Image please. 30 POINTS

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

Phosphorus atomic radius is smaller than magnesium atomic radius <br> True or false

stealth61 [152]

Answer:

True

Explanation:

Atomic radius can be defined as a measure of the size (distance) of the atom of a chemical element such as hydrogen, oxygen, carbon, nitrogen etc, typically from the nucleus to the valence electrons. The atomic radius of a chemical element decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon). Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).

<em>Hence, the atomic radius of phosphorus is smaller than the atomic radius of magnesium. Basically, the atomic radius of phosphorus is 98 pm while the atomic radius of magnesium is 145 pm.</em>

4 0

3 years ago

PLEASE HELP MEEEEEEEEEEEEEEEEE

RSB [31]

The answer is the second choice.

5 0

3 years ago

Read 2 more answers

How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant

schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

1 g = 1,000 mg
The molar mass of Cu is 63.55 g/mol
When 1 mole of Cu is deposited, 2 moles of electrons circulate.
The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
1 A = 1 C/s

The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

$336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s$

3 0

3 years ago

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

konstantin123 [22]

C3 H4 N1 ~~that’s what I think

8 0

3 years ago