Steady State Kinetics in Regulating Homeostasis of Cells

Learning Objectives

• This Exemplar will teach the following concepts from the ACS Examinations Institute General Chemistry ACCM:

• VII. A. 2. a. Chemical reactions may occur at a wide range of rates, and a key aspect of rate is related to the concentration of species involved in the reaction. 

• VII. E. 1. a. A catalyst is defined as an agent that increases the rate of the reaction while not being consumed by the reaction. Catalysts can be either homogeneous or heterogeneous and play an important role in applications, such as catalytic converters on cars, or environmental events, such as ozone depletion.

• VII. E.  2. b. A catalyst increases the rate of the reaction by providing a new reaction pathway with a lower activation energy. Catalysts provide an alternative reaction pathway that lowers this activation energy and affect both forward and reverse reactions.

Kinetics in Biology

Kinetics is a measure of the rate of a chemical reaction with respect to the concentrations of the reactants and products. In a certain type of kinetics common in biological systems called steady state kinetics, all chemical variables in the reaction maintain constant concentrations despite external forces acting upon them.⁶ For this to happen, there must be flow going through the system, or in other words things cannot just be at a standstill. A common example used is how the volume of water in a bathtub remains constant when the tap is on but the drain is also open. Flow is occurring in this system, but the volume or concentration of water inside the bathtub remains constant. Enzymes are typically involved in keeping steady state in biological systems, assisting in the flow process and ensuring that the rate of formation of a compound is balanced with its rate of destruction to ensure constant concentration.⁸

In the case where some variables of a system remain constant but not all, a flow system does not have to occur. Instead, there can exist a closed system, similar to that found in equilibrium reactions, where multiple chemical reactions take place to keep the concentration of certain variables steady. In this case, it is extremely uncommon to achieve an actual steady state, so instead steady state approximation is used.⁷ This is because variables endlessly fluctuate back and forth around the value of the theoretical steady state, nearing closer and closer but never actually reaching steady state. Therefore, an approximation is needed to estimate the rates of production and destruction at steady state.

 

Reaction Kinetics

The rate of a kinetic reaction is given by the equation:

Rate = k[A]^m[B]ⁿ

Where k is the rate constant for the specific reaction, [A] is the concentration of species A, m is the order of reaction with respect to species A, [B] is the concentration of species B, and n is the order of reaction with respect to species B. 

The order of reaction with respect to a species can be determined by holding the concentration of one species constant while altering the concentration of the other species. This will allow for analysis of the effect that alteration on the concentration of the latter species has on the rate of the reaction. An example is provided below.

Example \(\PageIndex{1}\)

Determining order of reaction. 

Concentration of A (M)	Concentration of B (M)	Rate (M/s)
0.5	0.37	5.00e-6
1.0	0.37	2.00e-5
0.5	0.74	1.00e-5

 

Here we see that when [B] is held constant, and [A] is doubled, the rate increases by a magnitude of 4. From this we know that the order of reaction with respect to A is 2, since the rate changes by the square of the concentration of A. 

When [A] is held constant, and [B] is double, the rate increases by a magnitude of 2. From this we know that the order of reaction with respect to B is 1, since the rate changes directly with the concentration of B. 

We can write this mathematically as:

m = log_R(x)

Where m is the order of reaction, R is the ratio of [B], and x is the ratio of the rates.

k can then be determined using the equation:

Rate = k[A]^m[B]ⁿ

k = Rate/([A]^m[B]ⁿ)

In this example, k = 5.00e-6/(0.52 * 0.371) = 5.4e-5 mol^-2L²s^-1.

Steady State Kinetics in Regulating Homeostasis of Cells

Steady-state kinetics is a process in which the concentration of one or several species is held constant. This is often used in the body to maintain homeostasis with various processes, such as with the enzyme complex Na+-K+ ATPase.⁵ The sodium-potassium pump helps maintain equilibrium in cells by moving sodium and potassium against their gradients, allowing for a constant high concentration of sodium extracellularly and constant high concentration of potassium intracellularly.^2,4 This sustained concentration is vital for cellular functions such as regulating cell volume, cell signaling, and maintaining the resting membrane potential of the cell.³ To maintain the constant concentrations at steady state equilibrium of sodium and potassium, active transport is used with the hydrolysis of ATP.¹

Figure 1. Diagram of Sodium-Potassium Pump⁹

Exercise \(\PageIndex{1}\)

Steady-State Kinetics Problems

Deduce the rate law of this reaction when [NO₃] is constant.

