Colloquium 1

Question 1

1. What effect will raising pH fom an acidic value to the physiological value of 7.4 have on the structural features shown in red in the diagram
2. At physiologic pH, what will be the charge on the side chain (R group) of free Asp? Of Lys?
3. Which 2 amino acids contain a side-chain hydroxyl group that can be glycosylated? The hydroxyl group can also be __________. Which amino acid contains a secondary group and its what forms a rigid ring?
4. What are 2 reasons why Val is not ionized when incorporated into a protein?

Answer 1

1. Deproteination(ionization) of the a-carboxyl group( pK2) to COO. The a-amino group (pK-9)will remain protonated
2. Charge on the R group of free Asp is negative, Lys is positive
3. Ser and Thr contain hydroxyl group that can be 0-glycosylated. The hydroxyl group can also phosphorylated. Pro has secondary amino group bandits a-aminoN and R group form a rigid ring
4. No Val is not ionized when incorporated into a protein because: -the a-amino acid and a-carboxyl groups are involved in peptide bonds and are unavailable for isolation -the side chain is nonpolar

Question 2

1. Based on the figure, Leu, a non polar amino acid woule be likely located where in a protein that spans the membrane?
2. Describe the hydrophobic effect
3. In sickle cell anemia(SCA) why does the replacement of a Glu by a Val on the suface of the deoxyHb moecule result in the association of the molecules?

Answer 2

1. Leu would be located within the hydrophobic membrane spanning domain of the protein, in the interior of a soluble protein.
2. Hydrophobic effect is the tendency of nonpolar molecules (o regions of molecules such as amino acid side chains) to cluster together in a polar environment such as aqueous solutions
3. The replcemnt of polar Glu by nonpolar Val created a hydrophobic region on the surface of the deoxyHb molecule that will interact with a hydrphobic region on other deoxyHb molecules. This interaction creates rigid polymers of deoxyHb that deform RBCs. Thus it is the hydrophic effect that drvies the association of deoxHb molcules in SCA.

Question 3

1. Which structure shown (A or B) represents L-Ala?
2. Which amino acid does not possess a chiral (asymmetric) carbon?
3. Which peptide is less soluble in an aqueous (polar) environment, Ala-Gly-Asn-Ser-Tyr or Gly-Met-Phe-Leu-Ala?

Answer 3

1. Structure A represents L-Ala. The L isomer of an amino acid has the -amino group on the left. The D isomer has the -amino group on the right. D and L isomers are mirror images of each other (enantiomers).
2. Gly, with its two H substituents, does not possess a chiral (asymmetric) carbon.
3. Because the Gly-Met-Phe-Leu-Ala peptide contains no charged or polar uncharged amino acids, it is less soluble than Ala-Gly-Asn-Ser-Tyr in an aqueous (polar) environment.

Question 4

1. What relationship is described by the Henderson–Hasselbalch equation shown (pH= pKa + log [A–]/[HA] )?
2. Is an acid with a large pKa stronger or weaker than one with a small pKa?
3. The pKa of acetic acid (CH3COOH) is 4.8. What is the pH of a solution containing acetic acid and its conjugate base (CH3COO) in a ratio of 10 to 1?
4. Physiologic buffers are important in resisting blood pH changes. Maximal buffering occurs when the pH is equal to the , while effective buffering can occur within .

Answer 4

1. The Henderson–Hasselbalch equation describes the relationship between the pH of a solution and the concentration of a weak acid [HA] and its conjugate base [A-].
2. An acid with a large pKa is weaker than one with a small pKa because the large pKa reflects less ionization (fewer H+ released). This is because pKa= -log Ka.
3. Because pH= pKa + log [A-]/[HA], when pKa is 4.8 and the ratio of the acid to its conjugate base is 10 to 1, the pH is equal to 4.8 + log of 0.1. Therefore,

pH = 4.8+ (-1) = 3.8.

4. Physiologic buffers are important in resisting blood pH changes. Maximal buffering occurs when the pH is equal to the pKa, while effective buffering can occur within +/-1 pH unit of the pKa

Question 5

1. Which FORM (I, II, or III) shown represents the isoelectric form of Ala?
2. Calculate the pI for Arg, which has three pKs: pK1 2.2, pK2 9.2, and pK3 12.5.
3. What will happen to the charge on His residues in a protein that moves from the cytoplasm (pH 7.4) to a lysosome (pH 5.0)?

Answer 5

1. The isoelectric form has no net charge. It is the zwitterionic (“two ion”) form. Therefore, FORM II is the isoelectric form of Ala.
2. The pI corresponds to the pH at which an amino acid is electrically neutral, that is, the average of the pKs on either side of the isoelectric form. For Arg, a dibasic amino acid with pK1 (most acidic group) = 2.2, pK2 = 9.2, and pK3 (least acidic group) = 12.5, the pI is 10.8 (the average of 9.2 and 12.5).
3. In a protein, the imidazole R group of His can be charged or uncharged depending on the local environment. It will be uncharged (deprotonated) at pH 7.4 and charged (protonated) at pH 5.0. [Note: In free His the pK of the R group is 6.0.]

Question 6

1. Based on the bicarbonate buffer system shown, what will happen to the availability of HCO3- when H + is lost, such as with emesis (vomiting)?
2. Use the Henderson–Hasselbalch equation to determine what will happen to pH when HCO3- is lost (e.g., with diarrhea) and when CO2 is increased (e.g., with pulmonary obstruction).
3. Aspirin (pKa = 3.5) is largely protonated and uncharged in the stomach (pH 1.5). What percentage of the aspirin will be in this lipid-soluble form at pH 1.5?

Answer 6

1. With emesis ( vomiting ), the loss of H+ (rise in pH) results in increased availability of HCO3- as the result of a compensatory rightward shift in the bicarbonate buffer system .
2. The Henderson–Hasselbalch equation is used to calculate how the pH of a system changes in response to changes in the concentration of an acid or its conjugate base. For the bicarbonate buffer system, pH = pK+ log [HCO3-]/[CO2]. Therefore, both the loss of HCO3- (base) with diarrhea and the increase in CO2 (acid) because of decreased elimination with pulmonary obstruction result in decreased pH.
3. pH = pK+log [Drug-]/[Drug-H]. Therefore, for aspirin in the stomach,
4. 5 = 3.5 + (- 2). Because the antilog of -2 is 0.01, the ratio of [Drug-]/[Drug-H] is 1/100. This means that 1 out of 100 (1%) of the aspirin molecules will be the Drug- form and 99 out of 100 (99%) will be the uncharged, lipid-soluble, Drug-H form.

Question 7

1. Which level of protein structure depicted can be correctly described as the “three-dimensional shape of a folded polypeptide chain”?
2. Mutations that insert, delete, or replace amino acids change this level of protein structure.
3. How many different isoforms of the tetrameric enzyme PK can be made from M and/or L subunits?
4. How many different tetrapeptides could be generated from three different amino acids?

Answer 7

1. The “three-dimensional shape of a folded polypeptide chain” describes a protein’s tertiary structure (No. 3 shown).
2. At a minimum, the primary structure (amino acid sequence) will change with mutations that insert, delete, or replace amino acids. [Note: Changes in the primary structure can also affect the higher levels of protein structure (No. 2 to 4 shown). Such changes frequently result in protein misfolding and can lead to loss of function, aggregation, or degradation.]
3. Five different forms of tetrameric PK can be made from M and/or L subunits: M4, M3L, M2L2 , ML3 , and L4 . Because PK is composed of more than one subunit, it has a quaternary structure.
4. There are 3^4 or 81 (where 3 the number of amino acids and 4 the chain length) different tetrapeptides that could be generated from three different amino acids.

Question 8

1. What is the name given to the bond outlined by the black box shown?
2. What are the characteristics of this bond?
3. With fever, why might proteins begin to unfold but not be hydrolyzed to peptides and free amino acids?

Answer 8

1. A peptide bond, a type of amide bond, is outlined by the black box. Peptide bonds link the amino acid residues in a peptide or protein by joining the a-amino group of one amino acid to the a-carboxyl group of the next as water is released.
2. The peptide bond has partial double-bond character, is rigid and planar, uncharged but polar, and almost always in the trans configuration that reduces steric interference by the R groups.
3. Peptide bonds are resistant to conditions (such as the heat from a fever ) that can denature proteins and cause them to unfold. However, they are susceptible to cleavage by enzymes known as proteases or peptidases. [Note: Strong acids or bases at high temperatures can nonenzymatically cleave peptide bonds.]

Question 9

1. Sequencing large polypeptides involves cleavage reactions, as shown. Which sites in a peptide are susceptible to cleavage by the endopeptidase trypsin? By cyanogen bromide?
2. What is the Edman degradation method?
3. What is the amino acid sequence of a nonapeptide if trypsin digestion yields three products (Asn, Met-Gln-Lys, and Ala-Gly-Met-Leu-Arg) and cyanogen bromide cleavage yields three products (Leu-Arg-Met, Gln-Lys-Asn, and Ala-Gly-Met)?

Answer 9

1. Trypsin, an endopeptidase, cleaves at the carboxyl side of Lys and Arg residues within a peptide. [Note: Exopeptidases remove the terminal amino acid.] Cyanogen bromide cleaves at the carboxyl side of Met residues.
2. The Edman degradation method chemically determines the sequence of amino acids through the sequential removal and identification of the N-terminal amino acids in the small peptides generated from a polypeptide by cleavage reactions.
3. Based on the overlapping amino acids in the products of the trypsin (Asn, Met-Gln-Lys, and Ala-Gly-Met-Leu-Arg) and the cyanogen bromide (Leu-Arg-Met, Gln-Lys-Asn, and Ala-Gly-Met) cleavage reactions, the amino acid sequence of the nonapeptide is Ala-Gly-Met-Leu-Arg-Met-Gln-Lys-Asn. [Note: The sequence of amino acids in a protein is always written from the N-terminal to the C-terminal amino acid.]

Question 10

1. Which type of secondary structure is illustrated at right?
2. How does the orientation of the hydrogen bonds differ between the a-helix and the B-sheet structures?
3. In proteins (e.g., the GPCRs for glucagon and the catecholamines) that contain several -helical membrane-spanning domains, why would Pro not be one of the amino acids found in these domains?

Answer 10

1. The figure illustrates an a-helix , a right-handed, helical, secondary structural element commonly encountered in both fibrous and globular proteins.
2. The hydrogen bonds in a coiled a-helix are intrachain bonds that are parallel to the polypeptide backbone, whereas those in a B-sheet (an extended structure) can be intra- or interchain bonds (depending on whether they form between sections of one polypeptide or between two polypeptides) that are perpendicular to the backbone. [Note: a-Helices and B-sheets may be components of supersecondary structures (motifs), such as a B-barrel.]
3. Pro contains a secondary amino group that is not compatible with the right-handed spiral of the a-helix because (1) it cannot participate in the hydrogen bonding and (2) it causes a kink in the protein. Consequently, Pro is not found in the membrane-spanning domains of proteins such as GPCRs. [Note: Amino acids with bulky or charged R groups can also disrupt formation of an a-helix.]

Question 11

1. What type of molecular interaction involved in stabilizing the tertiary structure of a protein is shown?
2. What type of interaction would likely occur between Asp and Lys?
3. The tertiary structures of proteins (such as albumin) that function in the extracellular environment are stabilized by the formation of covalent links between the oxidized side chains of which sulfur-containing amino acid(s)?

Answer 11

1. Shown are hydrophobic interactions between Ile and Leu, two amino acids with nonpolar R groups.
2. Ionic interactions ( salt bridges ) would likely occur between Asp (acidic R group) and Lys (basic R group).
3. Two sulfur-containing Cys residues, brought into close proximity by the folding of the peptide(s), are covalently linked through oxidation of their thiol side chains. The disulfide bonds formed stabilize the tertiary structure of the folded peptide, preventing it from becoming denatured in the oxidizing extracellular environment. [Note: Cys-containing albumin transports hydrophobic molecules (e.g., fatty acids and bilirubin) in the blood. Its levels are used as an indicator of nutritional status.]

Question 12

1. As illustrated, what secondary structural feature is enriched in the infectious form of a prion protein (PrP) as compared to the noninfectious form?
2. Why do most large denatured proteins not revert to their native conformations even under favorable environmental conditions?
3. What misfolded peptide formed by abnormal proteolytic cleavage is the dominant component of the plaque that accumulates in the brains of individuals with Alzheimer disease?

