Enzymes

Question 1

what is the transition state of an enzyme

Answer 1

high energy intermediate lying between S and P states. S is before reaction and P is after.

Question 2

What is the activation energy

Answer 2

minimum energy S (substrate) must have for a reaction to take place and P (product) to be formed.

Question 3

how to increase rate of reaction

Answer 3

temperature increases number of molecules with correct activation energy and increasing concentration increases chances of collision

Question 4

Why can’t you increase temp or conc in body to increase the rate of reaction in the body?

Answer 4

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

Question 5

how do enzymes increase rate of reaction and remain unchanged.

Answer 5

lower activation energy to facilitate formation of transition state.

Question 6

structure of active site and how it works

Answer 6

highly specific site found in a crevice on the inside of a enzymes so it excludes water as this may affect the reaction.

Question 7

induced fit hypothesis?

Answer 7

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.

Question 8

product-time graph

Answer 8

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

Question 9

what is the vmax found on velocity-subsrate conc graph

Answer 9

maximum theoretical velocity is all active sites filled with substrate (hyperbola flat part = vmax)

Question 10

what is the michaelias menten model

Answer 10

that the complex formed between the enzyme and substrate is a necessary intermediate in catalysis

Question 11

what is michaelias menten equation?

Answer 11

v0= vmax[s]/km(michaleaias constant)+[s]. rate therefore depends on substrate conc. shows a v0 versus substrate conc graph will be rectangular hyperbola

Question 12

how to work out Km from v0/[s] graph

Answer 12

find half vmax and use this to draw line across and down to find km

Question 13

low km=

high km=

Answer 13

low km= high affinity for substrate

high km= low affinity for substrate

Question 14

what is 1 unit of vmax or v0

Answer 14

the amount of enzyme the converts 1 micro mol of product per minute can be expressed in other units eg litres

Question 15

double amount of enzyme =

Answer 15

double the rate

Question 16

what is the lineweaver burk plot?

Answer 16

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.

Question 17

irreversible enzyme inhibition

Answer 17

bind tightly ( covalent bonds) e.g nerve gases. cannot dissociate

Question 18

reversible enzyme inhibition competitive and non competitive?

Answer 18

non covalent can freely dissociate. non competitive - binds another site (allosteric site). competitive - binds at active site.

Question 19

how competitive inhibition affects vmax and km?

Answer 19

vmax stays same because competitor can be outcompeted if conc increased enough. BUT km increases because competitor is competing for active site.

Question 20

how non- competitive inhibition affects vmax and km?

Answer 20

km stays same because inhibitor binds to site other than active site but vmax decreases because inhibitor cannot be outcompeted.