Bio Ch6

Front 1

At high [S], all E is ES so...

Back 1

-[ES]=[Etotal]
-plateu is due to satuaration effect

Front 2

Vmax

Back 2

-units: uM/min
-observed when virtually all E is present as ES complex
-max rate, extrapolate to at high [S]
-hyperbola
-record it per active site of E

Front 3

Kcat

Back 3

-turnover number
-units: sec-1
-describes the limiting rate of an enzyme (catalytic rxn at satuaration)
-concentration of S converted to P per unit time per enzyme site
-at high [S], reaction is independent of [S] b/c all E sites are filled

Front 4

Reach at which [S]=km

Back 4

-km is the substrate concentration that provides one-half the maximal rate
-it's [S] that provides 1/2 the rate
-50% of the sites are occupied by substrate (1/2 the rate)

Front 5

Consider [S] <<<<< Km

Back 5

-v is dependent upon substrate concentration and enzyme concentration
-v is dependent upon formation of ES -> P
-initial region where if double [S], rate doubles
-linear region

Front 6

Specificity constant

Back 6

Kcat/Km (uM-1sec-1)
-use to compare enzymes
-predicts how efficient enzyme is at binding S and going forward to P
-at extrememy E, it predicts the probability of forming a product (ES -> P)

Front 7

Designing an experiment regarding enzyme kinetics

Back 7

-take into account assumptions of steady state

Front 8

Experiment 1: [E] is very low relative to [S]

Back 8

-goal: measure initial period of reaction where P formation is linear
y=mx+b
-slope provides the rate K or V or Vo
-units: uM product produced per unit time (uM/s or uM x s-1)

Front 9

Experiment 2: function of [S]

Back 9

-increase [S]
-family of lines
y=mx+b
-report as uM x s-1 or s-1
-remember, low [S] have dramatic increase of rate with additional [S] (below slope line see this)

Front 10

High [S]

Back 10

-Km has no role when all E sites are satuarated
-slowest step (k+2) determines rate (rate limiting step)

Front 11

S=[Km]

Back 11

-Km is the [S] that provides 1/2 Kcat or Vmax
-50% sites ES/50% Efree

Front 12

[S] <<<<<<< Km

Back 12

-rate is directly proportional to [S]
-rate is linear
y=mx+b
-slope = Kcat/Km (uM-1s-1)
-Kcat tells how fast can go when all sites are satuarated
-Km predicts affinity of S for E

Front 13

kcat/km

Back 13

units: uM⁻1s⁻1
-predicts lower limit of K₊₁
-higher number means better substrate since it is lower limit predicotr
-higher # = weaker binding

Front 14

K₊₁

Back 14

-E colliding with S
-concentration dependence
units: uM-1 s-1

Front 15

K₋₁ and K₊₂

Back 15

units: s-1
-exponential functions
-concentration has nothing to do w/ ES ->P or ES backwards

Front 16

3 steady state parameters

Back 16

Kcat: s-1, turnover numerb
Km: uM, SS dissociation constant
Kcat/Km: uM-1s-1, specificity constant

Front 17

Are there enzymes that do not obey M-M kinetics?

Back 17

-yes, when have evidence of cooperativity or allosteric regulation
-look for sigmoidal behavior
-may not be able to detect sigmoidal behavior at the conditions of the expt.

Front 18

Linear Transformation of kinetic data

Back 18

-Lineweaver-Burk double reciprocal plot
-Eadie-Hofstee plot(v vs. v/[S])
-Hanes-Woolf plot([S]/v vs. [S])

Front 19

Lineweaver-Burk

Back 19

-used more often
-using really only low [S] data

Front 20

Weaknesses of Lineweaver-Burk

Back 20

-data the most difficult to obtain techinically (low[S]) given most weight in determining linear fit
-difficult to evaluate quality of data directly (scatter, satuaration)
-cannot determine Kcat (Vmax) or Km directly from the plot