As we discussed in class, alcohol dehydrogenase (ADH) will oxidize several alcoh

Question

As we discussed in class, alcohol dehydrogenase (ADH) will oxidize several alcohols. We also discussed when methanol is the substrate for ADH that the toxic compound formaldehyde is produced, and that ethanol can be used as a competitive inhibitor to prevent this problem. Assume a dog has accidently consumed 50 mL of windshield washer fluid (50% methanol), the owners knows that unless he corrects for this soon the dog will be in trouble. The owner has several bottles of wine that he recently purchased for a party (12 % ethanol). How much wine does the dog need to consume to reduce the activity of ADH for methanol to 5% of its normal Km?

ADH-ethanol Km = 1mM

ADH-methanol Km = 10mM

Assume the “working volume” of the dog is 15L

Assume the densities of both liquids are 0.9 g/mL

Explanation / Answer

Volume of windshield washer fluid = 50 mL

Since it has 50% methanol so volume of methanol = (50 / 100) * 50 = 25 mL

Mass of methanol = 25 mL * 0.9 g/mL = 22.5 g

Moles of methanol consumed by the dog = 22.5 g / 32 g/mol = 0.7 mol

Now he consumed 0.7 mol of methanol and diluted it in a working volume of 15 mL

Concentration = 0.7 mol / 15 L = 46.9 mM

Enzyme will function at Vmax * 46.9 mM / ( 10 mM + 46.9 mM) = 0.82 * Vmax

Now,

5 % of 0.82 Vmax = 0.041 Vmax

0.041 Vmax = Vmax * 46.9 mM / [ 10 mM ( 1 + ( [Ethanol] / 1 mM)) + 46.9 mM ]

[Ethanol] = 109 mM

In order to raise the alcohol concentration to 109 mM the dog must drink 109 mM * 15 L = 1.63 moles of ethanol

Mass of alcohol or ethanol to be consumed = 1.63 mol * 46 g/mol = 75 g ethanol.

Mass of ethanol = 75 / 12 / 100 = 625 g

Volume of wine to be consumed = 625 g / 0.9 g/mL = 694 mL wine.

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