Adv. Biochem. Book Questions

Q1: The type of intercellular signaling in which one cell can communicate with another over long distances is called
A. autocrine
B. endocrine
C. juxtacrine
D. paracrine
E. synaptic

A1: B. endocrine – A hormone is released into the blood and travels to the target tissue.

Q2: Intracellular receptors
A. usually bind hydrophobic ligands.
B. may be located either in the cytosol or nucleus in unbound state.
C. when bound to their ligand, regulate gene transcription.
D. when bound to their ligand, function as dimeric complexes binding to specific DNA sequences.
E. all of the above.

A2: E. all of the above – These describe different aspects of intracellular receptor function.

Q3: A cell surface receptor
A. reacts only with molecules too large to cross the plasma membrane.
B. when bound to its ligand, could result in activation of an enzymatic cascade.
C. always opens an ion channel when bound to its ligand.
D. must produce a second messenger when it binds to its ligand.
E. is usually also called GPCR.

A3: B. when bound to its ligand, could result in activation of an enzymatic cascade – Signal transduction often occurs this way.

Q4: Cells can terminate signal transduction by cell surface receptors by
A. reducing agonist availability in the vicinity of the target cell.
B. internalizing and degrading the receptor–agonist complex.
C. modifying the receptor so that it is inactive or desensitized.
D. all of the above.
E. none of the above.

A4: D. all of the above – These are all mechanisms to terminate signaling.

Q5: Calmodulin is
A. a nonspecific kinase.
B. a protein that binds Ca²⁺.
C. a second messenger.
D. an activator of nitric oxide synthase.
E. a protein channel that facilitates the influx of Ca²⁺.

A5: B. a protein that binds Ca²⁺ – It is a ubiquitous Ca²⁺-binding protein.

Q6: The guanylate cyclase that responds to NO
A. is in the catalytic domain of a membrane receptor.
B. is also activated when atrial natriuretic factor (ANF) binds to its receptor.
C. increases activity because of a conformational change when NO binds to its heme.
D. is a monomeric enzyme.
E. is found only in smooth muscle cells.

A6: C. increases activity because of a conformational change when NO binds to its heme – The soluble enzyme has low activity when heme has not bound a gaseous molecule.

Q7: The PI system begins with activation of phospholipase C, which initiates a sequence of events including all of the following except
A. activation of IP₃ by action of a phosphatase.
B. increase in intracellular Ca²⁺ concentration.
C. release of diacylglycerol (DAG) from a phospholipid.
D. activation of protein kinase C.
E. phosphorylation of certain cytoplasmic proteins.

A7: A. activation of IP₃ by action of a phosphatase – IP₃ is active; phosphatase action inactivates it.

Q8: Which of the following statements concerning G proteins is correct?
A. G proteins bind the appropriate hormone at the cell surface.
B. GTP is bound to G protein in the resting state.
C. α-Subunit may be either stimulatory or inhibitory because it has two forms.
D. Adenylate cyclase can be activated only if α- and β-subunits of G protein are associated with each other.
E. Hydrolysis of GTP is necessary for G protein subunits to separate.

A8: C. α-Subunit may be either stimulatory or inhibitory because it has two forms – It depends on the specific hormone and receptor interaction.

Q9: Elements leading to increased cyclic AMP in response to GHRH binding to its receptor include
A. activation of a monomeric G protein.
B. activation of adenylate cyclase by αs-subunit of a Gs protein.
C. activation of cyclic nucleotide phosphodiesterase.
D. activation of protein kinase A.
E. all of the above.

A9: B. activation of adenylate cyclase by αs-subunit of a Gs protein – Adenylate cyclase then converts ATP to cAMP.

Q10: Low GTPase activity in the mutated protein results in constitutive activation of Gs and adenylate cyclase because
A. GTP-bound α-subunit does not reform the αβγ trimer.
B. GTP-bound G protein binds more strongly to the membrane receptor.
C. GTP reacts directly with adenylate cyclase to activate it.
D. the trimeric form of the G protein is stabilized.
E. adenylate cyclase is phosphorylated more readily.

A10: A. GTP-bound α-subunit does not reform the αβγ trimer – α-GTP dissociates from the βγ dimer, preventing reformation.

