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Biochemistry-Inhibitor Question

  1. Mar 22, 2010 #1
    1. Question with all known variables

    The following question has got me completely confused. I know there was someone previous that had posted a question almost identical to mine however there are slight but significant changes.

    Ethanol in the body is oxidized to acetaldehyde by liver alcohol dehydrogenase (LADH). Other alcohols are also oxidized by LADH. For example, methanol, which is mildly intoxicating, is oxidized by LADH to the quite toxic product of formaldehyde. The toxic effects of ingesting methanol (a component of many commercial solvents) can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of 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 30 mL of methanol (a lethal dose), how much 80 proof whiskey (40% ethanol by volume) must be imbibed to reduce the activity of his LADH towards methanol to 1% of its original value? The adult human body contains ~40L of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. The densities of ethanol and methanol are both 0.79 g/cm3. Assume the KM values of LADH for ethanol and methanol to be 10-3 M and 10-2 M, respectively, and that Ki = KM for ethanol.

    The changes are the mL of methanol ingestion (30mL) and the proof of the whiskey (40% ethanol; 80 proof), and the percentage of reduction (1%).

    I applied the formulas from the last post in reference to this question but am completely lost. The formulas used last time were:

    2. Formulas Applied
    alpha=1 + ([etOH]/KetOH)

    (V[meOH]/V[etOH])= (Vmax*[meOH]/KmeOH+[meOH])/(Vmax*[meOH]/alpha*KmeOH+[meOH])

    which reduces to

    (V[meOH]/V[etOH])=(alpha*KmeOH+[meOH])/(KmeOH+[meOH])

    3. My attempt

    Molarity of methanol: 30mL; which equates to 23.7g of methanol; in 40L that is equal to 0.5925 g/L

    Dividing the molecular weight by 32.04g/mol I get 0.0184925 which is approximately 0.02M; Km is 0.01M

    Since the molar mass of methanol and ethanol are two fold, I can multiply the g/l by 4.

    However, unlike the previous problem, I cannot multiply by 2 because I do not have 50% EtOH, so because 40 is less than 50 I assume to multiply by 2.5 yielding:

    (30mL)(4)(2.5)=300mL

    300mL of EtOH to effectively reduce the Methanol to 1%?

    I'm completely lost.
     
  2. jcsd
  3. Mar 22, 2010 #2

    epenguin

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    Homework Helper
    Gold Member

    Yeah that's quite reasonable, I make it more than that. (But it's late and I'd like to do the calc. tomorrow or so.)

    You do well to ask if a number sounds reasonable. But consider, to reduce the MeOH binding to 1% of what it would be, the ethanol has got to occupy 99% of the sites. To do that if there were no methanol it would already have to be of the order of 100X its Ki, so about 0.1M. If the MeOH had comparable or more affinity than EtOH that would still not be enough, but as KMeOH is 10X more than KEtOH that is not bad for rough calculation.

    You swallowed 30ml of Methanol, you need 10X as many moles of ethanol by what I just said above, that's more than 10X the number of grams because the M.Wt. of ethanol is higher, and the meth was neat but the ethanol source is only 40% so I think it's in excess of 600 ml.

    Your kinetic calculation is a bit hard for me to follow. Maybe your have been taught a way of seeing things with this alpha; I don't find it helps transparency and maybe you have not been careful about the bracketing when you wrote it? Otherwise I think your formula is correct though.
    Personally for clarity I would write out the formula for vMeOH (Michaelis eqn.) , that in presence of what is for present purposes a competitive inhibitor which you call vEtOH, then divide one by the other, but I think that is what you did.
     
    Last edited: Mar 23, 2010
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