Lectures 58-82 Flashcards

Question

What is type 2 diabetes?

Answer 1

Type 2 diabetes is generally associated with increased resistance to the actions of insulin; insulin is produced but its effectiveness in enhancing utilization of glucose is decreased. Obesity coupled to lack of exercise promotes type 2 diabetes by mechanisms that are not clear.

Answer 2

Diabetes is thus associated with accumulation of high levels of glucose, especially in the blood. This causes metabolic problems: energy production from glucose is impaired; fat metabolism is elevated to provide energy but much of the fat is oxidized to ketone bodies (lecture 8) which are acidic and can cause ketosis; high levels of circulating glucose can react non-enzymatically with proteins and enzymes and glycosylate them to form modified proteins with altered functions; high levels of glucose can be reduced by aldose reductase to sorbitol, a sugar alcohol that can increase osmotic pressure, e.g., in the eye lens.

Answer 3

A convenient assay for measuring long-term blood glucose levels relies on the fact that hemoglobin reacts readily with high glucose to form a glycosylated hemoglobin called hemoglobin A1c. Diabetics are monitored for their hemoglobin A1c levels, as high levels (> 7 mg% or 7 mg/dl blood) are indicative of poor glucose control due to not taking insulin and other medications or not eating properly.

Answer 4

Glucagon promotes glucose production in the liver by two primary mechanisms: stimulation of glucose synthesis (gluconeogenesis) and stimulation of glycogen breakdown.

Answer 5

Glycogen is a polymer made up of many glucose residues, which functions as a storage form of glucose.

Answer 6

Glucagon increases fatty acid release from triglycerides stored in adipose tissue. The liver will oxidize fatty acids for its own energy requirements, since the liver will be providing glucose for fuel for other tissues, not for itself.

Answer 7

stimulating glycolysis, a glucose degradation pathway; stimulating glucose uptake in some tissues, e.g. muscle, adipose tissue; stimulating glycogen formation; stimulating fatty acids synthesis from glucose (lecture 9); and stimulating protein synthesis (several amino acids can be produced from glucose).

Answer 8

It's a tetramer made up of two alpha subunits and two beta subunits. The beta units span the membrane. When insulin binds to the alpha subunit the binding results in the autophosphorylation of several tyrosine residues on the beta subunits. (The beta subunit has tyrosine kinase activity when insulin binds to the alpha subunit.)

Answer 9

It can phosphorylate other proteins, the major ones being insulin receptor substrates IRS 1 and 2.

Answer 10

PIP3 kinase

Answer 11

The signaling molecule AKT. pAKT can activate proteins such as pho sphatases, PKC, mTOR, that subsequently carry out the actions of insulin.

Answer 12

Glucagon binds to its receptor found in liver and fat tissue, while the epi receptor is widely present. These two receptors are G-protein-coupled receptors with seven transmembrane spanning loops (GPCR). To review GPCRs: the ligand-bound receptor activates specific G proteins. Typically, G proteins are heterotrimers made up of α, β and γ subunits. In the inactive state, the α subunits bind GDP (hence “G” proteins). The activated glucagon or epi receptor catalyzes an exchange of GDP with GTP. This is followed by dissociation of the βγ subunits to form the active α-GTP G protein. This activates the enzyme adenylate cyclase to produce the critical second chemical messenger cyclic AMP:

Answer 13

The phosphorylation of enzymes by activating cAMP- dependent PKA.

Answer 14

As a general “rule,” insulin promotes the dephosphorylation of enzymes by activating certain phosphatases, e.g., protein phosphatase 2A.

Answer 15

The αGTP subunit has a GTPase activity, and with time the GTP is hydrolyzed to form the αGDP subunit. This reassociates with the βγ subunits that have been “hanging around” to reestablish the inactive GDP-αβγ G protein. With time, the ligands glucagon and epi dissociate from the receptor. Also, cAMP is hydrolyzed by the enzyme phosphodiesterase to AMP; hence, the cAMP-PKA signal is turned off.

Answer 16

phosphodiesterase activity

Answer 17

it must be transported into cells by carrier-mediated mechanisms.

Answer 18

Liver, RBC, brain, pancreas, and most cells carry out a passive carrier-mediated glucose transport via their glucose carriers GLUT1 and GLUT3. These carriers have a low Km for glucose – about 1 mM (serum glucose levels are typically 4 to 8mM) – so they catalyze basal glucose uptake. Liver and pancreas also contain GLUT2, which has a Km for glucose of 15-20 mM. This allows the pancreas to sense high glucose (to produce insulin) and the liver to utilize this high glucose.

Answer 19

Insulin has no effect on the rate of glucose uptake by GLUT 1, 2 and 3.

Answer 20

Muscle and fat cells.

Answer 21

Insulin strikingly elevates the number of GLUT 4 carriers on the plasma membrane of muscle and fat cells so that glucose uptake is elevated. This is a major mechanism by which insulin elevates glucose utilization in muscle and fat cells, i.e., increasing glucose uptake.

Answer 22

nsulin increases GLUT4 in the plasma membrane by stimulating transport of GLUT4 molecules sequestered in the golgi to the plasma membrane. The downstream insulin-activated target AKT plays a role in this transfer of GLUT4 from the golgi to the plasma membrane.

Answer 23

GLUT5 is present in the GI tract and kidney where it catalyzes active transport of glucoses. Active uptake of glucose is linked to Na+ transport by the gut and kidneys.

Answer 24

Glycolysis refers to the “lysis” (breakdown) of glucose. It occurs in the cytosol fraction of all living things.

Answer 25

Glucose+2NAD+ +2Pi +2ADP → 2pyruvicacid+2ATP+2NADH

Answer 26

The major function of glycolysis is to provide energy for cells. For some cells, e.g., RBCs, glycolysis is the only source of energy. For other cells, it is the primary source of energy, e.g., brain, embryonic tissue, exercising muscle. Technically speaking, glucose only produces 2 ATP/glucose when it is oxidized to pyruvate. However, the pyruvate and the NADH can provide more ATP when they are oxidized further in the mitochondria.

Answer 27

Another important function of glycolysis is to provide intermediates for other metabolic reactions. Examples of such intermediates are: α-glycerophosphate for triglyceride and phospholipid synthesis; 2,3 bis-phosphoglycerate in the RBC to regulate hemoglobin-oxygen affinity; acetyl CoA from the pyruvate, which can be used to synthesize fatty acids, cholesterol, ketone bodies, and steroids; amino acids such as serine, alanine, glycine.

Answer 28

Step 1 is catalyzed by hexokinase, which catalyzes the transfer of a phosphate from ATP to produce glucose 6-P (G6P) and ADP. Note there is an initial input of energy needed to activate the glucose. ** irreversible.

Answer 29

Hexokinases, which are not specific for glucose, are constitutive enzymes with a low Km for glucose, and are subject to product inhibition by G6P and ADP.

Answer 30

The liver and the pancreas have a special hexokinase isoform called glucokinase. Glucokinase has a high Km for glucose, is specific for glucose, is not subject to inhibition by G6P and ADP and is inducible at the transcriptional level by high carbohydrate diet and insulin, and repressed by glucagon.

Answer 31

Glucose+ ATP--> glucose 6-P (G6P) and ADP, catalyzed by hexokinase.

Answer 32

G6P --> F6P, catalyzed by phosphoglucoisomerase

Answer 33

Step 3 is conversion of F6P to fructose 1,6 bis-phosphate (F1,6 bis P) by phosphofructokinase (PFK): F6P+ATP → F1,6bisP+ADP The reaction is irreversible and is the rate-limiting step in glycolysis. Note the input of a second ATP. Since glycolysis produces 2 ATP/glucose, the subsequent steps will have to produce a total of 4 ATPs.

Answer 34

PFK is inhibited by ATP and by citrate (high energy signals)

Answer 35

Stimulated by AMP, Pi, NH4+ and, surprisingly, by F1,6 bis P, the product.

Answer 36

Thus PFK has 2 substrate binding sites (ATP, F6P), 2 allosteric inhibitory sites, and 4 allosteric activator sites.

Answer 37

It's a powerful activator of PFK.

Answer 38

PFK is ACTIVE when dephosphorylated and F2,6bisP is INACTIVE when dephosphorylated.

Answer 39

This occurs in the presence of insulin, i.e., insulin promotes the dephosphorylated states of the kinase (active) and F2,6 bis phosphatase (inactive). This makes sense because insulin wants to stimulate glycolysis and therefore activate RFK.

Answer 40

Glucagon and epinephrine, acting via cAMP-PKA, promote the phosphorylated state of the kinase (inactive) and F2,6 bis phosphatase (active), thus lowering levels of F2,6 bis P and inhibiting PFK.

Answer 41

aldolase cleaves the 6-carbon F1,6 bis P to 2 trioses, dihydroxyacetone P (DHAP) and glyceraldehyde 3-P (G3P) by cleaving between carbons 3 and 4. At this point: Glucose+2ATP → DHAP+G3P+2ADP **DHAP can be reduced to α-glycerophosphate, which is needed for triglyceride and phospholipid synthesis. For glycolysis, the DHAP is converted to G3P by triose phosphate isomerase (step 5). The above 2 steps are reversible. The net reaction therefore is now: Glucose+2ATP → 2G3P+2ADP.

Answer 42

In step 6, G3P is oxidized by glyceraldehyde 3-P dehydrogenase (GAPDH) G3P + NAD+ + Pi ↔ 1,3 bis phosphoglyceric acid (1,3 bis PGA) + NADH

Answer 43

1,3 bis PGA is a high-energy compound and produces ATP and 3-phosphoglyceric acid (PGA) in step 7, which is the phosphoglycerate kinase reaction (Fig. 3). 1,3bisPGA +ADP ↔ 3PGA+ATP Since 1 glucose yielded 2 G3P, we net 2 ATPs when 1 mole of glucose is metabolized to 2 moles of 3 PGA.

Answer 44

The energy associated with the oxidation of the aldehyde on C1 of G3P to the carboxylic acid on C1 of 3PGA was trapped as a high energy intermediate 1,3 bis PGA, which then produced ATP. Note, this occurs in the absence of mitochondria and oxygen.

Answer 45

In step 8, phosphoglyceromutase converts 3 PGA to 2 PGA, which is followed (step 9) by removal of H2O from 2 PGA by enolase to produce phosphoenolpyruvate (PEP), a high-energy compound.

Answer 46

In the final step of glycolysis (step 10), the high energy PEP can yield ATP as catalyzed by pyruvate kinase (Fig. 4), an irreversible reaction. PEP + ADP → Pyruvate + ATP 2ATPs per glucose will be produced at this step. Thus, we used 2 ATPs at the beginning (hexokinase, PFK) and we produced 4 ATPs (2 at the G3P 3PGA step and 2 at the pyruvate kinase step) for a net of 2 ATPs for each glucose → 2 pyruvates.

Answer 47

screenshot 15

Answer 48

Glucose+ ATP--> glucose 6-P (G6P) and ADP, catalyzed by hexokinase. F6P+ATP → F1,6bisP+ADP cat by phosphofructokinase (PFK) **rate limiting step PEP + ADP → Pyruvate + ATP Catalyzed by PK

Answer 49

Insulin stimulates glycolysis by increasing glucokinase levels and by activating the PFK and PK reactions via dephosphorylations

Answer 50

Glucagon inhibits glycolysis by decreasing glucokinase levels and inhibiting PFK and PK via cAMP-PKA- dependent phosphorylations.

Answer 51

High-energy signals, such as ATP and citrate, inhibit PFK while ATP, NADH, and acetyl CoA inhibit PK.

Answer 52

Thus, fructose enters glycolysis at the DHAP and G3P level, which continue on to pyruvate or in the reverse direction, to glucose.

