physikalische Grundlagen der Biochemie

\(A\xrightarrow[]{k} B\)

\(v=-\frac{d[A]}{dt}=k\cdot [A] \rightarrow\int^{[A]}_{[A]=A_0}\frac{d[A]}{[A]}=k\cdot \int^t_0dt \)

\(\rightarrow-\ln\frac{[A]}{A_0}=k\cdot t \rightarrow [A]=A_0\cdot e^{-k\cdot t}\)

\(t_{1/2}=\frac{\ln2}{k}=\frac{0.693}{k}=0.693\cdot \tau,\ \tau=\frac{1}{k}\) is the "time constant" (unit: s)

\(v_i=k\cdot[A_0]\)

\(A+B\xrightarrow[]{k}C,\ \frac{[A]}{dt}=-k\cdot[A]\cdot[B]\)

\(-\frac{d[A]}{dt}=k\cdot[A]^2 \rightarrow-\int^{[A]}_{[A]=A_0}\frac{d[A]}{[A]^2}=k\cdot\int^t_0dt\)

\(\rightarrow \frac{1}{[A]}-\frac{1}{[A_0]}=k\cdot t \leftrightarrow [A]=\frac{1}{k\cdot t+\frac{1}{[A_0]}}\)

\(\Delta G^{\circ '}=-n\cdot F\cdot \Delta E'_0\)

\(\Delta G_{Transport}=R\cdot T\cdot \ln \frac{c_2}{c_1}+z\cdot F\cdot\Delta V\)

\(\mathrm{A+B\rightarrow C;\ [A_0]\gg[B_0]}\)

\(k_{pseudo}=k[A_0]\ \mathrm{unit\ s^{-1}}\)

\([B]=[B_0]\cdot e^{-k_{psuedo}\cdot t}\)

\(t_{1/2}=\frac{\ln2}{k_{pseudo}}=\frac{\ln2}{k\cdot [A_0]}\)

small molecules: \(7\cdot 10^9 M^{-1}s^{-1}\)

protein-ligand interactions: \(10^8-10^9 M^{-1}s^{-1}\)

\(v=-\frac{d[A]}{dt}=k\)

\(k=A\cdot e^{-E_A/RT}\)

\(\frac{k_{cat}}{k_{uncat}}=e^{\frac{\Delta E_A}{RT}}\)

•  Enzymes accelerate chemical reactions by lowering the activation energy. - Enzymes leave the reaction they are catalysing in the same state in which they entered the reaction.
• Enzymes act in substoichiometric (catalytic) amounts because an individual enzyme catalyses the same reaction over and over again (multiple turnovers).
• Under conditions where the substrate concentration is lower than enzyme concentration ([S0]>>[E0]), the velocity of the catalyzed reaction increases linearly with enzyme concentration.
• Enzymes show saturation behaviour at high substrate concentrations: When all binding sites are occupied with substrate, the maximum reaction velocity (vmax) is reached and stays constant even when substrate concentration is further increased (zero order reaction conditions, see 1.4.).
• Enzymes accelerate the forward and reverse reaction by the same factor. Therefore, they do not change the equilibrium (energy difference) between substrate and product,
  but they accelerate the attainment of the equilibrium between substrate and product.

\(\mathrm{v_i=v_{max}\frac{[S]}{K_M+[S]},\ v_{max}=k_{cat}\cdot[E_0]}\)

\(\mathrm{K_M=\frac{k_{-1}+k_{cat}}{k_1} }\)

enzyme works with \(\mathrm{v_{max}/2}\)

typiccali in the range of \(10^{-3}-10^{-6} \mathrm{\ M}\)

\(\mathrm{k_{cat} }\)

• maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration \( \mathrm{\displaystyle [E_{T}]}\).
• first order rate constant, cannot be any greater than any first-order rate constant on the forward reaction pathway
• is the slowest unimolecular step in the catalytic cycle of an enzyme
• in range of \(\mathrm{10^5-10^6\ s^{-1}}\)
• \(\mathrm{\frac{1}{k_{cat}}=}\)time required to convert a single substrate molecule to product in the steady state at saturating [S]

\(\mathrm{ k_{cat}/K_M\ unit\ M^{-1}s^{-1} }\)

• how efficiently an enzyme converts substrates into products
• The higher the specificity constant, the more the enzyme "prefers" that substrate.
• second-order rate constant
• refers to the properties and the reactions of the free enzyme and free substrate
• used for quantifying the catalytic efficiency of an enzyme
• \(\mathrm{ k_{cat}/K_M \leq }\) second-order rate constant on the forward reaction pathway
• maximum theoretical value: \(\mathrm{ \sim 10^9\ M^{-1}s^{-1} }\) (value of a diffusion controlled second-order reaction)
• most efficient enzyme known is superoxide dismutase \(\mathrm{7\cdot 10^9 \ M^{-1}s^{-1}}\)

\(\mathrm{ v_i=k_{cat}\cdot[E_0]\cdot \left[\frac{[Glucose]}{K_{M\ Glucose}+[Glucose]} \right]\cdot\left[ \frac{[ATP]}{K_{M\ ATP}+[ATP]} \right] }\)

the enzyme only works with 0.25 \(\mathrm{ v_{max} }\) if \(\mathrm{ [Glucose]=K_{M\ Glucose} }\) and \(\mathrm{ [ATP]=K_{M\ ATP} }\).

If both binding sites are 50% occupied, only 25% of the enzyme molecules will be occupied with both substrates (requirement for catalysis).

The K_M values of both substrate must be determined independently!

• defines a simple binding equilibrium
• k_on for the association of the protein P and its ligand L to the protein-ligand complex PL
• k_off for the spontaneous dissociation of PL
• \(\mathrm{ P+L\overset{k_{on}}{\underset{k_{off}}{\rightleftharpoons}}PL }\)

\(\mathrm{k_{on}\cdot P\cdot L=k_{off}\cdot PL}\)

\(\mathrm{K_{Diss}=\frac{k_{off}}{k_{on}}=\frac{P\cdot L}{PL}=\frac{1}{K_a} \ unit: M}\)

K_a can be used to describe the strength of interaction of a protein-ligand complex (unit: M^-1)

• L will not decrease significantly even if all P molecules were occupied with L
• L after attainment of equilibrium is therefore equal to L_tot
• \(\mathrm{ K_{Diss}=\frac{P\cdot L_{tot}}{PL}\ or\ \frac{K_{Diss}}{L_{tot}}=\frac{P}{PL} }\)

\(\mathrm{y=\frac{PL}{P_{tot}}=\frac{(K_{Diss}+L_{tot}+P_{tot})-\sqrt{(K_{Diss}+L_{tot}+P_{tot})^2-4\cdot P_{tot}\cdot L_{tot}}}{2\cdot P_0}}\)

 \(\mathrm{K_{Diss}=\frac{P\cdot L_{tot}}{PL}\ if\ L_{tot}\gg P_{tot}}\)

\(\mathrm{\Delta G^{\circ} {'}=+\mathrm{\Delta G^{\circ} {'}}\ Produkt-\mathrm{\Delta G^{\circ} {'}}\ Edukt}\)