Q11: Which of the following statements about receptor tyrosine kinases is correct?
A. The catalytic domain is on the N-terminal end.
B. Activation of the kinase requires ATP.
C. Growth factor binding to the receptor triggers dimerization which activates the kinase activity.
D. Active tyrosine kinase can phosphorylate other proteins but not itself.
E. All of the above.

A11: C. Growth factor binding to the receptor triggers dimerization which activates the kinase activity – Dimerization is required for activation.

Q12: Ras protein is a critical regulator in cell proliferation, and its activity is enhanced by activated tyrosine kinase. Elements of its action include all of the following except
A. formation of cyclic GMP.
B. adaptor proteins binding to phosphorylated tyrosines on receptor tyrosine kinase.
C. recruitment and stimulation of Ras-activating protein.
D. exchange of GDP for GTP on the Ras protein.
E. initiation of a cascade in which several kinases are activated sequentially by phosphorylation.

A12: A. formation of cyclic GMP – Ras signaling does not involve cyclic GMP.

Q13: How does elevation of cyclic AMP in eukaryotic cells lead to altered transcription of certain genes?

A13: cAMP binds to protein kinase A, causing dissociation of catalytic from regulatory subunits, allowing catalytic subunits to translocate to the nucleus and phosphorylate CREB proteins, which regulate cAMP-sensitive genes.

Q14: How do excitatory and inhibitory neurotransmitters differ in their effects on ligand-gated ion channels?

A14: Excitatory neurotransmitters (e.g., acetylcholine, glutamate) bind cation-selective receptors, allowing Na⁺ influx and depolarization. Inhibitory neurotransmitters (e.g., GABA, glycine) bind anion-selective receptors, allowing Cl⁻ influx and hyperpolarization.

(Ch.22)
1. If a single gene contains information for synthesis of more than one hormone molecule, which is typically true?

A. All the hormones are produced by any tissue that expresses the gene.
B. All hormone molecules are identical.
C. Cleavage sites in the gene product are typically pairs of basic amino acids.
D. All peptides of the gene product have well-defined biological activity.
E. Hormones all have similar function.

Answer: C - One or more trypsin-like proteases catalyze the reaction, ensuring proper cleavage.

2. In the interaction of a hormone with its receptor, all of the following are true except:

A. More than one polypeptide chain of the hormone may be necessary.
B. More than one second messenger may be generated.
C. An array of transmembrane helices may form the binding site for the hormone.
D. Receptors have a greater affinity for hormones than for synthetic agonists or antagonists.
E. Hormones released from their receptor after endocytosis could theoretically interact with a nuclear receptor.

Answer: D - β-receptors bind isoproterenol more tightly than their natural hormones.

3. Some hormone–receptor complexes are internalized by endocytosis. This process may involve:

A. Binding of hormone–receptor complex to a clathrin-coated pit.
B. Recycling of receptor to the cell surface.
C. Degradation of receptor and hormone in lysosomes.
D. Formation of a receptosome.
E. All of the above.

Answer: E - A and D always occur; B or C happens depending on the receptor type.

4. Epinephrine:

A. Mediates its effects by binding to cytoplasmic receptors.
B. Is synthesized in the adrenal cortex.
C. Is synthesized from norepinephrine by a methyl transferase (PNMT).
D. Leads to the formation of cGMP.
E. Produces DAG and IP3.

Answer: C - This process is induced by cortisol under stress.

5. Binding of insulin to its receptor:

A. Occurs on the β-subunit.
B. Induces autophosphorylation.
C. Reduces binding of cytosolic substrate proteins.
D. Leads only to phosphorylation of proteins.
E. Does not lead to release of a second messenger.

Answer: B - Autophosphorylation occurs on tyrosine residues of the β-subunit.

6. Glucocorticoid receptors are in the cytoplasm. Which statement is incorrect?

A. The hormone must be in the free state to cross the cell membrane.
B. Cytoplasmic receptors may be associated with heat shock proteins.
C. The receptor–hormone complex is not activated/transformed until it is translocated to the nucleus.
D. In the nucleus, the activated receptor–hormone complex searches for specific sequences on DNA called HREs.
E. The activated receptor–hormone complex may either activate or repress transcription of specific genes.

