglobin exhibits greater changes in secondary and tertiary structure after oxygen binding (E) Hemoglobin binds proportionately more oxygen at low oxygen tension than does myoglobin

*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).

The active site of an enzyme will bind to which set of molecules indicated below?

*The answer is C.* Since rapidly multiplying cancer cells are dependent upon the synthesis of deoxythymidilate (dTMP) from deoxyuridylate (dUMP), a prime target in cancer therapy has been inhibition of dTMP synthesis. The anticancer drug fluorouracil is converted in vivo to fluorodeoxyuridylate (FdUMP), which is an analogue of dUMP. FdUMP irreversibly forms a covalent complex with the enzyme thymidylate synthase and its substrate N5,N10-methylene-tetrahydrofolate. This is a case of suicide inhibition, where an enzyme actually participates in the change of a substrate into a covalently linked inhibitor that irreversibly inhibits its catalytic activity.

The chemotherapy drug fluorouracil undergoes a series of chemical changes in vivo that result in a covalent complex such that it is bound to both thymidylate synthase and methylene-tetrahydrofolate. The inhibition of deoxythymidilate formation and subsequent blockage of cell division is due to (A) Allosteric inhibition (B) Competitive inhibition (C) Irreversible inhibition (D) Noncovalent inhibition (E) Noncatalytic inhibition

*The answer is B.* The Vmax of the enzyme in the presence of the modifier is the same compared to the absence of the modifier.

The data shown in the figure are based on measuring the enzyme kinetics of an enzyme in the absence (curve X) or presence (curve Y) of an allosteric modifier. The Vmax of the enzyme in the presence of the modifier is which ONE of the following as compared to the absence of the modifier? (A) Greater (B) The same (C) Less than (D) Insufficient data to make a determination

*The answer is B.* The allosteric modifier is acting as a competitive inhibitor of the enzyme. This is because the Km increases, but the Vmax is unaffected.

The data shown in the figure are based on measuring the enzyme kinetics of an enzyme in the absence (curve X) or presence (curve Y) of an allosteric modifier. The allosteric modifier is acting as which ONE of the following? (A) An activator of the enzyme (B) A competitive inhibitor of the enzyme (C) A non-competitive inhibitor of the enzyme (D) Insufficient data to make a determination

*The answer is B.* Proteins can be effective buffers of body and intracellular fluids. Buffering capacity is dependent upon the presence of amino acids having ionizable side chains with pKs near physiologic pH. In the example given, only histidine has an ionizable imidazolium group that has a pK close to neutrality (pK = 6.0). Valine and leucine are amino acids with uncharged, branched side chains. Lysine has a very basic amino group (pK = 10.5) on its aliphatic side chain that is positively charged at physiologic pH, and aspartic acid has a side chain carboxyl (pK = 3.8) that is negative at pH 7.

The greatest buffering capacity at physiologic pH would be provided by a protein rich in which of the following amino acids? (A) Lysine (B) Histidine (C) Aspartic acid (D) Valine (E) Leucine

*The answer is C.* This problem is best solved using the Michaelis Menton equation and comparing the velocity (as a function of maximal velocity) under fasting and nonfasting conditions. 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.

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 A.* The patient's enzyme has a lower Km than the normal enzyme and, therefore, requires a lower glucose concentration to reach ½Vmax. Thus, the mutation may have increased the affinity of the enzyme for glucose, but it has greatly decreased the subsequent steps of the reaction leading to formation of the transition-state complex, and thus Vmax is much slower. The difference in Vmax is so great that the patient's enzyme is much slower whether you are above or below its Km for glucose. You can test this by substituting 2 mM glucose and 4 mM glucose into the Michaelis-Menten equation, v = Vmax S/(Km + S) for the patient's enzyme and for the normal enzyme. The values are 0.0095 U/mg and 0.0129 U/mg for the patient's enzyme versus 23.2 U/mg and 37.2 U/mg for the normal enzyme, respectively (thus, B and C are incorrect). At near-saturating glucose concentrations, both enzymes will be near Vmax, which is equal to kcat times the enzyme concentration. Thus, it will take nearly 500 times as much of the patient's enzyme to achieve the normal rate (93 / 0.2), and so C is incorrect. E is incorrect because rates change most as you decrease substrate concentration below the Km. Thus, the enzyme with the highest Km will show the largest changes in rate.

The pancreatic glucokinase of a patient with MODY had a mutation replacing a leucine with a proline. The result was that the Km for glucose was decreased from a normal value of 6 mM to a value of 2.2 mM, and the Vmax was changed from 93 U/mg protein to 0.2 U/mg protein. Which of the following best describes the patient's glucokinase compared with the normal enzyme? (A) The patient's enzyme requires a lower concentration of glucose to reach ½ Vmax. (B) The patient's enzyme is faster than the normal enzyme at concentrations of glucose <2.2 mM. (C) The patient's enzyme is faster than the normal enzyme at concentrations of glucose >2.2 mM. (D) At near-saturating glucose concentration, the patient would need 90 to 100 times more enzyme than normal to achieve normal rates of glucose phosphorylation. (E) As blood glucose levels increase after a meal from a fasting value of 5 mM to 10 mM, the rate of the patient's enzyme will increase more than the rate of the normal enzyme.

*The answer is C.* The patient is "blowing off CO2" to reduce acid. The equation described in answer C is the conversion of carbon dioxide into a soluble form, then into bicarbonate. During an acidosis, the high levels of protons push the reaction described in answer C to the left, to the formation of water and carbon dioxide. As the carbon dioxide is exhaled, and the concentration of carbon dioxide decreases, more carbon dioxide is formed, thereby reducing the pool of free protons and raising the pH. The protonation of ammonia to form ammonium ion takes place in the kidney, and not the lungs. Its primary purpose is to alkalinize the urine if it is too acidic. The reaction described in answer B is the dissociation of a proton from β hydroxybutyrate (a ketone body) to form the anion of β hydroxybutyrate and a proton. This is the reaction that is occurring to bring about the ketoacidosis, and is not the compensatory respiratory alkalosis. Reaction D is the dissociation of water, which cannot buffer the acidosis.

The patient described in the previous question is hyperventilating to compensate for her metabolic acidosis. Which of the following reactions explains this partially compensating respiratory alkalosis? (A) H + NH₃ ↔ NH₄⁺ (B) CH₃CHOHCH₂COOH ↔ CH₃CHOHCH₂COO⁻ + H⁺ (C) CO₂ + H₂O ↔ H₂CO₃ ↔ H⁺ + HCO₃⁻ (D) H₂O ↔ H⁺ + HO⁻

*The answer is C.* The homeless man has developed scurvy owing to a lack of vitamin C in his diet. Vitamin C is a required cofactor for the hydroxylation of proline and lysine within the collagen molecule. The lack of hydroxyproline reduces the stability of the collagen because of reduced hydrogen-bonding capabilities within the collagen triple helix. The lack of vitamin C does not affect disulfide-bond formation, which is required to initiate triple-helix formation within the cell. Fibrillin is not altered by the lack of vitamin C; it is the protein mutated in Marfan's syndrome.

The question is based on the following case: A homeless man was seen at a clinic because of bleeding gums and loosening teeth. His dietary history revealed that he has been consuming only chocolate milk and fast-food hamburgers for the past 6 months. The patient is exhibiting these symptoms due to which one of the following? (A) Reduced synthesis of fibrillin (B) Reduced synthesis of collagen (C) Reduced hydrogen-bond formation in collagen (D) Increased hydrogen-bond formation in collagen (E) Reduced disulfide-bond formation in collagen (F) Increased disulfide-bond formation in collagen

*The answer is B.* Scurvy is due to a lack of vitamin C, which is obtained from citrus fruits, which have been lacking in the diet. The patient may also become deficient in the other vitamins listed, but the lack of those vitamins will not lead to the symptoms characteristic of scurvy.

The question is based on the following case: A homeless man was seen at a clinic because of bleeding gums and loosening teeth. His dietary history revealed that he has been consuming only chocolate milk and fast-food hamburgers for the past 6 months. The patient, due to his diet, had become deficient in which one of the following vitamins, which would lead to the symptoms observed? (A) Vitamin A (B) Vitamin C (C) Vitamin B1 (D) Vitamin B2 (E) Vitamin B6

*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.

This 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 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.

This 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.* Mad cow disease is a prion disorder, in which a misfolded prion protein in the brain forms aggregrates and precipitates, interfering with normal brain function. Prions can adopt a "stable" conformation, which consists primarily of α-helices, and an aggregation-prone conformation, which consists primarily of β sheets. Once in the aggregation-prone conformation, the protein aggregates, shifting the equilibrium between structure forms toward the aggregation-prone form. This feeds the aggregation-prone form until the precipitated protein begins to interfere with brain function, and will eventually lead to death. Prion disorders are not due to altered gene expression, viruses, proteolytic cleavage of a prion protein, or to the loss of the nuclear membrane.

Unbeknownst to its owners, a cow recently sacrificed for meat production had mad cow disease. The precipitating event in the cow's brain that led to this disease is which one of the following? (A) Altered gene expression (B) Infection of the brain with a virus (C) Proteolytic cleavage of an existing brain protein (D) An altered secondary and tertiary structures for an existing brain protein (E) Loss of the nuclear membrane

*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.

When the pH of a solution of a weak acid, HA, is equal to the pKa, the ratio of the concentrations 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 A.* An amino acid substitution from isoleucine to arginine will most likely cause a change in a protein's tertiary structure. This is because isoleucine is a nonpolar, uncharged, hydrophobic amino acid and arginine is a polar, charged, hydrophilic amino acid.

Which ONE of the following amino acid substitutions most likely causes a change in a protein's tertiary structure? (A) Isoleucine to arginine (B) Arginine to lysine (C) Valine to leucine (D) Threonine to serine (E) Asparagine to glutamine

*The answer is B.* The α-helical segments of proteins represent one of the most common secondary structures of proteins. The helical structure is composed of a spiraled polypeptide backbone core with the side chains of component amino acids extending outward from the central axis in order to avoid interfering sterically or electrostatically with each other. All the peptide bond carbonyl oxygens are hydrogen-bonded to a peptide linkage that is four residues ahead in the polypeptide. This leads to 3.6 amino acids per turn, spatially held together in the α-helical structure. Since the configuration of the α helix is compatible with being in the interior of proteins, amino acids with nonpolar, hydrophobic side chains predominate. Conversely, amino acids that are charged or have bulky side chains may interfere with the α-helical structure if present in large enough amounts. Proline and hydroxyproline are not at all compatible with the right-handed spiral of the α-helix. They insert a kink in the chain. Likewise, large numbers of charged amino acids such as lysine or histidine disrupt the helix by forming electrostatic bonds or by ionically repelling one another. In addition, amino acids with bulky side chains, such as tryptophan or isoleucine, also tend to disrupt the configuration of the α helix.

Which of the following amino acids is most compatible with an α-helical structure? (A) Tryptophan (B) Alanine (C) Lysine (D) Proline (E) Cysteine

*The answer is B.* A molecule of hemoglobin binds four molecules of oxygen, one per subunit. When an oxygen-saturation curve is plotted, instead of the hyperbolic curve seen with myoglobin, there is a sigmoid curve that is well to the right of that of myoglobin (Fig. 4.21). Higher oxygen concentrations are required to 50% saturate haemoglobin than is the case for myoglobin, indicating that myoglobin has a higher affinity for oxygen. Hemoglobin is an allosteric protein - a protein that has more than one binding site for ligands and in which binding of a ligand at one site influences the interaction of the protein and ligand at another site. Each of hemoglobin's four subunits is capable of binding an oxygen molecule. The oxygen-binding curve of haemoglobin is sigmoid because binding of the oxygen at low pressures when few of the sites are occupied is more difficult than when more are occupied. There is a progressive increase in affinity of hemoglobin for oxygen as site occupancy increases, so that at the higher oxygen pressures the affinity is increased 20-fold. Although the haem groups in hemoglobin are distant from one another, the initial binding of oxygen to one subunit facilitates the binding of further molecules of oxygen to the other subunits. This is known as a homotropic positive cooperative effect (homotropic, because the ligand bound at each site is the same i.e. oxygen). 2,3-Bisphosphoglycerate (BPG) plays an important physiological role in oxygen transport by lowering the affinity of haemoglobin in red blood cells for oxygen, and thus increasing the unloading of oxygen to the tissues; it moves the dissociation curve of oxyhaemoglobin to the right. BPG acts as a negative heterotropic allosteric modulator (or allosteric effector) of hemoglobin. Hemoglobin in the deoxygenated state has a higher binding affinity for protons than has oxyhaemoglobin. Put in another way, the oxygenated form of hemoglobin is a stronger acid than is the deoxygenated form, resulting in dissociation of protons from the molecule when oxygen binds, a phenomenon known as the Bohr effect. When oxygen is unloaded, hemoglobin takes up protons. This carries roughly half of the H+ ions generated by CO2 in the tissues, and helps to prevent a drop in the pH of the red blood cell to unphysiological levels.

Which of the following statements about hemoglobin is correct? (A) 2,3-Bisphosphoglycerate (BPG) increases the affinity of hemoglobin for oxygen. (B) Deoxygenated hemoglobin has a higher binding affinity for protons than has oxyhemoglobin. (C) Hemoglobin has a higher affinity for oxygen than does myoglobin. (D) One molecule of hemoglobin binds sixteen molecules of oxygen - four per subunit.

*The answer is B.* In an enzyme catalyzed reaction, any spontaneous reaction of substrate to product is considered negligible. A non-competitive inhibitor will decrease the Vmax of the enzyme, since the enzyme-substrate inhibitor complex turns over to product at a slower rate than the enzyme-substrate complex. Km with units of molarity, is a general measure of the inverse affinity of the enzyme for a substrate. Therefore a low Km would indicate a high affinity, and therefore an enzyme with lower Km would be more easily saturated.

Which of the following will be true regarding enzymes saturated with substrate? (A) Increasing the substrate concentration will appreciably increase the reaction rate (B) An enzyme with lower Km is more easily saturated than an enzyme with high Km (C) Any excess substrate will shift the equilibrium towards the product end of the reaction (D) At saturating levels of substrate, a competitive inhibitor will affect the reaction rate more than a non-competitive inhibitor

*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.

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 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.

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 D.* The patient has a form of osteogenesis imperfecta, which is due to a mutation in collagen, generating brittle bones. Mild trauma is sufficient to break the bones. Bisphosphonates decrease bone resorption by the osteoclasts, thereby strengthening the bone, even with the defective collagen molecule. Bisphosphonates do not affect the synthesis of collagen or fibrillin. Bone reabsorption is resorption of bone tissue, that is, the process by which osteoclasts break down the tissue in bones and release the minerals, resulting in a transfer of calcium from bone tissue to the blood.

While working an overnight shift in the emergency department you are called to see an 8-year-old boy who appears to have a fracture in his arm. Upon taking a history, you learn that this child has been to the ER multiple times for fractures, and the incidents that lead to the fracture would be described as mild trauma at best. X-rays indicate a number of healed fractures that the boy and his parents were unaware of (see example of arm X-ray below). Physical exam shows sky blue sclera. The parents then inform you that the child is taking bisphosphonates for his condition. The mechanism whereby the frequency of fractures is being reduced in this patient is which of the following? (A) Increased synthesis of collagen (B) Increased resorption of collagen (C) Decreased synthesis of collagen (E) Decreased resorption of collagen (F) Increased synthesis of fibrillin

*The answer is C.* The effect of modulators on enzyme activity is dependent upon cellular conditions, such that they may bind at high intracellular concentrations, or diffuse out at low concentrations. Allosteric enzymes bind modulators at specific sites, and are characterized by conformational changes upon binding. Covalent bonds are not readily reversible. Allosteric enzymes bind modulators in a reversible fashion.

With respect to the binding of regulatory compounds, what properties define an enzyme as being allosteric? (A) Irreversible, covalent binding of regulatory compounds (B) Irreversible, noncovalent binding of regulatory compounds (C) Reversible, noncovalent binding of regulatory compounds (D) Reversible, covalent binding of regulatory compounds\

*The answer is E.* Lane 3 represents an individual with sickle cell disease. The mutation in sickle cell disease is a valine for glutamate substitution at position 6 of the β-chain. This substitution removes a negative charge from the β-chain such that when the β-chain is migrated through an electric field it will not travel as far towards the positive pole as does the non-mutated protein. Thus, in the gel shown in the question, band X represents the hemoglobin S β-chain (since it does not migrate as far towards the positive pole), and band Y represents the non-mutated protein. The pattern shown in lane 1 is that of a carrier of HbS (one normal β-chain and one mutated β-chain). The pattern shown in lane 2 represents a person who does not carry a mutant allele (two normal alleles). Lane 3 represents someone with the disease (two mutant genes).

You order a hemoglobin electrophoresis on a patient suspected of having sickle cell disease. A blood sample was obtained and the red cells were isolated. Disruption of the red cells released the hemoglobin, which was run on a polyacrylamide gel. Following the electrophoresis, a Western blot was performed to locate the hemoglobin. The results of the Western blot are shown below. Which one of the following statements best represents the interpretation of the results? (A) The band marked as X refers to the wild-type hemoglobin protein (B) The band in lane 2 represents an individual with sickle cell disease. (C) A carrier of sickle cell disease is represented by the band in lane 3 (D) Lane 1 represents an individual with sickle cell disease (E) Lane 3 represents an individual with sickle cell disease

*The answer is C.* The patient is exhibiting methemoglobinemia, in which an increased percentage of his hemoglobin has the iron in the +3 oxidation state (normal is +2), which is a form that cannot bind oxygen. This condition can arise by a variety of mutations within hemoglobin which lead to destabilization of the iron in the heme ring. The red blood cell contains methemoglobin reductase, which will reduce the iron back to the +2 state (using NADPH as the electron donor), and mutations within the reductase can also lead to this condition. An acquired form of methemoglobinemia can be caused by exposure to oxidizing drugs or toxins (aniline dyes, nitrates, nitrites, and lidocaine) which exceed the reduction capacity of the red blood cells. Surprisingly, the majority of patients with this syndrome show no ill effects, other than the bluish discoloration of certain tissues. Excessive 2,3-bisphosphoglycerate in the erythrocyte would lead to increased oxygen delivery to the tissues as 2,3-bisphosphoglycerate stabilizes the deoxygenated form of hemoglobin (as would a mutant hemoglobin with an enhanced ability to bind 2,3-BPG). The E to V mutation at position 6 of the β-chain of hemoglobin leads to sickle cell disease.

You see a patient on an initial visit and are struck by the bluish coloring of the skin and mucous membranes. You ask the patient about this and you are told that it is a blood problem that the patient has had for his or her entire life. The patient's father had a similar condition, but not the mother. This condition could result from which one of the following changes within the erythrocyte? (A) An increase of 2,3-bisphosphoglycerate in the erythrocyte (B) An E to V mutation at position 6 of the β-chain of hemoglobin (C) Increased oxidation of heme iron to the +3 state (D) Enhanced oxygen binding to hemoglobin (E) A mutated hemoglobin which no longer exhibited the Bohr effect

*The answer is B.* The y-axis on the Lineweaver-Burk plot is 1/V.

The Lineweaver-Burk plot of the reciprocal of the Michaelis-Menten equation is shown below. It is used to graphically determine Km and Vmax. When V is the reaction velocity at substrate concentration S, the y axis experimental data are expressed as (A) V (B) 1/V (C) S (D) 1/S (E) V/Km

*The answer is D.* According to the Henderson- Hasselbalch equation, pH = pKa + log [base]/[acid]. (A useful mnemonic is that you need your "b/a" to remember it.) In the case of 0.2 M lactate and 0.02 M lactic acid, as presented in the question, pH = 3.9 + log 10 = 4.9.

A 0.22 M solution of lactic acid (pKa 3.9) is found to contain 0.20 M in the dissociated form and 0.02 M undissociated. What is the pH of the solution? a. 2.9 b. 3.3 c. 3.9 d. 4.9 e. 5.4

*The answer is B.* The patient shows many signs of vitamin C deficiency or scurvy, which is seen most frequently in infants, the elderly, and in alcoholic patients. Particularly indicative of vitamin C deficiency are the multiple small hemorrhages that occur under the skin (petechiae) and nails and surrounding hair follicles. Bleeding gums are a classic indicator of scurvy.

A 10-month-old white boy is being evaluated for weakness, pallor, hemorrhages under the fingernails, and bleeding gums. Radiographs indicate that bone near the growth plates shows reduced osteoid formation and grossly defective collagen structure. What would be the most effective treatment for this patient's condition? A. Oral vitamin A B. Oral vitamin C C. Exclusion of dairy products from the diet D. Oral iron supplementation E. Growth hormone treatment

*The answer is D.* The patient presents in an acute pain episode, or sickle cell crisis, which is the most common type of vasa-occlusive event in sickle cell disease. Hemoglobin S is the abnormal hemoglobin responsible for sickling of erythrocytes during an acute pain episode. Hemoglobin S has the same oxygen dissociation curve as normal hemoglobin, or hemoglobin A, in low concentrations. During a crisis, however, deoxygenated hemoglobin S forms a polymer that has less affinity for oxygen, and the curve is shifted to the right. Other conditions that shift the oxygen dissociation curve to the right include increased metabolism (ie, increased temperature, carbon dioxide, and increased hydrogen ions in tissue), increased 2,3-bisphosphoglycerate (2,3 BPG), and high altitude. This facilitates unloading of oxygen to tissues and is reflected by curve D in the image, A helpful mnemonic to keep in mind is that an increase in any factor (temperature, carbon dioxide, hydrogen ion concentration, and 2,3-BPG) causes a rightward shift of the hemoglobin-oxygen dissociation curve. (Remember~ though, that an increase in hydrogen ion concentration will lower the pH value.)

A 13-year-old African American female presents to the emergency department with severe pain in her right hip and back. She has experienced many similar episodes in the past and has a documented history of anemia. Referring to the graph above, if N (the solid curve) is the normal oxygen dissociation curve for hemoglobin, which curve represents the oxygen dissociation curve of the predominant form of hemoglobin in the patient's blood during this episode? A B C D

*The answer is A.* This patient is suffering from Kartagener syndrome, a variant of primary ciliary dyskinesia consisting of bronchiectasis, situs inversus, and chronic sinusitis. The underlying cause of the patient's disorder is a defect in the cilia in his airways, making them unable to beat appropriately Because embryonic cilia are also affected, body asymmetry is randomized such that 50% of patients will have situs inversus in addition to respiratory and sinusitis symptoms. Bronchiectasis develops over time; pat ients are not born with it. With chest physiotherapy, antibiotic treatment, and avoidance of smoking, patients can generally expect to have a normal lifespan.

