Background: Endogenous triacylglycerols (TAGS) represent the largest fuel reserve in the body.

Most TAGS are stored in adipose tissue and must be “mobilized” for use. Fatty acid (FA)

oxidation during endurance exercise permits sustained physical activity and delays the onset of

glycogen depletion and hypoglycemia. The use of adipose tissue FAs as a fuel requires: 1)

hydrolysis of TAGS (i.e., lipolysis) from adipose tissue, and 2) delivery of released FAs to

skeletal muscle mitochondria for oxidation.

Create a detailed narrative about how these processes work, in the order they would occur, e.g.

when beginning an endurance exercise bout, explaining each step from fatty acid mobilization in

adipose tissue through oxidation in skeletal muscle. In your narrative, be sure to discuss the

following information:

Question 1 3 pts

Discuss the primary hormone involved, where it’s produced, how it’s circulated, its receptor, and

its mechanism of action. Also, does the hormone take seconds, minutes, or hours to have an

effect on lipolysis?

Hormone involved: Epinephrine is the primary hormone involved in the induction of

lipolysis during endurance exercise.

Site of production/how it’s circulated: It is produced in the adrenal medulla and released

by the adrenal glands to circulate in the blood.

Mechanism of action:

o In the adipocytes, epinephrine binds to Beta-adrenergic receptors coupled to G-

proteins. Binding of epinephrine causes the G-protein to exchange GDP for GTP,

activating its alpha subunit. The activated alpha subunit dissociates from the

beta-gamma subunit of the G-protein and binds to the enzyme adenylyl cyclase,

which is responsible for converting ATP to cyclic AMP (cAMP). cAMP then

activates the enzyme protein kinase A (PKA), causing dissociation of the

regulatory subunits from the catalytic subunits. This dissociation allows the now

active PKA to phosphorylate perilipin-1, which causes the perilipin to dissociate

from CGI-58, a coactivator of the lipolytic enzyme adipose triglyceride lipase

(ATGL). CGI-58 is now able to activate ATGL to cleave triacylglycerols to

diacylglycerols. PKA is also responsible for the phosphorylation of hormone

sensitive lipase (HSL), which increases the enzyme’s activity 2-fold, creating a

50-100 fold increase in lipolysis. In addition, PKA phosphorylates perilipin-1,

allowing HSL to actually bind to the lipid droplets and cleave diacylglycerols to

monocylglycerols, causing a 100-fold increase in the enzyme’s activity. Fatty

acids then exit the adipocytes and are carried by the serum protein albumin to the

muscle for oxidation and the eventual production of ATP.

o Epinephrine also circulates to skeletal muscle, which has intra-myocellular lipid

droplets. Beta-adrenergic receptors coupled to G-proteins, adenylyl cyclase,

cAMP, and PKA are also involved in the phosphorylation of perilipins to enable

the enzymes ATGL and HSL to cleave fatty acids from glycerol. However,

perilipins 2 and 5 are the major perilipins that both block and enable lipolysis via

phosphorylation in skeletal muscle. In both adipocytes and skeletal muscle, the

enzyme monoglyceride lipase is also involved in cleaving monoacylglycerols to a

glycerol and a fatty acid; however, this enzyme does not seem to be regulated by

hormones such as epinephrine.

Time of effect on lipolysis: The effects of hormones are not incredibly rapid, so

epinephrine should take more than a few seconds to affect lipolysis; however, hormones

should also take less than an hour to exert their effects. Therefore, epinephrine should

begin to affect lipolysis in minutes.

Question 2 3 pts

List the three enzymes involved in lipolysis and briefly explain the reaction(s) catalyzed by each,

including ALL substrates and products. Please use words, not arrows, and spell out any terms at

least once before abbreviating them.

Adipose triglyceride lipase (ATGL) is responsible for the de-esterification of a

triacylglycerol to a diacylglycerol and a fatty acid. In adipose tissue, hormonal

stimulation causes the activation of protein kinase A (PKA), which phosphorylates

perilipin-1, causing the perilipin to dissociate from CGI-58, the coactivator of ATGL.

This dissociation enables CGI-58 to activate ATGL. In skeletal muscle, the major

perilipins involved in lipolysis are perilipins 2 and 5.

Hormone sensitive lipase (HSL) is responsible for the de-esterification of a

diacylglycerol to a monoacylglycerol and a fatty acid. Again, in adipose tissue, hormonal

stimulation causes the activation of PKA, which phosphorylates both perilipin-1 –

enabling HSL to bind to the lipid droplet – and HSL itself, increasing HSL activity.

