Metabolism Flashcards
What is glycolysis?
Glycolysis is the pathway for extracting energy from the breakdown of glucose.
Why is glucose such a central molecule in metabolism?
The free energy of the breakdown of glucose into CO2 + H2O is huge! ~2840 kJ/mol (although we can only extract a small part of this energy as much is converted to heat). Besides this glucose has many possible fates in a cell:
- Its a precursor to many biomolecules like nucleic acids, polysaccharides and ECM components.
- It can be stored as starch/glycogen/sucrose
- It can be oxidized to drive the synthesis of ATP
What is the net production of glycolysis?
Glucose → 2 pyruvate + 2 ATP + 2 NADH
So we get pyruvate that can be converted into acetyl CoA and go into TCA, ATP that can be used as energy and NADH that give reducing power in oxidative phosphorylation. We also get a bunch of intermediates that can be used as precursors in biosynthesis.
What happens to the pyruvate generated by glycolysis in aerobic vs anaerobic environments?
In aerobic conditions, the pyruvate is converted to Acetyl-CoA by the PDH complex and feeds into the citric acid cycle. It can also go into the pentose phosphate pathway.
In anaerobic conditions, the pyruvate undergoes fermentation into ethanol in yeast and lactate in some cells (eg muscle and erythrocytes) and some microorganisms. The fermentation process oxidizes NADH, which reforms NAD+ which is cycled back to glycolysis.
Glycolysis can be divided in two phases, which? What is different about them?
Glycolysis can be divided into the preparatory phase and the “pay off” phase.
- In the preparatory phase, 2 ATPs are invested (glucose → G6P and F6P → F1,6-BP) to activate the glucose and produce important intermediates. In the end of the investment phase we have 2 phosphorylated 3-C molecules instead of the 6-C start molecule.
- In the pay-off phase, 4 ATPs, 2 NADH and 2 pyruvates are generated, which make the reaction net positive.
What is the full reaction of glycolysis?
glucose + 2 NAD+ + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 2H+ + 2 H2O + 2 ATP
with a delta free energy of -85 kJ/M - very small compared to the free energy of glucose, but most of the energy is not extracted during glycolysis, but in the ETC later.
Go through the 10 steps of glycolysis and describe what kind of reaction each step is.
Investment phase:
1. Phosphorylation of carbon 6; glucose is first phosphorylated at the hydroxyl group on C-6.
- Isomerization of G6P to F6P (reversible); The glucose 6-phosphate formed is converted to fructose 6-phosphate
- Phosphorylation of F6P to F1,6-BP; this time at C-1, to yield fructose 1,6-bisphosphate. For both phosphorylations, ATP is the phosphoryl group donor.
- Fructose 1,6-bisphosphate is split to yield two different three carbon molecules, dihydroxyacetone phosphate (DHAP) and
glyceraldehyde 3-phosphate through the formation of a schiff base. This is the “lysis” step that gives the pathway its name - Isomerization of DHAP; The dihydroxyacetone phosphate is
isomerized to form a second molecule of glyceraldehyde 3-phosphate
Payoff phase:
6. Oxidation of G3P to 1,3-BPG (generating NADH); Each molecule of glyceraldehyde 3-phosphate is oxidized and phosphorylated by inorganic phosphate (via substrate level phosphorylation, high energy molecule P donor, not ATP) to form 1,3-bisphosphoglycerate.
- Phosphorylation of ADP to ATP with the phosphoryl being donated from 1,3-BPG (substrate level phosphorylation) which forms 3-PG. steps 6 & 7 are coupled and together favorable. Net, step / generates ATP and 3-PG.
- “mutase” reaction that is preparatory for next step 3-PG to 2-PG (switching places of the phosphate and OH group using His residues in active site).
- 2-PG to phosphoenol pyruvate (PEP) + H2O, which “locks” the PEP in a high energy state, if H2O didn’t leave, the molecule would immediately rearrange into a ketone.
- tautomerization of PEP (enol form) + ADP → pyruvate (keto form) + ATP practically irreversible, strongly exergonic.
Note that step 1, 3 and 10 are “irreversible” which gets very problematic when we try to go the other way around in gluconeogenesis.
During glycolysis, all of the nine glycolytic intermediates between glucose
and pyruvate is phosphorylated. What is the three functions of the phosphoryl group?
