Carbohydrate Metabolism (Exam II)

Question 1

What are the four carbohydrate metabolism processes?

Answer 1

1. Glycolysis
2. Lactate production (glucose → lactate)
3. Glycogen metabolism
4. Gluconeogenesis

Question 2

Epinephrine

Answer 2

Site of action: muscle, adipose & liver

Mechanism: c-AMP ↑Ca++ ↑

Action: glycogenolysis ↑ lipolysis ↑ glycolysis ↑

Question 3

Glucagon

Answer 3

Site of action: adipose & liver

Mechanism: c-AMP ↑

Action: lipolysis ↑ glycogenolysis ↑ gluconeogenesis ↑ glycolysis ↓

Question 4

Insulin

Answer 4

Site of action: muscle, adipose & liver

Mechanism: tyrosine kinase ↑ phosphatidylinositol-3,4,5-P3 ↑ c-AMP ↓

Action: glucose transport ↑ lipolysis ↓ fatty acid synthesis ↑ glycogen synthesis ↑

Question 5

Know the four GLUT isoforms, and be able to compare their expression in different tissues, their affinities for glucose, and their insulin dependence

Answer 5

GLUT1 (3 mM): red blood cells, fetal tissue
GLUT2 (20 mM): liver, β-cells, kidney, gut
GLUT3 (1-2 mM): brain
GLUT4 (5 mM): muscle & adipose (insulin-dependent)

Question 6

G-Protein Coupled Receptors (i.e., glucagon & epinephrine at β-receptors)

Answer 6

Agonist binds to receptor → adenylate cyclase exchanges ADP for ATP to phosphorylate cAMP → cAMP activates PKA → PKA phosphorylates a protein.

Question 7

Epinephrine at α-receptors

Answer 7

Agonist binds to receptor → PIP2 becomes IP3 + DAG → DAG activates PKC and IP3 released Ca2+ → PKC and Ca2+-Calmodulin Complex phosphorylate their respective proteins.

Question 8

Insulin receptor

Answer 8

Insulin binds to receptor → PIP2 becomes PIP3 → PIP3 activates PKB/Akt → PKB/Akt phoshporylates protein.

Question 9

Describe the process by which insulin signaling controls GLUT4 localization

Answer 9

1. binding of insulin to the α-subunits of its receptor activates a tyrosine kinase domain, resulting in autophosphorylation of tyrosine residues in the β-subunits.
2. The negative charge of the phosphates causes members of the IRS (insulin receptor substrate)
   family of proteins to bind to the β-subunits and be phosphorylated at two Tyr residues by the
   kinase activity of the activated insulin receptor
3. Phosphorylated-IRS dissociate from the receptor
   and bind to (and activate) proteins with a characteristic domain (SH2); PI-3-kinase
4. PI-3-kinase phosphorylates PIP2 (PI-4,5-
   bisphosphate) to PIP3 (PI-3,4,5-trisphosphate).
5. PIP3 activates PDK-1 (phosphoinositidedependent
   kinase) , initiating activation of such downstream effectors as Akt/PKB that results in movement of GLUT4 to the cell surface in adipose and muscle, increasing glucose uptake

Question 10

Know the reaction catalyzed by hexokinase, and be able to compare glucokinase activity with the activities of the other hexokinase isoforms

Answer 10

Hexokinase/Glucokinase phosphorylates glucose to glucose-6-phosphate.

Hexokinases: Low Km, product inhibition by G-6-P, dimeric, non-hepatic.

Glucokinase: High Km, no product inhibition, monomeric, hepatic

Question 11

Compare glucose uptake and utilization in brain, muscle, and liver

Answer 11

Glucose is transported across cell membranes by facilitated diffusion. I. e., transporters facilitate the movement of glucose down its concentration gradient.

Glucose-dependent tissues, such as brain, have low Km, insulin-independent glucose transporters.

In other tissues, such as muscle, the transporter activity has a low KM, but is insulin-dependent. In these tissues, insulin stimulates the movement of the transporter from internal stores to the plasma membrane (as opposed to stimulating transporter activity or promoting synthesis of new transporters).

In liver, glucose uptake occurs by a transporter that is insulin-independent, but which has a high KM and high Vmax. The effect, as in case of insulin-dependent transporters, is to limit glucose uptake to conditions when blood glucose levels are high. The high KM transporter allows these cells to act as “sensors” of high blood glucose levels.

Question 12

Know the net reaction accomplished by glycolysis

Answer 12

D-[Glucose] + 2 [NAD]+ + 2 [ADP] + 2 [P]i → [Pyruvate] + 2 [NADH] + 2 H+ + 2 [ATP] + 2 H2O

Question 13

Regulation of glucokinase in hepatocytes

Answer 13

In hepatocytes, glucokinase binds to GKRP (glucokinase regulatory protein), which is a competitive inhibitor of glucokinase.