Again, perilipins 2 and 5 are involved in skeletal muscle.

Monoglyceride lipase (MGL) is responsible for the de-esterification of a

monoacylglycerol to glycerol and a fatty acid. Unlike ATGL and HSL, the activity of

MGL is not known to be regulated by hormones.

Question 3 2 pts

Explain two ways covalent modification is involved in increasing the rate of lipolysis.

As described in the questions above, the rate of lipolysis is increased by the covalent

modification – phosphorylation – of both perilipins and HSL.

In adipocytes, the phosphorylation of perilipin-1 by PKA allows the dissociation of

ATGL’s coactivator CGI-58 from the perilipin and enables coactivator binding to ATGL,

increasing the cleavage of triacylglycerols to diacylglycerols, thus also increasing the rate

of lipolysis. In addition, phosphorylation of perilipin-1 allows HSL to bind to lipid

droplets, increasing the cleavage of diacyglycerols to monoacylglycerols and therefore

increasing lipolysis.

PKA also phosphorylates HSL, increasing the activity of the enzyme 2-fold, and thus

increasing lipolysis 50-100 fold.

Question 4 2 pts

List the two final products of lipolysis that are released into the blood from adipose tissue, their

primary destinations, and how they get there (e.g. bound/unbound, etc.).

The two final products of lipolysis are fatty acids and glycerol.

Primary destinations:

o Fatty acids are circulated to tissues that need energy, such as the skeletal muscle,

the cardiac muscle, and the liver where they will be oxidized for the eventual

production of ATP.

o Glycerol is circulated to tissues that contain the enzyme glycerol kinase, which is

capable of phosphorylating glycerol to glycerol-3-phosphate. Glycerol-3-

phosphate is needed to combine with fatty acyl-coA molecules to be re-esterified

to triacylglycerols when energy needs to be stored. Because adipose tissue does

not have glycerol kinase, glycerol is circulated to tissues that do have the enzyme,

such as the liver and kidney.

How products get to their destinations:

o Fatty acids are circulated bound to the serum protein albumin.

o Glycerol moves through the blood unbound.

Question 5 5 pts

How would a FA such as palmitate get from the blood, across the sarcolemma, and into the

mitochondrion of a skeletal muscle cell to get oxidized? Discuss enzymes, binding proteins,

and/or transport/transporters that help FAs move through the sarcolemma, the cytosol, and the

mitochondrial membranes, and each step involved.

Blood: Fatty acids are transported through the blood bound to the serum protein albumin.

Sarcolemma: Fatty acids then use plasma membrane fatty acid binding proteins to cross

the sarcolemma of myocytes.

Cytosol: Fatty acids are hydrophobic and need assistance to move through the cytosol as

well via cytosolic fatty acid binding proteins.

Mitochondria: Fatty acids must first be activated by the enzyme acyl-coA synthetase,

which converts fatty acids to acyl-coA in the presence of coenzyme A and two ATP

equivalents. Acyl-coA synthetase is located in the outer membrane of the mitochondria

or the endoplasmic reticulum. Although the fatty acid has been activated to acyl-coA, it

can only move across the outer membrane of the mitochondria. In the presence of

carnitine, the enzyme carnitine acyltransferase-I (CAT-I), located in the outer membrane

of the mitochondria, converts acyl-coA to acylcarnitine in the intermembrane space.

Acylcarnitine can be exchanged for carnitine and enter the mitochondrial matrix by way

of the transporter carnitine acylcarnitine translocase, located in the inner membrane of the

mitochondria. Once in the mitochondrial matrix, acylcarnitine can be converted back to

acyl-coA in the presence of coenzyme A by the enzyme carnitine acyltransferase-II

(CAT-II), generating carnitine. Carnitine generated by this enzyme will be used by the

translocase transporter to move more acylcarnitine into the mitochondrial matrix. In the

mitochondrial matrix, acyl-coA can be oxidized to acetyl-coA.

