colloquim 1- chapter summaries

Question 1

Sickle cell anemia, a sickling disease of red blood cells, results from the substitution of polar glutamate by nonpolar valine at the sixth position
in the β subunit of hemoglobin

Answer 1

Many extracellular proteins are stabilized by disulfide bonds. Albumin, a blood protein that functions as a transporter for a variety of molecules, is an example.

Question 2

Separation of plasma proteins by charge typically is done at a pH above the pI of the major proteins, thus, the charge on the proteins is negative.
In an electric field, the proteins will move toward the positive electrode at a rate determined by their net negative charge. Variations in the mobility pattern are suggestive of certain diseases.

Answer 2

Each amino acid has an α-carboxyl group and a primary α-amino
group (except for proline, which has a secondary amino group). At
physiologic pH, the α-carboxyl group is dissociated, forming the nega-
tively charged carboxylate ion (– COO–), and the α-amino group is protonated (– NH3+). Each amino acid also contains one of 20 distinctive side chains attached to the α-carbon atom. The chemical nature of this side chain determines the function of an amino acid in a protein, and provides the basis for classification of the amino acids as nonpolar, uncharged polar, acidic, or basic. All free amino acids, plus charged amino acids in peptide chains, can serve as buffers. The quantitative relationship between the pH of a solution and the concentration of a weak acid (HA) and its con jugate base (A–) is described by the Henderson-Hasselbalch equation. Buffering occurs within ±1pH unit of the pKa, and is maximal when pH = pKa, at which [A–] = [HA]. The α-carbon of each amino
acid (except glycine) is attached to four different chemical groups and is, therefore, a chiral or optically active carbon atom. Only the L-form of amino acids is found in proteins synthesized by the human body.

Question 3

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?

C. Point C represents the region where the net charge on glycine is zero. Correct answer = 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.

Answer 3

Which one of the following statements concerning the
peptide shown below is correct?
Gly-Cys-Glu-Ser-Asp-Arg-Cys

D. The peptide is able to form an internal disulfide bond. Correct answer = 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.

Question 4

Given that the pI for glycine is 6.1, to which electrode, positive or negative, will glycine move in an electric field at pH 2? Explain.

Correct answer = negative electrode. When the pH is less than the pI, the charge on glycine is positive because the α-amino group is fully pro-
tonated. (Recall that glycine has H as its R group).

Answer 4

The α-helix and β-sheet structures provide maximal hydrogen bonding for peptide bond components within the interior of polypeptides.

Proteins that bind to DNA contain a limited number of motifs. The helix-loop-helix motif is an example found in a number of proteins that function as transcription factors

Isoforms are proteins that perform the same function but have different primary structures. They can arise from different genes or from tissue-specific processing of the product of a single gene. If the proteins function as enzymes, they
are referred to as isozymes

Question 5

Central to understanding protein structure is the concept of the native
conformation (Figure 2.15), which is the functional, fully-folded protein
structure (for example, an active enzyme or structural protein). The
unique three-dimensional structure of the native conformation is deter-
mined by its primary structure, that is, its amino acid sequence.
Interactions between the amino acid side chains guide the folding of the
polypeptide chain to form secondary, tertiary, and (sometimes)
quaternary structures, which cooperate in stabilizing the native confor-
mation of the protein. In addition, a specialized group of proteins named
“chaperones” is required for the proper folding of many species of pro-
teins. Protein denaturation results in the unfolding and disorganization
of the protein’s structure, which are not accompanied by hydrolysis of
peptide bonds. Denaturation may be reversible or, more commonly, irre-
versible. Disease can occur when an apparently normal protein
assumes a conformation that is cytotoxic, as in the case of Alzheimer
disease and the transmissible spongiform encephalopathies (TSEs),
including Creutzfeldt-Jakob disease. In Alzheimer disease, normal pro-
teins, after abnormal chemical processing, take on a unique conforma-
tional state that leads to the formation of neurotoxic amyloid protein
assemblies consisting of β-pleated sheets. In TSEs, the infective agent
is an altered version of a normal prion protein that acts as a “template”
for converting normal protein to the pathogenic conformation

Answer 5

A peptide bond:
A. has a partial double-bond character.
Correct answer = A. The peptide bond has a partial double-bond character. Unlike its components—the α-amino and α-carboxyl groups—the–NH and –C=O of the peptide bond do not accept or give off protons. The peptide bond is
not cleaved by organic solvents or urea, but is labile to strong acids. It is usually in the trans configuration

Question 6

Which one of the following statements is correct?

C. β-Bends often contain proline. Correct answer = C. β-Bends often contain pro-
line, which provides a kink. The α-helix differs from the β-sheet in that it always involves the coiling of a single polypeptide chain. The β-sheet occurs in both parallel and antiparallel forms. Domains are elements of tertiary struc-
ture. The α-helix is stabilized primarily by hydrogen bonds between the –C =O and –NH– groups of peptide bonds.

Answer 6

Which one of the following statements about protein structure is correct?

E. The information required for the correct folding of a protein is contained in the specific sequence of amino acids along the polypeptide chain. Correct answer = E. The correct folding of a protein is guided by specific interactions between the
side chains of the amino acid residues of a polypeptide chain. The two cysteine residues that react to form the disulfide bond may be a great distance apart in the primary structure (or on separate polypeptides), but are brought into close
proximity by the three-dimensional folding of the polypeptide chain. Denaturation may either be reversible or irreversible. Quaternary structure requires more than one polypeptide chain. These chains associate through noncovalent interactions.

Question 7

An 80-year-old man presented with impairment of higher intellectual function and alterations in mood and behavior. His family reported progressive disorientation
and memory loss over the last 6 months. There is no family history of dementia. The patient was tentatively diagnosed with Alzheimer disease. Which one of the
following best describes the disease?

D. It is associated with the deposition of neurotoxic amyloid peptide aggregates.Correct answer = D. Alzheimer disease is associated with long, fibrillar protein assemblies consisting of β-pleated sheets found in the brain
and elsewhere. The disease is associated with abnormal processing of a normal protein. The accumulated altered protein occurs in a β-pleated sheet configuration that is neurotoxic. The Aβ amyloid that is deposited in the brain in
Alzheimer disease is derived by proteolytic cleavages from the larger amyloid precursor protein—a single transmembrane protein expressed on the cell surface in the brain and other tissues. Most cases of Alzheimer disease are sporadic, although at least 5–10% of cases are familial. Prion diseases, such as Creutzfeldt- Jakob, are caused by the infectious form (PrPSc )of a host-cell protein (PrPC).

Answer 7

Obtaining O2 from the atmosphere solely by diffusion greatly limits the size of organisms. Circulatory systems overcome this, but transport molecules such as hemoglobin are also required because O2 is only slightly soluble in aqueous
solutions such as blood.

Question 8

Hemoglobin A, the major hemoglobin in adults, is composed of four
polypeptide chains (two α chains and two β chains, α2β2) held together
by noncovalent interactions (Figure 3.24). The subunits occupy different
relative positions in deoxyhemoglobin compared with oxyhemoglobin.
The deoxy form of hemoglobin is called the “T,” or taut (tense) form. It
has a constrained structure that limits the movement of the polypeptide
chains. The T form is the low-oxygen-affinity form of hemoglobin. The
binding of oxygen to hemoglobin causes rupture of some of the ionic and
hydrogen bonds. This leads to a structure called the “R,” or relaxed form,
in which the polypeptide chains have more freedom of movement. The R
form is the high-oxygen-affinity form of hemoglobin. The oxygen
dissociation curve for hemo globin is sigmoidal in shape (in contrast to
that of myoglobin, which is hyperbolic), indicating that the subunits
cooperate in binding oxygen. Cooperative binding of oxygen by the four
subunits of hemoglobin means that the binding of an oxygen molecule at
one heme group increases the oxygen affinity of the remaining heme
groups in the same hemoglobin molecule. Hemoglobin’s ability to bindoxygen reversibly is affected by the pO2 (through heme-heme interactions), the pH of the environment, the pCO2, and the availability of 2,3-bisphosphoglycerate (2,3-BPG). For example, the release of O2 from Hb is enhanced when the pH is lowered or the pCO2 is increased (the Bohr effect), such as in exercising muscle, and the oxygen dissociation curve of Hb is shifted to the right. To cope long-term with the effects of chronic hypoxia or anemia, the concentration of 2,3-BPG in RBCs increases. 2,3-BPG binds to the Hb and decreases its oxygen affinity, and it, therefore, also shifts the oxygen-dissociation curve to the right. Carbon monoxide (CO) binds tightly (but reversibly) to the hemoglobin iron, forming carbon monoxy hemoglobin (Hb CO). Hemoglobinopathies are disorders caused either by
production of a structurally abnormal hemoglobin molecule, synthesis of insufficient quantities of normal hemoglobin subunits, or, rarely, both (Figure 3.25). The sickling diseases sickle cell anemia (Hb S disease) and hemoglobin SC disease, as well as hemoglobin C disease and the thalassemia syndromes are representative hemoglobinopathies that can have severe clinical consequences.

