3-\(\alpha \)-HSD Kinetic Model

\(\begin{array}{l}\frac{{d\left[ {3 - \alpha - d} \right]}}{{dt}} = \frac{{d{{\left[ {3 - \alpha - d} \right]}_{forward}}}}{{dt}} + \frac{{d{{\left[ {3 - \alpha - d} \right]}_{backward}}}}{{dt}} = \\ = \left[ {3 - \alpha - HSD} \right] \cdot \left( {{K_{ca{t_1}}} \cdot \frac{{\left[ {DHT} \right]}}{{{K_{{m_1}}} + \left[ {DHT} \right]}} - {K_{ca{t_2}}} \cdot \frac{{\left[ {3 - \alpha - d} \right]}}{{{K_{{m_2}}} + \left[ {3 - \alpha - d} \right]}}} \right)\end{array}\)

Background

All AKRs catalyze an ordered bi-bi reaction in which the cofactor binds first, followed by the binding of the steroid substrate. The steroid product is the first to leave, and the cofactor is the last. In this mechanism, \({K_{cat}}\) represents the slowest step in the kinetic sequence ^[2].

Approaches to modeling the process

1. Cofactor saturation assumption

We assume that the levels of the cofactors on the scalp are high enough so that they are always saturated in the enzymatic reaction. The advantage of this approach is that we can use the michaelis-menten reversible equation to describe the reaction. As we will explain later, this assumption may not be correct so we will offer other approaches. Another major disadvantage of this model is that it does not take the levels of cofactors into consideration, so it cannot help us to predict the system's behavior for different cofactor concentrations.

2. New Model Development

Taking cofactors into consideration, we can use principles from statistical mechanics in order to develop a completely new enzyme reaction function. The advantage of this approach is that it describes the kinetics of the enzyme in much more detail than michaelis-menten reversible, and can even offer some explanations to our wetlab results. The disadvantage of this approach is that there are no reaction constants available for it, so we will have to estimate them.

Approach 1 – cofactor saturation assumption

If we assume that the levels of cofactor in the enzyme's environment are high enough that they become saturated, the probability of finding an enzyme that is not connected to a cofactor is negligible. We also need to assume that the concentrations of both cofactors are almost equal, so the inhibition effect [need article] will not affect the reaction (as we will show later, a large ratio in favor of one cofactor will inhibit the other direction of the reaction). The new kinetic schematic is:

Since both levels of NADPH and NADP are saturated, we'll assume product inhibition occurs only with DHT and \(3\alpha diol\), so the reaction will resemble michaelis-menten reversible. Since the degradation rates of the hormones on the scalp are unknown, we will neglect them by assuming the degradation is slower by several orders of magnitude than the enzymatic reaction.

We can summarize the reactions by the following coupled differential equations:

\[\left( 1 \right)\left\{ {\begin{array}{*{20}{c}}{\frac{{d\left[ {3\alpha diol} \right]}}{{dt}} = \frac{{{V_{{m_f}}} \cdot \frac{{\left[ {DHT} \right]}}{{{k_s}}} - {V_{{m_r}}} \cdot \frac{{\left[ {3\alpha diol} \right]}}{{{k_p}}}}}{{1 + \frac{{\left[ {DHT} \right]}}{{{k_s}}} + \frac{{\left[ {3\alpha diol} \right]}}{{{k_p}}}}}}\\{\frac{{d\left[ {DHT} \right]}}{{dt}} = \frac{{{V_{{m_r}}} \cdot \frac{{\left[ {3\alpha diol} \right]}}{{{k_p}}} - {V_{{m_f}}} \cdot \frac{{\left[ {DHT} \right]}}{{{k_s}}}}}{{1 + \frac{{\left[ {DHT} \right]}}{{{k_s}}} + \frac{{\left[ {3\alpha diol} \right]}}{{{k_p}}}}}}\end{array}} \right.\]

Where:

• \({V_{{m_f}}}\) is the maximum forward reaction rate. Attained when all enzyme molecules are bound to the substrate (DHT).
• \({V_{{m_r}}}\) is the maximum backward reaction rate. Attained when all enzyme molecules are bound to the product(\(3\alpha Diol\)).
• \({K_s}\) is the substrate concentration at which the forward reaction rate is at half-maximum.
• \({K_p}\) is the product concentration at which the backward reaction rate is at half-maximum.

