Proteins

Q: What is a peptide bond?

A peptide bond is the amide link, -CO-NH-, formed when the carboxyl group of one α-amino acid condenses with the amino group of another and water is lost. It is planar because resonance gives the C-N bond partial double-bond character, so there is no free rotation about it.

Q: What is a zwitterion?

A zwitterion (dipolar ion) is the inner salt an amino acid forms when its carboxyl group donates a proton to its own amino group: H₃N⁺-CHR-COO⁻. It has both charges but is neutral overall, which explains the high melting points, water solubility and large dipole moments of amino acids.

Q: What is the isoelectric point of an amino acid?

The isoelectric point (pI) is the pH at which an amino acid exists mainly as the zwitterion, has no net charge and does not move in an electric field. For a neutral amino acid it is the average of its two pKa values; for glycine it is 5.97. Amino acids are least soluble at their pI.

Q: What are the four levels of protein structure?

Primary structure is the amino acid sequence, held by peptide bonds. Secondary structure is the α-helix or β-pleated sheet, held by hydrogen bonds. Tertiary structure is the overall folding, held by H-bonds, S-S bridges, ionic and hydrophobic forces. Quaternary structure is the arrangement of subunits, as in haemoglobin.

Q: What is denaturation of proteins?

Denaturation is the loss of a protein's natural shape and biological activity when heat, a change in pH or chemicals break the hydrogen bonds and other weak forces holding it. The secondary and tertiary structures are destroyed but the sequence is not. Boiling an egg and curdling milk are everyday examples.

Q: How do enzymes work?

Enzymes are protein catalysts. The substrate binds at the enzyme's active site to form an enzyme-substrate complex, like a key in a lock. This complex offers a path with a lower activation energy, so the reaction runs fast at body temperature, and the enzyme is released unchanged. Each enzyme is highly specific.

Q: Which protein topics are most important for NEET?

NEET follows NCERT closely: the zwitterion and amphoteric nature of amino acids, essential amino acids, the peptide bond, fibrous and globular proteins, the four levels of structure with their bonds, denaturation with examples, and how enzymes lower activation energy. One-line facts from these areas appear most often.

Q: How are amino acids and proteins tested in JEE?

JEE Main asks about zwitterions, the pI formula, the charge and migration of an amino acid at a given pH, and counting peptide bonds or possible sequences. JEE Advanced adds pI of acidic and basic amino acids, electrophoresis, Strecker and Gabriel syntheses, and sequencing puzzles using partial hydrolysis and Sanger's reagent.

ChemistryBiomoleculesFor JEE aspirants

Proteins are long chains of $α$ -amino acids joined by peptide bonds, $- CO - NH -$ . These notes on proteins start with the 20 amino acids, their zwitterions and isoelectric points, then cover peptides, the four levels of protein structure, denaturation, sequencing by Sanger's method, the tests for proteins and how enzymes work. Every structure and trend is drawn. The topic follows the NCERT Biomolecules chapter and is asked every year in NEET and JEE Main.

On this page1Amino acids2Zwitterion, pI3Preparation4Reactions5Peptides6Structure7Denaturation8Sequencing9Tests, enzymes10Examples

Key Formulas - Quick Reference

$α$ -Amino acid: $R - CH (N H_{2}) - COOH$ ; in water and in crystals it exists as the zwitterion $H_{3} N^{+} - CHR - CO O^{-}$ .
★ Must learn Isoelectric point (neutral amino acids): $pI = \frac{p K _{a 1} + p K _{a 2}}{2}$ ; glycine: $\frac{2.34 + 9.60}{2} = 5.97$ .
★ Must learn pH $<$ pI: net positive, moves to the cathode. pH $>$ pI: net negative, moves to the anode. pH = pI: no migration and least solubility.
★ Must learn Peptide bond $- CO - NH -$ : $n$ amino acids give $n - 1$ peptide bonds and lose $n - 1$ $H_{2} O$ ; $n$ different amino acids give $n!$ sequences.
Peptide unit is planar: C-N = 132 pm (normal C-N 147 pm) because of partial double-bond character; no free rotation.
$α$ -helix: right-handed, 3.6 residues per turn, H-bonds between C=O and N-H of the same chain. $β$ -sheet: chains side by side, H-bonds between chains.
★ Must learn 1° peptide bonds; 2° H-bonds; 3° H-bonds, S-S bridges, ionic, van der Waals and hydrophobic forces; 4° association of subunits.
★ Must learn Denaturation (heat, pH) destroys 2° and 3° structure but not the 1° sequence: boiled egg white, curdled milk.
★ Must learn Tests: biuret (violet, peptide bonds), ninhydrin (violet, amino group), Millon's (brick red, tyrosine).
Van Slyke: $RCH (N H_{2}) COOH$ + $HN O_{2}$ $\to$ $RCH (OH) COOH$ + $N_{2}$ + $H_{2} O$ ; Sanger's reagent (DNFB) tags the N-terminal residue.

1. Amino Acids

1.1 Structure

Amino acids are organic compounds that contain both an amino group, $- N H_{2}$ , and a carboxyl group, $- COOH$ . They are an important class of difunctional compounds because they are the building blocks (monomers) of proteins. The amino acids in proteins are $α$ -amino acids: both groups are attached to the same carbon, the $α$ -carbon, together with an H atom and a side chain R that differs from one amino acid to the next (Figure 1).

Figure 1: Every

α

-amino acid has

- N H_{2}

and

- COOH

on the same (

α

) carbon; only the side chain R changes. In the Fischer projection the

- N H_{2}

of natural amino acids sits on the left, as the OH of L-glyceraldehyde does.

Amino acids are also named by the position of the amino group relative to the carboxyl group: $α$ (on C-2), $β$ (on C-3), $γ$ (on C-4) and so on.

2-Aminoethanoic acid, $H_{2} NC H_{2} COOH$ (glycine): an $α$ -amino acid.
3-Aminopropanoic acid, $H_{2} NC H_{2} C H_{2} COOH$ : a $β$ -amino acid.
4-Aminobutanoic acid, $H_{2} NC H_{2} C H_{2} C H_{2} COOH$ : a $γ$ -amino acid.

1.2 The 20 amino acids of proteins

Several hundred amino acids occur in nature, but only 20 are found in proteins. They are known by trivial names and by three-letter (and one-letter) codes, which are used to write the sequences of peptides and proteins. Amino acids are also grouped as neutral, acidic or basic, by comparing the number of amino and carboxyl groups: one of each gives a neutral amino acid, an extra COOH an acidic one, and an extra basic group a basic one. Essential amino acids are marked *.

Amino acid	3-letter	1-letter	Side chain R	Type
Glycine	Gly	G	$- H$	non-polar (achiral)
Alanine	Ala	A	$- C H_{3}$	non-polar
Valine*	Val	V	$- CH (C H_{3})_{2}$	non-polar
Leucine*	Leu	L	$- C H_{2} CH (C H_{3})_{2}$	non-polar
Isoleucine*	Ile	I	$- CH (C H_{3}) C H_{2} C H_{3}$	non-polar
Methionine*	Met	M	$- C H_{2} C H_{2} SC H_{3}$	non-polar
Proline	Pro	P	$- C H_{2} C H_{2} C H_{2} -$ ring back to the N	non-polar
Phenylalanine*	Phe	F	$- C H_{2} C_{6} H_{5}$	non-polar, aromatic
Tryptophan*	Trp	W	$- C H_{2}$ -indolyl	non-polar, aromatic
Serine	Ser	S	$- C H_{2} OH$	polar
Threonine*	Thr	T	$- CH (OH) C H_{3}$	polar
Cysteine	Cys	C	$- C H_{2} SH$	polar
Tyrosine	Tyr	Y	$- C H_{2} C_{6} H_{4} OH$ (para)	polar, aromatic
Asparagine	Asn	N	$- C H_{2} CON H_{2}$	polar
Glutamine	Gln	Q	$- C H_{2} C H_{2} CON H_{2}$	polar
Aspartic acid	Asp	D	$- C H_{2} COOH$	acidic
Glutamic acid	Glu	E	$- C H_{2} C H_{2} COOH$	acidic
Lysine*	Lys	K	$- (C H_{2})_{4} N H_{2}$	basic
Arginine*	Arg	R	$- (C H_{2})_{3} NHC (= NH) N H_{2}$	basic
Histidine*	His	H	$- C H_{2}$ -imidazolyl	basic

Figure 2: Sort by the side chain R. An extra

- COOH

makes an amino acid acidic (pI about 3), an extra basic N makes it basic (pI 7.6 to 10.8); the starred ten are essential and must come from food.

