Question

Part C Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an ATP-requiring reaction, firefly luciferase can be used to test for the presence of ATP. In this way, luciferase can test for the presence of life. The coupled reactions are 1.2.luciferin+O2ATP⇌⇌oxyluciferin+lightAMP+PPiΔG∘=−31.6 kJ/mol If the overall ΔG∘ of the coupled reaction is -4.80 kJ/mol , what is the equilibrium constant, K, of the first reaction at 6 ∘C ?

Answer 1

K = 9.620 × 10⁻⁶

Step-by-step explanation:

From the given information:

Temperature T= 6° C

= (273 + 6)K

= 279 K

The correct and well presentation of the reactions are:

1.
`Luciferin + O_2` ⇆ oxyluciferin + light ΔG₁°

2. ATP ⇄ AMP + PP
`_i` ΔG₂° = -31.6 kJ/mol

The overall ΔG° = -4.80 kJ/mol

Let's first determine the ΔG₁° for the equation (1)

ΔG° = ΔG₁° + ΔG₂°

- ΔG₁° = - ΔG° + ΔG₂°

ΔG₁° = ΔG° - ΔG₂°

ΔG₁° = ( -4.80 - (-31.6) ) kJ/mol

ΔG₁° = 26.8 kJ/mol

Using the formula:

ΔG° = -RTIn K

`In \ K =(-\Delta G^0)/(RT) \\ \\ log \ K = -(\Delta G^0)/(2.303RT)`

`log \ K = -(\Delta G_1^0)/(2.303RT) \\ \\ log \ K = -(26.8 * 10^3 \ J/mol)/(2.303* 8.314 \ J/mol/K * 279 \ K) \\ \\ log \ K =- 5.017`

K = antilog (-5.017)

K = 9.620 × 10⁻⁶