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>.
Answer:
(C3H4O3) x 2 = C6H8O6, the molecular formula for Vitamin C.
Answer:
Explanation:
Hello there!
In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:
We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:
Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.
Regards!
Temperature will raise and water will evaporate
Answer: The mass of given amount of copper (II) cyanide is 462.4 g
Explanation:
To calculate the number of moles, we use the equation:
We are given:
Moles of copper (II) cyanide = 4 moles
Molar mass of copper (II) cyanide = 115.6 g/mol
Putting values in above equation, we get:
Hence, the mass of given amount of copper (II) cyanide is 462.4 g