Answer 12

1. The B -sheet secondary structure is enriched in the infectious PrP^Sc form of a PrP, which causes the transmissible spongiform encephalopathies, as compared to the noninfectious PrP C form that is a-helical rich. T
2. The folding of most large proteins is a facilitated process that requires the assistance of proteins known as chaperones and ATP hydrolysis.
3. AB is the misfolded peptide produced by abnormal proteolytic cleavage of amyloid precursor protein by secretases. ABeta forms an extended B-sheet and spontaneously aggregates to form fibrils that are the dominant component of the amyloid plaque that accumulates in the brains of individuals with Alzheimer disease. [Note: The B-sheets in AB have exposed hydrophobic amino acid residues. The hydrophobic effect drives the aggregation and precipitation of AB.]

Question 13

1. Which His residue (A or B), as shown, is the proximal His? What is its function? What is special about the location of this amino acid?
2. What type of secondary structure is most abundant in Mb? Does Mb have a quaternary structure?
3. Rhabdomyolysis (muscle destruction) caused by trauma, for example, is characterized by muscle pain, muscle weakness, and dark-colored urine. The dark color of the urine is the result of excretion of _____, a condition known as_____.

Answer 13

1. Choice A is the proximal His. It forms a coordination bond with the Fe 2+ in the heme prosthetic group. Polar His is located in the nonpolar crevice where heme binds.
2. Mb is rich in a-helices. Because it is a monomeric protein, Mb does not have a quaternary structure.
3. Rhabdomyolysis (muscle destruction) caused by trauma, for example, is characterized by muscle pain, muscle weakness, and dark-colored urine (shown). The dark color of the urine is the result of excretion of Mb , a condition known as myoglobinuria.

Question 14

1. Which form of Hb (deoxygenated or oxygenated) is referred to as the R form? What determines the equilibrium concentrations of deoxyHb and oxyHb?
2. How does the structure of Hb change as O2 binds to the heme Fe2+?
3. What condition, characterized by a “ chocolate cyanosis ,” results from the oxidation of Fe2+ to Fe3+ in Hb? Why might replacement of the distal His cause this condition?

Answer 14

1. The oxygenated, high-O2 -affinity form of Hb is referred to as the R form. The availability of O2 determines the equilibrium concentrations.
2. The binding of O2 to the heme Fe2 + pulls the Fe2+ into the plane of the heme. This causes salt bridges between the two aB dimers to rupture, thereby allowing movement that converts the T to the R form.
3. Methemoglobinemia, characterized by a “ chocolate cyanosis ” (dark-colored blood, bluish colored skin), results from the oxidation of Fe2+ to Fe3+ in Hb. Because the distal His stabilizes the binding of O2 to the heme Fe 2+ , its replacement with another amino acid will favor oxidation of Fe2+ to Fe3+ and decreased binding of O2 .

Question 15

1. Use the figure to determine the approximate amount of O2 that would be delivered by Mb and Hb when the pO2 in the capillary bed is = 26 mm Hg.
2. Why is the O2 -dissociation curve for Hb sigmoidal and that for Mb hyperbolic?
3. How might RBC production be altered to compensate for changes to Hb that result in an abnormally high affinity for O2 ?

Answer 15

1. At a pO2 of 26 mm Hg, Hb would have delivered 50% of its O 2 , while Mb would have delivered less than 10%. Hb has a lower O2 affinity at all pO 2 values and a higher P50 than does Mb, as shown. [Note: P 50 is that pO 2 required to achieve 50% saturation of the O2 -binding sites.]
2. Hb is a tetramer. The O2-dissociation curve for Hb is sigmoidal because the four subunits cooperate in binding O2 . The first O2 binds to Hb with low affinity. As subsequent subunits become occupied with O2 , the affinity increases such that the last O2 binds with relative ease. Because Mb is a monomeric protein, it does not show cooperativity. Consequently, its O2-dissociation curve is hyperbolic , not sigmoidal.
3. RBC production typically is increased (a process known as erythrocytosis ) to compensate for changes to Hb that result in an abnormally high affinity for O2 : more RBCs = more Hb = more O 2 carried.

Question 16

1. Which curve (A or B), as shown, represents the lower pH?
2. List two other allosteric effectors that, when increased, result in a rightward shift of the Hb O2 -dissociation curve. What does this shift reflect? Do these allosteric effectors stabilize the R or the T form of Hb?
3. How does the binding of CO 2 to Hb stabilize Hb’s deoxygenated form?

4, What is the Bohr effect?

Answer 16

1. Curve B represents the lower pH (higher H+ concentration).
2. Increased amounts of CO2 and 2,3-BPG also result in a rightward shift of the Hb O2 -dissociation curve. The shift reflects increased off-loading (delivery) of O2 to the tissues. These allosteric effectors stabilize the T ( deoxygenated ) form of Hb, enabling O2 delivery.
3. When CO2 binds to the amino termini of the four Hb subunits, forming carbaminohemoglobin, the negative charge is used to form a salt bridge that helps to stabilize Hb’s deoxygenated (T) form.

Hb – NH2 + CO2 ←→ Hb – NH –COO- + H+

4. The Bohr effect refers to the increase in O2 delivery when CO2 or H+ increases. In actively metabolizing tissue, Hb binds CO2 and H+ and releases O2 . The process is reversed in the lungs.

Question 17

1. How does the subunit composition of HbF, as illustrated, influence the O2 affinity of HbF?
2. What form of Hb replaces HbF, and when does this occur?
3. What form of Hb is measured to assess glycemic control in individuals with diabetes?

Answer 17

1. HbF contains 2a and 2y subunits. Relative to the B subunits, the y subunits have a reduced affi nity for 2,3-BPG. This results in HbF having an increased affinity for O2 . [Note: HbF is needed to obtain O2 from maternal HbA, and its increased affinity for O2 enables this process.]
2. HbF is the major Hb found in the fetus and the newborn but represents less than 2% of the Hb in most adults because it is replaced by HbA (2a and 2B subunits) by about 6 months after birth.
3. Nonenzymatically glycosylated ( glycated ) Hb, HbA1c, is measured because its concentration in the blood is a refl ection of the average blood glucose concentration over the previous 3 months. [Note: The goal value for HbA 1c in adults with diabetes is less than 6.5%.]

Question 18

1. How do the sickled RBCs illustrated cause infarction (cell/tissue death due to obstruction of blood flow)?
2. Which type of globin chain precipitates in B -thalassemia?
3. Is HbC disease a sickling or nonsickling disease? Why?

Answer 18

1. Sickled RBCs cause infarction because the rigid polymer of HbS makes the sickled cells less deformable than the nonsickled cells and, therefore, less able to move through blood vessels. This can cause a blockage that obstructs the delivery of O2 .
2. B -Thalassemia is a defect in the ability to make B-globin. Consequently, it is the excess a-globin chains that precipitate.
3. HbC disease is a nonsickling disorder because Lys (a polar amino acid) is substituted for polar Glu. In contrast, in HbS disease ( SCA ), nonpolar Val is substituted for Glu.

Question 19

A woman, age 70 years, activates her medical alert system and is transported to the hospital by ambulance. The patient tells you she has a headache , feels weak , and is nauseated and drowsy. She vomited several times at home over the last few hours. She thinks she has the fl u. Upon talking with her, you discover that she has been without power for 2 days due to a recent snowstorm and has been using a kerosene space heater to keep warm. You suspect CO poisoning, send a blood sample to the clinical laboratory for analysis, and begin O2 therapy .

Why would CO poisoning cause an affected individual to feel weak and drowsy?

Answer 19

CO is a colorless, odorless gas produced by incomplete combustion of hydrocarbons. CO binds reversibly to the Fe2 + of Hb, forming HbCO (known as carboxyhemoglobin ). CO competes with O2 and binds with a 200-fold higher affi nity. The bound CO stabilizes the R ( oxygenated ) form of Hb and shifts the O2 -saturation curve to the left. Because CO does not easily dissociate from Hb, O2 is not delivered. Treatment with O2 is required to displace the CO. The patient’s blood test revealed that HbCO accounted for 16% of her Hb (reference value, less than 2%; higher in smokers and urban dwellers). [Note: At higher concentrations of CO, use of hyperbaric O2 therapy (100% O2 under pressure) may be required to displace the CO.]

CO poisoning decreases O 2 delivery

Question 20

1. In the collagen chain shown, what amino acid (present at every third position) is represented by the black ball?
2. What is special about this amino acid? Which is/are descriptors of type 1 collagen? It is:

A. a fibrous protein.

B. an extracellular (secreted) protein.

C. a fibril-forming collagen.

D. composed of three -helical proteins.

E. found only in bone.

3. What targets the prepro- chains of collagen to the RER?

Answer 20

1. The black ball at every third position shown represents Gly. Gly is the smallest amino acid, having only an H for a side chain. Gly fi ts into the restricted space where the three chains come together.
2. Type 1 collagen is:

A. a fibrous protein. TRUE.

B. an extracellular (secreted) protein. TRUE.

C. a fi bril-forming collagen. TRUE.

D. composed of three -helical proteins. FALSE. Although the chains in collagen are called chains, the abundance of Pro in these chains prevents formation of the helix.

E. found only in bone. FALSE. It also is found in skin, blood vessels, tendon, and the cornea of the eye, and is the most abundant protein in the body.

3. An amino acid sequence at the N terminus ( N-terminal signal sequence ) targets proteins such as the prepro- chains of collagen to the RER.

Question 21

1. What enzyme catalyzes the reaction shown? Where does the reaction occur?
2. What is the function of the reaction in the formation of collagen?
3. Which other amino acid undergoes the same posttranslational reaction during collagen synthesis?
4. A deficiency in vitamin C ( ascorbate ), the coenzyme for the reaction shown, causes______ , a disease characterized by the production of collagen with decreased tensile strength

Answer 21

1. Prolyl hydroxylase, an enzyme of the RER, catalyzes the hydroxylation of Pro to Hyp.
2. Hydroxylation maximizes formation of the interchain H-bonds that stabilize the triple helical structure of collagen.
3. Lys also undergoes posttranslational hydroxylation to Hyl (which is formed by lysyl hydroxylase) during collagen synthesis. [Note: Hyl is a substrate for O-glycosylation.]
4. A deficiency in vitamin C ( ascorbate ), the coenzyme for the reaction shown, causes scurvy . Vitamin C is the coenzyme for both prolyl hydroxylase and lysyl hydroxylase . Without the additional stability provided by Hyp and Hyl, collagen has decreased tensile strength. Patients with scurvy may have bruise-like ecchymoses (shown) as a result of blood vessel fragility.

Question 22

1. Does the removal of the N- and C-terminal propeptides from procollagen (as shown) occur intracellularly or extracellularly? What is the fate of the triple-helical tropocollagen formed in the cleavage reaction?
2. Why are the disulfide bonds in the C-terminal propeptide domain of procollagen important for the formation of functional collagen?
3. What name is given to the group of diseases that may result from defects in processing events during collagen synthesis?

Answer 22

1. Removal of the terminal N- and C-propeptides from procollagen occurs extracellularly and is catalyzed by N-terminal and C-terminal procollagen peptidases . The triple-helical tropocollagen molecules formed in the cleavage reaction spontaneously associate to form collagen fibrils that are organized into an overlapping parallel array. The array is then cross-linked to produce collagen fibers. [Note: The spontaneous association of tropocollagen is an example of the hydrophobic effect.]
2. The disulfide bonds in the C-terminal propeptide domain bring the three chains into correct alignment for triple helix formation. They are important for the formation of functional collagen.
3. Collagenopathies are diseases that may result from defects in collagen-processing events, such as removal of the terminal propeptides and hydroxylation of proline.

Question 23

1. What enzyme catalyzes the oxidative deamination reaction shown?
2. Does the reaction occur within or outside of a cell? What is the function of the reaction?
3. Menkes syndrome is a disease of severe Cu 2+ deficiency. Why is connective tissue fragility characteristic of this syndrome?

