Exam 1 Flashcards
2009
The ΔG’º for hydrolysis of ATP is -30.5 kJ/mole. The ΔG’º for hydrolysis of PEP is -61.9 kJ/mole. The ΔG’º for the following reaction is:
PEP + ADP ⇔ Pyruvate + ATP
- -92.4 kJ
- +31.4 kJ
- -31.4 kJ
- +92.4 kJ
- cannot be determined from data provided
-31.4 kJ
( -61.9 - (-30.5) = -31.4 )
2009
For A ⇔ B, ΔG’º is negative.
- The rate of reaction of B to A is increased
- The rate of reaction of B to A is decreased
- The rate of reaction of B to A is dependent on the magnitude of ΔG’º
- The rate of reaction is not influenced by the sign of magnitude of ΔG’º
- There is no reaction at all in either direction becase it is at equilibrium
The rate of reaction is not influenced by the sign of magnitude of ΔG’º
2009
ln K (K equilibrium constant) is equal to
- RT ΔG’º
- RT ΔG’º
- ΔG’º/RT
- ΔG’º/RT
- ΔG’ºT/R
- ΔG’º/RT
( ΔG’º = -RT ln K)
2009
ΔG’º (under cellular conditions) for ATP hydrolysis is -50 kJ/mole and 4000 kJ our caloric intake (in 24 hours) is used for synthesis ATP. The amount of ATP formed by a human in 24 hours is (MW ATP ≈ 500)
- ≈ 6g
- 80g
- 0
- -40 Kg
- ATP is only hydrolized
-40 Kg

2009
In glycolysis ATP synthesis is catalyzed by
- 6-phosphofructo-1 kinase
- hexokinase
- phosphoglycerate kinase
- glyceraldehyde 3-phosphate dehydrogenase
- none of the above
phosphoglycerate kinase
2009
The irreversible reactions of glycolysis include that catalyzed by
- phosphoglycerate kinase
- fructose biphosphate aldolase
- 6-phosphofructo-1-kinase
- glyceraldehyde 3-phosphate dehygrogenase
- phosphoglucose isomerase
6-phosphofructo-1-kinase
2009
6-Phosphofructo-1-kinase activity can be decreased by all of the following EXCEPT
- low pH
- AMP
- decreased concentration of fructose 2,6-bisphosphate
- ATP at high concentrations
- citrate
AMP
2009
Which of the following supports gluconeogenesis?
- lysine degradation
- leucine degradation
- acetyl CoA + oxaloacetate + H2O ⇔ citrate + CoA
- pyruvate + ATP + HCO3 ⇔ oxaloacetate + ADP + Pi + H+
- α-ketoglutarate + aspartate ⇔ glutamate + oxaloacetate
pyruvate + ATP + HCO3 ⇔ oxaloacetate + ADP + Pi + H+
2009
In the Cori Cyle
- glucose is converted to pyruvate/lactate in anaerobic (i.e. red blood cells) tissues, and this pyruvate/lactate return to the liver, where it is converted to glucose
- nitrogen from alanine must be converted to urea, increasing the amount of energy required to drive the process
- only tissues with aerobic metabolism (i.e. mitochondria and O2) are involved
- the same amount of ATP is used in the liver to synthesize glucose as it is released during glycolysis, leading to no net efect on whole-body energy balance
- a three-carbon compound arising from glycolysis is converted to glucose at the expense of energy from fatty acid oxidation
glucose is converted to pyruvate/lactate in anaerobic (i.e. red blood cells) tissues, and this pyruvate/lactate return to the liver, where it is converted to glucose
2009
Gluconeogenic enzymes include all of the following EXCEPT
- phosphoenolpyruvate carboxykinase
- pyruvate carboxylase
- fructose 1,6-bisphosphatase
- phosphogucmutase
- glucose 6-phosphatase
phosphogucmutase
(only in glycolysis)
2009
Phospho-dephospho regulation of, 6-phosphofructo-2-kinase, and pyruvate kinase is an important regulatory mechanism in
- skeletal muscle
- liver
- intestine
- erythrocytes
- brain
liver
(where glycolysis occurs)
2009
Phosphorylation-dephosphorylation and allosteric activation of enzymes play roles in stimulating glycogen degradation. All of the following result in enzyme activation EXCEPT
- Phosphorylation of Phosphorylase kinase
- Phosphorylation of protein kinase A
- binding of AMP to Phosphorylase b
- Phosphorylation of Phosphorylase
- dePhosphorylation of glycogen synthase
Phosphorylation of protein kinase A
2009
Glucose 6-phosphatase, which is deficient in Von Glerke’s disease, is necessary for the production of blood glucose from
- fructose
- galactose
- amino acid carbon chains
- liver glycogen
- all of the above
all of the above
2009
Glucose labeled at C-1 is incubated with glycolytic enzymes and necessary cofactors. The pyruvate CH3COCOO- that is formed will be labeled at
- C-1
- C-2
- C-3
- All C atoms
- At none of the C atoms
C-3
2009
Removal of a glucose residue from glycogen (by phosphorylase) and passing it thorugh glycolysis to 2 pyruvates yields _____ ATP
- 1
- 2
- 3
- 4
- None
3
2009
Glucose labeled at C-3 is allowed to undergo fermentation to ethanol CH3CH2OH and CO2. The label will show in
- Ethanol at C-1 only
- Ethanol at C-2 only
- CO2 only
- Both A and C
- Both B and C
CO2 only
2009
The conversion of 1 mol of fructose 1,6-biphosphate to 2 mol of pyruvate by the glycolytic pathway results in a net formation of:
- 2 mol of NADH and 2 mol of ATP
- 1 mol of NADH and 1 mol of ATP
- 2 mol of NADH and 4 mol of ATP
- 1 mol of NAD+ and 2 mol of ATP
- 2 mol of NAD+ and 4 mol of ATP
2 mol of NADH and 4 mol of ATP
2009
Which of the following statements is true of muscle glycogen phosphorylase?
- It removes glucose residues from the reducing ends of the glycoen chains
- It exists in an active (a) form and an inactive (b) form that is allosterically regulated by AMP
- It degrades glycogen to form glucose 6-phosphate
- It catalyzes phosphorolysis of the α(1→6) bonds at the branch points of glycogen
- It degrades glycogen to glucose
It exists in an active (a) form and an inactive (b) form that is allosterically regulated by AMP
2009
Which of these enzymes is a control point, regulated by the cell’s energy supply?
- Phosphofructokinase-1
- Glycogen phosphorylase
- Pyruvate kinase
- All of the above are control points
- None of the above is a control point
All of the above are control points
2009
The main function of the pentose phosphate pathway is to
- Supply pentoses and NADPH
- Give the cell an alternate pathway should glycolysis fail
- Supply NADH
- Provide a mechanism for the utilization of the carbon skeletons of excess amino acids
- Supply energy
Supply pentoses and NADPH
2009
Which of the following describes fatty acid transport into the mitochondrial matrix?
- It is the rate-limiting step in fatty acid oxidation
- It is regulated by malonyl-CoA
- The cytosolic and matrix pools of CoA are distince and separate
- Once fatty acyl groups have entered the matrix, they are committed to oxidation to acetyl-CoA
- All of the above are true
All of the above are true
2009
Fatty acid synthase of mammalian cells
- uses a four-step sequence to lengthen a growing fatty acyl chain by one carbon
- contains ACP, which carries acyl groups attached through thioester linkages
- requires chemical energy in just one form, the reducing power of NADPH
- is activated by glucagon and epinephrine
- is activated by palmitoyl CoA, the principal product of fatty acid synthesis
contains ACP, which carries acyl groups attached through thioester linkages
2009
How are fatty acid oxidation and synthesis controlled so that futile cycling (when two opposing reactions/pathways occurring in the same location and at the same time, one forming an energy-rich compound, the other consuming an energy-rich compound) does not occur?
- A - they occur in different cellular compartments
- B - they employ different electron carriers
- C - the product of the first oxidation reaction inhibits the rate-limiting step of biosynthesis
- A and B
- A, B, and C
A and B
2009
In mammmalian cells, fatty acid synthesis occurs in the cytosol. The beginning subsrate for this series of reaction is acetyl-CoA, which is formed in the mitochondrial matrix. How does the acetyl-CoA move from matrix to cytosol?
- Acetyl-CoA reacts with oxaloacetate and leaves the matrix as citrate via the citrate transporter
- Acetyl-CoA combines with bicarbonate and is transported out of the matrix as pyruvate
- There is a specific transport protein for acetyl-CoA
- Acetyl-CoA is a nonpolar molecule and can diffuse across all membranse
- Inner and outer mitochondrial membranes are freely permeable to Acetyl-CoA due to the presence of transmembrane pore or channels.
Acetyl-CoA reacts with oxaloacetate and leaves the matrix as citrate via the citrate transporter
2010
Which of the following BEST describes the relationship between ΔG and the rate of a reaction?
- ΔG is linearly proportional to the rate
- ΔG is inversely proportional to the rate
- if ΔG > 0, the reaction is spontaneous
- if ΔG < 0, the reaction is at equilibrium
- ΔG provides no information about the rate
ΔG provides no information about the rate
2010
For the reaction A → B, the Keq is 104. If a reaction mixture originally contains 1 mmol of A and no B, which of the following is true? (R = 8.32 J mol-1K-1, T = 300ºK)
- At equilibrium, there will be far more A than B
- The reaction will definitely proceed toward B at a very high rate
- ΔGº’ for the reaction will be large and positive
- The rate of the reaction may be immeasurably small
- ΔGº’ will be approximately -23kJ/mole
2010: ΔGº’ will be approximately -23kJ/mole
2012, 2015: A. At equilibrium, there will be far more B than A
2010
In a skeletal muscle cell ΔGº’ for hydrolysis of ATP to ADP + Pi is -50 kJ/mole. The approximate ratio of [ATP]/ADP][Pi] found in the cells at 37°C is ΔGº’ for ATP hydrolysis -30.5 kJ/mole, R = 8.32 J mol-1K-1
- 5000/1
- 4000/1
- 2000/1
- 1000/1
- 200/1
2000/1
2010
Glucose 1-phosphate is converted to fructose 6-phosphate in two successive reactions?
