Answer:
The villi in the small intestine provide a large surface area with an extensive network of blood capillaries. This makes the villi well adapted to absorb the products of digestion by diffusion and active transport.
Explanation:
Answer:
<u>1. type IV hypersensitivity: delayed-type hypersensitivity. </u>
<u>2. type I hypersensitivity: IgE </u>
<u>3. type II hypersensitivity: IgG </u>
<u>4. type III hypersensitivity: immune complexes </u>
Explanation:
Type IV hypersensitivity is cell mediated hypersensitivity rather than antibodies as in all other types of hypersensitivities. It is also known as delayed type hypersensitivity because it usually respond in 2-3 days via T-Lymphocytes rather than antibodies like IgE or IgG. So we can say that in option 5 is mismatched as it relates type IV to IgG antibodies.
It can be caused by mutation, random mating, random fertilization, and recombination.
Hope this helps :)
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
The hydrologic cycle is another name for the water cycle.
