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3 years ago

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A typical deposit of cholesterol,C27H460,in an artery has a mass of 3.90mg.how many molecules of cholesterol are present in this

deposit?

1 answer:

love history [14]3 years ago

8 0

Answer:

6.074 X 10¹⁸ molecules.

Explanation:

Molecular mass(MM) of this compound C₂₇H₄₆O

= 27*(MM of C) + 46*(MM of H) + (MM of O)

= 27*(12.0107) + 46* (1.00784) + (15.999)

= 324.2889 + 46.36064 + 15.999

=386.64854 g

The mass of a mole of any compound is called it's molar mass. 1 molar mass has 6.022 X 10²³, or Avogadro's number, C₂₇H₄₆O molecules.

If 386.64854 g of C₂₇H₄₆O has 6.022 X 10²³ molecules of cholesterol, then

3.9mg or 0.0039g of C₂₇H₄₆O will have how many molecules?..

Using unitary method:

Number of molecules = $\frac{0.0039 * 6.022 *10^{23} }{386.64854}$

= 6.074 X 10¹⁸ molecules.

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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

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You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans

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Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

<u>Vi = 0.145 l or 145 ml</u>

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