Answer 53

Fructose-->F1P Uses 1 ATP cat by fuctokinase. F1P--> DHAP + GLyceraldehyde. DHAP-> G3P cat by adolase GLyceraldehyde-> G3P by triosephosphokinase After entering in glycolysis it can proceed in either direction, down to pyruvate or up to glucose.

Answer 54

Galactose-> galactose 1-P by galactokinase galactose 1-P-> glucose 1-P (G1P) by galactose 1-P uridyl transferase G1P-> G6P by phosphoglucomutase, which enters glycolysis or is converted to free glucose (mostly in liver).

Answer 55

Uridyl transferase is deficient, therefore high levels of galactose 1P accumulate, which, similar to F1P, is toxic to the liver. Treatment is obviously to limit lactose in the diet, a problem for nursing mothers.

Answer 56

it can be reduced to lactic acid (lactate) or it can enter the mitochondria for further metabolism by pyruvate dehydrogenase (PDH) to acetyl CoA. In liver, pyruvate can also be metabolized to oxalacetate by pyruvate carboxylase (next lecture). In RBCs, which lack mito, pyruvate must be reduced to lactate.

Answer 57

LDH catalyzes reduction of pyruvate to lactate. This reaction is reversible.

Answer 58

This reaction is critical to RBCs because NADH is reoxidized back to NAD+, which is necessary for glycolysis to continue at the glyceraldehyde 3-P dehydrogenase step. Tissues containing mito can oxidize the NADH, after it is transported into the mito, by the respiratory chain. The LDH reaction is also important to deoxidize NADH whenever glycolysis is very rapid, e.g., during muscle contraction, when high insulin levels are high, or in tumor cells.

Answer 59

Lactic acid leaves the RBC or muscle or tissue where it was produced and circulates in the blood. High levels alter blood pH (lacticacidemia), can cause cramps (e.g., in marathon runners), and can aggravate gout (will be discussed later). Most lactate enters tissues where it is reoxidized back to pyruvate, which then undergoes the pyruvate dehydrogenase reaction for energy production, or can be converted to glucose in the liver during gluconeogenesis.

Answer 60

LDH can be made up of 2 different peptide chains, H and M chains. LDH is a tetramer, and therefore five different LDH isoforms can be produced – H4, H3M, H2M2, HM3 and M4.

Answer 61

H4 and H3M

Answer 62

M4 and M3H

Answer 63

H4 has a higher binding affinity for Lactate and NAD and are expressed in tissues that would have use for holding on to those things. Meaning they have oxygen readily available and can further process lactate. RBCs and skeletal muscle often can only perform anaerobic respiration so they don't want an isotherm that will hold onto the lactate they want to get rid of it and get it out so they can make more.

Answer 64

A great thing happens to pyruvate in yeast cells, it is converted to ethanol (alcohol). Yeast contain an enzyme pyruvate decarboxylase (do not confuse with E1 of the PDH complex discussed next) which causes the decarboxylation of pyruvate to CO2 plus acetaldehyde Vitamin B1, thiamine pyrophosphate (TPP), is the cofactor for this reaction. Acetaldehyde then is reduced by alcohol dehydrogenase to ethanol. In this way, the NADH produced by glycolysis is used to produce ethanol and the net reaction of this fermentation is: Glucose+2ADP+2Pi  2CO2 +2Ethanol+2ATP

Answer 65

Decarboxylates pyruvate into Co2 plus acetaldehyde in yeast.

Answer 66

Vitamin B1, thiamine pyrophosphate (TPP)

Answer 67

In most tissues, in order to extract more energy from glucose, pyruvate is converted to acetyl CoA by PDH. The acetyl CoA can be further oxidized in the TCA cycle, generating NADH and FADH2 for energy production by the respiratory chain.

Answer 68

No. The PDH reaction is irreversible. Therefore acetyl CoA cannot be converted to pyruvate, which means that fatty acids which are primarily oxidized to acetyl CoA cannot produce pyruvate and therefore glucose.

Answer 69

E1, Thiamine pyrophosphate (TPP) (derivative of thymine, vitamin B1)

Answer 70

E2, Lipoamide

Answer 71

E3, FAD (derivative of riboflavin)

Answer 72

Pyruvate reacts with the carbanion of TPP on E1 to yield CO2 and hydroxyethyl-TPP.

Answer 73

The hydroxyethyl carbanion on TPP of E1 reacts with the lipoamide on E2. The acetate formed by oxidation of the hydroxyethyl moiety is linked to one of the thiols of the reduced lipoamide as a thioester (~).

Answer 74

The acetate is transferred from the thiol of lipoamide to the thiol of coenzyme A, yielding acetyl CoA.

Answer 75

It reacts using E3 catalyst with FAD+. This reaction gives the oxidized lipoamide and FADH2. That reacts with NAD+ to generate FAD+ and NADH +H

Answer 76

It is a central compound in metabolism. The "high energy" thirster linkage makes it an excellent donor of the acetate moiety. A few functions: input to the Krebs Cycle, where the acetate moiety is further degraded to CO2. donor of acetate for synthesis of fatty acids, ketone bodies, and cholesterol.

Answer 77

E1 is inhibited by ATP, high energy charge (makes sense?). E2 is inhibited by its product, acetyl CoA, and E3 is inhibited by its product, NADH. These are also signs of high energy.

Answer 78

PDH kinase phosphorylates E1 and inhibits its activity. * NOTE This phosphorylation is NOT mediated by cAMP- PKA and is not affected by glucagon or epinephrine.

Answer 79

PDH phosphatase dephosphorylates E1-P and activates E1. This phosphatase is strikingly increased by insulin.

Answer 80

Note: insulin stimulates glucose to pyruvate at several steps (GK, PFK, PK) and here stimulates PDH to use the pyruvate.

Answer 81

Note: the reoxidation of FADH2 (on E3) by NAD is the only case in metabolism in which FADH2 reduces NAD+ to NADH. Usually, e.g., in the respiratory chain, NADH reduces FAD to FADH2.

Answer 82

The CO2 which comes off in the PDH reaction can be traced back to carbons 3 and 4 of the glucose (Fig. 6). Carbons 1 and 6 of the original glucose form the CH3 of acetyl CoA, while carbons 2 and 5 form the C=O of the acetyl group. These come off as CO2 in the TCA cycle.

Answer 83

C6H12O6 + 2 NAD+ 2 ADP + 2 Pi→ 2CH3COCOOH (pyruvate or pyruvic acid)+ 2 ATP + 2 NADH. Irreversible, occurs in cytosol of all cells. Pyruvate will be further metabolized

Answer 84

NADH is REOXIDIZED back to NAD+ which is necessary to turn G3P into 1,3bisphosthoglycerate in glycolysis. Mito can deoxidize NADH but RBCs don't have mito.

Answer 85

Most lactate enters tissues where it is reoxidized back to pyruvate, which then undergoes the pyruvate dehydrogenase reaction for energy production, or can be converted to glucose in the liver during gluconeogenesis.

Answer 86

Most lactate enters tissues where it is reoxidized back to pyruvate, which then undergoes the pyruvate dehydrogenase reaction for energy production, or can be converted to glucose in the liver during gluconeogenesis.

Answer 87

Glycolysis/PDH, fatty acid great down, amino acid breakdown.

Answer 88

Occurs in the mitochondrial matrix compartment of all cells (one enzyme, succinic dehydrogenase, is in the inner mito membrane).

Answer 89

AcetylCoA+3NAD+ +FAD+GDP+Pi +2H2O→2CO2 +CoA+3NADH+FADH2 + GTP + 2H+ Acetyl CoA + 3 NAD + FAD+ GDP+ Pi+ 2 H2O→ 2 CO2 +3 NADH+FADH2+GTP+CoASH

Answer 90

F1,6 Bis P and insulin stimulated dephosph.

Answer 91

Pyr + NAD+ CoA→ Acetyl CoA + CO2+ NADH

Answer 92

CO2 via TCA Cycle, ATP; → citrate which →fatty acids → ketones bodies,cholesterol, steroids → Acetylation RX e.g histones

Answer 93

PDH. E1, E2, E3 etc.

Answer 94

FADH2 from FAD

Answer 95

NADH from NAD+

Answer 96

Isocitrate + NAD+ ---> Alphakg + Co2 +NADH + H+

Answer 97

akg--> succinylCoA.

Answer 98

While the equilibrium of the malate dehydrogenase reaction in vitro greatly favors OAA malate, in cells OAA is continually being removed by the citrate synthase reaction, keeping the concentration of OAA low and pulling the reaction in the direction toward OAA formation.

Answer 99

One FADH2 and 3 NADH + 3H+ are produced in the TCA cycle, which can yield 1.5 (FADH2) + 7.5 (3 NADH) or a total of 9 ATPs. One GTP is equivalent to one ATP. Hence, 1 acetyl CoA yields about 10 ATPs when oxidized in the TCA cycle.

Answer 100

about 32 ATPs/G6P: Glucose--> 2 pyruvate + 2 NADH 2ATP 2 NADH (mito) 5ATP 2 Pyruvate--> 2AcetylCoA and 2NADH 2 NADH 5ATP 2 Acetyl CoA 20 ATP

Answer 101

Of the 6 carbons in G, C3,4 → CO2 in PDH RX | C1,2,5,6→CO2 in TCA cycle

Answer 102

a) Respiratory control regulates oxidation of NADH and FADH2 by the mito respiratory chain. This is discussed in more detail in the bioenergetics lectures. b) Energy charge. ATP inhibits citrate synthase, isocitrate dehydrogenase, αKg dehydrogenase c) Concentration of OAA. OAA is pulled out of the cycle during gluconeogenesis (lecture 10) and reduced to malate when NADH levels are elevated, e.g., after alcohol consumption.

Answer 103

citrate synthase, isocitrate dehydrogenase, αKg dehydrogenase

Answer 104

Succinyl CoA -> heme synthesis OAA-> gluconeogenesis, aa production (aspartate asparigine) Citrate-> fatty acids αKg-> aa. snth: glutamate, glutamine, proline, ornithine

Answer 105

a) Pyruvate carboxylase (#1 in Fig. 6). This enzyme contains biotin, the vitamin B used to carry CO2 in enzymatic reactions. ATP is necessary for this carboxylation. Importantly, the enzyme is strikingly stimulated by acetyl CoA. Does this make sense? As will be discussed in the nitrogen/amino acid lectures later on, many amino acids are catabolized to TCA cycle intermediates: - glutamate  αKg (#2 in figure below) - valine or isoleucine or methionine or threonine  succinyl CoA (#3 in figure below) - phenylalanine or tyrosine  fumarate (#4 in figure below) - aspartate  OAA (#5 in figure below) ``` PEP carboxykinase (not shown in the figure). This enzyme converts the glycolytic intermediate PEP to OAA: CO2 +PEP+GDP ↔ OAA+GTP ``` Note that the high energy PEP can produce ATP in the pyruvate kinase reaction, and GTP (equivalent to ATP) in the PEPCK reaction. PEPCK is induced by glucagon under gluconeogenic conditions, and we will discuss this and pyruvate carboxylase in lecture 10.

Answer 106

Inhibits the glycolysis enzyme, phosphofructokinase. Therefore inhibiting isocitrate dehydrogenase would cause a buildup of citrate and negatively impact the rate of glycolysis.