Source used for points 2 & 3:

https://books.google.com/books?id=IYskSAWI9McC&pg=PA219&lpg=PA219&dq=how+do+f

atty+acids+cross+the+sarcolemma&source=bl&ots=-TuRkaLlcW&sig=ZlLOUgg9-

a7BjutuUHrqhY4ZaF0&hl=en&sa=X&ved=0CD8Q6AEwBGoVChMI1YKgpseJyQIVgkuICh0

9Mw-

T#v=onepage&q=how%20do%20fatty%20acids%20cross%20the%20sarcolemma&f=false

Question 6 5 pts

Explain Beta-oxidation of palmitate. Describe key steps, enzymes, products, the ATP produced

from each product, where the ATP is produced (process/cycle), and the net ATP produced from

palmitate (after it enters the myocyte).

Step 1: oxidation: In the first step of Beta-oxidation, two hydrogens are removed from

carbons 2 and 3 (alpha and beta carbons) of the acyl-coA carbon chain and transferred to

FAD, reducing it to FADH2, by the enzyme acyl-coA dehydrogenase. The oxidation of

these two carbons produces a double bond between carbons 2 & 3, forming the molecule

delta2-trans-enoyl-coA. FADH2 will form two ATP when it drops off its electrons at the

electron transport chain. Thus, step one of Beta-oxidation produces two ATP.

Step 2: hydration: In the second step of Beta-oxidation, water is added to the double bond

between carbons 2 & 3, putting an OH group on carbon 3 and a hydrogen on carbon 2.

This step is catalyzed by the enzyme delta2-enoyl-coA hydratase and forms 3-

hydroxyacyl-coA. As no electron carriers are produced in this step, no ATP is produced.

Step 3: oxidation: In the third step of Beta-oxidation, two hydrogens are removed from

the third carbon, oxidizing it as the two hydrogens are transferred to NAD+, reducing

NAD+ to NADH. This step is catalyzed by L(+)-3-hydroxyacyl-coA dehydrogenase and

forms 3-ketoacyl-coA. As this step generates an electron carrier, it produces ATP.

NADH drops off its electrons to a different protein in the electron transport chain, so it

will produce three ATP instead of two like FADH2.

Step 4: cleavage: In the final step of Beta-oxidation, the acyl-coA chain is split between

carbons 2 and 3 in the presence of coenzyme A by the enzyme thiolase, yielding acetyl-

coA. The acyl-coA chain is now two carbons shorter and can continue to go through

these same steps of Beta-oxidation until it has produced its maximum amount of acetyl-

coA molecules. Because palmitate is sixteen carbons long, it can produce eight two-

carbon acetyl-coA molecules.

Net ATP produced from palmitate:

o 2 ATP in FADH2 and 3 ATP in NADH = 5 ATP in electron carriers from one

cycle of Beta-oxidation.

o 16 carbon fatty acid = 7 splits to generate 8 two-carbon acetyl-coA molecules. 7

splits = 7 cycles x 5 ATP per cycle = 35 ATP from electron carriers for the

complete Beta-oxidation of palmitate.

o 8 acetyl-coA molecules are produced from Beta-oxidation. Each acetyl-coA

molecule will generate 12 ATP in the form of electron carriers (3 NADH = 3

ATP/NADH + 1 FADH2 = 2 ATP/FADH2 + 1 ATP) when it is completely

oxidized to CO2 in the tricarboxcylic acid cycle (TCA).

o 8 acetyl-coA molecules x 12 ATP per TCA cycle = 96 ATP from electron carriers

for the complete oxidation of palmitate to CO2 molecules.

o 96 ATP from TCA cycle + 35 ATP from Beta-oxidation = 131 total ATP

o However, activation palmitate to acyl-coA by the enzyme acyl-coA synthetase

requires 2 ATP, so 2 ATP must be subtracted from this total to get the net ATP.

o 131 total ATP – 2 ATP from activation of palmitate to acyl-coA = 129 net ATP

Question 7 10 pts

Describe the bi-functional enzyme and each reaction it catalyzes, including ALL substrates and

products and whether the reaction’s rate increases or decreases (and how/mechanisms) when the

bi-functional enzyme is phosphorylated. Describe all of this in the context of dietary

conditions/state. Be sure to include (some version of) each of the following terms in your

response:

Enzyme

Catalyzes

Active

Inactive

Domain

Glycolysis

Gluconeogenesis

Postprandial state

Fasted state

Bi-functional

Phosphorylated

Dephosphorylated

Phosphofructokinase 1

Phosphofructokinase 2

Fructose 1,6-

bisphosphatase

Fructose 2,6-

bisphosphatase

Allosteric

Activator

Inhibitor

Protein Kinase A

Regulatory molecule

The bi-functional enzyme phosphofructokinase-2/fructose 2,6-bisphosphatase (PFK-2/F-

2,6-pase) is responsible for regulating glycolysis and gluconeogenesis via the production

of fructose 2,6-bisphosphate, which allosterically affects the glycolytic enzyme

phosphofructokinase-1(which catalyzes the conversion of fructose-6-phosphate to

fructose 1,6-bisphosphate) and the gluconeogenic enzyme fructose 1,6-bisphosphatase

(which catalyzes the conversion of fructose 1,6-bisphosphate to fructose-6-phosphate).