Answer 23

1. Lysyl oxidase catalyzes the oxidative deamination of Lys to allysine.
2. The reaction occurs outside of a cell (extracellularly). It forms reactive aldehydes (such as allysine) that condense with Lys residues in neighboring collagen molecules to form the covalent cross-links characteristic of mature collagen. [Note: Two allysine residues can form cross-links via aldol condensation.]
3. Connective tissue fragility is characteristic of Menkes syndrome because lysyl oxidase is a Cu2+ -requiring enzyme. Decreased activity of this enzyme, as a consequence of decreased Cu 2+ , would impair the final step in collagen synthesis. [Note: X-linked Menkes syndrome ( kinky hair disease ) is the consequence of a defect in the transporter that moves dietary Cu 2+ out of intestinal cells. This decreases Cu 2+ availability for the rest of the body. In addition to lysyl oxidase, other Cu 2+ -requiring enzymes ( cytochrome c oxidase ,dopamine hydroxylase, superoxide dismutase , and tyrosinase ) are affected.]

Question 24

1. Which heritable collagen-based disease is characterized by stretchy skin , as shown? Which type of collagen is affected in this disease? Which type of collagen is mutated in the vascular form of this disease that is associated with potentially lethal arterial rupture?
2. Which heritable collagen-based disease is characterized by bone fragility (as shown) and is the most severe form of the disease? Which type of collagen is affected in this disease?

Answer 24

1. Classic Ehlers-Danlos syndrome ( EDS ) is characterized by stretchy skin. In classic EDS, type V collagen is affected. The vascular form of EDS, associated with potentially lethal arterial rupture, is caused by mutations to type III collagen.
2. Osteogenesis imperfecta ( OI ), a heritable collagen-based disease, is characterized by bone fragility. Shown at right is the most severe form of OI, which typically is lethal in the perinatal period. Type I collagen is affected in OI. [Note: A common mutation in OI results in the replacement of Gly, which prevents appropriate packing of the a chains in the triple helix.]

Question 25

1. In elastin, a connective tissue protein that forms an extensively interconnected network with rubberlike properties, the interconnections are formed by cross-links, as shown. What name is given to these cross-links?
2. What is the role of fibrillin in the production of elastin?
3. Why does a deficiency of the protease that normally destroys neutrophil elastase lead to lung pathology? Why might the liver also be affected?

Answer 25

1. Desmosine cross-links between three allysine side chains and one unaltered lysyl side chain provide the extensive interconnections that give elastin its mechanical ability to stretch.
2. Fibrillin is one of the glycoprotein microfibrils that functions as a scaffold onto which tropoelastin is deposited. [Note: Once deposited, tropoelastin undergoes the lysyl oxidase– mediated oxidative deamination required for cross-link formation.]
3. AAT is a protease inhibitor that normally destroys neutrophil elastase. The elastase , a protease , can destroy elastin in the walls of lung alveoli, thereby causing emphysema if unopposed by AAT. AAT deficiency in the lungs is the result of mutations that cause polymerization and retention of AAT in the liver, the primary site of its synthesis. Hepatic retention can damage the liver and result in cirrhosis.

Question 26

1. Which one of the six major classes of enzymes is illustrated by the reaction shown?
2. Why is NAD said to function as a coenzyme– cosubstrate (not a coenzyme–prosthetic group) in enzymatic reactions such as the one shown?

3. McArdle disease type V GSD is caused by a defi ciency in muscle glycogen phosphorylase ( myophosphorylase ), an enzyme of glycogen degradation. How will a decrease in P i affect the activity of this enzyme?

Answer 26

1. Shown is an enzyme that belongs to the class known as oxidoreductases (that most commonly function as dehydrogenases ).
2. NAD functions as a coenzyme–cosubstrate in enzymatic reactions because it is only loosely bound to the enzyme and leaves the enzyme in a changed form. [Note: FAD is an example of a coenzyme–prosthetic group. It is tightly bound to the enzyme and is returned to its original form on the enzyme.]
3. Based on its designation as a phosphorylase, myophosphorylase (deficient in McArdle disease ) uses Pi to cleave bonds in glycogen. Therefore, a decrease in Pi will decrease enzymatic activity. [Note: The enzyme cleaves the a (1→4) glycosidic bond in glycogen, thereby generating the phosphorylated product glucose 1-P.]

Question 27

1. Enzymes are protein______ that increase the__ ______of a chemical reaction. As shown, they contain an_____ ___ , which is a small___ __ _____ on the surface of the enzyme to which a specific_____ binds, forming an _________complex leading to product formation. Binding may cause a conformational change in the enzyme, a process known as __ _______.
2. What is the difference between a holoenzyme and an apoenzyme?
3. Elevated blood ALP suggests a pathology. ALP is found primarily in the liver as ALP-1 and in bone as ALP-2. Levels of the two forms can help differentiate between a liver and a bone pathology. What term is used to describe the tissue-specifi c forms of an enzyme?

Answer 27

1. Enzymes are protein catalysts that increase the rate ( velocity ) of a chemical reaction. As shown, they contain an active site , which is a small pocket ( or cleft) on the surface of the enzyme to which a specific substrate binds, forming an enzyme-substrate complex leading to product formation. Binding may cause a conformational change in the enzyme, a process known as induced fit. [Note: RNA catalysts are referred to as ribozymes.]
2. A holoenzyme is an enzyme with its nonprotein component, and an apoenzyme is missing the nonprotein component. The nonprotein component is required for enzymic activity.
3. Isozyme ( isoenzyme ) is the term used to describe the tissue-specific forms of an enzyme, such as ALP-1 and ALP-2 . Isozymes catalyze the same reaction but differ in their amino acid composition (primary structure).

Question 28

1. What name is given to that region of the curve shown at right marked by the asterisk (*)?
2. Which arrow (blue or red) represents the free energy of activation of the uncatalyzed reaction?
3. How do enzymes dramatically increase the reaction rate relative to the uncatalyzed reaction?
4. How do enzymes affect the G of a reaction?

Answer 28

1. The asterisk (*) marks the transition state.
2. The blue arrow represents the free energy of activation of the uncatalyzed reaction.
3. The lower the free energy of activation, the faster the reaction rate. Enzymes lower the free energy of activation by (1) providing an alternate, energetically favorable reaction pathway and (2) stabilizing the transition state of this pathway. Stabilization increases the concentration of the reactive intermediate that can be converted to product, thereby increasing the reaction rate. [Note: The turnover number (k cat ), the number of substrate molecules converted to product per second, is increased.]
4. Enzymes have no effect on the delta G of a reaction. Therefore, the free energies of the reactants and the products are the same in the catalyzed and uncatalyzed reactions

Question 29

1. What processes are shown at right? What effect do they have on the velocity of an enzymecatalyzed reaction?
2. What general name is given to the enzyme that catalyzes the forward reaction?
3. What other environmental factors infl uence the velocity of an enzyme-catalyzed reaction?
4. Tyrosinemia type 1 ( infantile tyrosinemia ) is caused by a defi ciency in fumarylacetoacetate hydrolase that catalyzes the last reaction in the degradation of Tyr. It is treated with a drug that inhibits an enzyme earlier in the pathway. What is the biochemical rationale for this therapy?

Answer 29

1. Phosphorylation and dephosphorylation , covalent modifi cations to proteins, are shown. Depending on the enzyme, these modifi cations may increase or decrease the velocity of an enzyme-catalyzed reaction. [Note: The change in enzyme activity is the result of a conformational change in the enzyme caused by the covalent modifi cation.]
2. Kinases catalyze phosphorylation reactions using ATP as the phosphate source. They are opposed by phosphatases.
3. Changes in the concentration of the enzyme, coenzyme, and substrate; temperature; and pH are additional factors that infl uence the velocity of an enzyme-catalyzed reaction.
4. Nitisinone is prescribed for infantile tyrosinemia because it decreases production of the substrate for the hydrolase , thereby decreasing the velocity of the reaction. Additionally, by preventing substrate accumulation, this substrate reduction therapy prevents entry of the substrate into side reactions that produce harmful products.

Question 30

1. Supply the missing terms in the Michaelis–Menten equation shown.
2. What is the steady state assumption?
3. True or false: When the [S] is much less than the K m , the V 0 is proportional to [S], and the reaction is said to be fi rst order.
4. If 1/V 0 and 1/[S] were plotted, what shape would result? What is the X intercept on this plot? The Y intercept?
5. If a mutation to the gene that codes for an enzyme results in a 12-fold increase in the K m of the enzyme for its physiologic substrate, what effect has the mutation had on the affi nity of the enzyme for the substrate?

Answer 30

1. [S] is the missing term in the Michaelis–Menten equation.
2. The steady state assumption is that the concentration of ES does not change with time. That is, the rate of formation of ES is equal to that of the breakdown of ES to E + S and E + P.
3. True: When [S] is much less than the Km , the V0 is proportional to [S], and the reaction is said to be first order, as shown.
4. A straight line would be seen if 1/V 0 and 1/[S] were plotted. The X intercept on this Lineweaver-Burk plot is -1/Km , and the Y intercept is 1/Vmax.
5. Increasing the Km of the enzyme for its physiologic substrate decreases the affi nity of the enzyme for the substrate.

Question 31

1. What type of inhibition is shown?
2. Which line represents the uninhibited enzyme? Which line represents the highest concentration of inhibitor?
3. What type of inhibition results in a decrease in the apparent V max ? Is K m also affected by the inhibitor?
4. Orlistat, a weight-loss drug, covalently bonds to lipases that hydrolyze dietary fat (TAGs) and inhibits their enzymic activity. Is this an example of reversible or irreversible enzyme inhibition?

Answer 31

1. Competitive inhibition is shown, in which the inhibitor and the S compete for the same binding site on the enzyme. As a result, the apparent Km increases because a higher [S] is required to achieve 1/2 V max .
2. The blue line represents the uninhibited enzyme. The black line represents the highest concentration of inhibitor.
3. Noncompetitive inhibition results in a decrease in apparent Vmax. K m is not affected. [Note: In noncompetitive inhibition, the inhibitor does not compete with the S and can bind the E and the ES complex (as shown).]
4. Covalent bonding of an inhibitor to an enzyme (as seen with orlistat ) irreversibly inhibits the enzyme. [Note: Covalent modifi cation of an enzyme, such as is seen with the acetylation of COX by aspirin, also causes irreversible enzyme inhibition.]

Question 32

1. Which curve shown represents an allosteric enzyme?
2. Will a positive allosteric effector that influences the K0.5 shift the V0 versus [S] plot to the left or to the right?
3. Sialuria (sialic acid in the urine) is a rare, AD condition caused by a mutation in the rate-limiting enzyme of sialic acid synthesis. The mutation decreases the enzyme’s ability to bind CMP–sialic acid, the end product of the pathway. Why does this mutation result in increased production (and excretion) of sialic acid?

Answer 32

1. The green curve, with its sigmoidal shape, represents an allosteric enzyme.
2. A positive allosteric effector that influences K0.5 will shift the V 0 versus [S] plot to the left (as shown), refl ecting a lower K 0.5 . [Note: K0.5 is that [S] required to achieve half maximal velocity.]
3. The CMP–sialic acid is a feedback inhibitor of the pathway. Loss of this allosteric inhibition (as the result of decreased binding to the regulated enzyme) results in overproduction of sialic acid and, consequently, sialuria.

Question 33

A 66-year-old female is seen in the emergency department in the late evening. She was driven to the hospital by her husband. The woman reports she has had a “pressure” on her chest for the last few hours. She denies overt chest pain; jaw, neck, shoulder, arm, or epigastric pain; shortness of breath ( dyspnea ); and sweating ( diaphoresis ). The pressure does not seem to increase with exertion. The patient’s history is remarkable for hyperlipidemia that is being treated with diet and drugs. There is no family history of heart disease. The patient’s heart and respiratory rates are elevated. Blood is drawn and sent to the clinical laboratory for measurement of cardiac biomarkers. Sublingual nitroglycerin is given. An EKG is performed, and the results are consistent with a myocardial infarction ( MI ). Cardiac biomarker measurements reveal elevated levels of CK , the CK- MB (the isoform found predominately in cardiac muscle) to total CK ratio, and cTnI (a cardiac troponin).

What would be the characteristics of an ideal cardiac biomarker?

Answer 33

CK-MB is one of three isoforms of CK , an intracellular enzyme. CK -MB is the form found virtually exclusively in the heart, with CK -MM found in the brain, and CK -BB found in skeletal muscle. Troponins ( Tn ) are nonenzymatic, intracellular, regulatory proteins involved in contractility of skeletal and cardiac muscle. Both CK -MB and cTnI become elevated in the blood as a result of tissue necrosis in an MI, but their patterns are different. Although both cardiac biomarkers rise early in an MI, cTnI remains elevated for up to 10 days, whereas CK -MB remains elevated for up to 3 days.

