Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) <u>Trypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) <u>Chymotrypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the <u>"Lis"</u> and <u>"Arg"</u> (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the <u>"Phe"</u> (See figure 2). The second amino acid that can be broken is <u>tyrosine</u>, but this amino acid is placed in the <u>C terminal spot</u>, therefore will not be involved in the <u>hydrolysis</u>.
A <span>heating curve having horizontal plateaus at certain points during the heating process means that the process occurs at increase heating. </span>
Temperature Decreases
Pressure Increases
Salinity Increases
Answer:
The correct option is;
D. Calcium chloride breaks into 3 ions, while sodium chloride only breaks into 2 ions
Explanation:
To help melt the ice on icy streets, calcium chloride (CaCl₂) is a more preferred salt type to sodium chloride (NaCl) as when it dissociates into ions in the ice, it forms three ions: a calcium ion, Ca²⁺, and two chloride ions, 2 Cl⁻, while sodium chloride forms two ions: one sodium ion, Na⁺, and one chloride ion, Cl⁻
According to the freezing point depression equation, we have;
ΔT = 
× m × i
Where;
ΔT = Change in freezing point temperature
= The freezing point depression constant
m = The solution's molality
i = van't Hoff factor
The calcium chloride salt will cause the most depression because the van't Hoff factor is 3 for calcium chloride and 2 for sodium chloride.
