quiz2_answer - 3 + <--> (R)-NH 2 + H...

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Quiz 2 Question A Phosphorylation of glucose in presence of ATP in the cell: Glucose + ATP <--> G6P (Glucose-6-Phosphate) + ADP 1) What is the mass action ratio (Q) for this reaction? [ADP][G6P]/[Glucose][ATP] 2) What values do you need to calculate ! G o ? Please also include the equation. [ADP], [Glucose], [G6P] and [ATP] at the equilibrium ! G o ’ = -RT ln K eq 3) ! G o ’ = 7 kJ•mol -1 . Can you predict in which direction the reaction would go? ( no full credit without explanation ) No, need to know ! G’ . ! G o > 0: non favorable regarding standard conditions , but to predict in the cel l, we also need to consider the initial conditions since Question B Let’s consider the following acid dissociation reaction ( pKa=8) : (R)-NH
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Unformatted text preview: 3 + &lt;--&gt; (R)-NH 2 + H + 1) At which pH would the charged form be predominant? ( no full credit without explanation ) pH &lt; pKa. Once pH &gt; pKa, the basic (deprotonated, neutral) form is predominant. or the acid form is charged, so the uncharged (basic form) is predominant pH &gt; 8 2) At pH=7, how would you calculate the ratio of base to acid? (only equation is needed) pH = pKa + log([base]/[acid]) 3) Lets assume (R)-NH 3 + is a drug. Its absorption requires passage through a membrane. At which pH will the drug be optimally absorbed? Why? pH &gt; 8 To go through a membrane, the drug needs to be deprotonated (neutral) because of the hydrophobic environment of the membrane. &quot; G = &quot; G ! + RT ln Q...
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