An ideal cardiac biomarker would be (1) released by injury, (2) cardiac specifi c, (3) elevated shortly after the injury and remaining so for an extended period, and (4) easily measured.

Question 34

1. Will the reaction shown proceed spontaneously in the forward (B➔A) or the reverse (A➔B) direction?
2. Is the G of this reaction positive or negative at equilibrium?
3. Compare and contrast G and G 0 .
4. Glutamine synthetase catalyzes the amidation of Glu to Gln. However, the reaction is endergonic. How is this problem solved in cells (e.g., skeletal myocytes) that synthesize Gln?

Answer 34

1. The reaction shown will proceed spontaneously in the reverse (A➔B) direction because the G of the forward direction is positive. The forward reaction is endergonic , and it will not proceed unless energy is provided.
2. At equilibrium , the G 0 (neither positive nor negative). [Note: Equilibrium is the point at which no net chemical change occurs. Therefore, for A➔B at equilibrium, the ratio of [B] to [A] is constant regardless of their actual concentrations.]
3. G is a measure of the capacity of a system to do work as it proceeds to equilibrium. G can be determined under standard conditions in which the concentration of the reactants and products is 1M (
   G 0 ), or it can be determined at any specifi ed concentrations (
   G ). Thus, G 0 is a constant (a reference value) and G is a variable. Their relationship is shown in the equation at bottom right.

4.The problem of glutamine synthesis being endergonic is solved by the production of a common intermediate that couples the glutamine synthetase reaction to the exergonic hydrolysis of ATP, such that the net G of the coupled reactions is negative.

Question 35

1. The ETC shown is an assembly of carriers that accept e from the reduced coenzymes and generated in processes. As e move through the ETC to , the terminal acceptor, they release that is used to pump across the mitochondrial membrane, thereby creating a that drives the phosphorylation of to .
2. What is transferred from NADH to the FMN prosthetic group of NADH dehydrogenase in Complex I?
3. What are the two relatively mobile e carriers of the ETC?
4. Primary CoQ defi ciency is an AR genetic condition that affects CoQ synthesis. What are the functional consequences of this defi ciency?

Answer 35

1. The ETC shown is an assembly of carriers that accept e from the reduced coenzymes NADH and FADH 2 generated in oxidative processes. As e move through the ETC to O 2 , the terminal acceptor, they release energy that is used to pump H across the inner mitochondrial membrane (into the intermembrane space), thereby creating a H + gradient that drives the phosphorylation of ADP to ATP.
2. NADH transfers a hydride ion and a proton (2 e 2 H ) to the FMN prosthetic group of NADH dehydrogenase in Complex I of the ETC. The e are subsequently transferred to CoQ via Fe-S proteins.
3. CoQ (a lipid-soluble component of the inner membrane) and cytochrome c (a protein in the intermembrane space) are relatively mobile e carriers of the ETC. [Note: CoQ accepts e from several mitochondrial dehydrogenases .]
4. Primary CoQ deficiency will impede e transfer from both Complexes I and II, decreasing the production of ATP. This will typically manifest as muscle weakness and exercise intolerance.

Question 36

1. How do the cytochromes shown transfer e ? Which one of the Complexes is also called cytochrome c oxidase ?
2. Trace the path through Complex II of the e derived from the oxidation of succinate to fumarate.
3. Cyanide poisoning causes a cytotoxic hypoxia in which cells are unable to use O 2 , even if it is plentiful. Will cyanide poisoning affect the activity of NADH dehydrogenase ?

Answer 36

1. The iron of the heme prosthetic group in cytochromes readily interconverts between the oxidized ferric form (Fe 3 ) and the reduced ferrous form (Fe 2 ), enabling cytochromes to transfer e . Complex IV is called cytochrome c oxidase because the e acceptor is O 2 and not the prosthetic group of a protein. Complex IV contains Fe (in the heme component) and Cu.
2. The e from succinate are fi rst transferred to the FAD prosthetic group of SD , reducing it to FADH 2 , and then to the Fe 3 of the Fe-S proteins, reducing it to Fe 2 as the FADH 2 is reoxidized. The e are picked up from the Fe 2 by CoQ , reducing it to CoQH 2 as Fe 2 is reoxidized. No H are pumped at Complex II.
3. Cyanide binds and inactivates Complex IV. By preventing transfer of e to O 2 , it causes the ETC to “back up,” resulting in accumulation of the reduced forms of its e carriers. Therefore, NADH dehydrogenase will be inhibited, and the NADH/NAD ratio in mitochondria will increase. [Note: ETC inhibition results in inhibition of ATP synthesis in coupled mitochondria because ATP synthase requires the H gradient.]

Question 37

1. Does the fi gure show NADH being oxidized or reduced? FMN?
2. If the two reactions shown below were coupled, in which direction would the e fl ow?
   Ubiquinone 2 H 2 e ➔ ubiquinol Eo 0.045V
   NAD 2 H 2 e ➔ NADH H E o 0.320V
3. What is the consequence of the incomplete reduction of O 2 to 2 H 2 O by the ETC as seen in reperfusion injury?

Answer 37

1. The NADH is being oxidized to NAD . The FMN is being reduced to FMNH 2 .
   * *Oxidation** is the loss of e and reduction the gain.

2.Ubiquinone 2 H 2 e ➔ ubiquinol E o 0.045V
NAD 2 H 2 e ➔ NADH H E o 0.320V
Because e fl ow is from the redox pair with the lowest E 0 to the pair with the highest, e will fl ow from NADH to ubiquinone (CoQ). [Note: Electrons subsequently will fl ow from ubiquinone to O 2 because the redox pairs involved in the transfer have increasingly positive E 0 values.]

3.Incomplete reduction of O 2 to 2 H 2 O by the ETC, as seen in reperfusion injury caused by the rapid return of O 2 (e.g., with thrombolytic therapy for an MI ), produces ROS (e.g., O 2
, H 2 O 2 , and OH ). ROS damage DNA and proteins and cause lipid peroxidation. [Note: Enzymes such as superoxide dismutase , catalase , and glutathione peroxidase protect cells from ROS.]

Question 38

1. How is the fl ow of e through the ETC coupled to ATP synthesis, as shown?
2. What will happen to e fl ow through the ETC in the presence of oligomycin?
3. A myocardial infarction ( MI ) is usually caused by occlusion of a coronary artery by a thrombus. What would be the immediate effects on the mitochondrial ETC in the event of an MI?

Answer 38

1. Flow of e through the ETC results in energy release used to pump H from the mitochondrial matrix to the intermembrane space at Complexes I (4 H), III (4 H), and IV (2 H), creating an electrochemical gradient. The energy of the gradient is used to drive the phosphorylation of ADP to ATP by Complex V ( ATP synthase , F 1 /F o ATPase ). Thus, the gradient is the common intermediate that couples the processes.
2. Oligomycin inhibits H fl ux through the F o domain of ATP synthase , thereby inhibiting ATP production at the F 1 domain. In coupled mitochondria, inhibition of ATP synthesis inhibits the ETC because of the diffi culty of pumping additional H against the steep gradient. Flow of e will eventually stop.
3. Flow of e through the ETC requires the reduction of O 2 to 2 H 2 O by Complex IV. As O 2 becomes limited during the MI , the ETC slows and stops. [Note: Thrombolysis allows rapid reperfusion.]

Question 39

1. What is the function of the malate-aspartate shuttle shown?
2. The glycerophosphate shuttle delivers e to the ETC via FADH 2 . Is this shuttle more or less effi cient than the malate-aspartate shuttle in generating ATP?
3. In mitochondrial myopathies , why are the mutations more likely to be in mtDNA rather than nuclear DNA?

Answer 39

1. The malate-aspartate shuttle moves reducing equivalents from the cytosol to the mitochondria because the inner mitochondrial membrane lacks a transporter for NADH. Cytosolic NADH is oxidized as OAA is reduced to malate, for which there is a membrane transporter. Mitochondrial malate is oxidized to OAA as NAD is reduced to NADH H . [Note: The OAA is transaminated to Asp.]
2. The NADH of the malate-aspartate shuttle is oxidized by Complex I, whereas the FADH 2 of the glycerophosphate shuttle is oxidized by CoQ. The P/O ratio for NADH is 3 and for FADH is 2. Therefore, more ATP will be generated using the malate-aspartate shuttle. [Note: The lower P/O ratio for FADH 2 refl ects the smaller number of H pumped (and, therefore, fewer ATP made) because FADH 2 is not oxidized by Complex I.]
3. Only a minority of the proteins required for OXPHOS is encoded by mtDNA , with most encoded by nuclear DNA. However, the mutational rate of mtDNA is about 10 times that for nuclear DNA.

Question 40

A 3-year-old girl is brought to the emergency department for suspected salicylate poisoning on the advice of the Poison Control Center her mother called after fi nding the child with an empty bottle of oil of wintergreen (used as a rub by the child’s grandfather to reduce his arthritic pain). At arrival, the child is hyperventilating and is shown to have a mixed acid–base disorder. Her breath has the odor of wintergreen. She is tachycardic and hyperactive. Her body temperature is elevated . The mother reports that the child vomited several times at home, was sweaty , and seemed to have trouble hearing. This constellation of signs and symptoms is suggestive of salicylate poisoning, and her blood level of salicylate was later determined to be 60 mg/dl (reference level is 0 mg/dl, with a therapeutic range 30 mg/dl). Salicylate ingestion is a common cause of poisoning in children. [Note: Oil of wintergreen is a non-aspirin source of salicylate and contains 7,000 mg of salicylate per teaspoon.]

Explain the finding of hyperthermia , given that salicylate can cause uncoupling of OXPHOS.

Answer 40

Salicylic acid (un-ionized) moves into the mitochondrial matrix where it ionizes to salicylate. [Note: When the pH is above the pK, the acid deprotonates. The pH is higher in the matrix because H are pumped out by the ETC.] By bringing H into the matrix, salicylic acid dissipates the H gradient. Because the phosphorylation of ADP to ATP by ATP synthase depends on this gradient, ATP synthesis stops, thereby preventing energy capture from e transport as ATP. Instead, the energy is lost as heat, causing hyperthermia. [Note: The gradient allows H fl ow through the F o domain of ATP synthase , causing rotation within the domain and resulting in conformational changes in the F 1 domain, which allow ADP phosphorylation.] Because the transport of e is not inhibited, O 2 utilization will continue. Salicylic acid, therefore, uncouples the ETC from OXPHOS. The drug 2,4-DNP works in a similar manner. UCP1 is a physiologic uncoupler found in BAT. It works in a manner different than exogenous uncouplers, and forms channels in the inner mitochondrial membrane that allow H to reenter the matrix, as shown. The resulting production of heat (nonshivering thermogenesis ) maintains body temperature in neonates.
With uncouplers , the energy from e transfer through the ETC is not captured as ATP but is lost as heat.

Question 41

1. What type of bond is displayed above the red question mark?
2. What enzyme joins monosaccharides to form a disaccharide?
3. Galactosemia is a rare metabolic disorder that inhibits an individual’s ability to metabolize the monosaccharide galactose. Why would those with galactosemia be advised to avoid dairy products?

Answer 41

1. It is a glycosidic bond between two hexoses. Specifi cally, it is a (1➔4) glycosidic bond.
2. A glycosyltransferase joins monosaccharides to form a disaccharide. [Note: Glycosyltransferases are also required for the formation of oligo- and polysaccharides.]
3. Dairy products contain lactose , which is a disaccharide of galactose and glucose. Because galactosemia is a defi ciency in the ability to metabolize galactose, individuals with the disorder are advised to avoid dairy products.

Question 42

1. Match the red letters shown with these three types of structural relationships: C-2 epimers, C-4 epimers, and isomers that are not epimers.
2. Define the term enantiomer.
3. Formation of which type of glycoside is impaired in individuals who cannot attach carbohydrate to selected Asn residues in a protein?

Answer 42

1. A C-4 epimers and B C-2 epimers. C isomers that are not epimers because they differ in confi guration at more than one carbon. [Note: Fructose has a keto group at the carbonyl C and is classifi ed as a ketose. The other sugars shown have an aldehyde group and are aldoses.]
2. An enantiomer is a special type of isomer in which the pair of structures is mirror images of one another, such as D- and L-glucose. [Note: Most sugars exist as D isomers, whereas most amino acids exist as L isomers.]
3. N-glycoside formation is impaired in individuals who cannot attach carbohydrate to the amide N of selected Asn residues in a protein, a process known as N-glycosylation. Impairments in this process are classified as congenital disorders of glycosylation ( CDGs ).

