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2 years ago

How did the nucleus evolve through time?

Chemistry

1 answer:

Tems11 [23]2 years ago

4 0

Answer:

The nucleus represents a major evolutionary transition. As a consequence of separating translation from transcription many new functions arose, which likely contributed to the remarkable success of eukaryotic cells. Here we will consider what has recently emerged on the evolutionary histories of several key aspects of nuclear biology; the nuclear pore complex, the lamina, centrosomes and evidence for prokaryotic origins of relevant players.

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The atomic weight of Ga is 69.72 amu. There are only two naturally occurring isotopes of gallium:

irga5000 [103]

Answer:

71 Ga has a naturally abundance of 36%

Explanation:

Step 1: Given data

Gallium has 2 naturally occurring isotopes: this means the abundance of the 2 isotopes together is 100 %. The atomic weight of Ga is 69.72 amu. This is the average of all the isotopes.

Since the average mass of 69.72 is closer to the mass of 69 Ga, this means 69 Ga will be more present than 71 Ga

Percentage 69 Ga> Percentage 71 Ga

<u>Step 2:</u> Calculate the abundance %

⇒Percentage of 71 Ga = X %

⇒Percentage of 69 Ga = 100 % - X %

The mass balance equation will be:

100*69.72 = x * 71 + (100 - x)*69

6972 = 71x + 6900 -69x

72 = 2x

x = 36 %

71 Ga has a naturally abundance of 36%

69 Ga has a naturally abundance of 64%

3 0

3 years ago

Which medical condition(s) can prevent molecules from entering the cell?

kvasek [131]

Answer:

Explanation:

any type of spreading disease that kills

6 0

2 years ago

Read 2 more answers

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

2 years ago

If three molecules of hydrogen react with one molecule of nitrogen, how many molecules of ammonia would be formed?

joja [24]

Answer:

The answer to your question is 2 molecules, or B(edg 2021).

Explanation:

edg 2021

8 0

2 years ago

The temperature and number of moles of a gas are held constant. Which of the following is true for the pressure of the gas?

Taya2010 [7]

Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K P2V2 = K

P1V1 = P2V2

3 0

3 years ago