1.3 Essential and non-essential amino acids

Essential amino acids cannot be made in the body and must come from the diet: valine, leucine, isoleucine, phenylalanine, threonine, tryptophan, methionine, lysine, arginine and histidine. Histidine and arginine are sometimes called semi-essential: the body makes them, but not fast enough for normal growth.
Non-essential amino acids are made in the body: glycine, alanine, serine, cysteine (and cystine, two cysteines joined by S-S), tyrosine, proline, hydroxyproline, aspartic acid and glutamic acid, among others.

1.4 Configuration

Except glycine (R = H), every $α$ -amino acid has a chiral $α$ -carbon and exists as two optically active forms. The amino acids in proteins all belong to the L-series: in the Fischer projection with COOH at the top, the $- N H_{2}$ group is on the left, just as the OH is on the left in L-glyceraldehyde (Figure 1). As with sugars, L is a configuration and says nothing about the sign of rotation.

1.5 Physical properties

Amino acids are colourless crystalline solids; many taste sweet.
They melt with decomposition at high temperatures, generally above 200 °C, instead of melting cleanly.
They dissolve in water but not in non-polar organic solvents.
They have large dipole moments.

All four facts point to an ionic structure, the zwitterion (Section 2).

Key idea

Every protein amino acid is an

α

-amino acid of the L-series (glycine is achiral); only the side chain R changes, and R decides whether it is neutral, acidic or basic.

2. Zwitterions and the Isoelectric Point

2.1 The zwitterion

An amino acid carries an acidic carboxyl group and a basic amino group in the same molecule, so it is amphoteric. In water, the carboxyl group loses a proton and the amino group gains one in an internal acid-base reaction. The product is an inner salt called a dipolar ion or zwitterion: it carries both a positive and a negative charge but is electrically neutral overall.

RCH (N H_{2}) COOH ⇌ RCH (N H_{3}^{+}) CO O^{-}

The highly polar zwitterions pack into strong crystal lattices, much like ionic compounds. That is why amino acids decompose instead of melting, dissolve in water but not in non-polar solvents, and have large dipole moments. In the zwitterion the acidic group is the substituted ammonium ion, $- N H_{3}^{+}$ , and the basic group is the carboxylate ion, $- CO O^{-}$ .

2.2 Behaviour in acid and in base

In strongly acidic solution (low pH) the carboxylate group takes back a proton, giving a cation with a net positive charge and two acidic sites ( $N H_{3}^{+}$ and COOH). In strongly basic solution (high pH) the $N H_{3}^{+}$ group loses its proton, giving an anion with a net negative charge and two basic sites ( $N H_{2}$ and $CO O^{-}$ ). Between the two, at some intermediate pH, the amino acid exists as the zwitterion with no net charge (Figure 3).

RCH (N H_{3}^{+}) COOH ⇌ p K_{a 1} RCH (N H_{3}^{+}) CO O^{-} + H^{+}

RCH (N H_{3}^{+}) CO O^{-} ⇌ p K_{a 2} RCH (N H_{2}) CO O^{-} + H^{+}

Figure 3: Acid turns the zwitterion into a cation (moves to the cathode); base turns it into an anion (moves to the anode). At the isoelectric point the net charge is zero.

The isoelectric point (pI or

p H_{i}

) is the pH at which an amino acid exists mainly as the zwitterion, carries no net charge, and does not migrate towards either electrode in an electric field.

2.3 Calculating the isoelectric point

A neutral amino acid has two acidic sites, so it has two $p K_{a}$ values: the more acidic site gives $p K_{a 1}$ and the less acidic site $p K_{a 2}$ . For glycine, with the cation written CA, the dipolar ion DI and the conjugate base CB:

H_{3} N^{+} C H_{2} COOH + H_{2} O ⇌ p K_{a 1} = 2.34 H_{3} N^{+} C H_{2} CO O^{-} + H_{3} O^{+}

H_{3} N^{+} C H_{2} CO O^{-} + H_{2} O ⇌ p K_{a 2} = 9.60 H_{2} NC H_{2} CO O^{-} + H_{3} O^{+}

K_{1} = \frac{[ DI ] [ H ^{+} ]}{[ CA ]} K_{2} = \frac{[ CB ] [ H ^{+} ]}{[ DI ]}

At the isoelectric point the amounts of cation and anion are equal, [CA] = [CB], so:

\frac{[ DI ] [ H ^{+} ]}{K _{1}} = \frac{K _{2} [ DI ]}{[ H ^{+} ]} \Rightarrow [H^{+}]^{2} = K_{1} K_{2} \Rightarrow pI = \frac{p K _{a 1} + p K _{a 2}}{2}

For glycine, $pI = \frac{2.34 + 9.60}{2} = 5.97$ . Compared with ethanoic acid ( $p K_{a} = 4.76$ ), the COOH of an amino acid is far more acidic ($\mathrm{p}K_{a1} \approx 2.3 $) b ec a u se t h e n e a r b y$ \mathrm{{-}NH_{3}^+}$ group withdraws electrons. The titration curve of glycine (Figure 4) shows both $p K_{a}$ values and the pI directly.

Amino acid	$p K_{a 1}$ (COOH)	$p K_{a 2}$ ( $N H_{3}^{+}$ )	pI
Glycine	2.34	9.60	5.97
Alanine	2.34	9.69	6.01
Valine	2.32	9.62	5.97
Leucine	2.36	9.60	5.98
Isoleucine	2.36	9.68	6.02
Methionine	2.28	9.21	5.74
Proline	1.99	10.60	6.30
Phenylalanine	1.83	9.13	5.48
Tryptophan	2.38	9.39	5.89
Asparagine	2.02	8.80	5.41
Glutamine	2.17	9.13	5.65
Serine	2.21	9.15	5.68
Threonine	2.09	9.10	5.60

Figure 4: Titrating glycine gives two buffer plateaus, at

p K_{a 1}

and

p K_{a 2}

. Halfway between them, at one equivalent, only the zwitterion is left:

pI = \frac{2.34 + 9.60}{2} = 5.97

The same two $p K_{a}$ values give the share of each form at any pH. The curves below are calculated, not sketched: the zwitterion dominates over a wide band around the pI.

Figure 5: Computed from the two

p K_{a}

values. At

pH = p K_{a 1}

cation and zwitterion are 50 : 50, at

pH = p K_{a 2}

zwitterion and anion are 50 : 50, and at the pI (5.97) the zwitterion is 99.95% of all glycine.

2.4 Acidic and basic amino acids

Amino acids with an acidic or basic side chain have a third $p K_{a}$ , which reflects the side-chain group. Their pI is the average of the two $p K_{a}$ values on either side of the neutral zwitterion: the two lowest for acidic amino acids (pI near 3) and the two highest for basic ones (pI near 10). Neutral amino acids have pI between about 5.5 and 6.3.

