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VikaD [51]
2 years ago
8

What happens to the cell membrane during exocytosis?

Chemistry
1 answer:
Cloud [144]2 years ago
8 0

Answer:

Endocytosis and Exocytosis: Differences and Similarities

ARTICLE Apr 28, 2020

by Nicole Gleichmann

Endocytosis and Exocytosis: Differences and Similarities

Endocytosis and exocytosis are the processes by which cells move materials into or out of the cell that are too large to directly pass through the lipid bilayer of the cell membrane. Large molecules, microorganisms and waste products are some of the substances moved through the cell membrane via exocytosis and endocytosis.

Why is bulk transport important for cells?

Cell membranes are semi-permeable, meaning they allow certain small molecules and ions to passively diffuse through them. Other small molecules are able to make their way into or out of the cell through carrier proteins or channels.

But there are materials that are too large to pass through the cell membrane using these methods. There are times when a cell will need to engulf a bacterium or release a hormone. It is during these instances that bulk transport mechanisms are needed.

Endocytosis and exocytosis are the bulk transport mechanisms used in eukaryotes. As these transport processes require energy, they are known as active transport processes.

Vesicle function in endocytosis and exocytosis

During bulk transport, larger substances or large packages of small molecules are transported through the cell membrane, also known as the plasma membrane, by way of vesicles – think of vesicles as little membrane sacs that can fuse with the cell membrane.

Cell membranes are comprised of a lipid bilayer. The walls of vesicles are also made up of a lipid bilayer, which is why they are capable of fusing with the cell membrane. This fusion between vesicles and the plasma membrane facilitates bulk transport both into and out of the cell.

What is endocytosis? Endocytosis definition and purposes

Endocytosis is the process by which cells take in substances from outside of the cell by engulfing them in a vesicle. These can include things like nutrients to support the cell or pathogens that immune cells engulf and destroy.

Endocytosis occurs when a portion of the cell membrane folds in on itself, encircling extracellular fluid and various molecules or microorganisms. The resulting vesicle breaks off and is transported within the cell.

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3) The relative concentrations of each gas must remain constant.

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The no. of moles of gases in each side is constant; there is 2 moles of gases at reactants side and 2 moles of gases at products side.

So, changing the volume will not affect on the equilibrium system.

<em>So, the right choice is:</em>

3) The relative concentrations of each gas must remain constant.

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1. What is the balanced chemical equation for the reaction that takes place between bromine and sodium iodine?
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What is the main difference between electron configuration and orbital notation?
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Answer:

Orbital Notation is more specific on where exactly the electron is placed.

Explanation:

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Glad I could help!

6 0
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An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
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Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

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2 years ago
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