Question 43

1. Which red letter(s) shown identifi es an anomeric carbon?
2. What chemical feature identifi es a sugar as a reducing sugar?
3. Are reducing sugars normally detected in urine?

Answer 43

1. Letters A and C identify anomeric carbons (carbonyl carbons now bound to two oxygens).
2. Reducing sugars contain a free hydroxyl group on the anomeric carbon of the cyclic form. In the acyclic form, the carbonyl carbon of the aldehyde group is oxidized (forming a carboxyl group) as a chromogenic agent (e.g., Benedict’s reagent ) is reduced. [Note: Fructose is a reducing sugar because its ketone group can isomerize to an aldehyde.]
3. Reducing sugars are not normally detected in urine, and their presence is indicative of a pathology (e.g., galactosemia ). If urine gives a positive result with a reducing-sugar test (e.g., Benedict’s reagent), the identity of the sugar can be determined using more specific tests (such as the glucose oxidase test for glucose).

Question 44

1. During mastication, acts briefly on dietary starch and glycogen, hydrolyzing random bonds.
2. Why are humans unable to digest cellulose, a plant carbohydrate?
3. In a rare, congenital disorder of carbohydrate digestion, affected individuals cannot break down the disaccharides sucrose and maltose. What protein is defi cient in this disorder?

Answer 44

1. During mastication, salivary -amylase acts briefl y on dietary starch and glycogen, hydrolyzing random (1➔4) bonds.
2. Humans do not make the (1➔4)-endoglucosidases that are necessary to digest cellulose.
3. The SI protein is defi cient in those who cannot break down sucrose and maltose. SI is synthesized and then cleaved into two functional subunits that form the sucrase–isomaltase (SI) enzyme complex, which hydrolyzes dietary sucrose, maltose, and isomaltose in the small intestine.

Question 45

1. What protein is responsible for the transport into the portal circulation of the three monosaccharides shown?
2. How are glucose and galactose transported into the intestinal mucosal cell?
3. How is fructose transported into the intestinal mucosal cell?
4. Individuals with gastroenteritis caused by giardiasis (infection with Giardia intestinalis ) commonly develop lactose intolerance that persists long after the infection has resolved. Why? What are the symptoms of lactose intolerance?

Answer 45

1. The GLUT-2 transporter allows for the transport of glucose, galactose, and fructose into the portal circulation.
2. Glucose and galactose are transported into the intestinal mucosal cell by SGLT-1.
3. Fructose enters the intestinal mucosal cell via GLUT-5.
4. Brush border cells are damaged by the gastroenteritis caused by giardiasis , resulting in disaccharide accumulation in the large intestine due to defi ciency of the disaccharidases made by these cells. Lactase , the brush border disaccharidas e that hydrolyzes lactose, is the last to recover in giardiasis. The movement of lactose into the large intestine will draw in water, resulting in bloating and an osmotic diarrhea (shown). Treatment of lactose intolerance is lactose avoidance.

Question 46

1. Which group of vitamins shown serves as coenzymes (or their precursors) for the enzymes of metabolism?
2. What is the product of the N 10 -formyl THF–dependent pathway shown?
3. Why do inadequate serum levels of folic acid (folate) cause a megaloblastic anemia?

Answer 46

1. The water-soluble vitamins ( B complex C ) are coenzymes (or their precursors). With the exception of vitamin B 12 , they are not stored.
2. Two molecules of N 10 -formyl THF are required as one-carbon donors in construction of the purine ring in purine nucleotide de novo synthesis.
3. Deficiency of water-soluble folate can be caused by increased demand (in pregnancy); decreased absorption (in alcoholism); and treatment with the DHFR inhibitor methotrexate , which increases folate excretion by preventing conversion to THF (shown). Folate defi ciency results in megaloblastic anemia because decreased purine nucleotide and dTMP synthesis impairs cell division but allows cell growth. The result is production of fewer, larger than normal RBC precursors. [Note: Women of childbearing age are advised to consume 0.4 mg/day of folate to reduce neural tube defect incidence.]

Question 47

1. What form of vitamin B 12 is required as the coenzyme for the mutase reaction shown?
2. What is the “folate trap,” and why does it lead to impaired DNA synthesis and cell division?
3. Follow-up blood tests on a patient with megaloblastic anemia reveal antiparietal cell Abs. What is the likely diagnosis?

Answer 47

1. Deoxyadenosylcobalamin is required to isomerize methylmalonyl CoA to succinyl CoA. Consequently, methylmalonic acid levels rise with B 12 defi ciency.
2. The “ folate trap ” links vitamin B 12 with folate. N 5 -methyl THF is the methyl group donor and B 12 is the initial acceptor (becoming methylcobalamin ) in the remethylation of Hcy to Met (shown). Because this is the only reaction in which THF carries and donates a methyl group, B 12 defi ciency will trap THF in a form that cannot be used in other reactions, preventing the N 10 -formyl THF formation required for purine synthesis and the N 5 ,N 10 -methylene THF required for dTMP synthesis. Consequently, DNA synthesis and cell division are impaired. Hcy levels rise with B 12 deficiency.
3. Megaloblastic anemia with antiparietal cell Abs is diagnostic for pernicious anemia. B 12 absorption by mucosal cells of the ileum requires IF , a glycoprotein made by gastric parietal cells. Autoimmune destruction of these cells prevents IF synthesis and, consequently, B 12 absorption, thereby causing pernicious anemia. [Note: Pernicious anemia also manifests with CNS effects if the disease is untreated. Because folate supplementation can mask B 12 defi ciency, thereby delaying diagnosis, the megaloblastic anemia is treated with both vitamins until the cause is determined.]

Question 48

1. In what types of reactions does TPP (generated from thiamine, as shown) serve as a coenzyme?
2. How is thiamine (vitamin B1) defi ciency diagnosed?
3. What is the biologically active form of vitamin B 6 (pyridoxine)?
4. What syndrome is endemic in regions that rely on polished rice as a main staple in their diet?
5. Why might vitamin C (ascorbic acid) defi ciency result in defective connective tissue?

Answer 48

1. TPP is the coenzyme in the transfer of a two-carbon unit (by transketolase [panel A]) and the oxidative decarboxylation of -keto acids by PDH and -KGD (panel B). [Note: BCKD also requires TPP.]
2. Thiamine defi ciency can be diagnosed by an increase in RBC transketolase activity upon TPP addition.
3. The biologically active form of vitamin B 6 ( pyridoxine ) is PLP , a coenzyme in most reactions involving amino acids. [Note: Aromatic amino acid hydroxylases require THB, not PLP.] Vitamin B 6 is the only water-soluble vitamin to show toxicity ( sensory neuropathy ) in excess.
4. Beriberi , endemic where polished rice is a dietary staple, is a severe thiamine-deficiency syndrome characterized by neurologic and cardiac dysfunction. In the United States, thiamine ( vitamin B 1 ) defi ciency is seen primarily with chronic alcoholism and its associated decreased intake or absorption of the vitamin. Wernicke-Korsakoff syndrome , characterized by confusion, ataxia, nystagmus, memory loss, and hallucinations, may develop.
5. Vitamin C ( ascorbic acid ) defi ciency may result in scurvy , a connective tissue disease, because it is the coenzyme for prolyl and lysyl hydroxylase , enzymes that contribute to the tensile strength of collagen , a fi brous ECM protein.

Question 49

1. What are the precursors for NAD (shown), a coenzyme in many oxidation-reduction reactions?
2. What are the biologically active forms of vitamin B 2 (ribofl avin)? What general name is given to proteins that bind these active, coenzyme forms?
3. What nutritional defi ciency disease results from inadequate vitamin B3 (niacin) intake? Why might niacin be included in the treatment of familial combined (type IIb) hyperlipidemia in which both VLDL and LDL are elevated?
4. Why does a biotin cycle defect result in multiple carboxylase defi ciency?

Answer 49

1. The precursors of NAD are vitamin B 3 ( niacin or nicotinic acid ), nicotinamide (which gets deaminated), and Trp (which gets metabolized to quinolinate ). However, Trp conversion to NAD is ineffi cient. [Note: NAD can be phosphorylated to NADP , and the oxidized forms reduced.]
2. The active, coenzyme forms of vitamin B 2 ( riboflavin ) are FMN(H 2 ) and FAD(H 2 ). They are prosthetic groups in flavoprotein dehydrogenases such as NADH dehydrogenase (contains FMN) of the ETC and succinate dehydrogenase (FAD) of the ETC and the TCA cycle.
3. Inadequate niacin intake results in pellagra , a nutritional disorder characterized by the four Ds: diarrhea , dermatitis , dementia , and (if untreated) death. Niacin is useful in type IIb hyperlipidemia treatment because high doses (100 times the RDA) strongly inhibit adipose lipolysis, which reduces FFA availability for use in the hepatic synthesis of TAGs and, therefore, results in decreased VLDL synthesis (and, consequently, LDL production). [Note: Niacin decreases Lp(a) and increases HDL.]
4. A biotin cycle defect results in multiple carboxylase deficiency because of an inability to add biotin (by holocarboxylase synthetase ) during carboxylase synthesis or remove it (by biotinidase ) during degradation. Dermatologic and neurologic signs are characteristic. Biotin supplementation is the treatment.

Question 50

1. Which of the retinoids shown is required for vision? Which mediates the other actions of vitamin A?
2. How does retinoic acid (RA) lead to changes in gene expression?
3. How are the effects of vitamin A defi ciency related to keratin, a fi brous protein that helps to prevent moisture loss in epithelia?

Answer 50

1. Retinal derived from retinol ( vitamin A ) oxidation is required for vision. The 11-cis form bound to the protein opsin forms rhodopsin , which, when bleached by light, triggers a nerve impulse from the optic nerve to the brain. Consequently, an early sign of defi ciency is night blindness. Prolonged defi ciency leads to an irreversible loss of visual cells. RA (from retinal oxidation) mediates other actions of the vitamin (e.g., epithelial tissue maintenance). [Note: -Carotene from plants can be cleaved to two molecules of vitamin A. The conversion is ineffi cient in humans, and most of our need is met by preformed vitamin A from animal sources.]
2. RA binds to a nuclear receptor (the RAR ) that functions as a STF. The RA–RAR complex binds to REs on DNA and alters expression of retinoid-responsive genes (e.g., the gene for keratin). The RAR is part of a family of nuclear receptors that includes those for steroid and thyroid hormones and calcitriol.
3. Expression of the gene for keratin is inhibited by vitamin A. Consequently, with vitamin A defi ciency, excess keratin is made, which can cause xerophthalmia (pathologic dryness of corneal epithelium) that can lead to blindness. [Note: With excess vitamin A, a toxicity syndrome (hypervitaminosis A) can develop. In the chronic form, skin and hair are affected as a result of decreased keratin synthesis.]

Question 51

1. In addition to the preformed dietary precursors of active vitamin D shown, how else do humans obtain a precursor?
2. How and where is the active form of vitamin D generated? How is the process regulated?
3. What is the primary function of vitamin D?
4. What causes rickets and what are its effects?

Answer 51

1. Humans can synthesize the cholecalciferol ( vitamin D 3 ) precursor of active vitamin D from 7-dehydrocholesterol (an intermediate in cholesterol synthesis) in skin exposed to sunlight.
2. The active form of vitamin D ( calcitriol ) is produced from cholecalciferol by two sequential hydroxylation reactions: (1) hepatic 25-hydroxylase produces 25-OH-D 3 ( calcidiol ), the predominant form of vitamin D in the plasma and the major storage form; and (2) renal 25-hydroxycalciferol 1-hydroxylase produces 1,25-diOH-D 3 (calcitriol). 1-Hydroxylase activity is tightly regulated by plasma phosphate and Ca 2 levels. Low phosphate stimulates the enzyme directly, and low Ca 2 stimulates it indirectly through direct stimulation of PTH secretion. PTH upregulates 1-hydroxylase expression.
3. The primary function of 1,25-diOH-D 3 is to maintain adequate plasma levels of Ca 2 . When blood Ca 2 is low, it increases Ca 2 uptake by the intestine, minimizes Ca 2 loss by increasing renal reabsorption, and stimulates bone resorption (demineralization).
4. Vitamin D defi ciency impairs intestinal uptake of Ca 2 causing net bone demineralization and resulting in nutritional rickets in children and osteomalacia in adults. Incomplete collagen mineralization results in soft, pliable bones. Insuffi cient exposure to sunlight is a risk factor for defi ciency. [Note: Human milk is low in vitamin D.]

