biochem unit 3

Write a balanced reaction of glycolysis.

Glucose+ 2 ADP + NAD+ + Pi ↔ 2 Pyruvate + 2 ATP + 2 NADH + 2 H2O

Define glycolysis and state the starting molecule and main products.

Glycolysis is a metabolic pathway that breaks down glucose to generate 2 pyruvates, 2 net ATP, and 2 NADH for energy

Overall, explain why each step of glycolysis is unique and important for the whole process

STEP 1 Hexokinase:

- metabolically irreversible step!

- transferase mechanism to add phosphate to glucose

o traps glucose inside cell

o maintains glucose concentration gradient

o binds to enzymes better

o helps break glucose into 2 Pi-containing products

- key word: investment regulation

STEP 2 Phosphoglucose isomerase:

- isomerase mechanism to change an aldose to a ketose

- sets up for reaction 3 and reaction 4

o reaction 3: makes it easier to add phosphate to free hydroxyls

o reaction 4: C-OH must be 2 carbons away from C=O

- key word: set up

STEP 3 Phosphofructokinase-1:

- main point of regulation/committed/rate limiting step in glycolysis!

- metabolically irreversible step!

- transferase mechanism to add another phosphate to fructose-6-P

- key word: investment regulation

STEP 4 Aldolase:

- lyase mechanism that breaks Fructose-1,6-bisP into 2 products:

Dihydroxyacetone-P + Glyceraldehyde 3-P

- key word: set up

STEP 5 Triose Phosphate Isomerase:

- isomerase mechanism that converts a ketose to aldose

- endergonic reaction that allows glycolysis to proceed by one pathway

- key word: set up

STEP 6 Glyceraldehyde-3-P Dehydrogenase:

- oxidoreductase mechanism that allows for a phosphate to be incorporated without

the need of investing ATP.

- 2 reactions done by enzyme:

o 1) oxidation

o 2) phosphorylation

- pushing the reactions together make this reaction possible

- produces NADH

- key word: pay off

STEP 7 Phosphoglycerate Kinase:

- transferase mechanism that produces 1st ATP (substrate level phosphorylation)!

- key word: pay off

STEP 8 Phosphoglycerate Mutase:

- isomerase mechanism that changes the location of the phosphate group

- key word: set up

STEP 9 Enolase:

- lyase mechanism that creates a double bond and water leaves

- PEP has an uncomfortable phosphate group that prevents from forming ketone

- key word: set up

STEP 10 Pyruvate Kinase:

- metabolically irreversible step!

- transferase mechanism that allows the production of 2nd ATP (substrate-level phosphorylation)!

- key word: pay off

Explain what is allowing each reaction to be energetically favorable. Make sure to have in depth explanations for steps 1, 2, 6, and 10.

STEP 1: energetically favorable by coupling with ATP hydrolysis

- phosphorylating glucose is unfavorable, so breaking the phosphoanhydride bond in ATP makes the reaction spontaneous

STEP 2: in general isomerization reactions are close to equilibrium and the direction of the reaction depends on the ratio of products and reactants.

- also Fructose-6-P is continuously consumed by step 3 → pulling mechanism

STEP 3: energetically favorable by coupling with ATP hydrolysis

- phosphorylating fructose-1-phosphate is unfavorable, so breaking the phosphoanhydride bond in ATP makes the reaction spontaneous

STEP 4: energetically favorable because step 10 pulls reaction forward by keeping

concentration of intermediates low.

STEP 5: in general isomerization reactions are close to equilibrium and the direction of the reaction depends on the ratio of products and reactants. Step 10 pulls reaction forward by keeping concentration of intermediates low.

STEP 6: energetically favorable because the unfavorable phosphorylation is coupled to the oxidation using NAD+

- oxidation reaction of glyceraldehyde-3-P and 3-phosphoglycerate is very favorable.

- phosphorylation reaction of 3-phosphoglycerate to 1,3-bisphosphoglycerate is very unfavorable.