A 13-year-old boy is referred to a specialist for evaluation of chronic cough and frequent pulmonary infections. Detailed medical history reveals that the child has suffered from constant "runny nose" and episodes of sinus congestion. The child has gained weight appropriately. Physical examinat ion of the patient 's nasal passages reveals nasal polyps. Heart sounds are heard predominantly on the right side of the chest. Which of the following cellular structures is defective in this patient's condition? A. Cilia B. Golgi body C. Lysosome D. Mitochondria E. Ribosome

*The answer is A.* The oxygen-hemoglobin dissociation curve describes the relationship between the partial pressure of oxygen in the blood and the oxygen saturation of hemoglobin. The partial pressure of oxygen at which hemoglobin is 50% saturated is known as the P50, this value is used as a conventional measure of hemoglobin's affinity for oxygen. The P50, is about 26 mm Hg in normal individuals. A P50 shift from 26 to 20 mm Hg indicates that the affinity of hemoglobin for oxygen is increased (left shift of the O2 dissociation curve). There are numerous high-oxygen-affinity hemoglobin mutations that reduce the ability of hemoglobin to release oxygen in the tissues (eg, hemoglobins Chesapeake and Kempsey). However, the low tissue oxygen levels stimulate the kidneys to increase erythropoietin synthesis, which results in a compensatory erythrocytosis that helps maintain normal oxygen delivery. Thus, patients with high-oxygen-affinity hemoglobins are typically asymptomatic. *Educational Objective:* P50 refers to the partial pressure of oxygen at which hemoglobin is 50% saturated. High-oxygen~affinity hemoglobin have a decreased P50 that is represented by a leftward shift of the oxygen-dissociation curve. High-oxygen-affinity hemoglobins have reduced ability to release oxygen within the peripheral tissues, leading to renal hypoxia, increased erythropoietin synthesis, and compensatory erythrocytosis.

A 13-year-old girl undergoing hematologic evaluation in found to have a hemoglobin abnormality that decreases the partial pressure of oxygen at which hemoglobin is 50% saturated from 26 to 20 mm Hg. Which of the following sequelae is this patient most likely to develop? A. Erythrocytosis B. Hypoxia-induced hemolysis C. Increased erythrocyte osmotic fragility D. Megaloblastic erythrocyte changes E. Oxidant-induced hemolysis

*The answer is D.* Hyperextensibility of skin and hypermobility of joints are hallmark features of Ehlers Danlos syndrome. The physical findings and history, especially the patient's tall, thin body, her joint and skin hyperextensibility and past medical history of dislocations, are consistent with a collagen disorder. Another inherited collagen disorder, osteogenesis imperfecta, is unlikely due to her tall stature and the absence of evidence of frequent fractures. Vitamin C deficiency affects collagen synthesis and structure but exhibits a different set of clinical findings (eg, hemorrhage).

A 14-year-old girl is brought to the emergency department with shoulder pain and immobility consistent with dislocation. She is tall and thin and exhibits marked flexibility of her skin and joints wrists, fingers, and ankles. There are no apparent cardiac abnormalities or vision problems. She has a past medical history of dislocation of both shoulders and her right hip, as well as easy bruising. Microscopic examination of a skin biopsy shows disorganized collagen fibers. What is the most likely diagnosis in this case? A. Scurvy B. Osteogenesis imperfecta C. Prolyl hydroxylase deficiency D. Ehlers-Danlos syndrome E. Vitamin C deficiency

*The answer is C.* This patient has Marfan Syndrome, which is a deficiency in fibrillin-1. This disease is characterized by long arms and legs, a dislocated lens, and and enlarged aortic root.

A 14-year-old girl is brought to the physician because of a recent growth spurt of 15 cm (6 in) during the past year. She also has had increasing fatigue and palpitations during this period. Her paternal aunt has a history of palpitations and severe myopia. She is at the 95th percentile for height and 50th percentile for weight. Physical examination shows a long, thin face. Ophthalmologic examination shows dislocated lenses. Cardiac examination shows a hyperdynamic precordium with early click and systolic murmur. Echocardiography shows an enlarged aortic root and mitral valve prolapse. Abnormal synthesis of which of the following proteins is the most likely cause of this patient's disorder? (A) Collagen, type I (B) Elastin (C) Fibrillin-1 (D) Fibroblast growth factor R3 (E) Laminin (F) Neurofibromin (G) PAX 6

*The answer is B.* Hemoglobin is a tetramer that consists of 2 pairs of globin chains (total of 4 chains per molecule). During the first few weeks of embryogenesis, hemoglobin is synthesized by the yolk sac and contains zeta or epsilon globin chains (Choices G and F). Thereafter, one pair of the globin chains should always be alpha and the other should be non-alpha. Fetal hemoglobin (Hb F) production begins around 8 weeks gestation and replaces all embryonic hemoglobin by 14 weeks gestation when erythropoiesis in the fetal liver and spleen is established. Hb F consists of 2 alpha and 2 gamma protein subunits (α2γ2). Production declines at birth, and Hb F comprises ~60-80% of all hemoglobin in a term newborn. Hb F is gradually replaced by adult hemoglobin (Hb A, α2β2) (Choice A) during the first 6 months of life, after which Hb A composes the vast majority of adult hemoglobin. Compared to red blood cells with Hb A, those with Hb F have a high oxygen affinity as Hb F binds to 2.3-bisphosphoglycerate poorly. The greater affinity of Hb F facilitates transplacental oxygen delivery from the maternal circulation to that of the fetus. (Choice C) Hemoglobin A2 (α2δ2) is a normal hemoglobin variant that makes up 2% -3% of hemoglobin in a healthy adult and is functionally similar to Hb A. Patients with beta-thalassemia major have impaired beta globin production, resulting in an excess of alpha globin chains (eg. Hb A2. Hb F) and no Hb A. (Choices D and E] Alpha-thalassemia results from a shortage of alpha globin chain. Hemoglobin H (β4) and hemoglobin Barts (y4) have a very high oxygen affinity and cannot release oxygen, resulting in tissue hypoxia. Hemoglobin H disease manifests as chronic hemolytic anemia. Hemoglobin Barts is incompatible with life (eg, hydrops fetalis) as normal fetal and adult hemoglobin cannot be produced. *Educational Objective:* Hemoglobin F (Hb F) is the predominant hemoglobin type in the second and third trimesters of pregnancy and during the first few months after birth. Hb F consists of 2 alpha and 2 gamma protein subunits (α2γ2) and has a high affinity for oxygen, which facilitates oxygen transport across the placenta to the fetus. Hb A (α2β2) is the major hemoglobin in adults.

A 2-day-old boy is being examined in the newborn nursery prior to discharge from the hospital. He was born at 38 weeks gestation by vaginal delivery. The pregnancy and delivery were uncomplicated, and the boy has been breastfeeding, stooling, and urinating normally. The patient's mother has beta-thalassemia trait. and his father has a normal hemoglobin electrophoresis. Vital signs and physical examination are normal. Which of the following hemoglobin compositions is most likely predominant in this infant? (A) α2β2 (B) α2γ2 (C) α2δ2 (D) β4 (E) γ4 (F) ζ2ε2 (F) ζ2γ2

*The answer is D.* Tachypnea in term infants may result from brain injuries or metabolic diseases that irritate the respiratory center. The increased respiratory rate removes ("blows off") carbon dioxide from the lung alveoli and lowers blood CO2, forcing a shift in the indicated equilibrium toward the left: CO2 + H2O ⇌ H2CO2 ⇌ H+ + HCO3− Carbonic acid (H2CO2) can be ignored because negligible amounts are present at physiologic pH, leaving the equilibrium: CO2 + H2O ⇌ H+ + HCO3− The leftward shift to replenish exhaled CO2 decreases the hydrogen ion (H+) concentration and increases the pH (-log10[H+]) to produce alkalosis (blood pH above the physiologic norm of 7.4). This respiratory alkalosis is best treated by diminishing the respiratory rate to elevate the blood [CO2], force the above equilibrium to the right, elevate the [H+], and decrease the pH. The newborn does not have acidosis, defined as a blood pH below 7.4, either from excess blood acids (metabolic acidosis) or from increased [CO2] (respiratory acidosis). The baby also does not have metabolic alkalosis, caused by loss of hydrogen ion from the kidney (e.g., with defective tubular filtration) or stomach (e.g., with severe vomiting).

A 2-day-old neonate becomes lethargic and uninterested in breastfeeding. Physical examination reveals tachypnea (rapid breathing) with a normal heartbeat and breath sounds. Initial blood chemistry values include normal glucose, sodium, potassium, chloride, and bicarbonate (HCO3−) levels; initial blood gas values reveal a pH of 7.53, partial pressure of oxygen (PO2) normal at 103 mmHg, and partial pressure of carbon dioxide (PCO2) decreased at 27 mmHg. Which of the following treatment strategies is indicated? a. Administer alkali to treat metabolic acidosis b. Administer alkali to treat respiratory acidosis c. Decrease the respiratory rate to treat metabolic acidosis d. Decrease the respiratory rate to treat respiratory alkalosis e. Administer acid to treat metabolic alkalosis

*The answer is C.* Kartagener syndrome, or immotile cilia, is caused by a defect in dynein that prevents effective movement of cilia. The full syndrome is characterized by sinusitis, bronchiectasis, situs inversus, and male infertility. Cilia play an important role in moving mucus along the airway and clearing debris; the absence of this function contributes to the pulmonary findings of the syndrome. Cilia are also very important for leukocyte movement and phagocytosis. Infertility is present in most patients due to immotile cilia.

A 2-year-old boy presents to the pediatrician with fever, facial tenderness, and a green, foulsmelling nasal discharge. The patient is diagnosed with sinusitis, and the physician notes that he has a history of recurrent episodes of sinusitis. X-ray of the chest is ordered because of the fever; it reveals some dilated bronchi and shows the heart situated on the right side of his body. A congenital disorder is diagnosed. Which other finding would this patient be most likely to have? (A) Defective chloride transport (B) Elevated blood sugar (C) Infertility (D) Reactive airway disease (E) Tetralogy of Fallot

*The answer is C.* Kartagener syndrome, or primary ciliary dyskinesia, is caused by a defect in dynein that prevents effective movement of cilia. Characteristic manifestations of the syndrome are sinusitis, bronchiectasis, situs inversus, and infertility in both males and females. Situs inversus can be defined as a congenital condition in which some major organs are reversed from their normal position. As an example, the heart will be on the right side as opposed to the left. A CT would show pronounced bronchiectasis in the bilateral lower lobes. Cilia play an important role in moving mucus along the airway and clearing debris; the absence of this function contributes to the pulmonary findings of the syndrome. Cilia are also very important for leukocyte movement and phagocytosis. Infertility is caused by immotile cilia in the fallopian tubes and immotile sperm.

A 2-year-old boy presents to the pediatrician with fever~ facial tenderness, and a green, foul-smelling nasal discharge. The patient is diagnosed with sinusitis, and the physician notes that he has a history of recurrent episodes of sinusitis. X-ray of the chest is ordered because of the fever; it reveals some dilated bronchi and shows the heart situated on the right side of his body, A congenital disorder is diagnosed. Which of the following is associated with this patient's condition? A. Defective chloride transport B. Elevated blood sugar C. Infertility D. Reactive airway disease E. Tetralogy of Fallot

*The answer is D.* Cyanide binds to a variety of iron-containing enzymes, the most important of which is the cytochrome a-a3 complex. This complex is critical for electron transport during oxidative phosphorylation. By binding to this molecule, minute amounts of cyanide can inhibit aerobic metabolism and rapidly result in death. The typical clinical syndrome present in cyanide poisoning is rapidly-developing cutaneous flushing, tachypnea, headache, and tachycardia, often accompanied by nausea/vomiting, confusion, and weakness. Respiratory distress and cardiac dysfunction may follow. Laboratory studies indicate severe lactic acidosis in conjunction with a lessened difference between arterial and venous O2 content (ie. the venous blood is still highly oxygenated). The antidotal effect of nitrites for cyanide poisoning has been recognized since the late nineteenth century. Nitrites are oxidizers, and act primarily in cyanide poisoning by inducing the formation of methemoglobin. This occurs when ferrous iron in hemoglobin is oxidized to ferric iron. Methemoglobin cannot carry oxygen, but it does have a high affinity for cyanide. Methemoglobin can bind and sequester cyanide in the blood, thereby keeping the poison away from mitochondrial (and other} enzymes where cyanide exerts its toxic effects. Sodium thiosulfate is also used for cyanide poisoning, it combines with cyanide to form the less-toxic thiocyanate, which is excreted in the urine.

A 21-year-old laboratory worker experiences rapid-onset breathing difficulty, palpitations, and flushed skin. He has no significant past medical history and takes only loratadine for seasonal allergies. The patient is suspected to have accidental poisoning. Amyl nitrite from a laboratory safety kit is immediately administered via inhalation. Amyl nitrite affects the affinity of hemoglobin for which of the following? A. Carbon dioxide B. 2,3-biphosphoglycarata C. Carbon monoxide D. Cyanide E. Iron F. Lead

*The answer is D.* A ruptured aorta should be suspected as the cause of death in this patient with easy bruisability and hyperextensible skin. An aort ic aneurysm is shown in the image. Aortic aneurysms (both abdominal and thoracic) can be sequelae of Ehlers-Danlos syndrome (EDS)- a group of disorders characterized by joint hypermobility, skin hyperelasticity, easy bruising, and tissue weakness. The disorders are caused by a genetic defect in collagen that results in joint laxity and poor tissue integrity. The most common type of EDS is the hypermobility type, which is not associated with a known single collagen type deficiency. However, a defect in the synthesis of type III collagen is the most frequently seen in the vascular type of EDS, which is depicted in this vignette. In the vascular form of EDS, skin hyperelasticity is less prominent, instead patients often exhibit thin/transparent skin and increased varicosity. The mode of inheritance and the severity of the disease varies.

A 25-year-old man is brought to the emergency department by his next-door neighbor after being found unconscious. No medical history is available. Initial examinat ion reveals blood pressure of 60/40 mm Hg, pulse of 150/min, and respiratory rate of 10/min. He is of normal height . Despite the efforts of emergency department staff, the man cannot be resuscitated. Autopsy fails to reveal any appreciable spinal abnormalities, but several large bruises (present before admission), as well as the appearance of thin/transparent skin and varicose veins, particularly seen on the chest and abdomen, were noted. This patient most likely suffered from defective synthesis of which of the following structural proteins? A. Elastin B. Fibrillin C. Type I collagen D. Type III collagen E. Type IV collagen

*The answer is D.* The boy has Duchenne muscular dystrophy, which is due to mutations in the protein dystrophin, found in the muscle sarcolemma (plasma membrane). The lack of dystrophin alters the permeability properties of the plasma membrane, eventually leading to cell death. The disorder is not found in mitochondria, the liver, or in the β-chain of hemoglobin. This disease also does not alter gene transcription. As the muscles weaken, their function is compromised, leading to the complications of this form of muscular dystrophy.

A 3-year-old boy is evaluated by the pediatrician as the child has trouble rising from a sitting position. Examination reveals calf hypertrophy and limb-girdle weakness. The inborn error in this patient is due to which of the following? (A) Defective muscle mitochondria (B) A mutation in the β-chain of hemoglobin (C) A defect in the structure of the hepatocyte membrane (D) A defect in the structure of the sarcolemma (E) A defect in the transcription of muscle-specific genes

*The answer is D.* Cyanide is a very toxic compound that can be formed in the high-temperature combustion of many materials, such as polyurethane, acrylonitrile, nylon, wool, and cotton, thus making cyanide poisoning common in the setting of smoke inhalation. The most common cause of cyanide poisoning in industrialized countries is household fires. Cyanide modifies the iron within cytochrome oxidase (cytochrome aa3 ) in the mitochondria, thereby abnormally interrupting the electron transport chain and halting cellular respiration. Tissues with the highest oxygen demands, such as the heart, brain, and liver are most significantly affected because cyanide prevents oxygen from binding to cytochrome oxidase and serving as the final electron acceptor in the chain. On physical examination, the retinal arteries and veins are bright red due to absent tissue oxygen extraction. Additionally, some patients have an acrid smell (bitter almonds) on the breath. These findings are not present with carbon monoxide inhalation, the other common toxin associated with smoke inhalation. Antidotal therapy includes direct cyanide binding (with hydroxocobalamin), Induction of methemoglobinemia (with nitrite), and administering a therapeutic sulfur donor (with thiosulfate).

A 30-year-old man presents to the emergency department complaining of shortness of breath, dizziness, nausea, and vomiting. He also says that his heart feels " like it is jumping out of my chest. " Two days ago, he passed a burning house and stopped to help the residents evacuate. He reports inhaling a significant amount of smoke but declined medical assistance at the scene because he had no symptoms. The patient reports feeling very fatigued the day prior to presentation and stayed in bed for most of the day. On physical examination, his pulse is 90/min, blood pressure is 100/60 mm Hg, and respiratory rate is 30/min with deep, gasping respirations. The rest of the examination is unremarkable with the exception of bright red vessels in both of his eyes and an unusual, acrid smell on his breath. As part of the treatment plan for his condition, he was given hydroxocobalamin. What is the mechanism of action of the toxic agent that resulted in this patient's symptoms? A. Cross-linking of IgE bound to the receptor Fc-epsilon-RI on mast cells B. Decreased blood oxygen-carrying capacity C. Inhibition of acetylcholinesterase D. Inhibition of cellular respiration E. Ribosome inhibition

*The answer is D.* The patient described in the vignette has a respiratory acidosis, likely due to decreased ventilation and carbon dioxide retention after ingestion of alcohol and opiates. Acidemia is a condition that shifts the oxygen-hemoglobin dissociation curve to the right; in other words, hemoglobin's affinity for oxygen is reduced. This reduction in affinity enables a greater percentage of the oxygen in the blood to be delivered to peripheral tissues, which compensates for the hypoxia caused by decreased ventilation. Other conditions that cause a right shift in the oxygen-hemoglobin dissociation curve include increased temperature, increased 2,3 bisphosphoglycerate (2,3-BPG), and increased carbon dioxide level. During exercise, lactic acid builds up in tissues and causes a metabolic acidosis that shifts the curve in the same way as a respiratory acidosis would, as described in this vignette. Conversely, decreased temperature, decreased 2,3-BPG, decreased levels of carbon dioxide, and high pH all lead to a left shift of the oxygen dissociation curve.

A 32-year-old man is found unconscious on the sidewalk without evidence of trauma. In the emergency department, a toxicology screen is positive for high levels of ethanol, as well as opiates. His temperature is 36.1°C (97°F), heart rate 50/min, blood pressure 120/80 mm Hg, respiratory rate 8/min, and oxygen saturation 93% (normal range 95%- 99%). An arterial blood gas sample is drawn and is notable for the following: pH 7.29, partial pressure of carbon dioxide 60 mm Hg, and partial pressure of oxygen 80 mm Hg. Which of the following conditions would change hemoglobin's affinity to bind oxygen in the same way it is altered in this patient? A. Decrease in hematocrit B. Decreasing the amount of 2,3-bisphosphoglycerate (2,3-BPG) in RBCs C. Inhalation of carbon monoxide D. Metabolic changes associated with exercise E. Metabolic changes associated with protracted vomiting

*The answer is D.* Fibrilin-1 is a major component of the microfibrils that form a sheath around elastin fibers. Microfibrils are abundantly present in blood vessels, periosteum and the suspensory ligaments of the lens. Fibrillin in the extracellular space acts as a scaffold for deposition of elastic extruded from connective tissue cells. Defects in the fibrillin-1 genes cause classic Marfan's syndrome. Clinically, Marfan's syndrome has three prominent features: 1. Long, thin extremities, loose joints, and long fingers (arachnodactyly). 2. Ocular abnormalities such as dislocation of lens (ectopia lentis). 3. Cardiovascular abnormalities including ascending aortic aneurysm, ortic dissection, and mitral valve prolapse. The patient is this question likely suffered from an aortic dissection. *Educational Objective:* Marfan's syndrome is due to a defect in fibrillin, an extracellular glycoprotein that is abundant in the zonular fibers of the lens, the periosteum and the aortic media. The different locations of fibrillin production explains the varied clinical manifestations of Marfan's syndrome.

A 34-year-old man who died of internal hemorrhage inherited a defect of an elastin-associated glycoprotein that is abundant in the zonular fibers of the lens, periosteum, and the aortic media. The patient most likely suffered from: A. Osteopetrosis B. Ankylosing spondylitis C. Osteogenesis imperfecta D. Marfan's syndrome E. Ehlers-Danlos syndrome F. Achondroplasia

*The answer is B.* The patient has myasthenia gravis, in which she generates antibodies against the acetylcholine receptor. Treatment with edrophonium chloride leads to a transient increase in acetylcholine levels (through the temporary inactivation of acetylcholinesterase) such that acetylcholine can bind to receptors (via competition with the antibodies). Normal levels of acetylcholine are too low for such competition to be successful. This disorder does not generate antibodies which lead to acetylcholine destruction, inhibition of acetylcholinesterase, inhibition of acetylcholine synthesis, or release of acetylcholine at the synapse.

A 37-year-old female has trouble keeping her eyes open and swallowing and is beginning to slur her speech. The patient has also noticed a weakness in her arms and legs. Treatment with edrophonium chloride results in a temporary relief of symptoms. The underlying etiology of this disorder involves auto-antibodies that do which of the following? (A) Destroy acetylcholine (B) Block acetylcholine receptors (C) Inhibit acetylcholinesterase (D) Inhibit acetylcholine synthesis (E) Stimulate acetylcholine release into the synapse

*The answer is B* This patient 's presentation is consistent with a thoracic aortic dissection, Approximately 50% of patients younger than age 40 with a thoracic aortic dissection have Marfan syndrome. Marfan syndrome is one of the most common inherited connective tissue disorders. It is an autosomal-dominant disorder associated with a defect in fibrillin synthesis. Marfan syndrome patients are typically tall and have long arms, arachnodactyly, aortic aneurysms, and lens defects (upward and outward lens subluxation, which commonly manifests as bi lateral blurry vision).

A 37-year-old woman presents to the emergency department complaining of "tearing" chest pain that began all of the sudden approximately 30 minutes ago. She states that her pain started in her left chest and radiated to her back. She denies any history of similar pain. Review of systems was only notable for blurry vision, which has been present for years. On physical examination, the patient is a tall, slender woman with long arms and long fingers. She appears uncomfortable and diaphoretic. Her blood pressure is 145/90 mm Hg, heart rate is 90/min, respiratory rate is 22/min, and oxygen saturation is 97%, and she is afebrile. She denies any family history of coronary artery disease or hypercholesterolemia. What is the most likely cause of this patient's disease? A. Clot in the patient's left pulmonary artery B. Defect in f ibrillin synthesis C. Defect in LDL cholesterol receptors D. Defect in type I collagen synthesis E. Defect in type III collagen synthesis

*The answer is B.* This patient has α1-antitrypsin deficiency, a genetic disease characterized by a deficiency in the serine protease inhibitor α1-antitrypsin. This protein normally functions to inhibit neutrophil elastase in the lung. When deficient, there is overabundant activity of elastase, which destroys elastin and collagen in the alveolar walls, progressing to emphysema. Most patients with α1-antitrypsin deficiency are homozygous for the Z allele. Clinically, α1-antitrypsin deficiency can affect the lung, liver, and less commonly the skin. In the lung, the most common manifestation is early onset panacinar emphysema, which is more prominent at the lung bases than apices. Slowly worsening dyspnea is the most common symptom, although patients may initially complain of cough, sputum production, or wheezing. As in this case, patients who present early complaining of episodes of wheezing and productive cough may be told they have asthma. Although treatment for asthma may initially improve symptoms, it does not slow the progression of the disease. Her x-ray of the chest shows a pattern typical for this disease; hyperinflated lungs, a flattened diaphragm, and hyperlucent lungs due to decreased lung markings (it is difficult to see at this resolution due to the overlying breast tissue, but we expect that the lung markings would be especially absent at the bases). α1-Antitrypsin deficiency can also cause cirrhosis of the liver and panniculitis of the skin.