Answer 8

Which one of the following statements concerning the hemoglobins is correct?

A. Fetal blood has a higher affinity for oxygen than does adult blood because Hb F has a decreased affinity for 2,3-BPG. Correct answer = A. Because 2,3-BPG reduces the affinity of hemoglobin for oxygen, the weaker interaction between 2,3-BPG and Hb F results in a higher oxygen affinity for Hb F rela-
tive to Hb A. In contrast, if both Hb A and Hb F are stripped of 2,3-BPG, they have a similar affinity for oxygen. Hb F consists of α2γ2. Hb A1c is a glycosylated form of Hb A, formed nonenzymically in red cells. Hb A2 is a minor component of normal adult hemoglobin, first appearing shortly before birth and rising to adult levels (about 2% of the total hemoglobin) by 6 months of age.

Question 9

Which one of the following statements concerning the ability of acidosis to precipitate a crisis in sickle cell anemia is correct?

A. Acidosis decreases the solubility of Hb S. Correct answer = A. Hb S is significantly less soluble in the deoxygenated form, compared with oxyhemoglobin S. A decrease in pH (acido- sis) causes the oxygen dissociation curve to shift to the right, indicating a decreased affinity
for oxygen. This favors the formation of the deoxy, or taut, form of hemoglobin, and can precipitate a sickle cell crisis. The binding of 2,3- BPG is increased, because it binds only to the deoxy form of hemoglobins.

Answer 9

Which one of the following statements concerning the binding of oxygen by hemoglobin is correct?

C. The oxygen affinity of hemoglobin increases as the percentage saturation increases. Correct answer = C. The binding of oxygen at one heme group increases the oxygen affinity of the remaining heme groups in the same
molecule. Carbon dioxide decreases oxygen affinity because it lowers the pH; moreover, binding of carbon dioxide to the N-termini stabilizes the taut, deoxy form. Hemoglobin binds one molecule of 2,3-BPG. Deoxyhemoglobin has a greater affinity for protons and, therefore, is a weaker acid.

Question 10

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?

B. It should have increased O2 affinity and, consequently, decreased delivery of O2 to tissues.Correct 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.

Answer 10

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?

C. Methemoglobinemia. Correct answer = 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

Question 11

Basement membranes are thin, sheet-like structures that provide mechanical support for adjacent cells, and function as a semipermeable filtration barrier to macromolecules in organs such as the kidney and the lung

Lysyl oxidase is one of several copper-containing enzymes. Others include cytochrome oxidase (see p. 76), dopamine hydroxylase (see p. 286),
superoxide dismutase (see p.148) and tyrosinase (see p. 273). Disruption in copper homeostasis causes copper deficiency (X-linked Menkes dis-
ease) or overload (Wilson disease).

Answer 11

Collagen and elastin are fibrous proteins (Figure 4.15). Collagen molecules contain an abundance of proline, lysine, and glycine, the latter occurring at every third position in the primary structure. Collagen also contains hydroxyproline, hydroxylysine, and glycosylated hydroxylysine, each formed by posttranslational modification. Collagen molecules typically form fibrils containing a long, stiff, triple-stranded helical structure, in which three collagen polypeptide chains are wound around one another in a rope-like superhelix (triple helix). Other types of
collagen form mesh-like networks. Elastin is a connective tissue protein with rubber-like properties in tissues such as the lung. α1-Antitrypsin (α1-AT), produced primarily by the liver but also by tissues such as monocytes and alveolar macrophages, prevents elastin degradation in the alveolar walls. A deficiency of α1-AT can cause emphysema and, in some cases, cirrhosis of the liver.

Question 12

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 pul-monary symptoms?

B. Deficiency of α1-antitrypsin. Correct answer = 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 (deficiency of proline hydroxylase, dietary vitamin c, increased collagenase activity) refer to collagen, not elastin.

Answer 12

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. Correct answer = 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.

Question 13

What is the differential basis of the liver and lung pathology seen in α1-AT deficiency?

With α1-AT deficiency, the liver cirrhosis that can result is due to polymerization and retention of α1-AT in the liver, its site of synthesis. The lung pathology is due to this retention-based deficiency in α1-AT ( a serin protease inhibitor or ser-
pin) such that elastase (a serine protease) is unopposed

Answer 13

Potentially confusing enzyme nomenclature: synthetase (requires ATP), synthase (no ATP required); phosphatase (uses water to remove phosphoryl group), phosphorylase (uses Pi to break a bond and generate a phosphorylated
product); dehydrogenase (NAD+/FAD is electron acceptor in redox reaction), oxidase (O2 is acceptor but oxygen atoms are not incorporated into substrate), oxygenase (one or both oxygens atoms are incorporated).

Question 14

The optimum temperature for most human enzymes is between 35 and 40°C. Human enzymes start to denature at temperatures above 40°C, but thermophilic bacteria found in the hot springs have optimum temperatures of 70°C.

Plasma is the fluid, noncellular part of blood. Laboratory assays of enzyme activity most often use serum, which is obtained by centrifugation of whole blood after it has been allowed to coagulate. Plasma is a physiologic fluid, whereas
serum is prepared in the laboratory.

Answer 14

Enzymes are protein catalysts that increase the velocity of a chemical
reaction by lowering the energy of the transition state (Figure 5.23).
Enzymes are not consumed during the reaction they catalyze. Enzyme
molecules contain a special pocket or cleft called the active site. The
active site contains amino acid side chains that participate in substrate
binding and catalysis. The active site binds the substrate, forming an
enzyme–substrate (ES) complex. Binding is thought to cause a confor-
mational change in the enzyme (induced fit) that allows catalysis. ES is
converted to enzyme-product (EP), which subsequently dissociates to
enzyme and product. An enzyme allows a reaction to proceed rapidly
under conditions prevailing in the cell by providing an alternate reac-
tion pathway with a lower free energy of activation. The enzyme does
not change the free energies of the reactants or products and, there-
fore, does not change the equilibrium of the reaction. Most enzymes
show Michaelis-Menten kinetics, and a plot of the initial reaction
velocity (vo) against substrate concentration ([S]) has a hyperbolic
shape similar to the oxygen dissociation curve of myoglobin. Any sub-
stance that can diminish the velocity of such enzyme-catalyzed reac-
tions is called an inhibitor. The two most commonly encountered types
of reversible inhibition are competitive (which increases the apparent
Km) and noncompetitive (which decreases the apparent Vmax). In con-
trast, the multi subunit allosteric enzymes frequently show a sigmoidal
curve similar in shape to the oxygen dissociation curve of hemoglobin.
They typically catalyze the committed step (often the rate-limiting or
slowest step) of a pathway. Allosteric enzymes are regulated by
molecules called effectors (also modifiers) that bind noncovalently at a
site other than the active site. Effectors can be either positive (acceler-
ate the enzyme-catalyzed reaction) or negative (slow down the reac-
tion). An allosteric effector can alter the affinity of the enzyme for its
substrate, or modify the maximal catalytic activity of the enzyme, or
both. Enzymes can also be regulated by covalent modification, and by
changes in the rate of synthesis or degradation. Enzymes have diag-
nostic and therapeutic value in medicine.

Question 15

in cases of ethylene glycol poisoning and its characteristic metabolic acidosis, treatment involves correction of the acidosis, removal of any remaining
ethylene glycol, and administration of an inhibitor of alcohol dehydrogenase (ADH, alcohol:NAD+ oxidoreductase), the enzyme that oxidizes ethylene glycol to the organic acids that cause the acidosis. Ethanol (grain alcohol) frequently is the inhibitor given to treat ethylene glycol poisoning; it works by competitively
inhibiting ADH. As a competitive inhibitor, ethanol:

A. increases apparent Km without affecting Vmax. Correct answer = A. In the presence of a competitive inhibitor, an enzyme appears to have a lower affinity for substrate, but as the substrate level is increased, the observed velocity
approaches Vmax. (See panel B of Figures 5.12 and 5.14 to compare effects of competitive and noncompetitive inhibitors.)