Answer 107

Phosphate carrier - exchanges Pi with OH 2. Dicarboxylate carrier - exchanges Pi or malate or succinate for each other 3. Tricarboxylate carrier - exchanges citrate, isocitrate, malate or PEP for each other 4. αKg carrier - exchanges αKG for malate 5. Pyruvate carrier - exchanges pyruvate for OH or ketone bodies 6. Glutamate carrier - exchanges glutamate for OH 7. Aspartate carrier - exchanges aspartate for glutamate 8. Adenine nucleotide carrier - exchanges ADP for ATP

Answer 108

Carrier 8 is obviously important in transporting ATP produced by oxidative phosphorylation out of the mito in exchange for cytosolic ADP, which was produced from ATP hydrolysis in the cytosol, e.g., glucose + ATP  G6P + ADP. Carrier 8 functions in conjunction with Carrier 1 to bring ADP plus Pi back to the mito for eventual synthesis of ATP. Remember carrier 1 exchanges Pi with OH

Answer 109

4 and 7 4. αKg carrier - exchanges αKG for malate 7. Aspartate carrier - exchanges aspartate for glutamate

Answer 110

2. Dicarboxylate carrier - exchanges Pi or malate or succinate for each other 3. Tricarboxylate carrier - exchanges citrate, isocitrate, malate or PEP for each other 4. αKg carrier - exchanges αKG for malate

Answer 111

Shuttles are critical for transporting reducing equivalents from NADH or NADPH into or out of the mito; for providing acetyl CoA for fatty acid or cholesterol synthesis and for providing carbon intermediates for gluconeogenesis.

Answer 112

The α-glycerophosphate (αGP) and the malate-aspartate (MA) shuttle.

Answer 113

DHAP, a product of glycolysis, reacts with NADH, also a product of glycolysis, to produce αGP + NAD+, as catalyzed by the cytosolic α GPDH. Thus, NAD+ is reoxidized. However, what do we do with the αGP, and how can we regenerate DHAP in order to continue the shuttle? Furthermore, if we keep pulling DHAP out of glycolysis, glycolysis will stop. αGP can react with the mitochondrial αGPDH, which is located on the outer surface of the mito inner membrane to regenerate DHAP. The mito αGPDH is linked to FAD, not NAD+ as was the cytosolic αGPDH, so FADH2 rather than NADH is produced in the mito. αGP + αGPDH-FAD  DHAP + αGPDH-FADH2  Respiratory Chain.

Answer 114

``` The NADH produced by glycolysis reacts with OAA in the cytosol to produce NAD+ and malate, as catalyzed by the cytosolic malate dehydrogenase (mdh; step 1). NAD+ has been regenerated, but unless OAA is regenerated (similar to the need to regenerate DHAP in the αGP shuttle), this shuttle will stop. ```

Answer 115

``` Malate can enter the mito (step 2) using 3 different carriers (2, 3, 4) and can be reoxidized by the mito mdh (same one as in the TCA cycle) to produce OAA and NADH (step 3). Note: in this shuttle, the reducing equivalents of cytosolic NADH wind up as mito NADH (not FADH2 as with the αGP shuttle). ```

Answer 116

OAA is converted to the amino acid aspartate by a transamination reaction with glutamate. (glutamate becomes alphakg) This is catalyzed by mitochondrial AST or GOT. Aspartate and αKg can exit the mito on carriers 7 and 4, respectively (step 5). Once in the cytosol, the transamination reaction is reversed, as catalyzed by the cytosolic AST or GOT (step 6).

Answer 117

4 (alpha keto gluterate for malate) | 7 (aspartate for malate)

Answer 118

The liver, adipose, mammary gland, steroidogenic tissues such as the adrenals, and RBCs. It;s activity is low in really active tissues such as muscle brain and heart.

Answer 119

The PPP is another pathway that utilizes glucose for: A) Production of NADPH – needed for synthesis of cholesterol, fatty acids, steroids, detoxification reactions by the cytochrome P450 system. **B) Production of ribose – needed to produce the nucleotides for RNA and DNA synthesis; for NAD, FAD, CoASH C) Reduction of oxidative stress

Answer 120

glucose6phosphate

Answer 121

G6P+2NADP+--> CO2 +Ribose5P+2NADPH

Answer 122

The pathway ends at oxidation and you get both.

Answer 123

3 ribulose 5P the rearrangement reactions will produce 2 fructose 6P plus one glyceraldehyde 3P 3 pentoses (ribulose 5P) have been converted to glyceraldehyde 3P and 2 fructose 6P . Fructose 6P is easily converted to G6P via phosphoglucoisomerase (second enzyme of glycolysis). Glyceraldehyde 3P (via gluconeogenesis) is equivalent to 0.5 G6P; thus, we have regenerated 2.5 G6P by rearrangement reactions.

Answer 124

6 NADP+ + 3 G6P --> 3 CO2 + 6 NADPH + 3 Ribulose 5P | irreversible

Answer 125

The point is that RBCs need NADPH and therefore need a PPP to reduce oxidized glutathione back to reduced glutathione in order to avoid oxidative stress, hemolysis and destruction. The O2 . can either spontaneously form peroxide (O2=) which is primarily hydrogen peroxide (H2O2) at pH 7, or there are enzymes called superoxide dismutases which catalyze the conversion of superoxide to H2O2 O2 .+O2 .+2H+O2+H2O2 H2O2 is a powerful oxidant and must be rapidly removed or it will damage the RBC (to be discussed further in the oxygen radical lectures). The RBC decomposes H2O2 by two mechanisms. Catalase, a heme-containing enzyme, decomposes H2O2 to O2 + H2O. The second enzyme, called glutathione peroxidase (GPx) uses a cellular biochemical called glutathione (GSH) to catalyze the following reaction: H2O2 + 2GSH 2H2O + GSSG GSH = reduced glutathione GSSG = oxidized glutathione It needs NADH to reconvert it back to the reduced GSH

Answer 126

Glycogen is a polymer made up of glucose units linked by direct bonds between C1 of one glucose to C4 of an adjacent glucose. There are also branch points in which C1 of a glucose is linked to C6 of another glucose (Fig. 1). Branching increases solubility and allows several available sites to be degraded simultaneously.

Answer 127

Particles containing glycogen and the enzymes involved in glycogen metabolism are found in glycogen granules, located in the cytosol of cells, with liver and muscle containing most of the body glycogen.

Answer 128

Muscle glycogen serves to provide the muscle with glucose in times of need, e.g., exercise, fight, flight. Liver glycogen produces glucose which is to be released from the liver to the blood to provide glucose to other tissues .

Answer 129

phosphorylase and the debranching enzyme

Answer 130

Phosphorylase catalyzes a phosphorolysis of glycogen to produce one molecule of G1P and a glycogen that is shortened by one glucose unit. This is irreversible.

Answer 131

In addition, sometimes phosphorolysis is preferable to hydrolysis (like in the breakdown of glycogen or starch, as in the example above) because glucose 1-phosphate yields more ATP than does free glucose when subsequently catabolized to pyruvate.

Answer 132

Removal of one glucose unit occurs repeatedly until about 4 residues from a branch point. The debranching enzyme moves 3 of the 4 residues to another chain. The fourth residue which is in a 1-6 glycosidic bond with the other chain is hydrolyzed by the 1-6 glucosidase activity of the debranching enzyme to produce glucose and the now straight polymer. Phosphorylase now continues to degrade the straight chain polymer to G1P until it reaches another branch point.

Answer 133

Glycogen is broken down during starvation, during low carbohydrate diets, with exercise, and in diabetes.

Answer 134

Glycogen is synthesized under conditions of high glucose availability, the fed state, and high energy state.

Answer 135

G6P is mutated into G1P by phosphoglucomutase. The G1P is activated using UTP to form UDP-glucose + pyrophosphate (PPi). The glycogen polymer is built around a small protein core called glycogenin where a tyrosine- OH is linked (via autoglycosylation) to one or more glucose residues. New glucose units from UDP-glucose are added to the core primer glucose in 1-4 glycosidic bonds. Glycogenin–Tyr–glucose + UDPG -->Glycogenin–Tyr–glucose-glucose + UDP This addition is catalyzed by glycogen synthase and is irreversible.** After building up a number of glucose links, a branching enzyme transfers 6-7 glucose residues from the end of this chain to the C6-OH group of a glucose residue in that chain (or another chain) to form the 1-6 branching points (Fig. 4, below).

Answer 136

Glycogen synthase (GS) is stimulated by high levels of G6P. Makes sense? Glycogen phosphorylase (Ph) is stimulated by AMP (low energy signal) and inhibited by G6P and ATP (high energy signals).

Answer 137

In the dephosphorylated state. What is this similar to? PFK! What activates this state? Insulin. What phosphorylates it slash deactivates it? eli/glucagon. This makes sense you would want to make glycogen and break down glucose all in states of high glucose.

Answer 138

In the phosphorylated state. This is activated by eli/glucagon and inhibited by insulin.

Answer 139

Is present in liver but not muscle, hydrolyses G6P to G + Pi. This enzyme is transcriptionally activated by Epi and glucagon, so note that these hormones stimulate liver glycogen G6P Glucose. Why doesn’t G6P made from glycogen in liver enter glycolysis like it does in muscle? Recall from glycolysis in liver that Epi and glucagon decrease levels of F2,6 bis P, thus blocking glycolysis while increasing G6Pase. So the G6P does not enter glycolysis but instead is hydrolyzed to glucose, which then exits the liver.

Answer 140

The F2,6 bis P kinase and phosphatase enzymes in muscle are different isoforms from the liver F2,6 bis P kinase and phosphatase enzymes and show opposite properties: while phosphorylation decreases the kinase activity and activates the phosphatase in liver, thereby lowering F2,6 bis P levels and thus PFK activity, phosphorylation increases the kinase activity and decreases the phosphatase activity in muscle, thus increasing F2,6 bis P levels and stimulating PFK activity and glycolysis in muscle. Such considerations emphasize the importance of isoforms.

Answer 141

Remember F26bisP stimulates RFK. in muscle... dephosphorylation of the kinase makes it active and dephosphorylation of the phosphotase makes it inactive. This increases F26bisP which stims RFK. Insulin promotes this. In the LIVER however, Phosphorylation of the kinase makes it inactive and phosphorylation of the phosphotase makes it active. This makes F26bisP levels go down therefore inhibiting RFK and glycolysis.

Answer 142

In liver, phosphorylase acts as a glucose sensor. In the absence of glucose, phosphorylase binds protein phosphatase-1, thus preventing the phosphatase from dephosphorylating phosphorylase and glycogen synthase, which are therefore active and inactive, respectively. This should make sense as glycogen will not be synthesized in the absence of glucose. In the presence of glucose, the phosphatase-1 is released from phosphorylase because of a conformational change that occurs when glucose binds to phosphorylase. The now free protein phosphatase-1 can dephosphorylate phosphorylase and glycogen synthase, making them inactive and active, respectively.

Answer 143

There are a variety of glycogen storage diseases which basically block glycogen breakdown. These can interfere with the liver’s ability to provide glucose and the muscle’s ability to carry out muscular function under stress conditions.

Answer 144

G→G6P→G1P→UDPG. Need exisiting primer- Glycogen (n) + UDPG→ Glycogen ( n+1) + UDP

Answer 145

Most common inborn error of metabolism Complete deficiency is embryonic lethal. Certain drugs ↓G6PDH, can↑ RBC hemolysis, anemia. Those with low G6PDH sensitive to oxidant stress produced by drugs, chemicals, foods (Fava beans), high O2, high altitudes Wernicke Korsakoff Syndrome- poor memory, orientation,gait, mental function/ neuropsyc. Disorders. Due to 10X↑ in Km for TPP by TK

Answer 146

TPP is a cofactor for TX which is an enzyme necessary to catalyze the PPP pathway. this pathway is essential to reducing the burden of oxidative stress.

Answer 147

Glucose is synthesized primarily by the liver and to a small extent by the kidney. It is synthesized under conditions of starvation, low carbohydrate diet, and diabetes (even though glucose levels may be high in diabetics).