Question 52

Vitamin Functions Summary Card*****IMPORTANT

Answer 52

Question 53

Vitamin Deficiencies and Toxicities****IMPORTANT

Answer 53

Question 54

1. What type of bond links the dNMP monomers in each polymeric strand of the dsDNA molecule shown? What holds the strands together?
2. By convention, which strand of DNA is this: TTAGCCG? What dNMP is at the 5 -end of the sequence?
3. What is the functional signifi cance of A- and T-rich DNA compared to G- and C-rich DNA?

Answer 54

1. The dNMP monomers are linked by 3 →5 phosphodiester bonds that join the 3 -OH group of the deoxyribose of one dNMP to the 5 -OH group of the adjacent dNMP through a phosphoryl group. The complementary and antiparallel strands are held by H-bonds between the AT and GC bps and by hydrophobic interactions between the stacked bases.
2. By convention, (1) if only one strand of dsDNA is shown, it is the coding strand, and (2) a nucleic acid sequence is written 5 to 3 , making T (dTMP) the 5 -end of the TTAGCCG sequence.
3. AT bps are held together by two H-bonds , whereas GC bps are held by three. Consequently, AT pairs denature (“melt”) at a lower temperature than do GC pairs. [Note: The site where DNA synthesis begins (the origin of replication) is rich in AT bps to allow the strands of the double helix to separate. Separation is maintained at the origin by SSB protein.]

Question 55

1. Use the figure to list the order in which the following enzymes of prokaryotic replication function: DNA pol I , DNA pol III , helicase , ligase , primase , and topoisomerase.
2. Describe the function of each of the listed enzymes.
3. What is the clinical signifi cance of DNA gyrase inhibitors?

Answer 55

1. The order is: 1. helicase 2. topoisomerase 3. primase 4. DNA pol III 5. DNA pol I 6. ligase
2. Helicase unwinds dsDNA, and topoisomerases relieve the supercoiling created by unwinding via the cutting and rejoining of one strand (by topoisomerase I ) or both (by II ). Primase makes the RNA primer , then DNA pol III (a processive enzyme) elongates the primer with DNA ( 5 →3 polymerase activity) and removes errors ( proofreading 3 →5 exonuclease activity). DNA pol I removes the primer ( 5 →3 exonuclease activity), replaces it with DNA, and proofreads. Ligase joins the DNA made by pol I and pol III on the lagging strand, which is synthesized discontinuously (as Okazaki fragments ) in the 5 →3 direction (away from the replication fork). [Note: Replication is semiconservative because one parental strand is retained in each new duplex.]
3. Fluoroquinolones are drugs that inhibit DNA gyrase (a bacterial topoisomerase II). They prevent the ATP-dependent neutralization of positive supercoils by gyrase , thereby inhibiting bacterial replication and causing bacterial death.

Question 56

1. Which one of the high-fi delity eukaryotic DNA pols shown initiates replication of eukaryotic nuclear DNA? Which one is recruited to elongate the leading strand? Which one replicates mtDNA?
2. In what phase of the eukaryotic cell cycle does scheduled replication occur?
3. Why is 2 ,3 -dideoxyinosine used in HIV infection treatment?

Answer 56

1. DNA pol alpha is a multisubunit, multifunctional enzyme that makes the RNA primer required for DNA synthesis and extends the primer with dNMPs. DNA pol epsilon further extends the DNA on the leading strand. DNA pol gamma replicates mtDNA.
2. Scheduled DNA synthesis (in which the entire genome is replicated) occurs in the S phase (synthesis phase) of the eukaryotic cell cycle shown. [Note: Unscheduled replication to fi ll in gaps occurs as part of DNA repair.]
3. 2 ,3 -Dideoxyinosine ( ddI ), a purine nucleoside analog, is converted to ddATP in the cell. Because replication requires a 3 -OH group for the formation of a 3 →5 phosphodiester bond with the incoming dNTP, absence of this group in a dideoxynucleotide terminates replication by HIV’s reverse transcriptase. [Note: Dideoxynucleotides are also used in sequencing DNA.]

Question 57

1. What is the function of the 3-UCCCAA-5 sequence shown as part of telomerase ? What are telomeres?
2. What is the basic structural unit of eukaryotic chromatin?
3. What is epigenetics?

Answer 57

1. The 3-UCCCAA-5 sequence is the RNA template used to synthesize telomeric DNA , which consists of tandem repeats of the noncoding hexameric sequence AGGGTT. The RNA is a component of telomerase and is used by the enzymatic component (a reverse transcriptase ) to synthesize DNA in the 5 →3 direction. Telomeres are complexes of DNA and proteins that protect the ends of linear chromosomes. They normally shorten with each cell division, causing the cell to either senesce or apoptose, because most cells do not express telomerase. [Note: In some cells (e.g., cancer cells), telomerase expression allows unlimited division.]
2. The basic structural unit of eukaryotic chromatin is the nucleosome in which DNA is wound nearly twice around an octameric core of basic histone (H) proteins. Nucleosomes are joined by linker DNA bound by H1.
3. Epigenetics refers to heritable changes in gene expression without altering the nucleotide sequence of DNA. It includes covalent modifi cations to chromatin (e.g., the methylation of C in DNA, which decreases expression, and the acetylation of Lys in histones, which increases expression) and nucleosome repositioning. [Note: Epigenetic changes have been linked to cancer (e.g., through silencing of tumor suppressor genes).]

Question 58

1. In BER of the deamination of C to U, what enzyme removes U?
2. In prokaryotic MMR, how is the “correct” strand identifi ed?
3. What two systems are used in the repair of ds breaks in DNA?
4. What unique DNA lesion is caused by UV radiation? What process repairs the lesion? Inability to perform this repair results in what rare genetic disease?

Answer 58

1. In BER , uracil N-glycosylase removes U by cleaving the glycosidic bond between the base and the phosphorylated sugar of the nucleotide. An endonuclease and a lyase remove the sugar, creating an AP site. DNA pol and ligase complete the repair process. [Note: If not repaired, U will pair with A (rather than the correct CG pairing) in the next round of replication, causing a permanent change in DNA sequence (a mutation).]
2. In prokaryotic MMR , the degree of methylation identifi es the “correct” strand. The parental strand is more highly methylated than the daughter strand immediately after replication and is assumed to be the correct strand. [Note: Mutations to human MMR proteins result in hereditary nonpolyposis colorectal cancer ( HNPCC ).]
3. NHEJ and HR are two systems used in the repair of ds breaks in DNA. NHEJ is error prone because DNA lost in the process is not replaced.
4. UV radiation causes pyrimidine dimers , unique DNA lesions repaired by NER. [Note: NER differs from BER in that an oligonucleotide, not just a single nucleotide, is removed.] Xeroderma pigmentosum ( XP ) is the disease caused by an inability to repair these dimers due to defects in any of several XP proteins required for NER. With XP, early and numerous skin cancers result (shown).

Question 59

Which of the RNAs shown:
1. accounts for the largest percentage of cellular RNA?

2. carries amino acids to ribosomes?
3. is extensively modified only in eukaryotes?
4. contains a high percentage of modified bases?
5. is a ribozyme in translation?
6. contains extensive intrachain base-pairing?

Why is prokaryotic mRNA described as “polycistronic”? How does RNA differ from DNA?

Answer 59

1. rRNA accounts for the largest percentage of cellular RNA.
2. tRNA carries amino acids (bound to the 3􏰀-A) to ribosomes.
3. mRNA is extensively modified only in eukaryotes.
4. tRNA contains a high percentage of unusual bases (e.g., dihydrouracil and pseudouracil).
5. rRNA of the large ribosomal subunit is a ribozyme (RNA with catalytic activity) in translation and
   forms the peptide bond between amino acids.
6. tRNA contains extensive intrachain base-pairing that leads to a characteristic secondary structure
   (cloverleaf).

Prokaryotic mRNA is polycistronic because it encodes more than one gene (or cistron). [Note: Eukaryotic
mRNA is monocistronic.]
RNA differs from DNA in that it is smaller, contains ribose (not deoxyribose) and U (not T), and is single stranded. Its synthesis is selective (not “all or none”). [Note: RNA is like DNA in that it is an unbranched polymer of NMPs linked by 3􏰀→5􏰀 phosphodiester bonds, synthesized in the 5􏰀→3􏰀 direction.]

Question 60

How does prokaryotic RNA pol recognize and bind the appropriate region of DNA to initiate transcription, as shown? Is a primer required in transcription?

What is a “consensus sequence?”

What is the transcription product (conventionally shown) of the sequence TAGC (also conventionally shown)?

Individuals with tuberculosis are typically treated with a multidrug regimen that includes rifampin. How does rifampin work?

Answer 60

Prokaryotic RNA pol initiates transcription through the binding of its (sigma) subunit to consensus sequences within an untranscribed region of DNA known
as the promoter . The Pribnow box (TATAAT), located upstream (toward the 5 -end) of the transcription start site, is an example. RNA pol does not require
a primer and does not appear to have 3 →5 exonuclease activity (proofreading).

A consensus sequence is one in which the nucleotide base shown is the one most frequently encountered at that position (e.g., TATAAT).

The product of the transcription of TAGC is UAGC. By convention, if only one strand of DNA is shown, it is the coding strand and is written 5 to 3 . The RNA
product of transcription is identical to the coding strand, with U replacing T, and is written 5 to 3 .

Rifampin binds the subunit of prokaryotic RNA pol , preventing chain extension beyond three nucleotides, thus having a bactericidal effect . [ Note: The core
RNA pol contains 5 subunits: 2 and 1 (enzyme assembly), 1 (template binding), and 1 (5 →3 polymerase activity). Addition of forms the holoenzyme.]

Question 61

Which step in transcription is facilitated by formation of the hairpin structure shown? How else can this step be facilitated in prokaryotes?

What is the significance of PPi formation in replication and transcription?

A 23-year-old man, recently diagnosed with testicular cancer , is started on a drug regimen that includes dactinomycin (actinomycin D). Why is this drug
cytotoxic?

Answer 61

Transcription termination is facilitated by the hairpin structure (GC-rich stem loop) formed when a sequence in the DNA template generates a self-complementary
sequence in the nascent RNA. Just beyond the hairpin, a series of Us weakly base-pair to As in the template, facilitating separation of the DNA and RNA.
This intrinsic termination is the norm. Termination can also be facilitatedby bifunctional rho protein that binds to and moves along the RNA ( ATPase activity). When rho reaches RNA pol paused at the termination site, it separates the DNA–RNA hybrid helix ( helicase activity). [Note: Termination in eukaryotes is
RNA pol-dependent. Pol II termination of mRNA transcription is linked to 3-end
polyadenylation.]

The PP i formed in replication and transcription is hydrolyzed to 2 P i by pyrophosphatase . The loss of product drives polymerization in the forward direction, making it essentially irreversible. This is a common theme in
biochemistry.

Dactinomycin ( actinomycin D ) has antibiotic and anti-tumor activity. It intercalates between CG bps in DNA and interferes with the movement of RNA pol in both prokaryotes and eukaryotes. Because of its high cytotoxicity, dactinomycin is not used clinically as an antibiotic but is used to treat a variety of cancers.

Question 62

What role do the reactions shown play in eukaryotic transcription?

What is the function of RNA pol II ?

Based on the description below, does the cortisol receptor–cortisol complex function as a GTF or a STF?

Answer 62

The reactions shown covalently modify specifi c Lys residues in the histone components of eukaryotic chromatin. HAT acetylates Lys and eliminates its positive charge, thereby decreasing the strength of the interactions between histones and negatively charged DNA. This decondenses the chromatin, allowing access to DNA for transcription. HAT, then, is a co-activator. HDAC deacetylates Lys and favors
chromatin condensation. [Note: Decondensed, transcriptionally active chromatin is euchromatin, whereas the condensed, inactive form is heterochromatin.]

