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BIOC 202 > MT 2 > Flashcards

MT 2 Flashcards

1
Q

Chymotrypsin:

[enzyme type] in [organ]

Can cleave [?] bonds on the [?] terminal side of [?] amino acids [give 4 examples]

Synthesized as [precursor] in the [organ]; it is activated in the [organ] by [?] first, then by [?] in a process called [?]

Active chymotrypsin is composed of [#] chains, connected together by [?] bonds

In the active site, it has conserved reactive [?] that can act as a [?]

A

Serine protease in small intestines

Can cleave peptide bonds on the carboxy terminal side of large hydrophobic amino acids (ie. Trp, Phe, Tyr, Met)

Synthesized as chymotrypsinogen in the pancreas; it is activated in the small intestine by trypsin first, then by itself (autolysis; π-chymotrypsin cleaves itself)

Active chymotrypsin is composed of 3 chains, connected together by disulfide bonds

In the active site, it has conserved reactive Ser (S195) that can act as a strong nucleophile

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2
Q

Protease

A

Enzymes that break peptide bonds

Cleave the peptide bond by hydrolysis (addition of water)

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3
Q

What is the purpose of a protease? (5)

A
  1. Protein turnover (ie. reuse proteins)
  2. Maintaining protein homeostasis (ie. degrading misfolded proteins)
  3. Digestion
  4. Activation or deactivation of proteins (enzymes)
  5. Formation of multiple proteins from one polypeptide chain
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4
Q

Why are peptide bonds stable?

A

Resonance structures

The carbonyl is less electrophilic due to resonance structures, so it is less reactive compare to regular carbonyl groups.

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5
Q

Overview of Mechanism of Serine Proteases (2 steps)

A
  1. The nucleophilic oxygen on serine attacks the electrophilic carbonyl of the peptide bond in a nucleophilic attack, cleaving the peptide bond and covalently forming an acyl-enzyme intermediate.
  2. The intermediate is cleaved by water to generate the new carboxyl group and free enzyme (S195 is regenerated)
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6
Q

Why is S195 so reactive?

A

S195 forms H-bond with imidazole ring of His57

This H-bond positions S195 and polarizes the hydroxyl group of S195 (it has alkoxide character)

His acts as a base catalyst (pulls away H+ from S195)

Asp102 also H-bonds with His57, making it a better proton acceptor

Asp102, His57, and Ser195 are known as catalytic triad (true for all serine proteases)

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7
Q

Step-By-Step Catalysis by Chymotrypsin

A
  1. Substrate binds to the active site; large hydrophobic amino acids fit to the hydrophobic pocket next to the active site
  2. The alkoxide ion on S195 does nucleophilic attack on the carbonyl carbon of the substrate. This causes carbon to transciently adopt a tetrahedral geometry. His57 acts as a base catalyst
  3. The tetrahedral intermediate rearranges to the acyl-enzyme intermediate. The carbonyl reforms and the peptide bond is cleaved. His57 acts as an acid catalyst, donating proton to the amine group. The “new” amino peptide fragment leaves the active site.
  4. Water molecule enters the active site; His57 H-bonds with water, making it hydroxide-like
  5. The nucleophilic oxygen of water acts as a nucleophile, attacking the carbonyl carbon. His57 acts as a base catalyst, stealing H from water. This forms another short-lived intermediate– tetrahedral oxyanion
  6. Tetrahedral intermediate rearranges, reforming the carbonyl and separating the product from Ser195. His57 acts as an acid catalyst (H donor to Ser195). A “new” carboxyl peptide is formed and free Ser195 are generated.
  7. The “new” carboxyl peptide leaves active site. Enzyme is ready to catalyze another reaction.
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8
Q

What are some examples of serine proteases? How do they differ from each other?

