# 6.021/Notes/2006-09-29

## 4 state model

• Simplify the model with assumptions
• $\displaystyle{ \alpha_1=\alpha_3, \beta_1=\beta_3 }$ (binding same on inside and outside)
• $\displaystyle{ \alpha_2=\alpha_4, \beta_2=\beta_4 }$ (ability for protein to translocate/flip is independent of solute)
• Binding fast relative to translocation
• Only care about the dissociation constant as it will always be in steady state
• Instead of concentrations (which is per volume), it is easier to think about $\displaystyle{ \mathfrak{N}_E }$ (per surface area) $\displaystyle{ \mathfrak{N}_E=c_E\cdot d }$ where $\displaystyle{ d }$ is the membrane thickness
• This leads to the simple symmetric four state carrier model
• The solution can be interpreted intuitively
• The enzyme is first partitioned into facing in or out depending on $\displaystyle{ \alpha, \beta }$
• Then it is partitioned into whether has substrate bound by $\displaystyle{ K }$ and $\displaystyle{ c_s }$
• The concentration difference between inside and outside is not important. All that matters is the concentration relative to K.

Solution to simple symmetric 4-state carrier model:

$\displaystyle{ \mathfrak{N}^i_{ES}=\frac{\beta}{\alpha+\beta}\frac{c^i_s}{c^i_s+K}\mathfrak{N}_{ET} }$

$\displaystyle{ \mathfrak{N}^i_{E}=\frac{\beta}{\alpha+\beta}\frac{K}{c^i_s+K}\mathfrak{N}_{ET} }$

$\displaystyle{ \mathfrak{N}^o_{ES}=\frac{\alpha}{\alpha+\beta}\frac{c^o_s}{c^o_s+K}\mathfrak{N}_{ET} }$

$\displaystyle{ \mathfrak{N}^o_{E}=\frac{\alpha}{\alpha+\beta}\frac{K}{c^o_s+K}\mathfrak{N}_{ET} }$

$\displaystyle{ \phi_s=(\phi_s)_{max}(\frac{c^i_s}{c^i_s+K}-\frac{c^o_s}{c^o_s+K}) }$; $\displaystyle{ (\phi_s)_{max}=\frac{\alpha\beta}{\alpha+\beta}\mathfrak{N}_{ET} }$