A 45-year-old white woman presents to her physician complaining of several months of worsening shortness of breath. Previously she was told she had asthma because she was having intermittent episodes of wheezing combined with a productive cough and difficulty catching her breath. She used to run two miles every morning but can no longer walk more than 10 city blocks without stopping. She has never smoked. On physical examination she is using her accessory muscles to assist with respiration. Pulmonary examination is notable for an increased decreased FEV1/FVC ratio, decreased air movement with each breath, and increased resonance upon percussion. X-ray of the chest is shown in the image. Which of the following is the most likely underlying cause for this patient's disease? (A) A genetic mutation resulting in deficient levels of a protease (B) A genetic mutation resulting in deficient levels of a protease inhibitor (C) A mutation in the p53 gene (D) A mutation of the CFTR gene, which encodes a regulated chloride channel (E) Airway inflammation, airflow obstruction, and bronchial hyperresponsiveness

*The answer is D.* Carbon monoxide (CO) is a colorless, odorless, nonirritant gas that is generated as a byproduct of incomplete hydrocarbon combustion. Carbon monoxide emission from automobiles can result in carbon monoxide poisoning in poorly ventilated spaces. The scenario described above is typical for carbon monoxide poisoning. Another classic source of carbon monoxide poisoning is a faulty home heater. Carbon monoxide has 220 times more affinity for hemoglobin than does oxygen. Inhaled carbon monoxide rapidly diffuses across the alveolar membrane and binds tightly with heme-bound iron in hemoglobin and other hemeproteins. Carbon monoxide binding to hemoglobin results in the formation of carboxyhemoglobin. Carbon monoxide decreases the oxygen content of the blood by occupying oxygen binding sites. Carbon monoxide also inhibits the release of oxygen from hemoglobin in tissues by altering hemoglobin conformation into the relaxed form that has a very high affinity for oxygen. This results in a leftward shift of the oxygen dissociation curve and tissue hypoxia via deficient unloading of oxygen. Treatment of carbon monoxide toxicity is with 100% or hyperbaric oxygen.

A 46-year-old male is brought to the emergency department by EMS. He has attempted suicide twice in the past, once by cutting his wrists and another time by taking an overdose of amitriptyline. His current medications include quetiapine and fluoxetine. He has no known drug allergies. This evening a neighbor found him in a closed garage with the car running. As you examine him he loses consciousness end begins to seize. The toxic substance causing this patient's condition effects hemoglobin by: A. Oxidation of the iron moiety B. Oxidation of the porphyry ring C. Covalent linking to heme D. Competitive binding to heme E. Altering the partial pressure of oxygen

*The answer is B.* This patient has methemoglobinemia, which is characterized by a high blood level of methemoglobin in which the oxygen-carrying iron is present in the oxidized (ferric Fe3+) state instead of the reduced (ferrous Fe2+) state. Methemoglobin cannot bind to oxygen. Moreover, any remaining ferrous hemes in the hemoglobin tetramer exhibit increased oxygen affinity (ie, the oxygen dissociation curve is shifted to the left). Thus signs and symptoms of methemoglobinemia reflect decreased blood oxygen content and an even greater decrease in oxygen delivery to tissues (ie, cellular hypoxia). Symptoms include headache, dizziness, nausea, shortness of breath, confusion, seizures, and coma. Because oxygen diffusion at the alveolar-arterial level is not impaired, the arterial partial pressure of oxygen will be normal. Blood may have a characteristic muddy color secondary to the oxidization state of iron. Methemoglobinemia may occur as an adverse effect of oxidizing agents such as sulfonamides, dapsone, and local anesthetics (eg, benzocaine), from hereditary hemoglobin abnormalities, or secondary to a hereditary deficiency of the reduced form of nicotinamide adenine dinucleotide. Methylene blue has been shown to increase the conversion of Fe3+ back to Fe2+. An alternative treatment for methemoglobinemia is ascorbic acid (vitamin C).

A 47-year-old man presents with a 4-day history of weakness, dizziness, and shortness of breath after starting treatment with dapsone for dermatitis herpetiformis. His lips and fingernails have a bluish tint. Blood gas analysis is ordered to look for perturbations in the patient's blood oxygen and carbon dioxide levels. It is determined that the patient 's arterial partial oxygen pressure is 100 mm Hg and arterial partial carbon dioxide pressure is 40 mm Hg. The nurse who drew blood for the gas analysis notes that the patient's blood appeared brown in the collecting tube. Which of the following is the most appropriate treatment for this condition? A. Hyperbaric oxygen B. Methylene blue C. N-Acetylcysteine D. Protamine E. Thiosulfate

*The answer is B.* This woman suffers from β-thalassemia major, the most severe form of β-thalassemia, in which the β-chain is absent. β-Thalassemia major manifests as ineffective erythropoiesis, causing severe hemolysis and anemia, Hemolysis occurs as a result of excessive α chains aggregating to form inclusion bodies, which cause oxidative damage and destruction of RBCs, both in the bone marrow and in the circulation. This hemolysis leads to elevations in unconjugated bilirubin. Patients with β-thalassemia major are transfusion dependent and frequently develop sequelae of iron overload. The consequences of iron overload due to transfusion dependency or secondary hemochromatosis are described in the vignette. These manifestations are due to iron deposition in various tissues, including the pancreas, heart, and skin. β-Thalassemia is more common among Mediterranean populations, whereas α-thalassemia is more common among Asian and African populations.

A 48-year-old woman of Mediterranean descent presents because of fatigue, arthralgias, discomfort in her right upper abdominal quadrant, and polyuria. Laboratory tests are remarkable for elevated glucose level, elevated bilirubin, low hemoglobin, elevated reticulocytes, and increased transferrin saturation. Cardiac testing shows moderate restrictive cardiomyopathy. She frequently has required blood transfusions throughout her life. Which hereditary disorder does this patient most likely have? A. Absence of the hemoglobin α-chain B. Absence of the hemoglobin β-chain C. Mutation resulting in increased absorption of dietary iron D. Mutations in the gene encoding ankyrin E. Mutations resulting in copper accumulation

*The answer is B.* The process of type 1 collagen synthesis begins with collagen α-chain translation in the cytosol. Shortly after translation begins, a hydrophobic signal sequence at the N-terminus of the protein directs the ribosome to the rough endoplasmic reticulum (RER), where it extrudes the growing polypeptide chain into the RER cisternae (Choices A and D). Within the RER, numerous post-translational modifications are made to the peptide, including cleavage of the signal sequence and hydroxylation and glycosylation of certain residues. Intrachain disulfide bonds then form at the N- and C-terminal globular regions, helping to stabilize them. Special interchain disulfide bonds also form at the C-terminus between 3 pro-α chains, aligning the helical domains into a conformation favorable for triple helix formation (Choice E). The resulting procollagen molecule then undergoes exocytosis, after which the N- and C-tenminal nonhelical regions are cleaved from the triple helix by procollagen peptidases, forming tropocollagen. Tropocollagen subunits then self-assemble into collagen fibrils that are subsequently crosslinked by lysyl oxidase. Ehlers-Danlos syndrome (EDS) is a group of rare hereditary disorders involving connective tissues found in skin, tendons, ligaments, muscles, and vasculature. EDS usually results in hypermobile joints, fragile, hyperelastic skin; and easy bruising due to decreased tissue strength and support. It is caused by mutations affecting the collagen genes or the enzymes involved in collagen synthesis, such as lysyl-hydroxylase or procollagen peptidase. In the case of EDS due to procollagen peptidase deficiency, impaired cleavage of the procollagen N- and C-termini causes the formation of more soluble collagen that does not properly crosslink with other collagen molecules. This results in joint laxity. loose skin, and easy bruisability. *Educational Objective:* Extracellular peptidases cleave disulfide-rich terminal extensions from the procollagen molecule. This results in the formation of water-insoluble triple helical collagen subunits (tropocollagen) that self-assemble and undergo crosslinking by lysyl oxidase to form mature collagen fibrils. Impaired cleavage of procollagen causes the formation of more soluble collagen that does not properly crosslink with other collagen molecules.

A 5-year-old boy is brought to the physician by his parents because of easy bruising. Physical examination reveals soft and loose skin as well as multiple ecchymoses over the forearm and pretibial regions. Histologic evaluation with electron microscopy shows type 1 collagen fibrils that are abnormally thin and irregular. Biochemical evaluation reveals the presence of disulfide-rich globular domains within a purified sample of mature collagen fibrils. Which of the following stages of collagen synthesis is most likely impaired in this patient? A. Amino acid incorporation into proteins B. Extracellular protein cleavage C. Lysine residue hydroxylation D. Signal peptide recognition E. Triple helix formation

*The answer is E.* In the presence of insulin deficiency, a shift to fatty acid oxidation produces the ketones such as acetoacetate that cause metabolic acidosis. The pH and bicarbonate are low, and there is frequently some respiratory compensation (hyperventilation with deep breaths) to lower the PCO2, as in choice e. A low pH with high PCO2 would represent respiratory acidosis (choices a and b—the low-normal bicarbonate values in these choices indicate partial compensation). Choice d represents respiratory alkalosis as would occur with anxious hyperventilation (high pH and low PCO2, partial compensation with high bicarbonate). Choice c illustrates normal values.

A 5-year-old girl displays decreased appetite, increased urinary frequency, and thirst. Her physician suspects new onset diabetes mellitus and confirms that she has elevated urine glucose ketones. Which of the following blood values would be most compatible with diabetic ketoacidosis?

*The answer is E.* All statin drugs are competitive inhibitors of HMG-CoA reductase, which means it will increase the Km while Vmax remains the same.

A 54-year old man suffered a myocardial infarction and was given atorvastatin as one of the treatments. The drug is used to inhibit the key regulatory enzyme that was involved in the generation of the fatty streaks that contributed to his myocardial infarction. Suppose that the Km for the substrate of this enzyme is 5 X 10-3 M and the Vmax is 9 X 102 mmol/h. In the presence of atorvastatin, which of the following values would be consistent with the drug's biochemical action?

*The answer is B.* The Michaelis-Menten constant (Km) of an enzyme such as DHFR reflects the enzyme's affinity for a particular substrate [S], such as methotrexate, in an inverse fashion. Therefore, if the Km is high, the enzyme will have low affinity for the substrate. The Km of an enzyme is the concentration of substrate required to achieve a reaction velocity equal to half of the maximum reaction rate (Vmax). A competitive inhibitor binds reversibly to the same site that the substrate would normally occupy and thus competes with the substrate for that site. Therefore, in the presence of a competitive inhibitor, the concentration of DHFR to achieve half of Vmax will be increased. It should also be noted t hat competitive inhibitors can be overcome by add ing more substrate.

A 54-year-old woman has had longstanding rheumatoid arthritis. Her rheumatologist recently stated her on methotrexate, a competitive inhibitor of dihydrofolate reductase (DHFR). What effect does methotrexate have on DHFR? A. Methotrexate acts on DHFR by decreasing its Michaelis-Menten constant B. Methotrexate acts on DHFR by increasing its Michaelis-Menten constant C. Methotrexate does not affect DHFR's Michaelis-Menten constant D. The maximum reaction rate is decreased E. The maximum reaction rate is increased

*The answer is D.* The answer is D: A misfolded form of a normal protein. The pathologist is showing early clinical signs of Creutzfeldt-Jakob disease, caused by a misfolded prion protein, leading to protein aggregates in the brain. The initial seed for the aggregation was obtained from a cadaver that the pathologist was working on. Prions can exist in two states, the normal, nonaggregated form and an alternative conformation that is prone to aggregation (see differences in structure below, where PrPc is the normal conformation, and PrPsc is the abnormal structure). Once the alternative form reaches a critical concentration, aggregation ensues and shifts the equilibrium between the normal and abnormal forms to produce more abnormal form, feeding the aggregation. The prion is not a truncated neuronal protein (its primary structure can be the same in both forms of the protein), nor is it a truncated extracellular protein. This disorder is not due to alterations in hemoglobin or fibrillin. This patient will probably die within 1 year. There is no current treatment for the disease. As the disease progresses, he will probably develop blindness, involuntary movements, and severe deterioration of mental function.

A 56-year-old pathologist was taken to his family doctor by his son for he was showing mood changes, minor loss of memory, and decreased motor skills. During the patient history, it became clear that over the course of his career he had, on occasion, cut himself using the instruments he had been using on the cadavers he had been working on. A potential explanation for his symptoms is abnormal aggregation of which of the following proteins? (A) Hemoglobin in the red blood cells (B) Fibrillin in the extracellular compartments of the brain (C) A truncated neuronal protein (D) A misfolded form of a normal protein (E) A truncated extracellular protein

*The answer is B.* Hemoglobin electrophoresis is used to analyze the different forms of hemoglobin in patients with suspected hemoglobinopathy. Normal hemoglobin consists primarily of hemoglobin A (HbA), which migrates rapidly toward the positive electrode (anode) because of its negative charge. Hemoglobin S (HbS) is an abnormal type of hemoglobin in which a nonpolar amino acid (valine) replaces a negatively charged amino acid (glutamate) in the beta globin chain. This amino acid replacement decreases the negative charge on the HbS molecule, which causes HbS to move more slowly toward the anode. Similarly, hemoglobin C (HbC) has a glutamate residue replaced by lysine in the beta globin chain. Because lysine is a positively charged amino acid, HbC has even less total negative charge than HbS and moves even more slowly toward the anode. Both HbC and HbS result from missense mutations, a type of mutation in which a single base substitution results in a codon that codes for a different amino acid. Patients with sickle cell disease have HbS mutations in both beta chains; those with HbC disease have HbC mutations involving both beta chains. Patients with hemoglobin SC disease have 1 HbS allele and 1 HbC allele and will have 2 hemoglobin bands on electrophoresis. This patient's electrophoresis results show a single band that migrates loss than the HbA and HbS bands, meaning that he has HbC disease. Patients with HbC disease are typically asymptomatic and often have mild hemolytic anemia and splenomegaly.

A 6-year-old African American boy is brought to the physician because of easy fatigability. Physical examination reveals splenomegaly, and his complete blood count shows mild anemia. Hemoglobin electrophoresis is performed at alkaline pH on a cellulose acetate strip individuals with normal hemoglobin end known sickle. Findlngs for the patient are shown below compared to individuals with normal hemoglobin end known sickle cell disease. A. Frameshift mutation B. Missense mutation C. Nonsense mutation D. Silent mutation E. Trinucleotide expansion

*The answer is B.* This patient is exhibiting signs and symptoms of sickle cell anemia. Sickle cell anemia is a hemoglobinopathy that typically affects patients of African ancestry. A point mutation in the 6th codon of the beta-globin gene, which causes the substitution of valine (hydrophobic) for glutamic acid (hydrophilic), is responsible. The incorporation of this abnormal beta-globin protein into hemoglobin results in the formation of hemoglobin S (HbS). HbS polymerizes at low oxygen tension, causing sickling and hemolysis of erythrocytes and resultant vascular occlusion. This patient's poor exercise tolerance and exertional dyspnea are due to anemia. His history of acute chest syndrome, abdominal pain, and bone pain are due to vasoocclusive events in the lungs. spleen and bone, respectively. (Choice A) A phenylalanine deletion (AF508) is the most common cause of cystic fibrosis, the most common fatal genetic decease of Caucasians. (Choice E) Early termination of polypeptide synthesis (nonsense mutation) will produce a truncated protein. *Educational Objective:* Exertional dyspnea, pneumonia resulting in life-threatening acute cheat syndrome, and recurrent abdominal and bone pain are clinical features of sickle cell anemia. Sickle cell anemia results from a point mutation that causes valine to substitute for glutamic acid in the sixth position of the β-globin chain of hemoglobin.

A 6-year-old African American male is brought to your office for a routine check-up. His mother remarks that he often seems uninterested in playing with his peers and appears to "run out of breath quickly." His medical records reveal that he has missed several pediatric vaccinations and has been hospitalized twice, once with a "chest infection" and once with abdominal pain. The patient mentions to you that occasionally his "bones hurt." Which of the following protein changes most likely accounts for this patient's condition? A. Phenylalanine deletion B. Valine substitution for glutamic acid C. Phenylalanine substitution for praline D. Valine substitution for lysine: E. Early termination of polypeptide synthesis

*The answer is D.* Isozymes are multiple forms of a given enzyme that occur within a given species. Since isozymes are composed of different proteins, analysis by electrophoretic separation can be done. Lactate dehydrogenase is a tetramer composed of any combination of two different polypeptides, H and M. Thus the possible combinations are H4, H3M1, H2M2, H1M3, and M4. Although each combination is found in most tissues, M4 predominates in the liver and skeletal muscle while H4 is the predominant form in the heart. White and red blood cells as well as brain cells contain primarily intermediate forms. The M4 forms of the isozyme seem to have a higher affinity for pyruvate compared with the H4 form. Following a myocardial infarction, the H4 (LDH1) type of lactate dehydrogenase rises and reaches a peak approximately 36 h later. Elevated LDH1 levels may signal myocardial disease even when the total lactate dehydrogenase level is normal.

A 65-year-old obese male presents with severe indigestion and chest pain after a spicy meal. A lactate dehydrogenase (LDH) level obtained to evaluate possible myocardial infarction is normal, but the laboratory recommends that LDH isozymes be performed. The managing physician knows that lactate dehydrogenase is composed of two different polypeptide chains arranged in the form of a tetramer. Assuming that all possible combinations of the different polypeptide chains occur, how many isozyme forms of lactate dehydrogenase must be measured? a. Two b. Three c. Four d. Five e. Six

*The answer is C.* Oxidation of the heme component of hemoglobin to the ferric (Fe3+) state forms methemoglobin. This may be caused by the action of certain drugs, such as nitrates. The methemoglobinemias are characterized by chocolate cyanosis (a brownish-blue coloration of the skin and mucous membranes), and chocolate-colored blood as a result of the dark-colored methemoglobin. Symptoms are related to tissue hypoxia, and include anxiety, headache, dyspnea. In rare cases, coma and death can occur.

A 67-year-old man presented to the emergency department with a 1 week history of angina and shortness of breath. He complained that his face and extremities had a "blue color." His medical history included chronic stable angina treated with isosorbide dinitrate and nitroglycerin. Blood obtained for analysis was chocolate-colored. Which one of the following is the most likely diagnosis? A. Sickle cell anemia B. Carboxyhemoglobinemia C. Methemoglobinemia. D. β-Thalassemia E. Hemoglobin SC disease

*The answer is E.* In sickle cell (hemoglobin s [HbS]) anemia, the nonpolar amino acid valine replaces the charged amino acid glutamate at position 6 of the 5 globin chain. This results in the alteration of a hydrophobic portion of the β globin chain that fits into a complementary site on the α globin chain of another hemoglobin molecule. As a result, hemoglobin molecules aggregate under anoxic conditions. After polymerization, HbS initially forms a gel and then a meshwork of fibrous polymers causing the red blood cells to distort into an abnormal sickle shape. Sickling is promoted by conditions associated with low oxygen levels, increased acidity, or low blood volume (dehydration). Sickled cells are not flexible enough to pass through microvasculature. As a result, they impede blood flow and cause microinfarcts in tissues and painful vasoocclusive crises. Organs in which blood moves slowly (eg, spleen, liver) are predisposed to lower oxygen levels or acidity. Organs with particularly high metabolic demands (eg, brain, muscles, placenta) promote sickling by extracting more oxygen from the blood (oxygen unloading The sickling process is complex and incompletely understood. *Educational Objective:* Hemoglobin S (HbS) aggregates in the deoxygenated state. HbS polymers form fibrous strands that reduce red blood cell membrane flexibility and promote tickling. Sickling occurs under conditions associated with anoxia including low pH and high levels of 2,3-bisphosphoglycerate. These inflexible erythrocytes predispose to microvascular occlusion and microinfarcts.

A 7-year-old African American boy is brought to the office by his parents. The parents state that the boy has been hospitalized several times for severe pains in his back and extremities. The patient is very active when he is not in pair but gets quite tired by the end of the day. He has no other medical problems and takes no medications except acetaminophen for pain control. On examination, the conjunctivae are pale. Blood count reveals a hemoglobin level of 7.8 mg/dL and a reticulocyte count of 16%. A valine for glutamic acid substitution at position 6 of the β globin chain of the hemoglobin molecule is suspected. This patient's hemoglobin would most likely aggregate upon which of the following? A. 2,3-bisphosphoglycerate depletion B. β globin chain folding C. Capillary pH values >7.4 D. Interaction with fetal hemoglobin E. Oxygen unloading

*The answer is D.*The CK isoenzyme pattern at admission showed elevated MB isozyme, indicating that the patient had experienced a myocardial infarction in the previous 12-24 hours. [Note: 48-64 hours after an infarction, the MB isozyme would have returned to normal values.] On day 2, 12 hours after the cardioconversions, the MB isozyme had decreased, indicating no further damage to the heart. However, the patient showed an increased MM isozyme after cardioconversion. This suggests damage to muscle, probably a result of the convulsive muscle contractions caused by repeated cardioconversion. Angina is typically the result of transient spasms in the vasculature of the heart, and would not be expected to lead to tissue death that results in elevation in serum creatine kinase.

A 70-year-old man was admitted to the emergency room with a 12 hour history of chest pain. Serum creatine kinase (CK) activity was measured at admission (day 1) and once daily (look at figure). On day 2 after admission, he experienced cardiac arrhythmia, which was terminated by three cycles of electric cardio-conversion, the latter two at maximum energy. [Note: Cardioconversion is performed by placing two paddles, 12 cm in diameter, in firm contact with the chest wall and applying a short electric voltage.] Normal cardiac rhythm was reestablished. He had no recurrence of arrhythmia over the next several days. His chest pain subsided and he was released on day 10. Which one of the following is most consistent with the data presented? A. The patient had a myocardial infarction 48 to 64 hours prior to admission. B. The patient had a myocardial infarction on day 2. C. The patient had angina prior to admission. D. The patient had damage to his skeletal muscle on day 2. E. The data do not permit any conclusion concerning myocardial infarction prior to, or after, admission to the hospital.

*The answer is B.* The man is acidotic as defined by the pH lower than the normal 7.4. His hyperventilation with Kussmaul respirations can be interpreted as compensation by the lungs to blow off CO2, lower PCO2, increase [HCO3−]/[CO2] ratio, and raise pH. The correct answer, therefore, includes a low PCO2, eliminating choices c through e. Using the Henderson Hasselbalch equation indicates that the pH minus the pK for carbonic acid (7.3 − 6.1 = 1.2) equals log [15]/[0.03 × 30 mmHg] or log [15/0.9]. These values correspond to those in choice b. The man has compensated his metabolic acidosis (caused by the accumulation of ketone bodies such as acetoacetic acid) by increasing his respiratory rate and volume.

A 72-year-old male with diabetes mellitus is evaluated in the emergency room because of lethargy, disorientation, and long, deep breaths (Kussmaul respirations). Initial chemistries on venous blood demonstrate high glucose at 380 mg/dL (normal up to 120) and a pH of 7.3. Recalling the normal bicarbonate (22 to 28 mM) and PCO2 (33 to 45 mmHg) values, which of the additional test results below would be consistent with the man's pH and breathing pattern? a. A bicarbonate of 5 mM and PCO2 of 10 mmHg b. A bicarbonate of 15 mM and PCO2 of 30 mmHg c. A bicarbonate of 15 mM and PCO2 of 40 mmHg d. A bicarbonate of 20 mM and PCO2 of 45 mmHg e. A bicarbonate of 25 mM and PCO2 of 50 mmHg

*The answer is B.* This patient is suffering from chronic obstructive pulmonary disease (COPD). His history reveals a significant tobacco history, chronic cough, and worsening dyspnea, and physical examination shows signs of hyperinflation (barrel chest), decreased breath sounds, and characteristic pursing of the lips when breathing. Arterial blood gas analysis reveals mild hypoxemia and hypercapnia, which can develop in severe cases of COPD. The excess carbon dioxide in this patient's lungs and blood is caused by the ineffective expirations and air trapping from his COPD. Carbon dioxide is a substrate for carbonic anhydrase, the enzyme in RBCs that catalyzes its conversion to bicarbonate.