Glucose-1-P → glucose-6-P ΔG’º = -1.7 kcal/mol
Glucose-6-P → fructose-6-P ΔG’º = -0.4 kcal/mol
What is ΔG’º for the overall reaction?
- -2.1 kcal/mol
- -1.7 kcal/mol
- -1.3 kcal/mol
- 1.3 kcal/mol
- 2.0 kcal/mol
-2.1 kcal/mol
2010
The E0’ of the NAD+/NADH half reaction is -0.32 V. The E0’ of the oxaloacetate/malate half reaction is -0.175 V. When the concentrations of NAD+, NADH, oxaloacetate, and malate are all 10-5 M, the spontaneous reaction is:
- NAD+ + oxaloacetate → NADH + H+ + malatate
- NAD+ + NADH + H+ → malatate + oxaloacetate
- malatate + NADH + H+ → oxaloacetate + NADH + H+
- malatate + NAD+ → oxaloacetate + NADH + H+
- oxaloacetate + NADH + H+ → malatate + NAD+
oxaloacetate + NADH + H+ → malatate + NAD+
2010
A person who cannot synthesize liver fructose 1,6-biphosphate would probably not be able to
- Synthesize fats when the diet contained excess carbohydrate
- Metabolize fructose
- Convert fructose 1,6-biphosphate into triose phosphates
- Resynthesize glucose from lactate produced during exercise
- None of the above are correct
Resynthesize glucose from lactate produced during exercise
2010
glucose can not be synthesized from which of the following noncarbohydrate precursors?
- fumarate
- alanine
- lactate
- palmitic acid
- glycerol
palmitic acid
2010
Starting from a glucose residue in glycogen, how many net ATP molecules will be formed in the glycolysis of the residue to pyruvate?
- 1
- 2
- 3
- 4
- 5
3
2010
If the C-1 carbon of glucose were labeled with 14C, which of the carbon atoms in pyruvate would be labeled after glycolysis?
- The carboxylate carbon
- The carbonyl carbon
- The methyl carbon
- none
- all carbon atoms will be labeled
The methyl carbon
2010
Phosphofructokinase activity is enhanced by which of the following?
- Increased ATP concentration
- Increased fructose 2, 6-biphosphate concentration
- Increased citrate concentration
- Decreased AMP concentration
- Increased [glucagon] serum
Increased fructose 2, 6-biphosphate concentration
2010
Place the following steps of the reaction cascade of glycogen metabolism in the proper sequence
- activation of protein kinase
- formation of cyclic AMP by adenylate cyclase
- phosphorylation of phosphorylase kinase
- hormone binding to target cell receptors
- phosporylaiton of glycogen synthase a and phosphorylase b
- 1,2,3,4,5
- 2,4,1,5,2
- 4,2,1,3,5
- 5,4,3,2,1
- 2,1,4,5,3
4,2,1,3,5
2010
Inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate, which of the following compounds would you expect to increase in concentration following the addition of fluoride?
- glucose
- phosphoenolpyruvate
- glyoxylate
- 3-phosphoglycerate
- pyruvate
3-phosphoglycerate
2010
The conversion of one mole of glucose-6-phosphate to two moles of lactate in glycolysis is accompanied by a net gain of:
- 3 moles of ATP
- 2 moles of ATP
- 1 mole of ATP
- 1 mole of NADH
- 2 moles each of ATP and NADH
3 moles of ATP
2010
In glycolysis, fructose 1,6-biphosphate is converted to two products with a standard free-energy change (ΔG°’) of 23.8 kJ/mol. Under what conditions encountered in a normal cell will the free-energy change (ΔG°’) be negative, enabling the reaction to proceed spontaneously to the right?
- 2010 / 2015: Under standard conditions, enough energy is released to drive the reaction to the right
- 2010 / 2015: The reaction will not go to the right spontaneously under any conditions because the ΔG°’ is positive
- 2010 / 2015: The driving force of this reaction is supplied by a coupled exergonic reaction, the hydrolysis of ATP
- 2015: When there is a high concentration of products relative to the concentration of fructose 1,6- bisphosphate.
- 2010: The reaction may proceed given appropriate concentrations of fructose-1,6-bisphosphate and the products
- 2015: When there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products
- 2010 / 2015: none of the above conditions is sufficient
- 2010: The reaction may proceed given appropriate concentrations of fructose-1,6-bisphosphate and the products
- 2015: When there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products
2010
If glucose labeled with 14C in C-3 were fed to yeast carrying out of the ethanol fermentation, where would the 14C label be in the products?
- C-1 (methyl group) of ethanol
- C-1 of ethanol only
- CO2 only
- C-2 of ethanol and CO2
- C-1 of ethanol and CO2
CO2 only
2010
Glyceraldehyde-3-phosphate labeled with 14C in C-1 (the carbonyl carbon) is mixed wiht unlabeled dihydroxyacetone phosphate and the enzymes triose phosphate isomerase and aldolase. After a period of incubation, fructose-1,6-bisphosphate is isolated from this system. It will be labeled with 14C in:
- C-1 only
- C-3 only
- C-1 and C-3
- C-1 and C-6
- C-3 and C-4
C-3 and C-4
2010
Galactosemia is a genetic error of metabolism principally associated with
- excessive ingestion of galactose
- inability to digest lactose
- deficiency of galactokinase
- deficiency of UDP-glucose: galactose 1-phosphate uridylyltransferase
- deficiency of UDP-glucose epimerase
deficiency of UDP-glucose: galactose 1-phosphate uridylyltransferase
2010
When blood glucose levels are low, glucagon is secreted. Which of the following are the effects of increased glucagon levels on glycolysis and related reactions in the liver?
- Phosphorylation of phosphofructokinase 2 occurs
- Dephosphorylation of phosphofructokinase 2/fructose bisphosphatase 2 occurs
- phosphofructokinase-1 is activated
- phosphofructokinase-1 is inhibited
- glycolysis is accelarated
- glycolysis is slowed down
- 1,3,6
- 2,3
- 1,4,6
- 2,5
- 2,3,5
1,4,6
2010
The oxidation of 3 moles of glucose by the pentose phosphate pathway may result in the production of
- 3 moles of pentose, 4 moles of NADPH, and 3 moles of CO2
- 2 moles of pentose, 8 moles of CO2, and 4 moles of NADPH
- 3 moles of CO2, 6 moles of NADPH, and 3 moles of pentose
- 6 moles of CO2, 6 moles of NADPH, and 4 moles of triose
- 18 moles of CO2, 18 moles of NADPH
3 moles of CO2, 6 moles of NADPH, and 3 moles of pentose
2010
Which of the following is true of the control of gluconeogenesis?
- glucagon stimulates adenylyl cyclase, causing the formation of cAMP
- Cyclic AMP stimulates the phosphorylation and thus increases the activity of FBPase-2
- FBPase-2 activity lowers the level of fructose 2,6-bisphosphate, thus increasing the rate of gluconeogenesis
- It is reciprocally linked to the control of glycolysis
- all of the above are true
all of the above are true
2010
Which of the following has the (immediate) effect of increasing the rate of glycogen breakdown?
- A- increased concentration of cAMP
- B- increase in the [AMP]/[ATP] ratio
- C- increased secretion of glucagon
- A and C
- A, B, and C
A, B, and C
2010
Which of the following correctly matches the glycolytic reaction with the gluconeogenic enzyme used in the corresponding bypass reaction?
- Glucose → glucose 6-phosphate; glucose 6-phosphatase
- Fructose 6-phosphate → fructose 1,6-bisphosphate; phosphofructokinase-2
- Fructose 1,6-bisphosphate → dihydroxyacetone phosphate + glyceraldehyde 3-phosphate; glyceraldehyde 3-phosphate dehydrogenase
- 2-Phosphoglycerate → phosphoenolpyruvate; phosphoglycerate kinase
- Phosphoenolpyruvate → pyruvate; pyruvate kinase
Glucose → glucose 6-phosphate; glucose 6-phosphatase
2010
Transport of fatty acids from the cytoplasm to the mitochondrial matrix requires:
- ATP, coenzyme A, and hexokinase
- ATP, carnitine and coenzyme A
- carnitine, coenzyme A, and hexokinase
- ATP, carnitine, and pyruvate dehydrogenase
- ATP, coenzyme A, and pyruvate dehydrogenase
ATP, carnitine and coenzyme A
2010
Lipoprotein lipase acts in
- intracellelar lipid breakdown of lipoproteins
- intestinal uptake of dietary fat
- lipoprotein breakdown to supply needed amino acids
- hydrolysis of triacylglycerols of plasma lipoproteins to supply fatty acids to various tissues
- none of the above
hydrolysis of triacylglycerols of plasma lipoproteins to supply fatty acids to various tissues
Fatty acids are activated to acyl-CoAs and the acyl group is further transferred to carnitine because
- acyl-CoAs easily cross the mitochondrial membrane, but the fatty acids themselves will not
- fatty acids cannot be oxidized by FAD unless they are in the acyl-carnitine form
- carnitine is required to oxidize NAD+ to NADH
- acyl-carnitines readily cross the mitochondrial inner membrane, but acyl-COAs do not
- None of the above is true
acyl-carnitines readily cross the mitochondrial inner membrane, but acyl-COAs do not
2010
Which of the following is (are) true of the oxidation of 1 mol of palmitate (a 16-carbon saturated fatty acid; 16:0) by the β-oxidation pathway, beginning with the free fatty acid in the cytoplasm?
