Lecture 7 Flashcards
What is the general strategy of oxidative metabolism?
To oxidize glucose to 6 molecules of CO2 which is highly favorable. The electrons removed during this oxidation are transferred to cofactors NAD+ and FAD which are stable electron carriers that shuttle the electrons into the electron transport chain where they can be used to reduce oxygen to water and produce ATP in the process.
What is the general goal of the preparation phase of glycolysis?
Of the payoff phase?
Preparation: to convert glucose to 2 molecules of GAP
Payoff: to convert each molecule of GAP to pyruvate which can be used to enter the kreb’s cycle or used to make intermediate for other biosynthetic pathways
What is the first step of glycolysis?
Hexokinase phosphorylates glucose at carbon 6 to produce Glucose 6 Phosphate. This causes the glucose molecule to be charged, thus trapping the molecule inside the cell. The phosphate from this reaction comes from ATP.
Why is an imine less electrophilic than an aldehyde/ketone?
Because the electronegativity of oxygen is greater than that of nitrogen, so the pi* c=o bond is lower in energy than the pic=n bond, which makes pic=o a better acceptor and therefore better electrophile
Why does nature form iminium ions and enamines in order to make aldol reactions more favorable?
Because this makes the carbon associated with the iminium a better electrophile and there are lysine residues in enzyme actives sites that are able to readily make this transformation happen. This promotes the aldol/retro-aldol reaction making it more favorable. Similarly, enamines are more nucleophilic than enols so their use drives the reaction forward.
What is step 4 of glycolysis?
Fructose 1,6 Bis-Phosphate is converted to dihyroxyacetone phosphate (DHAP) and Glyceraldehyde 3 Phosphate (GAP) via enzyme Fructose 1,6-Bisphosphate aldolase
What is step 5 of glycolysis?
DHAP is converted to GAP by Triosephosphate Isomerase (TIM)
What are the stereoelectronic requirements of TIM’s first step?
1) Lysine in active site forms ion-ion interaction with PO32- which stabilizes negative charge and anchors substrate in pocket
2) Histidine in active site forms hydrogen bond between H on nitrogen and carbonyl of DHAP, making he pKa of Histidine’s hydrogen lower
3) The sigma c-h bond that is alpha to the carbonyl is parallel to the pi*c=o of the carbonyl
4) Glutamate in the active site is positioned such that the negatively charged oxygen is in proximity to the alpha hydrogen that is deprotonated in the first step
What are the stereoelectronic requirements of TIM’s second step?
1) Sigma o-h is parallel to pi*c=c
2) glutamate is always positioned in the active site above the substrate, so the proton that is added in the second step to carbon 2 is always added on top so it is a very stereospecific reaction, always produces R product
How is the negative charge that is produced on Histidine during the TIM mechanism stabilized?
By the presence of a nearby macrodipole on an alpha helix portion of the enzyme that has a net positive charge on one end and net negative charge on the other. The dipole is oriented such that the net positive end is positioned near the Histidine residue so the negative charge is stabilized by the nearby alpha helix.
What is step 6 of glycolysis?
GAP –> 1,3 BPG (remove hydride, adds phosphate)
Catalyzed by GAPDH (utilizes deprotonated cysteine in active site and NAD+ and a molecule of inorganic phosphate Pi)
Why is the thioester that is formed in step 6 during GAPDH mechanism a good electrophile?
Because sulfur is a large atom, so the orbitals that contains its lone pairs have poor overlap with the pi* system of the adjacent carbonyl, meaning that the lone pair on sulfur does not easily donate into the carbonyl so the carbonyl is left fairly electron deficient making the thioester a good electrophile.
What is step 7 of glycolysis?
1,3 BPG is converted to 3 PG.
The phosphate group on carbon 1 of 1,3 BPG is chelated with MG2+ and then the terminal oxygen of ADP attacks the phosphate on carbon 1 of 1,3 BPG to steal the phosphate from 1,3 BPG.
Describe why steps 6 and 7 are coupled.
Step 6 is endergonic (positive delta G and unfavorable) and step 7 is exergonic (negative delta G and favorable). Coupling these 2 reactions helps drive step 6 forward since the net of the two reactions is still negative. Additionally, the product of step 6 (and thus substrate of step 7) is kept in low concentration so as soon as 1,3 BPG is produced it is consumed by step 7 to produce 3PG. According to Le Chatelier’s principle, this helps drive the reaction forward for step 6 which is the unfavorable step.
What is step 8 of glycolysis?
3 PG is converted to 2 PG by phosphoglycerate mutase. Utilizes 2 histidine residues in active site, one of which begins mechanism with a phosphate group attached.
