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Biomolecules question

2012 · Shift 1 · Q19
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Biomolecules question

2012 · Shift 1 · Q19

JEE AdvancedChemistryBiomoleculesNumerical+3 / −1

The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH = 7.0 ?

IIT-JEE 2012 Paper 1 Offline Chemistry - Biomolecules Question 11 English

Peptide R1{R_1}R1​ R2{R_2}R2​
I H H
II H CH3C{H_3}CH3​
III CH2COOHC{H_2}COOHCH2​COOH H
IV CH2CONH2C{H_2}CON{H_2}CH2​CONH2​ (CH2)4NH2{(C{H_2})_4}N{H_2}(CH2​)4​NH2​
V CH2CONH2C{H_2}CON{H_2}CH2​CONH2​ CH2CONH2C{H_2}CON{H_2}CH2​CONH2​
VI (CH2)4NH2{(C{H_2})_4}N{H_2}(CH2​)4​NH2​ (CH2)4NH2{(C{H_2})_4}N{H_2}(CH2​)4​NH2​
VII CH2COOHC{H_2}COOHCH2​COOH CH2CONH2C{H_2}CON{H_2}CH2​CONH2​
VIII CH2OHC{H_2}OHCH2​OH (CH2)4NH2{(C{H_2})_4}N{H_2}(CH2​)4​NH2​
IX (CH2)4NH2{(C{H_2})_4}N{H_2}(CH2​)4​NH2​ CH3C{H_3}CH3​

Numerical answer
View written solutionFree

Correct answer: 4

Step-by-step Derivation:

  1. Understand the Structure of the Peptides: The question describes nine dipeptides. A dipeptide is formed from two amino acids linked by a peptide bond. The general structure of a dipeptide with side chains R₁ and R₂ is:

    H2N−CHR1−CO−NH−CHR2−COOHH_2N-CHR_1-CO-NH-CHR_2-COOHH2​N−CHR1​−CO−NH−CHR2​−COOH

    This structure has several ionizable groups that determine its overall charge: the N-terminal amino group (-NH_2$), the C-terminal carboxyl group ($-COOH), and any ionizable groups within the side chains R₁ and R₂.

  2. Determine the Charge of Ionizable Groups at pH = 7.0: The charge of an ionizable group at a given pH depends on its pKa value. The general rules are:

    • If pH < pKa, the group is predominantly in its protonated (acidic) form.

    • If pH > pKa, the group is predominantly in its deprotonated (basic) form.

    • **N-terminal amino group (-NH_2$):** The pKa is approximately 9-10. Since pH 7.0 < pKa, it will be protonated to $-NH_3^+. Its charge is +1.

    • **C-terminal carboxyl group (-COOH$):** The pKa is approximately 2-3. Since pH 7.0 > pKa, it will be deprotonated to $-COO^-. Its charge is -1.

    The net charge of the peptide backbone at pH = 7.0 is therefore (+1)+(−1)=0(+1) + (-1) = 0(+1)+(−1)=0. This is the zwitterionic form. Consequently, the overall charge of the dipeptide is determined solely by the sum of the charges on the side chains R₁ and R₂.

  3. Analyze the Charges of the Side Chains (R₁ and R₂) at pH = 7.0: We need to classify the given R groups based on whether their side chains are acidic, basic, or neutral.

    • Acidic Side Chain:

      • -CH_2COOH$ (from Aspartic Acid): The side chain carboxyl group has a pKa of ~3.9. Since pH 7.0 > pKa, it deprotonates to $-CH_2COO^-, contributing a charge of -1.

    • Basic Side Chain:

      • -(CH_2)_4NH_2$ (from Lysine): The side chain amino group has a pKa of ~10.5. Since pH 7.0 < pKa, it protonates to $-(CH_2)_4NH_3^+, contributing a charge of +1.

    • Neutral Side Chains:

      • ‘−H‘`-H`‘−H‘ (from Glycine)
      • −CH3-CH_3−CH3​ (from Alanine)
      • −CH2CONH2-CH_2CONH_2−CH2​CONH2​ (from Asparagine)
      • −CH2OH-CH_2OH−CH2​OH (from Serine) These side chains are not ionizable near pH 7.0 and have a charge of 0.

  4. Calculate the Net Charge for Each Peptide: A peptide will be positively charged if the sum of the charges of its side chains (R₁ and R₂) is greater than zero.

    • Peptide I: R₁ = H (0), R₂ = H (0). Net Charge = 0 + 0 = 0.
    • Peptide II: R₁ = H (0), R₂ = CH3CH_3CH3​ (0). Net Charge = 0 + 0 = 0.
    • Peptide III: R₁ = CH2COOHCH_2COOHCH2​COOH (-1), R₂ = H (0). Net Charge = -1 + 0 = -1.
    • Peptide IV: R₁ = CH_2CONH_2$ (0), R₂ = $(CH_2)_4NH_2 (+1). Net Charge = 0 + 1 = +1. (Positively charged)
    • Peptide V: R₁ = CH_2CONH_2$ (0), R₂ = $CH_2CONH_2 (0). Net Charge = 0 + 0 = 0.
    • Peptide VI: R₁ = (CH_2)_4NH_2$ (+1), R₂ = $(CH_2)_4NH_2 (+1). Net Charge = +1 + 1 = +2. (Positively charged)
    • Peptide VII: R₁ = CH_2COOH$ (-1), R₂ = $CH_2CONH_2 (0). Net Charge = -1 + 0 = -1.
    • Peptide VIII: R₁ = CH_2OH$ (0), R₂ = $(CH_2)_4NH_2 (+1). Net Charge = 0 + 1 = +1. (Positively charged)
    • Peptide IX: R₁ = (CH_2)_4NH_2$ (+1), R₂ = $CH_3 (0). Net Charge = +1 + 0 = +1. (Positively charged)

  5. Count the Positively Charged Peptides: The peptides that have a net positive charge at pH = 7.0 are IV, VI, VIII, and IX.

Therefore, the total number of positively charged peptides is 4.

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