Answer 148

lactate, pyruvate, glucogenic amino acids (covered in a later lecture, but 18 of the 20 amino acids in proteins are glucogenic) and, to a lesser extent, glycerol from breakdown of triglycerides.

Answer 149

In the liver, lactate is oxidized to pyruvate by LH.

Answer 150

a) Pyruvate+ATP+CO2-->OAA+ADP+Pi b) OAA+GTP--> PEP+CO2 +GDP (a) is the pyruvate carboxylase reaction and (b) is the PEP carboxykinase reaction. The net reaction of both is Pyr+ATP+GTP-> PEP+ADP+Pi+GDP

Answer 151

It is a reversible reaction. When more substrate is needed for the TCA cycle, PEPCK can make PEP from OAA produced in the cycle and then pyruvate carboxylase can make pyruvate to eventually make acetyl CoA to put back in the cycle. On the other hand in need of glucose, Pyruvate can be made into OAA with pyruvate carboxylase which can then be turned into PEP with PEPCK.

Answer 152

The first step occurs in the mito and the second step occurs either in the mito or cytosol because PEPCK is in both. It takes two high energy bonds to bypass or reverse that PK step in glycolysis. Rest of steps in gluconeogenesis are in the cytosol.

Answer 153

OAA can leave as malate using carriers 2, 3 or 4 or via the malate-aspartate shuttle using carriers 4 plus 7 (Fig. 2).

Answer 154

Acetyl CoA.

Answer 155

Note that 2 PEPs are needed to form one F1,6 bis P, ATP is needed at the phosphoglycero- kinase step and NADH is needed at the G3PDH step. Three high-energy bonds are needed to convert pyruvate to G3P, and 2 pyruvates are needed per glucose, so 6 ATP equivalents are needed to synthesize glucose from 2 pyruvates. Remember, only 2 ATPs were produced when glucose was metabolized to pyruvate. This is the high cost of bypassing irreversible steps.

Answer 156

Since PFK is irreversible, F1,6 bis P is converted to F6P by the gluconeogenic enzyme F1,6 bisphosphatase (F1,6 bis phase) (see Fig. 1). This enzyme shows the opposite regulatory properties of PFK, i.e., it is stimulated by ATP and citrate and is inhibited by AMP, Pi and NH4+.

Answer 157

F2,6 bis P, the critical allosteric activator of PFK, is a critical allosteric inhibitor of F1,6 bis phase. Accordingly, glucagon and epi, which lower F2,6 bis P levels to inhibit PFK, will stimulate F1,6 bis phase and gluconeogenesis by lowering F2,6 bis P levels. Insulin, which elevates F2,6 bis P levels to stimulate PFK, will inhibit F1,6 bis phase and gluconeogenesis by elevating F2,6 bis P levels.

Answer 158

Component of the phospholipids and glycolipids of biological membranes Major source of energy for most tissues Major storage form of energy as triglycerides Produce signal transduction molecules, e.g. inositol phosphates, diacylglycerol Produce prostaglandins and leukotrienes

Answer 159

In the fasted state, fatty acids are the primary fuel for energy in most tissues. The hormone-sensitive lipase in adipose tissue is activated by glucagon, epinephrine or norepinephrine via cAMP-dependent PKA phosphorylation (in the fed state, insulin inhibits this lipase via dephospho rylation). The activated lipase hydrolyses stored triglycerides to glycerol plus fatty acids. The glycerol exits adipose tissue and goes to the liver, where it can enter gluconeogenesis or glycolysis at the G3P and DHAP level. The fatty acids also exit and bind to albumin in the blood for transport to various tissues. The albumin-fatty acid complex cannot cross the blood brain barrier, hence fatty acids do not reach the brain for oxidation (a major reason why brain requires glucose constantly for energy). The fatty acids are transported from albumin to fatty acid binding proteins for delivery inside the tissue.

Answer 160

The fatty acid-albumin complex can not cross the blood brain barrier. Therefore the brain relies solely on glucose.

Answer 161

The fatty acid is first activated to the fatty acyl CoA ester by Acyl CoA synthase

Answer 162

Most of the fatty acids in our diet are long-chain, saturated fatty acids. long-chain fatty acyl CoA cannot enter the mito. A carnitine shuttle mechanism is needed, converting the long-chain fatty acyl CoA to the long-chain fatty acyl carnitine which can enter the mito via a carnitine translocase carrier. Once inside, the fatty acyl carnitine is converted back to the fatty acyl CoA. These conversions are catalyzed by acyl carnitine transferases 1 and 2

Answer 163

The acyl CoA now undergoes a sequence of four enzyme- catalyzed reactions which split a 2-carbon fragment of the acyl CoA to produce acetyl CoA plus a new fatty acyl CoA shortened by two carbon atoms (Fig. 4). This process is called β-oxidation.

Answer 164

One NADH and one FADH2 are produced in steps 1 and 3 of the reaction.

Answer 165

Thus, β-oxidation is a spiral and is repeated until all the carbons are oxidized to acetyl CoA. Each spiral produces one FADH2 and one NADH. For example, a C18 fatty acid will undergo 8 spirals:

Answer 166

However, the final spiral will produce acetyl CoA + propionyl CoA rather than 2 acetyl CoAs

Answer 167

The propionyl CoA is metabolized to succinyl CoA by a 3-step process

Answer 168

Availability of fatty acids via activated hormone-sensitive lipase. - Availability of carnitine. There are some inborn errors of metabolism which lower carnitine synthesis. In mammals, carnitine can be synthesized from the amino acid lysine. - Malonyl CoA, the product of the rate-limiting enzyme of fatty acid synthesis, inhibits carnitine acyl transferases 1 and 2. Does this make sense to you? - Rate of the electron transport chain. Why?

Answer 169

For the C18 fatty acid described above, how many ATPS will be produced? 8 NADH-> 20 ATPs 8 FADH2-> 12 ATPs 9 acetyl CoAs-> 90 ATPs Two ATPs were used to activate the fatty acid, so a net of about 120 ATPs are produced from the β-oxidation of a C18 fatty acid coupled to the TCA cycle and the respiratory chain. WOW, pretty impressive compared to total oxidation of glucose (about 30 ATPs/glucose).

Answer 170

We cannot pile up acetyl CoA because this would deplete CoASH (all tied up as acetyl CoA) and therefore fatty acid oxidation could not continue. The liver (and to a small extent the kidneys) have a new pathway for metabolism of acetyl CoA, the formation of ketone bodies (ketogenesis).

Answer 171

Ketogenesis is a 3-step pathway (Fig. 6) in which 2 acetyl CoAs are converted to the ketone body acetoacetate. These steps are in the mitochondria. Mechanistically, a third acetyl CoA is used but it is regenerated.

Answer 172

2 Acetyl CoA-> Acetoacetyl CoA-> HMGCoA -> Acetoacetate (net production of 2 CoA, really 3 are used but one is regenerated)

Answer 173

Acetoacetate can be reduced by β-hydroxybutyrate dehydrogenase to the second ketone body, **β-hydroxybutyrate, and can spontaneously decarboxylate to acetone plus CO2. The smell of acetone is detectible in the breath of diabetics, fasting individuals, alcoholics.

Answer 174

Ketone bodies made in the liver are valuable sources of fuel for other tissues, especially muscle, heart and kidney cortex. The ketone bodies exit the mito via carrier # 5 and also leave the liver, circulate in the blood (detectable by paper strips used by diabetics), and enter tissues where they are converted back to acetyl CoA and oxidized in the TCA cycle (Fig. 7). Ketone bodies can be thought of as circulating forms of acetyl CoA. Very high levels of ketone bodies in blood (ketosis) are a problem because they are strong acids and alter blood pH. This is a major problem in uncontrolled diabetics and poorly nourished alcoholics (alcohol is oxidized mainly in the liver eventually to acetyl CoA).

Answer 175

The transferase (Fig. 7) which converts acetoacetate to acetoacetyl CoA is normally not present in brain, hence, brain does not oxidize ketone bodies well. However, this enzyme is induced in brain under stressful conditions e.g. starvation, which allows the brain to now oxidize ketone bodies. This is a major metabolic adaptation because it lowers the need for glucose utilization by brain (discussed later in a metabolic interrelationships lecture).

Answer 176

Fatty acids are synthesized when carbohydrate (glucose) and protein (amino acids) levels are high. The synthesis occurs in the cytoplasm of most cells, with liver the most reactive.

Answer 177

Most of the excess fatty acids synthesized in the liver are exported as triglycerides packaged in very low density lipoproteins (VLDL) to adipose tissue, where the triglycerides are stored.

Answer 178

Fatty acid synthesis requires ATP and NADPH and is elevated by insulin and decreased by glucagon (Fig. 1).

Answer 179

High glucose results in excess fatty acid synthesis. The glucose is metabolized by glycolysis and pyruvate dehydrogenase to acetyl CoA, which enters the TCA cycle to produce citrate. Some of this citrate continues through the TCA cycle to produce the ATP necessary for fatty acid synthesis. However, some of the citrate leaves the mito on the tricarbxylic acid carrier to the cytosol, where it produces acetyl CoA. Thus, citrate is transporting mito acetyl CoA to the cytosol. This cytosolic acetyl CoA then enters lipogenesis (synthesis of lipids, fatty acids) to produce palmitic acid, the major fatty acid produced by the fatty acyl synthase complex discussed below.

Answer 180

Stored as triglycerides, especially in liver and adipose tissue Forms phospholipids needed for membranes Can be elongated to produce C18, C20 fatty acids Can be desaturated to produce unsaturated fatty acids

Answer 181

Citrate shuttle Mitochondrial acetyl CoA reacts with OAA to produce citrate (first step of the Krebs cycle catalyzed by citrate synthase), which is transported out of the mito via the citrate shuttle (Fig. 1). Citrate leaves the mito on the tricarboxylic acid carrier and is cleaved by the ATP-dependent citrate cleavage enzyme (also called citrate lyase) to OAA + acetyl CoA. The OAA is reduced to malate, which either reenters the mito and is oxidized back to OAA or is oxidized in the cytosol by the malic enzyme: Malate + NADP+  Pyruvate + CO2 + NADPH (cf. Fig. 2) This reaction produces NADPH, which is important for fatty acid synthesis. The pyruvate enters the mito where it is converted to OAA by the anaplerotic enzyme pyruvate carboxylase. Thus, OAA is regenerated in the mito from either malate or pyruvate. The acetyl CoA now enters the lipogenesis pathway. These reactions are summarized in Fig. 2.

Answer 182

Fatty acid biosynthesis requires acetyl CoA and malonyl CoA: acetyl CoA + 7 malonyl CoA  palmitoyl CoA + 7 CoA

Answer 183

The synthesis of malonyl CoA from acetyl CoA is catalyzed by the enzyme acetyl CoA carboxylase, the rate-limiting **enzyme of fatty acid synthesis catalyzes CH3-C-SCoA + ATP + CO2--> -O-C-CH2-C-SCoA + ADP + Pi OOO

Answer 184

Acetyl CoA carboxylase (ACC) its cofactor is biotin. Acetyl CoA carboxylase (ACC) is stimulated by citrate (note the many important roles of citrate in lipogenesis) and inhibited by the end product palmitoyl CoA. The enzyme is inhibited by both AMP-activated kinase, which phosphorylates the enzyme, and by glucagon/cAMP-PKA- dependent phosphorylation, which phosphorylates the enzyme on another site. AMP kinase, a low-energy signal, inhibits fatty acid synthesis. Insulin promotes dephosphorylation of ACC and markedly stimulates this rate-limiting enzyme of fatty acid synthesis.