A

Serine proteases all use the same catalytic mechanism (ie. same catalytic triad), but they all have different specificity

  1. Chymotrypsin has a large hydrophobic pocket, so it binds amino acids with large, hydrophobic side chains
  2. Trypsin has Asp at the bottom of the pocket, so it binds positively charged amino acids
  3. Elastase has 2 Val residues that narrow down the pocket, so it binds amino acids with small R-groups (ie. Ala, Val)
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9
Q

List 4 types of proteases

A
  1. Cysteine proteases
  2. Aspartate proteases
  3. Metalloproteases
  4. Serine proteases
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10
Q

Cysteine proteases

A

Use cysteine as a nucleophile

Example: caspases in apoptosis

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11
Q

Aspartate proteases

A

Have two aspartates that make water a strong nucleophile

Example: HIV proteases

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12
Q

Metalloproteases

A

Use metal ion to make water a strong nucleophile

Example: MMP - matrix metalloproteinase; regulates tissue remodelling

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13
Q

Hemoglobin and Oxygen Transport

A

Hb Oxygen is soluble in water, but not enough to supply tissues

RBC (red blood cells) use Hb to carry oxygen from lungs (high [O2], low [CO2], pH=7.4) to tissues (low [O2], high [CO2], pH=7.2) Carry O2 from high [O2] to low [O2}

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14
Q

Hemoglobin structure

A

Tetramer (quaternary structure) - 2 alpha and 2 beta subunits

Alpha subunit - 144 amino acids; 7 helices

Beta subunit - 146 amino acids; 8 helices

a1β1-a2β2 interface is weaker, has fewer interactions and is less rigid

The other interface (between α1-β1 and β2-α1) is very stable and has many interactions

Each subunit has 1 heme group (4 in total); the O2 binds to heme

Heme has protoporphyrin ring with Fe2+ in the centre Fe2+ has 6 coordination sites (4 are nitrogens from the ring; 5th is from His below the ring; 6th is above the ring and binds oxygen)

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15
Q

How does heme bind O2?

A

When the 6th site is empty, the iron is a little outside the ring

When it binds to O2, it “pops” into the ring, because the electron density changes

The histidine (at the 5th site) is attached to a-helix, so in turn, the a-helix moves, causing carboxyl end to move

This disrupts and creates new ion pairs in the α1β1-α2β2 interface

This conformational change increases the other hemes ability to bind O2

H6 binds oxygen cooperatively

H6 with 3 O2 bound has a 20x greater affinity to O2 compared to when no O2 is bound – this is known as allosteric interaction (ie. change in shape of a protein that results form binding a molecule at a point other than the active site)

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16
Q

What are the two states of hemoglobin?

A

The untwisted deoxyHb (low O2 affinity) = T-state

Twisted oxyHb (high O2 affinity) = R-state

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17
Q

How does cooperativity help O2 transport?

A

If Hb was always in R-state, it would bind O2 in lungs but never release it in the tissues

If Hb was always in T-state, it would bind only small amounts of O2 in the lungs, which is not enough for metabolism

Because Hb binds oxygen cooperatively, it can exist in R-state in the lungs and become saturated with O2, and then travel to the tissues where it can convert to T-state and release O2

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18
Q

Why does it convert from R-state to T-state? List the two main ideas.

A
  1. Bohr effect
  2. 2,3-bisphosphoglycerate (2,3-BPG)
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19
Q

What is the Bohr Effect?

A

H+ and CO2 promote the release of O2 from oxyHb, lowering affinity

This enhances Hb’s ability to release O2 in the tissues and pick it up in the lungs

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20
Q

What is the [H+] and [CO2] in the tissues and lungs?

A

[H+] and [CO2] is high in the tissues, resulting in low affinity to oxygen due to the Bohr effect.

[H+] and [CO2] is low in the lungs, resulting in high affinity to oxygen due to the Bohr effect.

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21
Q

What is the mechanism of the Bohr effect?

A

H+ : some His residues in the α1β1-α2β2 interface have pKa’s close to 7. In conditions of low pH (high [H2]) the histidines become protonated and will interact with other residues, stabilizing the T-state

CO2 : can bind to the free terminal amino group; carbonate ions can form new ionic interactions that will stabilize the T-state

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22
Q

How does 2,3-bisphosphoglycerate impact the conversion of R-state to T-state?