This inhibition is strongly potentiated by fructose 6-phosphate (which is in equilibrium with glucose-6- phosphate; hence the indirect regulation).

The glucokinase-GKRP complex is translocated into the nucleus, where the glucokinase is held in an inactive state.

High concentrations of glucose (or fructose 1-phosphate) reverse the inhibition by triggering dissociation of the glucokinase-regulatory protein complex.

Question 14

Know that NAD+ is consumed in glycolysis, and be able to discuss how the cell continues to provide NAD+ to support glycolysis and how the cell utilizes the NADH that is produced by glycolysis

Answer 14

Under aerobic conditions the enzymes of oxidative phosphorylation can re-oxidize NADH back to NAD and utilize this latent energy. However, these enzymes are found inside the mitochondrion, and NADH in the cytoplasm cannot cross the mitochondrial inner membrane to reach them. The solution to this problem is that the entire NADH molecule is not transferred into
the mitochondrion—only its reducing equivalents are. Several shuttle systems exist that can accomplish this.

In the malate-aspartate shuttle, electrons are transferred from cytosolic NADH to oxaloacetate, forming malate. This malate traverses the mitochondrial inner membrane by means of a specific transporter. Once in the matrix, the malate is reoxidized by malate dehydrogenase and NAD+ to form OAA and NADH. The OAA cannot escape the matrix, since there is no transporter for it; instead, a transamination reaction (discussed later in the course) takes place that converts OAA to aspartate, which can be transported back to the cytosol. Here, the aspartate undergoes transamination to reform OAA, completing the cycle.

The glycerol 3-phosphate shuttle (also known as the α-glycerol phosphate shuttle) couples the cytosolic oxidation of NADH with the mitochondrial reduction of FAD. Cytoplasmic NADH is utilized by a cytoplasmic enzyme, glycerol-3-phosphate dehydrogenase, to convert dihydroxyacetone phosphate to glycerol-3-phosphate. The glycerol-3-phosphate is then converted
back to DHAP by a mitochondrial version of the dehydrogenase, reducing FAD to FADH2 in the
process. Electrons are then transferred by the reduced flavin to the electron carrier Q, which can enter the respiratory chain in its reduced form, QH2.

Under anaerobic conditions (e.g., muscle during vigorous activity), other mechanisms must be used to transform cytoplasmic NADH to NAD+. In particular, pyruvate can be converted into lactate via homolactic fermentation. The concomitant oxidation of NADH to NAD+ restores the redox balance in cytoplasm and enables glycolysis to continue. Excessive build-up of lactate in muscle can lead to cramping.

Question 15

Know the three major regulatory points affecting glycolysis

Answer 15

hexokinase (glucose → glucose-6-phosphate)

phosphofructokinase or 6-phosphofructo-1-kinase or PFK1 (β-D-Fructose 6-phosphate (F6P) → β-D-Fructose 1,6-bisphosphate (F1,6BP))

Pyruvate Kinase (phosphoenolpyruvate → pyruvate)

Question 16

Regulation of PFK1

Answer 16

Negative allosteric: ATP, 3-phosphoglycerate, phosphoenolpyruvate, citrate, phosphocreatine (muscle), lactate and H+ (7)

Positive allosteric: AMP, fructose-6-phosphate, and fructose 2,6-bisphosphate

Question 17

The role of fructose-2,6,-bisphosphate

Answer 17

Fructose-2,6-bisphosphate (fructose-2,6-bisphosphate) is synthesized and degraded by a multifunctional enzyme having both kinase and phosphatase activity. The kinase activity is referred to as a 6-phosphofructo-2-kinase or PFK 2. Fructose-2,6-biphosphate can promote synthesis of fructose-1,6-bisphosphate.

Question 18

Control of 6-phosphofructokinase-1 in the liver

Answer 18

↓ Glucagon ↓ cAMP-deP PK ↑PFK-2/F2,6Pase

ratio ↑ → PFK-1 active (glycolysis ↑)

Question 19

Tissue differences in the control of 6-phosphofructo-1-kinase

Answer 19

Liver: primary effector is F-2,6-P2. Physiological condition: hyperglycemia

Heart: primary effector is F-2,6-P2. Physiological condition: stress

Muscle: primary effector is AMP. Physiological condition: contraction

All tissues: primary effector is AMP. Physiological condition: anoxia

Question 20

Pyruvate kinase deficiency

Answer 20

Mature erythrocytes have no mitochondria—they rely
solely on glycolysis for ATP production. Hence, defects in glycolysis frequently manifest as anemias. One of the most common such defects is pyruvate kinase deficiency. Patients with mutations in the pyruvate kinase gene suffer chronic hemolytic anemia. These mutations can affect stability of the enzyme, substrate binding or turnover, or response to allosteric effectors.