Amino acid	$p K_{a}$ values	Averaged pair	pI
Aspartic acid (acidic)	1.88, 3.65 (side COOH), 9.60	1.88 and 3.65	2.77
Glutamic acid (acidic)	2.19, 4.25 (side COOH), 9.67	2.19 and 4.25	3.22
Histidine (basic)	1.82, 6.00 (imidazole), 9.17	6.00 and 9.17	7.59
Lysine (basic)	2.18, 8.95, 10.53 (side $N H_{3}^{+}$ )	8.95 and 10.53	9.74
Arginine (basic)	2.17, 9.04, 12.48 (guanidino)	9.04 and 12.48	10.76

2.5 Electrophoresis and solubility

When an amino acid solution is placed in an electric field, what happens depends on the pH:

In acidic solution (pH below pI) cations are in excess, so the amino acid moves towards the cathode.
In alkaline solution (pH above pI) anions are in excess, so it moves towards the anode.
At the pI, cations and anions are exactly balanced. Any molecule is a cation and an anion for equal lengths of time, so a small movement one way is cancelled by an equal movement back, and there is no net migration.

Different amino acids have different pI values, because of their different side chains, so a mixture can be separated by electrophoresis at a chosen pH (Figure 6). An amino acid is least soluble at its isoelectric point, where the concentration of the dipolar ion is highest; making the solution more acidic or more basic increases the concentration of the more soluble cation or anion. This is used to crystallise amino acids from solution at their pI.

Figure 6: Compare the buffer pH with each pI. Above its pI an amino acid is negative and moves to the anode; below its pI it is positive and moves to the cathode.

Exam Trick

"Low pH loads H $^{+}$ ": in acid the amino acid picks up $H^{+}$ and turns positive, so it runs to the cathode; in base it loses $H^{+}$ and runs to the anode. Opposite charges attract.

Every migration question follows the same three steps, shown as a flowchart:

Figure 7: Flowchart: compare pH with pI and the charge follows. Below the pI the amino acid carries an extra

H^{+}

(cation); above it, it has lost one (anion).

Key idea

pH below pI: cation, to the cathode. pH above pI: anion, to the anode. At the pI: no net charge, no migration, least solubility.

Quick Recall: tap to check

Which group is the acid in the glycine zwitterion?

- N H_{3}^{+}

; the base is

- CO O^{-}

pI of alanine (

p K_{a}

2.34 and 9.69)?

\frac{2.34 + 9.69}{2} = 6.01

(6.02 with unrounded values).

Lysine at pH 7: which electrode?

Cathode. Its pI (9.74) is above 7, so it is positive.

3. Preparation of $α$ -Amino Acids

3.1 From proteins

Refluxing a protein with dilute (or 6 M) hydrochloric acid hydrolyses it to a mixture of $α$ -amino acids. The mixture can be separated by:

fractional crystallisation;
fractional distillation of the esters, followed by hydrolysis (Fischer's method);
selective precipitation as salts with phosphotungstic acid and picric acid;
distribution between n-butanol and water (Dakin's method);
column, paper and gas chromatography, and ion-exchange chromatography in an amino acid analyser;
electrophoresis.

3.2 Amination of $α$ -halo acids

An $α$ -chloro or $α$ -bromo acid is heated with excess ammonia (about 50 °C); the ammonium salt is then acidified.

chloroacetic acid ClC H_{2} COOH + (excess) 3N H_{3} 50 °C H_{2} NC H_{2} COON H_{4} + N H_{4} Cl

H_{2} NC H_{2} COON H_{4} + H^{+} H_{2} NC H_{2} COOH + N H_{4}^{+}

2-bromopropanoic acid C H_{3} CH (Br) COOH + 3N H_{3} 50 °C C H_{3} CH (N H_{2}) COON H_{4} + N H_{4} Br

Acidifying the second product gives alanine.

3.3 Gabriel phthalimide synthesis

Potassium phthalimide reacts with ethyl chloroacetate. Acid hydrolysis of the product then frees glycine, phthalic acid and ethanol. Because the nitrogen is protected, only one alkyl group can attach.

potassium phthalimide C_{6} H_{4} (CO)_{2} NK + ethyl chloroacetate ClC H_{2} COO C_{2} H_{5} C_{6} H_{4} (CO)_{2} NC H_{2} COO C_{2} H_{5} + KCl

C_{6} H_{4} (CO)_{2} NC H_{2} COO C_{2} H_{5} + 3 H_{2} O H^{+}, Δ phthalic acid C_{6} H_{4} (COOH)_{2} + glycine H_{2} NC H_{2} COOH + C_{2} H_{5} OH

3.4 Strecker synthesis

An aldehyde is treated with ammonia and HCN; in practice a mixture of ammonium chloride and potassium cyanide is used in water, which supplies both ( $N H_{4} Cl + KCN$ $\to$ $N H_{4} CN + KCl$ ; $N H_{4} CN$ $\to$ $N H_{3} + HCN$ ). HCN adds to the C=O to form a cyanohydrin, ammonia replaces the OH to give an $α$ -aminonitrile, and acid hydrolysis of the CN group gives the amino acid.

formaldehyde HCHO + HCN cyanohydrin HOC H_{2} CN

HOC H_{2} CN + N H_{3} aminonitrile H_{2} NC H_{2} CN + H_{2} O

H_{2} NC H_{2} CN + 2 H_{2} O + H^{+} glycine H_{2} NC H_{2} COOH + N H_{4}^{+}

Acetaldehyde, $C H_{3} CHO$ , gives alanine in the same way.

JEE Advanced

Laboratory syntheses give racemic amino acids. Strecker synthesis adds HCN to a flat C=O group, and the Gabriel and $α$ -halo acid routes start from achiral or racemic material, so the new $α$ -carbon is made equally often in both configurations: the product is a (±) mixture, 50% D and 50% L. Proteins use only L, so a synthetic amino acid must be resolved (for example with an enzyme that acts only on the L-form) before it can be used. Glycine, with no chiral centre, needs no resolution.

4. Chemical Reactions of Amino Acids

Amino acids show reactions of the carboxyl group, of the amino group, and of both groups together.

4.1 Reactions of the carboxyl group

With bases. The $- N H_{3}^{+}$ group of the zwitterion gives its proton to the base, forming a salt:

glycine H_{3} N^{+} C H_{2} CO O^{-} + NaOH sodium aminoacetate H_{2} NC H_{2} COONa + H_{2} O

Esterification. HCl first converts the dipolar ion into the ammonium salt of the acid, which is then esterified; a base such as moist $A g_{2} O$ (AgOH) frees the ester:

H_{3} N^{+} C H_{2} CO O^{-} + HCl [H_{3} NC H_{2} COOH]^{+} C l^{-}

[H_{3} NC H_{2} COOH]^{+} C l^{-} + C_{2} H_{5} OH dry HCl [H_{3} NC H_{2} COO C_{2} H_{5}]^{+} C l^{-} + H_{2} O

[H_{3} NC H_{2} COO C_{2} H_{5}]^{+} C l^{-} + AgOH ethyl aminoacetate H_{2} NC H_{2} COO C_{2} H_{5} + AgCl + H_{2} O

Decarboxylation. Heating with barium hydroxide removes $C O_{2}$ and gives an amine: glycine gives methylamine and alanine gives ethylamine.

glycine H_{2} NC H_{2} COOH + Ba (OH)_{2} Δ methylamine C H_{3} N H_{2} + BaC O_{3} + H_{2} O

alanine C H_{3} CH (N H_{2}) COOH + Ba (OH)_{2} Δ ethylamine C H_{3} C H_{2} N H_{2} + BaC O_{3} + H_{2} O

Reduction. $LiAl H_{4}$ reduces the COOH group to $C H_{2} OH$ , giving amino alcohols: glycine gives 2-aminoethanol and alanine gives 2-aminopropan-1-ol.