The three functions of the phosphoryl group are:
1. Phosphorylated sugars can’t leave the cell (no transporters) so no further energy is necessary to retain phosphorylated intermediates in the cell, despite the large
difference in their intracellular and extracellular concentrations.
- Phosphoryl groups are essential components in the enzymatic conservation of metabolic energy. The bonds are high energy.
- Binding energy resulting from the binding of phosphate groups to the active sites of enzymes lowers the activation energy and increases the specificity of the enzymatic
reactions.
What is gluconeogenesis?
The pathway for synthesizing glucose from
noncarbohydrate precursors, like pyruvate, related three- and four-carbon compounds as well as certain amino acids to glucose.
Name three sources that can be converted into the starting material for gluconeogenesis.
- Lactate (animals)
- fatty acids (both)
- glucogenic amino acids like alanine. (both)
- 3PG from calvin cycle (plants)
Where does gluconeogenesis mainly occur in mammals?
In the liver! It also occur to some extent in the renal cortex and in the epithelial cells that line the small intestine.
Gluconeogenesis is not simply the reverse steps of glycolysis, why not? How is this solved?
Because three of the 10 reactions in glycolysis have a large negative energy change, which means that the reverse is not achievable.
This is solved by having three alternative enzymes catalyzing bypass reactions, reactions that are sufficiently exergonic to be effectively irreversible in the direction of glucose synthesis. Thus, both glycolysis and gluconeogenesis are irreversible processes in cells. In animals, both pathways occur largely in the cytosol, necessitating their reciprocal and coordinated regulation.
Explain the first bypass reaction in gluconeogenesis.
As glycolysis ends with the conversion of PEP to pyruvate, we need to convert pyruvate into PEP.
Pyruvate is transported into mitochondria or is generated from alanine within mitochondria by transamination, in which the α-amino group is transferred from alanine (leaving pyruvate) to an α-keto carboxylic acid (this is also a transport mechanism for pyruvate in the blood).
a) In the mitochondria, pyruvate carboxylase adds on a carboxyl group to the pyruvate (with biotin as coenzyme) to form oxaloacetate (OAA). The role of the biotin is to covalently bind the carboxylate ion HCO3- (created in a ATP driven reaction).
b) OAA is then transported out by interconvesion to malate and in the cytosol OAA is converted to PEP catalyzed by phosphoenolpyruvate carboxykinase by decarboxylation of the added CO2- group (from step a) that rearranges electrons so that the carbonyl O attacks the gamma phosphoryl of GTP which leaves as GDP. Since CO2 leaves as a gas, this reaction leads to an increase in entropy and thus is exergonic.
Explain the second and third bypass reactions in gluconeogenesis.
The second and third bypasses are simple
dephosphorylations by phosphatases (compared to the kinases that perform the corresponding steps in glycolysis)
- The F-1,6-BP to F6P reaction is catalyzed by Mg2+-dependent fructose 1,6-bisphosphatase (FBPase-1), which promotes the essentially irreversible hydrolysis of the C-1 phosphate (not phosphoryl group transfer to ADP)
Fructose 1,6-phosphate + H2O → fructose 6-phosphate + Pi - the G6P to glucose reaction is catalyzed by glucose 6-phosphatase which is a simple hydrolysis of a phosphate ester.
Glucose 6-phosphate + H2O →glucose + Pi
What is the total energetic cost of gluconeogenesis?
Formation of one molecule of glucose from pyruvate requires four ATP, two GTP, and two NADH; it is expensive.
Hexokinase, the enzyme catalyzing the first step of glycolysis (requiring ATP) exist as two different versions in mammals, Hexokinase I in muscle and Hexokinase IV in the liver. How does the regulation of these differ?
The hexokinase I (muscle) has a very low Km, it basically work at max speed all the time due to the high energy needs of muscle and is not tightly regulated.
The liver hexokinase on the other hand, have a much higher Km so it is regulated by substrate level and is further regulated by binding to regulatory proteins and transported into the nucleus, where it can’t act. The transport is favored by high levels of either glucose or F6P, as this indicate that the energy needs are met.
The rapid hormonal regulation of glycolysis and gluconeogenesis is mediated by fructose 2,6-bisphosphate, how?
F-2,6-BP is formed by phosphorylation of
fructose 6-phosphate, catalyzed by phosphofructokinase-2 (PFK2), a regulatory pathway in response to glucagon.
