Enzymes Flashcards
what is the transition state of an enzyme
high energy intermediate lying between S and P states. S is before reaction and P is after.
What is the activation energy
minimum energy S (substrate) must have for a reaction to take place and P (product) to be formed.
how to increase rate of reaction
temperature increases number of molecules with correct activation energy and increasing concentration increases chances of collision
Why can’t you increase temp or conc in body to increase the rate of reaction in the body?
because would disrupt homeostasis and denture proteins. therefore require enzymes which act as biological catalysts and speed up the rate of reaction by lowering activation energy
how do enzymes increase rate of reaction and remain unchanged.
lower activation energy to facilitate formation of transition state.
structure of active site and how it works
highly specific site found in a crevice on the inside of a enzymes so it excludes water as this may affect the reaction.
induced fit hypothesis?
active site has complimentary shape to subsrate and when substrate binds it induces changes onto the actibve site allowing it to bind by weak interactions (no covalent bonds) so can still leave. This is now a complementary active site to the substrate. allows catalysis of product.
product-time graph
hyperbola crave. as time increase product increases and then levels off. v0 = reaction velocity. extrapolate from t=0. Rate of reaction = change in conc/time
what is the vmax found on velocity-subsrate conc graph
maximum theoretical velocity is all active sites filled with substrate (hyperbola flat part = vmax)
what is the michaelias menten model
that the complex formed between the enzyme and substrate is a necessary intermediate in catalysis
what is michaelias menten equation?
v0= vmax[s]/km(michaleaias constant)+[s]. rate therefore depends on substrate conc. shows a v0 versus substrate conc graph will be rectangular hyperbola
how to work out Km from v0/[s] graph
find half vmax and use this to draw line across and down to find km
low km=
high km=
low km= high affinity for substrate
high km= low affinity for substrate
what is 1 unit of vmax or v0
the amount of enzyme the converts 1 micro mol of product per minute can be expressed in other units eg litres
double amount of enzyme =
double the rate
what is the lineweaver burk plot?
rearranged michelaias equation to give 1/v0 / 1/[s] to give an easier reading of km and half vmax. slop = km/vmax. km found on x axis. further away from 0 = higher km. y axis shows vmax. higher up y axis is lower vmax.
irreversible enzyme inhibition
bind tightly ( covalent bonds) e.g nerve gases. cannot dissociate
reversible enzyme inhibition competitive and non competitive?
non covalent can freely dissociate. non competitive - binds another site (allosteric site). competitive - binds at active site.
how competitive inhibition affects vmax and km?
vmax stays same because competitor can be outcompeted if conc increased enough. BUT km increases because competitor is competing for active site.
how non- competitive inhibition affects vmax and km?
km stays same because inhibitor binds to site other than active site but vmax decreases because inhibitor cannot be outcompeted.
