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!
"College" is not the subject. Please be more specific.
Yeah sorry, this is my first time here. I just noticed that after I posted it.
To solve this problem, we need to determine the amount of brew the dog needs to consume in order to lower the activity of ADH on methanol to 5% of its uninhibited value. We will do this by calculating the concentration of ethanol needed to inhibit methanol oxidation, and then converting it to the volume of brew the dog needs to consume.
Let's break down the steps:
Step 1: Calculate the concentration of ethanol needed to inhibit methanol oxidation by 95%.
We know that the Km value for ethanol is 1mM. Since we want to inhibit methanol oxidation to 5% of its uninhibited value, we can use the equation for competitive inhibition:
Km(apparent) = Km * (1 + [I] / Ki)
Where:
- Km(apparent) is the apparent Km in the presence of the inhibitor (ethanol).
- Km is the Km value for methanol (10mM).
- [I] is the concentration of the inhibitor (ethanol) needed to achieve the desired inhibition.
- Ki is the dissociation constant for the inhibitor (ethanol).
In this case, the Km(apparent) is 0.05 * 10mM (5% of 10mM), which is 0.5mM.
Plugging in the values, we get:
0.5mM = 1mM * (1 + [I] / Ki)
To solve for [I], rearrange the equation:
[I] / Ki = (0.5mM / 1mM) - 1
[I] / Ki = 0.5 - 1
[I] / Ki = -0.5
Since the Ki value for ethanol in its role as a competitive inhibitor is the same as its Km value, [I] / Ki = -0.5. Therefore, [I] (the concentration of ethanol) is -0.5 * 1mM = -0.5mM.
However, concentrations cannot be negative, so we take the absolute value:
[I] = 0.5mM
Step 2: Convert the concentration of ethanol to the amount of brew.
To do this, we need to calculate the amount of ethanol in the brew in grams, using the density of ethanol (0.79g/mL).
Mass of ethanol = Volume of brew * Concentration of ethanol * Density of ethanol
We know the concentration of ethanol in the brew is 4.2% v/v, which means 4.2 mL of ethanol per 100 mL of brew.
Converting 4.2 mL of ethanol to the volume in liters:
Volume of ethanol in brew = (4.2 mL / 100 mL) * 17 L
Volume of ethanol in brew = 0.0714 L
Now, let's calculate the mass of ethanol in the brew:
Mass of ethanol = 0.0714 L * 0.042 g/mL = 0.0029988 g
Step 3: Convert the amount of ethanol in the brew to the volume of brew the dog needs to consume.
First, we need to find the corresponding concentration of ethanol in the brew:
Concentration of ethanol = Mass of ethanol / Volume of brew
Plugging in the values, we get:
Concentration of ethanol = 0.0029988 g / Volume of brew
Since we want the concentration of ethanol to be 0.5mM (0.0005M), we can set up the equation:
0.0005M = 0.0029988 g / Volume of brew
To solve for Volume of brew, rearrange the equation:
Volume of brew = 0.0029988 g / (0.0005M)
Volume of brew = 5.9976 L
Therefore, the dog needs to consume approximately 5.9976 liters (or 5997.6 mL) of the brew in order to lower the activity of ADH on methanol to 5% of its uninhibited value.