The tetrapeptide, NRMD, exhibits four pKa values: pK1 = 1.88, pK2 = 3.65, pK3 = 8.80,...

The tetrapeptide, NRMD, exhibits four pK_a values: pK₁ = 1.88, pK₂ = 3.65,
pK₃ = 8.80, and pK₄ = 12.48.

What is the isoelectric point for this peptide?

Group of answer choices

pI = 2.8

pI = 5.3

pI = 6.2

pI = 8.1

pI = 10.6

Solutions

Expert Solution

Answer - 6.2

Explanation - The following net charge changes occur -

The isoelectric point (pI) is the pH where the net charge of the peptide is 0.

As can be seen from the diagram, that the net charge of peptide is between pK2 and pK3.

Thus, pI = (pK2 +pK3)/2

= (3.65 + 8.80)/2

=12.45/2

=6.225

Hence, isoelectric point for this peptide is 6.2.

gladiator answered 1 year ago

Next > < Previous

Related Solutions

PK1 and PK2 are protein kinases involved in a phosphorylation cascade. If either protein is inactive...

PK1 and PK2 are protein kinases involved in a phosphorylation cascade. If either protein is inactive due to a mutation the response is not observed. A different mutation causes PK1 to be permanently active. This activity induces a response even if PK2 is inactive. Which of the following conclusions is supported by the above information? A. PK1 activates PK2 B. PK2 activates PK1 C. PK2 is a transcription factor D. PK1 inhibits PK2 Ca2+ can act as a second messenger...

What is the pH of a 0.1M solution of monosodium succinate pk1= 4.19 pk2= 5.56

Consider the titration of 50.0ml of 0.0200M diprotic acid with pk1=4.00 and pk2=8.00 by 0.100M NaOH....

Consider the titration of 50.0ml of 0.0200M diprotic acid with pk1=4.00 and pk2=8.00 by 0.100M NaOH. Calculate the pH at the following points: 0ml, 2ml, 5ml, 7ml, 10ml, 15ml, 20ml, 25ml

125.0-mL aliquot of 0.124 M diprotic acid H2A (pK1 = 4.03, pK2 = 8.06) was titrated...

125.0-mL aliquot of 0.124 M diprotic acid H2A (pK1 = 4.03, pK2 = 8.06) was titrated with 1.24 M NaOH. Find the pH at the following volumes of base added: Vb = 0.00, 1.80, 6.25, 11.50, 12.50, 13.50, 18.75, 24.00, 25.00, and 28.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

For the triprotic acid, H+glutamic acid, pK1=2.23, pK2=4.42 and pK3=9.95. If one has 0.10L of a...

For the triprotic acid, H+glutamic acid, pK1=2.23, pK2=4.42 and pK3=9.95. If one has 0.10L of a 1.0E-2 M solution and titrates it with 0.255 L of a 1.0E-2 M KOH solution, what is the pH of the resulting solution? The answer is 10.04. I just have no clue how to get there.

A 25.00 mL aliquot of a 0.100 M maleic acid, HOOC-CH=CH-COOH (pK1=1.87) and (pK2 = 6.23)...

A 25.00 mL aliquot of a 0.100 M maleic acid, HOOC-CH=CH-COOH (pK1=1.87) and (pK2 = 6.23) was titrated with 0.1000 M NaOH. Construct a titration curve by calculating pH at a) initial pH b) after adding 5.00 mL of NaOH c) after adding 24.90 mL of NaOH d) at equivalence point e) after adding 25.01 mL of NaOH f) after adding 25.50 ml of NaOH g) after adding 49.90 mL NaOH h) after adding 50.00 mL NaOH i) after adding...

What is the isoelectric point (pI) of the polypeptide given the following pKa values: alpha-COOH pka=...

What is the isoelectric point (pI) of the polypeptide given the following pKa values: alpha-COOH pka= 9.67 alpha-NH3+ pKa= 2.17 guanidium R group pKa= 12.48 carboxylic acid R group pKa= 4.25

For a monoprotic acid, verify that: a) At pH values above the pKa, the concentation of...

For a monoprotic acid, verify that: a) At pH values above the pKa, the concentation of [HA] decreases by tenfold for each unit increase in pH (i.e. dpHA/dpH = 1) b) At pH values below the pKa, the concentation of [A-] decreases by tenfold for each unit decrease in pH (i.e. dpA/dpH = -1)

Using a cation exchange resin, a mixture of four amino acids, Asp (pKa 3.9), Arg (pKa...

Using a cation exchange resin, a mixture of four amino acids, Asp (pKa 3.9), Arg (pKa 12.5), Ser (pKa 13) and Lys(pKa 10.5) are separated using an elution gradient of increasing NaCl solution. What would be the correct elution sequence? Explain your reasoning. Draw the chromatogram would look like for this mixture if there was twice as much Asp and Arg in the sample than Ser and Lys.

Calculate the pH of a 0.228 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the...

Calculate the pH of a 0.228 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine ( H3NCH2CH2NH3 ) are 6.848 (pKa1) and 9.928 (pKa2). AND Calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

Subjects