NO_(g) + O_2(g) → NO_3(g)         fast

NO_(g) + NO_3(g) → 2 NO_2(g)    slow

Answer

The first things to note are that the rate determining step is step 2, and that NO₃ is an intermediate in the reaction.

The concentration of NO₃ is constant and does not change throughout the reaction, as given in the problem. This means that d[NO₃]/dt = 0.

From this it can be deduced that the rate formation of NO₃ is equal to the rate of consumption of NO₃, since the concentration is held constant. It is important to note that the rate of the reverse reaction in step 2 is negligible since this is the rate determining step, which is product favored. 

Formation rate: k₁[NO][O₂]

Consumption rate: k_-1[NO₃] and k₂[NO][NO₃]

This can be written out in an equation as

d[NO₃]/dt = k₁[NO][O₂] - k_-1[NO₃] - k₂[NO][NO₃]

0 = k₁[NO][O₂] - k_-1[NO₃] - k₂[NO][NO₃]

This can then be solved for [NO₃]:

0 = k₁[NO][O₂] - (k_-1 - k₂[NO])[NO₃]

(k_-1 - k₂[NO])[NO₃] = k₁[NO][O₂]

[NO₃] = k₁[NO][O₂]/(k_-1 - k₂[NO])

The rate determining step is 2, so 

R = k₂[NO][NO₃]

The intermediate [NO₃] cannot be left in the final rate law. Substituting the equation for [NO₃] gives us,

R = k₂k₁[NO]²[O₂]/(k_-1 - k₂[NO])

Exercise \(\PageIndex{2}\)

Steady-State Kinetics Problem #2

Deduce the rate law of this reaction when [NOCl₂] is constant.

2 NO(g) + Cl₂(g) → 2NOCl(g)   

––––––––––––––––––––––––

NO(g) + Cl₂(g) ⇄ NOCl₂(g)   

NOCl₂(g) + NO(g) → 2 NOCl(g)

Answer

First, finding the equation for the initial rate law of the second elementary step allows gives 

r = k₂[NO][NOCl₂]

where k2 is the rate constant of the second step of the reaction

Then, since NOCl₂ is an intermediary reactant whose concentration remains constant throughout the reaction, it follows that d[NOCl₂]/dt = 0.

This can be written out using all rate laws in the reaction

d[NOCl₂]/dt = 0 = k₁[NO][Cl₂] - k_-1[NOCl₂] - k₂[NO][NOCl₂]

Solving for [NOCl₂] we get

[NOCl₂] = k₁[NO][Cl₂]/(k_-1 + k₂[NO])

R = k₁k₂[NO]²[Cl₂]/(k_-1 + k₂[NO] )

Exercise \(\PageIndex{3}\)

Steady-State Enzyme-Substrate Problem

Determine [ES] in the enzyme substrate complex when [ES] is constant.

E + S ⇄ ES ⇄ E + P

Answer

The first thing in this problem is to note that the concentration of [ES] is constant and does not change throughout the reaction, as given in the problem.

This means that d[ES]/dt = 0. 

From this it can be deduced that the rate formation of ES is equal to the rate of consumption of ES, since the concentration is held constant. It is important to note that the rate of the reverse reaction in step 2 is nearly non-existent since the formation of substrate from product is not favored in biological reactions, so this should not be included under the formation of [ES].

Formation rate: k₁[E][S]

Consumption rate: k₁[ES] and k₂[ES]

This can be written out in an equation as

d[ES]/dt = k₁[E][S] - k_-1[ES] - k₂[ES]

0 = k₁[E][S] - k_-1[ES] - k₂[ES]

This can then be solved for [ES]:

0 = k₁[E][S] - (k_-1 - k₂)[ES]

(k_-1 - k₂)[ES] = k₁[E][S]

[ES] = k₁[E][S]/(k_-1 - k₂)

Bibliography

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(2)

Manzoor, N.; Haque, M. M.; Khan, L. A. Pre-Steady State Kinetics of ATP Hydrolysis by Na,K-ATPase. Cell Biochem Funct 2009, 27 (3), 135–141. https://doi.org/10.1002/cbf.1545.

(3)

Gorman, A. L. F.; Marmor, M. F. Steady-State Contribution of the Sodium Pump to the Resting Potential of a Molluscan Neurone. J Physiol 1974, 242 (1), 35–48.

(4)

Pirahanchi, Y.; Jessu, R.; Aeddula, N. R. Physiology, Sodium Potassium Pump. In StatPearls; StatPearls Publishing: Treasure Island (FL), 2022.

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(9)

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