Answer 15

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):

B. coenzyme-cosubstrate. Correct answer = 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.

Question 16

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 (Figure 5.24). 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?

D. The patient had damage to his skeletal muscle on day 2. Correct answer = D. The CK isoenzyme pattern at admission showed elevated MB isozyme, indi-
cating 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.

Answer 16

Incomplete reduction of oxygen to water produces reactive oxygen species (ROS), such as superoxide (O2–•), hydrogen peroxide (H2O2) and hydroxyl radi-
cals (OH•). ROS damage DNA and proteins, and cause lipid peroxidation. Enzymes such as superoxide dismutase (SOD), catalase, and glutathione per-
oxidase are cellular defenses against ROS.

Question 17

The change in free energy (ΔG) occurring during a reaction predicts
the direction in which that reaction will spontaneously proceed. If ΔG is
negative (that is, the product has a lower free energy than the sub-
strate), the reaction goes spontaneously. If ΔG is positive, the reac-
tion does not go spontaneously. If ΔG = 0, the reactions are in
equilibrium. The ΔG of the forward reaction (A → B) is equal in magni-
tude but opposite in sign to that of the back reaction (B → A). The ΔGs
are additive in any sequence of consecutive reactions, as are the stan-
dard free energy changes (ΔGos). Therefore, reactions or processes
that have a large, positive ΔG are made possible by coupling with
cleavage of adenosine triphosphate (ATP), which has a large, nega-
tive ΔG. The reduced coenzymes NADH and FADH2 each donate a pair
of electrons to a specialized set of electron carriers, consisting of FMN,coenzyme Q, and a series of cytochromes, collectively called the electron transport chain. This pathway is present in the inner mitochondrial membrane, and is the final common pathway by which electrons derived from different fuels of the body flow to oxygen, reducing it to water. The terminal cyto chrome, cytochrome oxidase, is the only cytochrome able to bind oxygen. Electron transport is coupled to the transport of protons (H+) across the inner mitochondrial membrane from the matrix to the intermembrane space. This process creates electrical and
pH gradients across the inner mitochondrial membrane. After protons have been transferred to the cytosolic side of the inner mitochondrial membrane, they reenter the mitochondrial matrix by passing through the Fo channel in
ATP synthase (Complex V), dissipating the pH and electrical gradients and causing conformational changes in F1 that result in the synthesis of ATP from ADP + Pi. Electron transport and phosphorylation are thus said to be
tightly coupled (Figure 6.17). Inhibition of one process inhibits the other. These processes can be uncoupled by uncoupling proteins found in the inner mitochondrial membrane, and by synthetic compounds such as 2,4-dini-
trophenol and aspirin, all of which increase the permeability of the inner mitochondrial membrane to protons. The energy produced by the transport of electrons is released as heat rather than being used to synthesize ATP. Mutations in mitochondrial DNA (mtDNA) are responsible for some cases of mitochondrial diseases, such as Leber hereditary optic neuropathy. The release of cytochrome c into the cytoplasm and subsequent activation of proteolytic caspases results in apoptotic cell death.

Answer 17

A muscle biopsy specimen from a patient with a rare disorder, Luft disease, showed abnormally large mitochondria that contained packed cristae when examined in the electron microscope. Basal ATPase activity of the mitochondria was seven times greater than normal. From these and other data it was concluded that oxidation and phosphorylation were partially uncou-
pled. Which of the following statements about this
patient is correct?

E. The patient shows hypermetabolism and elevated core temperature.Correct answer = E. When phosphorylation is partially uncoupled from electron flow, one
would expect a decrease in the proton gradient across the inner mitochondrial membrane and, hence, impaired ATP synthesis. In an attempt to compensate for this defect in energy capture, metabolism and electron flow to oxygen is
increased. This hypermetabolism will be accompanied by elevated body temperature because the energy in fuels is largely wasted, appearing as heat. The electron transport chain will still be inhibited by cyanide.

Question 18

Explain why and how the malate-aspartate shuttle moves NADH reducing equivalents from the cytosol to the mitochondrial matrix.

There is no transporter for NADH in the inner mitochondrial membrane. However, NADH can be oxidized to NAD+ by the cytoplasmic isozyme of malate dehydrogenase as oxaloacetate is reduced to malate. The malate is
transported across the inner membrane, and the mitochondrial isozyme of malate dehydrogenase oxidizes it to oxaloacetate as mitochondrial NAD+ is reduced to NADH. This NADH can be oxidized by Complex I of the electron transport chain, generating three ATP through the coupled processes of respiration
and oxidative phosphorylation.

Answer 18

CO binds to and inhibits Complex IV of the electron transport chain. What effect, if any, should this respiratory inhibitor have on oxidative phosphorylation?

Inhibition of the electron transport chain by respiratory inhibitors such as CO results in an inability to maintain the proton gradient. Oxidative phosphorylation is therefore inhibited, as are ancillary reactions, because they also require the proton gradient.

Question 19

A colorimetric test can detect a reducing sugar in urine. A positive result is indicative of an underlying pathology because sugars are not normally
present in urine, and can be followed up by more specific tests to identify the reducing sugar.

Sucrase and isomaltase are enzymic activities of a single protein which is cleaved into two functional subunits that remain associated in the cell membrane, forming the sucrase-isomaltase complex. Maltase forms a similar complex with an
exoglucosidase (glucoamylase) that cleaves α(1→4) glycosidic bonds in dextrins.

Answer 19

Monosaccharides (simple sugars, Figure 7.12) containing an alde- hyde group are called aldoses and those with a keto group are called ketoses. Disaccharides, oligosaccharides, and polysaccharides consist of monosaccharides linked by glycosidic bonds. Compounds with the same chemical formula are called isomers. If two monosaccharide isomers differ in configuration around one specific carbon atom (with the exception of the carbonyl carbon), they are defined as epimers of
each other. If a pair of sugars are mirror images (enantiomers), the
two members of the pair are designated as D- and L-sugars. When a sugar cyclizes, an anomeric carbon is created from the aldehyde group of an aldose or keto group of a ketose. This carbon can have two configurations, α or β. If the aldehyde group on an acyclic sugar gets oxidized as a chromogenic agent gets reduced, that sugar is a reducing sugar. A sugar with its anomeric carbon linked to another structure is called a glycosyl residue. Sugars can be attached either to an –NH2 or an –OH group, producing N- and O-glycosides. Salivary α-amylase acts on dietary polysaccharides (glycogen, amylose, amylopectin), producing oligosaccharides. Pancreatic α-amylase continues the process of polysaccharide digestion. The final digestive processes occur at the mucosal lining of the small intestine. Several disacchari-
dases [for example, lactase (β-galactosidase), sucrase, maltase, and isomaltase] produce monosaccharides (glucose, galactose, and fructose). These enzymes are secreted by and remain associated with the luminal side of the brush border membranes of intestinal mucosal cells. Absorption of the monosaccharides requires specific transporters. If carbohydrate degradation is deficient (as a result of heredity, intestinal disease, malnutrition, or drugs that injure the mucosa of the small intestine), undigested carbohydrate will pass into the large intestine, where it can cause osmotic diarrhea. Bacterial fermentation of the compounds produces large volumes of CO2 and H2 gas, causing abdominal cramps, diarrhea, and flatulence. Lactose intolerance, caused by a lack of lactase, is by far the most common of these deficiencies

Question 20

Which of the following statements best describes glucose?

B. It is a C-4 epimer of galactose. Correct answer = B. Glucose and galactose dif-
fer only in configuration around carbon 4, and so are C-4 epimers that are interconvertible by the action of an epimerase. Glucose is an aldose
sugar that typically exists as a pyranose ring in solution; fructose, however, is a ketose with a furanose ring. The D-isomeric form of carbohydrates is most typically the form found in biologic systems, in contrast to amino acids. Salivary
amylase does not produce monosaccharides. Homopolysaccharides of glucose include branched glycogen in which the glycosidic linkages are the α form, as well as unbranched cel- lulose that has β linkages.

Answer 20

A young black man entered his physician’s office complaining of bloating and diarrhea. His eyes were sunken and the physician noted additional signs of
dehydration. The patient’s temperature was normal. He explained that the episode had occurred following a birthday party at which he had participated in an ice
cream eating contest. The patient reported prior episodes of a similar nature following ingestion of a significant amount of dairy products. This clinical pic-
ture is most probably due to a deficiency in:

E. lactase.Correct answer = E. The physical symptoms suggest a deficiency in an enzyme responsible for carbohydrate degradation. The symptoms observed following the ingestion of dairy products suggest that the patient is deficient in lactase.

