3 edited body edited May 30 at 1:01 Karsten Theis 10.6k1313 silver badges5757 bronze badges Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein with $$m$$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $$m$$ inhibitors: $$\ce{ T_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens when there is competitive inhibition? It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{EL2I2}$$$$\ce{ES2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{EL4}$$$$\ce{ES4}$$ because not all active sites are turning over substrate. Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein with $$m$$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $$m$$ inhibitors: $$\ce{ T_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens when there is competitive inhibition? It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{EL2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{EL4}$$ because not all active sites are turning over substrate. Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein with $$m$$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $$m$$ inhibitors: $$\ce{ T_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens when there is competitive inhibition? It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{ES2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{ES4}$$ because not all active sites are turning over substrate. 2 added 54 characters in body edited May 29 at 21:25 Karsten Theis 10.6k1313 silver badges5757 bronze badges Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein R with $$m$$ subunits attached to the enzyme. RIt binds cooperatively binding to inhibitor, andwhich allosterically affectingaffects enzyme activity. ItThe regulatory protein would also have two statestates, T and R, either free or bound to $$m$$ inhibitors: $$\ce{ R_m + m I <=> R_mI_m}$$$$\ce{ T_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens forwhen there is competitive inhibition? It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{EL2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{EL4}$$ because not all active sites are turning over substrate. Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein R with $$m$$ subunits attached to the enzyme. R binds cooperatively binding to inhibitor, and allosterically affecting enzyme activity. It would also have two state, either free or bound to $$m$$ inhibitors: $$\ce{ R_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand, the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens for competitive inhibition It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{EL2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{EL4}$$ because not all active sites are turning over substrate. Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein with $$m$$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $$m$$ inhibitors: $$\ce{ T_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens when there is competitive inhibition? It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{EL2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{EL4}$$ because not all active sites are turning over substrate. 1 answered May 29 at 21:18 Karsten Theis 10.6k1313 silver badges5757 bronze badges Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$ The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them: $$\ce{T4 + 4 L <=> R4L4}$$ From this, you can write the law of mass action and the binding isotherms. If we apply this to an enzyme with a Hill coefficient of $$n$$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium: $$\ce{F_n + nS<=> E_nS_n}$$ For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $$v_\text{max}$$. Mechanism of cooperative inhibition In the OP's question, the Hill coefficient for inhibition $$m$$ is different from that for enzyme activity ($$n$$). So we could have a regulatory protein R with $$m$$ subunits attached to the enzyme. R binds cooperatively binding to inhibitor, and allosterically affecting enzyme activity. It would also have two state, either free or bound to $$m$$ inhibitors: $$\ce{ R_m + m I <=> R_mI_m}$$ When the allosteric inhibitor is bound to ligand, the enzyme lacks activity. Combination of the two Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $$v_\text{max}$$ times those two fractions. $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$ What happens for competitive inhibition It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $$\ce{EL2I2}$$) are populated; if they are, they would contribute to product formation, but less than $$\ce{EL4}$$ because not all active sites are turning over substrate.