- Activation of the free fatty acid requires the equivalent of two ATP
- Inorganic pyrophosphate (PPi) is produced
- Carnitine functions as an electron acceptor
- 8 mol of FADH2 are formed
- 8 mol of acetyl-CoA are formed
- there is no direct involvement of NAD+
- 1, 2, 6
- 1, 3, 5
- 1 and 5 only
- 5 only
- 1, 2, and 5
1, 2, and 5
2010
Place the following steps in lipid oxidation in their proper sequence
- a - Thiolysis
- b - Reaction of fatty acyl-ClA with carnitine
- c - Oxidation requiring NAD+
- d - Hydrolysis of triacylglycerol by lipase
- e - Activation of fatty acid by joining to CoA
- f - Hydration
- 5, 4, 3, 2, 1, 6
- 4, 5, 2, 3, 4, 1
- 4, 2, 5, 3, 6, 1
- 1, 4, 5, 2, 6, 1
- 5, 4, 2, 3, 1, 6
4, 5, 2, 3, 4, 1
d, e, b, c, f, a
- Hydrolysis of triacylglycerol by lipase
- Activation of fatty acid by joining to CoA
- Reaction of fatty acyl-ClA with carnitine
- Oxidation requiring NAD+
- Hydrolysis of triacylglycerol by lipase
- Thiolysis
2012
When the serum glucose concentrations falls below the normal level, which of the following will occur?
- A. Glycolysis in red blood cells and nerve cell slows down
- B. Glycolysis in these cells continues normally
- C. Skeletal muscle contraction stops so as to save glucose
- D. Glycolysis in liver is increased
- E. None of the above
B. Glycolysis in these cells continues normally
2012
Biological oxidation-reduction reactions always involve:
- A. direct participation of oxygen.
- B. formation of water.
- C. mitochondria.
- D. transfer of electron(s)
- E. transfer of hydrogens
D. transfer of electron(s)
2012
Which of the following statements concerning signal transduction by the insulin receptor is not correct?
- A. Activation of the receptor protein kinase activity results in the activation of additional protein kinases.
- B. Binding of insulin to the receptor activates a protein kinase.
- C. Binding of insulin to the receptor results in a change in its quaternary structure.
- D. The receptor protein kinase activity is specific for tyrosine residues on the substrate proteins.
- E. The substrates of the receptor protein kinase activity are mainly proteins that regulate transcription
E. The substrates of the receptor protein kinase activity are mainly proteins that regulate transcription
2012
Which of the following does not involve cyclic AMP?
- A. Regulation of glycogen synthesis and breakdown
- B. Regulation of glycolysis
- D. Signaling by epinephrine
- C. Signaling by acetylcholine
- E. Signaling by glucagon
C. Signaling by acetylcholine
2012
Hormone-activated phospholipase C can convert phosphatidylinositol 4,5-bisphosphate to:
- A. diacylglycerol + inositol triphosphate.
- B. diacylglycerol + inositol+ phosphate.
- C. glycerol + inositol + phosphate.
- D. glycerol + phosphoserine.
- E. phosphatidyl glycerol + inositol + phosphate.
A. diacylglycerol + inositol triphosphate.
2012
Gluconeogenesis occurs during intense exercise. This seems counter-intuitive.
- A. This is not true; two opposing pathways would result in net hydrolysis of ATP
- B. Intense exercise consumes fatty acids only,
- C. Skeletal muscle converts glucose residues to pyruvate and lactate. Rising level of lactate causes lactate to pass from muscle to blood to liver where the lactate can be converted (by gluconeogenesis) to glucose.
- D. During exercise, glucose is transported from blood into muscle. When serum glucose has been depleted, liver forms glucose by gluconeogenesis and glycogenolysis.
- E. Insulin activates the Glut 4 and glucose transport into muscle increases.
C. Skeletal muscle converts glucose residues to pyruvate and lactate. Rising level of lactate causes lactate to pass from muscle to blood to liver where the lactate can be converted (by gluconeogenesis) to glucose.
2012
In the coordinated control of glycolysis and gluconeogenesis in liver
- A. Citrate inhibits PFK and stimulates F-1 ,6-Bpase.
- B. Fructose 2,6-bisphosphate stimulates F-1,6-Bpase and inhibits PFK.
- C. Acetyl CoA inhibits PFK and stimulates F-1,6-Bpase.
- D. AMP inhibits PFK and stimulates F-1,6-Bpase.
- E. NADPH inhibits PFK and stimulates F-1,6-Bpase.
A. Citrate inhibits PFK and stimulates F-1 ,6-Bpase.
Avidin, a 70-kd protein in egg white has a high affinity for biotin. In fact, it is a highly specific inhibitor of biotin enzymes. Which of the following conversions would be blocked by the addition of avidin to a cell homogenate?
- A. Glucose → Pyruvate
- B. Pyruvate → Glucose
- C. Oxaloacetate → Glucose
- D. Malate →oxaloacetate
- E. Glyceraldehyde-3-P → Fructo 1 ,6-Bis P
B. Pyruvate → Glucose
2012
Glucose, labeled with 14C in different carbon atoms, is added to a crude extract of a tissue rich in the enzymes of the pentose phosphate pathway. The most rapid production of 14CO2 will occur when the glucose is labeled in:
- A. C-1
- B. C-3
- C. C-4
- D. C-5
- E. C-6
A. C-1
2012
Glucose labeled with 14C at C-6 is added to a solution containing the enzymes and cofactors of the oxidative branch of the pentose phosphate pathway. The radioactive label will be found in D-ribulose-5P at
- A. C-1
- B. C-2
- C. C-3
- D. C-4
- E. C-5
E. C-5
2012
When blood glucagon rises, which of the following hepatic enzyme activities FALLS?
- A. adenyl cyclase
- B. protein kinase A
- C. 6-phosphofructo-2-kinase
- D. fructose 1 ,6-bisphosphatase
- E. hexokinase
E. hexokinase
2012
Malignant hyperthermia is a genetic abnormality in which exposure to certain agents, especially the widely used general anesthetic halothane, produces a dramatic rise in body temperature, acidosis, hyperkalemia, and muscle rigidity. Death is rapid if the condition is untreated and may occur the first time a susceptible person is anesthetized. The defect causes an inappropriate release of Ca2+ from the sarcoplasmic reticulum of muscle. Many heat producing processes are stimulated in an uncontrolled fashion by the release of Ca2+, including glycolysis and glycogenolysis.
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.
A. activating phosphorylase kinase b, even in the absence of cAMP
2012
Malignant hyperthermia is a genetic abnormality in which exposure to certain agents, especially the widely used general anesthetic halothane, produces a dramatic rise in body temperature, acidosis, hyperkalemia, and muscle rigidity. Death is rapid if the condition is untreated and may occur the first time a susceptible person is anesthetized. The defect causes an inappropriate release of Ca2+ from the sarcoplasmic reticulum of muscle. Many heat producing processes are stimulated in an uncontrolled fashion by the release of Ca2+, including glycolysis and glycogenolysis.
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.
D. phosphorylation of protein kinase A.
2012
Malignant hyperthermia is a genetic abnormality in which exposure to certain agents, especially the widely used general anesthetic halothane, produces a dramatic rise in body temperature, acidosis, hyperkalemia, and muscle rigidity. Death is rapid if the condition is untreated and may occur the first time a susceptible person is anesthetized. The defect causes an inappropriate release of Ca2+ from the sarcoplasmic reticulum of muscle. Many heat producing processes are stimulated in an uncontrolled fashion by the release of Ca2+, including glycolysis and glycogenolysis.
Which one of the following is not a characteristic of phosphoprotein phosphorylase-1 (PP1)?
- A. PP1 can be phosphorylated by protein kinase A (PKA).
- B. PP1 can dephosphorylate glycogen phosphorylase, glycogen synthase, and phosphorylase kinase.
- C. PP1 is allosterically activated by glucose-6-phosphate.
- D. PP1 is inhibited by activated glycogen phosphorylase
- E. PP1 is phosphorylated by glycogen synthase kinase-3 (GSK3).
E. PP1 is phosphorylated by glycogen synthase kinase-3 (GSK3).
2012
The role of hormone-sensitive triacylglycerol lipase is to:
- A. hydrolyze lipids stored in the liver.
- B. hydrolyze membrane phospholipids in hormone-producing cells.
- C. hydrolyze triacylglycerols stored in adipose tissue.
- D. synthesize lipids in adipose tissue.
- E. synthesize triacylglycerols in the liver.
C. hydrolyze triacylglycerols stored in adipose tissue.
2012
Which of the following statements concerning the β oxidation of fatty acids is true?
- A. About 1,200 ATP molecules are ultimately produced per 20-carbon fatty acid oxidized.
- B. One FADH2 and two NADH are produced for each acetyl-CoA.
- C. The free fatty acid must be carboxylated in the β position by a biotin-dependent reaction before the process of β oxidation commences.
- D. The free fatty acid must be converted to a thioester before the process of β oxidation commences.
- E. Two NADH are produced for each acetyl-CoA.