A

DeoxyHb binds to 2,3-BPG, which stabilizes the T-state, lowering affinity to O2 It binds in a special pocket formed by the 2 β subunits

The pocket has a positive charge to interact with the negatively charged 2,3-BPG

Part of transition from T-state to R-state is the removal of 2,3-BPG from Hb

Conformational changes during T→R make the pocket too small for 2,3-bBPG to fit; thus, 2,3-BPG leaves, stabilizing the R-state.

The binding curve shifts, resulting in higher affinity.

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23
Q

Why do we need the protein part of hemoglobin? Why don’t we just have heme?

A
  1. Cooperativity (only works with protein)
  2. CO binds to heme 25,000x better than O2; Hb lowers the affinity for CO by shielding heme
  3. Prevents iron (Fe2+) from being oxidized (to Fe3+) because the protein has interactions that shield iron from being oxidized
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24
Q

Metabolism

A

Highly integrated network of reactions (chemical pathways) that enable a cell to extract energy from the environment and use this energy for biosynthesis

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25
Q

Catabolism

A

Reactions that use fuel to generate useful energy

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26
Q

Anabolism

A

Reactions that use energy to form complex structures

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27
Q

Cellular respiration equation

A

glucose + O2 → CO2 + H2O + energy

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28
Q

Thermodynamics and ATP

A

In order for a pathway to proceed:

  1. Reactions must be specific to its final product
  2. Reactions must be thermodynamically favoured
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29
Q

How do endergonic reactions proceed?

A

Couple with exergonic reactions

The overall free energy change for a chemically coupled series of reactions is equal to the sum of the individual steps

Common method: couple with ATP hydrolysis

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30
Q

What are the reactions and the net reaction for the conversion of glucose to G-6-P?

A
  1. Glucose + Pi → Glu-6-P + H2O
  2. ATP + H2O → ADP + Pi

Net reaction: Glucose + ATP → Glu-6-P + ADP

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31
Q

What is standard state?

A

25 degrees Celsius (298K), 1 atm pressure, [] = 1M, pH=7

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32
Q

ATP

A

Called the universal energy currency of the cell

Consists of ribose, adenine, and three phosphates

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33
Q

What structure is this?

A

Pi

Inorganic Phosphate

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34
Q

What structure is this?

A

Ppi

Pyrophosphate

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35
Q

Why is the hydrolysis of ATP so favourable?

A
  1. Relief of charge repulsion

At pH=7, ATP has a -4 charge, and ADP has a -3 charge

  1. Resonance stabilization of Pi
  2. Ionization of ADP

At pH=7, ADP-2 + e- → ADP-3 + H+

  1. Greater solvation of the product (ie. more H-bonds with H2O)
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36
Q

Does ATP hydrolysis drive reactions?

A

No, usually it is not ATP hydrolysis that drives reactions, even if we write it this way; rather, it is phosphoryl transfer from ATP

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37
Q

Phosphoryl transfer

A

Movement of phosphate group from one molecule to another; this is how reactions are coupled

Molecules other than ATP can transfer phosphates; all have negative -ΔG0’ (more negative means higher phosphoryl transfer capacity; something with more negative ΔG0’ can easily phosphorylate something less negative)

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38
Q

Creatine phosphate

A

creatine phosphate + ADP + H+ ⇌ ATP + creatine @ standard state (negative delta G)

Creatine phosphate is used to regenerate ATP in muscle during the first few seconds of heavy muscle contraction

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39
Q

Exercise Time vs Energy Source

A

1-3 seconds: ATP

3-9 seconds: Creatine-P

Up to 1 minute: Glycolysis

Up to 40 minutes: Glycogen

>40 minutes: Fats + glucose

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40
Q

Carbohydrates general chemistry

A

An aldehyde or ketone compounds with multiple hydroxyl-groups

General formula: (CH2O)n

Sugars can be drawn as linear aldehydes (aldoses) or ketones (ketoses) depending on where the carboxyl is placed

Sugars have chiral carbons, thus there exists different enantiomers

Sugars with more than 3 carbons have multiple chiral centers (they can exist as diastereomers)

Fischer projections help to view enantiomers

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41
Q

What are the simplest sugars?