Question 21

Routine examination of the urine of an asymptomatic pediatric patient showed a positive reaction with Clinitest (a copper reduction method of detecting reduc-
ing sugars), but a negative reaction with the glucose oxidase test. Which one of the following sugars is least likely to be present (assuming a single elevated
saccharide)?

C. Sucrose. Correct answer = C. Clinitest is a nonspecific test that produces a change in color if urine is positive for reducing substances, including reducing
sugars (glucose, fructose, galactose, xylulose,lactose), amino acids, ascorbic acid, and certain drugs and drug metabolites. Because sucrose is not a reducing sugar, it is not detected by Clinitest. Glucose oxidase method will not detect increased levels of galactose or other sugars in urine. It is therefore important that a copper
reduction method be used as a screening test. In those instances when the copper method is positive and the glucose oxidase method is negative, glucosuria is ruled out.

Answer 21

α-Glucosidase inhibitors such as acarbose and miglitol taken with meals are used in the treatment of diabetes. Explain. What effect should these drugs have
on the digestion of lactose?

α-Glucosidase inhibitors slow the production of glucose from dietary carbohydrates, thereby reducing the post-prandial rise in blood glucose and facilitating better blood glucose control in diabetics. These drugs have no effect on lactose digestion because the disaccharide lactose contains a β-glycosidic bond, not an α.

Question 22

Essentially all cells except mature erythrocytes can synthesize phospholipids, whereas triacylglycerol synthesis occurs essentially only in liver, adipose tissue, lactating mammary glands, and intestinal mucosal cells.

Lung maturity of the fetus can be gauged by determining the ratio of DPPC to sphingomyelin, usually written as the L (for lecithin)/S ratio, in amniotic fluid. A ratio of two or above is evidence of maturity, because it reflects the major shift from sphingomyelin to DPPC synthesis that occurs in the pneumocytes at about
32 weeks of gestation.

.

Answer 22

A ceramide containing a fatty acid 30 carbons long is a major component of skin, and regulates skin’s water permeability

When cells are transformed (that is, when their cell division and growth are dysregulated), there is a dramatic change in the glycosphingolipid composition of the plasma membrane.

Question 23

Phospholipids are polar, ionic compounds composed of an alcohol (for
example, choline or ethanolamine) attached by a phosphodiester bridge
to either diacylglycerol (producing phosphatidylcholine or phospha tidyl -
ethanolamine) or to sphingosine (Figure 17.25). The alcohol sphingosine
attached to a long-chain fatty acid produces a ceramide. Addition of a
phosphor ylcholine produces the phospholipid sphingomyelin, which is
the only significant sphingophospholipid in humans. Phospholipids are the
predominant lipids of cell membranes. Nonmembrane-bound phospho-
lipids serve as components of lung surfactant and bile. Dipalmitoylphos -
phatidylcholine (DPPC, also called dipalmitoyl lecithin, DPPL) is the
major lipid component of lung surfactant. Insufficient surfactant production
causes respiratory distress syndrome. Phospha tidylinositol (PI) serves
as a reservoir for arachidonic acid in membranes. The phosphorylation of
membrane-bound PI produces phosphatidylinositol 4,5-bisphosphate
(PIP2). This compound is degraded by phospholipase C in response to the
binding of a variety of neurotransmitters, hormones, and growth factors to
membrane receptors. The products of this degradation, inositol 1,4,5-tris -
phos phate (IP3) and diacylglycerol mediate the mobilization of intracellu-
lar calcium and the activation of protein kinase C, which act synergistically
to evoke cellular responses. Specific proteins can be covalently attached
via a carbohydrate bridge to membrane-bound phosphatidylinositol (glyco-
syl phosphatidy linositol, or GPI). A deficiency in the synthesis of GPI in
hematopoietic cells results in a hemolytic disease, paroxysmal nocturnal
hemoglobinuria. The degradation of phosphoglycerides is performed by
phospholipases found in all tissues and pancreatic juice. Sphingomyelin
is degraded to a ceramide plus phosphorylcholine by the lysosomal
enzyme sphingomyelinase. A deficiency in sphingomyelinase causes
Niemann-Pick (A + B) disease. Glycolipids (glycosphingolipids) are
derivatives of ceramides to which carbohydrates have been attached.
When one sugar molecule is added to the ceramide, a cerebroside is pro-
duced. If an oligosaccharide is added, a globoside is produced.

Answer 23

If an acidic N-acetylneuraminic acid (NANA) molecule is added, a ganglioside
is produced. Glycolipids are found predominantly in cell membranes of the
brain and peripheral nervous tissue, with high concentrations in the
myelin sheath. They are very antigenic. Glycolipids are degraded in the
lysosomes by hydrolytic enzymes. A deficiency of one of these enzymes
produces a sphingolipidosis, in each of which a characteristic sphin-
golipid accumulates. Prostaglandins (PG), thromboxanes (TX), and
leukotrienes (LT) are produced in very small amounts in almost all tissues,
act locally and have an extremely short half-life. They serve as mediators
of the inflammatory response. The dietary precursor of the eicosanoids is
the essential fatty acid, linoleic acid. It is desaturated and elongated to
arachidonic acid—the immediate precursor of prostaglandins—which is
stored in the membrane as a component of a phospholipid, generally
phosphatidy linositol. Arachidonic acid is released from the phospholipid by
phospholipase A2. Synthesis of the PG and TX begins with the oxidative
cyclization of free arachidonic acid to yield PGH2 by prostaglandin
endoperoxide synthase—an ER membrane protein that has two catalytic
activities: fatty acid cyclooxygenase (COX) and peroxi dase. Opposing
effects of PGI2 and TXA2 limit clot formation. There are two isozymes of
the synthase: COX-1 (constitutive) and COX-2. NSAIDs inhibit both. LT are
li n ear molecules produced by the 5-lipoxygenase pathway. They mediate
allergic response and are unaffected by NSAIDs.

Question 24

Aspirin-induced asthma (AIA) is a severe reaction to nonsteroidal anti-inflammatory drugs (NSAIDs) characterized by bronchoconstriction 30 minutes to several hours after ingestion. It is seen in as many as 20% of adults. Which of the following statements best explains the symptoms seen in patients with
AIA?

B. NSAIDs inhibit COX but not lipoxygenase, resulting in the flow of arachidonic acid to leukotriene synthesis. Correct answer = B. NSAIDs inhibit COX but not lipoxygenase, so any arachidonic acid available is used for the synthesis of bronchoconstricting-leukotrienes. NSAIDs have no effect on the CFTR protein, defects in which are the cause of cystic fibrosis. Steroids, not NSAIDs,
inhibit phospholipase A2. COX is inhibited by NSAIDs, not activated. NSAIDs have no effect on phospholipases.

Answer 24

An infant, born at 28 weeks of gestation, rapidly gave evidence of respiratory distress. Lab and x-ray results supported the diagnosis of infant respiratory
distress syndrome (RDS). Which of the following statements about this syndrome is true?

D. The concentration of dipalmitoylphosphatidyl-choline in the amniotic fluid would be expected to be lower than that of a full-term baby.Correct answer = D. Dipalmitoylphosphatidylcholine (DPPC, or dipalmitoyl lecithin) is the lung surfactant found in mature, healthy lungs. RDS can occur in lungs that make too little of
this compound. If the lecithin/sphingomyelin ratio in amniotic is greater than two, a new-born’s lungs are considered to be sufficiently mature—premature lungs would be expected to have a ratio lower than two. The RDS would not be due to too few Type II pneumocytes—these cells would simply be secreting sphingomyelin
rather than DPPC at 28 weeks of gestation

Question 25

A 25-year-old woman with a history that included hepatosplenomegaly with eventual removal of the spleen, bone and joint pain with several fractures of the femur, and a liver biopsy that showed wrinkled-looking cells with accumulations of glucosylcer -
amides was presented at Grand Rounds. The likely diagnosis for this patient is:

C. Gaucher disease. Correct answer = C. The adult form of Gaucher
disease causes hepatosplenomegaly, osteoporosis of the long bones, and the characteristic wrinkled appearance of the cytosol of cells. This is also the sphingolipidosis in which glucosylceramides accumulate. The deficient enzyme is β-glucosidase (a glucocerebrosidase).