D. The free fatty acid must be converted to a thioester before the process of β oxidation commences.
2012
The balanced equation for the degradation of CH3(CH2)10COOH via the β-oxidation pathway is:
- A. CH3(CH2)10COOH + 5FAD + 5NAD+ + 6CoA—SH + 5H2O + ATP → 6 Acetyl-CoA + 5FADH2 + 5NADH + 5H+ + AMP + PPi
- B. CH3(CH2)10COOH + 5FAD + 5NAD+ + 6CoA—SH + 5H2O → 6 Acetyl-CoA + 5FADH2 + 5NADH + 5H+
- C. CH3(CH2)10COOH + 6FAD + 6NAD+ + 6CoA—SH + 6H2O + ATP → 6 Acetyl-CoA + 6FADH2 + 6NADH + 6H+ + AMP + PPi
- D. CH3(CH2)10COOH + 6FAD + 6NAD+ + 6CoA—SH + 6H2O → 6 Acetyl-CoA + 6FADH2 + 6NADH + 6H+
A. CH3(CH2)10COOH + 5FAD + 5NAD+ + 6CoA—SH + 5H2O + ATP → 6 Acetyl-CoA + 5FADH2 + 5NADH + 5H+ + AMP + PPi
2013
Restriction enzymes
- A. used in Northern analysis
- B. used for protein processing
- C. recognize palindromic sequences of DNA
- D. recognize palindromic sequences of RNA
- E. cut randomly in the genome
C. recognize palindromic sequences of DNA
2013
PCR:
- A. stands for polymer chain response
- B. can only use DNA as a template
- C. can be used to amplify RNA sequences
- D. requires gel electrophoresis for detection
- E. is used in western analysis
B. can only use DNA as a template
2013
mRNA:
- A. is the most abundant RNA in the cell
- B. can be separated from other RNA species by polyA isolation
- C. can be analyzed by Southern analysis
- D. is not really important
- E. can be used as a template for PCR
B. can be separated from other RNA species by polyA isolation
2013
Generation of a knockout animal involves:
- A. homologous recombination
- B. random insertion of exogenous DNA
- C. targeted insertion of exogenous RNA
- D. use of siRNA to eliminate DNA expression
- E. competent rat species
A. homologous recombination
2013
Which of the following is a true statement about DNA:
- A. All DNA in a eukaryotic organism is in heterochromatin form unless activated
- B. Prokaryotic DNA is capped by telomere sequences
- C. 1.5% of eukaryotic DNA is considered genes
- D. All DNA in eukaryotic organisms is in euchromatin form unless activated
- E. 1.5% of all eukaryotic DNA is coding.
E. 1.5% of all eukaryotic DNA is coding.
2013
fMet :
- A. is the last amino acid added to the polypeptide chain
- B. is necessary for the elongation step of translation
- C. only exists in the tRNA structure
- D. has a specific tRNA separate from Met.
- E. has a different codon than Met
D. has a specific tRNA separate from Met
2013
Eukaryotic translation:
- A. is exactly the same as prokaryotic transcription
- B. requires a multiprotein complex that links the transcribed RNA to the translational machinery via the 5’ cap and poly A tail.
- C. does not use AUG as the start codon
- D. requires a shine-delgardo sequence to start translation.
- E. is dependant on the sigma factor
B. requires a multiprotein complex that links the transcribed RNA to the translational machinery via the 5’ cap and poly A tail.
2013
RNA polymerase II is distinct in that it
- A. Is only used to generate tRNAs
- B. Is only found in eukaryotic organisms
- C. Is effective on a variety of different promoters
- D. Is used to make primers for DNA replication
- E. Makes Okazaki fragments.
C. Is effective on a variety of different promoters
2013
In prokaryotes, RNA polymerase finds the start of transcription by:
- A. Specific sequences in the promoter region
- B. Identification of the cap structure
- C. Sliding along the DNA and looking for unwound segments
- D. Signaling the histone molecules to disconnect from the wound DNA
- E. Sliding along the DNA to find a specific stem-loop structure
A. Specific sequences in the promoter region
2013
Which of the following statements is FALSE:
- A. Many amino acids can be encoded by several codons
- B. Mitochondria have a slightly different “code” than that of the eukaryotic organism
- C. The wobble hypothesis comes from the finding that there is greater freedom in pairing in the first position of the codon
- D. Aminoacyl-tRNA synthetases function to add the amino acid onto the tRNA
- E. Aminoacyl-tRNA synthetases can proofread
C. The wobble hypothesis comes from the finding that there is greater freedom in pairing in the first position of the codon
2013
Splicing:
- A. can occur through ATP-dependant and –independent mechanism.
- B. Is critical to remove exons so introns can be joined together
- C. Depends on a 5’ pG and an internal branch site
- D. Generates a holiday junction
- E. A & C
2013 Exam options
- A. requires ATP
- B. Is critical to remove introns so exons can be joined together
- C. Depends on a 5’ pG and an internal branch site
- D. A & C
- E. B & C
E. A & C
- can occur through ATP-dependant and –independent mechanism.
- Depends on a 5’ pG and an internal branch site
2013 Exam Answers
- B. Is critical to remove introns so exons can be joined together
- C. Depends on a 5’ pG and an internal branch site
2013
Histone acetylation plays a critical role in:
- A. DNA replication
- B. RNA structure
- C. Activation of transcription
- D. The Lac operon
- E. Proofreading the degenerate code
C. Activation of transcription
2013
Protein translation in the endoplasmic reticulum:
- A. Only occurs in prokaryotes
- B. Requires a signal sequence
- C. Is necessary for all proteins in the cell
- D. A & C
- E. B & C
B. Requires a signal sequence
2013
Facilitated diffusion through a biological membrane is:
- A. driven by a difference of solute concentration.
- B. driven by ATP.
- C. Endergonic.
- D. generally irreversible
- E. not specific with respect to the substrate
A. driven by a difference of solute concentration.
2013
The acetylcholine receptor is a
- A. Steroid receptor.
- B. a membrane protein that permits a ligand to pass through the membrane only when opened by the appropriate ion.
- C. a membrane protein that permits an ion to pass through the membrane only when opened by the appropriate ligand.
- D. a molecule that binds to the membrane thereby allowing ions to pass through.
- E. always requires a second ligand to close the channel once it is opened.
C. a membrane protein that permits an ion to pass through the membrane only when opened by the appropriate ligand.
2013
The figure above is a hydropathy plot. What does this predict about bacteriorhodopsin?
- A. It is very water soluble
- B. It will bind to specific sequences in DNA as a transcription factor
- C. It is an integral membrane protein with seven membrane spanning domains.
- D. It is an integral membrane protein with five membrane spanning domains.
- E. It will fold into a beta-pleated sheet.

C. It is an integral membrane protein with seven membrane spanning domains.
2013
When a computer is looking for a potential open reading frame (ORF) it:
- A. finds ATG and stop codon.
- B. scans for translatable length higher than set threshold.
- C. repeats 3 times.
- D. A & B.
- E. A, B & C
D. A & B.
2013
Which one of the following is not involved in classical steroid hormone action?
- A. Cell surface receptors
- B. Hormone-receptor complexes
- C. Specific DNA sequences
- D. Transcription activation and repression
- E. Zinc fingers
A. Cell surface receptors
2013
The signal sequences that direct proteins to the nucleus are:
- A. always at the amino terminus of the targeted protein.
- B. cleaved after the protein arrives in the nucleus.
- C. glycosyl moieties containing mannose 6-phosphate residues.
- D. not located at the ends of the peptide, but in its interior.
D. not located at the ends of the peptide, but in its interior.
2013
What is the name of the above lipid:
- A. 18:1 (Δ9) cis-9-octodecenoic acid
- B. 18:1 (Δ10) cis-10-octodecenoic acid
- C. 17:1 (Δ9) cis-9-octodecenoic acid
- D. 17:1 (Δ10) cis-10-octodecenoic acid

A. 18:1 (Δ9) cis-9-octodecenoic acid
2013
In comparison to 20:4 (Δ5,8,11, 14), the above lipid would:
- A. Have a higher melting temperature
- B. Have a lower melting temperature
- C. more likely to be incorporated into membranes
- D. A & C
- E. A & B
B. Have a lower melting temperature
2013
In the picture to the left, the bottom of the gel is toward the cathode, and the top is toward the anode. Which of the below statements are true.
- A. Linking number is larger at the bottom of the gel, than at the top.
- B. Linking number is related to the number of bands in the lane.
- C. The DNA at the top of the gel is supercoiled
- D. The DNA at the bottom of the gel is supercoiled.
- E. The second lane is RNA.

D. The DNA at the bottom of the gel is supercoiled.
2013
Histone H1 is different from the other histones in that:
- A. It is associated with packaging RNA.
- B. It is found primarily in prokaryotes.
- C. It is not part of the histone core particle, but seals the nucleosome at the edge of the linker DNA.
- D. It is found only in eukaryotes.
- E. It is a pseudogene.
C. It is not part of the histone core particle, but seals the nucleosome at the edge of the linker DNA.
2013
Single strand repair:
- A. Occurs when there is a mismatch base in the DNA strand causing a distortion.
- B. Requires a RNA intermediate
- C. Requires ligase
- D. A & C
- E. A & B
D. A & C
2013
REC A is a protein that:
- A. Is an ATPase
- B. Is important for DNA recombination
- C. Is critical for binding to histone H2
- D. A & C
- E. A & B
E. A & B
2013
Given p(A)=0.5, p(C)=0.1, p(G)=0.2, and p(T)=0.2. Which string is most likely NON-RANDOM?
- A. AACTG
- B. AAAAA
- C. ATATA
- D. CCACG
- E. CCCCC
E. CCCCC
2013
Genome assembly REQUIRES
- A. Multiple runs to generate matching of overlapping read sequences
- B. Use of Sanger sequencing
- C. Use of pyrosequencing or Sanger sequencing
- D. Use of shotgun sequencing
- E. An understanding of intron exon boundaries
A. Multiple runs to generate matching of overlapping read sequences
2013
What do we lose when we view DNA as a string?
- A. Any useful information
- B. Histone binding and DNA accessibility information
- C. the ability to identify ORFs
- D. splicing information
B. Histone binding and DNA accessibility information
2013
What is the (3’-5’) DNA sequence encoded in this Sanger sequencing gel?
