Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b.
Glu-Ala-Phe-Gly-Ala-Tyr by chymotrypsin
1 answer:
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>.
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Answer:
Explanation:
Given that:
The flow rate Q = 0.3 m³/s
Volume (V) = 200 m³
Initial concentration = 2.00 ms/l
reaction rate K = 5.09 hr⁻¹
Recall that:
where;
Thus; the concentration of species in the reactant = 102.98 mg/l
b). If the plug flow reactor has the same efficiency as CSTR, Then:
The volume of the PFR is ≅ 140 m³
Answer:
Explanation:
There are few simple rules in organic chemistry which a person must learn.
1. In predicting a structure from ozonolysis, number of oxygen present in products must be counted. In this particular example, there are 2+3+1 = 6 oxygen atoms are present in three molecules. This means that there should be at least three confirmed double bonds must be present in the molecule of Dictyopterene.
2. There are in total of 3+5+3 = 11 carbon atoms present, making 2^11 = 2048 possible structures for this particular structure. But since there are three double bonds present, this will reduce the number of possible structures as well.
3. Questions also says that when fully hydrogenated, the molecule generates Butylcycloheptane. That means that our base molecule is Butylcycloheptane and this molecule contains at least three double bonds.
So, first we draw the molecule of butylcycloheptane. After this, we think of were double bonds could be present. We do have the products. Let's make their structures first. I have also mentioned the possible breakup where the ozonolysis has occured by color code. You can see them in the reference image attached.
Part 1 out of 2: There is only one chiral carbon present in the molecule and have two possible isomers. Cis and Trans. Only one stereogenic carbon is present, that's why two possible isomers. Image attached.