RNA pol II synthesizes nuclear pre-mRNA. It also synthesizes some small, noncoding RNAs such as snoRNA (for rRNA processing), snRNA (for mRNA
splicing), and miRNA (for RNAi). [Note: RNA pol I synthesizes rRNA, and RNA pol III synthesizes tRNA and 5S rRNA.]

Based on the description, the cortisol receptor in complex with cortisol, a steroid hormone, functions as a STF [Note: Cortisol mediates the stress response.].

Question 63

What are GTFs? What is the function of the GTF TFIID shown?
What is an enhancer?
Why is -amanitin (a cyclic peptide produced by some mushrooms) toxic to eukaryotic cells?

Answer 63

GTFs are trans-acting proteins that bind cis-acting consensus sequences in the core promoter and
initiate all eukaryotic gene transcription. TFIID , a GTF for RNA pol II , recognizes and binds promoter
elements (e.g., the TATA box), making it functionally analogous to prokaryotic . [ Note: TFIIF brings pol II
to the promoter. Phosphorylation of the polymerase by TFIIH allows it to escape the promoter and initiate
elongation.]
An enhancer is a DNA sequence involved in upregulating gene transcription in specifi c tissues under specifi c conditions. Relative to the gene, it can either be on the same strand of DNA or on the other, up- or downstream, and close or far away. Enhancers contain cis-acting REs to which trans-acting STFs bind.
Through looping of the DNA, the STFs can interact with GTFs bound to the promoter, as shown.

RNA pol II is highly sensitive to -amanitin , which binds the polymerase and interferes with its movement along the DNA, inhibiting eukaryotic mRNA synthesis.

Question 64

What four posttranscriptional processing events generate the functional tRNA molecule
(whose secondary structure is shown) from the primary tRNA transcript?

What three posttranscriptional processing events convert eukaryotic pre-mRNA to its functional form?

Why do mutations to DNA that change or delete the conserved dinucleotide sequences at
the 5 or 3 splice site end in the pre-mRNA for a Cu 2 transporter result in the most severe
phenotype of Menkes syndrome?

Answer 64

A functional tRNA molecule is generated from its primary transcript by (1) removal of nucleotides
from the 5 - and 3 -ends, (2) addition of CCA (orange boxes) to the new 3 -end, (3) removal
of an intervening sequence from the anticodon loop by endo - and exonucleases , and (4) base
modifi cations (yellow boxes).

Eukaryotic pre-mRNA is converted to functional mRNA by (1) addition of a methylated
guanosine cap to the 5 -end via an unusual 5→5 triphosphate link, (2) addition of a polyA
tail by polyadenylate polymerase to the 3 -end created by cleavage just past an AAUAAA
sequence, and (3) splicing (removal of noncoding introns and joining of coding [expressed]
exons ) by two transesterifi cation reactions mediated by snRNPs of the spliceosome . The fi rst
creates a lariat via an unusual 2→5 phosphodiester linkage between the branch site A and the
splice donor site G . The second cleaves the lariat and joins adjacent exons via a 3 →5 linkage.

Mutations that change or delete the conserved dinucleotide sequences at either the 5 splice
donor (GU) or 3 splice acceptor (AG) site have severe consequences because they reduce
functional mRNA generation and, therefore, its functional protein product such as the intestinal Cu 2
transporter in Menkes syndrome .

Question 65

The genetic code is a triplet code of four nucleotide bases (U, C, A, and G, as shown). Why do only 61 of the 64 triplets code for amino acids in the translation of mRNA?

What are the three possible consequences of changing a single nucleotide base in the coding region of a mRNA?

Is cystic fibrosis ( CF ) a triplet expansion disease?

Answer 65

Only 61 of the 64 triplets code for amino acids because 3 of them (UAA, UAG, and UGA) are stop ( termination ) codons , which are recognized and bound
by RFs that terminate mRNA translation.

Changing a single nucleotide base in the coding region of a mRNA (a point mutation ) results in a new codon that may (1) code for the same amino acid
because of the degeneracy of the code (a silent mutation), (2) code for a different amino acid (a missense mutation ), or (3) be a termination codon
(a nonsense mutation ).

CF is not a triplet expansion disease. The most common cause of CF is loss of the codon for Phe (F) at position 508 ( F508) in the CFTR protein. In contrast,
triplet expansion diseases are characterized by extra copies of a trinucleotide (e.g., Huntington disease , in which expansion occurs in the coding region,
resulting in extra Gln residues in the protein, and fragile X syndrome , in which it occurs in the 5 - UTR , resulting in gene silencing). [ Note: A frame-shift
mutation occurs if a number (not a multiple of 3) of nucleotides is added or deleted, altering the mRNA’s reading frame.]

Question 66

What enzymes generate a charged tRNA (shown)? What two activities do these enzymes
possess?

What are “isoaccepting tRNAs”?

What name is given to the ability of a tRNA to recognize and bind to more than one codon for
a specific amino acid?

What is the role of eukaryotic ribosomes that remain in the cytosol compared to those that
associate with the ER membrane?

Answer 66

Aminoacyl tRNA synthetases use ATP in the two-step process shown that generates a charged tRNA,
one in which an amino acid is esterifi ed to the 3 -A. The synthetases are also able to proofread and remove an incorrect amino acid from the enzyme or the tRNA.

Isoaccepting tRNAs are all the tRNAs that can be charged with the same amino acid. They have different anticodons .

A tRNA’s ability to recognize and bind more than one codon for a specifi c amino acid is known as wobble .
It results from nontraditional pairing between the third (3 ) base in the mRNA codon and the fi rst (5 ) base
in the tRNA anticodon. [Note: Recall that two strands of nucleic acid orient in an antiparallel manner.]

Eukaryotic ribosomes that remain in the cytosol synthesize proteins required at that site or in the nucleus, mitochondria, or peroxisomes. In contrast, those on the ER membrane synthesize proteins that will be secreted; incorporated into the plasma membrane; reside in the ER, Golgi, or lysosomes or be incorporated into their membranes. [Note: ER with ribosomes attached is termed RER.]

Question 67

Does the figure show the product of translation initiation in a prokaryote or eukaryote? Which
ribosome is depicted, a 70S or an 80S? To which site on the ribosome is the charged tRNA i
bound?

How is the start AUG distinguished from other AUGs in eukaryotic translation? Contrast with
prokaryotes.

What NTP gets hydrolyzed in all three steps of translation?

Vanishing white matter disease ( VWM ) is a severe AR neurodegenerative disorder caused by mutations to eIF2B, a GEF of translation. What is the function of GEFs?

Answer 67

fMet (made from N 10 -formyl-THF and Met after charging of the tRNAi ) is characteristic of
prokaryotic initiation, so a 70S prokaryotic ribosome (30S 50S subunits) is depicted. Eukaryotic
ribosomes are 80S (40S 60S). The charged tRNA iis bound to the P site of the ribosome and is the only tRNA that fi rst goes to this site. All others go to the A site .

In eukaryotes, the start AUG is distinguished by its proximity to the 5 cap, which is bound by
proteins of the eIF-4 family . The 40S subunit binds near the cap and scans the mRNA for the
fi rst AUG, an ATP-requiring process. In prokaryotes, in contrast, the purine-rich SD sequence
upstream of the start codon pairs with the 16S rRNA of the 30S subunit, positioning the subunit on
the mRNA at the start codon without scanning.

GTP is hydrolyzed to GDP in translation initiation, elongation, and termination.

GEFs reactivate guanine nucleotide–binding factors by removing GDP and allowing GTP to
bind. eIF-2-GTP recognizes the tRNA i and takes it to the P site of the 40S subunit. When the 60S
subunit joins, the GTP is hydrolyzed to GDP. The GEF for eIF-2-GDP is eIF-2B mutated in VWM .

Question 68

How is the elongation intermediate shown achieved in prokaryotes?

What role does a ribozyme play in elongation?

Why are two GTP hydrolyzed for every amino acid added to a peptide during elongation?

Which aspects of prokaryotic translation are inhibited by chloramphenicol, erythromycin,
streptomycin, and the tetracyclines?

Answer 68

The elongation intermediate is achieved in prokaryotes by EF-Tu-GTP , which brings all tRNAs (except tRNA i)
to the A site of the ribosome in response to codons on the mRNA, a process known as decoding . GTP is
hydrolyzed to GDP, and the GEF EF-Ts facilitates its removal.

rRNA (23S in prokaryotes and 28S in eukaryotes) of the large subunit is the catalyst that forms a peptide bond
by condensing the carboxyl end of the growing peptide at the P site with the amino group of the amino acid at the A site ( transpeptidation , shown). Therefore, the rRNA is a ribozyme . The catalytic activity is known as peptidyltransferase .

In addition to decoding and bond formation, elongation involves movement of the ribosome one codon in the 3 direction. This puts the peptidyl-tRNA in the P site, leaving the A site available, as the uncharged tRNA exits the E site . EF-G-GTP mediates movement in prokaryotes. GTP is hydrolyzed to GDP.

Prokaryotic translation is inhibited at the peptidyltransferase reaction by chloramphenicol , translocation by erythromycin , initiation by streptomycin , and elongation by the tetracyclines . [Note: Diptheria toxin covalently modifi es EF-2 and inhibits eukaryotic elongation.]

Question 69

Complete the chart by filling in the names of the factors that function in translation in a
eukaryotic cell (denoted by the blue E).

What is the primary regulatory step of eukaryotic translation? How is it regulated?

What is a “polysome”?

Individuals with Swyer syndrome are genotypically 46,XY and phenotypically female. In
some cases, the nuclear localization sequence (NLS) of the protein that functions as a STF
to initiate maleness is mutated. How would this result in a 46,XY female?

Answer 69

Initiation is the primary regulatory step of eukaryotic translation. It is inhibited by eIF-2 phosphorylation. Specific kinases phosphorylate the factor in response to the environmental stresses of unfolded proteins in the ER, amino acid deprivation, heme deficiency in erythroid cells, and viral infection. Inhibiting translation at initiation conserves energy (primarily as GTP).

A polysome ( polyribosome ) is a complex of more than one ribosome simultaneously translating one mRNA.

STFs bind to response elements in nuclear DNA. Mutation to the NLS of the STF that initiates maleness would prevent its posttranslational targeting from the cytosol to the nucleus, preventing transcription of the genes required for maleness. Consequently, the
fetus will develop as a 46,XY female.

Question 70

What two effects does phosphorylation (shown) have on protein activity?

What proteins of coagulation undergo -carboxylation?

In I-cell disease , the inability to phosphorylate mannose to mannose 6-P prevents the
cotranslational targeting of acid hydrolases . What is the effect on lysosomes? Where does
the mannose originate?

Answer 70

Phosphorylation , a covalent modifi cation, causes conformational changes in a protein that can lead to
(1) activation (e.g., glycogen phosphorylase and the STF CREB) and (2) inactivation (e.g., glycogen synthase ).

The clotting proteins FII, FVII, FIX, and FX and the anticlotting proteins C and S undergo vitamin K–
dependent -carboxylation of specifi c Glu residues. [ Note: Biotin-dependent carboxylation of Lys residues is
seen with carboxylases that use pyruvate, acetyl CoA, propionyl CoA, and methylcrotonyl CoA as substrates.]

In I-cell disease , lysosomes contain inclusion bodies of undegraded materials as a consequence of the
mistargeting of acid hydrolases because of a defect in the phosphotransferase required to generate the mannose 6-P lysosomal targeting signal. The mannose is part of an oligosaccharide made on dolichol (an ER membrane lipid) and transferred en bloc to the amide N of an Asn residue in the hydrolase as it moves through the ER lumen. The glycosidic linkage between Asn and GlcNAc of the oligosaccharide (shown) is formed by N-glycosylation .
[Note: In O-glycosylation, sugars are sequentially added to the –OH group of selected Ser, Thr, or Hyl residues.]

Question 71

How do proteins that function as trans-acting factors recognize and bind to cis-acting DNA elements, as shown?

Where are the genes that encode regulatory trans-acting factors located relative to the genes they regulate?

Give two examples of cis-acting elements and their trans-acting factors involved in regulation of the lac operon.

Why is the lac operon turned off in the presence of both glucose and lactose?

Answer 71

Trans-acting proteins contain structural motifs (e.g., the helix-turn-helix and zinc fi nger) that recognize and bind specifi c cis-acting DNA elements.

Genes that encode trans-acting molecules can be upstream or downstream of the regulated gene.
[Note: They can be on a different chromosome in eukaryotes.]

