Question

Methanol and ethanol are both oxidized by alcohol dehydrogenase (ADH). Methanol is poisonous because it is oxidized by ADH to form the highly toxic compound formaldehyde. The adult body has 40 L of water, throughout which these alcohols are rapidly and uniformly distributed. The densities of both alcohols are 0.79 g/mL. The Km of ADH for ethanol is 1.0x10-­‐3 M, and for methanol the Km is 1.0x10-­‐2 M. The molecular mass of methanol is 32 g/mol and that of ethanol is 46 g/mol. If an individual consumed 100 mL of methanol (a leathal amount), how much 100-­‐proof whiskey (50% ethanol by volume) must this person consume in order to reduce the activity of ADH toward methanol to 5%? Assume the Ki for ethanol = Km.

Answer 1

Step 1

Calculate the concentration of methanol

Molarity = moles of methanol / volume of solution in L

Moles = mass/ molar mass

Mass = density x volume

volume of methanol = 100 mL

mass of methanol = volume x density = 100 mL x 0.79 g /mL = 79.0 g

molarity = 79.0 g x 1 mol methanol / 32.04 g methanol ) /40.0 L

=0.06172

Now we know

Velocity of methanol without inhibitor is given as follow.

v = Vmax x [S]/ km +[S]

with using inhibitor

v2 =Vmax x [S] / km^app + [S]

we know V2 with inhibitor = 0.05 V

lets plug this value

0.05 x ( km [S] / Km + [S] ) = km [S]/ Km(app) + [S]

Km (app) = (Km+[S]/0.05 ) –[S]

We also know

Km (app) = km ( 1+ [I]/Ki)

Lets plug this value

km ( 1+ [I]/Ki) = (Km+[S]/0.05 ) –[S]

[I]= ki( 1/km ( km + [S]/ 0.05 – [S] ) – 1

Lets plug the value of substrate concentration ( methane) and ki (ethanol ) and km ( methanol)

[I] = (1.0 E-3 ) ( 1/ 1E-2) x ( 1E-2 + 0.0617) /0.05 ) – 0.0617)

[I]= 0.136 M

Calculate volume

Volume = molarity x volume x molar mass / density

= 0.136 M x 40.0 L x 46.07 g / 0.79 g per mL

= 317.6 mL

Volume needed for 100 percent

= 317.6 x 2 = 635 mL

The answer is 635 mL