Answer 25

Folic acid can partially reverse the hematologic abnormalities of B12 deficiency and, therefore, can mask a cobalamin deficiency. Thus, therapy of megaloblastic anemia is often initiated with folic acid and vitamin B12 until the cause of the
anemia can be determined.

Multiple carboxylase deficiency results from a defect in the ability to link biotin to carboxylases or to remove it from carboxylases during their
degradation. Treatment is biotin supplementation.

Populations consuming diets high in fruits and vegetables show decreased incidence of some chronic diseases. However, clinical trials have failed to show a definitive benefit from supplements of vitamins A, C, or E; multivitamins with
folic acid; or antioxidant combinations for the prevention of cancer or cardiovascular disease.

Question 26

Which one of the following statements concerning vitamin B12 is correct?

C. It requires a specific glycoprotein for its absorption. Correct answer = C. Vitamin B12 requires intrinsic factor for its absorption. A deficiency of vitamin B12 is most often caused by a lack of intrinsic factor. However, high does of the vitamin, given
orally, are sufficiently absorbed to serve as treatment for pernicious anemia. The cofactor forms are methylcobalamin and deoxyadenosylcobalamin. Vitamin B6, not vitamin B12, is involved in the transfer of amino groups. B12 is found in food
derived from animal sources.

Answer 26

Retinol:

B. in its ester form is transported from the intestine to the liver in chylomicrons.
Correct answer = B. Retinyl esters are incorporated into chylomicrons. Retinoic acid cannot be reduced to retinol. Retinal, the aldehyde form of retinol, is the chromophore for rhodopsin. Retinal is photoisomerized during the visual cycle.
Retinoic acid, not retinol, is the most important retinoid

Question 27

Which one of the following statements concerning vitamin D is correct?

A. Chronic renal failure requires the oral administration of 1,25-dihydroxycholecalciferol. Correct answer = A. Renal failure results in the
decreased ability to form the active form of the vitamin, which must be supplied. The vitamin is not required in individuals exposed to sunlight 1,25-Dihydroxycholecalciferol is the active form of the vitamin. Vitamin D and parathyroid hormone both increase serum calcium. A deficiency of vita-
min D decreases the secretion of calcitonin.

Answer 27

Vitamin K:

D. is synthesized by intestinal bacteria. Correct answer = D. Vitamin K is essential for clot formation, decreases coagulation time, and is present in low concentrations in milk. It is one of four fat-soluble vitamins.

Question 28

Deficiency of which vitamin results in: beri-beri; scurvy; pellagra; night blindness; rickets or osteoma-lacia; pernicious anemia; megaloblastic anemia;
bleeding?

Thiamin; vitamin C; niacin; vitamin A, vitamin D; vitamin B12 (due primarily to a deficiency ofintrinsic factor), both B12 and folate; vitamin K.

Answer 28

Transcription factors bind DNA through a variety of motifs, such as the helix-loop-helix, zinc fingers, and leucine zippers (see p. 18)

Systemic lupus erythematosus, an often fatal inflammatory disease, results from an autoimmune response in which individuals produce antibodies against their own nuclear proteins such as snRNPs.

Question 29

There are three major types of RNA that participate in the process of
protein synthesis: ribosomal RNA (rRNA), transfer RNA (tRNA), and
messenger RNA (mRNA) (Figure 30.20). They are unbranched poly-
mers of nucleotides, but differ from DNA by containing ribose instead of
deoxyribose and uracil instead of thymine. rRNA is a component of the
ribosomes. tRNA serves as an “adaptor” molecule that carries a specific amino acid to the site of protein synthesis. mRNA carries genetic
information from the nuclear DNA to the cytosol, where it is used as the
template for protein synthesis. The process of RNA synthesis is called
transcription, and its substrates are ribonucleoside triphosphates. The
enzyme that synthesizes RNA is RNA polymerase, which is a multisub-
unit enzyme. In prokaryotic cells, the core enzyme has five subunits—
2α, 1β, 1β’, and 1Ω—and possesses 5’→3’ polymerase activity that
elongates the growing RNA strand. This enzyme requires an additional
subunit—sigma (σ) factor—that recognizes the nucleotide sequence
(promoter region) at the beginning of a length of DNA that is to be tran-
scribed. This region contains characteristic consensus nucleotide
sequences that are highly conserved and include the Pribnow box and
the –35 sequence. Another protein—rho (ρ) factor—is required for
termination of transcription of some genes. There are three distinct
classes of RNA polymerase in the nucleus of eukaryotic cells. RNA
polymerase I synthesizes the precursor of large rRNA in the nucleolus.
RNA polymerase II synthesizes the precursors for mRNA and some
ncRNAs in the nucleoplasm, and RNA polymerase III produces the pre-
cursors of tRNA in the nucleoplasm. In both prokaryotes and eukary-
otes, RNA polymerase does not require a primer, and has no
proofreading activity

Answer 29

Promoters for genes transcribed by RNA poly-
merase II contain consensus sequences, such as the TATA or Hogness
box, the CAAT box, and the GC box. They serve as binding sites for
proteins called general transcription factors, which, in turn, interact
with each other and with RNA polymerase II. Enhancers are DNA
sequences that increase the rate of initiation of transcription by binding
specific transcription factors that serve as transcription activators.
Eukaryotic transcription requires that the chroma tin be accessible. A
primary transcript is a linear copy of a transcription unit—the segment
of DNA between specific initiation and termination sequences. The pri-
mary transcripts of both prokaryotic and eukaryotic tRNA and rRNA are
posttranscriptionally modified by cleavage of the original transcripts by
ribonucleases. rRNA of both prokaryotic and eukaryotic cells are syn-
thesized from long precursor molecules called preribosomal RNA.
These precursors are cleaved and trimmed by ribonucleases, producing
the three largest rRNA, and bases and sugars are modified. Eukaryotic
5S rRNA is synthesized by RNA polymerase III , and is modified sepa-
rately. Pro karyotic mRNA is generally identical to its primary transcript,
whereas eukaryotic mRNA is extensively modified co- and posttran-
scriptionally. For example, a 7-methylguanosine “cap” is attached to
the 5’-terminal end of the mRNA through a 5’→5’ linkage. A long poly-A
tail—not transcribed from the DNA—is attached to the 3’-end of most
mRNA. Most eukaryotic mRNAs also contain intervening sequences
(introns) that must be removed to make the mRNA functional. Their
removal, as well as the joining of expressed sequences (exons),
requires a spliceosome composed of small, nuclear ribonucleoprotein
particles that mediate the process of splicing. Eukaryotic mRNA is
monocistronic, containing information from just one gene. Prokaryotic
and eukaryotic tRNA are also made from longer precursor molecules. If
present, an intron is removed by nucleases, and both ends of the
molecule are trimmed by ribonucleases. A 3’-CCA sequence is added,
and bases at specific positions are modified, producing “unusual”
bases

Question 30

A 1-year-old male with chronic anemia is found to have β-thalassemia. Genetic analysis shows that one of his β-globin genes has a mutation that creates a new splice acceptor site 19 nucleo tides upstream of the normal splice acceptor site of the first intron. Which of the following best describes the new mRNA molecule that can
be produced from this mutant gene?

D. Exon 2 will be too long.Correct answer = D. Because the mutation cre-
ates an additional splice acceptor site (the 3’-end) upstream of the normal acceptor site of intron 1, the 19 nucleotides that are usually found at the 3’-end of the excised intron 1 lariat can remain behind as part of exon 2. Exon 2 can, therefore, have these extra 19 nucleotides at its 5’-end. The presence of these extra nucleotides in the coding region of the mutant mRNA molecule will prevent the ribosome from
translating the message into a normal β-globin protein molecule. Those mRNA for which the normal splice site is used to remove the first intron will be normal, and their translation will produce normal β-globin protein.

Answer 30

The base sequence of the strand of DNA used as the template for transcription is GATCTAC. What is the base sequence of the RNA product? (All sequences are
written 5’→ 3’ according to standard convention.)

E. GUAGAUC.
Correct answer = E. The RNA product has a sequence that is complementary to the template strand and identical to the coding strand of DNA. Uracil (U) is found in RNA in place of the thymine (T) in DNA. Thus, the DNA template 5’- GATCTAC-3’ would produce the RNA product 3’-CUAGAUG-5’ or, written correctly in the stan-
dard direction, 5’-GUAGAUC-3’.