A 75-year-old man goes to his physician because he has experienced increasing short ness of breath and a chronic cough over the past 10 years. He has a more than 100-pack-year history of smoking cigarettes. The patient appears to have a very large barrel-shaped chest and often purses his lips when he breathes. On examination, the physician hears diminished breath sounds. Arterial blood gas analysis shows a partial arterial oxygen pressure of 75 mm Hg and a partial arterial carbon dioxide pressure of 50 mm Hg. As a result of his disease, the substrate of which enzyme has built up in this patient's lungs? A. ATP synthase B. Carbonic anhydrase C. Cyclin-dependent kinases D. Nicotinamide adenosine dinucleotide phosphate oxidase E. Topoisomerase

*The answer is D.* Leucine and isoleucine have nonpolar methyl groups as side chains. As for any amino acid, titration curves obtained by noting the change in pH over the range of 1 to 14 would show a pK of about 2 for the primary carboxyl group and about 9.5 for the primary amino group; there would be no additional pK for an ionizable side chain. Recall that the pK is the point of maximal buffering capacity when the amounts of charged and uncharged species are equal (see answer to question 104). Aspartic and glutamic acids (second carboxyl group), histidine (imino group), and glutamine (second amino group) all have ionizable side chains that would give an additional pK on the titration curve. The likely diagnosis here is maple syrup urine disease, which involves elevated isoleucine, leucine, and valine together with their ketoacid derivatives. The ketoacid derivatives cause the acidosis, and the fever suggests that the metabolic imbalance was worsened by an infection.

A child presents with severe vomiting, dehydration, and fever. Initial blood studies show acidosis with a low bicarbonate and an anion gap (the sum of sodium plus potassium minus chloride plus bicarbonate is 40 and larger than the normal 20 to 25). Preliminary results from the blood amino acid screen show two elevated amino acids, both with nonpolar side chains. A titration curve performed on one of the elevated species shows two ionizable groups with approximate pKs of 2 and 9.5. The most likely pair of elevated amino acids consists of a. Aspartic acid and glutamine b. Glutamic acid and threonine c. Histidine and valine d. Leucine and isoleucine e. Glutamine and isoleucine

*The answer is C.* Glycosaminoglycans (mucopolysaccharides) are polysaccharide chains that may be bound to proteins as proteoglycans. Each proteoglycan is a complex molecule with a core protein that is covalently bound to glycosaminoglycans—repeating units of disaccharides. The amino sugars forming the disaccharides contain negatively charged sulfate or carboxylate groups. The primary glycosaminoglycans found in mammals are hyaluronic acid, heparin, heparan sulfate, chondroitin sulfate, and keratan sulfate. Inborn errors of glycosaminoglycan degradation cause neurodegeneration and physical stigmata described by the outmoded term "gargoylism." Glycogen is a polysaccharide of glucose used for energy storage, and has no sulfate groups. Collagen and fibrillin are important proteins in connective tissue. γ-aminobutyric acid is a γ-amino acid involved in neurotransmission.

A child stops making developmental progress at age 2 years and develops coarse facial features with thick mucous drainage. Skeletal deformities appear over the next year, and the child regresses to a vegetative state by age 10 years. The child's urine tests positive for glycosaminoglycans that include which of the following molecules? a. Collagen b. γ-aminobutyric acid c. Heparan sulfate d. Glycogen e. Fibrillin

*The answer is A.* The primary structure of collagen peptides consists of repeating tripeptides with a gly-X-Y motif, where gly is glycine and X and Y are any amino acid. The small CH2 group connecting the amino and carboxyl groups of glycine contrasts with the larger connecting groups and side chains of other amino acids. The small volume of glycine molecules is crucial for the α helix secondary structure of collagen peptides. This in turn is necessary for their tertiary helical structure and their assembly into quaternary tripeptide, triple helix structures. The most severe clinical phenotypes caused by amino acid substitutions in collagen peptides are those affecting glycine that prevent α helix formation. The child has a disorder called Stickler syndrome that exhibits autosomal dominant inheritance.

A child with tall stature, loose joints, and detached retinas is found to have a mutation in type II collagen. Recall that collagen consists of a repeating tripeptide motif where the first amino acid of each tripeptide is the same. Which of the following amino acids is the recurring amino acid most likely to be altered in mutations that distort collagen molecules? a. Glycine b. Hydroxyproline c. Hydroxylysine d. Tyrosine e. Tryptophan

*The answer is C.* The normal pKa for a histidine side chain is 6.0, meaning that at pH 6.0, 50% of the histidine side chains are protonated and 50% deprotonated. For the pKa to be raised to 8.2, there must be an environment which stabilizes the protonated form of the side chain (because now one has to reach a pH of 8.2 before 50% of the histidine side chains have lost their proton). A polar environment would stabilize histidine holding on to its proton, as compared to a hydrophobic environment, which would promote side chain deprotonation at a low pH. The core of globular proteins is usually composed of hydrophobic amino acids (such as phenylalanine, valine, and leucine), and in that environment, one would expect the pKa of the histidine side chain to be reduced. Surface-associated domains usually interact with water and are not where active sites are often found (it is too diffi cult to control the environment of the active site if water can freely enter the site). At a surface-associated domain, one would not expect much change in the histidine side chain pKa. Many enzymes catalyze reactions based on the ability of amino acid side chains to accept or donate protons, which will be a function of the pKa of the dissociable proton on the amino acid side chain.

A critical histidine side chain in an enzyme's active site displays a pKa value of 8.2. Which of the following best describes the local environment in which this histidine residue resides? (A) A surface-associated domain (B) A very hydrophobic environment (C) A very polar environment (D) Buried deep within the core of this globular protein (E) Surrounded by phenylalanine, valine, and leucine residues

*The answer is C.* Collagen is the most abundant protein in the human body and is synthesized by fibroblasts. osteoblasts, and chondroblasts. It consists of 3 polypeptide alpha chains hold together by hydrogen bonds, forming a rope~like triple helix (collagen molecule). Collagen molecules self assemble into Fibrils, which subsequently crosslink to form collagen fibers. The triple helical conformation of collagen molecules occurs due to the simple and repetitive amino acid sequence within each alpha chain, in which glycine (Gly) occupies every third amino acid position (Gly-X-Y). Glycine is the most abundant amino acid in collagen and, due to its small size, is the only amino acid that can fit into the confined space between individual alpha chains *Educational Objective:* Glycine is the most abundant amino acid in collagen. The triple helical conformation of collagen molecules occurs due to the repetitive amino acid sequence within each alpha chain, in which glycine (Gly) occupies every third amino acid position (Gly-X-Y).

A dermatology researcher is studying the role of different amino acids in wound healing. She cultures mature dermal fibroblasts in growth media. After several days, the fibroblasts begin synthesizing polypeptide chains then assemble into triple helical structures, followed by fibrils. The fibrillate proteins are hydrolyzed and separated into their constituent amino acids via paper chromatography. Which of the following amino acids is most likely to be found in highest quantity in these proteins? A. Alanine B. Cysteine C. Glycine D. Lecuine E. Lysine F. Proline

*The answer is D.* The sum of the major cations in blood (Na+ plus K+) minus the sum of the major anions (HCO3− plus Cl−) is called the anion gap (filled by phosphate ions, negatively charged proteins, etc.). An anion gap over 20 suggests the presence of an abnormal anion, such as acetoacetate, which occurs in diabetics. The anion gap in this teenager is elevated at 30 (140 + 4) − (103 + 11). Assuming that all of the gap is made up by acetoacetate anion, then 7 L × 30 mmol/L = 210 mmol of acetoacetate and 210 mmol of sodium excreted per hour. Even with acidosis and a pH of 7.1, virtually all of the acetoacetic acid (pK 4.8) is present as acetoacetate and less than 1% is excreted as acetoacetic acid. To calculate the exact amount of acetoacetic acid present, the Henderson-Hasselbalch equation rearranges to (7.1 − 4.8) = 2.3 =log[base]/[acid]; [CH3C=OCH2COO−]/[CH3C=OCH2COOH] = antilog 2.3 = 102. Less than 0.3 mmol of acetoacetic acid is excreted for each liter of blood filtered through the kidney.

A diabetic teenager is found to have a pH of 7.1 and normal electrolyte levels (Na+ = 140 mM, K+ = 4 mM, Cl− = 103 mM) except for a bicarbonate of 11 mM (normal 22 to 28 mM). The urine tests positive for ketone bodies, mostly due to acetoacetic acid and acetoacetate (CH3C=OCH2COOH and CH3C=OCH2COO−), which have a pK of 4.8. In this case, it is assumed that acetoacetate is the only significant anion in the blood besides chloride, and that each acetoacetate anion binds and removes one sodium cation during excretion by the kidney. Given that the patient has a normal glomerular filtration rate of about 7 L of blood per hour without any retention of acetoacetate/acetoacetic acid, the rates of sodium, acetoacetate, and acetoacetic acid loss will be a. 10 mmol/h of each species b. 50 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretion c. 100 mmol/h of sodium and acetoacetic acid, virtually no acetoacetate excretion d. 200 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretion e. 300 mmol/h of each species

*The answer is E.* With a pKa of 9.0, the weak base needs to lose a proton to enter cells in its uncharged form (this base is most likely −NH3 + below pH 9.0, and −NH2 above pH 9.0). Thus, the higher the pH, the greater the proportion of drug which is in its unionized form. At pH values less than 9.0, greater than 50% of the drug will be ionized, which will slow its entry into cells. At pH 9.2, less than 50% of the drug is ionized, and as the unionized form enters the cell, it will reduce the concentration of unionized drug in the circulation, thereby forcing a re-equilibration and generating more unionized drug. At the next highest pH value listed, 7.6, less than 8% of the drug is unionized, and the rate of transport would be much less than at pH 9.2.

A drug contains one ionizable group, a weak base with a pKa of 9.0. The drug enters cells via free diffusion through the membrane in its uncharged form. This will occur most readily at which of the following pH values? (A) 3.5 (B) 5.5 (C) 7.0 (D) 7.6 (E) 9.2

*The answer is D.* A left shift in the hemoglobin curve indicates an increase in oxygen affinity. This increased affinity also means oxygen dissociation from hemoglobin in tissues is decreased. A left shift can be seen with increased amounts of fetal hemoglobin, decreased amounts of 2,3-diphosphoglycerate (2,3-DPG), decreased temperature, increased pH, and decreased carbon dioxide. A good way to remember these last three factors is to think that changes seen with exercise (increase in temperature, decrease in pH, increase in carbon dioxide) all decrease oxygen affinity, so that the opposite increases oxygen affinity. Exposure to carbon monoxide leads to an allosteric change in the hemoglobin molecule, reducing the offloading of oxygen in the periphery. This leads to a left shift of the oxygen dissociation curve.

A given hemoglobin curve is shifted to the left. What could explain this shift? A. Decreased pH B. Increased 2,3-diphosphoglycerate levels C. Increased adult hemoglobin D . Increased carboxyhemoglobin levels E. Increased temperature

*The answer is C.* Sickle cell disease results from a missense mutation in the second nucleotide of the sixth codon, inherited in an autosomal recessive fashion, which results in a glutamic acid to valine substitution in the β-globin subunit of hemoglobin. The resulting hemoglobin S (HbS) molecule forms polymers that distort the cells and leave them poorly deformable. The thalassemias, including Hb Bart result from deletions or mutations that cause reduced or absent synthesis of hemoglobin subunits.

A glutamic acid to valine substitution in the beta-globin subunit of hemoglobin causes which one of the following? (A) α-thalassemia (B) β-thalassemia (C) sickle cell disease (D) Hb Bart hydrops fetalis syndrome

*The answer is B.* The oxygen-hemoglobin dissociation curve describes the relationship between the partial pressure of oxygen (x-axis) and the hemoglobin oxygen saturation (y-axis). Oxygen saturation increases in a sigmoidal fashion as the pO2 increases because of the increase in oxygen-binding affinity that occurs after the first oxygen molecule binds to hemoglobin. As more oxygen molecules bind to hemoglobin, the number of available binding sites decreases and the curve eventually flattens out. The partial pressure of oxygen in the blood at which hemoglobin is 50% saturated is known as the P50 (dotted black line in diagram above), this value is a standard measure of hemoglobin's affinity for oxygen and is about 26 mm Hg in normal individuals. A leftward shift of the oxygen hemoglobin dissociation curve occurs when hemoglobin has increased affinity for oxygen (je, a lower Pg). Because decreased temperatures help to stabilize the bonds between oxygen and hemoglobin, hypothermia increases hemoglobin's oxygen affinity and shifts the dissociation curve to the left.

A group of investigators is researching the changes in oxygen~hemoglobin binding that occur under various clinical conditions. They are especially interested in situations that after the shape and position of the oxygen-hemoglobin dissociation curve. Which of the following processes would most likely cause a shift from the blue curve to the red curve in the graph below? A. Chronic high-altitude adaptation B. Hypothermia C. Hypoventllation D. Severe anemia E. Strenuous exercise

*The answer is B.* Analysis of the data indicates that in the presence of the inhibitor, the Km of the enzyme is the same as in the absence of the inhibitor, but the Vmax is significantly reduced (the extrapolated lines intersect on the x-axis). These characteristics are the hallmark of noncompetitive inhibition; the inhibitor binds to a site distinct from the substrate binding site and alters the protein's conformation such that activity is reduced, but not substrate binding. A competitive inhibitor would demonstrate an increased Km, but an unaltered Vmax (line intersection on the y-axis). Activation of the enzyme would either decrease the Km or increase the Vmax, or both. Uncompetitive inhibitors are very rare in pharmacology. Such an inhibitor alters both the Km and Vmax such that parallel lines are seen on double-reciprocal plots. The basic concept behind this question is critical for an understanding of how drugs alter enzyme activities (the basis for pharmacology).

A kinetic analysis of the effect of a drug on an enzyme's activity was performed, and the results shown below were obtained. The drug would be best classified as which one of the following? (A) A competitive inhibitor (B) A noncompetitive inhibitor (C) An uncompetitive inhibitor (D) A competitive activator of the enzyme (E) A noncompetitive activator of the enzyme

*The answer is C.* Once acetylcholinesterase has been covalently modified by an inhibitor, it cannot be reactivated. The only way to regain this activity is by new synthesis of acetylcholinesterase, which would not have the covalent modification found in the inhibited enzyme. Since acetylcholine is released at nerve muscle junctions, once new acetylcholinesterase has been synthesized the released acetylcholine can be cleaved in order to allow relaxation of the muscle.

A laboratory worker was working with a potent organophosphorus inhibitor of acetylcholinesterase in the lab when a drop of the inhibitor flew into his eye. This resulted in a pin-point pupil in that eye that was nonreactive and unresponsive to atropine. He eventually (over a period of weeks) recovered from this incident. The reason for the long recovery period is which of the following? (A) Retraining of the ciliary muscles (B) Regrowth of neurons which were damaged by the inhibitor (C) Resynthesis of the inhibited enzyme (D) Induction of enzymes which take the place of the inhibited enzyme (E) Induction of proteases to reactivate the inhibited enzyme

*The answer is B.* The tendency for hydrophobic side chains to cluster is driven by the entropy of water. Water will form a cage around hydrophobic molecules, which requires a decrease in water entropy. The decrease in entropy will be minimized, however, if water only has to form one large cage around a cluster of hydrophobic molecules, rather than a large number of small cages around separate hydrophobic molecules. Thus, the driving force for the hydrophobic side chains to cluster is to minimize their interactions with water and to allow water to maximize its entropy. It is not related to disulfide bond formation (cysteine is not a hydrophobic residue) nor to hydrogen bond formation of the side chains (hydrophobic side chains do not participate in hydrogen bonding). The clustering of hydrophobic side chains may increase van der Waals interactions (just by placing these residues in close proximity), but it will not necessarily minimize them. Steric hindrance between side chains occurs as a protein folds, as the negative van der Waals interactions will prevent side chains from interfering with each other and the overall protein structure. The polymerization of sickle cell hemoglobin molecules is due to hydrophobic interactions between adjacent deoxygenated HbS molecules.

A major driving force for protein folding is the hydrophobic effect, in which hydrophobic amino acid side chains tend to cluster together, usually in the core of globular proteins. This occurs primarily due to which of the following? (A) Increasing hydrogen bond formation (B) Increasing the entropy of water (C) Increasing disulfide bond formation (D) Minimizing van der Waals interactions (E) Reducing steric hindrance between amino acid side chains

*The answer is B.* At 28 weeks, this neonate's lungs have not had the opportunity to fully develop and are deficient in dipalmitoyl phosphatidylcholine, which is the primary component of pulmonary surfactant Many lecithins make up pulmonary surfactant, with dipalmitoylphosphatidylcholine being one of the most important of them. Pulmonary surfactant, synthesized by type II pneumocytes, is essential in stabilizing air-expanded alveoli by decreasing surface tension, thus prevent ing lung collapse during expiration. Pulmonary surfactant is usually made most abundantly after the 35th week of gestation. The most common way to determine lung maturity is by the lecithin: sphingomyelin ratio of the amniotic fluid. If this ratio is >2.0, then the risk of developing neonatal respiratory distress syndrome is significantly decreased.

A neonate born at 28 weeks' gestation is having difficulty breathing. On physical examination, the neonate's heart rate is 120/min, blood pressure is 100/60 mm Hg, and respiratory rate is 55/min. He has nasal flaring and subcostal retractions. Which of the following components is deficient in this infant? A. γ-Aminobutyric acid B. Dipalmitoyl phosphatidylcholine C. Elastase D. Functional cilia E. Sphingomyelin

*The answer is A.* Given a pH of 7.1 in the cyanotic newborn, then 7.1 − 6.1 = 1 = log (10) = log [HCO3−]/[CO2] = log [HCO3−]/0.03 × PCO2. Since the [HCO3−] is 12 mM, the PCO2 × 0.03 must be 1.2 mM and the PCO2 40 mmHg. This normal calculated value for PCO2 means that the baby must have metabolic acidosis, a common accompaniment of hypoxia (low PO2) that can be treated by providing oxygen or administering alkali to ameliorate the acidosis. If the baby had respiratory acidosis, the PCO2 would be elevated; this would be treated by increasing the respiratory rate to blow off CO2. Renal treatment of acidosis would require increasing acid excretion or alkali retention. The lungs compensate acidosis with increased breathing rates or tidal volumes to blow off CO2 and increase pH, the kidneys by retaining HCO3−. The lungs can compensate alkalosis somewhat by decreasing breathing rates or volumes to retain CO2 (and decrease oxygenation within limits), the kidneys by increasing excretion of HCO3−.

A newborn with tachypnea and cyanosis (bluish color) is found to have a blood pH of 7.1. A serum bicarbonate is measured as 12 mM, but the blood gas machine that would determine the partial pressures of oxygen (PO2) and carbon dioxide (PCO2) is broken. Recall the pKa of 6.1 for carbonic acid (reflecting the HCO3−/CO2 equilibrium in blood) and the fact that the blood CO2 concentration is equal to the PCO2 in mmHg (normal value = 40 mmHg) multiplied by 0.03. Which of the following treatment strategies is indicated? a. Administer oxygen to improve tissue perfusion and decrease metabolic acidosis b. Administer oxygen to decrease respiratory acidosis c. Increase the respiratory rate to treat respiratory acidosis d. Decrease the respiratory rate to treat respiratory acidosis e. Administer medicines to decrease renal hydrogen ion excretion

*The answer is C.* In contrast to competitive inhibitors, noncompetitive inhibitors are not structural analogues of the substrate. Consequently, noncompetitive inhibitors bind to enzymes in locations remote from the active site. For this reason, the degree of inhibition is based solely upon the concentration of inhibitor and increasing the substrate concentrations do not compete with or change the inhibition. Therefore, unlike the increase in Km seen with competitive inhibition, in noncompetitive inhibition Vmax increases while Km usually remains the same. While competitive inhibitors can be overcome at sufficiently high concentration of substrate, noncompetitive inhibition is irreversible.

A noncompetitive inhibitor of an enzyme a. Increases Km with no or little change in Vmax b. Decreases Km and decreases Vmax c. Decreases Vmax d. Increases Vmax e. Increases Km and increases Vmax

*The answer is B.* After glucose is ingested in the form of a carbohydrate meal, it can only be utilized by cells once it is phosphorylated to glucose-6-phosphate. The phosphorylation of glucose traps glucose within cells, effectively removing it from the circulation and thereby lowering blood glucose levels. The body uses two similar enzymes to phosphorylate glucose: hexokinase and glucokinase. Hexokinase is an enzyme that functions in most tissues; glucokinase functions exclusively in hepatocytes and pancreatic islet cells. Compared with glucokinase, hexokinase has a greater affinity for glucose (a lower Km) but a lesser maximum enzymatic rate (a lower Vmax). Therefore, hexokinase can effectively bind small amounts of glucose and enhance phosphorylation, but the reaction rate is slow and the enzyme reaches its maximum velocity at a relatively low concentration of substrate. In contrast, glucokinase has a lower affinity for glucose (a high Km) but a greater maximum enzymatic rate (a greater Vmax)- Glucokinase, therefore, does not significantly function until there is a high concentration of glucose, at which time it can rapidly phosphorylate a large amount of substrate. Glucokinase therefore functions in the hepatocyte and in the pancreatic islet cell as a glucose "sensor/ in that it is act ivated in the presence of relative hyperglycemia. It is glucokinase's high Vmax (and high Km) that enables the body to rapidly restore normal blood glucose levels after a high carbohydrate load .

A nondiabetic man consumes a carbohydrate-rich meal. His fasting blood glucose is 80 mg/dl, and his postprandial blood glucose rises to 160 mg/dL What enzymatic characteristic enables this patient to maintain normal postprandial glucose levels despite ingesting a high carbohydrate load? A. The high Km of hexokinase B. The high Vmax of glucokinase C. The high Vmax of hexokinase D. The long half-life of glucokinase E. The long half-life of hexokinase F. The low Km of glucokinase

*The answer is C.* The normal PCO2 value coupled with a low bicarbonate value and pH of 7.32 indicates a metabolic acidosis due to shock arising from the trauma. This condition can be managed by administration of a solution of the conjugate base of a weak acid. Although it may seem that sodium bicarbonate would be the natural choice to rapidly increase blood pH and replenish bicarbonate, this treatment should be reserved for severe cases of acidosis because of its risk of kidney damage. The best treatment option is to administer sodium lactate, which helps replace fluid loss due to potential internal bleeding as well as buffer some of the acid. Sodium gluconate solution would be an alternative option. Both of these agents help buffer the acid and are better tolerated by the kidneys than bicarbonate. Sodium hydroxide is a strong base and highly toxic. Dextrose (glucose) would not affect blood pH in this case. Normal saline would be valuable for fluid replenishment but has no buffering capability.

