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Competitive inhibitors (biochemistry)

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data

    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 100 mL of methanol (a lethal dose), how much 100 proof whiskey (50% ethanol by volume) must be imbibed to reduce the activity of his LADH towards methanol to 5% 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 1.0 X 10-3 M and 1.0 X 10-2 M, respectively, and that Ki = KM for ethanol.


    2. Relevant equations

    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. The attempt at a solution

    after determining the concentration of methanol in the body to be 62mM the above equation is filled in with the values

    100%/50%= (alpha*1.0e-2+62e-3)/(1.0e-2+62e-3)

    20=(alpha*1.0e-2+62e-3)/(1.0e-2+62e-3)

    alpha=(20*7.2e-2-62e-3)/1.0e-2=137.8

    then, using the top equation and the determined and given values we get

    137.8=1+[etOH]/1.0e-3

    which gives [etOH] as 0.1368M

    But I can't remember how to convert M to L, and I'm not sure if this is even the right way to solve the problem, so any help on what is done, and where I go from concentration would be great! Thanks!
     
  2. jcsd
  3. Mar 17, 2009 #2
    Oh- I think I figured out how to get how many liters of EtOH are needed-

    0.1368 M * 40 L = 5.472 mol
    5.472mol * 46g/mol= 251.71g
    251.71g * 1mL/0.79g= 318.62mL

    If anyone knows how to do this, can you check my work and let me know if it's right. Thanks again!
     
  4. Mar 18, 2009 #3

    epenguin

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    One of us is out by about 1 order of magnitude.

    A ballpark figure would be good enough if this were a practical question. I'd do that to get basic understanding and clear up the trivial mistakes one of us has made and then maybe refine for academic purposes.

    You will not be far out if you say occupy 95% of the sites with ethanol because I worked out the methanol is about 2 mM quite below its Km, i.e. the enzyme far from saturated by this dose of methanol.



    Km is 10-3M.

    So to occupy 95% of sites you need around 1/0.05 = 20.Km ethanol.
    = 0.02 M ethanol.

    M. Wt. ethanol = 46
    so that's 0.92 g/l (nearly 1)you need in the blood.
    40 l blood, you need 36.8g

    you're told density ethanol 0.79 g/ml so you need 36.8/0.79 = 46.6 ml

    it's 50% by vol so drink 93 ml

    Feasible.

    All I had time for, I will come back later today or tomorrow
     
    Last edited: Mar 18, 2009
  5. Mar 18, 2009 #4
    I can see where you were going with it- we just are given certain equations to use and the textbook has got to be one of the worst for anything (I hate! Wiley text) and doesn't really explain the equations very well. Unfortunately, neither does my professor
    :( so we are forced to look at equations that contain the variables that we are using and have the values that we would be searching for, and wriggle them around to make them work. Sometimes it doesn't turn out so well.....

    the equations were Michaelis-Menten reactions for enzymes- one with no inhibitors, one with competitive inhibitors. So that's why the equation worked out the way it did....have to turn in my homework now, but thanks anyway!
     
  6. Mar 18, 2009 #5

    epenguin

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    Gold Member

    Back now. I did it before in a hurry and of course there seems to be a mistake and now I am nearer your answer. :blushing:

    I seem to have miscalculated the molarity of the methanol. 100 ml that's 79 g, in 40 l that's nearly 2 g/l.

    Divide by MW 32 I make it around 0.06 M. So with Km of 0.01 M the methanol instead of unsaturating is fairly saturating.

    So then we're into the opposite approximation to what I said. The ethanol has 10 X more affinity than methanol, so at equal concentrations will bind 10 X more ethanol than methanol. You want that to be about 20 X so ethanol has to be about twice the molarity of the methanol. The MW ethanol is very roughly twice that of methanol so you need 4X the g/l, 4X the g in fact .

    He drank pure methanol according to the problem, but you have 50% alcohol as whiskey so that's 8X the vol. 800ml :eek: . Actually a more refined calc with equations looking a bit like yours gives me 640 ml. Still :eek:.



    What is this book?
    This kind of enzyme kinetics is really not difficult, but if you don't get the basic ideas straight it will look like a lot of incomprehensible equations that all look the same; when you have it straight you can almost see the equation representing the mechanism.

    It boils down to the enzyme is distributed between different forms represented by equilibrium equations between them, which involve eq constants and subs. or inhibitor molarities. The rate depends on the amount of one or two of those forms. The mechanism easily tells you how. Write down how, like here v = k*[Enz-MeOH] (the complex). Write down how much that complex is as fraction of all of the forms - here [E]+[E-EtOH]+[Enz-MeOH], this is give by the equilibrium equations. But you need to think what is really happening and what you expect otherwise you just have confusing equations without intuition.

    Then in this problem they added a bit of what you have to do in real life like worry about molarities, volumes, Molecular weights etc. I made a mistake with that, I wonder if it's right now? :confused:
     
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