glycine H_{2} NC H_{2} COOH + 4 [H] LiAl H_{4} 2-aminoethanol H_{2} NC H_{2} C H_{2} OH + H_{2} O

4.2 Reactions of the amino group

With strong acids. The carboxylate group takes a proton, giving a salt such as glycine hydrochloride:

glycine H_{3} N^{+} C H_{2} CO O^{-} + HCl glycine hydrochloride [H_{3} NC H_{2} COOH]^{+} C l^{-}

Acylation. Acid chlorides and anhydrides acylate the amino group:

glycine H_{2} NC H_{2} COOH + acetyl chloride C H_{3} COCl N-acetylglycine C H_{3} CONHC H_{2} COOH + HCl

alanine C H_{3} CH (N H_{2}) COOH + (C H_{3} CO)_{2} O N-acetylalanine C H_{3} CH (NHCOC H_{3}) COOH + C H_{3} COOH

benzoyl chloride C_{6} H_{5} COCl + H_{2} NC H_{2} COOH base hippuric acid C_{6} H_{5} CONHC H_{2} COOH + HCl

With nitrous acid. The primary amino group is replaced by OH and nitrogen gas is given off: glycine gives glycolic acid and alanine gives lactic acid.

glycine H_{2} NC H_{2} COOH + HN O_{2} glycolic acid HOC H_{2} COOH + N_{2} ↑ + H_{2} O

alanine C H_{3} CH (N H_{2}) COOH + HN O_{2} lactic acid C H_{3} CH (OH) COOH + N_{2} ↑ + H_{2} O

Van Slyke method. This reaction is used to estimate amino acids. The $N_{2}$ is given off quantitatively (one N atom from the amino acid, one from $HN O_{2}$ ) and its volume is measured: one mole of amino acid gives one mole of $N_{2}$ .

With nitrosyl chloride. The amino group is replaced by Cl:

glycine H_{2} NC H_{2} COOH + NOCl chloroacetic acid ClC H_{2} COOH + N_{2} ↑ + H_{2} O

With 1-fluoro-2,4-dinitrobenzene (DNFB, Sanger's reagent). The free $- N H_{2}$ group displaces fluoride from the ring, giving a yellow 2,4-dinitrophenyl (DNP) amino acid. This is the basis of Sanger's end-group method (Section 9).

RCH (N H_{2}) COOH + DNFB F C_{6} H_{3} (N O_{2})_{2} DNP-amino acid RCH (COOH) NH C_{6} H_{3} (N O_{2})_{2} + HF

4.3 Reaction of both groups: effect of heat

On heating (about 200 °C), $α$ -amino acids lose water between the $- COOH$ of one molecule and the $- N H_{2}$ of another, twice over, to form a cyclic diamide called a diketopiperazine:

glycine 2 H_{2} NC H_{2} COOH Δ 2,5-diketopiperazine C_{4} H_{6} N_{2} O_{2} + 2 H_{2} O

5. Peptides

5.1 The peptide bond

Peptides are polymers in which $α$ -amino acids are joined by amide links formed from the carboxyl group of one amino acid and the amino group of the next, with loss of water. This amide link, $- CO - NH -$ , is the peptide bond (peptide linkage).

glycine H_{2} NC H_{2} COOH + alanine H_{2} NCH (C H_{3}) COOH Gly-Ala H_{2} NC H_{2} CONHCH (C H_{3}) COOH + H_{2} O

Figure 8: The peptide bond,

- CO - NH -

, forms when

- COOH

of one amino acid loses OH and

- N H_{2}

of the next loses H. The N lone pair delocalises onto O, so C-N is 132 pm (a normal C-N is 147 pm) and the whole unit stays flat.

Two amino acids give a dipeptide, three a tripeptide, four a tetrapeptide, and so on, whether the units are the same or different.
More than about ten amino acids give a polypeptide. A polypeptide with more than about a hundred residues and a molecular mass above 10 000 u is called a protein. The line is not sharp: insulin, with 51 residues, is still called a protein.
The amino acid with the free $- N H_{2}$ group is the N-terminal residue and is written on the left; the one with the free $- COOH$ is the C-terminal residue, written on the right. So Gly-Ala and Ala-Gly are different dipeptides.
The tripeptide glycylalanylphenylalanine is written Gly-Ala-Phe: glycine at the N-terminus and phenylalanine at the C-terminus.

Bradykinin is an important natural nonapeptide found in blood plasma. It helps to regulate blood pressure:

Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg

5.2 Geometry of the peptide bond

X-ray studies (Linus Pauling) show that the peptide unit is flat: the carbonyl C and O, the N and its H, and the two $α$ -carbons all lie in one plane, and the H of N-H is usually trans to the O of C=O. The C-N bond of the peptide link is 132 pm long, shorter than a normal C-N single bond (147 pm), because resonance gives it partial double-bond character (Figure 5). There is therefore no free rotation about this bond, which is why the peptide unit stays planar and why cis/trans forms exist (the trans form is favoured).

Key idea

n

amino acids give

n - 1

peptide bonds and release

n - 1

water molecules. The peptide unit is flat because C-N has partial double-bond character.

Quick Recall: tap to check

How many peptide bonds are in a pentapeptide?

Four (

n - 1

Why is there no free rotation about the C-N bond of a peptide?

Resonance puts the N lone pair into the C-N bond, giving it partial double-bond character (132 pm).

Which end of Gly-Ala carries the free

- N H_{2}

Glycine, the N-terminus, written on the left.

6. Proteins: Roles and Classification

Proteins are large polypeptides, from about 50 to more than 8000 amino acid units per molecule. They have many biological roles:

As enzymes and hormones (such as insulin) they catalyse and regulate the reactions in the body.
As skin and hair they form the body's outer covering, and as muscle they give movement.
As antibodies they protect against disease.
Haemoglobin carries oxygen from the air we breathe (and some $C O_{2}$ ).
Nucleoproteins in the genes help to carry and pass on genetic information during cell division, and many proteins give structural support together with other substances.

Proteins are classified by composition, by molecular shape and by function:

Basis	Type	Features	Examples
Composition	Simple proteins	give only $α$ -amino acids on hydrolysis	egg albumin, serum globulins, glutelin (wheat), oryzenin (rice)
Composition	Conjugated proteins	amino acids plus a non-protein part, the prosthetic group	casein (phosphate), haemoglobin (haem), nucleoproteins (nucleic acid), lipoproteins (lipid), glycoproteins (carbohydrate)
Shape	Fibrous proteins	long parallel chains held by H-bonds and S-S bridges; thread-like, stretch and contract; usually insoluble in water	keratin (hair, wool, nails), fibroin (silk), myosin (muscle), collagen
Shape	Globular proteins	chains coiled into compact spheres, held by H-bonds, S-S bridges and ionic (salt) bridges; usually soluble in water	egg albumin, insulin, haemoglobin, most enzymes
Function	Structural proteins	fibrous proteins that hold the body together	collagen (skin, cartilage, bone), keratin
Function	Blood proteins	keep osmotic pressure, carry oxygen, clot blood	albumin, haemoglobin, fibrinogen (in that order)

Fibrous proteinslong parallel chains, thread-like
H-bonds and S-S between chains
insoluble in water
keratin, myosin, silk fibroin

Globular proteinschains folded into compact spheres
H-bonds, S-S, salt bridges inside
soluble in water
albumin, insulin, most enzymes

7. Structure of Proteins

Proteins have definite three-dimensional shapes, and several factors decide the exact shape. Their structure is described at four levels (Figure 9).

7.1 Primary structure

The primary structure is the sequence in which the amino acids are joined in each polypeptide chain, together with the positions of any disulphide links. The chain is linear, and all its units are joined by peptide bonds. Frederick Sanger worked out the first complete sequence, that of insulin, in the early 1950s. Any change in the primary structure, even of one amino acid, gives a different protein.

