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Summary of exam notes for Metabolic Biochemistry covering Enzyme Kinetics and Inhibition, Nitrogen Metabolism, Glycolysis, PPP, Gluconeogenesis,
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Samenvatting biochemistry seventh edition Hfdst 11 tm 13 + 16 tm 19 + 22
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Metabolisme Samenvatting
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Samenvatting H8 "Biochemistry: A Short Course"
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Voorbeeld van de inhoud
Enzyme Kinetics & InhibitionL-B Graph & Tabulated Table:
- (mM) = (mM-1) (µmol s-1) = (µmol s-1)-1 (µmol min-1) = (µmol min-1)-1
- Ki Value (dissociation constant for the binding of the inhibitor to the enzyme) should be in 3.s.f. and in
mM e.g. 0.0309 mM or 30.9 µM. In drugs, the smaller the
Ki value is, the smaller amount of medication needed in order - Competitive Inhibition: Km Vmax
to inhibit the activity of that enzyme = most potent/best (Kmi > Km) and (Vmaxi = Vmax)
inhibitor - Uncompetitive Inhibition: Km Vmax
(Kmi< Km) and (Vmaxi< Vmax)
Title for LB: ‘’ Lineweaver-Burk Plot of 1/V versus 1/[S] for an - Non-competitive Inhibition: Km Vmax
enzyme-substrate complex in the presence/absence of a reversible (Kmi = Km) and (Vmaxi< Vmax)
inhibitor’’. (if name of inhibitor is stated then mention in title ie. ‘’LB- Inhibitor does not effect the equilibrium of I binding to
Plot … for the enzyme Pyruvate Kinase reaction in the [S] so Km remains the same
presence/absence of 2-phosphoglycerate’’).
Inhibitors are substances that work by binding directly to the enzyme and blocking the active site of the enzyme
thereby blocking binding of S and causing the enzyme-catalysed
reaction to occur more slowly.
‘’ Type of inhibition taking place is .. It is clear that Km has.. while V
max remains unchanged (within experimental error).’’
Intercepts on axes + Units: example
1/Vmax(i) = 0.565 (µmol s-1)-1 so Vmax(i)= 1.77 µmol s-1
-1/Km(i) = - 0.911 (mM-1) so Km(i)= 1.10 mM
Show that when V0=1/2 Vmax then [S]= Show that when [S]=Km, V0= 1/2 Show that when Km=[S], V0= 1/2
Overcoming/Relieving Inhibition by increasing [S]: Molecular level
Km Vmax
All cases of non-competitive inhibition and in some competitive inhibition (where I binds elsewhere than Vmax active site) are forms of allosteric
-M-M: V0= Vmax [S] -M-M: V0=
regulation. I binds to T state = allosteric inhibitor. Substrates bind to RVmax
state=[S]
allosteric activators -M-M: V0= Vmax [S]
(Km + [S]) (Km + [S]) (Km + [S])
Competitive Inhibition:
-Inhibitor binds to Free E only. Binding of I and S to enzyme is mutually exclusive
-A-Let V0 = ½ inhibitor
competitive Vmax, ½ Vmax=by
functions Vmax [S] with the substrate
competing -When [S] for = Km,
the V0=site
active Vmax Km
of the -When the
enzyme. Therefore, Km= rate[S], V0= Vmax
of catalysis [S] on the
depends
relative concentrations of the I and(KmS. + [S])
This means that increasing [S] will overcome Kminhibition
+ Km by decreasing the chance of the inhibitor[S] + [S]
binding to
-½
the Vmax (Km
enzyme. + [S])
Change = reflects
in Km Vmax [S]a decrease in affinity -V0 = enzyme
of the Vmax Km for substrate due to inability of S molecule
-V0 = Vmax to [S]
interact with enzyme mol
- ½ Vmax
bound to I. Km +1/2 Vmax [S] = Vmax [S] 2 Km 2 [S]
- ½ Vmax Km =Inhibition:
Non-Competitive ½ Vmax [S] V0= Vmax = ½ Vmax V0= Vmax = ½ Vmax
-A-Km=
non-competitive
[S] inhibitor reacts with E:S or free E or both respectively.
2 S and I do not compete for same binding2 site. It binds somewhere else
than active site but its able to change the conformation of the active site in a way that [S] cannot efficiently catalyse the reaction
-This means increasing [S] will NOT overcome inhibition because it does not bind to the [S] binding site and binding of S and I are independent of
each other. Level of inhibition depends on conc. of I= Vmax for an in enzyme in presence of non-c I will be less than under uninhibited conditions.
-They slow the rate of reaction to form the E:S product complex.
Uncompetitive Inhibition: aka anti-competitive inhibition
-Inhibitor binds to the E-S complex only. V max decreases in presence of I as some of the enzyme molecules will be ‘out of commission’. Km
decreases also because of some S is bound to ESI complexes therefore cannot be converted into product.
-This means increasing [S] will NOT relieve inhibition. Binding to E-S complex makes it more difficult for S to dissociate or be converted into product.
Michaelis-Momentum Equation:
Calculating Ki for Competitive Inhibition: Km (i) = Km (1+ [I])/Ki V0= Vmax [S]
Calculating Ki for Non-Competitive Inhibition= Vmax (i)= Vmax / (1 + [I]/Ki) (Km + [S])
Calculating Ki for Uncompetitive Inhibition = Vmax (i)= Vmax / (1 + [I]/Ki)
or Km (i) = Km/ (1+ [I]/Ki)
Nitrogen Metabolism
Nitrogen fixation
Is a process where inorganic molecular nitrogen (N2) from the atmosphere is converted to the biologically useful
ammonia (NH4+) by symbiotic bacteria (living in plants) or non-symbiotic bacteria (living in soil) using bacterial
nitrogenase enzyme complex. High ADP levels inhibit nitrogenase.
Nitrogen Cycle:
1) Nitrogen fixation converts N2 to NH4+. Nitrate can also be converted to NH4+.
2) Ammonia is transformed back to N2 by nitrification followed by denitrification.
3) Overall reaction: N2 + 10H+ + 8e- + 16ATP → 2NH4 + 16 ADP + 16Pi + H 2 (16 ATP hydrolysed per N 2
molecule reduced)
AA synthesis:
- 6 Biosynthetic AA families: Oxaloacetate, Pyruvate, Ribose-5-Phosphate, α-
ketoglutarate, 3-phosphoglycerate, Phosphoenolpyruvate, Erythrose 4-phosphate
Humans can synthesize 20 AA found in proteins. Essential AA cannot be synthesised from humans &
must be obtained from dietary sources. Conditional AA are required during illness/stress.
- 2 main reaction synthesis of AA ; 1) Animation 2) Transamination
Animation: Ammonia reacts with a-ketoglutaric acid to form glutamic acid via
glutamate dehydrogenase and glutamine synthetase. Combined actions of Glutamine
synthetase and glutamate synthase (NADPH) assimilates fixed nitrogen into an organic
compound α- ketoglutarate (a citric acid intermediate) to produce AA Glutamate.
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