Enzyme Reaction question using kinetics An enzyme reaction follows the enzyme-su

Question

Enzyme Reaction question using kinetics

An enzyme reaction follows the enzyme-substrate complex kinetics. The value for K_m is 5.0 tnmolL while the value for Vuvc is 3.0 mmolt-min for a solution having an enzyme concentration of 1 mg enzjine per liter. For an enzyme concentration of 1 mg/L how much time is required for the concentration of substrate to change front 16 mmol/l. to 8 mmol/l? For an enzyme concentration of 1 mg/L how much time is required for the concentration of substrate to change from 8 mmol/l, to 4 mmol/L? For an enzyme concentration of 1 mg/L, if the initial substrate concentration is 12 mmol/L, what concentration of substrate will remain after 3 minutes of reaction? For an enzyme concentration of 3 mg/L, if the initial substrate concentration is 12 mmol/L, what concentration of substrate will remain after 3 minutes of reaction?

Explanation / Answer

Enzyme kinetics

For a given [E]o , (initial enzyme concentration) and high value of [S],( initial substrate concentration,) the rate of product formation become independent of [So] reaching a maximum value ,(Vmax,)maximum velocity

Rate of reaction=r=d[P]/dt=rate of product formation=rate of substrate reaction or change(change in substrate concentration with time)

[S]=substrate concentration

r=Vmax[S]/km+[S]

given km=5.0mmol/L

Vmax=3.0mmol/L.min

1)For [S]o=16mmol/L

Rate=r=(3.0 mmol/L.min)(16mmol/L)/(5.0mmol/L)+ (16mmol/L)=48 (mmol/L)2.min/21 mmol/L=2.286mmol/L.min

Change of substrate concentration=[S]o-[S]=16-8=8 mmol/L

Time for change= Change of substrate /rate=8 mmol/L/(2.286mmol/L.min)=3.5 min

2) change of substrate=8mmol/L-4mmol/L=4mmol/L

Rate=(3.0 mmol/L.min)(8mmol/L)/(5.0mmol/L)+ (8mmol/L)=24(mmol/L)2.min/13 mmol/L=1.846 mmol/L.min

Time=4mmol/L/(1.846 mmol/L.min)=2.2 minutes

3)[S]o=12mmol/L

Rate==(3.0 mmol/L.min)(12mmol/L)/(5.0mmol/L)+ (12mmol/L)=36(mmol/L)2.min/17 mmol/L=2.117 mmol/L.min

Time=3min

Change of substrate concentration=rate*time=2.117mmol/L.min*3min=6.35 mmol/L

[S] after 3 min=12mmol/L-6.35mmol/L=5.65mmol/L

4) same as (3)=a

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