- A. …ATGGTGACA…
- B. …TACCACTGT…
- C. …ACAGTGGTA…
- D. …TGTCACCAT…
- E. None of the above
B. …TACCACTGT…
2013
Pyrosequencing differs from Sanger sequencing in that:
- A. It does not require a primer
- B. It generates longer reads (2,900 bp versus 400- 900 bp)
- C. It does not result in chain termination
- D. A & C
- E. A & B
C. It does not result in chain termination
2014
A signaling receptor such as is shown below would be expected to
- A. bind a broad range of hormones
- B. be unresponsive without an auxiliary signal
- C. bind only one kind signal molecule or only signals containing particular chemical groups
- D. generate a response without binding a signal molecule
- E. none of the above

C. bind only one kind signal molecule or only signals containing particular chemical groups
2014
As part of the β-adrenergic receptor system the protein Gscx dissociates from the receptor complex upon epinephrine binding and moves to adenylyl cyclase, exchanging a bound GDP for GTP in the process. This activates adenylyl cyclase, which produces c-AMP. After some time Gscx ceases to activate, and the production of c-AMP stops. Why?
- A. Cellular proteases attack Gscx, inactivating it.
- B. When the concentration of c-AMP rises, it binds to Gscx, thereby inhibiting it, an example of feedback inhibition.
- C. c-AMP stimulates phosphorylation of the receptor system, causing epinephrine to dissociate and ending the response
- D. Gscx slowly hydrolyzes its GTP, thereby losing its ability to bind to adenylyl cyclase, so it returns to become part of the receptor complex.
- E. none of the above
D. Gscx slowly hydrolyzes its GTP, thereby losing its ability to bind to adenylyl cyclase, so it returns to become part of the receptor complex.
2014
Cyclic AMP has what effect on protein kinase A?
- A. On binding to the catalytic subunits, it causes them to dissociate from the regulatory subunits. The resulting change in catalytic subunit conformation activates their ability to phosphorylate other proteins.
- B. On binding to the regulatory subunits, it causes the catalytic subunits to dissociate from the regulatory subunits. This makes the substrate binding cleft in the catalytic subunits available to phosphorylate other proteins.
- C. It stimulates hydrolysis of the regulatory subunits.
- D. It inhibits hydrolysis of the catalytic subunits
- E. c-AMP has no effect at all on protein kinase A.
B. On binding to the regulatory subunits, it causes the catalytic subunits to dissociate from the regulatory subunits. This makes the substrate binding cleft in the catalytic subunits available to phosphorylate other proteins.
2014
What of the following is/are true of calmodulin?
- A. It acts as a “sensor” for free calcium ion.
- B. When its Ca++ binding sites are occupied, it undergoes a conformational change leading it to bind to other proteins.
- C. On binding to other proteins it activates their phosphorylating function.
- D. Its calcium binding sites have the conformation of an “EF hand.”
- E. All of the above are true.
E. All of the above are true.
2014
- Which of the steps in the insulin cascade shown here would be expected to amplify directly the insulin binding signal?
- A. phosphorylation of IRS-1
- B. dissociation of Grb2 from SOS
- C. phosphorylation of MEK
- D. the catalytic activity of Raf-1 once Raf-1 has been activated
- E. more than one of the above

E. more than one of the above
2014
- In bacterial chemotaxis, attractant binding leads to phosphorylation of an asp on component 2, which binds to the flagellar motor thereby causing counterclockwise rotation and thus a “run.” Tumbling, caused by clockwise motor rotation, ends a run, allowing change of direction. What must happen for the transition from run to tumble to occur?
- A. Component 2 must dissociate from the motor.
- B. A phosphatase must act on component 2.
- C. The attractant must be consumed.
- D. A timer must exist to ensure that runs away from the attractant are shorter than those toward it.
- E. More than one of the above.
E. More than one of the above.
2014
The aldolase reaction in glycolysis
- A. converts fructose-6-phosphate to fructose-1,6-bisphosphate.
- B. converts dihydroxyacetone phosphate to glyceraldehyde-3 phosphate.
- C. converts pyruvate to lactate in active muscle.
- D. glyceraldehydes-3-phosphate to 1,3-bisphosphoglyceric acid.
- E. none of the above
E. none of the above
2014
How many carbon dioxide molecules are produced in the glycolytic conversion of glucose to pyruvate?
- A. none
- B. one
- C. two
- D. four
- E. more than 4
A. none
2014
Why is glucose-6-phosphate isomerized to fructose-6-phosphate in glycolysis?
- A. The fructose-6-phosphate is more resistant to spontaneous hydrolysis, which would cause loss of the phosphate.
- B. It allows phosphorylation at C1 to occur so that when the six carbon sugar is split into two three carbon pieces, each will have a phosphate on it.
- C. The fructose-6-phosphate is more likely to be in the open chain form, which makes it a good substrate for enzymes later in the path.
- D. The fructose-6-phosphate is less likely to be in the open chain form, which makes it a good substrate for enzymes later in the path.
- E. none of the above
B. It allows phosphorylation at C1 to occur so that when the six carbon sugar is split into two three carbon pieces, each will have a phosphate on it.
2014
What main purpose is served by the anerobic conversion of pyruvate to lactate?
- A. to provide a lactate reserve for conversion back to pyruvate
- B. to prevent pyruvate from forming acetyl CoA which would enter the TCA cycle
- C. to promote gluconeogenesis
- D. to regenerate NAD+ from NADH, without which glycolysis couldn’t continue
- E. none of the above
D. to regenerate NAD+ from NADH, without which glycolysis couldn’t continue
2014
The enzymes of glycolysis (in no particular order) are
- 1) enolase
- 2) phosphofructokinase-1
- 3) phosphoglycerate kinase
- 4) phosphohexose isomerase
- 5) pyruvate kinase
- 6) hexokinase
- 7) aldolase
- 8) glyceraldehyde 3-phosphate dehydrogenase
- 9) triose phosphate isomerase
- 10) phosphoglycerate mutase
Which of the following sets enzymes do not play a role in gluconeogenesis, the back reaction being catalyzed by other enzymes not listed here?
- A. 2, 5, 6
- B. 1, 3, 5
- C. 3, 4, 7
- D. 7, 9,
- 10 E. none of the above
A. 2, 5, 6
2014
Why, if all the intermediates between glucose and pyruvate are the same for glycolysis and gluconeogenesis, are different enzymes used for some of the steps in each direction?
- A. so that both routes can operate at the same time
- B. so that separate controls for each path can ensure that only one path operates at any given time
- C. to reduce the high energy phosphate consumption of gluconeogenesis so that the ATP from glycolysis is not wasted
- D. to preserve reducing power by using less NADH in gluconeogenesis
- E. all of the above
B. so that separate controls for each path can ensure that only one path operates at any given time
2014
α-keto acids can be converted by transamination to amino acids for protein synthesis, and when energy is needed, proteins can be broken down and amino acids can be transaminated to α-keto acids so their carbon skeletons can be oxidized. Taking glycolysis and the TCA cycle as a whole, how many of the intermediates can be transaminated directly to amino acids?
- A. none
- B. one
- C. two
- D. three
- E. more than three
D. three
2014
Although molecular oxygen does not participate directly in any of the reactions in the citric acid cycle, the cycle only operates when oxygen is present. This is because O2:
- A. is necessary as an activator for several enzymatic dehydrogenations (oxygenations) in the cycle.
- B. is necessary for producing water, which is crucial for all cellular processes.
- C. accepts electrons from the electron transfer chain, allowing reoxidation of NADH to NAD+ .
- D. removes toxic by-products of the citric acid cycle.
- E. has all of the above functions.
C. accepts electrons from the electron transfer chain, allowing reoxidation of NADH to NAD+ .
2014
Which one of the following statements about gluconeogenesis is false?
- A. For starting materials, it can use carbon skeletons derived from certain amino acids.
- B. It consists entirely of the reactions of glycolysis, operating in the reverse direction.
- C. It employs the enzyme glucose 6-phosphatase.
- D. It is one of the ways that mammals maintain normal blood glucose levels between meals.
- E. It requires metabolic energy (ATP or GTP).
B. It consists entirely of the reactions of glycolysis, operating in the reverse direction.
2014
A crude tissue preparation capable of carrying out the reactions of the citric acid cycle given acetyl-CoA uniformly labeled in both carbons with 14C. After one turn of the cycle, oxaloacetate would have 14C in:
- A. all four carbon atoms.
- B. no pattern that is predictable from the information provided.
- C. none of its carbon atoms.
- D. the keto carbon and one of the carboxyl carbons.
- E. the two carboxyl carbons.
A. all four carbon atoms.
2014
In the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase complex to yield acetyl-CoA, the first step is removal of CO2 and attachment of the acetyl group to thiamine pyrophosphate. The subsequent steps involve redox reactions. Which of the following gives them in the order in which they occur?
- A. reduction of NAD+ , oxidation of reduced lipoic acid, reduction of lipoic acid with attachment of the acetyl to lipoate, reduction of FAD
- B. reduction of FAD, oxidation of reduced lipoic acid, reduction of NAD+ , reduction of lipoic acid with attachment of the acetyl to lipoate
- C. reduction of lipoic acid with attachment of the acetyl to lipoate, oxidation of reduced lipoic acid, reduction of FAD, reduction of NAD+
- D. oxidation of reduced lipoic acid, reduction of NAD+ , reduction of lipoic acid with attachment of the acetyl to lipoate, reduction of FAD
- E. none of the above
C. reduction of lipoic acid with attachment of the acetyl to lipoate, oxidation of reduced lipoic acid, reduction of FAD, reduction of NAD+
2014
Which of the TCA cycle reactions yields reducing power directly, i.e. NADH or FADH2?
- A. the citrate synthase reaction
- B. conversion of succinate to fumarate
- C. the aconitase reaction
- D. conversion of fumarate to malate
- E. more than one of these
B. conversion of succinate to fumarate
2014
If succinyl-CoA labeled with 14C in C1 (which carries the CoA) is converted to fumarate, the label will appear in fumarate
- A. equally in C1 and C2
- B. equally in c2 and c3
- C. only in c1
- D. only in c4
- E. equally in c1 and c4
equally in c1 and c4
2014
Several steps of the TCA cycle are regulated allosterically. Which of the following combinations would be expected to slow or stop the cycle from operating?