If we just look at the reaction of F6P -> F-1,6-BP and the bypass reaction, PFK-1 is allosterically activated by F-2,6-BP, while allosterically inhibitory for FBPase-1 (catalyzing the bypass reaction).
When the blood glucose level decreases, the hormone glucagon signals the liver to produce and release more glucose and to stop consuming it for its own needs. The pancreas release glucagon which in the liver triggers an increase in cAMP which in turn activates PKA, which phosphorylates PFK-2 (that catalyzes formation of F-2,6-BP). With less F-2,6-BP we no longer have inhibition of FBPase-1, and thus favor gluconeogenesis. (p. 1973)
What is glycogen?
Glycogen is the storage molecule of glucose in animals.
Explain how glycogen is synthesized in short.
- To start glycogen synthesis, the glucose 6-phosphate is converted to glucose 1-phosphate in the phosphogluco-mutase reaction:
Glucose 6-phosphate ⇌ glucose 1-phosphate - The product is then converted to UDP-glucose by the action of UDP-glucose pyrophosphorylase, in a key step of glycogen biosynthesis:
Glucose 1-phosphate + UTP → UDP-glucose + PPi
Notice that one ATP equivalent is the cost of the formation of each UDP-glucose.
- UDP-glucose is the immediate donor of glucose residues in the reaction catalyzed by glycogen synthase, which promotes the
transfer of the glucose residue from UDP-glucose to a nonreducing end of a branched glycogen molecule, forming an (α1 → 4) linkage. - A “branching enzyme” transfers 6-7 glycogen units from the non reducing end to a C-6 of another unit, creating (α1 → 6) branches.
Explain the process of glycogen degradation.
Glycogen degradation is catalyzed by glycogen phosphorylase, which breaks the (α1 → 4) bonds with Pi as nucleophile to reform G1P, which shortens the glycogen chain by one residue. The G1P is then either phosphorylated into G6P and used in glycolysis or into glucose and exported where it’s needed.
A debranching enzyme transfers branches to the non-reducing end which is in turn broken down by the glycogen phosphorylase reaction described above.
Pyruvate is a central metabolite that links glycolysis and the citric acid cycle. Explain the reaction catalyzed by the PDH complex in short.
The overall reaction of the PDHc is a oxidative decarboxylation, pyruvate is first decarboxylated, and then linked to CoA in a transacetylation reaction through a thioester linkage, where the acetyl group is considered “activated” as the hydrolysis of thioesters have a large negative free energy.
The reduced enzyme components are reoxidized by NAD+ which gets reduced to NADH with reducing power that can be utilized elsewhere in the cell.
What is the overall production of the citric acid cycle?
In the TCA, acetyl-CoA undergoes a series of redox reactions that harvests the bond energy in the form of NADH, FADH2 and ATP molecules.
Overall, one turn of the citric acid cycle releases two carbon dioxide molecules and produces three NADH, one FADH2 and one ATP or GTP. The citric acid cycle goes around twice for each molecule of glucose that enters cellular respiration because there are two pyruvates—and thus, two acetyl-CoAs made per glucose. So the net production per glucose is 6 NADH, 2FADH, 2ATP/GTP.
Note: since the reaction catalyzed by PDHc also produces 1 NADH per pyruvate, you could consider those too.
Three of the eight steps of the citric acid cycle are tightly regulated, why these? What kind of regulation is the most common?
The three tightly regulated steps (1,3 and 4) are the irreversible ones. Irreversible reactions are generally the ones being regulated, as energy is wasted if irreversible reactions that are not needed take place. Reversible reactions don’t require an energy investment.
The regulation is mostly allosteric, where molecules indicating a low energy state activates (ADP, Ca2+) and products and molecules indicating a well fed state (ATP) inhibits. For the first step (Ac-CoA to citrate), high concentrations of later products in the cycle inhibits as well, such as succinyl-CoA, as this indicates that the cycle is running fine.
The first reaction of the citric acid cycle is the reaction it’s named after. Explain the reaction mechanism in short.
The first step, catalyzed by citrate synthase, is a Claisen condensation. In the active site, the two catalytically active residues are a deprotonated Asp and a His.
- When Acetyl-CoA binds, the Asp takes up a hydrogen from the methyl group and with electron delocalization, the carbonyl oxygen oxidizes the His residue and forms an enol intermediate stabilized by hydrogen bonding with the His.
