Question

4) (7 marks) 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/ ml.)

Answer 1

The methanol is converted to formaldehyde inside the body by the action of ADH. It follows the Michaelis Menten equation of the enzyme-catalyzed reaction. The ethanol is administrated into the body of the dog as a competitive inhibitor. There are equation to deduce the velocity of the same reaction in the presence of a competitive inhibitor. The volume of ethanol administrated is to be calculated so that the velocity of methanol conversion is reduced to 5% of the normal conversion rate. We will start considering the unknown volume of MGD as V. The detailed calculation is shown in the following attachments step by step.

① According to Michaelis - Menten equation of enzyme catalysed reactions Vmax · [s] en + [3] km where, ve velocity of Vmax - maximum possible velocity of reaction substrate [5] - Concentration of km = Michaelis constant. ADH here. For the substrate of Methanol is which, - km z 10 mm km (meth) Vmax are ;) 36 ml don't need to Vmax (meth) /we know its value [s] - Concentration of methanol in body & we to calculate this value. of windshield washer fluid has been consumed, which contains 50 % (v/w) methanol. it (36 * )= 18 mil pure methanol is taken in volume massa densi (mass - density x volume) density of methanot (provided) = 0.79 q fmt methanol (0.79 x 18) methanol g A .. 18 ml 14.22 g

ix) 14.22 fluid. methout is dissolved in 17L body so, concentration in [m] no. Of moles volume in L methanol, - 14.22 v] 14.22 14.227 molecular weight mol 32 of methanol = 32 Cs 12 ) CH₃ ottaa Atomic t = 1 = 12 + (143) + 16 +1 reight O 16 32 vid Methanot cone. 2 14.22/32 m 17 3 4.22 x 10 MM 32 17 7.7573 mm. a. I's] = 7.7573 mm 3 To block. the ADH mediated reaction ethanol administracted, as competitive inhibitor. ► The equation for competitive inhibition is Vmax [5] ks (1+ (2) + [5] eqe ki

where, reaction relocity in the presence of inhibitor. Vmax maximum reaction rebcity posible in the absence of inhibitor Substrate concentration Chere methand) inhibitor concentration Chere ethanol) [1] ks = km for substrate km of ethanol Ki here equals to const. inhibitor) 04 (in general affinity are the values So, as follows- not needed. Vmax exact value is methanol [s] concentration of 7.7573 mm (already calculated) methanol 10 mm kg km for 1 mm ethanol ki a km for concentration of ethanol administrated. will have to calculate. о we V mh i say MGD is administrated. of MGD contain 4.2 w/ ethanol. mh ethanol. (1004.) 4.2 V So, ramt MGD = 100

13 density of ethanol 0.79 g/ml So, mass ethanol of volume density & 4.2 V 0.79 X g 100 inf Molecular meight of ethanol 2, 46 CH₃ CH₂ olt (1242) + (146) + 16 24 + 6 + 16 2 mass in) No. of noles of ethanol mw 0.79 x 4.2 x V nol. 46 x 100 of mol y Conc. of ethanol vol of body fluid 0.79 x 4.2 x v M 46 x 100 x 17 0.79 x 4.2 XV x 10 103 2 м М 46 x 100 x 17 2 0.04243 v mM So, [1] - 0.04243 V mm

5 From eg Vmax X 7.7573 enit without inhibitor 10 t 7.7573 from et 2 7.7573 7.7573 + V. Vmax with inhibitor 10 (14 0:04213 ) According to the problem, 5 % of V without inhibitor 2 withand inhibitor 5 * 7.7543 X 7.7873 Vmax 100 = Vimax 7.7573 10 (1+0.0423 v) +7.7573 10 4 7.7573 20 = ) 1fox 17.7573 5 ( 10+ 0.423 V + 7-7573) 355. 146 5*(17-7573 + 0.423 v => 0.423 Va 337.3887 -> V2 797.61 MGD So, 797.61 ml of is to beat administerated to reduce the conversion rate of methansh of original. 15 y.