3. The amino acid histidine has 3 pKa values: pKi 1.8, pK2-9.2, pK, \"6.0. An al

Question

3. The amino acid histidine has 3 pKa values: pKi 1.8, pK2-9.2, pK, "6.0. An aliquot of fully protonated histidine (pH 1) is titrated against NaoH. At 1.5 equivalents oH, what species of histidine are present in the solution? COO COO Hi-CH Hy-CH Species A Species B Species D Species C A. Species A only B. Species B only C. Equal amounts of species A & Species B D. Equal amounts of Species B & Species C E. Species A, B, and Care all present in significant quantities. 4. The components of poison ivy and poison oak that produce the characteristic itchy rash are catechols substituted with long-chain alkyl groups (Figure 1, below). If you were exposed to poison ivy, which of the treatments below would best be applied to the affected area? A. Wash the area with cold water. B. Wash the area with dilute vinegar, lemon juice, or urine C. Wash the area with soap and water. D. Wash the area with soap, water, and baking soda (sodium bicarbonate). E. Do nothing to the affected area. (CH2)14CH3 OH pKa "8 Substituted catechol in poison ivy.

Explanation / Answer

a) Suppose 1 mole of histidine is taken and for titration 1.5 moles of NaOH is used.

1 mole of NaOH will first deprotect the most acidic proton (COOH proton) and 1 mole of form B is produced. Rest 0.5 mole of NaOH will deprotect the next acidic proton (ring NH proton) and will produce 0.5 mole of form C from 1 mole of B. So, 0.5 mole of C is produced with 0.5 moles of B remains.

So equal amount of B and C are produced.

B) The poison is catechol derivative containing acidic phenolic OH functional groups. The treatment of this poison is to neutralise the catechols with suitable base.

So the treatment is wash the area with water, soap and baking soda ( sodium bicarbonate)

because soap and sodium bicarbonate are basic salts and produces strong base (NaOH and KOH) in aqueous solution which neutralise the catechol molecules.

C) At pH 7 , terminal N has +1 charge, phenolic OH has 0 charge, terminal carboxylic acid group has -1 charge, other side chain carboxylic acid group has -1 charge.

Net charge is 0+1-1-1= -1

D) in size exclusion chromatography smaller proteins (low molecular weight) are eluted last as they enter into the pores of the column and hence the elution time becomes longer. Whereas larger proteins (high molecular weight) can not e get into pores and they are eluted very fast.

Protein 3 will be eluted first (as it has the highest molecular weight) then protein 1 and then protein 2.

Order of emergence is protein 3, protein 1, protein 2

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