Liver alcohol dehydrogenase (ADH) is a nonspecific enzyme. Its normal substrate is ethanol but it will oxidize other primary alcohols like methanol. Methanol produces formaldehyde which is quite toxic and can lead to blindness. A dog ingested about 36 ml of windshield fluid which is an aqueous solution of 50% v/v methanol. Methanol will be excreted if the oxidation of formaldehyde is blocked. Ethanol can act as a competitive inhibitor of methanol oxidation by ADH. The dog is offered a a brew that contains 4.2% v/v ethanol. How much brew must the dog consume in order to lower the activity of the ADH on methanol to 5% of its uninhibited value, if the Km values for canine are 1mM for ethanol and 10mM for methanol? (Assume the Ki for ethanol in its role as a competitive inhibitor os methanol oxidatino is the same as its Km. Both methanol and ethanol will quickly distribute throughout the dog's 17 L of body fluids. The densities of both methanol and ethanol are 0.79g/mL).

Please help. Anything will be greatly appreciated!

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To solve this problem, we need to calculate the amount of brew (ethanol) the dog must consume in order to lower the activity of ADH on methanol to 5% of its uninhibited value.

1. Calculate the concentration of methanol in the windshield fluid:
- The windshield fluid is 50% v/v methanol, which means 50 ml methanol in 100 ml of fluid.
- Therefore, the concentration of methanol in the windshield fluid is (50 ml / 100 ml) * 0.79 g/ml = 0.395 g/ml.

2. Calculate the amount of methanol ingested by the dog:
- The dog ingested 36 ml of windshield fluid, so the amount of methanol ingested is 36 ml * 0.395 g/ml = 14.22 g.

3. Calculate the concentration (in mM) of methanol in the dog's body fluids:
- Convert the mass of methanol to moles: 14.22 g / 32.04 g/mol = 0.444 mol.
- Convert moles to mmol: 0.444 mol * 1000 = 444 mmol.
- Divide mmol by the total volume of body fluids (17 L): 444 mmol / 17 L = 26.12 mM.

4. Calculate the concentration (in mM) of methanol that would inhibit ADH activity by 95%:
- ADH activity needs to be reduced to 5% of its uninhibited value.
- Therefore, the concentration of methanol should be 5% of its Km value for canine, which is 10 mM.
- 5% of 10 mM = 0.05 * 10 mM = 0.5 mM.

5. Calculate the concentration of ethanol (brew) needed to achieve the desired inhibition:
- The Km value for ethanol is given as 1 mM.
- The concentration of brew needed can be calculated using the equation for competitive inhibition:

Ki = [E] [I] / [EI]

- Rearranging the equation to solve for [I], we get:

[I] = (Ki * [EI]) / [E]

- Substituting the known values, we get:

[I] = (1 mM * 0.5 mM) / 10 mM = 0.05 mM.

6. Convert the concentration of ethanol (brew) to the amount (in ml) needed to achieve the desired inhibition:
- The concentration of brew is given as 4.2% v/v, which means 4.2 ml ethanol in 100 ml of brew.
- Therefore, the concentration of ethanol (brew) is (4.2 ml / 100 ml) * 0.79 g/ml = 0.03318 g/ml.
- Convert the mass of ethanol to moles: 0.03318 g / 46.07 g/mol = 0.000719 mol.
- Convert moles to mmol: 0.000719 mol * 1000 = 0.719 mM.
- Divide the desired concentration ([I]) by the concentration of ethanol (brew):

Amount of brew = [I] / Concentration of brew
= 0.05 mM / 0.719 mM
= 0.0696 ml.

Therefore, the dog must consume approximately 0.0696 ml (or 69.6 μl) of the brew to lower the activity of ADH on methanol to 5% of its uninhibited value.