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:
eukaryotes that ensures that the number of chromosomes will not double from parent to offspring when gametes fuse during fertilisation. Homologous pairs of chromosomes separate in meiosis I, so the gametes are haploid (n), and each gamete receives only one member of each chromosome pair.
Explanation:
I hope this helps!
Answer: option D.
Phenotype
Explanation:
Theory of natural selection shows that different organism survive, adapt and reproduce in their environment due to their different phenotypic variation. The phenotypic variation exist among individual organisms and it is heritable which is transfer from one generation to another.
Phenotype are traits or physical characteristics or observable characteristics of organism e.g organisms appearance, behavior e.t.c.
Answer:
You can say:
Enzymes are soluble.
Enzymes are proteinous in nature.
Enzymes are sensitive to temperature.
Enzymes are sensitive to the activity or alkalinity of their environment.
Enzymes can br inactivated by inhibitors.
Hi Mark
The answer is : D
A hox gene likely affected
I hope that's help:)