Question 31

A 4-year-old child who becomes easily tired and has trouble walking is diagnosed with Duchenne muscular dystrophy, an X-linked recessive disorder. Genetic anal-
ysis shows that the patient’s gene for the muscle protein dystrophin contains a mutation in its promoter region. Of the choices listed, which would be the most likely effect of this mutation?

A. Initiation of dystrophin transcription will be defective.Correct answer = A. Mutations in the promoter typically prevent formation of the RNA polymerase II transcription complex, resulting in a decrease in the initiation of mRNA synthesis. A
deficiency of dystrophin mRNA will result in a deficiency in the production of the dystrophin protein. Capping, splicing and tailing defects are not a consequence of promoter mutations. They can, however, result in mRNA with decreased stability (capping and tailing defects), or a mRNA in which too many or too few introns
have been removed (splicing defects)

Answer 31

A mutation to this sequence in eukaryotic mRNA will
affect the process by which the 3’-end poly-A tail is
added to the mRNA.

D. AAUAAA. Correct answer = D. An endonuclease cleaves mRNA just downstream of this polyadenylation signal, creating a new 3’-end to which the pol A polymerase adds the poly-A tail using ATP as the substrate in a template-independent process.
CAAT, GGGGCGT, and TATAAA are sequences found in promoters for RNA polymerase II. CCA is added to the 3’-end of tRNA by nucleotidyl
transferase.

Question 32

The small ribosomal subunit binds mRNA and is responsible for the accuracy of translation by ensuring correct base-pairing between the codon in the mRNA and the anticodon of the tRNA. The large ribosomal subunit catalyzes formation of the peptide bonds that link amino acid residues in a protein.

One important difference is that translation and transcription are coupled in prokaryotes, with translation starting before transcription is completed. Coupling is a consequence of the lack of a nuclear membrane in prokaryotes.

Answer 32

Codons are composed of three nucleotide bases presented in the mRNA language of A, G, C, and U. They are always written 5’→3’. Of the 64 possible three-base combinations, 61 code for the 20 common amino acids and three signal termination of protein synthesis (translation). Altering the nucleotide sequence in a codon can
cause silent mutations (the altered codon codes for the original amino acid), missense mutations (the altered codon codes for a different amino acid), or nonsense mutations (the altered codon is a termination codon). Characteristics of the genetic code include specificity, universality, and degeneracy, and it is nonoverlapping
and commaless. Requirements for protein synthesis include all the amino acids that eventually appear in the finished protein, at least one specific type of tRNA for each amino acid, one aminoacyl-tRNA synthetase for each amino acid, the mRNA coding for the protein to be synthesized, fully competent ribosomes, protein fac-
tors needed for initiation, elongation, and termination of protein synthesis, and ATP and GTP as energy sources. tRNA has an attachment site for a specific amino acid at its 3’-end, and an anticodon region that can recognize the codon specifying the amino acid the tRNA is carrying. Ribosomes are large complexes of protein
and rRNA. They consist of two subunits. Each ribosome has three binding sites for tRNA molecules—the A, P, and E sites that cover three neighboring codons. The A-site codon binds an incoming aminoacyl-tRNA, the P- site codon is occupied by peptidyl-tRNA, and the E site is occupied by the empty tRNA as it is about to exit the
ribosome. Recognition of an mRNA codon is accomplished by the tRNA anticodon. The anticodon binds to the codon following the rules of complementarity and antiparallel binding. (Nucleotide sequences are ALWAYS assumed to be written in the 5’ to 3’ direction unless otherwise noted.) The “wobble” hypothesis states that the
first (5’) base of the anticodon is not as spatially defined as the other two bases. Movement of that first base allows nontraditional base-pairing with the last (3’) base of the codon, thus allowing a single tRNA to recognize more than one codon for a specific amino acid. For initiation of protein synthesis, the components of the trans-
lation system are assembled, and mRNA associates with the small ribosomal subunit. The process requires initiation factors. In prokaryotes, a purine-rich region of the mRNA (the Shine-Dalgarno sequence) base-pairs with a complementary sequence on 16S rRNA, resulting in the positioning of the small subunit on the mRNA so that translation can begin. The 5’-cap on eukaryotic mRNA is used to position the small subunit on the mRNA. The initiation codon is AUG; N-formylmethionine is the initiating amino acid in prokaryotes, whereas methion-
ine is in eukaryotes. The polypeptide chain is elongated by the addition of amino acids to the carboxyl end of its growing chain. The process requires elongation factors. The formation of the peptide bond is catalyzed by peptidyltransferase, which is an activity intrinsic to the rRNA of the large subunit. Following peptide bond formation, the ribosome advances along the mRNA in the 5’→3’ direction to the next codon (translocation). Because of the length of most mRNAs, more than one ribosome at a time can translate a message, forming a polysome.
Termination begins when one of the three termination codons moves into the A site. These codons are recognized by release factors. The newly synthesized protein is released from the ribosomal complex, and the ribosome is dissociated from the mRNA. Initiation, elongation, and termination are driven by the hydrolysis of GTP;
initiation in eukaryotes also requires ATP for scanning. Numerous antibiotics interfere with the process of protein synthesis. Many polypeptide chains are covalently modified during or after translation. Such modifications include removal of amino acids, phosphorylation, which may activate or inactivate the protein, glycosylation,
which plays a role in protein targeting, or hydroxylation such as that seen in collagen. Proteins must fold to achieve their functional form. Folding can be spontaneous or facilitated by chaperones. Proteins that are defec-
tive, for example misfolded, or destined for rapid turnover are marked for destruction by the attachment of chains of a small, highly conserved protein called ubiquitin. Ubiquitinated proteins are rapidly degraded by a cytosolic complex known as the proteasome.

Question 33

A 20-year-old anemic man is found to have an abnormal form of β-globin (Hemoglobin Constant Spring) that is 172 amino acids long, rather than the 141
found in the normal protein. Which of the following point mutations is consistent with this abnormality?

A. UAA → CAA.Correct answer = A. Mutating the normal stop codon for β-globin from UAA to CAA causes the ribosome to insert a glutamine at that point. It
will continue extending the protein chain until it comes upon the next stop codon further down the message, resulting in an abnormally long protein. A change from UAA to UAG would simply change one termination codon for another and would have no effect on the protein. The replacement of CGA (arginine) with UGA (stop)
would cause the protein to be too short. GAU and GAC both encode aspartate and would cause no change in the protein. Changing GCA (alanine) to GAA (glutamate) would not change the size of the protein product.

Answer 33

A pharmaceutical company is studying a new antibiotic that inhibits bacterial protein synthesis. When this antibiotic is added to an in vitro protein synthesis
system that is translating the mRNA sequence AUGUUUUUUUAG, the only product formed is the dipeptide fMet-Phe. What step in protein synthesis is most likely inhibited by the antibiotic?

D. Ribosomal translocation. Correct answer = D. Because fMet-Phe is made,
the ribosomes must be able to complete initiation, bind Phe-tRNA to the A site, and use peptidyltransferase activity to form the first peptide bond. Because the ribosome is not able to proceed any further, ribosomal movement (translo-
cation) is most likely the inhibited step. The ribosome is, therefore, frozen before it reaches the termination codon of this message.

Question 34

A tRNA molecule that is supposed to carry cysteine (tRNAcys) is mischarged, so that it actually carries alanine (ala-tRNAcys). Assuming no correction
occurs, what will be the fate of this alanine residue during protein synthesis?

B. It will be incorporated into a protein in response to
a cysteine codon.
Correct answer = B. Once an amino acid is attached to a tRNA molecule, only the anticodon of that tRNA determines the specificity of incorporation. The mischarged alanine will, therefore, be incorporated in the protein at a position determined by a cysteine codon.

Answer 34

In a patient with cystic fibrosis caused by the ΔF508 mutation, the mutant cystic fibrosis transmembrane conductance regulator (CFTR) protein folds incorrectly. The patient’s cells modify this abnormal protein by attaching ubiquitin molecules to it. What is the fate of this modified CFTR protein?

D. It is degraded by the proteasome. Correct answer = D. Ubiquitination usually
marks old, damaged, or misfolded proteins for destruction by the proteasome. There is no known cellular mechanism for repair of damaged proteins.

Question 35

Many antimicrobials inhibit protein translation. Which of the following antimicrobials is correctly paired with its mechanism of action?

E. Erythromycin binds to the 50S ribosomal subunit.Correct answer = E. Erythromycin and clindamycin both bind the 50S ribosomal subunit. Tetracyclines inhibit the 30S ribosomal subunit (chloramphenicol inhibits peptidyl transferase).
Diphtheria toxin binds EF-2, inactivating it through ADP ribosylation. Puromycin has structure similar to that of aminoacyl-tRNA. It is incorporated into the growing chain, and inhibits further elongation of the peptide chain.

