Question

Alcohol dehydrogenase (ADH) will oxidize several alcohols. 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
person has accidently consumed 100 mL of methanol. How much 100 proof
whiskey (50% alcohol) does the person need to consume to reduce the activity of
ADH for methanol to 5%?
ADH-ethanol Km = 1mM
ADH-methanol Km = 10mM
Assume the “working volume” of the person is 40L
Assume the densities of both liquids are 0.79 g/mL

Answer 1

Ans. Part 1: Alcohol dehydrogenase (ADH) kinetics for methanol as substrate.

Mass of methanol consumed = Volume x density = 100.0 mL x (0.79 g/ mL) = 76.0 g

Moles of methanol consumed = Mass / Molar mass

                                                = 79.0 g / (32.04216 g/ mol)

                                                = 2.4655017 mol

Molarity of methanol = Moles / Volume of solution in liters

                                    = 2.4655017 mol / 40.0 L

                                    = 0.06163754 M                                           ; [1 M = 10³⁺ mM]

                                    = 61.638 mM

So, [S] = [methanol] = 61.638 mM

Now, Velocity of enzyme catalysis is given by MM equation-

Vo = Vmax [S] / (Km + [S])

Or, Vo = (Vmax x 61.638 mM) / (10 mM + 61.638 mM) = 0.8604 Vmax

            Hence, Vo = 0.8604 Vmax

Part 2: To reduce ADH activity methanol to 5%, the remaining 95% ADH activity must be dedicated to ethanol.

So, calculate [ethanol] required to attain reaction velocity 95% of Vmax.

Or,

Required reaction velocity in presence of ethanol = 95% of Vmax in presence of methanol

                                                            = 95 % of 0.8604 Vmax

                                                            = 0.8174 Vmax

Therefore calculate [ethanol] to attain the reaction velocity of 0.8174 Vmax.

# Part 3: Using MM equation foe ethanol-

            0.8174 Vmax = (Vmax x [S]) / (1 mM + [S])

            Or, (1 mM + [S]) = (Vmax x [S]) / 0.8174 Vmax = 1.2234 [S]

            Or, 1 mM = 1.2234 [S] - [S] = 0.2234 [S]

            Or, [S] = 1 mM / 0.2234 = 4.476 mM

Therefore, required [ethanol] in body = 4.476 mM = 4.476 x 10^-3 M

# Part 4: Calculate required quantity of pure ethanol

Required moles of ethanol = Molarity x volume of body fluid in liters

                                    = (4.476 x 10^-3 M) x 40.0 L

                                    = 0.1791 mol

Required moles of ethanol = Moles x Molar mass

                                    = (0.1791 mol) x (46.06904 g/ mol)

                                    = 8.2487 g

Volume of ethanol required = Mass / Density

                                    = 8.2487 g / (0.79 g/ mol)

                                    = 10.4414 g

# Part 5: Required quantity of whisky

Pure ethanol content of 100 proof whiskey = 50 % ethanol (wt/ v)

                                    = 50.0 g ethanol/ 100 mL

Now,

Required amount of whiskey =

Required amount of pure ethanol / ethanol content of whiskey

= 10.4414 g / (50.0 g/ 100 mL)

= 20.88 mL

Therefore, required volume of whiskey = 20.88 mL