Question 21

Pyruvate kinase regulation

Answer 21

Pyruvate kinase is a homotetramer.

ATP tends to inhibit the active (unphosphorylated) enzyme. The ATP doesn’t completely inactivate the enzyme, but it makes it less active than it would be in the absence of ATP.

PEP and F-1,6-P2 can allosterically drive the phosphorylated form (which is normally highly inactivated) toward a partially activated form

Question 22

Liver isoform of pyruvate kinase

Answer 22

The liver isoform of pyruvate kinase contains a serine residue in the N domain that is not found in the other isoforms. This residue is a substrate for the cAMP-dependent protein kinase. The result of the phosphorylation is an increase in KM and sensitivity to negative allosteric regulators. Looking at the monomer structure, it is difficult to see how phosphorylation at this
serine would affect activity, since it lies on the opposite side of the molecule from the active site.

Question 23

Why are there four isoforms of pryvuate kinase?

Answer 23

L and R result from alternate promoters which result in different amino termini. This explains why L contains a serine at residue 11, which R lacks

M1 and M2 are the result of alternate splicing.

Understand that isoenzymes may result from different genes or different transcriptional or post-transcriptional modifications.

Question 24

Know the overall pathway and function of the hexose monophosphate pathway and the major products of this pathway.

Answer 24

The primary function of this pathway is the generation of NADPH. A secondary function for this pathway is to serve as a source of ribose-5-phosphate synthesis.

Oxidative phase: irreversible

1. Glucose 6-phosphate + NADP+ → 6-phosphoglucono-δ-lactone + NADPH (Enzyme: glucose 6-phosphate dehydrogenase)
2. 6-phosphoglucono-δ-lactone + H2O → 6-phosphogluconate + H+
3. 6-phosphogluconate + NADP+ → ribulose 5-phosphate + NADPH + CO2 (Enzyme: 6-phosphogluconate dehydrogenase)

Non-Oxidative Phase: reversible reactions.
1. ribulose 5-phosphate → ribose 5-phosphate (Ribulose 5-Phosphate Isomerase) OR xylulose 5-phosphate (Ribulose 5-Phosphate 3-Epimerase)

2. xylulose 5-phosphate + ribose 5-phosphate → glyceraldehyde 3-phosphate + sedoheptulose 7-phosphate (transketolase + TPP)
3. sedoheptulose 7-phosphate + glyceraldehyde 3-phosphate → erythrose 4-phosphate + fructose 6-phosphate (transaldolase)
4. xylulose 5-phosphate + erythrose 4-phosphate → glyceraldehyde 3-phosphate + fructose 6-phosphate (transketolase + TPP)

Major products: ribulose 5-phosphate + 2 NADPH + 2 H+ + CO2 (oxidative phase)

Major productS: fructose-6-phosphate + glyceraldehyde-3-phosphate (non-oxidative phase)

Question 25

Control of hexose monophosphate pathway

Answer 25

Control of the hexose monophosphate pathway is via regulation of the glucose-6-phosphate dehydrogenase—

NADPH is a competitive inhibitor of the enzyme, so an
increased NADPH / NADP+ ratio inhibits G6PD.

Insulin can up-regulate the production of the glucose-6-phosphate dehydrogenase enzyme

Question 26

Be able to describe the effects of G6PD deficiency upon the cell’s response to oxidative stress

Answer 26

These pathologies are the result of the inability to maintain the tripeptide glutathione in the reduced form (GSH). Glutathione is critical in reducing disulfides in proteins and hydrogen peroxides formed at high oxygen levels in the RBC. Glutathione and the
glutathione peroxidase system is the principal antioxidant defense system in mammalian cells. Defects in this system lead to decreased ability to remove H2O2 and other reactive oxygen species (ROS) and hence leads to more fragile RBCs that break down prematurely.

Enzyme deficiency cuts off NADPH supply which increases ratio of GSSG/GSH.

Question 27

Know the principle mechanisms by which fructose is metabolized in liver. Be able to describe the cellular bases of essential fructosuria and hereditary fructose intolerance, and explain why the latter presents a more severe pathology than the former.

Answer 27

1. Fructokinase phosphorylates fructose at the 1 position. No mechanism exists to isomerize fructose-1-phosphate to glucose-1-phosphate.
2. Aldolase B, the liver isozyme of aldolase, will split both F-1-P and F-1,6-P2.

