Answer:
A.
Explanation:
The <u>tertiary structure </u>of proteins is related to the interactions between the amino acids of the <u>primary structure</u>. Thus, these interactions give it a specific three-dimensional configuration which is very sensitive to <u>functionality</u>.
For example, <u>allosteric inhibitions</u> are related to this concept. When the <u>inhibitor</u> changes the tertiary structure of the protein it loses all <u>activity</u> and for the catalysis of the reaction.
Thus, the primary structure (which is related to the specific <u>sequence of amino acids</u>) will determine the tertiary structure since the chain folds will be a consequence of<u> intra-amino acid interactions</u>.
Answer: Adenine and guanine are the two purines and cytosine, thymine and uracil are the three pyrimidines. The main difference between purines and pyrimidines is that purines contain a sixmembered nitrogencontaining ring fused to an imidazole ring whereas pyrimidines contain only a sixmembered nitrogencontaining ring. They both are types or categories of nitrogen containing bases present in nuclei acids of DNA and RNA.
Purines are 2 Ring or Carbon Ring, Nitrogen containing bases. That consist of these 2 rings next placed next to each other. These examples include - Adenine and Guanine.
Pyrimidines are 1 or single Ring Nitrogen containing structures. There are 3 nitrogenous bases that are categorized as pyrimidines. Cytosine, Thymine, and Uracil.
There are no states in the picture, but Na should have a "(s)" after it, and Cl2 should have a "(g)" after it. NaCl should have an "(s)". Chlorine is a diatomic element so it has a "2" subscript on it.
Hope this helped! :)
It’s the 3d one
Good luckkkkkkkk
Answer:
pH = 7.8
Explanation:
The Henderson-Hasselbalch equation may be used to solve the problem:
pH = pKa + log([A⁻] / [HA])
The solution of concentration 0.001 M is a formal concentration, which means that it is the sum of the concentrations of the different forms of the acid. In order to find the concentration of the deprotonated form, the following equation is used:
[HA] + [A⁻] = 0.001 M
[A⁻] = 0.001 M - 0.0002 M = 0.0008 M
The values can then be substituted into the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008M/0.0002M) = 7.8