Answer 185

First two steps of the PPP pathway. | - Malic enzyme under lipogenic conditions - Transhydrogenase

Answer 186

8Acetyl CoA + 7ATP + 14NADPH + 14H+ + 7H2O  | Palmitoyl CoA + 7CoA + 14NADP+ + 7ADP + 7Pi

Answer 187

Citrate->acetylCoA->malonylCoA->palmitate->palmytolCoA

Answer 188

Malonyl CoA, needed for fatty acid synthesis, is a powerful inhibitor of the acyl carnitine transferases which bring fatty acids into the mito for β-oxidation (lecture 11). Hence, fatty acid synthesis and oxidation do not occur at the same time.

Answer 189

Cholesterol is component of biological membranes, usually providing some order and rigidity to membrane structure. In liver, cholesterol is a precursor of the bile salts which are necessary to emulsify and digest fats. In endocrine tissues, cholesterol is a precursor of the steroid hormones, e.g., glucocorticoids, mineralocorticoids, sex hormones; vitamin D.

Answer 190

Cholesterol is synthesized from acetyl CoA; hence all nutrients – glucose, fatty acids, amino acids (even alcohol) – can provide carbon atoms for cholesterol biosynthesis.

Answer 191

Cholesterol synthesis occurs in the cytosol and the endoplasmic reticulum

Answer 192

Acetyl CoA->Acetoacetyl CoA-> HMGCoA->Mevalonate->isopentyl pyrophosphate-> squalene-> lanosterol-> cholesterol.

Answer 193

HMGCoA reductase, the rate-limiting enzyme in cholesterol synthesis. This enzyme is the site of inhibition of the statin drugs, e.g., Lipitor, Simvastatin, which are very commonly prescribed to lower cholesterol levels.

Answer 194

Isopentyl PP can produce other important products such as ubiquinone (Q10), the carbohydrate dolichol found in membrane glycolipids, and vitamins E and A.

Answer 195

This transcription factor binds to the sterol regulatory element of the HMGCoA reductase gene and activates the gene when cholesterol and other sterols are low; i.e., SREBP-1 senses cellular cholesterol and sterol levels. SREBP-1 is strongly activated by insulin. Hence, insulin increases cholesterol synthesis by activating transcription of HMGCoAR and by maintaining the HMGCoAR enzyme in the active dephosphorylated state. Cholesterol, besides inhibiting HMGCoA reductase synthesis, also promotes degradation of the HMGCoA reductase enzyme (decreases the half-life). Therefore, high levels of cholesterol will decrease its own synthesis by deceasing HMGCoAR transcription and protein stability. Conversely, low cholesterol will elevate its synthesis by not down regulating HMGCoAR transcription or stability. Cholesterol itself will feed back inhibit transcription of the HMGCoA reductase gene. It does this by preventing the processing and nuclear translocation of the transcription factor SREBP-1.

Answer 196

Physicians test for their patient’s LDL and HDL levels as a rough index of risk of atherosclerosis. HDL cholesterol is the “good” cholesterol, as HDLs transport cholesterol from the blood to the liver for metabolism. LDL cholesterol is the “bad” cholesterol since LDLs contain high amounts of cholesterol. Current guidelines recommend total cholesterol level in blood of less than 200 mg/dl, LDL levels less than 130 (preferably less than 100) mg/dl, and HDL levels more than 40 mg/dl, the higher the better. Exercise, the B vitamin niacin and small amounts of alcohol generally elevate HDL levels for reasons that are not clear.

Answer 197

``` PFK (active) Pyruvate Kinase (active) Glycogen Synthase (active) PDH (active) Acetyl CoACarboxylase (active) ``` ``` Phosphorylase Kinase (inactive) Glycogen phosphorylase (inactive) ``` This leads to ACTIVE Glycolysis Fatty Acid Synthesis Glycogen Synthesis

Answer 198

``` Glycogen phosphorylase (active) Phosphorylase Kinase (active) ``` ``` PFK (inactive) Pyruvate Kinase (inactive) Glycogen Synthase (inactive) PDH (inactive) Acetyl CoACarboxylase (inactive) ``` This leads to ACTIVE Glycagen breakdown Beta oxidation of fatty acids Gluconeogenesis

Answer 199

Insulin can maintain HMGCoA reductase in a dephosphorylated state, keeping it active. Glucagon phosphorylates the enzyme therefore deactivating it. Insulin also activates the transcription factor SREPB-1 which activates transcription of HMGCoA reductase

Answer 200

An additional problem is that accumulated LDLs in the blood are subjected to oxidation of their phospholipids by oxygen radicals (to be discussed later) to produce oxidized LDLs. These oxidized LDLs are recognized as “foreign” by macrophages of the immune system, which take them up via macrophage scavenger receptors. The accumulation of oxidized LDLs in macrophages results in swelling and engorgement with cholesterol and other lipids and conversion of macrophages to foam cells. Foam cells clog up the endothelial space of blood vessels producing fatty streaks and atherosclerotic plaques, which are among the first signs of atherosclerosis.

Answer 201

Resins alter bile salt binding causing more bile salts to be excreted from the body. this drives the process in the direction of producing more bile salts which removes some of the unneeded cholesterol.

Answer 202

They serve to inhibit HMGCoA reductase, the rate limiting step in cholesterol synthesis.

Answer 203

The mitochondrial electron transport chain (respiratory chain) oxidizes NADH and FADH2 via a series of enzymes embedded in the inner membrane. Electrons are transferred (blue arrows, Fig. 2) via ubiquinone (coenzyme Q) and cytochromes to molecular O2, which is reduced to H2O. Energy from the oxidation of NADH and FADH2 by O2 is used to produce ATP from ADP + Pi by the ATP synthase (FOF1 ATPase) complex associated with the respiratory chain.

Answer 204

This pumping of protons through complexes I (4H+ pumped per electron passing through), III (2H+ pumped/e) and IV (4 H+ pumped/e) generates an electrochemical gradient (ΔP) across the membrane. The ΔP is composed of a membrane potential because of pumping positive charges out of the mito and a proton gradient because of pumping protons out of the mito.

Answer 205

The ATP synthase contains a proton pore through the inner membrane and a catalytic head-piece that protrudes into the matrix.

Answer 206

As protons that were pumped out during electron transport to O2 (respiration) are re-entering via the proton pore of the ATP synthase complex, there is a conformational change in the catalytic headpiece which releases ATP bound to one site, while catalyzing the formation of a new ATP from ADP plus Pi at another site. This newly formed ATP will be transferred to the bound site and will be released by the next proton entering the proton pore.

Answer 207

In this model, the transfer of electrons and pumping of protons is coupled to the synthesis of ATP, i.e., oxidative phosphorylation. The rate of respiration (transfer of electrons, reduction of O2) is regulated by the rate of ATP synthesis, by the ΔP, or by other bioenergetic work functions which utilize the ΔP (discussed below). If the ΔP is not utilized, the rate of electron transfer or O2 uptake will decrease because of back pressure exerted on the respiratory chain by the high accumulated ΔP. The faster the ΔP is utilized for ATP synthesis or bioenergetic work functions, the faster is the rate of electron transfer in order to keep generating the ΔP.

Answer 208

If the mito membrane becomes damaged and leaky to protons, or if certain chemicals called uncoupling agents carry protons in and out of the mito, a ΔP will be immediately dissipated as the pumped-out protons non-specifically re-enter the mito rather than enter through the proton pore of the ATP synthase.

Answer 209

In fact, there is a group of uncoupling proteins that form channels through the mito inner membrane for protons to pass from the intermembrane (cytosolic) space to the matrix, thus short-circuiting the ATP synthase. Uncoupling protein 1 (UCP1) is associated with heat production in brown adipose tissue, important for infants to maintain body temperature and for animals that hibernate. UCP3 in skeletal muscle has been a drug target for obesity, the idea being to find activators of UCP3 to “waste” the ΔP as heat rather than use it to support biosynthetic reactions such as fat synthesis.

Answer 210

NADH dehydrogenase complex. NADH is oxidized by complex I, the NADH dehydrogenase complex, which contains iron-sulfur (FeS) proteins and FMN (flavin mononucleotide). Electrons are passed from NADH  FMN  FeS Centers  ubiquinone (Q) to produce reduced Q, QH2.

Answer 211

It pumps out 4 H+ protons and produces coenzyme Q

Answer 212

Complex II is succinate dehydrogenase, the TCA cycle enzyme, and contains FAD. Electrons from succinate are passed to FAD to Q, producing QH2. Similarly, FADH2 produced by β-oxidation of fatty acids (lecture 11) and αGP (αGP shuttle) are passed on to Q via an electron transfer flavoprotein.

Answer 213

Complex III is called the b:c1 complex and it pumps protons. QH2 passes one electron at a time to Cyt b (Fe 3+) to produce reduced cyt b (Fe2+), which in turn reduces the FeS, which reduces cyt c1 which then reduces Cyt c (Fe3+) to Cyt c (Fe2+). Protons are pumped out by the b:c1 complex. QH2 -> cyt b -> FeS -> Cyt c1 -> Cyt c. 2 H+ protons are pumped out. Just a note-cytochome C is a peripheral protein on the outside of the inner mito membrane.

Answer 214

Complex IV is the cytochrome oxidase complex, which is made up of cytochromes a and a3 and two major copper centers, Cu A and Cu B. Reduced cyt C passes one electron at a time through Cyt oxidase, reducing Cu A, then Cyt a, then CuB and then Cyt a3. Four electrons are needed to reduce molecular O2 to H2O: O2 + 4H+ + 4e-  2H2O. Note that four electrons pass through Complex IV to reduce O2 to H2O, and that protons are pumped by the cut oxidase complex. Cyt c→CuA→Cyt a→CuB→Cyt a3→OXYGEN

Answer 215

Rotenone. A ΔP cannot be maintained so ATP will not be synthesized.

Answer 216

Antimycin A. With antimycin A, QH2 accumulates and Cyt c is not reduced.

Answer 217

cyanide and carbon monoxide are inhibitors of Complex IV. reduced Cyt C accumulates.

Answer 218

As mentioned above, respiratory control refers to the control of oxygen uptake by the ΔP. Basically, it refers to the rate of O2 uptake in the presence of a respiratory chain substrate plus ADP and Pi, relative to the rate in the presence of substrate only . O2 uptake in the presence of substrate alone is usually low since the ΔP that is generated is not being used, whereas O2 uptake in the presence of substrate plus ADP + Pi is rapid because the ΔP is being used to produce ATP.

Answer 219

In ETC, the FADH2 & NADH are oxidized to produce H+ which are combined with Oxygen to produce H2O & the energy produced is utilized to produce ATPs. Uncouplers let the oxygen combine with H+ without the formation of ATP, so there is no ATP production but all the other reactions keep on going. Dec. ATP production will also inc. the rate of reactions of Glycolysis & TCA cycle

Answer 220

The product of the previous irreversible step builds up.

Answer 221

If it's a PDH deficiency that can be any of the enzymes. It it's an E1 deficiency TPP can help.

Answer 222

Pyruvate is a very important enzyme. It is the first step of gluconeogenesis and also allows OAA replenishment for the TCA cycle through the PEP pathway. You could treat with a high glucose diet so they don't rely on gluconeogenesis. A high aspartate diet could also be tried because that will generate OAA.

Answer 223

Caffeine is a PDE inhibitor. PDE's cleave the cAMP--PKA complex. Inhibiting this would render it constitutively active.

Answer 224

Glycogen synthase is allosterically stimulated by G6P. Therefore glycogen synthase is activated when elevated glucose leads to increased intracellular G6P. In the liver, the cAMP PKA pathway leads to phosphorylation/inactive Glyc. Syn. which in turn leads to glucose to be released into the blood. Eventually that glucose will lead to increased cytosolic G6P. The binding of that G6P changes it's structure a little to allow for dephospho rylation. this mechanism essentially serves as a way to maintain blood glucose homeostasis.

