If I decrease koff by a factor of three, what is my new reaction rate
- In reality, the fall in neurotransmitter in the synaptic cleft following the action potential results from two processes – reuptake into the presynaptic cell and breakdown by enzymes. Let’s analyze just the enzyme portion. Imagine the enzyme that breaks down the neurotransmitter needs to decrease the concentration from 10 uM to zero in 1 msec. The Km = 10 nM, the kcat = 1000 s-1 (a very fast enzyme). What concentration of enzyme would you need to achieve an initial rate that would break down 10 uM in 1 msec? Watch your units! (In reality the rate would go down as the neurotransmitter concentration goes below the Km, but we will ignore that here.)
- a) For a reaction that is governed by M-M kinetics, if I have a Vmax of 1 μM/s and a KM
of 1 μM, what is the reaction rate if I my substrate concentration is 0.1 μM?
b) If I decrease koff by a factor of three, what is my new reaction rate?
c) If I instead phosphorylate 80% of my enzyme and completely inactivate them, what is my new reaction rate?
3) Imagine that a cell surface receptor for some hormone ligand has an extracellular ligand binding domain, a transmembrane domain, and an intracellular domain that is an enzyme that phosphorylates other proteins. When no ligand is bound, the enzyme is turned off and when ligand is bound the enzyme is fully active. Which will lead to the largest increase in the phosphorylation rate in the cell: doubling the number of receptors on the cell, doubling the hormone concentration, or doubling the affinity of the receptor for its ligand? Which will lead to the smallest increase in the phosphorylation rate? Explain your answer.