Minerals will be the correct answer because they are needed to start everything
Answer:
The correct answer is "It blocks translocation of the large ribosomal subunit, preventing the movement of peptidyl-tRNA from the A (acceptor) site to the P (peptidyl) site of the ribosome".
Explanation:
Many antibiotics work by inhibiting bacterial protein synthesis, and they act at different levels of the synthesis process. The mechanism of action of antibiotics could be assessed using different strategies on the laboratory. In this case an antibiotic with the synthetic polynucleotide 5’-AUGUUUUUUUUU resulting in disrupting peptide synthesis to Met-Phe would block translocation of the large ribosomal subunit, preventing the movement of peptidyl-tRNA from the A (acceptor) site to the P (peptidyl) site of the ribosome. The A site is the point of entry of the tRNA and where the first and second amino acid are added, whereas the P site is where the whole polypeptide is synthesized. This antiobiotic does not allow that protein synthesis continues to the P site of the ribosome, therefore the proteins synthesized only have two amino acids.
Explanation:
Receptors are specific molecular components of a biological system with which drugs interact to produce changes in body function. This fundamental event of drug-receptor interaction initiates communication, through signal molecules (first messenger) of great structural and functional diversity, with receptors coupled to ionic channels, to G-proteins, catalytic or regulators of DNA transcription, as well as The integrated support of second messengers. It also mentions certain diseases that arise due to dysfunction of these receptors of the receptor-effector systems, such as the appearance of aberrant receptors or as products of "oncogenes".
Answer:
The correct option is;
a. 11.21
Explanation:
The chemical equation for the reaction is;
NH₄OH ↔ NH₄⁺ + OH⁻
At the start of the reaction, we have;
NH₄OH ↔ NH₄⁺ + OH⁻
I 0.15 0 0
C 0.15 - x +x +x
The = 1.78 × 10⁻⁵
Therefore, we have;
1.78 × 10⁻⁵ = [NH₄⁺][OH⁻]/NH₄OH = x²/(0.15 - x)
Which gives;
(0.15 - x) 1.78 × 10⁻⁵ = x²
x² + 1.78 × 10⁻⁵x - 2.67× 10⁻⁶ = 0
Factorizing the above equation online, gives;
(x + 0.00164294)(x - 0.00162514) = 0
Therefore;
x = -0.00164294, or x = 0.00162514
We use the positive result, which is x = 0.00162514 = 1.62514 × 10⁻³ = [OH⁻]
Therefore;
pOH = -log[OH⁻] = -log(0.00162514) = 2.789
pH + pOH = 14
Therefore;
pH = 14 - pOH = 14 - 2.789 ≈ 11.21089
The pH of a 0.15 M NH₄OH solution with = 1.78 × 10⁻⁵ = 11.21.
The pH of a 0.15 M NH₄OH solution with = 1.78 × 10⁻⁵ is approximately 11.21