7.2 Secondary structure

The secondary structure is the fixed local shape of the polypeptide backbone: how the chain is folded or bent, and which bonds hold that shape. It arises because the C=O and N-H groups of peptide bonds form hydrogen bonds. There are two main types:

$α$ -Helix. The chain coils into a right-handed spiral, like a screw. The spiral is held by H-bonds between the N-H of one amino acid and the C=O of the amino acid on the next turn (four residues along), which lie vertically above one another. X-ray studies show about 3.6 amino acid units per turn. Such proteins are elastic: on stretching, the weak H-bonds break and the chain extends like a spring, and the bonds re-form when the tension is released. Examples: myosin, keratin ( $α$ -keratin of hair and wool) and tropomyosin.
$β$ -Pleated sheet. The chains are stretched out almost fully and laid side by side, held together by many H-bonds between the C=O of one chain and the N-H of the next. The sheet is pleated, and sheets can stack on each other. Example: the fibroin of silk.

α

-Helixone chain coiled, right-handed
H-bond: C=O of residue

n

to N-H of

n + 4

3.6 residues per turn
keratin, myosin

β

-Pleated sheetchains stretched side by side
H-bonds between neighbouring chains
pleated, stackable sheets
silk fibroin

7.3 Tertiary structure

The tertiary structure is the overall folding of the whole polypeptide chain, including its helices and sheets, into a particular shape: long fibres or compact globules. It is held by H-bonds, disulphide (S-S) bridges, ionic (salt) bridges, van der Waals forces and hydrophobic interactions between side chains. Examples are the coat protein of tobacco mosaic virus (TMV) and myoglobin.

7.4 Quaternary structure

Some proteins are made of two or more polypeptide chains, called subunits, held together by forces other than covalent (peptide and disulphide) bonds. The spatial arrangement of these subunits is the quaternary structure, the final three-dimensional shape of the protein. Haemoglobin, with two $α$ and two $β$ chains, is the classic example.

Figure 9: Each level builds on the one before: the sequence (1°) folds into helices and sheets (2°), these pack into a 3D shape (3°), and some proteins join several chains (4°).

Level	What it describes	Held by	Examples
Primary (1°)	sequence of amino acids in each chain	peptide (covalent) bonds; S-S bridges fix where chains or loops join	insulin (sequenced by Sanger)
Secondary (2°)	local shape of the chain: $α$ -helix or $β$ -pleated sheet	H-bonds between C=O and N-H groups	keratin, myosin ( $α$ -helix); silk fibroin ( $β$ -sheet)
Tertiary (3°)	overall folding of the whole chain into fibres or globules	H-bonds, S-S bridges, ionic (salt) bridges, van der Waals and hydrophobic forces	myoglobin, TMV coat protein, globular enzymes
Quaternary (4°)	arrangement of two or more chains (subunits)	the same non-covalent forces between subunits	haemoglobin (2 $α$ + 2 $β$ )

Exam Trick

"Sequence, Shape, Fold, Team": 1° is the sequence, 2° the local shape (H-bonds only), 3° the fold of the whole chain (S-S, ionic, H-bonds, hydrophobic), 4° the team of separate chains, as in haemoglobin.

8. Denaturation of Proteins

A protein found in a living system, with its own structure and biological activity, is a native protein. When it is exposed to a change in temperature or pH (or to heavy-metal salts, alcohol, detergents or strong shaking), the hydrogen bonds and other weak forces that hold its shape are disturbed. The globules unfold and the helices uncoil, and the protein loses its biological activity. This is denaturation (Figure 10).

The secondary and tertiary structures are destroyed, but the primary structure (the peptide bonds and sequence) stays intact.
Boiling an egg coagulates the soluble globular protein of egg white into an insoluble mass.
Milk curdles when bacteria in it make lactic acid, which lowers the pH and denatures the milk proteins.

Figure 10: Denaturation breaks the weak forces that hold the folded shape, so the protein unfolds and stops working, but its amino acid sequence is untouched.

Denaturationheat, pH, heavy-metal salts
2° and 3° structure lost
peptide bonds intact
boiled egg white, curdled milk

Hydrolysis6 M HCl, 110 °C, 24 h (or enzymes)
peptide bonds broken
gives free amino acids
used to find the composition

Key idea

Denaturation destroys the shape (2°, 3°) and the activity, never the sequence (1°).

Quick Recall: tap to check

Which bonds hold the secondary structure?

Hydrogen bonds between C=O and N-H of peptide groups.

Is haemoglobin an example of tertiary or quaternary structure?

Quaternary: two

α

and two

β

subunits.

Does boiling an egg break peptide bonds?

No. It denatures the albumin; the sequence is unchanged.

9. Finding the Primary Structure

To work out the structure of a peptide or protein, chemists find which amino acids are present, in what amounts and in what order, and where any disulphide links are.

Break the disulphide bridges. Peroxymethanoic (performic) acid, $HCOOOH$ , oxidises each S-S bridge to two sulphonic acid groups, separating the linked chains.

disulphide bridge R - S - S - R^{'} + 5 [O] + H_{2} O HCOOOH RS O_{3} H + R^{'} S O_{3} H

Find the composition. Heating with 6 M HCl at about 110 °C for 24 hours hydrolyses every peptide bond. The amino acid mixture is separated by ion-exchange chromatography in an amino acid analyser, using the acid-base properties of the amino acids. This gives the amino acids present and their relative amounts.
Find the sequence. Partial hydrolysis (dilute acid or enzymes) breaks the chain into smaller peptides whose overlaps show the order, and end-group analysis identifies the terminal residues.

9.1 Partial hydrolysis

Partial hydrolysis of a tetrapeptide made of Ala, Gly, Phe and Val gave the tripeptide Gly-Phe-Val and the dipeptide Ala-Gly. The dipeptide shows that Ala is joined to Gly, and Gly starts the tripeptide, so the sequence is Ala-Gly-Phe-Val. Cutting the tetrapeptide at bond a (Ala-Gly) gives the tripeptide, and cutting at bond b (Gly-Phe) gives the dipeptide (Figure 11).

9.2 End-group analysis

By convention a peptide is written with the amino end on the left and the carboxyl end on the right: the N-terminus and the C-terminus. Every amino group except the N-terminal one is used in a peptide bond, so only the N-terminal amino group is free to act as a nucleophile.

Sanger's method. The free $- N H_{2}$ group reacts with 1-fluoro-2,4-dinitrobenzene (DNFB), which tags the N-terminal nitrogen with a 2,4-dinitrophenyl (DNP) group. Complete acid hydrolysis then cleaves every peptide bond, giving the yellow DNP-labelled N-terminal amino acid and a mixture of unlabelled amino acids; the labelled one is identified by chromatography. For Val-Phe-Gly-Ala, the products are DNP-Val, Phe, Gly and Ala (Figure 11). The drawback is that the whole chain is destroyed after tagging only one residue.

Edman degradation. A more useful method removes only the tagged terminal amino acid and leaves the rest of the chain intact, so it can be tagged again. The sequence is read one amino acid at a time.

Figure 11: Overlapping fragments give the order of residues; Sanger's reagent (DNFB) labels the free

- N H_{2}

end, so the labelled amino acid is the N-terminal one.

10. Tests for Proteins

Test	Reagent and procedure	Positive result	Shows
Biuret test	dilute $CuS O_{4}$ added to an alkaline (NaOH) protein solution	violet colour	peptide bonds (two or more)
Millon's test	Millon's reagent: mercury dissolved in nitric acid (mercuric and mercurous nitrates); warm	white precipitate that turns brick red	tyrosine (phenolic OH)
Ninhydrin test	boil with dilute ninhydrin solution	violet (Ruhemann's purple)	free $α$ -amino groups: all proteins and amino acids

The biuret test is named after biuret, $H_{2} NCONHCON H_{2}$ , which gives the same violet colour with alkaline copper sulphate. Amino acids and dipeptides do not give it, because they have fewer than two peptide bonds.

