5) Methanol and ethanol are both oxidized by alcohol dehydrogenase (ADH). Methan

Question

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

Explanation / Answer

solution:

The enzyme catalyzed reaction that takes place is:

E + S ---->ES ---->E + P

where E is enzyme alcohol dehydrogenase(ADH), S is substrate(methanol) and P is the product(formaldehyde). Ethanol is used as an inhibitor (I).

According to Michaelis-Menten kinetics, the reaction velocity of methanol conversion is given as:

v = Vmax[S] / Km+[S]

when inhibitor (I) is added, the velocity becomes

v2 = Vmax[S] / K'm+[S] where K'm is a new value for Km.

In order to reduce the activity of ADH towards methanol to 5%, velocity of methanol reaction should be 0.05 times its original value.

so, 0.05*v = v2

sustitute for v and v2 we get:

0.05*Vmax[S]/Km+[S] = Vmax[S]/K'm+[S]

Now we want an expression for K'm. So, divide both sides of above equation by Vmax[S] and then multiply both sides by K'm+[S], we get

0.05(K'm+[S]/Km+[S]) = 1

then multiply both sides by Km+[S] and divide by 0.05,

K'm+[S] = Km+[S]/0.05

Subtract [S] from both sides and thus we get the expression for K'm:

K'm = Km+[S]/0.05 - [S] ..............(1)

According to the Michaelis-Menten model for competitive inhibition, we get value of K'm :

K'm = Km*(1+[I]/Ki) .......................(2)

Ki is the dissociation constant of ethanol.

Equating equations (1) and (2), we get

Km*(1+[I]/Ki) = Km+[S]/0.05 - [S] .......................(3)

now, we have to calculate the concentration of ethanol that is [I].

So, get the expression for [I]

Divide equation (3) by Km and subtract 1 from both sides,

[I]/Ki = 1/Km*(Km+[S]/0.05 - [S]) - 1

now, multiply both sides by Ki,

[I] = Ki*(1/Km*(Km+[S]/0.05 - [S]) - 1)

We have, Km for ethanol = 1.0*10-3 M

and Km = Ki for ethanol, so Ki = 1.0*10-3 M

Km for methanol = 1.0*10-2 M

calculate substrate concentration as follows:

[S] = 100 mL methanol/40L * 0.79g/mL(density of methanol) * 1mol methanol/32g/mol = 0.0617 mol/L

now substitute these values in equation for [I],

[I] = 1.0*10-3 M*(1/1.0*10-2 M(1*10-2 M+0.0617 M/0.05 - 0.0617 M) - 1 )

[I] = 0.13623 M

from [I] we can calculate the volume of ethanol,

0.13623 mol/L * 40L * 46g/mol * 1ml/0.79g = 317.3mL

Since, whiskey is only 50% ethanol, volume of ethanol = 2*317.3mL = 634.6 mL

So, 634.6 mL of whiskey must be consumed.

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