Help please it’s due soon just calculate the atomic mass of lithium that’s all i need

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

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

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

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 $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

7 0

3 years ago

Can someone help me with number 1 and 2 plz!

GalinKa [24]

Answer:

1) 0 N

2) 8 N

Explanation:

The net force is the sum of all of the forces acting on the object.

For question 1, we can see that there is a force of 5 N acting to the right and 5 N acting to the left. If we define the right to be positive and the left to be negative, then the net force equals:

Fnet = 5N - 5N = 0 N

Therefore, the net force in question 1 is 0 N.

For question 2, the process is very similar. We want to find the sum of the forces acting on the object. In this case, there are forces of 3 N and 5 N acting to the right.

Fnet = 3 N + 5 N = 8 N

Therefore, the net force in question 2 is 8 N.

Hope this helps!

3 0

2 years ago

4Fe + 30₂ ⇒ Fe₂0₃

DaniilM [7]

Answer:

A

The nuber of each one should be same

5 0

2 years ago

Which of the following is not equal to 42.7 millimeters?

Stella [2.4K]

The answer is: C. 0.00427 m.

A) 1 km = 1000000 mm.

d = 0.0000427 km · 1000000 mm/km.

d = 47.7 mm.

B) 1 hm = 100000 mm.

d = 0.000427 hm · 100000 mm/hm.

d = 42.7 mm.

C) 1 m = 1000 mm.

d = 0.00427 m · 1000 mm/m.

d = 4.27 mm.

D) 1 cm = 10 mm.

d = 4.27 cm · 10 mm/cm.

d = 42.7 mm.

Millimeter (abbreviated: mm, a thousandth part of metar) is an unit of distance in the metric system.

8 0

3 years ago

what is the molecular formula for a compound that is 39.99% carbon, 6.73% hydrogen, and 53.28% oxygen and has a molar mass of 18

4vir4ik [10]

Answer:

IDEK

Explanation:

5 0

3 years ago