- A. High AMP concentration and high NAD+ concentration
- B. High calcium concentration and high ADP concentration
- C. High ATP concentration and high NADH concentration
- D. High CoASH concentration and high ADP concentration
- E. all of the above
C. High ATP concentration and high NADH concentration
2014
Although vertebrates cannot convert acetyl-CoA to carbohydrates, plants can by using the reactions of the glyoxylate cycle. For this bypass to work they must pass an intermediate that they make in peroxisomes or glyoxisomes to the TCA cycle in mitochondria. Which of the following is the intermediate?
- A. isocitrate
- B. malate
- C. acetyl-CoA
- D. succinate
- E. none of these
D. succinate
2014
In the reaction of acetyl-CoA and oxaloacetate to form citrate, the cellular concentration of oxaloacetate is very low. The reaction is nevertheless essentially irreversible. Why?
- A. Citrate builds up, pushing the reaction to the right
- B. The standard free energy of the reaction is so strongly negative that the effect of the low OAA concentration is overcome.
- C. Citrate synthase is a very active enzyme with a high kcat.
- D. Allosteric interactions force the reaction to the right.
- E. none of the above
B. The standard free energy of the reaction is so strongly negative that the effect of the low OAA concentration is overcome.
2014
The iron in FeS clusters usually participates in reactions by
- A. reducing disulfides to SH groups.
- B. oxidizing SH groups to disulfides
- C. gaining or losing an electron according the reaction Fe++ → Fe+++ + e
- D. binding or releasing sulfur atoms
- E. consuming radical oxygen species
C. gaining or losing an electron according the reaction Fe++ → Fe+++ + e
2015
What is the cost (in ATP equivalents) of transforming glucose to pyruvate via glycolysis and back again to glucose via gluconeogenesis?
- A. 1
- B. 2
- C. 3
- D. 4
- E. 5
D. 4
2015
Starting with fructose 1,6 bisphosphate and incubating with all glycolytic enzymes and cofactors what is the yield of ATP?
- A. 1
- B. 2
- C. 3
- D. 4
- E. 5
D. 4
2015
The conversion of 1 mol of fructose 1,6-bisphosphate to 2 mol of pyruvate by the glycolic pathway results in a net formation of:
- A. 1 mol of NAD+ and 2 mol of ATP.
- B. 1 mol of NADH and I mol of ATP.
- C. 2 mol of NADH and 2 mol of ATP.
- D. 2 mol of NADH and 4 mol of ATP.
- E. 2 mol of NAD+ and 4 mol of ATP.
D. 2 mol of NADH and 4 mol of ATP.
2015
Oxaloacetate uniformly labeled with 14C (i.e., with equal amounts of 14C in each of its carbon atoms) is condensed with unlabeled acetyl-CoA. After a single pass through the citric acid cycle back to oxaloacetate, what fraction of the original radioactivity will be found in the oxaloacetate?
- A. all
- B. 1/2
- C. 1/3
- D. 1/4
- E. 3/4
B. 1/2
2015
Which of the following statements apply (applies) to the β oxidation of fatty acids?
- The process takes place in the cytosol of mammalian cells.
- Carbon atoms are removed from the acyl chain one at a time.
- Before oxidation, fatty acids must be converted to their CoA derivatives.
- NADP+ is the electron acceptor.
- The products of β oxidation can directly enter the citric acid cycle for further oxidation.
- A. 1, 2 and 3
- B. 1, 2 and 5
- C. 3 and 5 only
- D. l and 3 only
- E. 4 only
C. 3 and 5 only
2015
Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue preparation capable of carrying out the reactions of the citric acid cycle. After one turn of the cycle, oxaloacetate would have 14C in:
- A. all four carbon atoms.
- B. no pattern that is predictable from the information provided.
- C. none of its carbon atoms.
- D. the keto carbon and one of the carboxyl carbons.
- E. the two carboxyl carbons.
A. all four carbon atoms.
2015
The two moles of CO2 produced in the first turn of the citric acid cycle have their origin in the:
- A. carboxyl and methylene carbons of oxaloacetate
- B. carboxyl group of acetate and a carboxyl group of oxaloacetate.
- C. carboxyl group of acetate and the keto group of oxaloacetate.
- D. two carbon atoms of acetate.
- E. two carboxyl groups derived from oxaloacetate.
E. two carboxyl groups derived from oxaloacetate
2015
Which of the enzymatic steps of the TCA cycle listed below is irreversible under cellular conditions?
- A. aconitase
- B. isocitrate dehydrogenase
- C. succinate dehudrogenase
- D. fumarase
- E. None of these is irreversible under cellular conditions
B. isocitrate dehydrogenase
2015
If electron transport is operating,
- A. the pH in the mitochondrial matrix will be lower than in the intermembrane space.
- B. the pH in the mitochondrial matrix will be higher than in the intermembrane space.
- C. the pH in the mitochondrial matrix will be the same as in the intermembrane space.
- D. the pH in the two compartments will differ only when electron transport starts.
- E. cannot tell from data givem.
B. the pH in the mitochondrial matrix will be higher than in the intermembrane space.
2015
The normal physiological role of ATP synthase is
- A. to pump protons out of the mitochondrion.
- B. to catalyze ATP hydrolysis.
- C. to use the proton electrochemical gradient to phosphorylate ADP.
- D. to use ATP to drive metabolic reactions.
- E. more than one of the above.
C. to use the proton electrochemical gradient to phosphorylate ADP.
2015
Glycolysis produces two molecules of pyruvate from one molecule of glucose. If each pyruvate is then converted to acetyl-CoA, and if each acetyl-CoA undergoes one turn of the TCA cycle, how many electrons will enter the electron transport chain?
- A. None
- B. Four
- C. Six
- D. Twenty
- E. none of these is correct
D. Twenty
2015
Which of the hemes shown below cannot be removed from its protein even if the protein is fully unfolded?
- A. Protoporphryn IX
- B. Heme C
- C. Heme A
- D. All can be removed upon unfolding the protein.
- E. None can be removed by unfolding the protein.

B. Heme C
2015
How many molecules of NADH would be necessary to convert a 4Fe-4S iron sulfur center from a fully oxidized state to a fully reduced state?
- A. None. NADH cannot interact directly or indirectly with 4Fe-4S centers.
- B. One
- C. Two
- D. Four
- E. None of the above
C. Two
2015
Yeast cells grow in the absence of oxygen, consuming glucose for energy and producing ethanol. If exposed to air, they consume much less oxygen and cease to make ethanol. Why?
- A. Oxygen inactivates the enzyme that converts pyruvate to ethanol.
- B. Oxygen slows the growth of the culture so less glucose is needed, and the cells use the ethanol instead.
- C. Oxygen is an allosteric inhibitor of glycolytic enzymes, reducing their activity.
- D. Oxygen allows the electron transport chain to operate, so that each glucose molecule yields more energy, thereby reducing the amount of glucose required for growth and avoiding diversion of pyruvate to ethanol.
- E. More than one of the above
D. Oxygen allows the electron transport chain to operate, so that each glucose molecule yields more energy, thereby reducing the amount of glucose required for growth and avoiding diversion of pyruvate to ethanol.
2016
How many of the TCA cycle reactions involve oxidations, i.e. production of NADH or FADH2?
- A. none
- B. one
- C. two
- D. three
- E. more than three
E. more than three
2016
How many complete turns of the TCA cycle would be required to produce four molecules of CO2?
- A. none
- B. one
- C. two
- D. three
- E. more than three
C. two
2016
How many net water molecules are produced when aconitase converts citrate to isocitrate?
- A. none
- B. one
- C. two
- D. three
- E. more than three
- A. none
2016
Lipoic acid is a key part of the pyruvate dehydrogenase complex. It participates in the oxidative decarboxylation of pyruvate, becoming reduced in the process, and is, in turn, reoxidized in preparation for another catalytic cycle. Which of the following is true of lipoic acid?
- A. It contains a nicotinamide group like that in NAD+ .
- B. It contains a flavin like that in FAD.
- C. It interacts directly with NAD+ when it is reoxidixed.
- D. It swings from the active site where it is oxidized to the site where it is reduced.
- E. More than one of these.
D. It swings from the active site where it is oxidized to the site where it is reduced.
2016
Which of the following is NOT involved in oxidative decarboxylations such as those catalyzed by αketoglutarate dehydrogenase, pyruvate dehydrogenase, and the enzyme catalyzing oxidation of isoleucine?
- A. CoA-SH is produced
- B. NADPH is produced
- C. NAD+ is produced
- D. CO2 is produced
- E. All the above are correct
B. NADPH is produced
2016
Which of the following reactions would cause the position of a radioactively labeled carbon atom to be scrambled so its origin in acetyl-CoA could no longer be accurately determined?
- A. Citrate synthase
- B. succinate dehydrogenase
- C. fumarase
- D. Succinyl-CoA synthetase
- E. None of these
D. Succinyl-CoA synthetase
2016
In muscle tissue, Ca++ signals contraction, and it also stimulates energy-yielding metabolism to replace the ATP consumed by contraction. The effect of calcium on α-ketoglutrate dehydrogenase activity would be expected to
- A. shift the sigmoid Vo versus [S] curve to the right.
- B. shift the sigmoid Vo versus [S] curve to the left.
- C. inhibit the enzyme
- D. reduce the amount of NADH produced in the reaction.
- E. None of the above
B. shift the sigmoid Vo versus [S] curve to the left.
2016
In plants, some invertebrates and some microorganisms, carbohydrate biosynthesis is made possible by
- A. the TCA cycle
- B. glycolysis and gluconeogenesis
- C. the glyoxylate cycle
- D. all of these combined
- E. none of these
D. all of these combined
2016
Which statement is not true of phosphofructokinase-1
- A. It is inhibited by fructose 2,6-bisphosphate.
