Answer:
Air-sea exchange is a physio-chemical process and is important for the cycling of gases such as carbon dioxide, methane, nitrous oxide, dimethylsulfide and ammonia
Answer: Four amino acids.
Explanation:
RNA (ribonucleic acid) and DNA (deoxyribonucleic acid) are polymers made up of long chains of nucleotides. Thus, a nucleotide is the basic building block of nucleic acids and consists of a sugar molecule (ribose in RNA or deoxyribose in DNA) linked to a phosphate group and a nitrogenous base. <u>The nitrogenous bases can be Adenine (A), Thymine (T), Cytosine (C), Guanine (G) or Uracil (U) replacing T in RNA</u>. DNA is the molecule that stores the genetic information to synthesize polypeptides or proteins (set of amino acids). When proteins need to be created, DNA is transcribed into RNA and this RNA is translated in ribosomes to create polypeptides.
Complementarity is the ability to combine with their complement. A and T (or U) are complementary, while C and G are complementary to each other. Complementarity is the principle of replication and transcription, because it is a propery of both DNA and RNA sequences. Their the nucleotide bases at each position of the sequences are complementary, allowing cells to copy information.
Nucleotides in RNA are grouped into codons, which consist of groups of 3 nucleotides where each codon codes for an amino acid. Therefore, <u>the sequence of bases in nucleic acids determines which protein will be synthesized</u>. Protein synthesis begins with a start codon (AUG, which codes for the amino acid methionine) and ends with a stop codon (UGA, UAG and UAA). So, if there are 15 nucleotides, there are 15 bases. Since they are grouped in groups of 3, we will have a polypeptide of 4 amino acids.
<u>3 nucleotides form a codon, so 12 nucleotides form 4 codons giving 4 amino acids. The last codon, is formed by the last 3 nucleotides and form the stop codon that stop protein synthesis</u>.
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