Answer: C - Dissociation of the heat shock protein in the cytosol activates the complex.

7. All of the following are normal events leading to secretion of aldosterone except:

A. Renin is released by the kidney in hypovolemia.
B. Angiotensinogen binds to membrane receptors.
C. The PI cycle is activated, producing IP3 and DAG.
D. Ca²⁺ levels in the cell rise.
E. Aldosterone is secreted into the blood.

Answer: B - Angiotensinogen must be cleaved by renin and further processed.

8. Once ovulation occurs, which of the following statements is correct?

A. FSH, via cAMP as a second messenger, stimulates the follicle to release 17β-estradiol.
B. Blood levels of progesterone fall as pregnancy progresses as the corpus luteum dies.
C. Inhibin produced by the follicle prevents release of LH.
D. The primary influence for the corpus luteum to produce progesterone and estradiol is FSH.
E. All of the above.

Answer: A - cAMP-dependent protein kinase stimulates estradiol synthesis.

9. If the stalk between the hypothalamus and anterior pituitary is severed, which hormone would still be released?

A. ACTH
B. Estradiol
C. Oxytocin
D. Testosterone
E. Thyroxine

Answer: C - Oxytocin is released from the posterior pituitary.

10. In hypopituitarism, the ovarian cycle would be affected because:

A. FSH and LH are synthesized in the anterior pituitary.
B. GnRH from the hypothalamus stimulates the release of FSH and LH.
C. Inhibins would be secreted in much larger than normal amounts.
D. The corpus luteum would be maintained.
E. GnRH would bind to receptors to activate protein kinase A.

Answer: B - GnRH stimulates FSH and LH release from the anterior pituitary.

11. The PI cycle begins with activation of phospholipase C, initiating all of the following except:

A. Activation of IP3 by action of a phosphatase.
B. Increase in intracellular Ca²⁺ concentration.
C. Release of diacylglycerol (DAG).
D. Activation of protein kinase C.
E. Phosphorylation of certain cytoplasmic proteins.

Answer: A - IP3 is active; phosphatase inactivates it.

12. Which of the following statements concerning G proteins is correct?

A. G proteins bind the appropriate hormone at the cell surface.
B. G proteins interact with cytoplasmic receptors.
C. G proteins are second messengers.
D. G protein causes the regulatory subunits of the protein kinase to dissociate from the catalytic subunits.
E. Activated G protein may either activate or inhibit the production of second messengers.

Answer: E - G proteins have both stimulatory and inhibitory subunits.

13. How is the iodine present in thyroid hormones conserved by the body?

Answer: Iodinated tyrosyl residues in thyroglobulin are taken up by endocytosis. Lysosomal hydrolysis releases T3 and T4, while mono- and diiodothyronines are deiodinated, and iodide is reused.

14. What is the relationship between 7-dehydrocholesterol and 1α,25-dihydroxycholecalciferol?

Answer: 7-dehydrocholesterol in the skin is converted by UV light to cholecalciferol (Vitamin D₃), then hydroxylated in the liver to 25-hydroxycholecalciferol and in the kidney to the active form, 1,25(OH)₂D₃.

(Ch. 14)
1. A bond may be high energy for any of the following reasons except
A. Products of its cleavage are more resonance stabilized than the original compound.
B. The bond is unusually stable, requiring a large energy input to cleave it.
C. Electrostatic repulsion is relieved when the bond is cleaved.
D. A cleavage product may be unstable, tautomerizing to a more stable form.
E. The bond may be strained.

Answer: B - High energy does not mean high stability; it refers to high free energy of hydrolysis.

2. All of the following tricarboxylic acid cycle intermediates may be added or removed by other metabolic pathways except
A. Citrate
B. Fumarate
C. Isocitrate
D. α-Ketoglutarate
E. Oxaloacetate

Answer: C - Most TCA intermediates have multiple metabolic roles; isocitrate is not directly utilized elsewhere.

3. The inner mitochondrial membrane contains a transporter for
A. NADH
B. Acetyl CoA
C. GTP
D. ATP
E. NADPH

Answer: D - ATP and ADP are transported in opposite directions; NADH and acetyl CoA require shuttles.

