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

According to the question

Given,

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

Then,

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,

Velocityof enzyme catalysis is given by MM equation-

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

Vo = (Vmax x 61.638 mM) / (10 mM + 61.638 mM)

=0.8604 Vmax

Hence,

Vo= 0.8604 Vmax

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

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

= 95 % of 0.8604 Vmax

=0.8174

Therefore ,

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

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

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

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

[S] = 1 mM / 0.2234 = 4.476 mM

Therefore,

Required [ethanol] in body =4.476 mM =4.476*10^-3

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

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