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BB10235 > Glycolysis > Flashcards

Glycolysis Flashcards

1
Q

What is the basis of glycolysis

A
  1. The anaerobic breakdown of glucose to pyruvate
  2. Occurs in the cell cytoplasm
  3. Little ATP formed
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2
Q

What happens if there is not enough oxygen once glycolysis is produced

A
  1. If the concentration of pyruvate is too high for it to be further oxidised in the mitochondria, it is converted to lactate (lactic acid)
  2. More ATP is produced in the mitochondria when pyruvate is further oxidised in aerobic respiration
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3
Q

Describe what glucose needs to enter the cell

A
  1. Glucose is very hydrophilic so cannot enter the cell by itself
  2. There are two families of cell membrane glucose transporter proteins:
  3. Facilitated diffusion transporters: the GLUT family (for absorbing glucose from the blood into cells to produce ATP)
  4. Sodium-glucose linked transporters: the SGLT family (active transport - for absorbing glucose against a concentration gradient e.g. absorption in the small intestine/kidneys)
  5. Distribution of different transporter proteins makes a difference as to which carbohydrates can enter which tissues (e.g. GLUT3/4/5 in different places)
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4
Q

Describe what affect insulin action has

A
  1. Insulin receptor signalling stimulates Glut4 exocytosis: 2. Glut4 protein is transported to the cell surface membrane and allows increased influx of glucose from the blood.
  2. Increased insulin causes increased glucose transport
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5
Q

What is the first reaction of glycolysis

A
  1. Transfer of phosphoryl group from ATP to glucose to form glucose-6-phosphate (G6P)
  2. Catalysed by hexokinase
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6
Q

What is hexokinase

A
  1. A transferase enzyme: EC 2.7.1.1
  2. Mg2+ is an enzyme cofactor
  3. Chelation of the terminal phosphates to the Mg2+ ion allows for nucleophilic substitution reaction by glucose C6 OH
  4. Glucose binds to the active site and then the enzyme changes conformation to fold around the glucose molecule
  5. Hexokinase has epimer selectivity for glucose
    (different hexokinase isoforms have different specificities e.g. for fructose)
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7
Q

Describe the Hexokinase Ⅰ,Ⅱ, Ⅲ affinity

A
  1. High affinity for glucose so enables cells to take up glucose for phosphorylation from the bloodstream.
  2. Human hexokinase Ⅰ has a regulatory domain and a kinase domain: the product glucose 6 phosphate can bind to allosteric site on regulatory domain which inhibits the kinase domain
  3. ADP and ATP can also bind to allosteric sites to alter catalytic activity
  4. This regulates the amount of G6P produced
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8
Q

Describe Glucokinase Hexokinase Ⅳ affinity

A
  1. Low affinity for glucose so takes up glucose at high concentrations (e.g. excess glucose in the liver, this results in equilibrium between glucose in blood and liver)
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9
Q

Describe the energy of the addition of phosphate to glucose

A
  1. Addition of phosphate group to glucose is endergonic (+ve ΔG) so it must be coupled with another reaction for it to occur
  2. It is coupled with the hydrolysis of ATP which is exergonic (-ve ΔG) and releases a phosphate to be added to glucose
  3. This is irreversible so glucose is trapped as glucose-6-phosphate which is impermeable and therefore unable to leave the cell
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10
Q

Describe the biochemical effects of glucose phosphorylation

A
  1. The phosphate gives the glucose molecule a negative charge so it can no longer leave the cell by diffusion
  2. Phosphate group is very electronegative so it has an electron-withdrawing effect.
  3. This increases reactivity of the saccharide as elsewhere on the molecules is more susceptible to nucleophilic attack
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11
Q

What is the second step of glycolysis

A
  1. The conversion of G6P to fructose-6-phosphate

2. Catalysed by phosphoglucose isomerase

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12
Q

Describe the process of the conversion of G6P to fructose-6-phosphate

A
  1. The ring form must first be broken- an acid catalyses this
  2. Base catalysis occurs to cause isomerization- removes the acidic proton from C2
  3. THe proton is replaced on C1 in an overall proton transfer
  4. The ring is closed to form the product- released to yield free enzyme
  5. Enzyme is so efficient meaning the reaction is so fast that the rate is diffusion limited
  6. This reaction is reversible
  7. This enzyme has a separate independent function as an extracellular signalling molecule
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13
Q

What is the third step of glycolysis

A
  1. Conversion of F-6P to fructose-1,6-bisphosphate

2. Catalysed by phosphofructokinase

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14
Q

Describe the process of the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate

A
  1. The active site of the enzyme binds to the sugar and an ATP
  2. Each binding site consists of two different subunits which close around either side of the sugar molecule
  3. The terminal OH group of the fructose molecule acts as a nucleophile and can attack the phosphorus atom of the terminal ATP because of the electron withdrawing effect of the oxygens around it.
  4. This therefore cleaves the terminal phosphate group from ATP, transferring it to the fructose molecule
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15
Q

What inhibits the activity of phosphofructokinase enzyme

A
  1. High levels of ATP

2. Has same Vmax but higher concentrations of substrate F6P needed to reach it

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16
Q

Describe the structure of phosphofructokinase enzyme

A
  1. Bacterial phosphofructokinase is a homotetramer (consists of 4 identical subunits)
  2. Human PFK is even more complicated
  3. It has an active state (R-state) and an inactive state (T-state)
17
Q

What stabilises the R-state of PFK

A
  1. The regulatory site on the enzyme stabilises the state of the enzyme
  2. R-state is stabilised by ADP binding to the allosteric site
18
Q

What happens to PFK if the concentration of ATP increases

A
  1. If concentration of ATP increases, it will displace ADP from the allosteric site and enzyme will be inactivated as it will be in the T-state (enzyme changes conformation so that active sites are closed off)
  2. This means that the enzyme is able to be activated when concentration of ATP is low so that glycolysis can continue to produce more ATP, but it is inactivated when ATP is not needed.
  3. The state of one subunit will influence the state of the other subunits in the tetramer (due to changes in conformation of the subunit in their different states)
19
Q

Why is it important to regulate the activity of PFK

A
  1. Regulation of this enzyme is important because after this step of glycolysis is completed, the molecule is committed to complete breakdown (the product cannot be used for other cellular processes)
  2. Citrate also acts as a regulatory molecule of phosphofructokinase - it is produced in the citric acid cycle which uses acetyl CoA
  3. In aerobic conditions production of ATP and citrate increase, reducing glycolysis due to allosteric inhibition
  4. Different isoforms of the PFK unit consist of different combinations of subunits which means regulation of ATP production can differ in different tissues e.g. muscles
20
Q

What is the fourth step of glycolysis

A
  1. Conversion fructose-1,6-bisphosphate to dihydroxyacetone phosphate (DHAP) and Glyceraldehyde-3-phosphate (GAP)
  2. Catalysed by Aldose
21
Q

Describe the process of the conversion of fructose-1,6-bisphosphate to dihydroxyacetone phosphate (DHAP) and Glyceraldehyde-3-phosphate (GAP)

A
  1. In the active site, lysine is held next to the fructose substrate
  2. The key catalytic amino acid residues involved in the reaction are lysine and tyrosine
  3. Lysine covalently binds and stabilizes the intermediates - acts as a nucleophile to attack the carbonyl group of the fructose molecule (bacteria may use magnesium ions in place of the lysine)
  4. This displaces the oxygen and allows a keto-enol rearrangement
  5. Tyrosine acts as an efficient hydrogen acceptor which allows cleavage of the C-C bond to form 2 3C molecules
22
Q

Describe the structure of aldose

A
  1. Aldolase is a homo-tetramer each with its own active site
23
Q

What is the fifth step of glycolysis

A
  1. Dihydroxyacetone phosphate is converted to Glyceraldehye-3-phosphate
  2. Catalysed by Triose phosphate Isomerase
24
Q

Describe the process of the conversion of dihydroxyacetone phosphate (DHAP) to Glyceraldehyde-3-phosphate (GAP)

A
  1. Only GAP can continue along glycolytic pathway after step 4
  2. But DHAP and GAP are ketose-aldose isomers like F6P and G6P.
  3. Occurs by base catalysis (same mechanism as phosphoglucose isomerase)
  4. There is an enediol intermediate (double bond, alcohol)
  5. 1 molecule of glucose therefore produces two molecules of glyceraldehyde-3-phosphate
25
Q

Describe triose phosphate isomerase

A
  1. Transition state analogue can be used as a transition state inhibitor in order to study enzyme structure
  2. This is a rapid and efficient enzyme
26
Q

What is the classification for phosphoglucose isomerase

A

Isomerase- EC 5.3.1.9

27
Q

What is the classification for hexokinase

A

Transferase- EC 2.7.1.1

28
Q

What is the classification for phosphofructose kinase

A

Transferase- EC 2.7.1.11

29
Q

What is the classification for aldose

A

EC 4.1.2.13

30
Q

What is the classification for triosephosphate isomerase

A

Isomerase- EC 5.3.1.1

BB10235 (16 decks)

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