Exam Trick

"Bonds, amino, phenol": biuret needs two or more peptide bonds, ninhydrin needs a free amino group (so all amino acids give it), and Millon's needs the phenol of tyrosine.

11. Enzymes

Life needs a large number of reactions to take place in an orderly way. Enzymes are the biocatalysts that make this possible: they are made by living cells, and almost all of them are globular proteins. Each enzyme is very specific; it catalyses one reaction (or one type of reaction) of one substrate.

Enzymes are usually named after the compound or class of compounds they act on, with the ending -ase: maltase hydrolyses maltose. Enzymes that oxidise one substrate while reducing another are called oxidoreductases.
They work in mild conditions: body temperature and near-neutral pH.

Enzyme	Reaction it catalyses
Invertase (sucrase)	sucrose $\to$ glucose + fructose
Maltase	maltose $\to$ 2 glucose
Urease	$CO (N H_{2})_{2} + H_{2} O$ $\to$ $C O_{2} + 2N H_{3}$
Pepsin, trypsin	proteins $\to$ peptides and amino acids
Carbonic anhydrase	$H_{2} C O_{3}$ $⇌$ $H_{2} O + C O_{2}$
Nucleases	RNA, DNA $\to$ nucleotides

11.1 How enzymes work

The substrate binds to a region of the enzyme called the active site and forms an enzyme-substrate complex, in the way a key fits a lock. The complex offers a path of lower activation energy, so the reaction is much faster; the products then leave, and the enzyme is free to act again (Figure 12).

E + S ⇌ ES ⟶ E + P

NCERT gives an example: the activation energy of the acid hydrolysis of sucrose is quoted as 6.22 kJ mol $^{- 1}$ , but only 2.15 kJ mol $^{- 1}$ when the enzyme sucrase catalyses it.

Figure 12: The substrate binds the active site as a key fits a lock. The enzyme-substrate complex offers a path with a much lower activation energy, so the reaction runs fast at body temperature.

12. Uses of Proteins

Proteins are an essential part of food; meat, eggs, fish, pulses and cheese supply them.
Casein, a milk protein, is used to make artificial wool and silk.
Amino acids for medicines and feeding experiments are prepared by hydrolysing proteins.
Gelatin is used in desserts, salads, sweets and bakery goods.
Leather is made by tanning the proteins of animal hides.
Haemoglobin carries oxygen (and some $C O_{2}$ ) in the blood; protein hormones such as insulin control body processes; enzymes catalyse specific biological reactions.

13. The Whole Topic on One Page: Mind Map

The mind map ties the page together: amino acids join by peptide bonds, the chain folds into four levels of structure, and enzymes are the working result.

Figure 13: Mind map: from one amino acid to a working enzyme. Every exam question sits on one branch; the pI and the four levels of structure are asked most.

14. Solved Examples

Solved Example 1

Which of these statements are true for glycine? 1. It exists in crystalline form. 2. It is optically active. 3. It is soluble in water. 4. It can form a zwitterion.
(A) 1, 2 and 3
(B) 1, 2 and 4
(C) 1, 3 and 4
(D) 2, 3 and 4

Solution:

Answer: (C). Glycine exists as the ionic zwitterion $H_{3} N^{+} C H_{2} CO O^{-}$ , so it is a crystalline solid and dissolves in water. Its acidic and basic groups in one molecule form this internal salt. It is not optically active, because its $α$ -carbon carries two H atoms (R = H).

Solved Example 2

Glycine is made by the Gabriel synthesis: potassium phthalimide + X

\to

Y, then Y with

H_{3} O^{+}

on heating

\to

Z + glycine. X, Y and Z are
(A) chloroacetic acid; N-(carboxymethyl)phthalimide; phthalic acid
(B) bromoacetic acid; N-(carboxymethyl)phthalimide; phthalic anhydride
(C) ethyl chloroacetate; ethyl phthalimidoacetate; phthalic acid
(D) all are correct

Solution:

Answer: (C). The $α$ -halo ester $ClC H_{2} COO C_{2} H_{5}$ (X) is used, because a free acid would simply protonate the phthalimide anion. X alkylates the nitrogen to give ethyl phthalimidoacetate, $C_{6} H_{4} (CO)_{2} NC H_{2} COO C_{2} H_{5}$ (Y). Acid hydrolysis breaks both amide bonds and the ester, giving phthalic acid (Z), glycine and ethanol.

Solved Example 3

In acidic medium, amino acids are present as
(A)

H_{2} NCHRCOOH

(B)

H_{3} N^{+} CHRCO O^{-}

(C)

H_{3} N^{+} CHRCOOH

(D)

H_{2} NCHRCO O^{-}

Solution:

Answer: (C). In acid the carboxylate group of the zwitterion takes a proton, leaving the cation $H_{3} N^{+} CHRCOOH$ , which moves to the cathode.

Solved Example 4

Consider these statements. 1. At the isoelectric point,

α

-amino acids do not migrate towards either electrode. 2. p-Aminobenzenesulphonic acid exists as a dipolar ion, but p-aminobenzoic acid does not. 3. Sulphanilic acid dissolves in aqueous base but not in aqueous acid. 4.

H_{3} N^{+} C H_{2} COOH

(

p K_{a}

2.4) is more acidic than

RC H_{2} COOH

(

p K_{a}

4-5). Which are correct?
(A) 1 and 2 only
(B) 1, 2 and 3 only
(C) 2, 3 and 4 only
(D) all four

Solution:

1 is true: at the pI the cation and anion concentrations are equal, so there is no net migration.
2 is true: $- S O_{3} H$ is strong enough to protonate the weakly basic aryl amino group; ArCOOH is not.
3 is true: in sulphanilic acid, $p - H_{3} N^{+} C_{6} H_{4} S O_{3}^{-}$ , the $- N H_{3}^{+}$ gives its proton to a base, forming the soluble anion $p - H_{2} N C_{6} H_{4} S O_{3}^{-}$ . The $- S O_{3}^{-}$ group is too weakly basic to accept a proton from an acid.
4 is true: the electron-withdrawing $- N H_{3}^{+}$ group raises the acidity of the COOH.

Answer: (D).

Solved Example 5

The secondary structure of proteins is due to
(A) peptide bonds
(B) hydrogen bonds
(C) covalent bonds
(D) coordinate bonds

Solution:

Answer: (B). The $α$ -helix and $β$ -pleated sheet are held by hydrogen bonds between C=O and N-H groups of peptide bonds.

Solved Example 6

Write the structure of the tripeptide Val-Phe-Ser.

Solution:

Valine is the N-terminal residue (free $N H_{2}$ ) and serine the C-terminal residue (free COOH); two peptide bonds join the three units:

H_{2} N - CH (C H_{3})_{2} CH - CO - NH - C H_{2} C_{6} H_{5} CH - CO - NH - C H_{2} OH CH - COOH

Solved Example 7

For valine,

(C H_{3})_{2} CHCH (N H_{2}) COOH

p K_{a}

of the carboxyl group is 2.31 and of the amino group 9.69. Calculate the isoelectric point and draw the form of valine present at the pI, at pH 2 and at pOH 2.

Solution:

For a neutral amino acid, $pI = \frac{2.31 + 9.69}{2} = 6.00$ .

At pH 6.00 (the pI): the zwitterion, $(C H_{3})_{2} CHCH (N H_{3}^{+}) CO O^{-}$ , net charge 0.
At pH 2 (below $p K_{a 1}$ ): the cation, $(C H_{3})_{2} CHCH (N H_{3}^{+}) COOH$ , net charge +1.
At pOH 2, that is pH 12 (above $p K_{a 2}$ ): the anion, $(C H_{3})_{2} CHCH (N H_{2}) CO O^{-}$ , net charge −1.

