Explanation:
<u>-101 mRNA codons</u>
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In the genetic code, an amino acid is encoded by 3 nucleotides, while there are just 4 bases
. Each amino acid is specifically encoded by a codon- a triplet sequence of nucleotides...
Thus 99 amino acids= 99 codon sequences.
...along with a start and stop codon, this would require 101 mRNA codons
Further Explanation:
The nucleic acids are comprised of smaller units called nucleotides and function as storage for the body’s genetic information. These monomers include ribonucleic acid (RNA) or deoxyribonucleic acid (DNA). They differ from other macromolecules since they don’t provide the body with energy. They exist solely to encode and protein synthesis.
Basic makeup: C, H, O, P; they contain phosphate group 5 carbon sugar does nitrogen bases which may contain single to double bond ring.
Codons are three nucleotide bases encoding an amino acid or signal at the beginning or end of protein synthesis.
RNA codons determine certain amino acids so the order in which the bases occur within in the codon sequence designates which amino acid is to be made bus with the four RNA nucleotides (Adenine, Cysteine and Uracil) Up to 64 codons (with 3 as stop codons) determine amino acid synthesis. The stop codons ( UAG UGA UAA) terminate amino acid/ protein synthesis while the start codon AUG begins protein synthesis.
Learn more about transcription at brainly.com/question/11339456
Learn more about DNA and RNA brainly.com/question/2416343?source=aid8411316
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Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
Radiated would be the answer
Answer:
Explanation:
Hello!
In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:
Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:
Ka1 > Ka2 > Ka3
Best regards!
Dipole-dipole interactions, and London dispersion interactions
