Plant cells have a cell wall. They also have a larger water vacuole.
Answer:
The right answers are mentioned in the picture.
A base pair (bp) is the pairing of two nucleobases located on two complementary strands of DNA or RNA. This pairing is carried out by hydrogen bridges. There are four types of nucleic bases: A-T-C-G, these letters Adenine, Thymine, Cytosine, and Guanine. A with T and C with G.
It is also necessary to take into account the antiparallel character of the DNA strands. If a strand is in the 5 '3' direction, its complete strand is in the 3 '5' direction.
Explanation:
Carbohydrates are not important as a stimulus in the gastric phase of gastric secretion, whereas distension, peptides, and low acidity are.
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
Some organisms observed by Charlis Darwin are as follow:
1. Beetles
2. Giant tortoise
3. Mockingbird
4. Rhea
5. Pigeon
6. Sand lady slipper
