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Zepler [3.9K]
3 years ago
15

A protein has the amino acid sequence

Chemistry
1 answer:
kherson [118]3 years ago
6 0

Answer:

Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments

Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments

Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)

Explanation:

Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL

Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL.  All produced in three cut sites.

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A negative feedback is one in which the process that produces the feedback is interrupted. That is, the process is stopped as a result of the feedback received.

In this case, CO2  which leads to global warming dissolves in the ocean which serves a large sink for the gas. However, as the increase in ocean temperatures causes decrease in solubility of CO2, more CO2 is released leading to further temperature rise. This is an example of a negative feedback loop.

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2 years ago
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3 years ago
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3 years ago
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A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the
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Answer:

(a) 102.6g/mol

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}

Solving for the molarity of base we've got:

M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L

Now, we can compute the moles of the base as:

n_{base}=0.425mol/L*0.1L=0.0425mol

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(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

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Best regards.

8 0
3 years ago
Complete them with correct formulas<br> Then balance them
otez555 [7]

Answer:

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2 years ago
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