A patient arrives in the trauma center suffering from unknown internal injuries as a result of a traffic accident. She is semiconscious with a blood pressure of 64/40 mm Hg and appears to be going into shock. Blood gases reveal a PCO2 of 39 mm Hg (normal = 40 mm Hg) and a bicarbonate of 15 mM (normal = 22-30 mM), with pH = 7.22. The best course of action to manage this patient's acidosis would be to start intravenous administration of a solution of: A. Sodium bicarbonate B. 5% dextrose C. Sodium lactate D. Sodium hydroxide E. Normal saline

*The answer is B.* The patient is suffering from Huntington disease, which is transmitted in an autosomal dominant pattern in which a triplet repeat is expanded within the Huntington disease gene. This triplet repeat codes for a polyglutamine tract in the mature protein, which leads to its eventual failure and disease symptoms. Huntington's is not caused by an exonic deletion or a nonsense mutation. Splicing is normal for the gene, and the mature mRNA is stable.

A patient has midlife onset of the following symptoms: abnormal, involuntary jerking body movements, an unsteady gait, personality changes, and chewing and swallowing difficulty, which has led to a gradual weight loss. The patient's father had similar symptoms before his death at the age of 45. Cellular analysis indicated precipitated proteins in the nucleus. This disease has, at its origins, which biochemical problem? (A) An exonic deletion (B) A polyglutamine tract in an exon of the defective gene (C) A nonsense mutation leading to the production of a truncated protein (D) A splicing mutation, leading to the insertion of intronic sequences into the mature protein (E) Production of an unstable mRNA, leading to reduced protein production

*The answer is D.* Cystic fibrosis patients have a thickening of the pancreatic duct, leading to nutrient malabsorption, as pancreatic enzymes have difficulty reaching the intestinal lumen. Lipid malabsorption syndromes frequently lead to deficiencies in fat-soluble vitamin uptake (vitamins E, D, K, and A). Vitamin K is required for the carboxylation of glutamic acid side chains on blood clotting proteins. This provides a means for these proteins to chelate calcium, and to bind to platelet surfaces. In the absence of gamma -carboxylation of glutamate, the clotting complexes cannot form, and a clotting disorder is observed. Vitamin C is required for proline hydroxylation, and as vitamin C is a water-soluble vitamin, lipid malabsorption does not affect vitamin C uptake. Transaminations require vitamin B6, another water-soluble vitamin. Dolichol can be synthesized in the body, so its absorption is not an issue under these conditions. Endogenous fatty acids will provide energy for protein synthesis in individuals with lipid malabsorption problems.

A patient, who was recently diagnosed with cystic fibrosis, displays an increased blood clotting time. This is most likely due to which of the following? (A) Lack of proline hydroxylation (B) Inability to catalyze transaminations (C) Lack of dolichol and an inability to glycosylate serum proteins (D) Inability to carboxylate glutamic acid side chains (E) Reduction in the synthesis of blood clotting factors due to lack of lipids for energy production

*The answer is A.* Arginine and lysine are basic amino acids that are positively charged at physiological PH. DNA has a negatively charged phosphate backbone and therefore has a strong affinity for arginine end lysine due to their positive charge. Of note, arginine and lysine are overexpressed in histones since histones are needed to bind negatively-charged DNA.

A researcher is looking to publish a significant paper on amino acids- He intends to engineer a specific protein sequence that will strongly bind DNA, with hopes of developing new pharmaceutical drugs for rare, neglected diseases. He is aware that amino acids have different properties that are useful in specific clinical scenarios, such as electrochemical charge at a certain Based on their electrochemical charges, which of the following amino acid sequences will most strongly bind DNA? A. Arginine--Lysine--Lysine B. Glutamate--Leucine--Leucine C. Histidine--Phenylalanine--Methionine D. Tryptophan--Tyrosine--Glycine E. Valine--Threonine--Isoleucine

*The answer is D.* This question is using a theoretical scientific experiment to ask about the distribution and biochemical characteristics of the glucose transporters (GLUTs), GLUT-2 is the isoform present in liver cells,. pancreatic~ cells, renal tubular cells, and small intestinal epithelial cells. It has several important properties, including bidirectional transport and low affinity for glucose. The low affinity prevents the transporter from becoming saturated at higher glucose levels. Since Km is inversely proportional to substrate affinity, GLUT-2 has a relatively high Km. GLUT-2 activity is not dependent upon insulin.

A scientist is studying the uptake of glucose by different human cell types. She designs an experiment to measure the amount of glucose uptake via glucose transporters (GLUTs) in these different cell types. She decides to perform the tests in the presence and absence of insulin. When studying hepatocytes, she would be studying a GLUT with which of the biochemical characteristics shown in the table above?

*The answer is A.* Vmax is defined as the rate of reaction when an enzyme is fully saturated with its substrate. The relationship between the reaction rate and concentration of substrate depends in large part on the affinity of the enzyme for its substrate. Given the data collected by the scientist, Vmax cannot be determined. Her results demonstrate that the enzyme has not been saturated, since there is still a linear, first-order relationship between reaction velocity (V) and substrate concentration ([ S]). As increasing concentrations of substrate are added to the same amount of enzyme in each experiment, reaction velocity is seen to increase. Reaction velocity never reaches a stable value for the substrate concentrations tested. The table simply does not extend to concentrations high enough to demonstrate a stabilizing enzymatic velocity, and therefore the researcher's data are inconclusive and a Vmax for the KC1989 enzyme cannot be determined.

A scientist working in a laboratory recently discovered a new enzyme, KC 198, which is located in the human intestinal brush border. KC1989 has been purified and its substrates, CAR and ALC, known. The researcher decides to perform an experiment in order to determine the Vmax for KC 1989. She adds KC 1989 to a reaction vessel and measures the rate of product formation over 1 minute. She repeats the experiment six times, increasing the concentration of CAR/ ALC added by 10 mM each time. Her data are shown in the table. She is planning to present her results to her supervisor soon and hopes to determine the value Vmax of KC1989. Are her current data conclusive or inconclusive? If conclusive, what is the Vmax for KC1989? A. Data are inconclusive, Vmax undetermined B. Data are conclusive, Vmax = 0.30 mol/s C. Data are conclusive, Vmax = 0.50 mol/s D. Data are conclusive, Vmax = 0.80 mol/s E. Data are conclusive, Vmax = 0. 70 mol/s

*The answer is B.* G-protein coupled receptors exist in an equilibrium between their active and inactive states that is dependent on whether ligand is present, and the affinity of the ligand for the receptor. When active, these receptors catalyze guanine-nucleotide exchange (GTP for guanosine diphosphate) of their associated G proteins. The Michaelis-Menten constant (Km) for any enzyme-catalyzed reaction is inversely proportional to the affinity of the enzyme for its substrate. Therefore, the Km for compound A will be lower than that for compound B because compound A has a higher affinity for the receptor than compound B. The maximum rate of reaction (Vmax) will be reached at a lower concentration of A than it would for B, although the Vmax is unchanged.

A scientist working in a research laboratory has been examining different agonists of serotonin receptor 1B (5-HT1B), a G-protein-coupled receptor. Compound A has a much higher affinity for 5-HT1B than compound B. Both compounds have a higher affinity for the receptor than serotonin. Which of the following describes the relationship between compound A and compound B when considering the guanine-nucleotide exchange activity of 5-HT1B? (A) Km for the exchange reaction with compound A is higher than that with compound B (B) Km for the exchange reaction with compound A is lower than that with compound B (C) Km values with compounds A and B are the same (D) The maximum reaction rate with compound A is greater than that with compound B (E) The maximum reaction rate with compound B is greater than that with compound A

*The answer is B.* Hb Bart is the most severe form of α-thalassemia and results from deletion or dysfunction of all four α-globin alleles.

A severe form of α thalassemia that results in death in neonates is which one of the following? (A) hemoglobin H disease (B) Hb Bart (C) thalassemia major (D) thalassemia minor

*The answer is B.* The dissociation of acids in water can be described by the equilibium HA + H2O H3O+ + A−, or, more simply, HA ⇌ H+ + A−. The ratio of [H+] [A−]/[HA] is constant (called the dissociation constant Ka or Ka′) for each acid. The resulting equation of [H+] [A−]/[HA] = Ka can be rearranged to [H+] = Ka [HA]/[A−] or 1/[H+] = 1/Ka + [A−]/[HA]. Taking logarithms of both sides, and noting that pH is defined as −log [H+], one derives the Henderson-Hasselbalch equation of pH = −log K (pK) + log [A−]/[HA]. Strong acids dissociate completely in water, so that [H+] and [A−] are equivalent to the amount of HA added to solution (Ka becomes infinite and meaningless). Since pure water has an [H+] concentration of 1.0 × 10−7 M, pH = −log 10−7 = −(−7) = 7.0. For hydrochloric acid, the dissociation of HCl H+ + Cl− in water is shifted entirely to the right, leaving virtually no HCl. The dilution of 2 mM hydrochloric acid with an equal volume of water produces an [H+] concentration of 1.0 × 10−3 M and a pH of 3. Note that the preexisting 10−7 [H+] of water is negligible compared to that resulting from the addition of HCl.

A solution of acid is prepared for cleaning surgical instruments by adding 0.5 L of 2 mM hydrochloric acid (HCl) to 0.5 L of pure water, which has a hydrogen ion concentration of 10−7 M. The initial pH of the pure water, then the pH after adding the HCl, are a. 7, then 3 b. 7, then 4 c. 7, then 1 d. 14, then 3 e. 14, then 4

*The answer is B.* To reach pH 7.0, approximately 100% of the α-carboxyl group (pKa1 = 2.19) and 90% of the γ-carboxyl group (pKa2 = 4.25) of glutamic acid must be dissociated. At that pH, approximately twice the amount of NaOH as glutamic acid molecules has been utilized to titrate the two carboxyl groups. Since each milliliter of a 1 M NaOH solution contains 1 mmol of OH− ion, about 3 mmol of the amino acid is present.

A solution of glutamic acid is titrated from pH 1.0 to 7.0 by the addition of 5 mL of a solution of 1 M NaOH. What is the approximate number of millimoles of amino acid in the sample (pKa1 = 2.19, pKa2 = 4.25, pKa3 = 9.67)? a. 1.5 b. 3.0 c. 6.0 d. 12.0 e. 18.0

*The answer is D.* The patient is in the midst of diabetic ketoacidosis, in which the production, but nonuse, of ketone bodies (which are acids), results in a significant lowering of blood pH. This patient will be creating a respiratory alkalosis to attempt to compensate for a metabolic acidosis. Under conditions of an acidosis, the proton concentration of the blood needs to be reduced. Due to the presence of carbonic anhydrase in the red blood cell, as carbon dioxide is exhaled, protons are removed from solution. As the concentration of carbon dioxide is reduced, bicarbonate (HCO3−) reacts with a proton (H+) to form carbonic acid, which then dissociates to form water (H2O) and carbon dioxide (CO2). Thus, as carbon dioxide is exhaled, the proton concentration decreases, and the acidosis is reduced. The exhalation of oxygen or nitrogen will not affect the proton levels in the blood, nor will the loss of nitrous oxide or superoxide.

A type 1 diabetic is brought to the emergency department due to lethargy and rapid breathing. Blood measurements indicated elevated levels of glucose and ketone bodies. Blood pH was 7.1. The patient was exhibiting enhanced breathing to exhale which one of the following gases in order to correct the abnormal blood pH? (A) Oxygen (B) Nitrogen (C) Nitrous oxide (D) Carbon dioxide (E) Superoxide

*The answer is C.* Based on the graph, when the substrate is present, Tamiflu results in the same Vmax and higher Km compared to the line when no inhibitor added. These are hallmarks of competitive inhibitors of enzymes, which Tamiflu is. Noncompetitive inhibitors result in decreased Vmax and the same Km with no inhibitor added, which is shown by the Relenza line in the graph.

A worldwide pandemic of influenza caused by human-adapted strains of avian influenza or bird flu is a serious health concern. One drug for treatment of influenza, Tamiflu (oseltamivir), is an inhibitor of the influenza viral neuraminidase required for release of the mature virus particle from the cell surface. Recent reports have raised concerns regarding viral resistance of Tamiflu compelling the search for alternative inhibitors. Another drug, Relenza (zanamavir), is already FDA approved for use in a prophylactic nasal spray form. The graph below show kinetic data obtained for viral neuraminidase activity (measured as the release of sialic acid from a model substrate) as a function of substrate concentration in the presence and absence of Relenza and Tamiflu. Based on the kinetic data, which of the following statements is correct? A. Both drugs are competitive inhibitors of the viral neuraminidase. B. Both drugs are noncompetitive inhibitors of the viral neuraminidase. C. Tamiflu increases the Km value for the substrate compared to Relenza D. Relenza increases the Vmax value for the substrate compared to Tamiflu. E. Relenza is not an inhibitor of neuraminidase, but inhibits another viral enzyme.

*The answer is A.* This question describes hydrops fetalis, a clinical manifestation of the most severe form of α-thalassemia. The α-globin chain is encoded by two genes on chromosome 16, for a total of four alleles (two from each parent). Different phenotypes are based on the number of mutated α-globin alleles. Remember, because of poor a-globin chain production, there will be an overproduction of β-globin in the adult or γ-globulin in the newborn to form unstable, harmful tetramers (hemoglobin H and hemoglobin Barts, respectively). When only one of the four alleles is mutated (to form nonfunctional globin chains), there is no gross clinical phenotype, but people with such mutations may be silent carriers. If two alleles are deleted, a mild microcytic hypochromic anemia occurs (often confused with iron-deficiency anemia). The double mutation may be (1) cis (on one chromosome, both alleles are bad, whereas, on the other chromosome, both alleles are normal) or (2) trans (one allele on both chromosomes is abnormal). The former is seen in populations of Asian origin, whereas the latter is more common in people of African origin. A trans deletion cannot lead to hydrops fetalis because a normal copy of the gene exists on each chromosome. In this vignette, each parent must be a carrier of the α-thalassemia trait in a cis fashion, such that the offspring may inherit two chromosomes from both parents without any α-globin gene (effectively a quadruple α-globin mutant). Because no functional α-chains are made, the fetus is able to make only the hemoglobin y4-tetramer~ also called hemoglobin Barts (Hb Barts). The high oxygen affinity of Hb Barts results in poor oxygen delivery to peripheral tissues and ultimately congestive heart failure, anasarca, and intrauterine fetal death.

A young Chinese couple is seeking genetic counseling following their third spontaneous pregnancy loss. Autopsies were performed after each stillbirth. Laboratory testing on the parents demonstrates low hemoglobin level with low mean corpuscular volume. Which of the following fetal genetic abnormalities is responsible for the recurrent pregnancy loss? A. Mutation that results in the absence of four of the α-globin genes B. Mutation that results in the absence of the β-globin chains C. Mutation that results in the absence of three of the α-globin genes D. Mutation that results in the substitution of glutamic acid with valine in the β -chain E. Mutat ion that results in the underproduction of the β-globin chain

*The answer is E.* Many enzymes interact to regulate blood clotting. Plasmin is activated by proteolytic cleavage of its zymogen, plasminogen. The activating protease is called tissue plasminogen activator (tPA). Plasmin hydrolyzes fibrin clots to form soluble products, and is used to dissolve clots in coronary arteries that cause myocardial infarction. Platelets, thrombin, and fibrinogen promote clotting through the intrinsic pathway and would be contraindicated in myocardial infarction. Platelets form a plug at the site of bleeding and bind prothrombin to facilitate its conversion to thrombin. Fibrinogen is the substrate acted upon by thrombin to yield the fibrin mesh of blood clots. Heparin is a mucopolysaccharide that terminates clot formation by interfering with a number of steps in the coagulation cascade. Heparin inhibits the formation of clots, but cannot dissolve clots that have already formed.

A young man with hypercholesterolemia is rushed to the hospital with crushing chest pain radiating to his left arm and a probable heart attack. Which of the following treatments should be considered? a. A platelet transfusion b. Heparin infusion c. Thrombin infusion d. Fibrinogen infusion e. Tissue plasminogen activator infusion

*The answer is E.* This patient is suffering from methemoglobinemia, in which symptoms typically develop as the percentage of circulating hemoglobin that is instead methemoglobin rises above 3%. Methemoglobin contains Fe3+ (as opposed to the Fe2+ of circulating hemoglobin) and is formed by nonoxygen oxidizing agents. One such agent known to cause this is dapsone, which is used for toxoplasmosis prophylaxis in HIV-positive patients. The Fe3+ of methemoglobin is unable to bind oxygen, and so cannot deliver it to the cells of the body and the patient becomes cyanotic. Nicotinamide adenine dinucleotide reductase (NADH) converts Fe3+ into Fe2+ by simultaneously oxidizing NADH to the oxidized form of nicotinamide adenine diphosphatase, and so oxygen-carrying capacity is restored. This system may be overwhelmed in the presence of oxidizing agents such as dapsone. Of note, the arterial blood gas of patients with methemoglobinemia will show a normal partial oxygen pressure of because the level of dissolved oxygen is normal; it is only the level of hemoglobin- bound oxygen that is reduced, hence, reduced oxygen saturation. The patient may be acidotic secondary to lactic acidosis from an oxygen deficit at the tissue level.

A young woman currently being treated for HIV is brought to the emergency department because of a headache and cyanosis of her nail beds and lips. She also reports feeling dizzy. The resident on call immediately places her on supplemental oxygen and draws blood for arterial blood gas analysis. While drawing the blood, he notes that the arterial blood has a dark brown color. Blood gas analysis reveals a pH of 7.39, partial oxygen pressure of 96 mm Hg, partial carbon dioxide pressure of 35 mm Hg, and oxygen saturation of 82% on room air. What enzyme is primarily responsible for preventing this condition in the normal adult? (A) ATPase (B) Flavin adenine dinucleotide reductase (C) GTPase (D) Lactase (E) Nicotinamide adenine dinucleotide reductase (F) Pyruvate kinase

*The answer is B.* Coenzymes-cosubstrates are small organic molecules that associate transiently with an enzyme and leave the enzyme in a changed form. Coenzyme-prosthetic groups are small organic molecules that associate permanently with an enzyme and are returned to their original form on the enzyme. Cofactors are metal ions. Heterotropic effectors are not substrates.

ADH requires NAD+ for catalytic activity. In the reaction catalyzed by ADH, an alcohol is oxidized to an aldehyde as NAD+ is reduced to NADH and dissociates from the enzyme. The NAD+ is functioning as a (an): A. apoenzyme. B. coenzyme-cosubstrate. C. coenzyme-prosthetic group. D. cofactor. E. heterotropic effector.

*The answer is E.* Immune responses involving the soluble antibody or humoral system are initiated first in IgM class. Long-term immunity is mediated by IgG molecules that circulate in the plasma. Acute allergic responses frequently involve increased levels of IgE molecules.

After first-time exposure to ragweed pollen, an initial immune response occurs followed by long-term sensitization to recurrent exposures to ragweed. Analysis for antibodies specific for the ragweed pollen would show immunoglobulins of which of the following classes at each stage of the immune response?

*The answer is D.* The child is exhibiting the symptoms of hereditary spherocytosis, a defect in spectrin in the erythrocyte membrane. This membrane problem leads to an abnormal shape of the red blood cell, such that the spleen removes them from circulation (hence, the large spleen), leading to an anemia due to a reduction of red blood cells in circulation. This defect is not due to a loss of hemoglobin or glucose-6-phosphate dehydrogenase (a lack of glucose-6-phosphate dehydrogenase will lead to red cell damage and cell fragments on peripheral smear under oxidizing conditions, conditions not observed with this patient). A lack of iron in the erythrocyte can lead to an anemia (due to insufficient oxygen binding to hemoglobin and reduced oxygen delivery to the tissues), but it would not lead to an altered cell shape. A loss of methemoglobin reductase would lead to increased levels of methemoglobin, which cannot bind oxygen, but would also not lead to a cell shape change.

An 8-month-old infant exhibits jaundice and lethargy. Physical exam detects splenomegaly. Blood work displays a microcytic anemia with abnormal erythrocytes (see picture below) under all conditions. This defect is most likely due to a hereditary mutation in which of the following? (A) Hemoglobin (B) Glucose-6-phosphate dehydrogenase (C) Iron transport into the erythrocyte (D) Spectrin (E) Methemoglobin reductase

*The answer is A.* Ectopia lentis (a displaced lens) is a common finding in Marfan's syndrome (occurs in 50%-80% of cases). Other typical findings include an arm span exceeding height (ratio > 1.05), reduced upper-to-lower body segment ratio, arachnodactyly, scoliosis greater than 20 degrees, dural ectasia, and aortic involvement, including dilation of the aortic root (leading to aortic regurgitation), aortic dissection, and mitral valve prolapse. Marfan 's syndrome is a connective tissue disease, most typically due to a mutation of the fibrillin gene, located on chromosome 15. Fibrillin is one of the major components of microfibrils, and these microfibrils are critical components of elastin in the aorta, suspensory ligaments of the lens of the eye, and other connective tissues.

An 8-year-old boy presents to his ophthalmologist on recommendation from his teacher. On examination, the ophthalmologist notices an upwardly displaced lens. Genetic testing reveals a mutation on chromosome 15, and physical examination reveals an abnormally large arm span. What is the most likely cardiovascular complication to be seen in this patient? A. Aortic regurgitation B. Aortic stenosis C. Hypertension D. Ischemic stroke E. Venous thrombosis

*The answer is E.* The patient has sickle cell anemia (HbSS), a disease of autosomal recessive inheritance. Sickle cell hemoglobin, or hemoglobin S (HbS), is generated when glutamic acid at the 6th position of the beta-globin gene is replaced with valine. HbS occurs most frequently in populations previously exposed to Plasmodium fa/ciparum malaria (ie, in Africa, India, the Mediterranean, Saudi Arabia). When oxygen tension is low, HbS is prone to crystallize, causing the RBCs to malform into a "sickle" shape. Clinical problems result from microvascular occlusive events. Sickling is facilitated by increased temperature (fever), decreased pH (acidosis), and high mean corpuscular hemoglobin concentration (dehydration).

An African-American patient with a history of lifelong pain crises involving the back and chest has a peripheral blood smear similar to that in the image. He states that the crises are often precipitated by overexertion or dehydration. What is the genetic abnormality associated with this condition? A. Deletion of glutamic acid B. Frameshift mutation C. Loss of promoter region D. Substitution of lysine for glutamic acid E. Substitution of valine for glutamic acid

*The answer is C.* In the technique of polyacrylamide gel electrophoresis (PAGE), the distance that a protein is moved by an electrical current is proportional to its charge and inversely proportional to its size. Patients with normal hemoglobin A have two α-globin and two β-globin chains, each encoded by a pair of normal globin alleles. Mutation in one α- or β-globin allele alters the primary amino acid sequence of the encoded globin peptide. If the amino acid change alters the charge of the peptide, then the hemoglobin tetramer assembled with the mutant globin peptide has a different charge and electrophoretic migration than the normal hemoglobin tetramer. The electrophoresis of native (undenatured) hemoglobin therefore produces two species (two bands) rather than one, each retaining its heme molecule and red color. If the hemoglobins were first denatured into their α-globin and β-globin chains as with SDS-polyacrylamide gel electrophoresis, then the similar size of the α- or β-globin peptides would cause them to move closely together as two colorless bands. Identification of these peptides as globin would require use of labeled antibody specific for globin (western blotting). Since the sodium dodecyl sulfate (SDS) detergent covers the protein surface and causes all proteins to be negatively charged, the distance migrated is solely dependent (inversely proportional) to protein size. High-performance liquid chromatography (HPLC) uses ionic resins to separate proteins by charge. The columns are run under high pressure, rapidly producing a series of proteins that are separated from most negative to most positive (or vice versa, depending on the charge of the ionic resin). A mutant hemoglobin with altered charge should produce a second red protein in the pattern. In dialysis, semipermeable membranes allow smaller proteins to diffuse into the outer fluid, but not larger proteins such as hemoglobin.

