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VikaD [51]
3 years ago
8

What happens to the cell membrane during exocytosis?

Chemistry
1 answer:
Cloud [144]3 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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0.033g

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg

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Procedure for the experiment.
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A solid residue of limestone and some gypsum.

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The sulfur dioxide (so2) stack-gas concentration from fossil-fuel combustion is 12 ppmv. determine the stack-gas so2 concentrati
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The concentration of SO_{2} in the stack gas = 12 ppmv

That means 12 L of  SO_{2} is present per 10^{6} L gas

The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,

PV = nRT

(1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K)

V = 22.4 L

1 mol SO_{2} occupies 22.4 L

Moles of SO_{2} = 12 L *\frac{1 mol}{22.4 L} = 0.5357 mol SO_{2}

Mass of  SO_{2} =0.5357 mol *\frac{64.06 g}{1 mol}  = 34.32 g SO_{2}  *\frac{10^{6} microgram}{1 g} =3.432 *10^{7}μg

Converting 10^{6} L to m^{3}:

10^{6} L *\frac{1 m^{3}}{1000 L} = 10^{3}   m^{3}

Calculating the concentration in μg/m^{3}:

\frac{3.432 * 10^{7} microgram}{10^{3} L}    = 3.432 * 10^{4}  microgram/m^{3}

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