One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3
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:
How may we help kind sir
Explanation:
and if this was for points thanks
Answer:
The volume for the ideal gas is: 4647.5 Liters
Explanation:
Formula for the Ideal Gases Law must be applied to solve this question:
P . V = n . R . T
We convert the T° to K → 100°C + 273 = 373 K
We convert pressure value from kPa to atm.
2 kPa . 1atm/101.3 kPa = 0.0197 atm
We replace data in the formula.
V = ( n . R . T) / P → (3 mol . 0.082 . 373K) / 0.0197 atm =
The volume for the ideal gas is: 4647.5 Liters