An adult with mild, chronic anemia does not respond to iron supplementation. Blood is drawn and the red cell hemoglobin is analyzed. Which of the following results is most likely if the patient has an altered hemoglobin molecule (hemoglobinopathy)? a. Several proteins but only one red protein detected by high performance liquid chromatography (HPLC) b. Two proteins detected in normal amounts by western blotting c. Several proteins and two red proteins separated by native gel electrophoresis d. Two labeled bands a slight distance apart after SDS-gel electrophoresis and reaction with labeled antibody to α- and β-globin e. A reddish mixture of proteins retained within a dialysis membrane

*The answer is C.* The light chain and part of the heavy chain at the amino terminal contain the antibody-combining site in the "hypervariable regions." These regions are all contained in fragment A, which is known as Fab. Fragment B is known as Fc. The Fc fragment contains a site for binding of complement. The Fab fragments mediate complement fixation. Each of the fragments can be further dissociated into two subunits by breaking its disulfide bridge with mercaptoethanol or some other reducing agent.

An immunoglobulin (see the figure below) is hydrolyzed by papain to form two A fragments and one B fragment. It is true of fragment A that it a. Contains the constant regions b. Is the heavy chain c. Contains the light chain d. Is not functional as an antibody-combining site e. Cannot be further dissociated by mercaptoethanol

*The answer is A.* In addition to its function as a carrier of O2 and CO2, hemoglobin buffers sudden additions of acid or base to the blood by virtue of the histidine 146 on each β chain. However, protonation of the imidazole of histidine causes deoxygenation of hemoglobin. Thus, decreased binding of O2 occurs in the high-pH conditions of acidosis. 2,3-bisphosphoglycerate (BPG) binds specifically to deoxyhemoglobin; that is, BPG cross-links positively charged residues on the β chain, thereby decreasing oxygen affinity and stabilizing the deoxygenated form of hemoglobin. The addition of each O2 molecule to deoxyhemoglobin requires the breakage of salt links, such as those formed by 2,3-BPG. Each subsequent O2 molecule requires the breakage of fewer salt links. Thus, initial O2 binding actually results in an increased affinity for subsequent O2 binding, which in turn results in a cooperative allosteric binding mechanism. CO2 reacts reversibly with the amino acid terminals of hemoglobin to create carbaminohemoglobin, which is negatively charged and which forms salt bridges stabilizing deoxyhemoglobin. Hence, CO2 binding lowers the affinity of hemoglobin for O2.

An increased affinity of hemoglobin for O2 may result from which of the following? a. Initial binding of O2 to one of the four sites available in each deoxyhemoglobin molecule b. High pH c. High CO2 levels d. High 2,3-bisphosphoglycerate (BPG) levels within erythrocytes e. Acidosis

*The answer is C.* In adults, hemoglobin A is the predominant form of hemoglobin. It is a tetramer consisting of two alpha and two beta chains. Normally the synthesis of alpha and beta chains is tightly regulated such that one α chain is synthesized for every β chain. Hemoglobin formation begins within a few weeks of conception. The initial hemoglobin formed by a fetus in utero is called embryonic hemoglobin (Gower). This hemoglobin is composed of two zeta (ζ) and two epsilon (ε) chains (ζ2ε2) and is produced in the embryonic yolk sac. Within a few weeks, the fetal liver starts synthesizing hemoglobin F (fetal hemoglobin). This form of hemoglobin is composed of two alpha and two gamma chains (α2γ2). HbF is the major hemoglobin in the fetus during last few months of gestation and in infants during first few weeks of postnatal life. HbA synthesis starts during the final month of gestation and gradually replaces HbF during postnatal life. Knowing the chronology of fetal hemoglobin formation and the gradual transition to adult hemoglobin (HbA) is important in understanding the relationship between clinical manifestations and postnatal age in beta thalassemia. Thalassemias are hereditary hemolytic anemias resulting from defective synthesis of globin chains. As described above, the synthesis of alpha and beta globin chains is very coordinated. In patients with thalassemia, the synthesis of either alpha or beta chains is defective. Beta thalassemia is caused by defective synthesis of beta chains. There are two copies of the beta globin gene (one from each parent). If only one gene is defective the patient will have beta thalassemia trait (beta thalassemia minor) and lack significant anemia. A defect in both beta globin genes results in severe hemolytic anemia known as beta thalassemia major. In this disease, alpha chains are produced normally but they cannot form stable tetramers due to the lack of beta globin chains. This failure to form stable hemoglobin leads to precipitation of alpha globin chains and premature lysine of red blood cells. Beta thalassemia cannot become symptomatic as long as there are significant amounts of gamma chains present because gamma chains make up for the absence of HbA beta chains in forming tetramers. Thus, in late gestation and early postnatal life. the expression of hemoglobin A is offset by gamma chain production (Choice C). As gamma chain production wanes, patients will become symptomatic.

An infant born to a Greek. immigrant appears healthy at birth but develops transfusion-dependent hemolytic anemia by the age of 8 months. His erythrocytes contain insoluble aggregates of hemoglobin subunits. The child developed normally in utero because at that time he produced high quantities of: A. α-globin B. β-globin C. γ-globin D. δ-globin E. ζ-globin

*The answer is E.* When interpreting Lineweaver-Burk plots, remember that the axes represent reciprocal values of velocity (1/v , y-axis) and substrate (1/[S], x-axis). The [S] is highest at the origin and becomes lower as it moves away from the origin on the x axis. The velocity increases as it moves down the y -axis, with 1/vmax represented by the point at which the line crosses the y -axis (infinite [S]}. If there were a mutation that decreased the amount of enzyme, the velocities. including Vmax, would decrease (although the Km would remain the same). Therefore 1/v, including 1 1/vmax would increase and the line representing this situation would be higher than the normal control in the right-hand quadrant and would intercept the y -axis at a higher point, Line X fits these criteria (choices E and C). The point at which the lines cross the x -axis determines the Km for the enzymes. in the sense that the value of the intercept (a negative number) = -1/Km. It is easiest to just drop the negative signs on each side of the equation and define the absolute value of the intercept as uKm , The closer this intercept gets to the origin of the graph the higher the Km value becomes. In this graph, line Y Fits this criterion (choices A and E) Therefore, the only choice that fits both mutations correctly is choice E.

An investigator is evaluating the kinetics of triose phosphate isomerase (TPI) in leukocytes from two patients. Dihydroxyacetone phosphate is used as the substrate and the reaction is coupled to the glyceraldehyde dehydrogenase (GAD) reaction by using excess GAD, inorganic phosphate, and NAD+. This both drives the reaction toward completion and permits the rate to be followed by the increased absorption at 340 nm due to the formation of NADH. To obtain results on an equal cellular basis, the aliquots used are adjusted so that an equal amount of leukocyte DNA is added in all cases. The following data are obtained: Which of the following choices best explain these data? 1. Normal enzyme control 2. Mutation producing increased Km 3. Mutation decreasing amount of functional enzyme A. Line X Line Y Line Z B. Line Y Line X Line Z C. Line Y Line Z Line X D. Line Z Line X Line Y E. Line Z Line Y Line X

*The answer is E.* One can estimate the Michaelis-Menten constant (Km) on this graph by taking the concentration of substrate at one-half of the enzyme's maximal velocity (Vmax). When one does this, B (glucokinase) has a higher Km than A (hexokinase). This property of glucokinase allows it to regulate glycogen production in the liver as well as glucose-dependent insulin secretion in the pancreatic β-cells .

An investigator is interested in comparing carbohydrate metabolism in hepatocytes and other body t issues. She isolates an enzyme from myocytes (A) and from hepatocytes (B) involved in glucose metabolism. She then compares their enzymatic activities, which are illustrated in the image. Which of the following best characterizes the relationship between these enzymes? A. A Is the same enzyme as B but in the presence of a competitive inhibitor B. A is the same enzyme as B but in the presence of a non-competitive Inhibitor C. At maximal substrate concentrations, both enzymes display first- order kinetics D. B has a greater affinity for glucose than A E. B has a higher Michaelis-Menten constant (Km) than A

*The answer is D.* At physiological pH, lysine would have a net charge of +1. This is because the amino group would be protonated (+1), the carboxyl group would be deprotonated (-1), and the side chain would be protonated (+1). +1 + +1 + -1 = +1.

At physiological pH, lysine (pK amino group = 9.0; pK carboxylate group = 2.2; pK of ionizing side-chain = 10.5) would exist mainly as a species with a net charge of which ONE of the following? (A) -2 (B) -1 (C) 0 (D) +1 (E) +2

*The answer is B.* Primary protein structures denote the sequence of amino acids held together by peptide bonds (carboxyl groups joined to amino groups to form amide bonds). The types of amino acids then determine the secondary structure of peptide regions within the protein, sometimes forming spiral α helices or flat pleated sheets. These regional peptide secondary structures then determine the overall three-dimensional tertiary structure of a protein, which is vital for its function. Amino acid substitutions that alter the charge of an amino acid side chain, like the change from glutamic acid (charged carboxyl group) to valine (nonpolar methyl groups) in choice b, are most likely to change the secondary and tertiary protein structure. A change in hemoglobin structure can cause instability, decreased mean cellular hemoglobin concentration (MCHC), and anemia. A change from glutamic acid to valine at position 6 in the β-hemoglobin chain is the mutation responsible for sickle cell anemia.

Blood is drawn from a child with severe anemia and the hemoglobin protein is degraded for peptide and amino acid analysis. Of the results below, which change in hemoglobin primary structure is most likely to correlate with the clinical phenotype of anemia? a. ile-leu-val to ile-ile-val b. leu-glu-ile to leu-val-ile c. gly-ile-gly to gly-val-gly d. gly-asp-gly to gly-glu-gly e. val-val-val to val-leu-val

*The answer is B.* Calcium ions are the regulators of contraction of skeletal muscle. Calcium is actively sequestered in sarcoplasmic reticulum by an ATP pump during relaxation of muscle. Nervous stimulation leads to the release of calcium into the cytosol and raises the concentration from less than 1 mM to about 10 mM. The calcium binds to troponin C. The calcium-troponin complex undergoes a conformational change, which is transmitted to tropomyosin and causes tropomyosin to shift position. The shift of tropomyosin allows actin to interact with myosin and contraction to proceed.

Contraction of skeletal muscle is initiated by the binding of calcium to a. Tropomyosin b. Troponin c. Myosin d. Actomyosin e. Actin

*The answer is E.* At the substrate concentration ([S]) of 3.00 mM, the reaction rate approaches V max and the velocity of the reaction changes minimally with changes in [S]. At Vmax, the rate will vary in direct proportion to changes in enzyme concentration ([E]} and thus this is a suitable substrate concentration for conducting assays of enzymes.

During an investigational study, an assay is set up to determine the activity of a hypothetical enzyme in biological samples. Enzyme activity is measured by adding increasing amounts of substrate to an otherwise optimal incubation system. The data obtained are shown Which of the following substrate concentrations is most suitable for conducting the desired enzyme assays? A. 0.30 mM B. 1.20 mM C. 1.50 mM D. 2.00 mM E. 3.00 mM

*The answer is A.* Globin chains of the hemoglobin (Hb) tetramer are compactly folded due to nonpolar hydrophobic residues in the interior and charged polar residues on the surface. In sickle cell disease (SCD) with HbS, the usual acidic (negatively charged) glutamic acid (glu) residue at the sixth position on the β-globin chain is replaced by a nonpolar (neutral charge) valine (Val) residue. This single glu → val substitution leads to the alteration of a region on the β-globin surface that interacts with a complementary site on another Hb molecule. As a result of the charge difference, the hydrophobic interactions that occur cause aggregation of Hb molecules (under anoxic conditions) and subsequent erythrocyte sickling (distortion and inflexibility), which is promoted by low oxygen levels, increased acidity, or dehydration. In patients with HbC, another common Hb variant, glu is replaced by a basic polar (positively charged) lysine (lys) residue. Because lys is charged (although it has . opposite posterity to glu), hydrophobic interactions between Hb molecules do not occur. The presence of lys causes HbC to have decreased mobility on electrophoresis. *Educational Objective:* Hemoglobin S (HbS) contains valine in place of glutamic acid at the sixth amino acid position of the β-globin chain. This promotes hydrophobic interaction among Hb molecules and results in HbS polymerization and erythrocyte sickling.

Hemoglobin C (HbC) disease is caused by a single amino acid substitution (glutamic acid → lysine) at position S in the β-globin chain of the Hb molecule. Patients homozygous for HbC have mild chronic hemolytic anemia, whereas those with hemoglobin (HbS) generally have a more severe condition. Which of the following properties of HbS best explains why HbS disease is associated with more pronounced clinical manifestations than HbC disease? A. HbS allows hydrophobic interaction among hemoglobin molecules B. HbS decreases β-globin interaction with 2,3-diphosphoglycerate C. HbS impairs oxygen binding to the heme moiety D. HbS impairs proper folding of the α-helix in the β-globin chain E. HbS stabilizes the iron moiety at the ferric (Fe3+) state

*The answer is B.* 2,3-Bisphosphoglycerate (BPG) promotes oxygen unloading by binding to a pocket formed by the two a subunits. It can bind only when they are close together, such as in the taut form. It is essential to add inosine to stored blood for transfusions to prevent the loss of 2,3-BPG. Any change that enhances the taut form of hemoglobin will decrease hemoglobin's affinity for oxygen.

Hemoglobin consists of four polypeptide subunits: two α subunits and two β subunits. The arrangement of these subunits shifts between a taut and relaxed conformation, resulting in changes in hemoglobin's oxygen affinity. At a given partial pressure of oxygen, which of the following will decrease hemoglobin's affinity for oxygen? (A) Decreasing the partial pressure of carbon dioxide (B) Increasing the amount of 2,3-bisphosphoglycerate in RBCs (C) Increasing the number of oxygen molecules bound to a hemoglobin from one to three (D) Increasing the pH by moving from peripheral tissue to lung (E) The presence of excess carbon monoxide

*The answer is D.* Immunoglobulin G is composed of pairs of light chains and heavy chains attached by disulfide bridges. If the reducing agent mercaptoethanol is used to break the disulfide bridges and urea is used to disrupt noncovalent interactions, two identical light subunits (25 kd) and two identical heavy chains (50 kd) per protein can be resolved with electrophoresis. A small amount of carbohydrate is also present. In contrast, the proteolytic enzyme papain cleaves the heavy chains, which results in two Fab molecules consisting of the entire light chain attached to the amino terminal half of each heavy chain and two Fc molecules consisting of the carboxyl terminal half of each heavy chain. Other proteolytic enzymes are nonspecific. Levels rise and fall in the serum dependent upon specific induction by antigen.

Immunoglobulin G molecules can be characterized by which of the following statements? a. They are maintained at a constant level in the serum b. They contain nucleic acids c. They contain mostly carbohydrate d. They can be separated into subunits with a reducing agent and urea e. They can be separated into subunits with a proteolytic enzyme and urea

*The answer is A.* Regular arrangements of groups of amino acids located near each other in the linear sequence of a polypeptide are the secondary structure of a protein. The α helix, β sheet, and β bend are the secondary structures usually observed in proteins. In both the α helix and the β sheet, all the peptide bond components participate in hydrogen bonding. That is, the oxygen components of the peptide bond form hydrogen bonds with the amide hydrogens. In the case of the α helix, all hydrogen bonding is intrachain and stabilizes the helix. In the case of β sheets, the bonds are interchain when formed between the polypeptide backbones of separate polypeptide chains and intrachain when the β sheet is formed by a single polypeptide chain folding back on itself. While the spiral of the α helix prevents the chain from being fully extended, the chains of β sheets are almost fully extended and relatively flat. The chains of β sheets can be either parallel or antiparallel. When the N-terminals of chains run together, the chain or segment is considered parallel. In contrast, when N-terminal and C-terminal ends of the chains alternate, the β strand is considered antiparallel.

In comparing the secondary structure of proteins, which description applies to both the α helix and the β-pleated sheet? a. All peptide bond components participate in hydrogen bonding b. N-terminals of chains are together and parallel c. The structure is composed of two or more segments of polypeptide chain d. N-terminal and C-terminal ends of chains alternate in an antiparallel manner e. The chains are almost fully extended

*The answer is A.* Glucose is immediately phosphorylated to form glucose 6-phosphate (G6P) after being transported into all extrahepatic cells by the enzyme hexokinase. Hexokinase has a very low Km and thus extremely high affinity for glucose, virtually trapping glucose into entering the metabolic pathways of most tissues. G6P is the starting point for glycolysis, glycogenesis, and the pentose phosphate pathway. In contrast, the liver's hexokinase is called glucokinase and it is only active following a meal, when blood glucose levels are above about 5 mM. It has a relatively low affinity and high Km. In this manner, the liver sequesters and stores glucose as glycogen for later distribution to tissues only when it is in excess.

In the figure below, which letter designates an enzyme that has an extremely low Km in most but not all tissues? a. Letter A b. Letter B c. Letter C d. Letter D e. Letter E

*The answer is C.* Allosteric enzymes, unlike simpler enzymes, do not obey Michaelis-Menten kinetics. Often, one active site of an allosteric enzyme molecule can positively affect another active site in the same molecule. This leads to cooperativity and sigmoidal enzyme kinetics in a plot of [S] versus V. The terms competitive inhibition and noncompetitive inhibition apply to Michaelis-Menten kinetics and not to allosteric enzymes.

In the study of enzymes, a sigmoidal plot of substrate concentration ([S]) versus reaction velocity (V ) may indicate a. Michaelis-Menten kinetics b. Myoglobin binding to oxygen c. Cooperative binding d. Competitive inhibition e. Noncompetitive inhibition

*The answer is A.* Fetal hemoglobin (Hb F) is the predominant form of hemoglobin in prenatal life and persists at high levels for some time after birth. Hb F does not contain β-globin subunits so it is unaffected by the mutation that causes sickle cell disease. Hb A is the normal adult hemoglobin. HbS and HbD are both abnormal hemoglobins caused by the common sickle cell mutation and coinheritance of another mutation in the same gene respectively.

Individuals with sickle cell disease appear healthy at birth. This is due to which one of the following? (A) presence of HbF (B) presence of HbS (C) presence of HbA (D) presence of HbD

*The answer is D.* Inhibitor X represents a competitive inhibitor, while inhibitorY represents a noncompetitive inhibitor. The binding of a competitive inhibitor lowers the affinity of the enzyme for the substrate (increased Michaelis-Menten constant, or Km), but the maximum velocity (Vmax) of the reaction remains unchanged. Increasing the concentration of substrate can overcome the effects of a competitive inhibitor. Binding of a noncompetitive inhibitor does not change the affinity of the enzyme for the substrate (Km is unchanged), but it does decrease the maximum velocity, The best answer is that competitive inhibitor X will effect no change in Vmax and that the noncompetitive inhibitorY will decrease Vmax·

Inhibitor X and inhibitor Y bind an enzyme. Inhibitor X reversibly binds the same site as the substrate, whereas inhibitor Y interacts with the enzyme at a different location. How is the maximum velocity of the reaction between the enzyme and substrate affected when inhibitors X and Y bind the enzyme separately?

*The answer is B.* Lysine's side chain is in its protonated below the pH of 10.5. Glutamate side chain is deprotonated above the pH of 4.0. So these amino acids are able to maintain the salt bridge between the pH of 4.0 and 10.5.

Ionic bonds between positively and negatively charged amino acid side chains, sometimes called salt bridges, often stabilize protein tertiary structures. Given the salt bridge: Over what range of pH will this salt bridge be stable? (Assume that the side chains must be at least 50% charged in order for the salt bridge to be stable and the pK for -NH3 is ~10.5 and the pK for -COO- is ~4.0). Choose the ONE best answer. (A) between pH 3.0 and 9.5 (B) between pH 4.0 and 10.5 (C) between pH 5.0 and 11.5 (D) between pH 4.0 and 11.5 (E) between pH 3.0 and 10.5

*The answer is D.* Weak acids like lactic acid never completely dissociate in solution and are thus defined by the property that at least some of the protonated (undissociatedmacid) form and the unprotonated (conjugate base) form of the acid are present at all concentrations and pH conditions. The indicated pKa of 5.2 is consistent with the idea that the lactate anion retains a strong affinity for protons, a hallmark of a weak acid. The lactate anion is highly water soluble. All weak acids obey the Henderson-Hasselbalch equation.

Lactic acid is considered to be a weak acid because: A. It is insoluble in water at standard temperature and pressure. B. It fails to obey the Henderson-Hasselbalch equation. C. Little of the acid form remains after it dissolves in water. D. The equilibrium between the acid and its conjugate base has a pKa of 5.2. E. The lactate anion has minimal tendency to attract a proton.

*The answer is D. In order to make a xenobiotic more soluble, a hydrophilic group needs to be added to the xenobiotic. Of the possible answer choices, only glucuronic acid (glucose with a carboxylic acid at position 6 instead of an alcohol group) is a hydrophilic molecule. Glucuronic acid is added to the xenobiotic at position 1, using the activated intermediate UDP-glucuronate. Once added to the xenobiotic, the highly soluble glucuronate confers enhanced solubility to the adduct. Phenylalanine contains a hydrophobic side chain, and palmitate, linoleate, and cholesterol are all very hydrophobic molecules. Their addition to a xenobiotic would decrease, rather than increase, its solubility.

Liver catabolism of xenobiotic compounds, such as acetaminophen (Tylenol), is geared toward increasing the solubility of such compounds for safe excretion from the body. This can occur via the addition of which compound below in a covalent linkage with the xenobiotic? (A) Phenylalanine (B) Palmitate (C) Linoleate (D) Glucuronate (E) Cholesterol

*The answer is E.* The transition from point 1 to point 2 on the graph above represents the loading of O2 onto partially deoxygeneted hemoglobin. At very low pO2, the hemoglobin molecule is fully deoxygenated and binding of the first O2 molecule is relatively difficult (as indicated by the early flatness of the curve). As PO2 increases, O2 binds to 1 of the 4 heme moieties on the hemoglobin molecule. causing the oxygen-binding affinity of the other hemoglobin subunits to increase (steepening of the curve). Additional O2 molecules bind as the oxygen partial pressure increases. As hemoglobin becomes saturated with oxygen, very little additional binding occurs, and the curve levels out. In the peripheral tissues, the release of O2 from hemoglobin is enhanced by increased pCO2 and the resultant decrease in pH (Bohr effect). This effect occurs due to the histidine side chains found on the alpha and beta hemoglobin subunits. As the tissues release CO2, the majority is converted by erythrocyte carbonic anhydrase to bicarbonate and H+. While bicarbonate is shifted out of the erythrocytes in exchange for chloride ions found in the plasma, the hydrogen ions remain within the erythrocytes. These hydrogen ions are buffered by the histidine residues on hemoglobin and in the process, help stabilize the deoxygeneted form of hemoglobin and decrease its affinity for oxygen.

Molecular biologists studying the properties of hemoglobin are investigating the structural changes associated with oxygen loading and unloading. During the transition from point 1 to point 2 on the graph shown below, hemoglobin molecules are most likely to release which of the following? A. Chloride B. Heme C. Oxygen D. Phosphate E. Protons

*The answer is D.* Native collagen is composed almost entirely of the triple helix structure.

