<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
Answer:
Apparently 0 is the answer which makes sense
Explanation:
https://www.tiger-algebra.com/drill/(x-5)(x_1)/3(x_1)=x-5/3/
False politics & economics have to do with government not the environment but that’s pretty self explanatory
Answer:
D) In case 1, both PS I and PS II completely lose function; in case 2, a proton gradient is still produced.
Explanation:
The light dependent reaction of photosynthesis, which produces the ATP and NADPH needed in the light independent stage of the process, includes complexes of proteins and pigments called PHOTOSYSTEMS. These photosystems (I and II) are key to the functionality of the light dependent reactions in the thylakoid.
The major pigment present in both photosystems is CHLOROPHYLL A, which absorbs light energy and transfers electrons to the reaction center. Chlorophyll B is only an accessory pigment meaning it can be done without. Hence, if all of the chlorophyll A is inactivated in the algae but leaves chlorophyll B intact as in case 1, both PS I and PS II will lose their function because Chlorophyll A is the major pigment that absorbs light energy in both photosystems.
In case 2, if PS I is inhibited and PS II is unaffected, a PROTON GRADIENT WILL STILL BE PRODUCED because the splitting of water into protons (H+) and electrons (e-) occurs in PSII. Hence, H+ ions can still be pumped into the inner membrane of the thylakoid in order to build a proton gradient even without the occurrence of PS I.