Answer 225

Catalyze the dysmutation convertion of two superoxides into H2O2.

Answer 226

Fe2+ +H2O2 → Fe3+ +OH- +.OH | essentially an Fe catalyzed HW reaction. Fe is replenished with help of superoxide.

Answer 227

ROS, especially the hydroxyl radical, oxidize the SH group of cysteine residues of proteins to the disulfides (Fig. 5) or to the sulfoxide (SO) or the sulfonic acid (SOOH). Enzymatic activity dependent on cysteine will be lost or protein tertiary structure will be disrupted. The CH3S of methionine is readily oxidized to the methionine sulfide – CH3SO – interfering with functions of methionine such as methylation reactions ROS oxidize the aromatic rings of Phe and Tyr or the indole ring of Try, open up the rings, form protein carbonyls (Fig. 7) and cause loss of enzymatic activity and structure. ROS can cleave the peptide linkages holding amino acids together in proteins disrupting and cleaving proteins.

Answer 228

ROS cleave the phosphodiester bonds holding bases in RNA and DNA together and break the chain structure of RNA and DNA. ROS oxidize the purine and pyrimidine bases and prevent appropriate base pairing. ROS can cause deaminations, e.g., remove the amino group from adenine or guanine to form hypoxanthine (Fig. 9) or xanthine, respectively, or remove the amino group from cytosine to form uracil.

Answer 229

Polyunsaturated fatty acids are a major target for oxidation by ROS, a process called lipid peroxidation (LP), which disrupts normal membrane structure. Rancidity of foods is largely due to LP as they age. LH +.OH→ L.+ H2O The hydrogens on the carbon 11 methylene (CH2) group are very sensitive to oxidation and abstraction by ROS, resulting in the production of a lipid radical (L.) on carbon 11: CH3-(CH2)4-CH=CH-CH.-CH=CH-(CH2)7-COOH While several reactions may now occur, a most favorable one is reaction of the L. with O2 to produce the lipid alkoxyl radical LO2.. The lipid alkoxyl radical can interact with a second polyunsaturated fatty acid, e.g., linoleic or linolenic or arachidonic acid or others, to abstract an H and form the lipid hydroperoxide and a new second lipid radical: LOO. +“LH→LOOH+“L.. The second lipid radical can, like the first, react with O2 to form a lipid peroxyl radical (“LOO.) and repeat the above. These are called the propagation steps of LP (Fig. 10B). Note that many molecules of polyunsaturated fatty acids can be oxidized by just one initiating hydroxyl radical. Fig. 10A

Answer 230

Vitamin E (α-tocopherol) is the answer, and is the major antioxidant promoting termination of LP (Fig. 10D).

Answer 231

Remarkably, there is a cell type which is designed to produce ROS to protect against infection by invading organisms and tumor cells. Phagocytes, a type of WBC, contain the enzyme NADPH oxidase, which is activated when the phagocytes come into contact with bacteria and other cells, e.g., tumor cells. This activation is called the respiratory burst, since oxygen uptake is increased because the NADPH oxidase catalyzes the following reaction: NADPH + O2→ NADP+ + O2.- The rapid and large production of superoxide and other ROS kills the invaders. (The PPP pathway is activated during the respiratory burst Why?)

Answer 232

Catalase, present in the peroxisomal fraction of the cell (as are many oxidase enzymes which produce H2O2), catalyzes removal of two moles of H2O2 (Fig. 12). Catalase, a heme-containing enzyme, has a relatively high Km for H2O2 and a very high Vmax. 2H2O2 catalase 2H2O + O2 The glutathione peroxidase system is found in the cytosol fraction and in the mito, and functions in conjunction with GSH, NADPH and glutathione reductase (Fig. 13). 2 GSH + H2O2 → GSSG + 2H2O Fig. 13 To regenerate GSH, GSSG is reduced by NADPH and glutathione reductase: GSSG+ NADPH→ 2GSH + NADP

Answer 233

One reason for excessive ROS formation is that during ischemia, the mito electron transfer chain is in a highly reduced state because of the lack of oxygen. Once reperfusion is initiated and oxygen re-introduced, there is a burst of ROS from, e.g., reduced ubisemiquinone, which may overwhelm tissue antioxidant defense. A second mechanism for reperfusion injury involves the enzyme xanthine dehydrogenase (XDH), usually called xanthine oxidase (XO). This enzyme is important for purine breakdown (to be discussed); it oxidizes the purine bases xanthine and hypoxanthine (which are derived from AMP and GMP) to uric acid for excretion. Normally, the enzyme is in the dehydrogenase form (XDH) and uses NAD+ (and not oxygen) for this oxidation, producing NADH. Under certain conditions, there is a conversion of the enzyme from the dehydrogenase form to the oxygenase form, xanthine oxidase (XO). This form can use oxygen, in preference to NAD, to oxidize xanthine and hypoxanthine to uric acid, but now the other product is superoxide, not NADH. During ischemia, XDH is converted to the XO form because certain calcium-activated proteases are increased, since calcium floods the ischemic tissue due to a lack of ATP production that prevents calcium from being pumped out of the cell. Also, ATP is not made due to the lack of oxygen, and it breaks down to hypoxanthine while GTP breaks down to xanthine. So we now have accumulation of these substrates and of the XO form of the enzyme. What is missing? The other substrate for XO: oxygen. Reperfusion provides oxygen and again the tissue is flooded with a burst of ROS via the XO reaction (Fig. 15).

Answer 234

Basically the medium/blood used to reperfuse the ischemic tissue is loaded with antioxidants, especially GSH and vitamins E and C. It also contains chelators of iron such as desferrioxamine (why?). It also contains an inhibitor of XO and XDH called allopurinol (to be discussed in our gout lecture). This knowledge on reperfusion injury has been of tremendous benefit in clinical medicine for treatment of infarcts, 13 stroke, and transplantation of organs. Preservation fluids for organs to be used for transplants are loaded with the antioxidants mentioned above, which highly improves organ function and graft survival.

Answer 235

Healthy adults are in nitrogen balance, meaning that the amount of nitrogen consumed, mainly in the form of protein, is equivalent to the amount of nitrogen excreted, mainly as urea (Table 2). Positive nitrogen balance reflects conditions of growth and development, where nitrogen intake exceeds nitrogen excretion. Negative nitrogen balance occurs when nitrogen excretion exceeds nitrogen intake. This occurs during illness, infection and catabolic stress, inadequate intake of protein in the diet, or intake of a protein diet deficient in essential amino acids.

Answer 236

Transaminations allow removal of the alpha amino group from an amino acid, or allow transfer of an alpha amino group from an amino acid to an alpha-keto acceptor. This results in production of a new amino acid from the alpha keto acceptor and a new alpha keto acceptor from the original amino acid.

Answer 237

GOT reaction: glutamate + oxaloacetate → αKG + aspartate

Answer 238

glutamate + pyruvate → αKG + alanine

Answer 239

All transaminases, and many other enzymes involved in amino acid metabolism, use pyridoxal phosphate (PLP) as the cofactor. Pyridoxal P is derived from vitamin B6, pyridoxine (Fig. 5).

Answer 240

Excess amino acids are metabolized to pyruvate, TCA cycle intermediates, and acetyl CoA. The alpha amino groups of the various amino acids are collected into one common pool, glutamate, by the transamination reactions discussed in the preceding lecture: R-CH-COOH + αKG --> R-CH-COOH + glutamate NH2 O

Answer 241

Glutamate will then release these amino groups as ammonia, and the ammonia will be converted to urea in the liver by the urea cycle pathway. Most amino acid metabolism occurs in the muscle and liver. The glutamate produced by transamination in the muscle will leave muscle and circulate to the liver, where it mixes with glutamate produced by liver transamination reactions. In the glutamate dehydrogenase reaction (Fig. 1), which uses NAD+ or NADP+ as a cofactor, glutamate is oxidized to αKG + NH3 + NAD(P)H.

Answer 242

Glutamate + NAD(P)+ --> αKG + NAD(P)H + NH3.

Answer 243

Besides transamination, some amino acids such as serene, histidine, threonine and cysteine, can break down by other pathways that ultimately produce NH3.

Answer 244

Ammonia produced in muscle must also get to the liver. Free circulation of ammonia in the blood is a no-no, since NH3 is toxic to the brain. Instead, the muscle ammonia reacts with glutamate to produce glutamine, by the irreversible glutamine synthase reaction (Fig. 2) Glutamate + ATP + NH3 → Glutamine + ADP + Pi In liver, the enzyme glutaminase hydrolyzes glutamine back to glutamate plus NH3 (Fig. 3), and this NH3 enters the urea cycle.

Answer 245

Steps 1 and 2 occur in the mitochondria: formation of carbamyl phosphate (CP) and formation of citrulline from CP and ornithine, an amino acid not found in proteins. The citrulline leaves the mito, and steps 3-5 occur in the cytosol. Arginine is eventually produced from the interaction between citrulline and aspartate, with argininosuccinate as an intermediate. In the last step, arginase hydrolyzes arginine to ornithine and urea. The urea is excreted in the urine, and ornithine, used in step 2 to produce citrulline, is regenerated – hence, urea cycle – and enters the mito for the next round.

Answer 246

ammonia-->CP-->Ornithine-->Citrulline-->aspartate-->arginosuccinate-->fumarate + arginine--> Urea + ornithine.

Answer 247

catalyzes the formation of CP in the urea cycle. It is activated by N-acetylglutamate, NAGA, whose synthesis from glutamate plus acetyl CoA is activated by arginine (Fig. 6).

Answer 248

Thus, four high energy bonds are needed to produce one mole of urea (two for CPSI and two for formation of argininosuccinate).

Answer 249

Note: The aspartate needed for step 3 can also come from glutamate via the GOT reaction: OAA + glutamate aspartate + αKG. Hence, the alpha amino group of almost any amino acid can be collected via transamination as glutamate, then wind up in aspartate via GOT, then wind up in the second NH2 of urea. In other words, aspartate is the direct donor of the second NH2 of urea, but aspartate could have gotten its alpha amino group from any other amino acid via GOT.

Answer 250

The urea cycle can be linked to the TCA cycle. The fumarate produced in step 4 can enter the TCA cycle and be converted to malate and then OAA. The OAA can then be converted to aspartate via GOT, as discussed above. Thus, aspartate is also regenerated

Answer 251

There are inborn deficiencies of each of the five enzymes of the urea cycle. The most common is deficiency of OTC wh. Deficiency of a urea cycle enzyme is associated with ammonia toxicity as NH3 cannot be converted to urea for excretion. Symptoms include GI-tract irritability, nausea, vomiting, lethargy, and, depending on the extent of the deficiency, neurological disturbances, mental retardation, seizures, coma and death. It is critical for the pediatrician to assay for blood ammonia levels early in life.

Answer 252

Among the reasons are: interference with blood pH (since NH3 is basic), depletion of αKG out of the TCA cycle via the glutamic dehydrogenase reaction, and formation in excess of the excitatory neurotransmitter glutamine via the glutamine synthase reaction.

Answer 253

- Low protein diets to minimize excess amino acids. - Clean out the gut – remove bacteria and yeast which produce lots of NH3. - Gene therapy, especially for OTC deficiency, is a hot area of research. - Use drugs such as benzoate or phenylacetate which can react with glycine or glutamine and remove these amino acids

Answer 254

liver function/urea cycle function.

Answer 255

kidney function since the kidney filters and removes urea.

Answer 256

Blood GOT and GPT levels are a general indication of tissue damage as these very active transaminases leak out of the injured tissue to the blood. They are not very tissue-specific, although they are present at highest levels in the liver.

Answer 257

methylating agent is not methionine but S- Adenosylmethionine (SAM), which is produced in the methionine adenosyl transferase reaction. it's the major biological methylating agent. basically methionine plus ATP

Answer 258

Epi, creatine, melatonin, etc.

