Question 15 of 20 correct Map General Chemistr sity Science Books presented by S

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Question 15 of 20 correct Map General Chemistr sity Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly The protein catalase catalyzes the reaction described by and has a Michaelis-Menten constant of KM 25 mM and a turnover number of 4.0 x 107 st The total enzyme concentration is 0.015 AM and the initial substrate concentration is 6.08 pM. Catalase has a single active site. First, convert all concentrations to mM. Calculate the value of Rmax (often written as Vmax) for this enzyme Recall that 106 AM 1 M 103 mM. Then use this formula to solve for Rmax. Number where [Elt is the total 0.06 x 10 mM. s k, enzyme concentration and [E] k2 is the turnover number Calculate the initial rat R (often written as Vo), of this reaction. Number RJ 0.005629 x 10 mM. S-1 Previous ® Give Up & View solution Try Again Next Exit

Explanation / Answer

Given that  Km = 25 mM

  turnover number Kcat = 4.0 × 107s–1

total enzyme concentration [E] = 0.015 M = 0.015 x 10-3 mM

initial substrate concentration [S] = 6.08 M = 6.08 x 10-3 mM

1) Vmax =  Kcat  [E]

= (4.0 × 107 s–1 )( 0.015 x 10-3 mM)

=  0.06 × 104 mM s–1

Therefore, Vmax = 0.06 × 104 mM s–1

2) Vo = Vmax [S] / {Km+[S]}

= (0.06 × 104 mM s–1) (6.08 x 10-3 mM) / {25 mM +  6.08 x 10-3 mM}

= 0.146 mM s–1

Therefore, Vo = 0.38 mM s–1

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