Complete Question
The complete question is shown on the first uploaded image
Answer:
[S]<<KM | [S]=KM | [S]>>KM | Not true
____________ | Half of the active | Reaction rate is | Increasing
[tex][E_{free}][/tex] is about | sites are filled of | independent of | [tex][E_{Total}][/tex] will
equal to [tex][E_{total}][/tex]. | | [S] | lower KM
_____________________________________________|____________
[ES] is much | | Almost all active
lower than | | sites are filled
[tex][E_{free}][/tex] | |
Explanation:
Generally the combined enzyme[tex][ES][/tex] is mathematically represented as
[tex][ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)[/tex]
for Michaelis-Menten equation
Where [tex][S][/tex] is the substrate concentration and [tex]K_M[/tex] is the Michaelis constant
Considering the statement [tex][S] < < K_M[/tex]
Looking at the equation [tex][S][/tex] is denominator so it can be ignored(it is far too small compared to [tex]K_M[/tex]) hence the above equation becomes
[tex][ES] = \frac{[E_{total}][S]}{K_M}[/tex]
Since [tex][S][/tex] is less than [tex]K_M[/tex] it means that [tex]\frac{[S]}{K_M} < < 1[/tex]
so it means that [tex][ES] < < [E_{total}][/tex]
What this means is that the number of combined enzymes[tex][ES][/tex] i.e the number of occupied site is very small compared to the the total sites [tex][E_{total}][/tex] i.e the total enzymes concentration which means that the free sites [[tex]E_{free}][/tex] i.e the concentration of free enzymes is almost equal to [tex][E_{total}][/tex]
Considering the second statement
[tex][S] = K_M[/tex]
So this means that equation one would now become
[tex][ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}[/tex]
So this means that half of the active sites that is the total enzyme concentration are filled with S
Considering the Third Statement
[tex][S] >>K_M[/tex]
In this case the [tex]K_M[/tex] in the denominator of equation 1 would be neglected and the equation becomes
[tex][ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}][/tex]
This means that almost all the sites are occupied with substrate
The rate of this reaction is mathematically defined as
[tex]v =\frac{V_{max}[S]}{K_M [S]}[/tex]
Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and [tex]V_{max}[/tex] is he maximum velocity of the reaction
In this case also the [tex]K_M[/tex] at the denominator would be neglected as a result of the statement hence the equation becomes
[tex]v = \frac{V_{max}[S]}{[S]} = V_{max}[/tex]
So it means that the reaction does not depend on the concentration of substrate [S]
For the final statement(Not True ) it would match with condition that states that increasing [tex][E_{total}][/tex] will lower [tex]K_M[/tex]
This is because [tex]K_M[/tex] does not depend on enzyme concentration it is a property of a enzyme