Answer 259

When SAM gives up its methyl group, it is converted to S-Adenosyl homocysteine (SAH) and SAH is hydrolyzed to adenosine plus homocysteine.

Answer 260

Homocysteine can be converted back to methionine by accepting a methyl group from the vitamin cofactors N5 -methyl tetrahydrofolate and B12 (cobalamins). This reaction is inhibited by methionine.

Answer 261

Tetrahydrofolate functions as a single carbon carrier in this reaction (next lecture). In this case, it is carrying a methyl group on nitrogen 5 of its structure, N5-CH3 THF. The N5-CH3 THF first transfers the methyl group to vitamin B12 (cobalamine) to produce methyl cobalamine. The methyl cobalamine then transfers the methyl group to homocysteine to regenerate methionine.

Answer 262

If sufficient methionine is available, there is little need to regenerate more methionine from homocysteine. Instead, homocysteine can enter the transsulfuration pathway whereby the amino acid cysteine can be synthesized. Homocysteine reacts with serine to form cystathionine as catalyzed by cy- tathionine synthase, a PLP requiring enzyme. This enzyme is feedback inhibited by high levels of cysteine. Cystathionine is then hydrolyzed by cystathionase, also a PLP-requiring enzyme, to cysteine plus α- ketobutyrate. Basically, serine is picking up the sulfur of homocysteine, which was originally derived from methionine, to now become cysteine.

Answer 263

Methionine, valine, isoleucine and threonine. Odd chain fatty acids are also degraded to propionyl CoA.

Answer 264

Vitamin B12.

Answer 265

Conversion of homocysteine to methionine and conversion of propionyl CoA eventually to succinyl CoA are the two mammalian reactions requiring vitamin B12

Answer 266

succinyl CoA which can be further oxidized in the TCA cycle for energy, or metabolized to oxaloacetate and eventually to glucose during gluconeogenesis, or can produce heme

Answer 267

Homocystinuria is a metabolic inborn error of metabolism in which large amounts of homocysteine accumulate. This causes neurological disorders, failure to grow and to thrive, skeletal abnormalities, and cardiovascular abnormalities. Normal serum levels of homocysteine are about 10 μM. Levels rise to about 20 to 40 μM with deficiencies of vitamins B6 or B12 or THF, and are markedly elevated (> 200 μM) with deficiencies in cystathionine synthase or homocysteine methyl transferase. Part of the toxic effects of high levels of homocysteine are related to elevated oxidative stress and to smooth muscle cell proliferation. What are some reasons why homocysteine may accumulate

Answer 268

Methionine synthase (homocysteine->met) N5N10Methylene FH4 reductase (can't turn homo back to met either, basically can't transfer the methyl group) Cystathione-B-synthase (Homocys->cysteine)

Answer 269

Note that cysteine is one of the three amino acids that make up GSH. Thus, large amounts of cysteine are needed not only for protein synthesis but also for GSH synthesis. The transsulfuration pathway is thus important as a protective pathway against oxidative stress via cysteine formation.

Answer 270

Glycoprotein, Intrinsic factor (IF), needed for absorption of B12. ↓Intrinsic factor → pernicious anemia, essentially due to disturbance of folate metabolism as accumulate N5CH3 THF if no B12.

Answer 271

MMA- GI irritation, vomit, CNS disturbances, convulsions, mental retardation, death Due to↓MMA mutase, ↓B12 ↓IF, ↓transferase which makes deoxyadenosyl B12 Treatment -↓Met, Val, The, Isol in diet- all essential aa so hard.

Answer 272

Folic acid, derived from the diet or ingested as a vitamin, is reduced by dihydrofolate reductase to dihydrofolate (DHF or FH2), and DHF is reduced by dihydrofolate reductase to tetrahydrofolate (THF or FH4; Fig. 1), which is the active cofactor form. THF functions in one-carbon metabolism as it transfers single carbon atoms in several critical biochemical reactions. The carbon atoms used are the methyl group (-CH3), the methylene group (-CH2), and the methenyl group (=CH). The intermediate N5-CH2-N10 methylene THF can be used directly, or it can be reduced to the N5-CH3 THF (by N5-N10 methylene THF reductase), or it can be oxidized to N5-HC=N10 methenyl THF. N5-CH3THF  N5-CH2- N10THF N5-CH=N10 THF The N5-CH=N10 methenyl THF can be used directly or reduced to the N5-CH2- N10 methylene THF. However, note that the N5CH3-THF cannot be oxidized to can be reduced or oxidized These four are all the same oxidation state – single H around the carbon atom. the N5-CH2-N10 THF. The significance of this to the problem of pernicious anemia will be discussed in lecture.

Answer 273

serine, glycine and histidine.

Answer 274

Note that reactions (a) and (b) above show the interrelationship between serine and glycine with each other as mediated by THF, i.e., serine can produce glycine (reaction a to the right) or glycine can produce serine (reaction a to the left). Also, one serine can produce two N5-CH2-N10 THF by combining reactions (a) and (b).

Answer 275

PLP acts as a coenzyme in all transamination reactions, and in certain decarboxylation, deamination, and racemization reactions of amino acids.

Answer 276

- N5CH3THF regenerates methionine from homocysteine, as discussed in the previous lecture. - N5-HC=N10 methenyl THF (or N5 formyl THF) is used to provide two of the carbon atoms in the synthesis of purine rings (discussed later). - N5-CH2-N10 methylene TNF is used to synthesize dTMP from dUMP. Recall that DNA contains thymidine while RNA contains uracil. In pyrimidine synthesis pathway (discussed in a later lecture), UMP is produced. Fig. 4 shows how UMP (the deoxy form is called dUMP) uses N5-CH2-N10 methylene THF to produce TMP (dTMP). During this reaction, THF is oxidized to DHF: dUMP + N5–CH2-N10 THF → dTMP + DHF Since the active form of folic acid is the THF form, DHF has to be reduced back to THF. This is accomplished by NADPH plus DHF reductase. DHF + NADPH → THF + NADP+ During this reaction, THF is oxidized to DHF: dUMP + N5–CH2-N10 THF → dTMP + DHF Since the active form of folic acid is the THF form, DHF has to be reduced back to THF. This is accomplished by NADPH plus DHF reductase. DHF + NADPH → THF + NADP+

Answer 277

purine synthesis starts with an activated form of ribose called 5- phosphoribosyl-1-pyrophosphate (PRPP), which is synthesized from ribose 5-P and ATP as catalyzed by PRPP synthase (or ribose P pyrophosphokinase). Recall that ribose 5-P is produced from G6P in the pentose cycle. PRPP synthase is feedback inhibited by IMP, AMP, and GMP.

Answer 278

The rate-limiting step of purine biosynthesis is formation of 5-phosophoribosyl-1-amine from PRPP + glutamine, as catalyzed by amido phosphoribosyltransferase (Fig. 2). This enzyme is product inhibited by IMP, AMP, and GMP.

Answer 279

IMP is the first purine to be produced by this de novo purine synthesis pathway. IMP contains the base hypoxanthine, and of course is not found normally in RNA or DNA.

Answer 280

AMP can inhibit step 1 in its synthesis, thereby driving IMP to GMP formation. Similarly, GMP can inhibit step 2 in its synthesis, thereby driving IMP to AMP. Also note that AMP synthesis requires GTP as a source of energy while GMP synthesis requires ATP. These regulatory considerations allow a balance between AMP and GMP synthesis to occur.

Answer 281

With the addition of ATP and glutamine it makes xanthylate which eventually becomes GMP. This process is inhibited by GMP.

Answer 282

With the addition of GTP and aspartate this gives rise to adenylosuccinate which eventually gives rise to AMP. This is inhibited by AMP.

Answer 283

Purine bases not salvaged are excreted as uric acid. Adenine is ultimately converted to hypoxanthine, which is then oxidized by xanthine oxidase to xanthine. Guanine is ultimately also converted to xanthine. Xanthine is then oxidized to uric acid by xanthine oxidase, and uric acid is excreted into the urine. Xanthine oxidase, when the enzyme is in its oxidase form, produces superoxide radical and H2O2 during these reactions. Recall our discussion on ischemia-reperfusion injury in the oxygen radical lecture.

Answer 284

Excess uric acid in the serum will form crystalline, insoluble deposits of calcium and sodium urate in the joints, which are very painful and cause inflammation. This condition is called gout. It occurs with diets high in purines, e.g., liver, in individuals with deficiencies in the salvage enzymes (why?), and can be precipitated by alcohol and by ketoacidosis.

Answer 285

Basic treatments are to inhibit xanthine oxidase with the drug allopurinol (Fig. 7), an analogue of hypoxanthine. This causes hypoxantine and xanthine, which are much more soluble than uric acid, to be excreted instead of uric acid. Drugs like colchicine and methotrexate are also used since they block inflammatory reactions and the immune system which become activated in response to the high levels of uric acid.

Answer 286

The amino acids aspartate and glutamine are required for pyrimidine (UMP, CMP) synthesis. Glutamine is used to produce carbamyl phosphate via CPSII, which is product inhibited by UMP or UTP. Aspartate reacts with carbamyl P to form N-carbamyl aspartate as catalyzed by aspartate transcarbamylase, the rate-limiting enzyme in pyrimidine synthesis.

Answer 287

Aspartate + CP --> N carbamyl aspartate this is catalyzed by aspartate transcarbamylase. This enzyme is product inhibited by CTP and stimulated by ATP (a purine, which stimulates pyrimidine synthesis – note the balance). Note orotic acid (not erotic acid), an intermediate in the pathway. High levels of orotic acid in blood can be due to decreases in the urea cycle enzyme ornithine transcarbamylase. Why?

Answer 288

Glutamine+Co2->CP (by way of carbonyl phosphate) + aspartate-----> orotate---> UMP->UDP->UTP... goes to RNA. Can then move on to synthesize DNA (dCTP and dTTP) from CTP.

Answer 289

CTP->CDP->dCDP (using rr) dCDP can either become dCTP or move to dUMP->dUMP->dTMP->dTDP->dTTP

Answer 290

Ribonucleotide reductase contains several binding sites, the active site where catalysis takes place, an activity site, and a substrate Fig.3 specificity site (Fig. 3). When ATP binds the activity site, the enzyme is activated; when deoxy ATP binds this site, the enzyme has low activity. Thus, ATP, a ribonucleotide, is stimulating synthesis of deoxyribonucleotides, while dATP, a deoxyribonucleotide, is preventing more synthesis of deoxynucleotides.

Answer 291

The enzyme also contains a substrate specificity site, which dictates which ribonucleotide diphosphate will be converted to a deoxyribonucleotide diphosphate at the enzyme’s active site. Consider: When ATP is high, the enzyme is turned on. If ATP, a purine, binds to the substrate specificity site, this will specify that pyrimidine substrates should be converted to the deoxypyrimidines: CDP  dCDP and UDP dUDP. dCDP and dUDP is eventually metabolized to dCTP + dTTP by nucleoside diphosphate kinase. As the deoxypyrimidines build up, dTTP displaces ATP from the substrate specificity site and now specifies formation of the deoxypurine dGDP: GDP → dGDP. dGDP is converted to dGTP, which slows down formation of deoxypyrimidines: CDP dCDP; UDP dUDP; dTTP Once enough dGTP is produced, it displaces dTTP from the substrate specificity site and now specifies that the other deoxypurine should be produced: ADP dADP, and dATP and dGTP production should be decreased:  GDP dGDP  dGTP. Once enough dATP accumulates, it is a signal that enough deoxynucleotides have been produced. The dATP displaces ATP from the overall activity site and shuts ribonucleotide reductase down. Note the exquisite balance between formation of deoxypyrimidines and deoxypurines.