Answer 35

Translation of a synthetic polyribonucleotide containing the repeating sequence CAA in a cell-free protein-synthesizing system produces three homopolypeptides: polyglutamine, polyasparagine, and polythreonine. If the codon for glutamine and
asparagine are CAA and AAC, respectively, which of the following triplets is the codon for threonine?

E. ACA.Correct answer = E. The synthetic polynucleotide sequence of CAACAACAACAA . . could be read by the in vitro protein synthesizing sys-
tem starting at the first C, the first A, or the second A. In the first case, the first triplet codon would be CAA, which codes glutamine; in the second case, the first triplet codon would be AAC, which codes for asparagine; in the last case, the first triplet codon would be ACA, which codes for threonine.

Question 36

Why is the genetic code described as both degenerate and unambiguous?

A given amino acid can be coded for by more than one codon (degenerate code), but a give codon codes for just one particular amino acid (unambiguous code).

Answer 36

α1-Antitrypsin (A1AT) deficiency can result in emphysema, a lung pathology, because the action of elastase, a serine protease, is unopposed. Deficiency of
A1AT in the lungs is the consequence of its retention in (rather than secretion from) the liver, the site of its synthesis. Proteins such as A1AT that are designed
to be secreted are best characterized by which of the following statements?

D. They are produced from translation products that have an N-terminal hydrophobic signal sequence.Answer = D. Synthesis of secreted proteins is begun on free (cytosolic) ribosomes. As the N- terminal hydrophobic signal sequence of the
peptide emerges from the ribosome, it is bound by the signal recognition particle, taken to the rough endoplasmic reticulum (RER), threaded into the lumen, and removed as translation continues. The proteins move through the RER and
the Golgi, and undergo processing such as N- glycosylation (RER) and O-glycosylation (Golgi). In the Golgi, they are packaged in secretory
vesicles and released from the cell. The smooth endoplasmic reticulum is associated with synthesis of lipids, not proteins, and has no ribo-
somes attached. Phosphorylation at carbon 6 of terminal mannose residues in glycoproteins targets these proteins (acid hydrolases) to lyso-
somes. The N-terminal methionine is removed from most proteins during processing.

Question 37

Transcriptional attenuation can occur in prokaryotes because translation of an mRNA begins before its synthesis is complete. In eukaryotes this does not occur because, as a result of having a membrane-bound nucleus, transcription
and translation are spatially and temporally separate processes.

Over 60% percent of the approximately 30,000 genes in the human genome undergo differential splicing. The use of alternative polyadenylation and transcription start sites is also seen in many genes. This explains, at least in part, how 30,000
genes can give rise to over 100,000 proteins.

Answer 37

Gene expression results in the production of a functional gene product (either RNA or protein) through the processes of replication, transcription, and translation (Figure 32.19). Genes can be either constitutive (always expressed, housekeeping genes) or regulated (expressed only under certain conditions in all cells or in a subset of cells). The ability to appropriately express (positive regulation) or repress (negative regulation) genes is essential in all organisms. Regulation of gene expression occurs primarily at the level of transcription in both prokaryotes and eukaryotes, and is mediated through the binding of trans-acting proteins to cis-acting regulatory elements on the DNA. In eukaryotes, regulation also occurs through modifications to the DNA, as well as through posttranscriptional and posttranslational events. In prokaryotes such as E. coli, the coordinate
regulation of genes whose protein products are required for a particular metabolic pathway is achieved throug operons (groups of genes sequentially arranged on the chromosome along with the regulatory elements that determine their transcription). The lac operon contains the Z, Y, and A structural genes, the protein products of
which are needed for the catabolism of lactose. When glucose is available, the operon is repressed by the binding of the repressor protein (the product of the lacI gene) to the operator, thus preventing transcription. When only lactose is present, the operon is induced by an isomer of lactose (allolactose) that binds the repressor protein, preventing it from binding to the operator. In addition, cAMP binds the CAP protein, and the complex binds the DNA at the CAP site. This increases promoter efficiency and results in the expression of the structural genes through the production of a polycistronic mRNA. When both glucose and lactose are present, glucose prevents formation of cAMP and transcription of these genes is negligible. The trp operon contains genes needed for the synthesis of tryptophan, and like the lac operon, it is regulated by positive and negative control. Unlike the lac operon, it is also regulated by attenuation, in which mRNA synthesis that escaped repression by trp is terminated before completion. Transcription of rRNA and tRNA is selectively inhibited in prokaryotes by the stringent response to amino acid starvation. Translation is also a site of prokaryotic gene regulation: when ribosomal
proteins are in excess, they bind the Shine-Dalgarno sequence on their own polycistronic mRNA, preventing ribosomes from binding. Gene regulation is more complex in eukaryotes. Operons are not present, but coordinate regulation of the transcription of genes located on different chromosomes can be achieved through the binding of trans-acting proteins to cis-acting elements. In multicellular organisms, hormones can cause coordinated regulation, either through the binding of the hormone receptor–hormone complex itself to the DNA (as with
steroid hormones) or through the binding of a protein that is activated in response to a second messenger (as with glucagon). In each case, binding to DNA is mediated through structural motifs such as the zinc-finger. Co-and posttranscriptional regulation is also seen in eukaryotes, and includes splice-site choice, polyA-site
choice, mRNA editing, and variations in mRNA stability as seen with transferrin receptor synthesis and with RNA interference. Regulation at the translational level can be caused by the phosphorylation (inhibition) of eIF-2. Gene expression in eukaryotes is also influenced by availability of DNA to the transcriptional apparatus, the amount of DNA, and the arrangement of the DNA.

Question 38

32.1 Which of the following mutations is most likely to result in reduced expression of the lac operon?

C. Cya– (no adenylyl cyclase made)
Correct answer = C. In the absence of glucose, adenylyl cyclase makes cAMP, which forms a complex with the CAP protein. The cAMP–CAP complex binds the CAP site on the DNA, causing RNA polymerase to bind more efficiently to the lac operon promoter, thus increasing expression of the operon. With cya– mutations, adeny-
lyl cyclase is not made, and so the operon is unable to be turned on even when glucose is absent and lactose is present. The absence of a repressor protein or decreased ability of the repressor to bind the operator results in constitu-
tive (constant) expression of the lac operon.

Answer 38

Which of the following is best described as being trans-acting?

D. Repressor. Correct answer = D. A repressor is a molecule that transits to the DNA, binds, and reduces the expression of that DNA; it is said to be trans-acting. The CAP site, operator, and promoter are part of the DNA itself, and so are said to be
cis-acting

Question 39

Which of the following is the basis for the intestine-specific expression of apoprotein B-48?

D. RNA editing Correct answer = D. A repressor is a molecule that transits to the DNA, binds, and reduces the expression of that DNA; it is said to be trans- acting. The CAP site, operator, and promoter are part of the DNA itself, and so are said to be cis-acting.

Answer 39

Which of the following is most likely to be true in hemochromatosis, a disease of iron accumulation in the body?

B. The mRNA for the TfR is not bound by IRPs, and
is rapidly degraded.
C. The mRNA for apoferritin is not bound by IRPs at
its 5’ stem-loop IRE, and is translated
E. Both B and C

Correct answer = E. When iron levels in the body are high, as is seen with hemochromatosis, there is increased synthesis of the iron-storage
molecule, apoferritin, and decreased synthesis of the transferrin receptor (TfR) that mediates iron uptake by cells. These effects are the result of trans-acting iron regulatory proteins (IRPs) binding iron rather than binding cis-acting iron-
responsive elements (IREs), resulting in degradation of the mRNA for TfR, and increased translation of the mRNA for apoferritin.

Question 40

After several weeks of chemotherapy with methotrexate, a cancer patient’s tumor begins to show signs of resistance to treatment. Which of the following mech-
anisms is most likely to explain this resistance to methotrexate?