A

Simplest sugars: trioses (3C’s)

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42
Q

How do you differentiate between different enantiomers of sugars? Which is which?

A

When looking at a Fischer projection of a sugar, if the chiral center for the furthest carbonyl carbon has hydroxyl (~OH) group on the right, it is a “D” ; if the ~OH is on the left, it is an “L” form

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43
Q

Which enantiomer is present in living cells?

A

Usually “D” form in living cells

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44
Q

Hexose

A

6 carbon sugar

16 different aldoses; 8 different ketoses

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45
Q

Epimers

A

Sugars that differ only at one chiral carbon

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46
Q

Why do sugars exist as ring structures?

A

Aldehyde and ketones can react with alcohols to form hemiacetals and hemiketals

ie. C5-OH of aldohexose (6-C sugar) can attack C1 carbonyl, forming a 6 membered ring called pyranose
ie. C5-OH of ketohexose can attack C2 carbonyl, forming a 5-membered ring known as furanose
ie. C-O-OH group can also attack C-2 carbonyl, forming a pyranose ring

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47
Q

Pyranose

A

6-membered sugar ring

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48
Q

Furanose

A

5-membered sugar ring

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49
Q

Anomeric carbon

A
  • The new chiral center that is formed when the cyclic form is created; it is where the carbonyl carbon was previously
  • it is reactive
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50
Q

Anomers

A

Alpha/Beta

If the hydroxyl group of the anomeric carbon is below the ring, it is α

If the hydroxyl group of the anomeric carbon is above the ring, is it β

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51
Q

In the cell, glucose can exist as…

A

1/3 α-ring anomer

2/3 β-ring anomer

<1% is open chain

*note, there is spontaneous conversion (“mutarotation”) between anomers

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52
Q

Mutarotation

A

the change in the optical rotation because of the change in the equilibrium between two anomers, when the corresponding stereocenters interconvert cyclic sugars show mutarotation as α and β anomeric forms interconvert

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53
Q

What structure is this?

A

ATP

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54
Q

What structure is this?

A

Ribose

(Beta-D Ribofuranose)

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55
Q

What structure is this?

A

Adenine (purine base)

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56
Q

What are these bonds called?

A

Phosphodiester bonds

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57
Q

Draw D-Glucose (open-chain)

A
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58
Q

Draw D-Fructose (open-chain)

A
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59
Q

Draw D-Galactose (open-chain)

A
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60
Q

Convert this to ring form (pyranose)

A
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61
Q

Convert this to ring form (furanose)

A
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62
Q

Draw α-D-Glucopyranose

A
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63
Q

Draw α-D-Fructofuranose

A
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64
Q

Draw β-D-Glucopyranose

A
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65
Q

Draw β-D-Fructofuranose

A
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66
Q

What structure is this?

A

Pyran

67
Q

What structure is this?

A

Furan

68
Q

Draw α-D-glucopyranosyl-(1→4)-D-glucopyranose

What’s another name for this molecule?

A
69
Q

What structure is this?

A

β-D-galactopyranosyl-(1→4)-β-D-glucopyranose

Lactose (β form)

70
Q

What structure is this?

A

α-D-glucopyranosyl-(1→4)-D-glucopyranose

Maltose

71
Q

Draw β-D-galactopyranosyl-(1→4)-β-D-glucopyranose

What’s another name for this molecule?

A
72
Q

Draw β-D-fructofuranosyl α-D-glucopyranoside

What’s another name for this molecule?

A
73
Q

Draw α-D-glucopyranosyl-(1→1)-α-D-glucopyranoside

What’s another name for this molecule?

A
74
Q

What structure is this?

A

β-D-fructofuranosyl α-D-glucopyranoside

Sucrose

75
Q

What structure is this?

A

α-D-glucopyranosyl-(1→1)-α-D-glucopyranoside

Trehalose

76
Q

What can the anomeric carbon react with?

A
  • proteins (glycosylation)
    ie. serine, threonine, tyrosine → O-glycosylation (-OH group)
    ie. arginine, asparagine → N-glycosylation (-N)
  • other sugars → form disaccharides, polysaccharides
77
Q

Draw the condensation reaction between α-D-glucose and β-D-glucose.

