Question

Consider the formation of double-stranded (DS) DNA from two complementary single strands (S and S') through the following mechanism involving an intermediate helix (IH): \\[ \begin{array}{l} \mathrm{S}+\mathrm{S}^{\prime} \frac{k_{1}}{\mathrm{k}_{-1}} \mathrm{IH} \\ \mathrm{IH} \stackrel{k_{2}}{\longrightarrow} \mathrm{DS} \end{array} \\] a. Derive the rate law expression for this reaction employing the preequilibrium approximation. b. What is the corresponding rate-law expression for the reaction employing the steady state approximation for the intermediate IH?

Short answer

The rate law expressions for the given reaction after employing the preequilibrium approximation and the steady state approximation for the intermediate IH are as follows: a. Preequilibrium approximation: \[ \mathrm{rate}=k_{2}K[\mathrm{S}][\mathrm{S}^{\prime}] \] b. Steady state approximation: \[ \mathrm{rate}=k_{2}\left(\frac{k_{1}[\mathrm{S}][\mathrm{S}^{\prime}]}{k_{-1}+k_{2}}\right) \]

Step-by-step solution

Write the equilibrium expression

The reaction reaches a rapid equilibrium before the slow step, so we have: \[ \begin{array}{l} \mathrm{S}+\mathrm{S}^{\prime} \longleftrightarrow \mathrm{IH} \end{array} \] The reaction has an equilibrium constant K, which can be expressed as: \[ K=\frac{[\mathrm{IH}]}{[\mathrm{S}][\mathrm{S}^{\prime}]} \]

Write the rate law for the slow step

The slow step is the formation of DS from IH: \[ \mathrm{IH} \stackrel{k_{2}}{\longrightarrow} \mathrm{DS} \] The rate law for the slow step can be written as: \[ \mathrm{rate}=k_{2}[\mathrm{IH}] \]

Substitute the equilibrium expression into the rate law

To obtain the rate law in terms of the reactants' concentrations, substitute the equilibrium expression into the rate law: \[ \mathrm{rate}=k_{2}[\mathrm{IH}]=k_{2}K[\mathrm{S}][\mathrm{S}^{\prime}] \] This is the rate law expression for the reaction employing the preequilibrium approximation. #b. What is the corresponding rate-law expression for the reaction employing the steady state approximation for the intermediate IH?#

Write the rate of formation and consumption of IH

For the steady-state approximation, the rate of formation of IH is equal to its rate of consumption. The rate of formation of IH is given by: \[ \mathrm{Formation\:rate}=k_{1}[\mathrm{S}][\mathrm{S}^{\prime}] \] The rate of consumption of IH is given by: \[ \mathrm{Consumption\:rate}=k_{-1}[\mathrm{IH}]+k_{2}[\mathrm{IH}] \]

Use the steady state approximation

Set the formation rate equal to the consumption rate: \[ k_{1}[\mathrm{S}][\mathrm{S}^{\prime}]=k_{-1}[\mathrm{IH}]+k_{2}[\mathrm{IH}] \] Rearrange the equation to solve for [IH]: \[ [\mathrm{IH}]=\frac{k_{1}[\mathrm{S}][\mathrm{S}^{\prime}]}{k_{-1}+k_{2}} \]

Write the rate law for the steady-state case

The rate law for the steady-state case is the rate of formation of DS: \[ \mathrm{rate}=k_{2}[\mathrm{IH}] \] Substitute the expression for [IH] we obtained using the steady-state approximation: \[ \mathrm{rate}=k_{2}\left(\frac{k_{1}[\mathrm{S}][\mathrm{S}^{\prime}]}{k_{-1}+k_{2}}\right) \] This is the rate-law expression for the reaction employing the steady state approximation for the intermediate IH.

Key concepts

Preequilibrium Approximation

Understanding the preequilibrium approximation is vital when analyzing certain chemical reactions, including the formation of double-stranded DNA from single strands. The preequilibrium approximation assumes that the initial steps of a reaction reach an equilibrium much faster than the subsequent steps. This leads to a temporary steady state where the concentrations of the reactants and the intermediate product remain relatively constant.

In the context of DNA formation, when single strands S and S' combine, they form an intermediate helix (IH) rapidly and then, more slowly, transition into a double-stranded DNA (DS). Because the formation of IH is fast and reversible, it is assumed to be at equilibrium. By equating the forward and reverse reaction rates at this stage, we derive an equilibrium constant, K, which relates the concentration of the intermediate to the concentration of the initial strands, expressed as \[K = \frac{[\mathrm{IH}]}{[\mathrm{S}][\mathrm{S}']}\].

Applying preequilibrium to the rate law for the reaction's slower step, the formation of DS, allows us to express the rate in terms of reactants' concentrations, leading to a simplified rate law for the overall process. This approximation is invaluable for understanding reaction dynamics and calculating rate laws for complex mechanisms where intermediate species are involved.

Steady State Approximation

The steady state approximation is another insightful approach used in chemical kinetics, particularly for reactions involving unstable intermediates. This method assumes that after a short initial period, the rate of formation of an intermediate species equals its rate of consumption, resulting in a steady (unchanged) concentration of the intermediate over time.

In the case of double-stranded DNA formation, the intermediary is the IH. The rates of both its formation and consumption are considered. By setting these rates equal to each other, \[k_{1}[\mathrm{S}][\mathrm{S}'] = k_{-1}[\mathrm{IH}] + k_{2}[\mathrm{IH}]\], the concentration of IH can be deduced. This allows for the expression of the rate law in terms of only the observable species—S and S'.

The steady state approximation simplifies the mathematics of complex reaction mechanisms by eliminating the need to consider time-dependent changes in the intermediate. It's particularly useful when dealing with reactions where intermediate species are difficult to detect or measure directly.

Rate Law Expression

Understanding the rate law expression is crucial for predicting how fast a reaction will occur under various conditions. A rate law expresses the rate of a reaction in terms of the concentration of its reactants, often raised to some power which corresponds to their reaction order.

In the formation of double-stranded DNA, the rate law can be derived using either the preequilibrium or steady state approximation. The preequilibrium approximation gives us a rate law that reflects the initial, fast equilibrium, resulting in a rate proportional to the product of the concentrations of the single strands S and S'. On the other hand, the steady state approximation leads to a more nuanced rate law that accounts for the formation and consumption of the intermediate, resulting in the formula \[\mathrm{rate} = k_{2}\left(\frac{k_{1}[\mathrm{S}][\mathrm{S}']}{k_{-1}+k_{2}}\right)\].

By employing the correct approximation and deriving the rate law, chemists can make accurate predictions about reaction rates, which is essential for process control and optimization in industrial applications and laboratory experiments.

Double-Stranded DNA Formation

Double-stranded DNA formation is a fundamental biological process that lies at the heart of genetics and molecular biology. It involves the pairing of two complementary single DNA strands to form the familiar double helix structure. This pairing is mediated by hydrogen bonds between the complementary bases on each strand.

The reaction mechanism for double-stranded DNA formation can be described by the association of single strands (S and S') to form an intermediate helix (IH), which then converts into the double-stranded DNA (DS). The application of kinetic approximations like preequilibrium and steady state helps us understand and model this critical biophysical process, providing insight into the rate at which genetic information is stabilized into its double-helical form.

Grasping the process of double-stranded DNA formation is critical not only for academic purposes but also for practical applications, such as DNA replication, repair, and the biotechnological processes that underpin modern molecular biology and genetic engineering.