A. Overproduction of dihydrofolate reductase. Correct answer = A. Methotrexate interferes with folate metabolism by acting as a competitive inhibitor of the enzyme dihydrofolate reductase. This starves cells for tetrahydrofolate, and makes them unable to synthesize purines and dTMP. This is especially toxic to rapidly growing
cancer cells. Overproduction of dihydrofolate reductase, usually caused by amplification of its gene, can overcome the inhibition of the enzyme at the metho trexate concentrations used for chemotherapy, and can result in resis-
tance of the tumor to treatment by this drug

Answer 40

The ZYA region of the lac operon will be efficiently transcribed if:

C. the operator is mutated and can’t be bound by the repressor. Correct answer = C. The lac operon is negatively regulated by the repressor protein binding
to the operator region and preventing RNA polymerase from transcribing the Z, Y, and A genes of the operon. If glucose is present, the operon is off as a result of catabolite repression. It is only when glucose is gone, cAMP levels are increased, and lactose is available that the operon is maximally induced. If the inducer
can’t bind the repressor, the repressor binds the operator and represses transcription.

Question 41

In bacteria, restriction endonucleases “restrict” the expression of non-bacterial DNA through cleavage. Bacterial DNA is protected by methylation of bases at the restriction site

Recombinant DNA technology has allowed the production, in nonhuman systems, of human enzymes that have been modified for appropriate targeting to lysosomes. Such recombinant enzyme replacement therapy (ERT) is used in the treatment of lysosomal storage diseases such as Pompe, Gaucher, and Hurler.

PCR can generate 100 billion identical copies of a specific DNA sequence in an afternoon, using a highly automated system.

Answer 41

Restriction endonucleases are bacterial enzymes that cleave double-
stranded DNA into smaller fragments. Each enzyme cleaves dsDNA at
a specific four to eight-base-long nucleotide sequence, producing DNA
segments called restriction fragments. The sequences that are recog-
nized are palindromic. These enzymes form either staggered cuts
(sticky ends) or blunt-end cuts on the DNA. A DNA sequence that is
recognized by a restriction enzyme is called a restriction site.
Bacterial DNA ligases can anneal two DNA fragments from different
sources if they have been cut by the same restriction endonuclease.
This hybrid combination of two fragments is called a recombinant DNA
molecule. Introduction of a foreign DNA molecule into a replicating cell
permits the amplification (production of many copies) of the DNA—a
process called cloning. A vector is a molecule of DNA to which the
fragment of DNA to be cloned is attached. Vectors must be capable of
autonomous replication within the host cell, and must contain at least
one specific nucleotide sequence recognized by a restriction endonu-
clease. It must also carry at least one gene that confers the ability to
select for the vector, such as an antibiotic resistance gene.
Prokaryotic organisms normally contain small, circular, extrachromoso-
mal DNA molecules called plasmids that can serve as vectors. They
can be readily isolated from the bacterium, joined with the DNA of
interest, and reintroduced into the bacterium which will replicate, thus
making multiple copies of the hybrid plasmid. A DNA library is a col-
lection of cloned restriction fragments of the DNA of an organism. A
genomic library is a collection of fragments of double-stranded DNA
obtained by digestion of the total DNA of the organism with a restric-
tion endonuclease and subsequent ligation to an appropriate vector. It
ideally contains a copy of every DNA nucleotide sequence in thegenome. In contrast, complementary DNA (cDNA) libraries contain only those DNA sequences that are complementary to messenger RNA (mRNA) molecules present in a cell, and differ from one cell type to another.

Question 42

Because cDNA has no intervening sequences, it can be cloned into an expression vector for the synthesis of eukaryotic proteins by bacteria. Cloned, then purified, fragments of DNA can be sequenced, for example, using the Sanger dideoxy method. A probe is a small, single-stranded piece of DNA (usually labeled with a radioistope, such as 32P, or another recognizable compound, such as biotin) which has a nucleotide sequence complementary to the DNA molecule of interest (target DNA). Probes can be used to identify which clone of a library or which band on a gel contains the target DNA. Southern blotting is a technique that can be used to detect specific sequences present in DNA. The DNA is cleaved using a restriction endonuclease, and the pieces are separated by gel electrophoresis and are denatured and transferred (blotted) to a nitrocellulose membrane for analysis. The fragment of interest is detected using a probe. The human genome contains many thousands of
polymorphisms (DNA sequence variations at a given locus) that do not affect the phenotype of the individual. Polymorphisms can arise from single base changes and from tandem repeats. A polymorphism can serve as a genetic marker that can be followed through families. A restriction fragment length polymorphism is a genetic
variant that can be observed by cleaving the DNA into restriction fragments using a restriction enzyme. A base substitution in one or more nucleotides at a restriction site can render the site unrecognizable by a particular restriction endonuclease. A new restriction site also can be created by the same mechanism. In either case,
cleavage with the endonuclease results in fragments of lengths differing from the normal that can be detected by DNA hybridization. This technique can be used to diagnose genetic diseases early in the gestation of a fetus. The polymerase chain reaction (PCR), a test tube method for amplifying a selected DNA sequence, does not rely on the biologic cloning method. PCR permits the synthesis of millions of copies of a specific nucleotide sequence in a few hours. It can amplify the sequence, even when the targeted sequence makes up less than one part in a million of the total initial sample. The method can be used to amplify DNA sequences from any source. Applications of the PCR technique include: 1) efficient comparison of a normal cloned gene with an uncloned mutant form of the gene, 2) detection of low-abundance nucleic acid sequences, 3) forensic analysis of DNA
samples, and 4) prenatal diagnosis and carrier detection, for example, of cystic fibrosis. The products of gene expression (mRNA and proteins) can be measured by techniques such as the following: Northern blots are very similar to Southern blots except that the original sample contains a mixture of mRNA molecules that are separated by electrophoresis, then hybridized to a radiolabeled probe. Microarrays are used to determine the differing patterns of gene expression in two different types of cells—for example, normal and cancer cells. Enzyme-linked immunosorbent assays and Western blots (immunoblots) are used to detect specific proteins. The goal of gene therapy is the insertion of a normal gene to replace a defective gene. Insertion of a foreign gene into an animal creates a transgenic animal that can produce therapeutic proteins or serve as a model for human diseases.

Answer 42

HindIII is a restriction endonuclease commonly used to cut human DNA into pieces before inserting it into a plasmid. Which of the following is most likely to be
the recognition sequence for this enzyme?

D. AAGCTT.Correct answer = D. The vast majority of restriction endonucleases recognize palindromes, and AAGCTT is the only palindrome among the

choices. Because the sequence of only one DNA strand is given, one must determine the base sequence of the complementary strand. To be a palindrome, both strands must have the same sequence when read in the 5’→3’ direc-
tion. Thus, the complement of 5’-AAGCTT-3’ is also 5’-AAGCTT-3’.

Question 43

An Ashkenazi Jewish couple brings their 6-month-old son to you for evaluation of listlessness, poor head control, and a fixed gaze. You determine that he has
Tay-Sachs disease, an autosomal recessive disorder. The couple also has a daughter. The diagram below shows this family’s pedigree, along with
Southern blots of an RFLP very closely linked to the hexosaminidase A gene, which is defective in Tay- Sachs. Which of the statements below is most accu-
rate with respect to the daughter?

E. She is homozygous normal. Correct answer = E. Both the father and mother
must be carriers for this disease. The son must have inherited a mutant allele from each parent. Because he shows only the 3-kb band on the Southern blot, the mutant allele for this disease must be linked to the 3-kb band for both parents.
The normal allele must be linked to the 4-kb band in both parents. Because the daughter inherited the 4-kb band from both parents, she must be homozygous normal for the hexosaminidase A gene.

Answer 43

A physician would like to determine the global patterns of gene expression in two different types of tumor cells in order to develop the most appropriate
form of chemotherapy for each patient. Which of the following techniques would be most appropriate for this purpose?

E. Microarray. Correct answer = E. Microarray analysis allows the determination of mRNA production (thus, gene expression) from thousands of genes at
once. A Northern blot only measures mRNA production from one gene at a time. Western blots and ELISA measure protein production (also gene expression), but only from one gene at a time. Southern blots are used to analyze DNA,
not gene expression

Question 44

A 2-week-old infant is diagnosed with a urea cycle defect. Enzymic analysis showed no activity for ornithine transcarbamoylase (OTC). Molecular analysis revealed that the mRNA product of the gene for OTC was identical to that of a control. Which of the techniques listed below was most likely used to analyze the size and amount of the mRNA?

Answer 44

B. Northern blot. Correct answer = B. Northern blot allows analysis of the mRNA present (expressed) in a particular cell or tissue. Southern blot is used for DNA analysis, whereas Western blot is used for pro- tein analysis. Dideoxy chain termination is used to sequence DNA. Polymerase chain reaction (PCR) is used to generate multiple, identical copies of a DNA sequence in vitro.