A
78
Q

Polysaccharide

A

Several molecules of monosaccharides connected by glycosidic bonds

Sugars in polysaccharides are in ring structures (pyranose or furanose)

79
Q

What are the bonds between sugar molecules called?

A

Glycosidic bonds

80
Q

Glycolysis

A

The sequence of reactions that break down 1 molecule of glucose into 2 molecules of pyruvate with the net production of 2 ATPs.

81
Q

Is glycolysis an aerobic or anaerobic process?

A

Anaerobic; doesn’t require oxygen

82
Q

What can pyruvate be converted into?

A
  1. H2O + CO2 (via Krebs cycle and oxidative phosphorylation)
  2. Lactate (lactic fermentation; doesn’t require oxygen)
  3. Ethanol (alcohol fermentation; doesn’t require oxygen)
83
Q

The pathway of glycolysis can be broken down to 2 stages. What are they?

A
  1. Preparatory stages: trapping and destabilization of glucose; cleavage of phosphofructose
  2. Payoff stage: ATP and pyruvate are generated
84
Q

How does glucose enter the cell?

A

Through facilitated diffusion via glucose transporters

85
Q

Glycolysis: Stage 1

A

Glucose is phosphorylated using ATP by hexokinase on C6, yielding glucose-6-phosphate (G-6-P)

This traps glucose in the cell, because it lowers [glucose] and G-6-P also cannot go back through the glucose transporters due to its charge

This also destabilizes the molecule of glucose This reaction is irreversible

86
Q

Draw the reaction for step 1 of glycolysis

A

[Insert picture]

87
Q

Glycolysis: Stage 2

A

G-6-P is converted to F-6-P by phosphohexose isomerase

Multistep process, converting G-6-P to open ring structure before converting it to F-6-P

This reaction is reversible

88
Q

Draw the reaction for step 2 of glycolysis

A
89
Q

Glycolysis: Stage 3

A

F-6-P is phosphorylated by phosphofructokinase-1 (PFK-1) to fructose-1,6-bisphosphate (F-1,6-BP)

This reaction is reversible

This is an important regulation point, because it is the committed step; bottleneck step

Once this step occurs, it is committed to glycolysis; previous step to form G-6-P is also irreversible, but G-6-P can be used for things other than glycolysis, where as F1,6-BP cannot)

This is also the rate-determining step of glycolysis

90
Q

Draw the reaction for step 3 of glycolysis

A
91
Q

Glycolysis: Stage 4

A

Aldolase cleaves F-1,6-BP into DHAP and G3P

Reversible reaction

You now have two 3C fragments

Only G3P goes further in glycolysis, so DHAP is converted to G3P in the next reaction

92
Q

Draw the reaction for step 4 of glycolysis

A
93
Q

Glycolysis: Stage 5

A

DHAP is converted to G3P by triose phosphate isomerase

This reaction is reversible

94
Q

Draw the reaction for step 5 of glycolysis

A

[insert picture]

95
Q

Glycolysis: Stage 6

A

G3P is oxidized and phosphorylated to 1,3-BPG by GAPDH

Requires inorganic phosphate and NAD+ 1,3-BPG has higher phosphoryl transfer potential

Reaction is reversible

96
Q

We can imagine stage 6 of glycolysis as happening in 2 steps to understand the mechanism (in reality, this is not true).

What are these 2 steps?

A
  1. Oxidize G3P to a carboxylic acid using NAD+ (NAD+ + 2e- + 2H+ → NADH + H+)

This is very favourable; ΔG<0

  1. The carboxylic group is attached to orthophosphate

Not favourable; ΔG>0

Note: these two reactions are coupled by thioester (S-C=O) intermediate, so that the first reaction drives the 2nd reaction (free acid is never formed)

97
Q

GAPDH Mechanism

A
  1. Form hemiacetal with a cysteine residue in the active site
  2. NAD+ oxidizes hemiacetal to form thioester; in the process, NAD+ is converted to NADH

3-4. When NADH is replaced by a fresh NAD+, inorganic phosphate can attack the carbonyl and cleave the reaction intermediate from cysteine, forming the product