4. During the transfer of electrons to O2 via the mitochondrial electron transport chain:
A. The energy released is used to translocate protons across the inner membrane.
B. A proton gradient is generated with the matrix now being more positive than the intermembrane space.
C. Pumping of protons across the membrane occurs each time electrons are moved.
D. No charge gradient develops because an OH- moves each time a proton does.
E. The energy is used directly in the addition of Pi to ADP to form ATP.

Answer: A - Energy is used to drive proton translocation, establishing a proton gradient.

5. ATP synthase (Complex V) consists of two domains, F1 and Fo.
A. Both are integral membrane protein complexes of the outer membrane.
B. F1 domain provides a channel for proton translocation.
C. F1 binds ATP but not ADP.
D. F1 domain catalyzes the synthesis of ATP.
E. Only the Fo domain contains more than one subunit.

Answer: D - The F1 domain is responsible for ATP synthesis.

6. All of the following statements are correct except
A. Reactive oxygen species (oxygen radicals) result when there is a concerted addition of four electrons at a time to O2.
B. Superoxide anion (O2-) and hydroxyl radical (·OH) are two forms of reactive oxygen.
C. Superoxide dismutase protects against damage by converting O2- to H2O2.
D. Reactive oxygen species damage phospholipids, proteins, and nucleic acids.
E. Glutathione protects against H2O2 by reducing it to water.

Answer: A - Reactive oxygen species arise from stepwise, not concerted, electron transfer.

7. The active form of pyruvate dehydrogenase is favored by all of the following except
A. Low [Ca2+].
B. Low acetyl CoA/CoASH.
C. High [pyruvate].
D. Low NADH/NAD+.

Answer: A - High Ca2+ activates pyruvate dehydrogenase via phosphatase activate

8. If a mutant pyruvate dehydrogenase has poor binding of its prosthetic group, increasing dietary intake of which may help?
A. Lipoic acid
B. Niacin (for NAD)
C. Pantothenic acid (for CoA)
D. Riboflavin (for FAD)
E. Thiamin (for TPP)

Answer: E - TPP is the essential cofactor for the first reaction in pyruvate

9. Complex I
A. Transfers electrons directly from NADH to ubiquinone.
B. Can transfer electrons from FADH2 of succinate dehydrogenase as well as from NADH.
C. Does not contain FeS centers.
D. Transfers electrons coupled to the transport of four protons across the membrane.
E. Is a mobile electron carrier traveling along the outer face of the inner membrane.

Answer: D - Complex I transfers electrons while pumping protons across the membrane.

10. An inability to reoxidize NADH due to Complex I defects would
A. Inhibit isocitrate dehydrogenase and slow the TCA cycle.
B. Force the oxaloacetate–malate equilibrium toward oxaloacetate.
C. Promote the γ-glycerol phosphate shuttle for NADH transport.
D. Allow NADH to diffuse freely from mitochondria to the cytosol.
E. Force the succinate–fumarate equilibrium toward succinate.
Answer: A - NADH accumulation inhibits TCA cycle enzymes.

Answer: A - NADH accumulation inhibits TCA cycle enzymes.

11. Cyanide
A. Minimally inhibits the electron transport chain.
B. Inhibits mitochondrial respiration but not energy production.
C. Also binds copper in cytochrome oxidase.
D. Binds to Fe3+ of cytochrome a3.
E. Poisoning could be reversed by increasing O2 concentration.

Answer: D - Cyanide binding to Fe3+ in cytochrome a3 blocks electron transport.

12. If cyanide is added to mitochondria actively oxidizing succinate:
A. 2,4-Dinitrophenol will cause ATP hydrolysis.
B. 2,4-Dinitrophenol will restore succinate oxidation.
C. Electron flow ceases but ATP synthesis continues.
D. Electron flow ceases but ATP synthesis is restored by 2,4-DNP.
E. 2,4-DNP and oligomycin will cause ATP hydrolysis.

Answer: A - Uncouplers like 2,4-DNP permit ATP hydrolysis by ATP synthase.