Native collagen is composed almost entirely of which of the following types of structures? (A) α-Helix (B) β-Pleated sheet (C) Random coils (D) Triple helix (E) Two peptides connected by a disulfide bond

*The answer is B.* This is the silent carrier state for α-thalassemia, and in many cases, the hematologic findings are normal.

Normal hematologic findings would be expected with which on of the following? (A) α⁰-thalassemia (B) α⁺-thalassemia (C) thalassemia major (D) β-thalassemia

*The answer is A.* β-Thalassemia is defined by the absence or reduced synthesis of the β-globin subunits of hemoglobin.

Reduced or absent synthesis of β-globin subunits of hemoglobin results in which one of the following? (A) β-thalassemia (B) α⁰-thalassemia (C) α⁺-thalassemia (D) sickle cell anemia

*The answer is D.* Rotenone competitively inhibits reduced nicotinamide adenine dinucleotide dehydrogenase (NADH), As a result, the Michaelis-Menten constant increases because rotenone competes with NADH at the enzymatic active site. However, the maximum reaction rate for a given concentration of enzyme does not change, because it can eventually be achieved by increasing the amount of substrate (ie, NADH) available to the enzymes.

Rotenone is a naturally occurring chemical that is used primarily as a fish poison. It is a reversible competitive inhibitor of reduced nicotinamide adenine dinucleotide dehydrogenase, the first complex in the electron transport chain. Application of rotenone will halt cellular respiration at this stage. Which of the following combinations describes the effects of rotenone on the enzyme kinetics of NADH dehydrogenase? A. The Michaelis-Menten constant decreases while the maximum reaction rate decreases. B. The Michaelis-Menten constant decreases while the maximum reaction rate remains unchanged C. The Michaelis-Menten constant increases while the maximum reaction rate decreases D. The Michaelis-Menten constant increases while the maximum reaction rate remains unchanged E. The Michaelis-Menten constant remains the same while the maximum reaction rate decreases

*The answer is A.* Hemeproteins such as hemoglobin and myoglobin contain heme groups that are used to reversibly bind oxygen for transportation and storage. Hemoglobin A (the major form of hemoglobin in adults) is a tetramer consisting of 2 alpha and 2 beta chains. Each hemoglobin subunit is associated with a heme moiety, so each hemoglobin molecule has 4 heme groups. After binding to 1 oxygen molecule, the oxygen affinity of other heme molecules increases, this heme-heme interaction is responsible for the characteristic sigmoid shape of the oxygen~hemoglobin dissociation curve. In contrast to hemoglobin, myoglobin is a monomeric protein and the primary oxygen-storing protein in skeletal and cardiac muscle tissue, it is only found in the bloodstream after muscle injury. The partial pressure of oxygen at which 50% of myoglobin molecules are oxygen saturated (P) is only 1 mm Hg, which is much lower than the P, of hemoglobin (26 mm Hg). Myoglobin also has only a single heme group and so does not experience heme-heme interactions, therefore, its oxygen-dissociation curve is hyperbolic. The secondary and tertiary structures of myoglobin and the hemoglobin beta subunit are almost identical (the o-subunits are also very similar to myoglobin). Because individual hemoglobin subunits are structurally similar to myoglobin, their oxygen-binding behavior is also similar to he individual subunits will have a hyperboic oxygen dissociation curve (Choice A). *Educational Objective:* The individual subunits of the hemoglobin molecule are structurally analogous to myoglobin. If separated, the monomeric subunits will demonstrate a hyperbolic oxygen-dissociation curve similar to that of myoglobin.

Scientists studying the principles behind oxygen-hemoglobin dissociation have discovered a way to successfully separate hemoglobin tetramers into individual alpha end beta subunits. During an experiment, a solution is created that contains only monomeric beta-hemoglobin subunits under physiologic conditions. If measured, the oxygen dissociation curve of the dissolved beta subunits will most likely resemble which of the following lines?

*The answer is C.* Cysteine has a sulfhydryl group in its side chain. Although methionine has a sulfur in its side chain, a methyl group is attached to it.

Several complexes in the mitochondrial electron transport chain contain non-heme iron. The iron in these complexes is bound tightly to the thiol group of which amino acid? A. Glutamine B. Methionine C. Cysteine D. Tyrosine E. Serine

*The answer is C.* Substitution of alanine for lysine removes from each β subunit a positive charge that is important for making a salt bridge with BPG. BPG should still bind but just not as well as it would to normal adult hemoglobin and the affinity would be decreased. Because BPG binding stabilizes the deoxy form of hemoglobin, reduced BPG binding affinity would make the deoxy-to-oxy transition occur at lower PO2 values, ie, affinity of the mutant hemoglobin for O2 would be increased.

Some patients with erythrocytosis (excess RBCs) have a mutation that converts a lysine to alanine at amino acid 82 in the β subunit of hemoglobin. This particular lysine normally protrudes into the central cavity of deoxyhemoglobin, where it participates in binding 2,3-bisphosphoglycerate (BPG). Which of the following effects would you predict this mutation to have on the affinity of hemoglobin for BPG and O2, respectively, in such patients? A. Increase, Decrease B. Increase, Increase C. Decrease, Increase D. Decrease, Decrease E. No effect on either binding function

*The answer is B.* When an enzyme obeys classic Michaelis-Menten kinetics, the Michaelis constant (Km) and the maximal rate (Vmax) can be readily derived. By plotting a reciprocal of the Michaelis-Menten equation, a straight-line Lineweaver-Burk plot is produced. The y intercept is 1/Vmax, while the x intercept is −1/Km. Thus, a reciprocal of these absolute values yields Vmax and Km.

The Vmax of an enzyme with the kinetic data of a Lineweaver-Burk plot is a. Reciprocal of the absolute value of the intercept of the curve with the x axis b. Reciprocal of the absolute value of the intercept of the curve with the y axis c. Absolute value of the intercept of the curve with the x axis d. Slope of the curve e. Point of inflection of the curve

*The answer is D.* A free sulfhydryl group in the drug would be able to form a disulfide bond with the protein (-CH2-S-S-CH2), which is an oxidation-reduction reaction. This would render the disulfide resistant to acid or base-catalyzed hydrolysis. Forming a bond with the other groups listed would lead to relatively easy hydrolysis reactions, rendering the inhibitory bond unstable. Since the inhibition is stable, the best choice is a sulfhydryl group. The drug is a proton pump inhibitor and reduces acid secretion by the chief cells in the stomach, thereby alleviating symptoms of acid refl ux in the patient.

The activated form of the drug omeprazole (used to treat pepticulcer disease) prevents acid secretion by forming a covalent bond with the H+, K+-ATPase, thereby inhibiting the enzyme's transport capabilities. Analysis of the drug-treated protein demonstrated that an internal cysteine residue was involved in the covalent interaction with the drug. Further analysis indicated that the bond was not susceptible to acid or base catalyzed hydrolysis. Based on this information, one would expect the drug to contain which of the following functional groups that would be critical for its inhibitory action? (A) A carboxylic acid (B) A free primary amino group (C) An imidazole group (D) A reactive sulfhydryl group (E) A phosphate group

*The answer is E.* Because the apparent Vmax is near 100 mmol/sec, Vmax/2 equals 50 mmol/sec. The substrate concentration giving this rate is 0.50 mM.

The activity of an enzyme is measured at several different substrate concentrations, and the data are shown in the table below. Km for this enzyme is approximately A. 50.0 B. 10.0 C. 5.0 D. 1.0 E. 0.5

*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).

The bonds labeled A and B in the compound shown are best described as which one of the following?

*The answer is C.* Ingestion of an acid or excess production by the body, such as in diabetic ketoacidosis, may induce metabolic acidosis, a condition in which both pH and HCO3− become depressed. In response to this condition, the carbonic acid-bicarbonate system is capable of disposing of the excess acid in the form of CO2. The equilibrium between bicarbonate and carbonic acid shifts toward formation of carbonic acid, which is converted to CO2 and H2O in the RBC catalyzed by carbonic anhydrase, an enzyme found mainly in the RBC. The excess CO2 is then expired by the lungs as a result of respiratory compensation for the acidosis. The main role of the kidneys in managing acidosis is through excretion of H+ rather than CO2.

The composite pKa of the bicarbonate system, 6.1, may appear to make it ill-suited for buffering blood at physiologic pH of 7.4. Nevertheless, the system is very effective at buffering against additions of noncarbonic acids. Changes in the bicarbonate/carbonic acid ratio in such cases can be regulated by: A. Recruitment of bicarbonate reserves from the peripheral tissues. B. Conversion of carbonic acid to CO2 and excretion in the urine. C. Conversion of carbonic acid to CO2 followed by removal by the lungs. D. Reaction of excess carbonic acid with the amino termini of blood proteins. E. Binding of carbonic acid by hydroxide ions from the fluid phase of blood.

*The answer is A.* The Km of the enzyme in the presence of the modifier is greater compared to the absence of the modifier.

The data shown in the figure are based on measuring the enzyme kinetics of an enzyme in the absence (curve X) or presence (curve Y) of an allosteric modifier. The Km of the enzyme in the presence of the modifier is which ONE of the following as compared to the absence of the modifier? (A) Greater (B) The same (C) Less than (D) Insufficient data to make a determination

*The answer is C.* The figure in the question shows the titration curve of glycine, an amino acid with two dissociable protons—one from the α-carboxyl group and the other from the α amino group. The maximum buffering capacity of any ionizable function is at the pH equivalent to the pKa of the dissociation, as represented by points A and B on the graph. The curve clearly demonstrates two ionizable functions, one that ionizes at high pH (∼9) that would be atypical of α or side chain carboxyl groups. Point A would suggest the ionization of a relatively strong acid like a carboxyl group rather than of a base like the α-amino group.

The graph below shows a titration curve of a common biochemical compound. Which of the following statements about the graph is true? a. The compound has one ionizable function b. The compound has three ionizable side chains c. The maximum buffering capacity of the compound is represented by points A and B on the graph d. Point A could represent the range of ionization of an amino function e. Points A and B represent the respective pKs of α and side chain carboxyl groups

*The answer is D.* Keratins are a type of intermediate filament that comprises a large portion of many epithelial cells. The characteristics of skin, nails, and hair are all due to keratins. Keratins contain a large amount of the disulfide amino acid cystine. Approximately 14% of the protein composing human hair is cystine. This is the chemical basis of depilatory creams, which are reducing agents that render keratins soluble by breaking the disulfide bridges of these insoluble proteins. The basic structure of intermediate filament proteins is a two- or three-stranded α-helical core 300 amino acids in length.

The highest concentration of cystine can be found in a. Melanin b. Chondroitin sulfate c. Myosin d. Keratin e. Collagen

*The answer is C.* C represents the isoelectric point or pI, and as such is midway between pK1 and pK2 for this monoamino monocarboxylic acid. Glycine is fully protonated at Point A. Point B represents a region of maximum buffering, as does Point D. Point E represents the region where glycine is fully deprotonated.

The letters A through E designate certain regions on the titration curve for glycine (shown below). Which one of the following statements concerning this curve is correct? A. Point A represents the region where glycine is deprotonated. B. Point B represents a region of minimal buffering. C. Point C represents the region where the net charge on glycine is zero. D. Point D represents the pK of glycine's carboxyl group. E. Point E represents the pI for glycine.

*The answer is D.* The structure of myoglobin is illustrative of most water-soluble proteins. Globular proteins tend to fold into compact configurations with nonpolar cores. The interior of myoglobin is composed almost exclusively of nonpolar, hydrophobic amino acids like valine, leucine, phenylalanine, and methionine. In contrast, polar hydrophilic residues such as arginine, aspartic acid, glutamic acid, and lysine are found mostly on the surface of the water-soluble protein.

The oxygen carrier of muscle is the globular protein myoglobin. Which one of the following amino acids is highly likely to be localized within the interior of the molecule? a. Arginine b. Aspartic acid c. Glutamic acid d. Valine e. Lysine

*The answer is C.* In any fluid, maximum buffering action is achieved by the acid whose pKa most nearly approximates the pH of the fluid. Physiologic pH is about 7.4, so that among those buffers listed in the question, NaH2PO4 is the most effective.

The pH of body fluids is stabilized by buffer systems. Which of the following compounds is the most effective buffer at physiologic pH? a. Na2HPO4, pKa5 12.32 b. NH4OH, pKa5 9.24 c. NaH2PO4, pKa5 7.21 d. CH3CO2H, pKa5 4.74 e. Citric acid, pKa5 3.09

*The answer is B.* Arginine is the most basic of the amino acids (pI~11) and would have the largest positive charge at pH 7.

The peptide ala-arg-his-gly-glu is treated with peptidases to release all of the amino acids. The solution is adjusted to pH 7, and electrophoresis is performed. In the electrophoretogram depicted below, the amino acid indicated by the arrow is most likely to be A. Glycine B. Arginine C. Glutamate D. Histidine E. Alanine

*The answer is B.* The presence of a noncompetitive inhibitor leads to a decrease in the observed Vmax. The Km is not affected.

The presence of a noncompetitive inhibitor (A) Leads to both an increase in the Vmax of a reaction and an increase in Km (B) Leads to a decrease in the observed Vmax (C) Leads to a decrease in Km and Vmax (D) Leads to an increase in Km without affecting Vmax

*The answer is E.* Regulatory proteins must bind with great specificity and high affinity to the correct portion of DNA. Several structural motifs have been discovered in DNA regulatory proteins: the zinc finger, the leucine zipper, and the helix-turn-helix (found in homeotic proteins). Due to the uniqueness of these structural arrangements, their presence in a protein indicates that the protein might bind to DNA. The β sheet, β bend, and α helix are secondary structures found in polypeptide chains, and the triple helix is a tertiary structure composed of three polypeptides as in collagen.

The presence of which of the following structural arrangements in a protein strongly suggests that it is a DNA binding, regulatory protein? a. β sheet b. Triple helix c. α helix d. β bend e. Zinc finger

*The answer is A.* Of the amino acid choices listed only histidine has a side chain which could conceivably buffer in the range of 7.2 to 7.4. The imidazole group of histidine has a pKa of 6.0, but this can be altered by the local environment of the protein. Aspartic acid and glutamic acid have side chain carboxylic acids, each of which has a pKa about 4.0 and would not be able to contribute to buffering at neutral pH. Both lysine and arginine have basic side chains, with pKa values about 9.5, and those too will not be able to buffer near neutral pH.

The protein albumin is a major buffer of the pH in the blood, which is normally kept between 7.2 and 7.4. Which of the following is an amino acid side chain of albumin that participates in this buffering range? (A) Histidine (B) Aspartate (C) Glutamate (D) Lysine (E) Arginine

*The answer is A.* The negative ∆G⁰ value indicates the reaction is thermodynamically favorable (irreversible), requiring a different bypass reaction for conversion of F1, 6BP to F6P in the gluconeogenic pathway.

The reaction catalyzed by hepatic phosphofructokinase-1 has a ∆G⁰ value of -3.5 kcal/mol. This value indicates that under standard conditions this reaction A. is reversible B. occurs very slowly C. produces an activator of pyruvate kinase D. is inhibited by ATP E. has a low energy of activation F. will decrease in activity as the pH decreases G. cannot be used for gluconeogenesis H. shows cooperative substrate binding I. is indirectly inhibited by glucagon J. is stimulated by fructose 2,6-bisphosphate

*The answer is C.* The calculation of pH requires substitution of the concentrations of acid and base into the Henderson-Hasselbalch equation. For the given ratio of acetate and acetic acid, the pH equals the pK of 4.8 plus the log of the concentration of base/acid (2 M acetate/0.2-M acetic acid). This simplifies to: pH = 4.8 + log 10 = 4.8 + 1 = 5.8

The relationship between the ratio of acid to base in a solution and its pH is described by the Henderson-Hasselbalch equation pH = pK + log [base]/[acid] The pK of acetic acid is 4.8. What is the approximate pH of an acetate solution containing 0.2 M acetic acid and 2 M acetate ion? a. 0.48 b. 4.8 c. 5.8 d. 6.8 e. 10.8

*The answer is B.* Substrate concentrations are usually expressed in terms of molarity, e.g., M = moles per liter, mM = millimoles per liter, μM = micromoles per liter. Km, the Michaelis constant, is expressed in terms of substrate concentration. Each unit of enzyme activity is described as the amount of enzyme that converts a specific amount of substrate to a product within a given time. The standard units of activity are micromoles of substrate per minute. Specific activity relates the units of enzyme activity to the amount of protein present in the reaction, expressed as units of enzyme activity per milligram of protein. If the enzyme is pure (no proteins except the assayed enzyme are present), then the specific activity is maximal and constant for that particular enzyme (units of activity per milligram of enzyme). The specific activity is a useful measure of enzyme purity that should increase during enzyme purification.

The specific activity of an enzyme would be reported in which of the following units of measure? a. Millimoles per liter b. Units of activity per milligram of protein c. Micromoles per minute d. Units of activity per minute e. Milligrams per micromole

*The answer is C.* Proteins can be separated on the basis of their overall charge at a given pH by ion exchange chromatography. At low pH all proteins have an overall positive charge because carboxyl groups are protonated. Thus, proteins tend to bind to a cation exchange column that has immobilized the negative charges. Usually, negatively charged sulfonic polystyrene resin is used, and Na charges are exchanged for the positively charged protein groups. Once binding has occurred, the pH and NaCl concentration of the eluting medium are increased, and proteins that have a low density of net negative charge emerge first, with those having a higher density of negative charge following. The only information that can be obtained from the information given in the question is that the enzyme has been purified over 100-fold. The turnover rate of the enzyme cannot be deduced. Likewise, the yield, which is the amount of original enzyme protein recovered, cannot be determined. The structure of the enzyme is not revealed by the information given.

The specific activity of glycogen phosphorylase increases from 2.5 U/mg homogenate protein to 325.5 U/mg protein after being bound to and eluted from a cation exchange column at pH 2.7. What can you conclude from this information? a. The yield of enzyme is greater than 80% b. The enzyme is negatively charged at pH 2.7 c. The enzyme is purified over 100-fold d. The enzyme is globular in structure e. The enzyme is in an activated state

*The answer is A.* The carboxyl of glutamate at position 6 on the β chain of normal hemoglobin is dissociated and negatively charged at pH 7.0. Substitution of uncharged valine for glutamate by mutation produces sickle cell hemoglobin, which is less negatively charged and has an increased electrophoretic mobility. Polymerization of the deoxygenated form of sickle hemoglobin occurs owing to the alteration of primary structure caused by the valine substitution. The insoluble, polymerized hemoglobin causes the erythrocyte to lose flexibility and to become rigid and sickle-shaped. The brittle cells produce anemia and block capillaries.

The substitution of valine for glutamate at position 6 on the two β chains in sickle cell hemoglobin causes which of the following? a. Increased electrophoretic mobility at pH 7.0 b. Increased solubility of deoxyhemoglobin c. Decreased polymerization of deoxyhemoglobin d. Unchanged primary structure e. More flexible red blood cells

*The answer is B.* When a modifier binds at the allosteric site, it affects the active site by altering Vmax and Km. The substrate binds to the active, or catalytic, site, where it is modified. Binding of both substrate and modifier is, of course, concentration-dependent. The velocity of an allosteric enzyme reaction depends on the concentration of both the substrate and the modifier.

The velocity-substrate curve below characterizes an allosteric enzyme system. The curve demonstrates that a. A modifier changes the binding constant for the substrate but not the velocity of the reaction b. A modifier binding to the allosteric site can also affect the catalytic site c. Binding of the substrate is independent of its concentration d. Binding of the modifier is independent of its concentration e. Binding of substrate to the allosteric site displaces modifier

*The answer is A.* The ratio of conjugate base to its acid for a physiologic buffer helps determine the pH of a solution according to the terms of the Henderson-Hasselbalch equation. When the concentration of base equals that of the acid form, the ratio is 1.0 and the pH = pKa. In this case, a ratio of acid to base of 100:1 inverts to a base to acid ratio of 1:100 and calculates pH = 1.86. Such a highly acidic condition is never actually achieved within muscle cells because other weak acids, including those provided by inorganic phosphates and proteins, help buffer the solution by binding excess protons arising from dissociation of the lactic acid.

The weak organic acid, lactic acid, has a pKa of 3.86. During strenuous exercise, lactic acid can accumulate in muscle cells to produce fatigue. If the ratio of the conjugate acid form lactate to the conjugate basic form of lactic acid in muscle cells is approximately 100 to 1, what would be the pH in the muscle cells? A. 1.86 B. 2.86 C. 3.86 D. 4.86 E. 5.86

*The answer is C.*In blood and other solutions at physiologic pH (approximately 7.0), only terminal carboxyl groups, terminal amino groups, and ionizable side chains of amino acid residues in proteins have charges. The basic amino acids lysine, arginine, and histidine have positive charges (protonated amines). The acidic amino acids aspartate and glutamate have negative charges (ionized carboxyls). Glutamine possesses an uncharged but hydrophilic side chain.

Under normal conditions in blood, which of the following amino acid residues of albumin is neutral? a. Arginine b. Aspartate c. Glutamine d. Glutamate e. Histidine

*The answer is B.* Water molecules have a dipole nature and dissolve salts because of attractions between the water dipoles and the ions that exceed the force of attraction between the oppositely charged ions of the salt. In addition, the latter force is weakened by the high dielectric constant of water. Nonionic but polar compounds are dissolved in water because of hydrogen bonding between water molecules and groups such as alcohols, aldehydes, and ketones.

Water, which constitutes 70% of body weight, may be said to be the "cell solvent." The property of water that most contributes to its ability to dissolve compounds is the a. Strong covalent bond formed between water and salts b. Hydrogen bond formed between water and biochemical molecules c. Hydrophobic bond formed between water and long-chain fatty acids d. Absence of interacting forces e. Fact that the freezing point of water is much lower than body temperature

*The answer is E.* At pH 10.5, the amino group, the carboxyl group attached to the alpha carbon, and the side chain of glutamate will be deprotonated because the pH is above their pKa.

Which ONE of the following ionic species of glutamate is predominant at pH 10.5?

*The answer is D.* Each of the 20 unique amino acids coded for by DNA is composed of an α-carbon atom bonded to a hydrogen, a carboxyl group, an amino group, and a side chain R group. The α-carbon is so named because it is adjacent to the carboxyl group. The distinctive side chains of each different amino acid allow variation in charge, shape, size, and reactivity. Although glutamine is often referred to as an acidic amino acid, in fact it is an uncharged polar amino acid with no ionizable group. It is an amide derivative of glutamate, which is an acidic amino acid with an ionizable carboxyl group. Aliphatic amino acids with large side chains, such as leucine, isoleucine, and valine, are hydrophobic in nature. Their hydrophobicity forces them to sequester together away from water in the interior of proteins. The three-dimensional structure of proteins is highly dependent on the hydrophobic side chains of aliphatic amino acids forming the interior of proteins. In contrast to aliphatic amino acids, which have no ionizable side chains, basic amino acids have ionizable amino groups that are positively charged at neutral pH. These include lysine and arginine, which have a pK of pH 10 and pH 12, respectively, and histidine, with an ionizable imidazole ring and a pK of 6.5. The different characteristics of the side chains of amino acids are responsible for the different qualities of the proteins into which they are incorporated.

Which of the amino acids below is the uncharged derivative of an acidic amino acid? a. Cystine b. Arginine c. Tyrosine d. Glutamine e. Proline f. Serine g. Leucine

*The answer is A.* All α-amino acids have an asymmetric α carbon atom to which an α-carboxyl group, an α-amino group, and an α-side chain are attached. Levorotary (L) isomers of amino acids compose proteins in nature. Because glycine has a hydrogen as its side chain, with two hydrogens, an amino group, and a carboxyl group on the α-carbon, it is the only optically inactive amino acid. Side chains contribute the distinctive properties at physiological pH to each amino acid (and hence proteins), which include: basic (positive); acidic (negative); neutral polar; neutral nonpolar; sulfur-containing (thiol); hydroxyl-containing; aromatic; hydrophobic; hydrophilic; branched; or straight-chained.

