Question

a protein in your cells (lactxase) converts lactate to another product lactate-x. the enazyme is specific for lactate (ch3-choh-coo-) and cannot use lactic acid ( ch3-coh-cooh) as its substrate. given the dissociation constant (Ka) is .001, the cell PH is 6.5 and glycolosis has just produced 10nM lactic acid, indicate how much lactate is available to lactxase.

Answer 1

Ans. Given,

I. Acid dissociation constant, Ka = 0.001 for the reaction-

                                                                        Lactic acid + H₂O --------> Lactate + H₃O⁺

            II. pH = 6.5

            III. [Lactic acid] = 10 nM = 10 x 10^-9 M = 1.0 x 10^-8 M

# Step 1: Calculate pKa of lactic acid-

            pKa = -log Ka = -log 0.001 = 3.0

# Step 2: Calculate [Lactate]

Henderson- Hasselbalch equation -

            pH = pKa + log ([A^-] / [AH])                    - equation 1

                        where, [A^-] = lactate            

[AH] = lactic acid

Note: all concentrations at equilibrium

Putting the values in equation 1-

6.5 = 3.0 + log ([A^-] / [AH])

Or, ([A^-] / [AH]) = antilog (6.5 -3.0) = antilog (3.5)

Or, ([A^-] / [AH]) = 3162.28

Or, [A^-] = 3162.28 [AH]                             - equation 2

# As explained in step 1 reaction, some of lactic acid dissociates into lactate ion. So, the sum of [A^-] and [AH] at equilibrium must be equal to the initial [AH] of 1.0 x 10^-8 M.

So,

            [A^-] + [AH] = 1.0 x 10^-8 M                                    - equation 3

# Putting the value of [A^-] from equation 2 into equation 3-

            3162.28 [AH] + [AH] = 1.0 x 10^-8

            Or, [AH] = (1.0 x 10^-8) M / 3163.28

            Hence, [AH] = 3.168 x 10^-12 M

Therefore,

At equilibrium [AH] = [lactic acid] = 3.168 x 10^-12 M

# Now, [Lactate] = [A^-] at equilibrium = 3162.28 [AH]

            Or, [Lactate] = 9.997 x 10^-9 M

Therefore,

            Available [Lactate] = 9.997 x 10^-9 M = 9.997 nM