Answer: D
Explanation:
x+10x = Tx*+ Water: water molecules move | Potato x+10Water = Potato slice in pure water
so the Answer is D
Answer:
a. + glucose, + lactose = On
b. - glucose, - lactose = Off
c. + glucose, - lactose = Off
d. - glucose, + lactose = On
Explanation:
Lac operon has both types of control, repressible and inducible.
Whenever glucose level is low in the cell, an enzyme known as adenylyl cyclase raises the level of cAMP which forms a dimer with CRP protein and they both act as activator of lac operon and cause expression.
Apart from this, when lactose is present in the cell, β-galactosidase enzyme metabolizes lactose to form allolactose which causes allosteric repulsion in the lac repressor and causes its removal from the operator. As soon as repressor is removed lac operon gets activated.
In wild type lac operons, the expression of lac operon occurs when glucose level is low in the cell and lactose is present but in this mutant presence or absence of glucose will not make a difference because CRP will bind Plac promoter independent of cAMP level i.e. activator CRP will work even in high glucose concentration. If lactose is present then lac operon will always express so in option 'a & d' lac operon will express but in option 'b & c' it will not express.
Exoskeleton
Exoskeleton is a evolutionary innovation. This is operated by muscles attached to it and acts as an armor. Other than support this is an adaptive mechanism to provide protection against predators. In vertebrates, it is considered as the dermal bone that developed via membranous ossification. Fishes retained their exoskeleton in the form of their scales. However, in some organisms terrestrial evolution allows the loss of exoskeleton and formed a more mobile endoskeleton,
Answer:
I think that the people that should have access to the information is your closest friends, family and your spouse
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
