Answer:
(a) (Glu)zo or(Phe-Met)3 at pH 7.0
O (Glu)zo ✔
O (Phe-Met)s ❌
(b) (Gly) zo or (Lys-Ala)3 at pH 7.0:
O (Gly12) ❌
O (Lys-Ala)✔
(c) (Ala-Asp-Gly)s or (Asn-Ser-His)s at pH 3.0:
O (Asn-Ser-His)s ✔
O (Ala-Asp-Gly)s ❌
(d) (Ala-Ser-Gly)s or (Asn-Ser-His)s at pH 6.0:
O(Ala-Ser-Gly)s ❌
O (Asn-Ser-Hish)s ✔
Explanation:
Polypeptides that has polar or charged side chains are more soluble than polypeptides with nonpolar side chains.
(a) At ph 7.0
(Glu)20 is negatively charged at pH 7 and more soluble
(Phe-Met)3 is observed to be less polar and less soluble
(b)At ph 7.0
(Lys-Ala)3 is positively charged (polar) and more soluble
(Gly)20 is uncharged as only the amino- and carboxyl-terminal groups are charged as its less polar and less soluble too.
(c) At pH 6.0
(Asn-Ser-His)5 has polar Asn side chains and partially protonated His side chains and it's more soluble unlike the (Ala-Asp-Gly)s at that pH.
(d) At pH 3.0
(Asn-Ser-His)s as partially protonated carboxylate groups of Asp residues and it is also neutral but the imidazole groups of His residues are fully protonated and positively charged. Hence it's more soluble than the (Ala-Ser-Gly)s at that particular pH.
C. must be controlled to coordinate with the cell cycle
Hope this helps :)
Answer:
The most obvious result of the removal of the top predators in an ecosystem is a population explosion in the prey species. ... When prey becomes more scarce, the predator population declines until prey is again more abundant. Therefore, the two balance each other. When the predators are removed, prey populations explode.
They would most likely also become extinct, since they require them to "Fix the nitrogen" during photosynthesis, they have a "You need me, I need you" kind of thing. One can't live without the other.