Note: We generate 2 molecules of 1,3-BPG per 1 molecule of glucose, so 2 molecules of NADH are generated

98
Q

Draw the reaction for step 6 of glycolysis

A
99
Q

Glycolysis: Stage 7

A

1,3-BPG phosphorylates ADP, forming ATP and 3-phosphoglycerate (3-PG) x2

This is another example of substrate level phosphorylation

Catalyzed by phosphoglycerate kinase

Reaction is reversible

100
Q

Draw the reaction for step 7 of glycolysis

A
101
Q

Glycolysis: Stage 8

A

Phosphoglycerate mutase converts 3-PG to 2-PG (2-phosphoglycerate) x2

Reaction is reversible

Requires cofactor: 2,3-bisphosphoglycerate (2,3-BPG)

102
Q

Mutase

A

An enzyme that causes intramolecular rearrangement (isomerization)

103
Q

Kinase

A

An enzyme that catalyzes the transfer of phosphate groups from high-energy, phosphate-donating molecules to specific substrates

104
Q

Phosphatase

A

An enzyme that uses water to cleave a phosphoric acid monoester into a phosphate ion and an alcohol.

105
Q

Phosphoglycerate mutase mechanism

A
  1. Enz-His + 2,3-BPG ⇌ (phosphatase activity) ⇌ Enz-His-P + 2-phosphoglycerate (product)
  2. Enz-His-P + 3-phosphoglycerate (substrate) ⇌ (kinase activity) ⇌ Enz-His + 2,3-BPG (cofactor)
106
Q

Draw the reaction for step 8 of glycolysis

A
107
Q

Glycolysis: Stage 9

A

Enolase removes water from 2-PG forming an enol: phosphoenolpyruvate (PEP) x2

Reversible reaction

PEP is another molecule with high phosphoryl transfer capacity

Note: Enol form of PEP exist in tautomer form

Phosphate traps PEP in enol form

108
Q

Draw the reaction for step 9 of glycolysis

A
109
Q

Glycolysis: Step 10

A

Pyruvate kinase transfers a phosphate from PEP to ADP, forming ATP x2 Irreversible reaction

110
Q

Draw the reaction for step 10 of glycolysis

A
111
Q

Net Glycolysis Reaction

A

Glucose + 2ADP + 2Pi + 2NAD+ → 2 pyruvate + 2 ATP + 2 NADH + 2H+ + 2H2O

112
Q

Fermentation

A

The process used in bacteria and year to convert pyruvate from glycolysis to either CO2 + ethanol (alcohol fermentation) or to lactate (lactic fermentation) to regenerate NAD+ from NADH. Humans can also do lactic fermentation - this happens in skeletal muscle during intensive exercise.

113
Q

Oxidation

A

Loss of e-

114
Q

Reduction

A

Gain of e-

115
Q

In biological systems, redox reactions are usually accompanied by addition/removal of _______.

A

Pairs of hydrogen ions (H+)

116
Q

What is the final electron acceptor in aerobic sysems?

A

Oxygen

2H+ + 2e- + 1/2 O2 → H2O

However, the highly energetic electrons are not transferred to oxygen directly; carriers are used to transfer electrons (and H+) to electron transport chain or to biosynthetic reactions that need them

117
Q

What structure is this?

A

NAD+ (oxidized form)

118
Q

What structure is this?

A

NADH (reduced form)

119
Q

What structure is this? [insert picture]

A

NADP+ (oxidized form)

120
Q

What structure is this?

A

NADPH (reduced form)

121
Q

What are NAD+ / NADH / NADP+ / NADPH derived from?

A

ATP and niacin (vitamin B3)

122
Q

What kinds of reactions are NAD+ / NADH used in?

A

Catabolic reactions

123
Q

What kinds of reactions are NADP+ / NADPH used in?

A

Anabolic reactions

124
Q

What structure is this? [insert picture]

A

FAD (oxidized form)

125
Q

What structure is this?

A

FADH2 (fully reduced form)

126
Q

What structure is this?