- The enol intermediate rearranges to attack the carbonyl carbon of OAA, with His positioned to retake the proton it previously donated. The carbonyl oxygen on OAA is protonated by another His residue and the condensation generated citroyl-CoA.
- The thioester linkage is then hydrolyzed so that CoA leaves and citrate is formed.
(Step two is a isomerization catalyzed by aconitase (FeS cluster as cofactor) to form Isocitrate (switching places of the H and OH group) to facilitate the first decarboxylation (step 3)).
What is the products and enzyme in the third step of the TCA cycle?
Step 3 is catalyzed by isocitrate dehydrogenase, where isocitrate is first oxidized by NAD(P)+ which forms NAD(P)H + H+, then decarboxylated (CO2 leaves) facilitated by Mg+ and then the enol intermediate is rearranged to generate α-Ketoglutarate.
The fourth step of the TCA cycle is the last irreversible step, describe it in short.
Step four is catalyzed by the α-ketoglutarate dehydrogenase complex in a reaction homologous to PDHc, a decarboxylative oxidation of α-ketoglutarate, which forms Succinyl-CoA, CO2 and NADH. (ΔG′° ≈−33.5 kJ/mol)
Step five of the TCA cycle is the ATP/GTP generating step. What drives the synthesis and how does the reaction work in short?
Hydrolysis of the thioester bond in Succinyl-CoA have a large negative free energy (ΔG′° ≈−35 kJ/mol) which drives the synthesis of a phosphoanhydride bond in in either GTP or ATP, with succinyl being formed in the process.
First, an inorganic phosphate is added in the place of CoA, then the Pi is taken up by a His residue in the active site, to then be transferred to GDP or ADP to form GTP or ATP.
Step 6-8 of the citric acid cycle are a series of reactions with a specific purpose, what? What is generated in the process?
The purpose of step 6-8 is to regenerate the OAA from succinate so that a new cycle can begin. Its a oxidation sequence (succinate - fumarate - L-malate -> oxidation to OAA) that generates an FADH2 (step 6) and one NADH (step 8).
Name the three steps of the TCA cycle where NADH is formed.
- Step 3, Isocitrate to α-ketoglutarate.
- Step 4, α-ketoglutarate to Succinyl-CoA
- Step 8, L-malate to OAA.
All in all, only two ATP (or GTP) are formed from one glucose (one from each pyruvate/acetyl-CoA) in the citric acid cycle, but the number of ATPs ultimately formed from the glycolysis-PDHc-TCA is much more, how many?
30-32 ATPs!
The citric acid cycle is not only a process that harvests energy from pyruvate, but also functions as a hub of metabolism, with catabolic pathways leading in and anabolic pathways leading out. Name tree key intermediates in the CA cycle, and what they can be used for.
- Citrate is a precursor for fatty acids and sterols.
- both α-ketoglutarate and Succinyl-CoA are precursors for amino acid synthesis, heme and nucleotides (purines).
- OAA is also a precursor for amino acid synthesis and nucleotides (pyrimidines)
Because so many of the intermediates in the citric acid cycle are used in biosynthesis, the withdrawal could potentially cause the cycle to stop running or slow down. What has the cell done to solve this?
When the withdrawal of cycle intermediates for use in biosynthesis lowers the concentrations of citric acid cycle intermediates enough to slow the cycle, the intermediates are replenished by anaplerotic reactions (Greek, “to refill”). These are for example:
- pyruvate to OAA in liver/kidney
- PEP to OAA in heart/skeletal muscle
- PEP to OAA in plants
- Pyruvate to malate in bacteria and eukaryotes.
The energy harvested through oxidation in glycolysis, the conversion of pyruvate to Acetyl-CoA and the citric acid cycle culminates in oxidative phosphorylation in the electron transport chain. Where is it located in eukaryotes and bacteria?
Oxidative phosphorylation happens in the inner membrane of mitochondria in eukaryotes and the inner membrane of bacteria. The outer membrane of mitochondria is porous so “small” molecules and ions can diffuse through, while the inner membrane is very tight and contains transporters for everything that should be taken in/out.
All the processes of energy catabolism (glycolysis, PDHc, the citric acid cycle and oxidative phosphorylation) take place in the mitochondria besides glycolysis, which takes place in the cytosol.