13. Equilibrium ratio of B/A for the reaction A → B (∆G°' = -29.7 kJ/mol at 37°C).

Answer: 100,000 - ∆G°' = -RT ln K; solving for K gives B/A = 100,000

14. Labeling of carbon in TCA cycle using 14C-labeled pyruvate

Answer: The labeled carbon appears in oxaloacetate and is released as CO2 in the second cycle.

(Ch.15)
1. Glucokinase

A. has a S₀.₅ greater than the normal blood glucose concentration.
B. is found in muscle.
C. is inhibited by glucose 6-phosphate.
D. is also known as the GLUT-2 protein.
E. has glucose 6-phosphatase activity as well as kinase activity.

Answer: A – Blood glucose is ~5 mM, and the S₀.₅ of glucokinase is ~7 mM. It is hepatic, not found in muscle, and is not inhibited by glucose 6-phosphate.

2. 6-Phosphofructo-1-kinase activity can be decreased by all of the following except

A. ATP at high concentrations.
B. citrate.
C. AMP.
D. low pH.
E. decreased concentration of fructose 2,6-bisphosphate.

Answer: C – AMP is an allosteric regulator that relieves inhibition by ATP, while citrate, low pH, and decreased fructose 2,6-bisphosphate inhibit the enzyme.

3. In the Cori cycle,

A. only tissues with aerobic metabolism (i.e., mitochondria and O₂) are involved.
B. a three-carbon compound arising from glycolysis is converted to glucose at the expense of energy from fatty acid oxidation.
C. glucose is converted to pyruvate in anaerobic tissues, and this pyruvate returns to the liver, where it is converted to glucose.
D. the same amount of ATP is used in the liver to synthesize glucose as is released during glycolysis, leading to no net effect on whole-body energy balance.
E. nitrogen from alanine must be converted to urea, increasing the amount of energy required to drive the process.

Answer: B – The liver uses energy from fatty acid oxidation to drive gluconeogenesis.

4. When blood glucagon rises, which of the following hepatic enzyme activities falls?

A. Adenyl cyclase
B. Protein kinase
C. 6-Phosphofructo-2-kinase
D. Fructose 1,6-bisphosphatase
E. Hexokinase

Answer: C – Increased glucagon activates adenyl cyclase, raising cAMP and activating protein kinase, which inactivates 6-Phosphofructo-2-kinase.

5. Glucose 6-phosphatase, which is deficient in Von Gierke disease, is necessary for the production of blood glucose from

A. liver glycogen.
B. fructose.
C. amino acid carbon chains.
D. lactose.
E. all of the above.

Answer: E – To enter the blood, glucose must be dephosphorylated. This enzyme is required for glucose release from glycogen, fructose metabolism, and gluconeogenesis.

6. UDP-glucose

A. is formed by the reaction between UTP and glucose.
B. is an intermediate in both the synthesis of glycogen and its degradation.
C. formation is irreversible because it generates pyrophosphate.
D. adds glucose units to preexisting glycogen in α-1,6 linkages.
E. is the substrate for the branching enzyme.

Answer: C – Hydrolysis of pyrophosphate makes UDP-glucose formation irreversible. It is involved in glycogen synthesis but not degradation.

7. Alcohol metabolism produces large amounts of NADH, which inhibits gluconeogenesis by

A. shifting the pyruvate–lactate equilibrium toward lactate.
B. favoring the production of oxaloacetate from malate.
C. preventing the movement of phosphoenolpyruvate from mitochondria to the cytosol.
D. inhibiting the electron transport chain.
E. inhibiting the malate–aspartate shuttle.

Answer: A – Excess NADH shifts the equilibrium toward lactate, reducing pyruvate availability for gluconeogenesis.

8. Insulin promotes hypoglycemia by a variety of mechanisms, including all of the following except

A. inactivating a transcription factor for key gluconeogenic enzymes.
B. inhibiting lipolysis in adipose tissue, thus decreasing the energy supply for gluconeogenesis in the liver.
C. decreasing levels of cAMP.
D. increasing activation of phosphoprotein phosphatase.
E. increasing the activity of cAMP-response element binding protein (CREB) and its binding to the cAMP-response element (CRE).