Solved Example 8

Account for the fact that 2-aminoethanoic acid (glycine) exists as a dipolar ion, as does p-aminobenzenesulphonic acid, but p-aminobenzoic acid does not.

Solution:

A zwitterion forms only when the acid group is strong enough to protonate the amino group. In glycine the aliphatic $- N H_{2}$ is basic enough to take $H^{+}$ from COOH, giving $H_{3} N^{+} C H_{2} CO O^{-}$ . In the aromatic acids the $- N H_{2}$ is only weakly basic (its lone pair is spread into the ring). $- S O_{3} H$ is a very strong acid and still protonates it, giving $p - H_{3} N^{+} C_{6} H_{4} S O_{3}^{-}$ ; $- COOH$ is too weak, so p-aminobenzoic acid stays as $p - H_{2} N C_{6} H_{4} COOH$ .

Solved Example 9

Sulphanilic acid has both an acidic and a basic group, yet (i) it dissolves in alkali but not in mineral acid, and (ii) it does not dissolve in organic solvents. Explain.

Solution:

(i) Sulphanilic acid exists as the zwitterion $p - H_{3} N^{+} C_{6} H_{4} S O_{3}^{-}$ . The weakly acidic $- N H_{3}^{+}$ gives $H^{+}$ to $O H^{-}$ , forming the soluble salt $p - H_{2} N C_{6} H_{4} S O_{3}^{-} N a^{+}$ . The $- S O_{3}^{-}$ group is too weakly basic to accept $H^{+}$ from strong acids, so no soluble cation forms in acid.

(ii) Its ionic (zwitterionic) character makes it insoluble in non-polar organic solvents.

Solved Example 10

The structure of glycine is

H_{3} N^{+} C H_{2} CO O^{-}

(a zwitterion). Which statement is correct?
(A) Glycine and other amino acids are amphoteric.
(B) The acidic group in amino acids is

- N H_{3}^{+}

.
(C) The basic group in amino acids is

- CO O^{-}

.
(D) All the statements are correct.

Solution:

Answer: (D). Amino acids contain both an acidic and a basic group, so they are amphoteric. In the zwitterion the proton donor is $- N H_{3}^{+}$ and the proton acceptor is $- CO O^{-}$ .

Solved Example 11

The force of attraction between neighbouring peptide chains is
(A) van der Waals force
(B) covalent bond
(C) hydrogen bond
(D) peptide linkage

Solution:

Answer: (C). Neighbouring chains are held by hydrogen bonds between the C=O of one chain and the N-H of the next, as in the $β$ -pleated sheet.

Solved Example 12

Peptides on hydrolysis give
(A) ammonia
(B) amines
(C) amino acids
(D) hydroxy acids

Solution:

Answer: (C). Peptides are condensation products of $α$ -amino acids, so hydrolysis of their peptide bonds gives back the $α$ -amino acids.

Solved Example 13

How many different tripeptides can be made from glycine, alanine and valine if each is used once? How many peptide bonds and water molecules are involved in each?

Solution:

The three amino acids can be arranged in $3! = 6$ orders (Gly-Ala-Val, Gly-Val-Ala, Ala-Gly-Val, Ala-Val-Gly, Val-Gly-Ala, Val-Ala-Gly). Each tripeptide has $3 - 1 = 2$ peptide bonds, and forming it releases 2 molecules of $H_{2} O$ .

Solved Example 14

Aspartic acid has

p K_{a}

values 1.88 (

α

-COOH), 3.65 (side-chain COOH) and 9.60 (

α

N H_{3}^{+}

). Lysine has 2.18, 8.95 (

α

N H_{3}^{+}

) and 10.53 (side-chain

N H_{3}^{+}

). Find the pI of each.

Solution:

Write the charge of each form. For aspartic acid the forms run +1, 0, −1, −2 as the protons leave, so the neutral form lies between the first two $p K_{a}$ values: $pI = \frac{1.88 + 3.65}{2} = 2.77$ .

For lysine the forms run +2, +1, 0, −1, so the neutral form lies between the last two: $pI = \frac{8.95 + 10.53}{2} = 9.74$ .

Solved Example 15

A mixture of aspartic acid (pI 2.77), glycine (pI 5.97) and lysine (pI 9.74) is placed at the centre of a paper strip wetted with a pH 6.0 buffer, and a voltage is applied. Predict the movement of each.

Solution:

At pH 6.0: aspartic acid (pH $>$ pI) is negative and moves to the anode; glycine (pH $\approx$ pI) has almost no net charge and stays near the start; lysine (pH $<$ pI) is positive and moves to the cathode (Figure 6).

Solved Example 16

0.15 g of glycine is treated with excess nitrous acid. What volume of nitrogen is given off at 273 K and 1 atm (molar volume 22.4 L)?

Solution:

Moles of glycine $= \frac{0.15}{75} = 0.0020$ mol. One mole of amino acid gives one mole of $N_{2}$ (Van Slyke reaction), so $V = 0.0020 \times 22400 = 44.8$ mL.

Solved Example 17

Glycine (

p K_{a 1} = 2.34

p K_{a 2} = 9.60

) is dissolved in a buffer of pH 2.34. The percentage of glycine present as the cation is
(A) 25%
(B) 50%
(C) 75%
(D) 100%

Solution:

Answer: (B). At $pH = p K_{a 1}$ , $[DI] / [CA] = 1 0^{pH - p K_{a 1}} = 1$ , so the cation and the zwitterion are 50 : 50. The anion is negligible ( $1 0^{2.34 - 9.60} \approx 5 \times 1 0^{- 8}$ of the zwitterion), as the species plot shows (Figure 5).

Solved Example 18

At pH 7.0, which amino acid moves towards the cathode during electrophoresis?
(A) aspartic acid
(B) glutamic acid
(C) glycine
(D) lysine

Solution:

Answer: (D). Lysine's pI is 9.74. At pH 7.0 (below its pI) it carries a net positive charge and moves to the cathode. Aspartic (2.77) and glutamic (3.22) acids are negative at pH 7 and move to the anode; glycine (5.97) is slightly negative and moves only a little.

Solved Example 19

Four different amino acids are joined, each used once, into a straight-chain tetrapeptide. How many peptide bonds does each tetrapeptide have, how many water molecules are released in making it, and how many different tetrapeptides are possible?

Solution:

Each tetrapeptide has $4 - 1 = 3$ peptide bonds, and 3 molecules of water are lost. The four residues can be ordered in $4! = 24$ ways, and each order (read from the N-terminus) is a different tetrapeptide.