Fructose 2,6-bisphosphate allosterically activates phosphofructokinase-1 by lowering its

Km and thus increasing its affinity for its substrate, fructose-6-phosphate; however,

fructose 2,6-bisphosphate allosterically inhibits fructose 1,6-bisphosphatase by increasing

its Km and thus lowering its affinity for its substrate, fructose 1,6-bisphosphate.

PFK-2/F-2,6-pase is under regulation itself. The bi-functional enzyme has two

portions/domains: the kinase portion, known as phosphofructokinase-2, and the

phosphatase portion, known as fructose 2,6-bisphosphatase. The kinase portion is

responsible for forming fructose 2,6-bisphosphate via phosphorylation of fructose-6-

phosphate using ATP and generating ADP. The phosphatase portion is responsible for

acting in the reverse direction by removing a phosphate from fructose 2,6-bisphosphate to

form fructose-6-phosphate using water and generating an inorganic phosphate.

o Substrate regulation of bi-functional enzyme:

Fructose-6-phosphate is responsible for the allosteric regulation of PFK-

2/F-2,6-pase. It allosterically activates the kinase portion of the enzyme

and allosterically inhibits the phosphatase portion of the enzyme.

o Hormonal regulation of bi-functional enzyme:

The hormone glucagon can cause the covalent modification of PFK-2/F-

2,6-pase via phosphorylation. Phosphorylation inactivates the kinase

portion and activates the phosphatase portion of the enzyme. When

glucagon binds to its receptor, it causes adenylyl cyclase to convert ATP

to cyclic AMP (cAMP), which can then activate protein kinase A (PKA).

PKA then phosphorylates PFK-2/F-2,6-pase, which causes inactivation of

the kinase and activation of the phosphatase. Therefore, fructose 2,6-

bisphosphate is dephosphorylated to fructose-6-phosphate by the action of

the F-2,6-pase portion of the enzyme.

The bi-functional enzyme in the postprandial state and its effect on

glycolysis/gluconeogenesis:

o In the postprandial state, blood glucose will be elevated, causing an increased flux

of glucose through glycolysis. This will generate more glucose-6-phosphate

which will be converted to fructose-6-phosphate as glucose moves through the

steps of glycolysis. As described above, fructose-6-phosphate allosterically

activates the kinase portion of the enzyme, phosphofructokinase-2, causing an

increase in the conversion of fructose-6-phosphate to fructose 2,6-bisphosphate.

The increased production of fructose 2,6-bisphosphate then allosterically activates

the glycolytic enzyme phosphofructokinase-1, thus increasing the rate of

glycolysis. However, the increased production of fructose 2,6-bisphosphate also

allosterically inhibits the gluconeogenic enzyme fructose 1,6-bisphosphatase, thus

inhibiting the rate of gluconeogenesis.

The bi-functional enzyme in the fasting state and its effect on

glycolysis/gluconeogenesis:

o In the fasting state, blood glucose will be low, causing the release of glucagon

from the pancreas. Upon binding to a receptor on a liver cell, glucagon will cause

the activation of adenylyl cyclase, which will convert ATP to cAMP. cAMP will

then activate PKA, and PKA will phosphorylate PFK-2/F-2,6-pase, causing

inactivation of the kinase portion and activation of the phosphatase portion.

Fructose 2,6-bisphosphate will then be dephosphorylated to fructose-6-phosphate

by the fructose 2,6-bisphosphatase portion of the enzyme. This decrease in

fructose 2,6-bisphosphate removes the allosteric inhibition of the gluconeogenic

enzyme fructose 1,6-bisphosphatase, and removes the allosteric activation of the

glycolytic enzyme phosphofructokinase-1, therefore increasing the rate of

gluconeogenesis and decreasing the rate of glycolysis.

In this way, PFK-2/F-2,6-pase regulates flux through glycolysis and gluconeogenesis

according to the physical state – postprandial or fasting – of the body, so that both

processes are not occurring at the same time and wasting valuable energy.