BIOL 430 EXAM

Catabolism

produces energy

generates monomers

produces reduced coenzymes and (e-)

overall- OXIDATIVE

LEO- lose electrons oxidate!

to be most oxidized means to have the most oxygens and the fewest Hs

ex. carbohydrates → pyruvate

Anabolism

uses energy

uses monomers

uses reduced coenzymes and (e-)

overall- REDUCTIVE

GER- gain electrons reduce!

to be most reduced means to have the fewest oxygens and the most Hs

ex. pyruvate → carbohydrates

Heavily reduced molecules serve as _____ while heavily oxidized molecules serve as _____

Reduced → long term energy

Oxidized → short term energy

Metabolism

totality of chemical reactions within the cell

metabolic pathways possess both near and far from equilibrium reactions

Far reactions

essentially irreversible under cellular conditions

very very negative Δ G

Near reactions

easily reversible under cellular conditions

slightly negative or positive Δ G

Metabolic pathways possess both near and far rxns because…

energetically expensive to maintain (replicate and transcribe) all 10 enzymes so if near rxns didn’t exist you would need even more energy because near enzymes are shared between glycolysis and glucleogenesis (7 enzymes shared)

Thermodynamic coupling

Typically occurs in one enzyme’s active site, but endergonic and exergonic reactions occur

ex. ATP → ADP -30.5 kJ/mol (Energy required from reaction 1 that goes towards ATP hydrolysis)

G+P → G6P + ADP +13.8 kJ.mol (This is the amount of energy used in reaction 1)

total Δ G° = -30.5+13.8= -16.7 kJ/mol

Biochemical reactions Δ G° and K_eq

Need to account for cellular reactants and products to accurately determine Δ G

Δ G_cellular= Δ G°’ +RT ln(Q) where Q= prod/react

NTPs and other high energy molecules serve as “chemical currency” inside cells

Hydrolysis products (gamma phosphate clipped) are lower energy

Hydrolysis leads to more conformational states for products (resonance, reduce electrostatic

repulsion)

Δ G° and Keq for ATP+CDP → ADP+CTP

Δ G°~0 kJ/mol

so K_eq=1

NAD⁺_(ox) + 2e^- + 2H⁺ → NADH_(red) + H⁺

FAD_(ox) + 2e^- + 2H⁺ → FADH_2(red)

Both NAD⁺ and FAD have 2 electron acceptors, so why is it NADH and FADH₂ at the end?

because of the respective strucuture of each molecule- only room for one H on NAD⁺

Glycolysis def

catabolic- breaking down of glucose into 2 pyruvate (+2ATP generated)

Gluconeogenesis def

anabolic- reverse of glycolysis, pyruvate to glucose