Answer: E – CREB increases gluconeogenic enzyme expression, but insulin promotes dephosphorylation, reducing its activity.

9. Ca²⁺ increases glycogenolysis by

A. activating phosphorylase kinase b, even in the absence of cAMP.
B. binding to phosphorylase b.
C. activating phosphoprotein phosphatase.
D. inhibiting phosphoprotein phosphatase.
E. protecting cAMP from degradation.

Answer: A – Phosphorylase kinase has a calmodulin subunit that is activated by Ca²⁺, even in the absence of cAMP.

10. Phosphorylation–dephosphorylation and allosteric activation of enzymes play roles in stimulating glycogen degradation. All of the following result in enzyme activation except

A. phosphorylation of phosphorylase kinase.
B. binding of AMP to phosphorylase b.
C. phosphorylation of phosphorylase.
D. phosphorylation of protein kinase A.
E. dephosphorylation of glycogen synthase.

Answer: D – Protein kinase A activates glycogen breakdown but is itself activated by cAMP, not phosphorylation.

11. The first step in liver metabolism of fructose is

A. isomerization to glucose.
B. phosphorylation to fructose 1,6-bisphosphate by ATP.
C. phosphorylation to fructose 1-phosphate by ATP.
D. phosphorylation to fructose 6-phosphate by ATP.
E. cleavage by aldolase.

Answer: C – Fructokinase phosphorylates fructose to fructose 1-phosphate.

12. The products initially produced by aldolase action on the substrate formed from fructose are

A. two molecules of dihydroxyacetone phosphate.
B. two molecules of glyceraldehyde 3-phosphate.
C. two molecules of lactate.
D. dihydroxyacetone phosphate and glyceraldehyde 3-phosphate.
E. dihydroxyacetone phosphate and glyceraldehyde.

Answer: E – Aldolase cleaves fructose 1-phosphate into dihydroxyacetone phosphate and glyceraldehyde.

13. If a cell is forced to metabolize glucose anaerobically, how much faster would glycolysis have to proceed to generate the same amount of ATP as it would get if it metabolized glucose aerobically?

Answer: – Anaerobically, glycolysis produces 2 ATP per glucose, while aerobic metabolism yields 32 ATP. Therefore, glycolysis must proceed 16 times faster anaerobically to generate the same ATP.

14. The alanine cycle requires more ATP per glucose molecule formed than the Cori cycle. Why is this?

Answer: – Both cycles require ATP to convert pyruvate to glucose. However, in the alanine cycle, nitrogen must be converted to urea, which requires 4 ATP per urea molecule, making it more energy-demanding than the Cori cycle.

(Ch.16)
Q1: All of the following interconversions of monosaccharides require a nucleotide-linked sugar intermediate except:
A. Galactose 1-phosphate to glucose 1-phosphate
B. Glucose 6-phosphate to mannose 6-phosphate
C. Glucose to glucuronic acid
D. Glucuronic acid to xylose
E. Glucosamine 6-phosphate to N-acetylneuraminic acid

A1: B. The glucose and mannose phosphates are interconverted via phosphohexose isomerases without a nucleotide intermediate.

Q2: All of the following are true about glucuronic acid except:
A. It is a charged molecule at physiological pH
B. As a UDP derivative, it can be decarboxylated to a component used in proteoglycan synthesis
C. It is a precursor of ascorbic acid in humans
D. Its formation from glucose is under feedback control by a UDP-linked intermediate
E. It can ultimately be converted to xylulose 5-phosphate and enter the pentose phosphate pathway

A2: C. Humans do not synthesize ascorbic acid from glucuronic acid.

Q3: NADPH/NADP+ is maintained at a high level in cells primarily by:
A. Lactate dehydrogenase
B. The combined actions of glucose 6-phosphate dehydrogenase and gluconolactonase
C. The action of the electron transport chain
D. Exchange with NAD+/NADH
E. The combined actions of transketolase and transaldolase

A3: B. Glucose 6-phosphate dehydrogenase initiates the oxidative phase of the pentose phosphate pathway, ensuring high NADPH levels.