- B. It is activated by AMP.
- C. It is inhibited by citrate.
- D. It is inhibited by ATP.
- E. ATP increases the K0.5 (the effective Km) for fructose 6-phosphate
A. It is inhibited by fructose 2,6-bisphosphate.
2016
Which of the following statements about photosystem II is correct?
- A. It is located in the inner mitochondrial membrane.
- B. It contains the electron carrier with the most negative E’o in the entire photosynthetic system.
- C. It contains a Mn-containing complex which splits water.
- D. Its final electron acceptor is NADPH.
- E. Water is the only electron donor capable of regenerating P680 from P680*.
C. It contains a Mn-containing complex which splits water.
2016
Which of the following statements about cyclic photophosphorylation and non-cyclic photophosphorylation is correct? (from a posted exam)
- A. Cyclic photophosphorylation involves only photosystem II and produces only ATP; non-cyclic photophosphorylation involves photosystems I and II and also produces only ATP.
- B. Both pathways liberate oxygen.
- C. Both pathways involve photosystems I and II.
- D. Cyclic photophosphorylation reduces NADP+ and liberates oxygen; non-cyclic photophosphorylation reduces NADP+ but does not liberate oxygen.
- E. Non-cyclic photophosphorylation reduces NADP+ and liberates oxygen; cyclic photophosphorylation produces ATP but does not liberate oxygen.
E. Non-cyclic photophosphorylation reduces NADP+ and liberates oxygen; cyclic photophosphorylation produces ATP but does not liberate oxygen.
2016
Some photosynthetic prokaryotes use H2S, hydrogen sulfide, instead of water as their photosynthetic hydrogen donor. How does this change the ultimate products of photosynthesis?
- A. Carbohydrate (CH2O) is not produced.
- B. H2O is not produced.
- C. Oxygen is not produced.
- D. ATP is not produced.
- E. The products do not change.
C. Oxygen is not produced.
2016
What is the difference between inhibitors and uncouplers of oxidative phosphorylation?
- A. Inhibitors stop electron transport but allow evolution of oxygen.
- B. Uncouplers stop electron transport but allow evolution of oxygen.
- C. Inhibitors allow electron transport but stop evolution of oxygen.
- D. Uncouplers allow electron transport but stop evolution of oxygen.
- E. None of the above is correct.
E. None of the above is correct.
The structure of ubiquinone is shown below. The ring part undergoes oxidation/reduction reactions. What is the function of the isoprenoid side chain?
- A. It is a left over from the biochemical synthesis of ubiquinone.
- B. It serves as a precursor in steroid biosynthesis.
- C. It helps control the rates of oxidation and reduction.
- D. It keeps the molecule within the membrane bilayer.
- E. none of the above
D. It keeps the molecule within the membrane bilayer.
2016
Which of the proteins involved in electron transport is also a TCA cycle enzyme?
- A. NADH dehydrogenase
- B. cytochrome oxidase
- C. succinate dehydrogenase
- D. malate dehydrogenase
- E. none of these
C. succinate dehydrogenase
2016
How many molecules of cytochrome c are required to fully oxidize reduced ubiquinone (QH2) to Q via catalysis by ubiquinone: cytochrome c oxidoreductase?
- A. None. Cytochrome c serves a catalytic role in this complex and undergoes no net change.
- B. One
- C. Two
- D. Four
- E. None of these answers is correct.
C. Two
2016
How many ATP molecules does it take to move one long chain fatty acyl-CoA across the inner mitochondrial membrane?
- A. None
- B. One
- C. Two
- D. Four
- E. None of these answers is correct.
A. None
2016
The enzymes of fatty acid catabolism are listed below. In which order do they act to remove acetyl-CoA from a fattyacyl-CoA?
- enoyl CoA hydratase
- acyl-CoA dehydrogenase
- acyl-CoA acetyltransferase (thiolase)
- β-hydroxyacyl-CoA dehydrogenase
- A. 1, 2, 3, 4
- B. 2, 1, 4, 3
- C. 4, 2, 3, 1
- D. 3, 3, 1, 4
- E. None of these sequences is correct.
B. 2, 1, 4, 3
2016
Why are auxiliary enzymes needed for conversion of unsaturated fatty acids to acetyl-CoA?
- A. They shift double bonds to allow for β-oxidation.
- B. They reduce double bonds which would otherwise obstruct β-oxidation.
- C. They convert cis double bonds to trans, as the β-oxidation enzymes cannot handle cis structures.
- D. They activate the normal β-oxidation enzymes so those enzymes can handle unsaturation.
- E. more than one of the above
E. more than one of the above
2015
In fatty acid biosynthesis acyl-CoA carboxylase is activated by a phosphatase and inactivated by c-AMP dependent protein kinase A (PKA). Which of the following statements about its regulation is correct?
- A. PKA activity is stimulated by high blood glucose.
- B. The phosphatase activity is stimulated by low blood glucose.
- C. When blood glucose is high, the carboxylase produces malonyl-CoA, which shuts off transfer of fatty acids to the mitochondrion.
- D. When the carboxylase is activated, fatty acid catabolism is turned off. E. more than one of the above
- E. more than one of the above
E. more than one of the above
2016
What do lipoic acid, the acyl carrier protein and biotin have in common?
- A. All are found at the ends of long swinging arms that move them from one active site to another.
- B. They are attached non-covalently to their proteins.
- C. All involve SH groups or disulfides in their functions.
- D. They have carbohydrate attached to them.
- E. They have nothing at all in common
A. All are found at the ends of long swinging arms that move them from one active site to another.
2016
In which of the steps of fatty acid oxidation do mitochondria and peroxisomes differ?
- A. the first oxidation → c=c double bond
- B. hydration of the double bond
- C. the second oxidation to the β-ketacyl-CoA
- D. the thiolytic cleavage that yields an acetyl-CoA
- E. more than one of the above
E. more than one of the above
2016
The CO2 required in fatty acid biosynthesis
- A. becomes part of the growing carbon chain.
- B. serves as the main buffer in the local environment as bicarbonate ion.
- C. is first used and then removed, never entering the final product.
- D. activates the synthetic system.
- E. deactivates the synthetic system
C. is first used and then removed, never entering the final product
2016
Why do plants and bacteria add double bonds beyond carbon nine, while animals do not?
- A. They arose earlier in evolution.
- B. They are metabolically more complex than animals.
- C. Because they can’t make or use cholesterol.
- D. The extra double bonds make adaptation to temperature changes possible. E. none of the above
D. The extra double bonds make adaptation to temperature changes possible. E. none of the above
2019
Glycolysis ends with _____ and feeds its end product to the TCA cycle as _____.
- A. phosphoenol pyruvate, ATP
- B. glyceraldehyde 3-P, dihydroxyacetone phosphate
- C. 1,3-bisphosphoglycerate, 2-phosphoglycerate
- D. pyruvate, acetyl-CoA
- E. none of the above
D. pyruvate, acetyl-CoA
2019
How many glycolysis intermediates or products yield amino acids upon direct transamination?
- A. None
- B. One
- C. Two
- D. Three
- E. more than three
B. One
2019
Which of the following glycolytic intermediates/products does NOT contribute to the carbon skeleton of aromatic amino acids?
- A. Pyruvate
- B. Phosphoenolpyruvate
- C. 3-Phosphoglyceric acid
- D. Acetyl CoA E.
- None of the above contribute
D. Acetyl CoA E.
2019
How many complete turns of the TCA cycle would be required to produce six molecules of CO2?
- A. none
- B. one
- C. two
- D. three
- E. more than three
D. three
2019
Which of the following is produced in oxidative decarboxylations such as those catalyzed by α-ketoglutarate dehydrogenase, pyruvate dehydrogenase, and the enzyme catalyzing oxidation of isoleucine?
- A. CoA-SH
- B. NADPH
- C. NAD+
- D. CO2
- E. All these are produced
D. CO2
2019
What is the order of reactions in the β-oxidation of fatty acids?
- formation of a carbonyl group
- formation of a carbon-carbon double bond
- hydration
- consumption of a CoASH
- A. 1, 2, 3, 4
- C. 4,3,2,1
- B. 2, 3, 1, 4
- D. 4,2,3,1
- E. none of these
B. 2, 3, 1, 4
2019
Which of the following is/are consumed in the β-oxidation of palmitoyl-CoA to acetyl-CoA?
- A. CoASH
- B. FADH2
- C. NAD+
- D. more than one of these
- E. none of these
D. more than one of these
2019
How many protons does complex II (succinate dehydrogenase) pump across the inner mitochondrial membrane for each succinate oxidized to fumarate?
- A. none
- B. two
- C. four
- D. eight
- E. none of these
A. none
2019
Sodium azide acts on the mitochondrion the same way as cyanide. What will be its effect on energy metabolism by the mitochondrion?
- A. There will be no effect at all; everything will proceed as normal.
- B. The flow of electrons will cease.
- C. Electrons will flow, but ATP will not be produced.
- D. Electrons will flow, but oxygen will not be consumed.
- E. None of these is correct
B. The flow of electrons will cease
2019
The principal biochemical benefit of ubiquinone as an electron carrier is that
- A. has a very high reduction potential, higher than the other electron carriers.
- B. it aids both aerobic and anerobic metabolism.
- C. it is ubiquitous throughout the cell, being found in all intracellular spaces.
- D. it can handle one electron at a time, unlike NADH and FADH2, which can only transfer two at a time.
- E. none of these is the main benefit.
D. it can handle one electron at a time, unlike NADH and FADH2, which can only transfer two at a time.
2019
What did Dorothy Crowfoot Hodgkin win the Nobel prize for?
