Liver alcohol dehydrogenases (ADH) is relatively nonspecific. Its normal substrate is ethanol, however, it will oxidize other primary alcohols, such as methanol, to their corresponding aldehydes. In the case of methanol this produces formaldehyde, which is quite toxic and can lead to blindness. Mistaking it for the cheap wine she usually prefers, my dog Lulu ingested about 36 mL of windshield washer fluid, which is an aqueous solution of 50% v/v methanol. I knew that methanol would be eventually excreted by Lulu's kidneys if its oxidation to formaldehyde could be blocked; I also knew that ethanol could act as a competitive inhibitor of methanol oxidation by ADH. Not wanting to become a seeing-eye human to a blind pooch, I decided to offer my pooch some her favorite hooch, Miller Genuine Draft (MGD), a tasteless brew that contains 4.2% v/v ethanol. a. How much MGD, in mL, must my Lulu consume in order to lower the activity of her ADH on methanol to 5% of its uninhibited value, if the Km values for canine ADH are 1 mM for ethanol and 10 mM for methanol? (Assume the Kl for ethanol in its role as a competitive inhibitor of methanol oxidation is the same as its Km. Both methanol and ethanol will quickly distribute throughout Lulu's 17 L of body fluids. The densities of both methanol and ethanol are 0.79 g/ mt";

volume of methanol consumed= 36 x 0.5 = 18 mL

(molar mass of methanol = 32 g/mol; molar mass of ethanol = 46 g/mol)

number of mol of methanol consumed = (0.79 x 18) / 32 = 0.02469 mol

concentration of methanol in 17L body fluid = 0.02469 / 17 = 1.4524 x 10-3 mol/L

without consuming ethanol, rate of methanol metabolism, v₁ = (v_max [methanol]) / ([methanol] + K_m,methanol)

after ingesting ethanol as competitive inhibitor, rate of methanol metabolism,

v₂ = (v_max[methanol]) / {[methanol]+K_m,methanol(1+[ethanol]/K_I,ethanol)} ,where as given in the question, K_I,ethanol = K_{m, ethanol}

for v₂ / v₁ = 5% = 0.05

v₂ / v₁ = {[methanol] + K_m,methanol(1 + [ethanol]/K_I,ethanol)} / ([methanol] + K_m,methanol)

0.05 = {[1.4524 x 10^-3]+(10 x 10^-3)(1 + [ethanol]/(1 x 10^-3)} / {[1.4524 x 10^-3]+(10 x 10^-3)}

[ethanol] = 0.02176 mol/L

For 17L body fluid, number of moles of ethanol needed = 0.02176 x 17 = 0.3699 mol

volume of ethanol needed = 0.3699 x 46 / 0.79 = 21.5384 mL

volume of MGD needed = 21.5384 / 0.042 = 512.84mL

Ans: 512.84mL of MGD is needed.

Step-by-step explanation

http://www1.lsbu.ac.uk/water/enztech/inhibition.html

http://www-plb.ucdavis.edu/courses/bis/105/lectures/EnzKinetics2.pdf

https://chem.libretexts.org/Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/08%3A_Transport_and_Kinetics/8.4%3A_Enzyme_Inhibition/Competitive_Inhibition

Michaelis-Menton Equation with competitive inibitor:

please see the attached file for the complete solution.