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iogann1982 [59]
3 years ago
12

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

Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

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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Hydrogen sulfide gas is reacted with dioxygen gas to produce gaseous sulfur dioxide and water. True or false?.
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Hydrogen sulfide gas is reacted with dioxygen gas to produce gaseous sulfur dioxide and water. It is false.

When hydrogen sulfide gas reacts with dioxygen gas, it produces water and solid Sulphur and not gaseous sulfur. The chemical equation of the given substances is :

                   2H2S +  O2 → 2S + 2H20

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If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of
navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

M=\frac{n}{V}

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

2.60g*\frac{1mol}{102.89g} =0.0253mol

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

M=\frac{0.0253mol}{0.160L}=0.158M

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L

That in milliliters turns out to be:

V=0.211L*\frac{1000mL}{1L}=211mL

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