Question

A cell-free extract of a lactobacillus bacterium that is known to contain all of the enzymes required for lactate fermentation is incubated anaerobically in 100 mL of a medium initially known to contain 200 mM glucose, 20 mM ADP, 40 mM ATP, 2 mM NADH, 2 mM NAD^+, and 20 mM inorganic phosphate (P_i). Assuming that lactate is removed continuously from the medium as soon as it is formed, what is the maximum amount of lactate (in millimoles) that can be formed if, in addition to the above components, an ATPase was added to the reaction mixture. Assume the ATPase catalyzes the following reaction: ATP + H_2O rightarrow ADP + P_i

Answer 1

$Glucose + 2 NAD^{+} + 2 ADP \rightarrow^{x_{1}} 2 Pyruvate + 2H_{2}O +2NADH + 2ATP$

$2 Pyruvate + 2 NADH \rightleftharpoons ^{x_{2}}_{x_{3}} 2 Lactate + 2 NAD^{+}$

$2ADP + 2P_{i} \rightarrow^{x_{4}} 2 ATP$

The above three equations summerise the given system. Where x_1, x₂,x₃, x₄ are rate constants of the given equations

Also, as Lactate is periodically being removed it cannot be converted back and it can be safely assumed that x₃ to be zero. Also it is clear that the product formation is non-growth associated.

Material balances can be written as

d[NADH]/dt = x1 - x2 -x3 = 0 --(1) As NADH remains constant in the system with out being generated or consumed.

d[Lactate]/dt = x2 - x3 = x2 --(2) as Lactate is continuosly being removed from the system.

Substituting (2) in (1) d[NADH]/dt = d[Lactate]/dt --(3)

From first Chemical reaction d[NADH]/dt = 2 d[Glucose]/dt --(4)

Therefore the amount of Lactate formed twice the rate of consumption of Glucose.

Assuming ideal conditions when all glucose is converted to Lactate, the amount of Lactate formed is maximum.

==> 200mM Glucose = 400mM Lactate.

In a general case the amount of glucose consumed is dependent on the rate constants of the reactions. As they are not mentioned in the problem they are idealized.