In a 50µl reaction containing 0.25 pmol of a Michaelis enzyme with Km = 4.0 x...

In a 50µl reaction containing 0.25 pmol of a Michaelis enzyme with Km = 4.0 x 10-7 M and substrate concentration of 2 x 10-7 M, product was formed at an initial velocity of 5.0 x 10-5 Mmin-1 .

a. If the enzyme concentration is doubled to 0.5 pmol in the reaction, how much will the velocity increase? Show mathematically

Expert Solution

Ans. Reaction velocity is directly proportional to enzyme concertation in the reaction mixture. That is, increasing the enzyme concertation, while keeping other variables constant, also increases the reaction velocity by same factor.

So,

New (increased) reaction velocity = (New [enzyme] / Original [enzyme]) x Original Vo

= (0.50 pmol / 0.50 pmol) x (5.0 x 10^-5 M min^-1)

= 1.0 x 10^-4 M min^-1

# Therefore, reaction velocity increases to 1.0 x 10^-4 M min^-1, i.e. 2 times that of the original reaction velocity.

gladiator answered 1 year ago

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