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USPshnik [31]
3 years ago
10

In an animal cell, what is the function of the lysosomes?

Biology
2 answers:
Alik [6]3 years ago
8 0
<span><span><span>Release enzymes outside of the cell (exocytosis)</span>
which may serve the purpose of destroying materials around the cell.</span><span><span>Break-down 'digestion' of materials from inside the cell (autophagy)</span> 
i.e. by fusing with vacuoles from inside the cell. 
This could include digesting worn-out organelles so that useful chemicals locked-up in their structures can be re-used by the cell.</span><span><span>Break-down 'digestion' of materials from outside the cell (heterophagy)</span>
i.e. by fusing with vacuoles from outside the cell.
This could include breaking-down material taken-in by phagocytes, which include many types of white blood cells - also known as leucocytes. Specific mechanisms of heterophagy can be:<span><span>phagocytic - by which cells engulf extracellular debris, bacteria or other particles - only occurs in certain specialized cells</span><span>pinocytic - by which cells engulf extracellular fluid</span><span>endocytic - by which cells take-up particles such as molecules that have become attached to the outer-surface of the cell membrane.</span></span></span><span><span>Recycle the products of biochemical reactions that have taken place following materials being brought into the cell by endocytosis (general term for this 'recycling' function: biosynthesis) </span>
Different materials (chemicals) are processed in different ways, e.g. some structures may be processed/degraded within lysosomes and others are taken to the surface of the cell.</span><span>Completely break-down cells that have died (autolysis)</span></span>

In general, the functions of lysosomes involve breaking-down i.e. processing to 'make safe' or make use of, or removing from the cell e.g. by exocytosis, useless and potentially harmful materials such as old worn-out parts of the cell or potential threats such bacteria. Lysosomes can therefore be thought of as the rubbish disposal units within cel

tatyana61 [14]3 years ago
5 0
<span><span>Release enzymes outside of the cell (exocytosis)
which may serve the purpose of destroying materials around the cell.</span><span>Break-down 'digestion' of materials from inside the cell (autophagy) 
i.e. by fusing with vacuoles from inside the cell. 
This could include digesting worn-out organelles so that useful chemicals locked-up in their structures can be re-used by the cell.</span><span>Break-down 'digestion' of materials from outside the cell (heterophagy)
i.e. by fusing with vacuoles from outside the cell.
This could include breaking-down material taken-in by phagocytes, which include many types of white blood cells - also known as leucocytes. Specific mechanisms of heterophagy can be:<span>phagocytic - by which cells engulf extracellular debris, bacteria or other particles - only occurs in certain specialized cellspinocytic - by which cells engulf extracellular fluidendocytic - by which cells take-up particles such as molecules that have become attached to the outer-surface of the cell membrane.</span></span><span>Recycle the products of biochemical reactions that have taken place following materials being brought into the cell by endocytosis (general term for this 'recycling' function: biosynthesis) 
Different materials (chemicals) are processed in different ways, e.g. some structures may be processed/degraded within lysosomes and others are taken to the surface of the cell.</span>Completely break-down cells that have died (autolysis)</span>

In general, the functions of lysosomes involve breaking-down i.e. processing to 'make safe' or make use of, or removing from the cell e.g. by exocytosis, useless and potentially harmful materials such as old worn-out parts of the cell or potential threats such bacteria. Lysosomes can therefore be thought of as the rubbish disposal units within cell

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\boxed {\boxed {\sf 121.905 \ J  }}

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We are asked to find the energy given mass, specific heat, and change in temperature. Therefore, we must use this formula;

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The mass is 15 grams and the specific heat is 0.129 J/(g×°C). Let's calculate the change in temperature.

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Now we know all the values:

m= 15 \ g \ \\c= 0.129 \ J / (g* \textdegree C) \\\Delta T= 63 \ \textdegree C

Substitue the values into the formula.

q= (15 \ g)( 0.129 \ J / (g* \textdegree C)) ( 63 \ \textdegree C)

Multiply the first numbers together. The grams will cancel.

q= (1.935 \ J/ \textdegree C) ( 63 \ \textdegree C)

Multiply again, this time the degrees Celsius cancels.

q= 121.905 \ J

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