Which of the characteristics below apply to the amino acid glycine? a. Optically inactive b. Large molecular diameter interfering with α helix formation c. Hydrophilic, basic, and charged d. Hydrophobic e. Hydrophilic, acidic, and charged

*The answer is E.* Pure metabolic acidosis (choice c) or pure metabolic alkalosis exhibits abnormal bicarbonate and normal lung function. Pure respiratory acidosis (choice d) or alkalosis (choice a) is associated with normal renal function (and normal blood acids) with a normal bicarbonate and abnormal PCO2. Thus choices b and e must involve compensation, since both the PCO2 and bicarbonate are abnormal. Choice e must represent compensated metabolic alkalosis since the PCO2 is high—if it were compensated respiratory acidosis with a high PCO2, the pH would be low.

Which of the combinations of laboratory results below indicates compensated metabolic alkalosis? a. Low PCO2, normal bicarbonate, high pH b. Low PCO2, low bicarbonate, low pH c. Normal PCO2, low bicarbonate, low pH d. High PCO2, normal bicarbonate, low pH e. High PCO2, high bicarbonate, high pH

*The answer B.* Sigmoidal control curve with ATP inhibiting and shifting curve to the right is needed.

Which of the diagrams illustrated below best represents the effect of ATP on hepatic phosphofructokinase-1 (PFK-1)?

*The answer is B.* Except for terminal amino acids, all α-amino groups and all α-carboxyl groups are utilized in peptide bonds. Thus only amino acids with side chains may be considered. Of these, 7 of the 20 common amino acids have easily ionizable side chains. These are the basic amino acids lysine, arginine, and histidine; the acidic amino acids aspartate and glutamate; and tyrosine and cysteine. Leucine, valine, and alanine have hydrocarbon side chains.

Which of the following amino acids is ionizable in proteins? a. Leucine b. Histidine c. Valine d. Alanine e. Glycine

*The answer is B.* The α-helical segments of proteins represent one of the most common secondary structures of proteins. The helical structure is composed of a spiraled polypeptide backbone core with the side chains of component amino acids extending outward from the central axis in order to avoid interfering sterically or electrostatically with each other. All the peptide bond carbonyl oxygens are hydrogen-bonded to a peptide linkage that is four residues ahead in the polypeptide. This leads to 3.6 amino acids per turn, spatially held together in the α-helical structure. Since the configuration of the α helix is compatible with being in the interior of proteins, amino acids with nonpolar, hydrophobic side chains predominate. Conversely, amino acids that are charged or have bulky side chains may interfere with the α-helical structure if present in large enough amounts. Proline and hydroxyproline are not at all compatible with the right-handed spiral of the α-helix. They insert a kink in the chain. Likewise, large numbers of charged amino acids such as lysine or histidine disrupt the helix by forming electrostatic bonds or by ionically repelling one another. In addition, amino acids with bulky side chains, such as tryptophan or isoleucine, also tend to disrupt the configuration of the α helix.

Which of the following amino acids is most compatible with an α-helical structure? a. Tryptophan b. Alanine c. Lysine d. Proline e. Cysteine

*The answer is E.* Nonregulatory enzymes, such as lactate dehydrogenase, typically exhibit a hyperbolic saturation curve when initial velocity is plotted against substrate concentration. Enzymes at key points in metabolic pathways are typically allosteric—their velocities at a given substrate concentration may be altered due to effects of metabolites in the pathway. Allosteric enzymes typically exhibit sigmoidal kinetics. Examples of allosteric enzymes include aspartate transcarbamoylase, which is inhibited by cytidine triphosphate (CTP); phosphofructokinase, which is inhibited by adenosine triphosphate (ATP) and activated by fructose 2,6 bisphosphate; hexokinase, which is inhibited by glucose-6-phosphate; and pyruvate kinase, which is inhibited by ATP. Allosteric enzymes produce sigmoidal kinetics when substrate concentration is plotted against reaction velocity. In contrast, hyperbolic plots are observed with Michaelis-Menten enzymes. The binding of effector molecules, such as end products or second messengers, to regulatory subunits of allosteric enzymes can either positively or negatively regulate catalytic subunits.

Which of the following enzymes exhibits a hyberbolic curve when initial reaction velocity is plotted against substrate concentration? a. Aspartate transcarbamoylase b. Phosphofructokinase c. Hexokinase d. Pyruvate kinase e. Lactate dehydrogenase

*The answer is D.* Mutations that alter the balance of α- and β-globin synthesis from their respective loci on chromosomes 16 and 11 produce thalassemias. Since these loci are autosomal, mutations at both homologous loci are required to produce severe thalassemia, as with the β thalassemias that involved altered RNA splicing at both β-globin loci. Methemoglobin is hemoglobin with the iron oxidized from the ferrous (Fe++) to the ferric (Fe+++) state. Methemoglobin cannot bind oxygen, so there is a specific enzyme (methemoglobin reductase) and reducing substances like glutathione in red cells that maintain hemoglobin iron in its reduced state. Point mutations that cause amino acid substitutions produce an abnormal hemoglobin rather than imbalance chain synthesis. Sickle cell anemia is one example, in which both β-globin chains have a valine replacing glutamine. The mutant β-globin chains in hemoglobin S have a "sticky patch" on their surface that is particularly adhesive when hemoglobin S is deoxygenated. For this reason, individuals with sickle cell anemia are prone to thrombotic crises (strokes, heart attacks, ischemic extremities) when they become dehydrated (increased hemoglobin concentration) or hypoxic (more deoxyhemoglobin S). There are also mutant hemoglobins (hemoglobin M, etc.) that predispose to oxidation of the iron group in heme, producing higher concentrations of methemoglobin and cyanosis (bluish color of the lips and fingertips).

Which of the following mutations would produce a severe thalassemia? a. Deletion of one α-globin locus b. Deletion of one β-globin locus c. Oxidation of heme groups to produce methemoglobin d. Altered RNA processing at both β-globin loci e. Sickle cell anemia

*The answer is E.* Cyclin B is targeted for degradation via ubiquitination.

Which of the following post-translational modifications is most likely to be found on a cyclin B protein that is targeted for degradation? (A) Acetylated lysine residues (B) Phosphorylated serine residues (C) Phosphorylated threonine residues (D) Phosphorylated tyrosine residues (E) Ubiquitinated lysine residues

*The answer is A.* At neutral pH, amino acids in solution are zwitterions (i.e., dipolar ions) containing both a protonated amino group (pK approximately 9.5) and a dissociated carboxyl group (pK approximately 2). At pH 7.4, the pH − pK from the Henderson-Hasselbalch equation is ∼5 for the carboxy group, predicting a ratio of base (carboxyl anion) to acid (carboxylic acid) of 10^5. Similarly, the pH − pK for the amino group is about −2, predicting a ratio of base (amino group) to acid (protonated ammonium ion) of less than 10−2. Amino acids with ionizable side chains may have charges in addition to those of the amino and carboxyl groups.

Which of the following statements about solutions of amino acids at physiologic pH is true? a. All amino acids contain both positive and negative charges b. All amino acids contain positively charged side chains c. Some amino acids contain only positive charges d. All amino acids contain negatively charged side chains e. Some amino acids contain only negative charges

*The answer is D.* Five classes of immunoglobulins are known: IgG, IgA, IgM, IgD, and IgE. The difference between each class is due to the variations in the constant chains from one class to another. The respective heavy chains corresponding to each class of immunoglobulin are γ and IgG; α and IgA; μ and IgM; δ and IgD; and ε and IgE. In contrast, the light chains are the same in each class: either κ or λ. The different biologic characteristics are due to the unique heavy chains. IgM is the first class of antibodies to be observed in the plasma following antigenic stimulation. IgG is the major antibody produced in serum at 10 days following antigenic stimulation. IgA acts against bacteria and viruses and is observed in external secretions such as mucus, tears, and saliva. While the undesirable effects of IgE in allergic reactions are known, its possible benefits are not understood. Likewise, the role of IgD is not known.

Which of the following statements concerning immunoglobulin is most accurate? a. The distinctiveness of the light chains gives the different classes of immunoglobulins their unique biologic characteristics b. IgE is the principal antibody in the serum c. The heavy chains are similar in each class of immunoglobulin d. The constant regions of the heavy chains are the same in each class of immunoglobulin e. IgE is the major immunoglobulin found in external secretions

*The answer is B.* There are two antigen-binding sites per antibody molecule, each defined by the N-terminals of the light and heavy chain from one subunit. Each of the two subunits have one light chain and one heavy chain, and all are held together by intra- or interchain disulfide linkages. The N-terminal peptides of heavy and light chains comprise the variable regions that recognize many different antigens, whereas the C-terminal peptides comprise the constant region that triggers responses to antigenantibody complexes. Antigen-binding sites are determined prior to encounter of specific antigens. A large repertoire of virgin B cells are produced after differentiation from stem and pre-B cells. Each virgin B cell carries a unique immunoglobulin M (IgM) molecule on its surface. The surface IgM is determined by one DNA rearrangement of variable-joiningconstant (V-J C) segments to produce a unique light chain and two DNA rearrangements of variable-diversity-joining-constant (V-D-J C) segments to produce a unique heavy chain. Those unique virgin B cells that do not encounter antigens die, while those that do encounter antigens become activated as plasma (immunoglobulin-producing) or memory B (immunity) cells. A spectrum of cells with specific antibodies are thus made before antigen is contacted, allowing antigen experience to guide the clonal selection process and determine immunoglobulin production/immune status. Note that some mature antibodies have multiple subunits, like the five subunits in IgM. Mature IgM antibodies thus have 10 binding sites in total.

Which of the following statements correctly describes immunoglobulins? a. Polypeptide chains composing immunoglobulins are held together by hydrogen bonds b. Each immunoglobulin has two antigen-binding sites per molecule c. Heavy immunoglobulin chains have constant N-terminal regions d. Light immunoglobulin chains have variable C-terminal regions e. Antigen-binding sites of all different antibodies are determined after encountering specific antigens

*The answer is B.* Each of the techniques listed separates proteins from each other and from other biologic molecules based upon characteristics such as size, solubility, and charge. However, only affinity chromatography can use the high affinity of proteins for specific chemical groups or the specificity of immobilized antibodies for unique proteins. In affinity chromatography, a specific compound that binds to the desired protein—such as an antibody, a polypeptide receptor, or a substrate—is covalently bound to the column material. A mixture of proteins is added to the column under conditions ideal for binding the protein desired, and the column is then washed with buffer to remove unbound proteins. The protein is eluted either by adding a high concentration of the original binding material or by making the conditions unfavorable for binding (e.g., changing the pH). The other techniques are less specific than affinity binding for isolating proteins. Dialysis separates large proteins from small molecules. Ion exchange chromatography separates proteins with an overall charge of one sort from proteins with an opposite charge (e.g., negative from positive). Gel filtration chromatography separates on the basis of size. Electrophoresis separates proteins on the principle that net charge influences the rate of migration in an electric field.

Which of the following techniques for purification of proteins can be made specific for a given protein? a. Dialysis b. Affinity chromatography c. Gel filtration chromatography d. Ion exchange chromatography e. Electrophoresis

*The answer is E.* Adult hemoglobin, or hemoglobin A, is composed of four polypeptide chains. Two of the chains are α chains and two are β chains. The chains are held together by noncovalent interactions. The hemoglobin tetramer can best be represented as being composed of two dimers, each containing the two different polypeptides. Thus the designation (α1-β1) (α2-β2), which refer to dimers 1 and 2, respectively, is the most correct way to refer to the quaternary structure of adult hemoglobin. Hydrophobic interactions are thought to be the main noncovalent interactions holding all four polypeptides together.

Which of the hemoglobin designations below best describes the relationship of subunits in the quaternary structure of adult hemoglobin? a. (α1-α2)(β1-β2) b. α1-α2-α3-α4 c. β-β-β-α d. (β1-β2-β3-α1) e. (α1-β1)-(α2-β2)

*The answer is E.* The major macromolecular components of ground substance are proteoglycans, which are made up of polysaccharide chains attached to core proteins. The polysaccharide chains are made up of repeats of negatively charged disaccharide units. This polyanionic quality of proteoglycans allows them to bind water and cations and thus determines the viscoelastic properties of connective tissues. Collagen is the other major component of connective tissue besides ground substance. The cornified layer of epidermis derives its toughness and waterproof nature from keratin. Keratins are disulfide-rich proteins that compose the cytoskeletal elements known as intermediate filaments. Hair and animal horns are also composed of keratin. Troponin is a component of muscle and fibrin of blood clots.

Which of the substances below is primarily found in ground substance (extracellular matrix)? a. Collagen b. Troponin c. Keratin d. Fibrin e. Proteoglycan

*The answer is A.* Collagens are insoluble proteins that have great tensile strength. They are the main fibers composing the connective tissue elements of skin, bone, teeth, tendons, and cartilage. Collagen is composed of tropocollagen, a triple stranded helical rod rich in glycine, proline, and hydroxyproline residues. Troponin is found in muscle, fibrillin in heart valves, blood vessels, and ligaments (it is defective in Marfan's syndrome). Fibrin is a component of blood clots and fibronectin is a component of extracellular matrix.

Which of the substances below is primarily found in tendons? a. Collagen b. Troponin c. Fibrillin d. Fibrin e. Fibronectin

*The answer is E.* Aspartate transcarbamoylase, which controls the rate of pyrimidine synthesis in mammals, is negatively inhibited by the allosteric effector cytidine triphosphate, an end product of pyrimidine synthesis. The allosteric modulation occurs via the binding of effectors at the regulatory site of the enzyme. Noncovalent bonds are formed during the binding between effector and enzyme. In contrast, all the other enzymes are activated or deactivated by covalent modification. Chymotrypsinogen is secreted as an inactive proenzyme (zymogen) in pancreatic juice and is irreversibly activated by trypsin cleavage of a specific peptide bond. Glycogen phosphorylase is reversibly activated by phosphorylation of a specific serine residue. At the same time, glycogen synthase is reversibly deactivated by phosphorylation of a specific serine residue, thereby preventing a futile cycle of breakdown and resynthesis of glycogen. Pyruvate dehydrogenase also is reversibly inactivated by phosphorylation of a specific serine residue. In all four enzymes, a single, discrete, covalent modification leads to conformational changes that allow the switching on or off of enzyme activity.

Which one of the following enzymes is regulated primarily through allosteric interaction? a. Chymotrypsin b. Pyruvate dehydrogenase c. Glycogen phosphorylase d. Glycogen synthase e. Aspartate transcarbamoylase

*The answer is B.* Two kinds of interacting protein filaments are found in skeletal muscle. Thick filaments 15 nm in diameter contain primarily myosin. Thin filaments 7 nm in diameter are composed of actin, troponin, and tropomyosin. The thick and thin filaments slide past one another during muscle contraction. Myosin is an ATPase that binds to thin filaments during contraction. α-actinin can be found in the Z line.

Which one of the following proteins is found in the thick filaments of skeletal muscle? a. α-actinin b. Myosin c. Troponin d. Tropomyosin e. Actin

*The answer is D.* The two cysteine residues can, under oxidizing conditions, form a disulfide bond. Glutamine's 3-letter abbreviation is Gln. Proline (Pro) contains a secondary amino group. Only one (Arg) of the seven would have a positively charged side chain at pH 7.

Which one of the following statements concerning the peptide shown below is correct? Gly-Cys-Glu-Ser-Asp-Arg-Cys A. The peptide contains glutamine. B. The peptide contains a side chain with a secondary amino group. C. The peptide contains a majority of amino acids with side chains that would be positively charged at pH 7. D. The peptide is able to form an internal disulfide bond.

*The answer is A.* The binding of an effector to the regulatory subunit of an allosteric enzyme causes a conformational change that either increases or decreases the activity of the enzyme's separate catalytic site. Only in some allosteric molecules, such as hemoglobin, does positive cooperativity occur. A positive effector increases substrate binding. This is the case with cyclic AMP-dependent protein kinase of the glycogen phosphorylase cascade. Cyclic AMP binds the regulatory subunit that dissociates from the catalytic subunit and thereby activates it. In the absence of cyclic AMP, the regulatory subunit tightly binds the catalytic subunit and inactivates the enzymes. Many allosteric enzymes are often placed at the first, or committed, step of a metabolic pathway. The end product of the pathway then acts as a negative effector of the enzyme. This is called feedback inhibition. An allosteric enzyme does not obey Michaelis-Menten kinetics.

Which one of the following statements correctly describes allosteric enzymes? a. Effectors may enhance or inhibit substrate binding b. They are not usually controlled by feedback inhibition c. The regulatory site may be the catalytic site d. Michaelis-Menten kinetics describe their activity e. Positive cooperativity occurs in all allosteric molecules except hemoglobin

*The answer is E.* Unlike Michaelis-Menten enzymes, allosteric enzymes exhibit sigmoidal plots when reaction velocity is plotted against substrate concentrations. The enzyme contains both a catalytic site and a regulatory site. The binding of regulatory molecules to the regulatory site alters enzyme activity. The binding of one substrate molecule can affect the binding of substrate to other catalytic sites.

Which one of the following statements correctly describes allosteric enzymes? a. Regulatory molecules bind the active site b. Regulatory molecules alter equilibrium but not activity c. Regulatory molecules do not affect activity or equilibrium d. Hyperbolic plots are obtained when reaction velocity is plotted against substrate concentration e. Binding of substrate to one site can affect other sites

*The answer is D.* Hemoglobin is a tetrameric hemoprotein whose oxygen saturation curves exhibit sigmoidal kinetics because of cooperative interactions among the four binding sites. Oxygen is bound to hemoglobin without changing the redox state of the iron from the ferrous state. Carbon monoxide and cyanide both bind to hemoglobin more tightly than does oxygen itself. O2 is released to tissues and exchanged with CO2 since increased CO2 levels in capillaries lead to decreased affinity of hemoglobin for O2.

Which one of the following statements correctly describes transport of O2 by hemoglobin? a. O2 binds to hemoglobin more avidly than does CO b. The binding of O2 to hemoglobin causes a valence change in the iron of the heme moiety c. Each of the four heme moieties binds O2 independently d. The plot of percentage of O2 bound versus O2 pressure is sigmoidal in shape e. Increased CO2 concentrations increase O2 affinity

*The answer is A.* The carbon next to a carboxyl (C=O) group may be designated as the α carbon, with subsequent carbons as β, γ, δ, etc. α-amino acids contain an amino group on their α carbon, as distinguished from compounds like γ-aminobutyric acid, in which the amino group is two carbons down (γ-carbon). In α-amino acids the amino acid, carboxylic acid, and the side chain or R group are all bound to the central α-carbon, which is thus asymmetric (except when R is hydrogen, as for glycine). Amino acids are classified as acidic, neutral hydrophobic, neutral hydrophilic, or basic, depending on the charge or partial charge on the R group at pH 7.0. Hydrophobic (water-hating) groups are carbon-hydrogen chains like those of leucine, isoleucine, glycine, or valine. Basic R groups, such as those of lysine and arginine, carry a positive charge at physiologic pH owing to protonated amide groups, while acidic R groups, such as glutamic acid, carry a negative charge owing to ionized carboxyl groups. Threonine with its hydroxyl side chain is neutral at physiologic pH. Neutral hydrophilic side chains have uncharged but polar or partially charged groups.

Which one of the following structures may be classified as a hydrophobic amino acid at pH 7.0? a. Isoleucine b. Arginine c. Aspartic acid d. Lysine e. Threonine

*The answer is A.* The patient is exhibiting the symptoms of primary amyloidosis, which is a protein folding disease in which immunoglobulin light chains are improperly processed and cannot be degraded. These proteins then form fibrils in tissues, which are insoluble. This disrupts the normal function of the tissue, and many tissues can accumulate these fibrils. Primary amyloidosis does not occur with abnormal deposits of collagen, fibrillin, albumin, or serum transaminases.

You are visited by a 40-year-old female patient complaining of weight loss, numbness in the hands and feet, fatigue, and difficulty swallowing. Physical exam notes an enlarged tongue, enlarged liver, a rubbery feeling around the joints, and bruising around the eyes. A bone marrow biopsy shows an abnormal staining of denatured protein (see below). These denatured proteins are most likely to be which of the following? (A) Antibody light chains (B) Collagen (C) Fibrillin (D) Albumin (E) Transaminases

*The answer is C.* HbA1c is glycosylated hemoglobin, reflecting the level of blood glucose over the lifetime of the erythrocyte (120 days). The higher the concentration of HbA1c, the more poorly controlled blood glucose levels are (normal is about 5.5% HbA1c). The glycosylation primarily occurs on the N-terminal valine residues of the β chains (which contain a free amino group). The amino acid sequences of hemoglobin and HbA1c are the same, there is no fatty acid addition (acylation) to the hemoglobin, the red cell contains no intracellular organelles for compartmentation to be an issue, and the rate of degradation of nonmodified hemoglobin and HbA1c are the same.

Your diabetic patient has a hemoglobin A1c (HbA1c) of 8.8. HbA1c differs from unmodified hemoglobin by which one of the following? (A) Amino acid sequence (B) Serine acylation (C) Valine glycosylation (D) Intracellular location (E) Rate of degradation

*The answer is A.* Phospholipids contain a very hydrophobic backbone and a "head group" that is primarily hydrophilic. The hydrophobic portion of the phospholipid remains embedded in the membrane while the hydrophilic head group faces the aqueous environment of the cell. The glycerol portion (or ceramide portion, which contains sphingosine) of the phospholipid, as well as the fatty acids, remains embedded in the membrane while only the head group (R) faces the aqueous environment. Thus, when a phospholipid flip-flops across the membrane, the head group will always end up facing the aqueous environment.

Your patient has a mechanical heart valve and is chronically anemic due to damage to red blood cells as they pass through this valve. One of the signals that target damaged red blood cells for removal from the circulation is the presence of phosphatidylserine in the outer leaflet of the red cell membrane. Phosphatidylserine is an integral part of cell membranes and is normally found in the inner leaflet of the red cell membrane. This flip-fl op of phosphatidylserine between membrane leaflets exposes which part of the phosphatidylserine to the environment? (A) The head group (B) Fatty acids (C) Sphingosine (D) Glycerol (E) Ceramide

*The answer is B.* Substitution of lysine by methionine decreases the ability of negatively charged phosphate groups in 2,3-BPG to bind the β subunits of hemoglobin. Because 2,3BPG decreases the O2 affinity of hemoglobin, a reduction in 2,3-BPG should result in increased O2 affinity and decreased delivery of O2 to tissues. The R form is the high-oxygen-affinity form of hemoglobin. Increased O2 affinity (decreased delivery) results in a left shift in the O2 dissociation curve. Decreased O2 delivery is compensated for by increased RBC production.

β-Lysine 82 in hemoglobin A is important for the binding of 2,3-BPG. In Hb Helsinki, this amino acid has been replaced by methionine. Which of the following should be true concerning Hb Helsinki? A. It should be stabilized in the T, rather than the R, form. B. It should have increased O2 affinity and, consequently, decreased delivery of O2 to tissues. C. Its O2 dissociation curve should be shifted to the right relative to Hb A. D. It results in anemia