3. The glyceraldehyde derived from F-1-P can be phosphorylated to glyceraldehyde-3-P by
triose kinase (triokinase), or it can be reduced to glycerol and then phosphorylated. Both phosphorylated products can be converted to DHAP.

Essential fructosuria is due to a deficiency of fructokinase and is relatively benign. It leads to accumulation of fructose in the urine.

Hereditary fructose intolerance (HFI) is an autosomal recessive disorder that results from the absence of aldolase B and has much more serious consequences (including vomiting, hypoglycemia, jaundice, metabolic acidosis, coma). Treatment involves strict attention to diet so as to minimize fructose intake.

Problems associated with aldolase B deficiency are due to the accumulation of F-1-P in liver, which depletes levels of ATP and Pi. Low Pi levels inhibit glycogenolysis,
and low ATP levels inhibit gluconeogenesis; both cause hypoglycemia. The enzyme AMP deaminase, found primarily in muscle, is activated by the decrease in
Pi. This results in an increase in purine catabolism and hyperuricemia, with the potential of gout. Additionally, low Pi prevents the phosphorylation of ADP. Adenylate kinase will convert 2 ADP to ATP +AMP, and AMP is degraded to urate.

Question 28

Know the principle mechanisms by which galactose is metabolized in liver. Be able to describe the cellular basis of classic galactosemia.

Answer 28

Galactose metabolism requires galactokinase to produce galactose-1-P.

This compound is then transformed into UDP-galactose by the action of galactose-1-phosphate uridylyltransferase (GALT), utilizing one equivalent of UDP-glucose in the process.

The uridylyl transferase switches the sugars, replacing the glucose on UDP-glucose with galactose

The UDP-galactose produced by the uridylyltransferase reaction may then be reversibly interconverted to UDP-glucose by the enzyme UDP-glucose 4-epimerase.

Uridylyltransferase (GALT) deficiency is the most common cause of galactosemia (“classic galactosemia”). In galactosemia, problems include failure to thrive, liver
damage, bleeding, and sepsis. If a diet restricted in galactose is provided within the first 10 days, the most severe complications (neonatal death, liver failure, intellectual disability) can be avoided. However, children with galactosemia remain at risk for developmental delays and problems with speech and motor function. Galactosemic females are at increased risk for premature ovarian failure. Cataracts are an important long-term complication associated with classic galactosemia, due to the conversion of the accumulated galactose to galactitol, a sugar alcohol, which leads to cataract formation

Question 29

Know the reactions specific to gluconeogenesis, as well as their locations and regulation.

Answer 29

The first two reactions of gluconeogenesis are required to bypass the highly exergonic reaction catalyzed by pyruvate kinase. In these reactions, pyruvate carboxylase and phosphoenolpyruvate carboxykinase (PEPCK) are coupled by the carboxylation and subsequent decarboxylation of the oxaloacetate intermediate. This requires one equivalent each of ATP and a GTP—in other words, two high-energy bonds.

Pyruvate carboxylase is a biotin-containing multifunctional protein, as are most carboxylases. Pyruvate carboxylase is allosterically activated by acetyl CoA.

The second irreversible reaction of glycolysis that must be bypassed in gluconeogenesis is the phosphofructokinase (PFK1) reaction. This is accomplished by the simple hydrolysis of the phosphate ester, catalyzed by the enzyme fructose-1,6-bisphosphatase.

Fructose-1,6-bisphosphatase is inhibited by the negative allosteric effectors F-2,6-P2 and AMP.

The last irreversible step that must be bypassed in order to produce glucose is the hexokinase reaction. Glucose-6-P is hydrolyzed by glucose-6-phosphatase, a membrane-bound enzyme located in the endoplasmic reticulum.

Glucose-6-phosphatase is present only in liver and kidney; therefore other tissues (for example, muscle) are unable to release free glucose from glucose-6-phosphate. While there appears to be no direct control of the phosphatase activity, the enzyme has a high KM
for G-6-P and therefore will function only when G-6-P levels rise.

Pyruvate carboxylase is a mitochondrial enzyme. The other gluconeogenic enzymes, however, are largely cytoplasmic. Hence, the transport of intermediates across the mitochondrial membrane is critical. The use of the OAA-malate shuttle appears.

Question 30

Be able to compare regulation of glycolysis and gluconeogenesis.

Answer 30

Effectors of the phosphoenolpyruvate substrate cycle:
Pyruvate kinase (+F-1,6-P2, -ATP)
Pyruvate carboxylase  (-ADP, +Acetyl CoA)

Effectors of the fructose phosphate substrate cycle:
6-phosphofructo-1-kinase (PFK1) (+ AMP +F-2,6-P2)
fructose-1,6-bisphosphatase (- AMP -F-2,6-P2)