A

FADH (semiquinone)

127
Q

What structure is this? [insert picture]

A

FMN (oxidized form)

128
Q

What structure is this?

A

FMNH+ (semiquinone)

129
Q

What structure is this?

A

FMNH2 (fully reduced)

130
Q

What is FAD / FADH2 / FMN / FMNH2 derived from?

A

ATP and riboflavin (vitamin B2)

131
Q

What is the equation for the conversion of FAD to FADH2?

A

FAD + 2H+ + 2e- ⇌ FADH2

132
Q

FAD/FMN vs NADH/NADPH in terms of movement

A

FAD/FMN is usually bound to something, unlike NADH/NADPH which is free to move around

133
Q

Why must NAD+ be regenerated?

A

The pool of NAD+/NADH is small, so NAD+ must be regenerated in order for glycolysis to run

134
Q

Fate of Pyruvate: Aerobic systems

A

NAD+ is regenerated from NADH by the electron transport chain

Oxygen is the final electron acceptor, and water is the final electron carrier

Pyruvate is converted into acetyl-CoA and oxidized further in the Kreb’s cycle

135
Q

Fate of Pyruvate: Anaerobic systems

A

There is no oxygen to accept electron, so fermentation occurs

136
Q

Fermentation in mammals

A

Enzyme lactate dehydrogenase reduces pyruvate to lactate and in the process converts NADH to NAD+, which allows glycolysis to continue

Pyruvate is a final electron acceptor and lactate is the final electron carrier

Happens in skeletal muscle during high intensity exercise

137
Q

Draw the fermentation reaction that occurs in skeletal muscles of mammals.

A
138
Q

Fermentation in yeast

A

Pyruvate is first decarboxylated by pyruvate decarboxylase, forming acetaldehyde

Acetaldehyde is reduced to ethanol by alcohol dehydrogenase, converting NADH back to NAD+

Note: NAD+ is generated at the second step

Acetaldehyde is the electron acceptor

Ethanol is the final electron carrier

139
Q

Regulation of glycolysis

A

Happens on the cellular level

Occurs at the three irreversible steps

  1. Step #1 - hexokinase
  2. Step #3 - PFK-1
  3. Step #10 - pyruvate kinase

All of these enzymes are regulated allosterically (in most cases)

140
Q

PFK-1 regulation

A

Most important rate-limiting step

Increased [ATP] inhibits PFK-1

Increased [AMP] activates PFK-1

Increased [fructose-2,6-bisphosphate] stimulates PFK-1 (this regulated by hormones and is not an intermediate of glycolysis)

Increased [citrate] inhibits PFK-1 (this is a Kreb’s cycle intermediate)

Increased [H+] inhibits PFK-1 (ie. lactic fermentation results in H+ production)

141
Q

Hexokinase regulation

A

Reaction #1

It is allosterically inhibited by its product, G-6-P

This is an example of product inhibition; prevents excessive glucose phosphorlyation

142
Q

Pyruvate kinase regulation

A

Reaction #10

Increased [ATP] inhibits pyruvate kinase

Increased alanine inhibits pyruvate kinase

Increased [fructose-1,6-bisphosphate] activates pyruvate kinase (this is an example of feed forward stimulation)

Regulated by hormones

143
Q

What is the logic that explains why alanine inhibits pyruvate kinase?

A

Alanine is made from pyruvate, so an excessive presence of alanine is indicative of an excessive production of pyruvate (ie. by pyruvate kinase)

144
Q

Feed forward stimulation

A

Upstream build-up results in downstream activation

145
Q

Draw a diagram that summarizes regulation of glycolysis in the cell.

A
146
Q

Conversion of pyruvate to acetyl-CoA

Done by multienzyme complex called (?)

Occurs in (?)

Involves (?) of pyruvate to (?), followed by formation of (?)

Free (?) never forms, but it helps us to understand the mechanism

A

Done by multienzyme complex called pyruvate dehydrogenase complex (PDH complex)

Occurs in the mitochondrial matrix

Involves decarboxylation/oxidation of pyruvate to acetate, followed by formation of acetyl-CoA (thioester)

Free acetate never forms, but it helps us to understand the mechanism

147
Q

Enzyme complex of PDH - What are the enzymes and cofactors involved?