Oxidative phosphorylation begins with the entry of electrons into the series of electron carriers called the respiratory chain. Most of these electrons arise from the action of dehydrogenases that collect electrons from catabolic pathways and funnel them into universal electron acceptors — nicotinamide
nucleotides (NAD+ or NADP+) or flavin nucleotides (FMN or FAD). Explain the mechanism of action for NAD+ and FAD.
NAD+-linked dehydrogenases remove two hydrogen atoms from their substrates. One of these is transferred as a hydride ion (:H−) to NAD +, and the other is released as H+ in the medium.
The oxidized FAD can accept either one electron (yielding the semiquinone/radical form) or two (yielding FADH2). Electron transfer occurs because the flavoprotein the FAD is attached to has a higher reduction potential than the compound oxidized.
In addition to FAD and NAD+, three other electron carriers are present in the respiratory chain, which? Describe them briefly.
- Ubiquinone (also called coenzyme Q,
or simply Q) can accept one electron to become the semiquinone radical (∙QH) or two electrons to form ubiquinol (QH2). It’s a small lipid soluble molecule that can move freely between the two lipid layers that constitute the inner membrane. - Cytochromes: Proteins with bound heme groups (a & b bound tightly but not covalently, c bound covalently). The iron in the heme groups are redox active in their oxidative state Fe2+ or Fe3+.
- Iron-sulfur proteins: proteins containing FeS clusters that almost exclusively exist to transfer electrons. Many different compositions of Fe and S.
The electron transfers happen in protein complexes embedded in the inner membrane, name them.
Complex I: NADH dehydrogenase (Fe-S)
Complex II: Succinate dehydrogenase (Fe-S)
Complex III: U:C oxidoreductase + Cyt. c (Heme and Fe-S)
Complex IV: Cytochrome oxidase (Heme and Cu)
All have different structure and size, but similar function, transferring electrons trough electron carriers to finally reduce O2.
Describe the reaction mechanism in complex I of the ETC.
Complex I catalyzes the transfer of a hydride ion (:H−) from NADH to FMN (analog to FAD). From the FMN, two electrons pass through a series of Fe-S centers to ubiquinone, which forms QH2, which diffuses into the lipid bilayer.
The complex is a proton pump which transfers 4H+ from the N side (matrix), to the P side (intermembrane space). This reaction is very energy demanding as the proton flux produces an electrochemical potential across the inner mitochondrial membrane. The free energy from the redox reaction facilitates this.
In short: 2 electrons donated from NADH as a hydride ion to Ubiquinone, the redox driving 4H+ to be pumped out into the intermembrane space.
Describe the reaction mechanism in complex II in the ETC.
Complex II catalyzes the oxidation of Succinate to fumarate, and the electrons is taken up by FAD which is reduced to FADH2, the two electrons is then transferred from FADH2 to Q through Fe-S clusters to finally reduce ubiquinone (Q) to QH2. This complex doesn’t directly contribute to the electrochemical potential across the membrane (not a H+ pump)
Describe the reaction mechanism in complex III in the ETC.
Complex III catalyzes the transfer of electrons from QH2 to cytochrome C, one at a time, each resulting in 2 H+ being transferred to the P side. At this stage we have moved 8 H+ over to the P side.
To ensure that the semiquinone radical (∙QH) doesn’t cause damage, “Q cycling” is utilized.
Describe the reaction mechanism in complex IV in the ETC.
Complex IV is also a proton pump (2H+ net (two H+ added to matrix by 2 NADH)). The 4 electrons from Cyt. C comes into the Cu-A center, goes through a heme center and moves to the Cu-B center in which the electrons are used to reduce 1/2 O2 to 1 H20 (two electrons for each oxygen)
Some problematic intermediates are generated in this reaction, ∙OH- and H2O2, which can form ∙OH (ROS) that mitochondria needs to deal with.
What is the summarized reaction of the electron transport chain?
NADH + 11 H+(N) + 1/2 O2 –> NAD+ + H2O + 10 H+(P)
So we have moved 10 protons from the N side to the P side, creating a charge difference of 20 units. The H+ transfer creates a gradient, both concentration (ΔpH) and charge (inside negative) and according to the chemiosmotic model, this creates a proton motive force (PMF) which drives APT synthesis.