- coupling reactions together make this step possible.

- The reaction in real life conditions is mainly driven by the ratio of NADH/NAD+

- Step 10 pulls reaction forward by keeping concentration of intermediates low.

STEP 7: energetically favorable by coupling the hydrolysis of a mixed anhydride bond with the formation of a phosphoanhydride bond in ATP. Step 10 pulls reaction forward by keeping concentration of intermediates low.

STEP 8: in general isomerization reactions are close to equilibrium and the direction of the reaction depends on the ratio of products and reactants. Step 10 pulls reaction forward by keeping concentration of intermediates low.

STEP 9: There reaction favor the formation of an enol, which is slightly unfavorable. Step 10 pulls reaction forward by keeping concentration of intermediates low.

STEP 10: energetically favorable because highly exergonic phosphate transfer to ADP

- Forming a phosphoanhydride bond in ATP is unfavorable, coupling this with the

tautomerization of PEP to pyruvate makes it favorable.

- substrate level phosphorylation (making ATP)

- PEP is unstable enol-phosphate, phosphate is a good leaving group.

- PEP turns to pyruvate, pyruvate is in keto form, which is more stable than enol form

Classify each step based on their closeness to equilibrium and state if they’re reversible or irreversible

step 1: far from equilibrium/irreversible

step 2: close to equilibrium/reversible

step 3: far from equilibrium/irreversible

step 4: close to equilibrium/reversible

step 5: close to equilibrium/reversible

step 6: close to equilibrium/reversible

step 7: close to equilibrium/reversible

step 8: close to equilibrium/reversible

step 9: close to equilibrium/reversible

step 10: far from equilibrium/irreversible

State the fates of glucose-6-P.

1) glycolysis and convert to 2 pyruvates

2) glycogenesis to make glycogen

3) pentose phosphate pathway to make pentoses

4) gluconeogenesis to make glucose

Calculate the net ATP yield by stating which step is using or producing ATP in glycolysis.

step 1: uses ATP (-1 ATP)

step 3: uses ATP (-1 ATP)

step 7: makes ATP*2 (+2 ATP)

step 10: makes ATP*2 (+2 ATP)

NET ATP calculation: -1 + -1 +2 +2 = 2 NET ATP

Predict the effect on the direction of enzymes and concentration of intermediates when different steps of glycolysis are inhibited.

It depends on the reversibility of steps

Explain why it is important to regulate steps 1, 3 and 10 of glycolysis.

they all are very exergonic and commitment points in glycolysis!

- step 1 must be regulated to make sure all cells have similar glucose levels

o uses feedback inhibition

- step 3 is the committed/rate limiting/main point of regulation step of glycolysis

o uses feedback inhibition and activation

- step 10 is the pull needed to make steps 4-9 stable

State the intermediates that inhibit or activate Step 1, 3 and 10 of glycolysis

step 1: hexokinase

- inhibitor: glucose-6-P

- activators: AMP and ADP

step 3: phosphofructokinase-1

- off pathway metabolites:

o activators: fructose-2,6-bisphosphate

- in pathway metabolites:

o inhibitor: PEP, citrate, and ATP

o activators: ADP and AMP

step 10: pyruvate kinase

- inhibitor: ATP

- activators: fructose-1,6-bisphosphate, AMP, and ADP

State the fates of pyruvate.

1) gluconeogenesis to make glucose

2) alcoholic fermentation to make ethanol and CO2

3) lactic fermentation to make lactate

4) pyruvate decarboxylation and CAC to make CO2

Describe the significance of lactic fermentation in cells that do glycolysis under anaerobic conditions.

In the absence of oxygen, cells cannot send pyruvate to mitochondria for oxidative phosphorylation. Instead, it undergoes lactic fermentation. This is the process where pyruvate and NADH convert to lactate and NAD+ via lactate dehydrogenase. The production of NAD+ ensures that glycolysis can continue