Q4: Fucose and sialic acid:
A. Are both derivatives of UDP-N-acetylglucosamine
B. Are parts of the carbohydrate chain that are covalently linked to the protein
C. Can be found in the core structure of certain O-linked glycoproteins
D. Are transferred to a carbohydrate chain when it is attached to dolichol phosphate
E. Are the repeating units of proteoglycans

A4: C. Core structures also contain galactose and N-acetylgalactosamine.

Q5: Glycosaminoglycans:
A. Are the carbohydrate portion of glycoproteins
B. Contain large segments of a repeating unit typically consisting of a hexosamine and a uronic acid
C. Always contain sulfate
D. Exist in only two forms
E. Are bound to protein by ionic interaction

A5: B. Glycosaminoglycans differ from glycoproteins by having repeating units.

Q6: All of the following are true of proteoglycans except:
A. Specificity is determined, in part, by the action of glycosyltransferases
B. Synthesis is regulated by UDP-xylose inhibition of UDP-glucuronic acid formation
C. Synthesis involves sulfation of carbohydrate residues by PAPS
D. Synthesis of core protein is balanced with polysaccharide moieties
E. Degradation is catalyzed in the cytosol by nonspecific glycosidases

A6: E. Proteoglycan degradation occurs in lysosomes, not the cytosol.

Q7: The most severe form of galactosemia:
A. Is a genetic deficiency of a uridylyltransferase that exchanges galactose 1-phosphate for glucose on UDP-glucose
B. Results from a deficiency of an epimerase
C. Is insignificant in infants but a major problem in later life
D. Is a defect in galactokinase
E. Would interfere with fructose metabolism

A7: A. The most severe form results from a deficiency in galactose 1-phosphate uridylyltransferase.

Q8: UDP-galactose:
A. Must be formed from galactose 1-phosphate
B. Is usually the first sugar linked to dolichol phosphate
C. Is used in the synthesis of chondroitin sulfate
D. Could not lead to sugar derivatives like glucuronic acid or xylose
E. Is the direct precursor of N-acetylgalactosamine

A8: C. UDP-galactose is involved in proteoglycan synthesis.

Q9: The role of phosphomannose isomerase is the interconversion of:
A. Mannose 6-phosphate and mannose 1-phosphate
B. Glucose 6-phosphate and mannose 6-phosphate
C. Fructose 6-phosphate and mannose 6-phosphate
D. Fructose 1-phosphate and mannose 1-phosphate
E. Glucose 6-phosphate and mannose 1-phosphate

A9: C. The enzyme catalyzes the conversion between fructose 6-phosphate and mannose 6-phosphate.

Q10: In Type Ic CDG, a defect in an enzyme transferring a glucosyl residue to a high-mannose dolichol pyrophosphate precursor results in a carbohydrate structure that is part of a(n):
A. N-linked glycoprotein
B. O-linked glycoprotein
C. Proteoglycan
D. Glycosaminoglycan
E. Complex lipid

A10: A. N-linked glycoproteins are synthesized via dolichol-linked intermediates.

Q11: The disorder described with skeletal abnormalities and normal blood sulfate levels is most likely caused by a defect in:
A. The ability to generate mannose 6-phosphate
B. Lysosomal glycosidase activity
C. The gene coding for a sulfate transporter
D. PAPS synthetase
E. A sulfatase

A11: D. Defects in PAPS synthetase disrupt sulfation in proteoglycan synthesis.

Q12: Sulfation is an important component of the synthesis of:
A. Most proteoglycans
B. Hyaluronate
C. Most glycoproteins
D. Conjugated bilirubin
E. All of the above

A12: A. Proteoglycans, except hyaluronate, require sulfation.

Q13: What is the role of transaldolase and transketolase in glucose metabolism?

A13: They facilitate the reversible conversion of pentose phosphates (ribose 5-phosphate + 2 xylulose 5-phosphate) into 2 fructose 6-phosphate and glyceraldehyde 3-phosphate via carbon transfers.

Q14: Essential fructosuria is a defect in fructokinase, while fructose intolerance is a defect in fructose 1-phosphate aldolase. Which disease leads to severe hypoglycemia after fructose ingestion, and why?

A14: Fructose intolerance leads to severe hypoglycemia because fructose 1-phosphate accumulates, sequestering phosphate and inhibiting ATP generation.