The lac repressor / operator site and the CAP / CAP site are trans-acting factors and cis-acting ele-
ments that regulate the lac operon . Prokaryotic operons contain the protein-coding genes (sequentially ordered) required for a pathway (e.g., the use of lactose and the synthesis of Trp) and the regulatory
elements that control their transcription. Operons are not found in eukaryotes. [ Note: Transcription is
the primary site for regulation of gene expression.]

In the presence of lactose, allolactose (its isomer) binds to the repressor protein and prevents it from
binding the operator. However, if glucose is also present, cAMP is not available to form a complex with
CAP protein and bind the CAP site, which prevents effi cient transcription initiation and turns off the
operon.

Question 72

How does the trp operon detect the elevated levels of Trp that result in attenuation, as shown?

Why is the lac operon described as “inducible,” whereas the trp operon is “repressible”?

How do prokaryotes coordinate both transcriptional and translational responses to environmental stresses (e.g., amino acid starvation)?

Answer 72

The 5 -end of the mRNA from the trp operon contains two consecutive codons for Trp. If charged tRNA Trp
is plentiful, the ribosomes move past these codons as they translate the polycistronic mRNA, allowing a
hairpin structure to form that attenuates transcription (stops it before completion). However, if charged
tRNA Trp is in short supply, the ribosomes stall, an alternative structure forms, and transcription continues.
[Note: Attenuation is possible because prokaryotic translation begins during transcription.]

Lactose prevents the repressor binding to the operator, which relieves inhibition of the lac operon and allows it to be induced (turned on) when glucose is unavailable. In contrast, Trp (a corepressor ) facilitates repressor binding, which maximizes trp operon inhibition and causes it to be repressed (turned off).

In amino acid starvation, an uncharged tRNA at the ribosomal A site activates stringent factor ( RelA ), which synthesizes ppGpp (an alarmone ) and results in inhibition of rRNA and tRNA synthesis. Conse-
quently, ribosomes cannot be made, and excess ribosomal proteins bind their polycistronic mRNAs at the SD sequences, blocking translation. This keeps ribosomal protein and rRNA synthesis in balance.

Question 73

How does the binding of the hormone glucagon to its cell-membrane GPCR with production of the second messenger cAMP (as shown) alter the cell’s transcriptional profile?

How do STFs mediate combinatorial control of transcription?

How does hydrocortisone, a glucocorticoid, decrease infl ammation ?

Answer 73

Glucagon binds to and activates its cell-membrane GPCR, upregulating production of cAMP by
adenylyl cyclase and, consequently, activating PKA . The kinase phosphorylates and activates CREB ,
a trans-acting STF. CREB binds to cis-acting CREs throughout the genome, affecting the expression of
glucagon-responsive genes (e.g., PEPCK ), thereby changing the cell’s transcription profile.

STFs bound to DNA recruit proteins to their activation domain, forming a multiprotein complex that
activates transcription. Regulation is achieved by the specifi c combinations of proteins in the complex,
a process known as combinatorial control .

Hydrocortisone (cortisol) enters the cell and binds to its cytosolic receptor, which induces a conformational change in the receptor that allows the receptor–hormone complex to enter the nucleus. The complex, a STF, binds to GREs throughout the genome, altering gene expression. Hydrocortisone induces expression of genes that code for proteins that have an anti-infl ammatory effect and represses pro-inflammatory genes.

Question 74

The eukaryotic gene shown contains fi ve exons. How can this one gene give rise to more than one protein product?

How do apo B-100 (made in the liver) and apo B-48 (made in the intestine) illustrate the use of one mature mRNA to produce different proteins in different tissues?

Why would inactivation of a gene encoding an IRP lead to decreased intracellular iron levels?

Answer 74

A single gene can give rise to more than one protein by alternative splicing , in which the use of different splice sites gives rise to different mRNAs in different tissues, greatly increasing the number of proteins that can be made from the 21,000 protein-coding genes in the human genome.

In the liver, 100% of the mature mRNA is expressed, giving rise to apo B-100 for VLDL synthesis. In the intestine, the mRNA undergoes an additional (and uncommon) modifi cation, RNA editing , in which deamination of a C to a U converts a sense to a nonsense codon. Creation of this stop codon allows translation of only the fi rst 48% of the mRNA and production of apo B-48 for CM synthesis.

TfR mRNA has cis-acting IREs at the 3 -end. When the iron levels in cells are low, trans-acting IRPs bind the
IREs and stabilize the mRNA, resulting in increased TfR synthesis that allows cells to endocytose transferrin-bound iron from the blood. Inactivation of the gene for an IRP would result in less TfR being synthesized (due to increased mRNA degradation), in turn, causing decreased intracellular iron levels.

Question 75

How does RNAi mediated by miRNA (as shown) regulate gene expression?

How can chemotherapy with methotrexate alter gene expression?

Why are methyltransferase inhibitors being explored as treatment for hemoglobinopathies that affect the
chain of Hb?

Answer 75

RNAi mediated by miRNA (a noncoding regulatory RNA) silences gene expression. Nuclear miRNA is
processed as shown and sent to the cytosol as ds miRNA. One strand (the guide strand) associates with RISC and hybridizes to the target mRNA, causing either decreased mRNA translation or increased mRNA degradation by an endonuclease of RISC. [Note: Synthetic siRNAs also can trigger RNAi and have clinical potential.]

Methotrexate inhibits DHFR , an enzyme required in the synthesis of dTMP and, consequently, DNA.
In response to methotrexate, amplification of the gene for DHFR occurs in some individuals, increasing DHFR production and leading to drug resistance.

Hypermethylation of C in CpG islands in the 5 regulatory portion of many genes silences expression, whereas hypomethylation increases it. Hypermethylation of the gene for the chain of Hb is involved in the switch to -chain expression shortly before birth. In hemoglobinopathies that affect the chain, methyltransferase inhibitors could increase expression of the chains and improve the clinical picture.

Question 76

How are the cohesive (“sticky”) ends shown created in dsDNA? Why are they useful in forming recombinant DNA?

What are the characteristics of a restriction endonuclease cleavage site?

Cloning of the gene for human insulin allowed for the production of the eukaryotic protein in a prokaryotic system. How can a DNA sequence be cloned?

Answer 76

The sticky ends are created by specifi c restriction endonucleases that cleave DNA in a staggered manner, generating single-stranded ends. Two pieces of DNA (from different sources) cut with the same restriction enzyme will have complementary “sticky” ends that can anneal and be covalently joined by DNA ligase , creating recombinant DNA . Some restriction enzymes generate blunt ends (shown).

Restriction endonucleases recognize short (4–8 bps), specific stretches of dsDNA that are palindromes with two-fold rotational symmetry (i.e., the sequence is the same on both strands if read in the same direction).

To clone a DNA sequence, a vector (e.g., a plasmid) is used that (1) is capable of autonomous
replication in the host, (2) contains a unique restriction site (also present in the DNA of interest),
and (3) contains antibiotic resistance genes. The plasmid and the DNA are cleaved by the same
restriction enzyme and ligated. The newly constructed plasmid is introduced into a bacterial host
( transformation ), which is grown in the presence of antibiotics to select for the recombinant
form of the plasmid. After the DNA has been amplifi ed ( cloned ), it can be released from the vector
and isolated [Note: Introduction of recombinant DNA into eukaryotic cells is termed “transfection”.]

Question 77

What specific type of plasmid vector is shown?

How is cDNA made?

Contrast a genomic library with a cDNA library. Which would be used to study mutations to promoter elements in the gene for beta globin?

Answer 77

An expression vector that allows a eukaryotic protein to be made in a prokaryotic cell is shown. The plasmid contains a bacterial promoter (for transcription) and a SD sequence (for translation by bacterial ribosomes). Eukaryotic cDNA is inserted downstream of the promoter and within a gene expressed by the bacterium. The mRNA product contains some codons for the bacterial protein and all the codons for the eukaryotic protein, which enhances translation
efficiency and results in a fusion protein .

cDNA synthesis requires a mRNA template, an oligo-dT primer, reverse transcriptase , and the four dNTPs.
The mRNA in the hybrid product is cleaved (e.g., by alkali or RNase H ) then removed and replaced by DNA pol . The resulting fragments are joined by DNA ligase , producing ds cDNA (shown).

A genomic library is a collection of fragments representing the entire genome of an organism, including introns and regulatory sequences, whereas a cDNA library is a collection of only the expressed sequences. As a dsDNA copy of mRNA, cDNA does not include introns and regulatory sequences. Because promoters and other regulatory elements are not expressed, a genomic library would be used to study them.

Question 78

What is the value of developing a labeled probe, as shown?

What is the biochemical basis of the technique used to sequence cloned DNA?

How can a synthetic nucleotide probe be used to determine whether or not an individual carries the sickle cell mutation?

Answer 78

A probe (a short piece of labeled ssDNA or RNA) is a screening tool designed to hybridize with the DNA of interest. It is used to identify which band on a gel or clone in a library contains the target DNA.

Sequencing of cloned, ssDNA involves annealing a primer to the 3 -end of the target DNA, adding DNA pol , all four dNTPs, and a limiting amount of
the four ddNTPs linked to different fl uorescent dyes. DNA pol elongates the chain until a ddNMP is incorporated. The absence of a 3 -OH group
terminates elongation and produces strands of every length, with the shortest representing the 5 -end. Separation of the products by size via gel electrophoresis followed by visualization of the fluorescent labels will yield a pattern of bands from which the base sequence of the complementary strand
can be read.

Under stringent hybridization conditions, a synthetic ASO probe will detect the sickle cell mutation in the gene for globin because it will hybridize only to the S sequence, as shown.

Question 79

What technique could be used to identify polymorphisms (variations in a particular DNA sequence) in two individuals, as shown?

What test would be performed to determine if a gene is being transcribed and the relative abundance of
product?

How could a family with a child that has phenylketonuria (PKU) determine if their fetus will be born with the disease?

Answer 79

RFLP analysis is used to study polymorphisms . An RFLP is a genetic variant observable if the length of the
restriction fragment is altered by creating or abolishing a restriction site or varying the number of short, tandem nucleotide repeats (shown). [Note: In RFLP analysis, a restriction digest of the DNA from each individual is separated by gel electrophoresis, denatured, and transferred (by blotting) to a membrane. A labeled probe is used to detect the presence (or absence) of the target sequence. The technique of digesting, separating, blotting, and detecting DNA is known as Southern blotting.]

Northern blotting would be used to determine if a gene is being transcribed and the relative abundance of its
RNA product. The technique is similar to Southern blotting except that it detects RNA, not DNA.

PAH is deficient in PKU . Mutations in the gene for PAH usually do not affect restriction sites. To use RFLP
analysis as a diagnostic screen for PKU, DNA from family members with and without PKU must be obtained to identify RFLPs closely linked to the gene ( indirect diagnosis ). [Note: Direct diagnosis is rare because only a few diseases (e.g., sickle cell anemia ) are caused by the mutation that causes the RFLP.]

Question 80

How might the transcriptional profile of a cancer cell (shown) be compared to that of a normal cell?

What are transgenic animals?

What is PCR?

How might PCR be used in genetic testing for cystic fibrosis (CF) caused by the F508 mutation in the CFTR protein?

Answer 80

The transcriptional (expression) profi le of a cancer cell and a normal cell can be compared by microarray analysis using gene chips , glass slides spotted (“arrayed”) with DNA fragments that represent specific coding regions. The mRNAs from each cell are converted to cDNAs, fluorescently labeled, and exposed to a gene chip. The amount of fluorescence bound to each spot is a measure of mRNA
abundance in the sample.

Transgenic animals are those that have been genetically altered to carry a foreign gene ( transgene ) in their genome through use of recombinant DNA techniques.
PCR is a technique to amplify targeted sequences of DNA. It requires short primers designed to be complementary to the 5 and 3 sequences fl anking the target, a heat-stable DNA pol , and dNTPs. The mixture is heated to form ssDNA, then cooled to allow the primers to anneal and be extended. The process (a repetitive cycle of denature, anneal, and extend) leads to an exponential increase in target DNA amount.

The F508 mutation in the gene for the CFTR protein causes a three-nucleotide deletion. Separation by size of the PCR product of the appropriate region of DNA can distinguish between homozygous normal, heterozygous ( carrier ), and homozygous mutant (CF-affected) individuals.