A

Enzymes: E1, E2, E3

Cofactors:

TPP is bound to E2

FAD is bound to E3

NAD+ is free

CoASH is free

148
Q

Coenzyme A

A

aka CoASH, CoA

Derived from ADP, vitamin B5 (pantothenate) and B-mercaptoethylamine

Carrier of an acyl-group

Forms high energy thioester bonds

149
Q

Draw acetyl-CoA

A
150
Q

Write out the reaction from acetyl-CoA to CoASH and the corresponding delta G.

A

Acetyl-CoA + H2O ⇌ acetate + CoASH

Delta G at standard state is -31 kJ/mole

151
Q

TPP

A

Thiamine pyrophosphate

Derived from vitamin B1 (thiamine)

Forms a reactive carbanion easily

152
Q

Draw the conversion of pyruvate to acetyl-CoA reaction

A
153
Q

What part of CoA is derived from vitamin B5?

A

Pantothenic acid component

154
Q

What structure is this?

A

Coenzyme A

155
Q

What structure is this?

A

Thiamine pyrophosphate (TPP)

156
Q

What structure is this?

A

Hydroxyethyl thiamine pyrophosphate

157
Q

Lipoic acid

aka (?)

Attached to (?) via (?), forming (?)

(?) (?) to (?) groups, resulting in the (?) group being attached to one of the (#) (?) atoms (breaking the (?))

Acts as a (?)

A

aka Lipoyllysine

Attached to E2 via lysine side chain, forming lipoyllysine

Oxidizes hydroxyethyls to acyl groups, resulting in the acyl group being attached to one of the 2 sulphur atoms (breaking the disulphide bridge)

Acts as a robotic arm

158
Q

5 Step Mechanism of PDH complex

A

1) Pyruvate enters E1 TPP attacks pyruvate and decarboxylates it to form the reaction intermediate: hydroxyethyl-TPP
2a) The lipoyllysine arm swings towards E1
2b) Hydroxyethyl group is oxidized to acetyl group and transferred to the lipoyllysine arm
3a) The arm swings into E2
3b) The acetyl group is transferred to a free CoASH (coenzyme A), forming acetyl CoA. The lipoyllysine is reduced in the process.
3c) Acetyl CoA leaves the PDH complex
4a) The arm swings to E3
4b) Here, the lipoyllysine arm is oxidized by FAD. FAD is reduced to FADH2.
5) NAD+ enters E3 and re-oxidizes FADH2 back to FAD. In the process, NAD+ is reduced to NADH + H+, which leaves E3.

Now we’re back to step #1, and another molecule of pyruvate can enter the PDH complex

159
Q

Regulation of PDH complex - E2 (dihydrolipoyl transacetylase)

A

Increased [acetyl CoA] inhibits E2 (product inhibition)

160
Q

Regulation of PDH complex - E3 (dihydrolipoyl dehydrogenase)

A

Increased [NADH] inhibits E3 (product inhibtion)

161
Q

Regulation of PDH complex - E1 (pyruvate dehydrogenase)

A

Majority of PDH complex regulation happens on E1 through phosphorylation (ie. of serine, tyrosine, threonine)

1) PDH-kinase phosphorylates PDH-E1, which deactivates it.

Increased [acetyl CoA] activates PDH-kinase

Increased [NADH] activates PDH-kinase

Increased [ATP] activates PDH-kinase

Increased [NAD+] inhibits PDH-kinase

Increased [ADP] inhibits PDH-kinase

2) PDH-phosphatase removes phosphate from PDH-E1, which re-activates it.

Increased [insulin] activates PDH-phosphatase

Increased [Ca2+] activates PDH-phosphatase

Note: Ca2+ is released during muscle contraction, thus suggestive is high energy consumption

In general, PDH is switched off when energy is high, and switched on when energy is low.

162
Q

Draw the tautomerization of pyruvate

A
163
Q

Fill in the standard free energies of hydrolysis of these phosphorylated compounds and acetyl-Co-A (kJ/mol)

A

BIOC 202 (4 decks)

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