While we couldn't find the kinetic constants for human scalp, we found an article which measured them on rat skin ^[5]. We will assume the constants are on the same order of magnitude as on the human scalp.

Time domain simulation

We simulated the system described above. The simulation has been done using the following constants:

In order to understand the breakdown process of DHT by the enzyme, we simulated the system for three initial concentrations of the hormone:

From the simulation we can see that for an initial concentration that is low by an order of magnitude from \({K_s}\), the time for the system to reach stable state is almost the same as for an initial concentration of \({K_s}\). However for a value that is higher by an order of magnitude, the time increases significantly.

The following video shows a parameter scan for different initial concentration values:

From this video we can learn that the time to break down initial concentrations that are larger by an order of magnitude from \({K_s}\) increase significantly for every small increment. By looking at equation (1) we can find two possible reasons for this:

• DHT saturation - for every increment in the initial concentration, the value of the derivative approaches \({V_{{{\mathop{\rm m}\nolimits} _f}}}\). For values that are larger by an order of magnitude than \({K_s}\), large increments in the initial concentration do not change much the value of the derivative.
• Product inhibition - most of the DHT molecules are converted to \(3\alpha - diol\). Larger concentration of the product decreases the breakdown rate of the substrate.

Percentage of DHT breakdown

Let C be the initial concentration of DHT, and the initial concentration of \(3\alpha - diol\). The relation between the substrate and the product is:
(2) \(\left[ {DHT} \right] + \left[ {3\alpha - diol} \right] = C\)

At stable state, \(\frac{d}{{dt}} = 0\), so from equation (1) we get: \[\begin{array}{l}\frac{{{V_{{m_f}}} \cdot \frac{{{{\left[ {DHT} \right]}_{final}}}}{{{k_s}}} - {V_{{m_r}}} \cdot \frac{{{{\left[ {3\alpha diol} \right]}_{final}}}}{{{k_p}}}}}{{1 + \frac{{{{\left[ {DHT} \right]}_{final}}}}{{{k_s}}} + \frac{{{{\left[ {3\alpha diol} \right]}_{final}}}}{{{k_p}}}}} = 0\\{V_{{m_f}}} \cdot \frac{{{{\left[ {DHT} \right]}_{final}}}}{{{k_s}}} - {V_{{m_r}}} \cdot \frac{{{{\left[ {3\alpha diol} \right]}_{final}}}}{{{k_p}}}\mathop = \limits^{(2)} {V_{{m_f}}} \cdot \frac{{{{\left[ {DHT} \right]}_{final}}}}{{{k_s}}} - {V_{{m_r}}} \cdot \frac{{C - {{\left[ {DHT} \right]}_{final}}}}{{{k_p}}} = 0\\{\left[ {DHT} \right]_{final}} \cdot \left( {\frac{{{V_{{m_f}}}}}{{{k_s}}} + \frac{{{V_{{m_r}}}}}{{{k_p}}}} \right) = \frac{{{V_{{m_r}}} \cdot C}}{{{k_p}}}\\{\left[ {DHT} \right]_{final}} = C \cdot \frac{{{V_{{m_r}}}}}{{\frac{{{k_p}}}{{{k_s}}} \cdot {V_{{m_f}}} + {V_{{m_r}}}}}\\ \Rightarrow \left[ \% \right] = \left( {1 - \frac{{{{\left[ {DHT} \right]}_{final}}}}{C}} \right) \cdot 100 = \left( {1 - \frac{{{V_{{m_r}}}}}{{\frac{{{k_p}}}{{{k_s}}} \cdot {V_{{m_f}}} + {V_{{m_r}}}}}} \right) \cdot 100\mathop = \limits^{from\,\,\,table} 0.717\% \end{array}\]

Problems with the approach