Specific chemicals are bound by carrier proteins and transferred on one side of the membrane. The conformational changes they go through next enable the molecule to cross the membrane and exit on the other side.
How carrier protein facilitate the diffusion?
When a molecule diffuses, it usually moves from a high concentration location to a low concentration area until the concentration is the same everywhere in the space.
Contrary to channel proteins, another form of membrane transport protein that is less selective in the molecules it transports, carriers are proteins that move a particular material through intracellular compartments, into the extracellular fluid, or across cells. Carrier proteins are found in lipid bilayer cell structures such cell membranes, mitochondria, and chloroplasts, just like other membrane transport proteins.
Therefore, carrier proteins can facilitate the diffusion of glucose or other substances into the cell.
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Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
Answer:
Water's high specific heat allows it to absorb a large amount of heat without changing much in temperature, keeping a relatively constant temperature in oceans and lakes which is beneficial to marine life. If water did not have a high specific heat, then it would increase in temperature very quickly and oceans and lakes would have drastic changes in temperature in very short periods of time. This could harm or kill the organisms living in them.
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hope this was helpful :)</h3>
The answer would be 0.26 centimeters.