Answer 292

ADA deficiency is due to a lack of the enzyme adenosine deaminase. This deficiency results in an accumulation of deoxyadenosine,[5] which, in turn, leads to: a buildup of dATP in all cells, which inhibits ribonucleotide reductase and prevents DNA synthesis, so cells are unable to divide. Since developing T cells and B cells are some of the most mitotically active cells, they are highly susceptible to this condition. an increase in S-adenosylhomocysteine since the enzyme adenosine deaminase is important in the purine salvage pathway; both substances are toxic to immature lymphocytes, which thus fail to mature. Because T cells undergo proliferation and development in the thymus, affected individuals typically have a small, underdeveloped thymus.[6] As a result, the immune system is severely compromised or completely lacking.

Answer 293

Since amino acids cannot be stored, excess amino acids are degraded by all tissues but especially the liver and muscle. The carbon products of these degradations are familiar and include acetyl CoA, acetoacetate, pyruvate, and the TCA cycle intermediates OAA, αKG, succinyl CoA, fumarate. These products are how the amino acids provide energy. The amino groups are of course converted to urea.

Answer 294

All but one amino acid, leucine, and to a large extent lysine are glucogenic.

Answer 295

leucine, lysine, phenylalanine, tyrosine, isoleucine and tryptophan (and to a limited extent threonine).

Answer 296

phenylalanine, isoleucine, tryptophan, and tyrosine.

Answer 297

3 family includes five amino acids – alanine, serine, glycine, cysteine, and tryptophan – which break down to pyruvate.

Answer 298

The C4 family consists of two amino acids: Aspartate is converted to OAA via the GOT transaminase reaction. Asparagine is hydrolyzed to aspartate plus ammonia by asparaginase.

Answer 299

The C5 family contains five amino acids: glutamate, glutamine, histidine, proline and arginine. These are all metabolized to αKG as shown. The latter four amino acids are converted first to glutamate. Glutamate is oxidized to αKG by the glutamate dehydrogenase reaction:

Answer 300

valine, isoleucine, threonine and methionine, These 4 amino acids are eventually metabolized to propionyl CoA. The propionyl CoA is then converted to D- and then L-methylmalonyl CoA, which produces succinyl CoA in the methylmalonyl CoA mutase reaction. This was covered in the methylation/B12 lecture. Recall the inborn error of metabolism, methylmalonyl aciduria, which is due to a deficiency of the mutase or of B12 or of the deoxyadenosyl transferase.

Answer 301

The acyl CoA dehydrogenase is deficient in the disease known as maple syrup urine disease. As a result, the original three branched-chain amino acids accumulate, as do the alpha keto acids, which were produced by the transamination reaction (some of these are excreted and the urine has an odor like maple syrup). These compounds are very toxic to the developing nervous system and cause severe neurological disturbances, failure to thrive, coma, and, if not treated, death within the first 2 years of life. How would you treat a child diagnosed with maple syrup urine disease?

Answer 302

Tyrosine itself or tyrosine produced from phenylalanine is eventually metabolized to fumarate and acetoacetate (Fig. 6). Hence these two amino acids are glucogenic and ketogenic.

Answer 303

Tryptophan is metabolized to alanine, acetyl CoA and ammonia, so it is glucogenic and ketogenic (Fig 9).

Answer 304

The cytosol of all tissues.

Answer 305

Glutamic acid plays the key role in the synthesis of many amino acids. It can be synthesized from αKG + NH3 + NAD(P)H by the reversible glutamic dehydrogenase reaction, first discussed in the urea cycle lectures (Table 1, #1). Note that this reaction incorporates free ammonia into the amino acid pool.

Answer 306

Can be synthesized from aspartate and glutamine via asparagine synthase (Table 1, #4). Note how glutamate can be used to produce aspartate and glutamine, which can react to produce asparagine.

Answer 307

Glycine can be produced from serine (Table 1, #8a) or it can be synthesized de novo from CO2 + ammonia + methylene THF (Table 1, #8b). The methylene could come from histidine or another serine. Note the relationship between serine and glycine as mediated via the folic acid pool. Reaction 8b is the basis of using benzoic acid to treat ammonia intoxication, as ammonia gets incorporated into glycine which then conjugates with benzoic acid to form a very soluble hippuric acid that is excreted.

Answer 308

Synthesis begins with the tyrosine hydroxylase reaction, the rate-limiting enzyme in the pathway. This enzyme is similar to phenylalanine hydroxylase, which converts phenylalanine to tyrosine. Note the use of THB as cofactor and oxygen, as the enzyme is a mixed function oxidase with one atom of oxygen incorporated into the product and one into water. The product is 3,4 dihydroxyphenylalanine, DOPA. DOPA is decarboxylated by DOPA decarboxylase to dopamine plus CO2. Dopamine is a critical neurotransmitter, helping to control coordination, motor control, and mood. Its deficiency, especially in the substantia nigra area of the brain, results in Parkinson’s disease. Treatment is to provide DOPA, which can cross the blood brain barrier, rather than dopamine, which, being positively charged at neutral pH, cannot. Dopamine becomes hydroxylated by dopamine β-hydroxylase to produce norepinephrine or 3,4 dihydroxy-phenethanolamine. Norepi is a critical vasodilator and regulator of cardiovascular tone and blood pressure. Vitamin C, ascorbate, is the cofactor for this copper-containing enzyme. Finally, norepi is methylated by SAM to produce epinephrine, as catalyzed by phenylethanolamine N-methyl transferase. Note the many cofactors utilized in catecholamine synthesis: THB, PLP, ascorbate, SAM.

Answer 309

THB, PLP, Ascorbate, SAM.

Answer 310

Glutathione (GSH) is produced from three amino acids – glutamic acid, glycine, and cysteine

Answer 311

seratonin and melatonin

Answer 312

glycine and SAM and part of arginine.

Answer 313

produced in one step from arginine by nitric oxide synthase

Answer 314

succinyl CoA and glycine,

Answer 315

Recall that succinyl CoA can be produced from four amino acids: valine, methionine, isoleucine and threonine via the methylmalonyl mutase-B12 reactions discussed previously.

Answer 316

Heme biosynthesis (Fig. 3) is a compartmentalized pathway, occurring both in the mito and in the cyto. The first step, the synthesis of ALA by ALA synthase (ALAS), occurs in the mito. This is the rate-limiting step of heme synthesis. Heme itself inhibits translation of ALA mRNA, activity of ALAS, and transfer of ALAS into the mito. ALA leaves the mito, and in the cytosol two moles of ALA condense to form a ring structure called porphobilinogen (PBG). The 5-membered ring structure shown (with 4 carbon atoms and one nitrogen) is called a pyrrole ring. This reaction is catalyzed by ALA dehydratase, a zinc-containing enzyme which is very sensitive to inhibition by the heavy metal lead. Four molecules of porphobilinogen condense to form the first porphyrin structure, uroporphrinogen III, as catalyzed by the enzymes porphobilinogen deaminase and uroporphyrinogen III synthase. The next two steps involved modifications of the side chain attachments to the porphyrin ring, mainly decarboxylations of the acetates to methyl groups to form coproporphyrinogen III and of two of the propionate side chains to vinyl groups to form protoporphyrinogen IX. The latter enters the mito where it is oxidized to protoporphyrin IX (note the alternating double bonds in the ring and the carbon bridge holding the ring together). The final step (yeah!) is to insert ferrous iron into protoporphyrin to produce heme (heme IX). This is catalyzed by ferrochelatase, an enzyme also very sensitive to lead. Thus, two steps in the heme biosynthesis pathway are inhibited by lead, making lead poisoning a major cause of anemia especially in children who may ingest lead paint chips. Heme is a potent product inhibitor of ferrochelatase.

Answer 317

heme breakdown, by the enzyme heme oxygenate, to biliverdin, iron and CO. Biliverdin is reduced by biliverdin reductase to bilirubin, much of which is conjugated to glucuronic acid and excreted in the bile (Fig. 6). We will discuss this in some detail, including the diagnostic significance of bilirubin.

Answer 318

Fuel Storage: We typically have about 1600 Kcal of energy stored as glycogen in liver (400 Kcal) and muscle (1200 Kcal), about 135,000 Kcal of energy stored as triglycerides in adipose tissue (could be lots more in today’s society), and about 24,000 Kcal of energy stored as protein, largely skeletal muscle protein (Table 1). We typically expend and consume about 2000-3000 Kcal of energy per day.

Answer 319

Amino acids (#14 in Fig 2) are used to synthesize new proteins to replace proteins lost by degradation and turnover. Some amino acids are used to synthesize purines, pyrimidines, heme, GSH, etc. Some amino acids are oxidized to produce about 15 to 20% of our necessary energy requirements. These reactions occur in all tissues but predominate in liver and muscle. Triglycerides (#2 in Fig. 2), present in the diet as chylomicrons, are hydrolyzed to glycerol and free fatty acids. Much of the glycerol and fatty acids are localized to adipose tissue where they resynthesize triglycerides which are stored (#13). Small amounts of fatty acids are oxidized in liver, heart and muscle for energy. In liver, the fatty acids are completely oxidized in the TCA cycle, with little ketogenesis occurring in the fed state. Glucose is the primary source of energy for almost all tissues and is oxidized to lactate by the RBC (#9) and to CO2 by brain (#8), muscle (#11) and liver (#5). Excess glucose is stored as glycogen in liver and muscle (#6 and #11). Excess glucose is used to produce fatty acids by liver ( #7) and by adipose tissue ( #13), which are then stored as triglycerides. Liver triglycerides leave the liver as very low density lipoproteins and are stored in the adipose tissue.

Answer 320

Triglycerides stored in adipose tissue are a primary source of energy for most, but not all, tissues, especially the brain. Fatty acids are released from their triglyceride storage form, circulate, and are oxidized by muscle (#6), liver (#7), heart, and other tissues (not shown). Muscle derives energy from stored glycogen and from stored protein. Glucose remains the primary fuel for the brain (#3) and some other tissues, and the only fuel for the RBC (#4). How is glucose being generated under these conditions? Liver glycogen is being degraded as a major source of body glucose (#2). Some gluconeogenesis starts to occur as a second source of glucose, with lactate produced by the RBC (#11) and gluconeogenic amino acids derived from muscle protein breakdown (#9) serving as the primary substrates for liver glucose production. Fatty acid oxidation by the liver is not complete, as OAA is pulled out of the TCA cycle to produce PEP on the way to glucose. Hence, acetyl CoA accumulates and, in order to continue fatty acid oxidation for energy, CoASH is regenerated while acetyl CoA enters ketogenesis. Ketone bodies are a preferred fuel for heart and muscle (#8). Because amino acid metabolism is increasing to provided carbon sources for gluconeogenesis, ammonia is produced, which has to be detoxified by the urea cycle (#10). Glycerol, produced in the adipose tissue when triglycerides are being degraded, can leave the fat cell and circulate to the liver where it can enter gluconeogenesis (#12).

Answer 321

Basic pathways are similar to the basal state just described except that: A) muscle and liver glycogen have become depleted. Liver gluconeogenesis becomes the only source for producing glucose. B) because of A, ketone bodies become the primary product of fatty acid oxidation by the liver, since OAA is depleted from the TCA cycle and large amounts of protein are initially broken down to provide the gluconeogenic amino acids. To minimize body protein breakdown and ketoacidosis, the brain adapts to using ketone bodies as fuel by activation of the succinyl CoA transferase gene.

Answer 322

``` Restrict substrate – Provide cofactors – Provide product – Replace enzyme – Provide alternate routes of elimination – Treat secondary effects ```

Answer 323

Replenishes the NAD and FAD The ETC does

Lectures 58-82 Flashcards

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