- A. Determining the three dimensional structure of insulin
- B. Determining the role of biotin in fatty acid metabolism
- C. Determining the three dimensional structure of vitamin B12
- D. Working out the path of fatty acid degradation
- E. none of the above
C. Determining the three dimensional structure of vitamin B12
2019
Which of the following describes photorespiration?
- A. Rubisco oxygenates ribulose 1,5-bisphosphate.
- B. Photorespiration generates glycolate, which is toxic.
- C. Oxygen is consumed and CO2 is produced as in mitochondrial respiration.
- D. Carbon, first fixed by photosynthesis into carbohydrate, is then lost.
- E. all of the above
E. all of the above
2019
In the research presented by South et al. in Science Magazine, biochemical engineering was used to circumvent photorespiration. What did they do?
- A. In Alternate Pathway 1 (AP1) genes from E. coli that convert glycolate to glycerate were introduced into the chloroplast.
- B. Participation in photorespiration of peroxisomal and mitochondrial pathways was avoided.
- C. In AP2 genes were introduced into the chloroplast that converted glycolate to malate.
- D. In AP3 undesirable genes were deleted
- E. more than one of the above
E. more than one of the above
2019
Of the three biochemically engineered constructs created by South et al. and introduced into tobacco plants
- A. all showed the same biomass increases in field trials.
- B. one had an approximately 40% increase in biomass.
- C. two failed to show any biomass increase
- D. the greenhouse results were radically different from the field trial results.
- E. none of the above
B. one had an approximately 40% increase in biomass
2019
In the best of South et al.’s tobacco modifications
- A. photosynthesis operated at a higher rate than in wild type.
- B. more ATP was generated by photosynthesis than in wild type.
- C. little or no fixed carbon was lost to photorespiration.
- D. the plants became intolerant of high light levels.
- E. more than one of the above
C. little or no fixed carbon was lost to photorespiration.
2020
What would be the net result of the phosphofructokinase-1 and fructose bisphosphatese-1 reactions?
A. loss of ATP by hydrolysis
B. increased gluconeogenesis
C. the synthesis of high energy phosphate bonds
D. no net change; their actions cancel one another.
E. none of the above
A. loss of ATP by hydrolysis
2020
How many NADH molecules are generated when one molecule of glucose is converted to pyruvate?
- A. none
- B. one
- C. two
- D. three
- E. more than three
C. two
2020
Imagine that one molecule of glucose is converted to two molecules of pyruvate by glycolysis and then back to glucose by gluconeogenesis. What will be the net direct change in the number of high energy phosphate bonds (i.e. ATP equivalents) if NAD+ and NADH are ignored?
- A. +2
- B. +4
- C. -2
- D. -4
- E. zero
D. -4
2020
When sedoheptulose 7-phosphate is produced from ribose 5-phosphate and xylulose 5-phosphate in the pentose phosphate pathway, what other product is produced in the reaction?
- A. glyceraldegyde 3-phosphate
- B. erythrose 4-phosphate
- C. fructose 6-phosphate
- D. glucose 6-phosphate
- E. none of these
A. glyceraldegyde 3-phosphate
2020
Why does a mutation in which people have less glucose 6-phosphate dehydrogenase (G6PD) than normal cause them to be resistant to malaria?
- A. The NADPH generated by the enzyme kills Plasmodium falciparum, the malaria parasite, directly.
- B. P. falciparum is very sensitive to damage by radical oxygen. People who have less glucose 6-phosphate dehydrogenase than normal generate less NADPH, which, if in sufficient quantity, would destroy radical oxygen that would otherwise kill the parasite.
- C. The red cells of people with G6PD deficiency don’t sickle.
- D. The pentose phosphate pathway in G6PD deficient people does not generate enough nucleic acid precursors for the parasite to survive.
- E. none of these
B. P. falciparum is very sensitive to damage by radical oxygen. People who have less glucose 6-phosphate dehydrogenase than normal generate less NADPH, which, if in sufficient quantity, would destroy radical oxygen that would otherwise kill the parasite.
2020
When glycogen is hydrolyzed by glycogen phosphorylase, the product is
- A. glucose
- B. glucose 6-phosphate
- C. pyruvate
- D. glucose 1-phosphate
- E. none of these
D. glucose 1-phosphate
2020
If one glucose molecule is subjected to glycolysis and if both the resulting acetyl-CoA molecules pass through the TCA cycle (two turns of the cycle), how many pairs of electrons are produced?
- A. two
- B. four
- C. six
- D. eight
- E. none of these
E. none of these
2020
In the pyruvate dehydrogenase reaction the disulfide of lipoic acid is reduced and must be oxidized back to the disulfide form if the enzyme is to catalyze action on another pyruvate molecule. What supplies the oxidizing power?
- A. NAD+
- B. NADH
- C. CoA-SH
- D. another molecule of lipoic acid
- E. none of these
A. NAD+
2020
At which of the following steps of the TCA cycle does oxidative decarboxylation occur?
- A. succinate dehydrogenase
- B. fumarase
- C. a-ketoglutarate dehydrogenase
- D. aconitase
- E. more than one of these
C. a-ketoglutarate dehydrogenase
2020
Why would AMP or ADP allosterically stimulate some of the TCA cycle enzymes?
- A. They fit into binding sites into which ATP cannot fit.
- B. Their presence at elevated concentrations implies that the cell needs ATP, so the enzymes are “turned on.”
- C. Their presence at elevated concentrations implies that the cell’s NAD+ concentration is too low.
- D. The statements is incorrect; it is calcium that stimulates the enzymes.
- E. None of the above
B. Their presence at elevated concentrations implies that the cell needs ATP, so the enzymes are “turned on.”
2020
Which TCA cycle intermediate yields aapartic acid in a one-step reaction?
- A. pyruvate
- B. citrate
- C. a-ketoglutarate
- D. oxaloacetate
- E. more than one of these
D. oxaloacetate
2020
The following fatty acid, in which the indicated carbon is labeled with 14C but the other carbons are 12C, is fed to an animal:
14CH3(CH2)9COOH
After allowing 30 minutes for fatty acid b oxidation, the label would most likely be recovered in:
- A. acetyl-CoA.
- B. beta-hydroxy butyryl-CoA.
- C. both acetyl-CoA and propionyl-CoA.
- D. palmitoyl-CoA.
- E. propionyl-CoA.
E. propionyl-CoA.
2020
In the oxidation of linoleic acid (18:2(cis-D9,cis-D12)), the double bonds require the action of enoyl CoA isomerase and dienoyl CoA reductase, which yield trans- D2-decanoyl-CoA, shown below. The process is completed by four b-oxidation cycles. In the first of these cycles, one step in the usual b-oxidation is skipped. Which step?
- A. enoyl-CoA hydratase
- B. thiolase
- C. b-hydroxyacyl-CoA dehydrogenase
- D. acyl-CoA dehydrogenase
- E. none of these: no step is skipped

D. acyl-CoA dehydrogenase
2020
What is the role of acetyl-CoA carboxylase (ACC) in the regulation of fatty acid metabolism?
- A. When activated by a phosphatase whose activity is stimulated by insulin when blood glucose is high it catalyzes formation of malonyl-CoA, leading to fatty acid synthesis.
- B. When blood glucose is low, it is inactivated by cAMP-dependent protein kinase A (PKA), which is stimulated by glucagon. Malonyl-CoA is therefore not formed, and fatty acid synthesis is prevented.
- C. It activates the carnitine shuttle.
- D. A and B
- E. It has no role in fatty acid metabolism.
D. A and B
2020
Cyclic electron flow in chloroplasts produces:
A. ATP and O2, but not NADPH.
B. ATP, but not NADPH or O2 .
C. NADPH, and ATP, but not O2 .
D. NADPH, but not ATP or O2 .
E. O2, but not ATP or NADPH.
A. ATP and O2, but not NADPH.
2020
Why does ubiquinone diffuse freely within the mitochondrial inner membrane while not escaping to the matrix or intermembrane space?
- A. It is covalently anchored to a type of molecule which is a diffusible membrane component.
- B. Its unusual distribution of oxygen atoms renders it lipid soluble.
- C. It has a large hydrophobic tail.
- D. Its interactions with protein electron carriers are so fast it doesn’t have time to escape but slow enough to allow diffusion.
- E. none of the above
C. It has a large hydrophobic tail.
2020
How many molecules of cytochrome c are required to fully oxidize reduced ubiquinone (QH2) to Q via catalysis by ubiquinone: cytochrome c oxidoreductase?
- A. None. Cytochrome c serves a catalytic role in this complex and undergoes no net change.
- B. One
- C. Two
- D. Four
- E. None of these answers is correct.
C. Two
2020
2,4-dinitrophenol, shown below, has a pKa of 4.0 for proton dissociation of the hydroxyl group. If we represent it as DNP-OH, the acid dissociation is
DNP-OH ⇌ DNP-O- + H+.
On the P side of the mitochondrial matrix, the DNP is in the protonated form. It passes easily through the membrane to the N side, where it dissociates and passes back through the membrane to the P side, and this cycle is repeated over and over. What will be the effect on energy metabolism by the mitochondrion?
- A. There will be no effect at all; everything will proceed as normal.
- B. The flow of electrons will cease.
- C. Electrons will flow, but ATP will not be produced.
- D. Electrons will flow, but oxygen will not be consumed.
- E. None of these is correct
C. Electrons will flow, but ATP will not be produced
2020
Ribulose 1 ,5-bisphosphate carboxylase/oxygenase is arguably the most important enzyme on earth because nearly all life is dependent, ultimately, on its action. The reactions catalyzed by this enzyme are influenced by:
- A. pH
- B. substrate concentration.
- C. Mg2+ concentration.
- D. temperature.
- E. all of the above
E. all of the above
2020
Of all the components of the “Z scheme,” which has the lowest E’° (i.e., is the best reducing agent)?
- A. P700
- B. P700*
- C. O2
- D. H2O
- E. NADPH
B. P700*