This one is long and has lots of stops. Save it for last. (Remember, the final v

Question

This one is long and has lots of stops. Save it for last. (Remember, the final volume has to be not too unreasonable. You wouldn't want the person have to drink 500L of vodka! Ethanol in the body is oxidized to acetaldehyde by live alcohol dehydrogenate (LADH). Other alcohols are also oxidized by LADH. For example, methanol, which is mildly intoxicating, is oxidized by LADFI to the quite toxic product formaldehyde The toxic effects of ingesting methanol (a component of many commercial solvents) can by reduced by administering ethanol The ethanol acts as a competitive inhibitor of the methanol by displacing it from LADH This provides sufficient time for the methanol to be harmlessly excreted by the kidneys If an individual has ingested 100 mL of methanol (a lethal dose), how much 100 proof vodka (50% ethanol by volume) must she imbibe to reduce the activity of her LADH towards methanol to 5% of its original value? The adult human body contains about 40 L of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. The densities of ethanol and methanol are both 0.79 g/mL. Assume the K_m values of LADH for ethanol and methanol to be 1.0 times 10^3 M and 1.0 times 10^-2 M. respectively, and that K_l = K_m for ethanol Figure 5.13 Invariant residues in the alpha and beta chains of mammalian hemoglobin. The colored dots, indicating the positions of invariant residues, lint the home pockets as well as the crucial alpha, beta interface. The invariant residues have been found in about 60 species. There are 43 invariant positions in the hemoglobin molecule.

Explanation / Answer

Step1- Calculate concentration of [CH3OH]

Given - 100 ml of methanol in 40 L of fluid, and the density of methanol is 0.79 g g/ml. From this we can solve for the concentration of methanol, or the substrate concentration [S].

Molar mass of methanol = 32 g/mol

V = Vmax [S] / (KM + [S]) where S here is methanol.

The amount ingested is (100 mL CH3 OH) (0.79 g/mL) / (32 g/mol) = 2.469 mol CH3OH

That's diluted in 40 L of solution for a concentration of [CH3OH] = 2.469 / 40 L = 0.0617 M

Step2- Calculate the concentration of ethanol inhibitor [I]

To solve for the concentration of ethanol (I) we need to meet the condition that the reaction rate of methanol conversion is reduced by 95%.

Given Km for ethanol to be 1.0 x 10^-3M, and according to the problem we assume KI = Km for ethanol, so Ki = 1.0 x 10^-3M.

With the values for Km, Ki, and [S], we can solve for the concentration of ethanol inhibitor [I] required by following equation.

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

I = 1.0 x 10^-3M (1/1.0 x 10^-2M (1.0 x 10^-3M +0.0617 M /0.05 - 0.0617 M) – 1)

I = 0.136M

Step3- Calculate the amount of Vodka-

Molar mass of ethanol = 46.07 g/mol

Given the concentration of ethanol inhibitor, we can find out how much volume of ethanol to add

(0.136 mol/L) x 40L x (46.07 g/mol) x 1/0.79 g/mL = 317.5ml

Since we are using vodka, this is only 50% ethanol. Therefore, we need to use

317.5ml x 2 = 635 ml 100 proof vodka is required

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