Triose phosphate isomerase has a rate enhancement of 1.0x109. If the enzyme-catalyzed rate is 1500 nmol s-1, what would the reaction rate be without the enzyme? Convert your answer to nmol day-1 before entering. (Pay very close attention to units.)
Well, well, well, looks like we have a good ol' enzyme question to tackle! Triose phosphate isomerase, huh? Quite a mouthful!
Now, you're telling me this enzyme has a rate enhancement of 1.0x10^9, and the enzyme-catalyzed rate is 1500 nmol s^-1. Alrighty then! Let's do some math magic.
If the enzyme-catalyzed rate is 1500 nmol s^-1, and the rate enhancement is 1.0x10^9, we can calculate the reaction rate without the enzyme by dividing the enzyme-catalyzed rate by the rate enhancement, like so:
1500 nmol s^-1 / 1.0x10^9 = 1.5x10^-6 nmol s^-1
But hold your horses! The question asked for the reaction rate in nmol day^-1. So we need to convert our answer. There are 24 hours in a day, and 60 minutes in an hour, and 60 seconds in a minute, so there are 24 x 60 x 60 = 86,400 seconds in a day.
1.5x10^-6 nmol s^-1 x 86,400 s day^-1 = 0.1296 nmol day^-1
So, without the enzyme, the reaction rate would be approximately 0.1296 nmol day^-1.
Phew! That was quite the journey, wasn't it? I hope I could shed some light on this enzyme's antics for you!
To find the reaction rate without the enzyme, we can use the rate enhancement factor provided.
The rate enhancement of 1.0x10^9 means that Triose Phosphate Isomerase (TPI) increases the reaction rate by a factor of 1.0x10^9 compared to the uncatalyzed reaction.
Given the enzyme-catalyzed rate of 1500 nmol s^-1, we can calculate the reaction rate without the enzyme.
1 nmol s^-1 = (1 nmol s^-1) * (1 min/60 s) * (60 min/1 h) * (24 h/1 day) = 1440 nmol day^-1
So, the reaction rate without the enzyme will be:
Reaction rate without the enzyme = Reaction rate with the enzyme / Rate enhancement factor
= 1500 nmol s^-1 / 1.0x10^9
Now we can convert the obtained result to nmol day^-1:
Reaction rate without the enzyme = (1500 nmol s^-1 / 1.0x10^9) * (1440 nmol day^-1)
= 2.16 nmol day^-1
Therefore, the reaction rate without the enzyme is 2.16 nmol day^-1.
To determine the reaction rate without the enzyme, we need to divide the enzyme-catalyzed rate by the rate enhancement value.
Rate without enzyme = Enzyme-catalyzed rate / Rate enhancement
Given:
Enzyme-catalyzed rate = 1500 nmol s^(-1)
Rate enhancement = 1.0 x 10^9
Substituting the given values:
Rate without enzyme = 1500 nmol s^(-1) / (1.0 x 10^9)
We need to convert the unit from seconds to days to match the answer format. There are 24 hours in a day and 3600 seconds in an hour:
Rate without enzyme = (1500 nmol s^(-1) / (1.0 x 10^9)) * (3600 s h^(-1)) * (24 h day^(-1))
Now, let's calculate the rate without the enzyme:
Rate without enzyme = (1500 nmol s^(-1) / (1.0 x 10^9)) * (3600 s h^(-1)) * (24 h day^(-1))
= 1500 nmol / (1.0 x 10^9) * 3600 * 24 nmol day^(-1)
= (1500 * 3600 * 24) / (1.0 x 10^9) nmol day^(-1)
= 12,960,000 nmol day^(-1)
Therefore, the reaction rate without the enzyme would be 12,960,000 nmol day^(-1).