Practice Questions

The acidic group in glycine is (A) $- COOH$ (B) $- CO O^{-}$ (C) $- N H_{2}$ (D) $- N H_{3}^{+}$ Answer: (D). In the zwitterion the proton donor is $- N H_{3}^{+}$ .
Can an amino acid be isolated at its isoelectric point?Answer: Yes. Its solubility in water is lowest at the pI, so it crystallises out at that pH.
(a) Name the linkages responsible for the primary and secondary structures of proteins. (b) On electrolysis, $α$ -amino acids move to the cathode in acid and to the anode in alkali. Explain. (c) What are essential and non-essential amino acids? Give two examples of each.Answer: (a) Primary: peptide bonds; secondary: hydrogen bonds. (b) In acid they exist as cations ( $H_{3} N^{+} CHRCOOH$ ), in alkali as anions ( $H_{2} NCHRCO O^{-}$ ). (c) Essential: not made in the body (leucine, lysine); non-essential: made in the body (glycine, alanine).
Glycine exists as a dipolar ion, while anthranilic acid (2-aminobenzoic acid) does not. Explain.Answer: In anthranilic acid the $- N H_{2}$ on the benzene ring is too weakly basic, and $- COOH$ too weakly acidic, for the proton transfer; in glycine the aliphatic $- N H_{2}$ is basic enough.
What is a peptide linkage? Describe the geometry and bond lengths of a peptide unit.Answer: The amide link $- CO - NH -$ . It is planar, with O and H trans; C-N is 132 pm (normal 147 pm) because of resonance, so there is no free rotation about it; C=O is 124 pm.
Lysine and glutamine form a dipeptide linkage. What are the two possible dipeptides?Answer: Lys-Gln, $H_{2} N (C H_{2})_{4} CH (N H_{2}) CO$ - $NHCH (COOH) C H_{2} C H_{2} CON H_{2}$ , and Gln-Lys, $H_{2} NCOC H_{2} C H_{2} CH (N H_{2}) CO$ - $NHCH (COOH) (C H_{2})_{4} N H_{2}$ : each joins the $α$ -COOH of one to the $α$ - $N H_{2}$ of the other (open chains, not rings).
Write the structure of alanine at pH 2 and at pH 10.Answer: pH 2: $C H_{3} CH (N H_{3}^{+}) COOH$ (cation); pH 10: $C H_{3} CH (N H_{2}) CO O^{-}$ (anion).
Aspartame, $H_{2} NCH (C H_{2} COOH) CO$ - $NHCH (C H_{2} C_{6} H_{5}) COOC H_{3}$ , is an artificial sweetener. (i) Name its four functional groups. (ii) Write its zwitterion. (iii) Which amino acids does hydrolysis give? (iv) Which of them is more hydrophobic?Answer: (i) Amine, carboxylic acid, amide (peptide bond) and ester. (ii) $H_{3} N^{+} CH (C H_{2} CO O^{-}) CO$ - $NHCH (C H_{2} C_{6} H_{5}) COOC H_{3}$ . (iii) Aspartic acid and phenylalanine (plus methanol from the ester). (iv) Phenylalanine, because of its non-polar benzyl group.
The peptide bond is (A) $- CO - NH -$ (B) $N H_{2} - CO - NH - R$ (C) $R - CO - NH - R$ (D) $- CON H_{2}$ Answer: (A).
If $K_{a 1}$ and $K_{a 2}$ are the ionisation constants of $H_{3} N^{+} CHRCOOH$ and $H_{3} N^{+} CHRCO O^{-}$ , the pH at the isoelectric point is (A) $p K_{a 1} + p K_{a 2}$ (B) $(p K_{a 1} + p K_{a 2})^{1/2}$ (C) $(p K_{a 1} \times p K_{a 2})^{1/2}$ (D) $(p K_{a 1} + p K_{a 2}) /2$ Answer: (D).
The simplest amino acid is (A) glycine (B) alanine (C) guanine (D) all of theseAnswer: (A).
The main structural unit of proteins is the (A) ester linkage (B) ether linkage (C) peptide linkage (D) all of theseAnswer: (C).
The pH at which an amino acid does not migrate in an electric field is its (A) isoelectric point (B) isoelectronic point (C) neutralisation point (D) none of theseAnswer: (A).
At the isoelectric point, amino acids are present as (A) $H_{2} NCHRCOOH$ (B) $H_{3} N^{+} CHRCO O^{-}$ (C) $H_{3} N^{+} CHRCOOH$ (D) $H_{2} NCHRCO O^{-}$ Answer: (B).
The biuret test is used to detect (A) sugars (B) fats (C) proteins (D) saturated oilsAnswer: (C).
Sanger's reagent is (A) 1-fluoro-2,4-dinitrobenzene (B) 5-(dimethylamino)naphthalene-1-sulphonyl chloride (C) 2,4,6-trinitrophenol (D) none of theseAnswer: (A). Option (B) is dansyl chloride, another N-terminal reagent.

Common Mistakes to Avoid

Watch out

Writing amino acids as neutral $H_{2} N - CHR - COOH$ in solids and water. They exist as zwitterions, which is why they decompose above 200 °C instead of melting.
Calling $- COOH$ the acidic group of the zwitterion. In $H_{3} N^{+} CHRCO O^{-}$ the acid is $- N H_{3}^{+}$ and the base is $- CO O^{-}$ .
Reversing the direction of migration. Below the pI the amino acid is a cation and moves to the cathode (−); above the pI it is an anion and moves to the anode (+).
Averaging the first and last $p K_{a}$ of an acidic or basic amino acid. Average the two values on either side of the neutral form (Asp: 1.88 and 3.65).
Saying glycine is optically active. It has two H atoms on the $α$ -carbon; all the other protein amino acids are L, and L does not mean laevorotatory.
Thinking denaturation breaks peptide bonds. It destroys the 2° and 3° structure; the sequence (1°) stays.
Expecting amino acids or dipeptides to give the biuret test. It needs at least two peptide bonds.
Counting peptide bonds as equal to the number of residues. A chain of $n$ amino acids has $n - 1$ peptide bonds.
Treating Gly-Ala and Ala-Gly as the same. Sequences are written from the N-terminus (left) to the C-terminus (right).

Frequently Asked Questions

What is a peptide bond?

A peptide bond is the amide link, $- CO - NH -$ , formed when the carboxyl group of one $α$ -amino acid condenses with the amino group of another and water is lost. It is planar because resonance gives the C-N bond partial double-bond character, so there is no free rotation about it.

What is a zwitterion?

A zwitterion (dipolar ion) is the inner salt an amino acid forms when its carboxyl group donates a proton to its own amino group: $H_{3} N^{+} - CHR - CO O^{-}$ . It has both charges but is neutral overall, which explains the high melting points, water solubility and large dipole moments of amino acids.

What is the isoelectric point of an amino acid?

The isoelectric point (pI) is the pH at which an amino acid exists mainly as the zwitterion, has no net charge and does not move in an electric field. For a neutral amino acid it is the average of its two $p K_{a}$ values; for glycine it is 5.97. Amino acids are least soluble at their pI.

What are the four levels of protein structure?

Primary structure is the amino acid sequence, held by peptide bonds. Secondary structure is the $α$ -helix or $β$ -pleated sheet, held by hydrogen bonds. Tertiary structure is the overall folding, held by H-bonds, S-S bridges, ionic and hydrophobic forces. Quaternary structure is the arrangement of subunits, as in haemoglobin.

What is denaturation of proteins?

Denaturation is the loss of a protein's natural shape and biological activity when heat, a change in pH or chemicals break the hydrogen bonds and other weak forces holding it. The secondary and tertiary structures are destroyed but the sequence is not. Boiling an egg and curdling milk are everyday examples.

How do enzymes work?

Enzymes are protein catalysts. The substrate binds at the enzyme's active site to form an enzyme-substrate complex, like a key in a lock. This complex offers a path with a lower activation energy, so the reaction runs fast at body temperature, and the enzyme is released unchanged. Each enzyme is highly specific.

Which protein topics are most important for NEET?

NEET follows NCERT closely: the zwitterion and amphoteric nature of amino acids, essential amino acids, the peptide bond, fibrous and globular proteins, the four levels of structure with their bonds, denaturation with examples, and how enzymes lower activation energy. One-line facts from these areas appear most often.

How are amino acids and proteins tested in JEE?

JEE Main asks about zwitterions, the pI formula, the charge and migration of an amino acid at a given pH, and counting peptide bonds or possible sequences. JEE Advanced adds pI of acidic and basic amino acids, electrophoresis, Strecker and Gabriel syntheses, and sequencing puzzles using partial hydrolysis and Sanger's reagent.

Previous year questions on Proteins

26 questions from past papers, each with a step-by-step solution.

Show all 26 questions

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1. Amino Acids

1.1 Structure

1.2 The 20 amino acids of proteins

1.3 Essential and non-essential amino acids

1.4 Configuration

1.5 Physical properties

2. Zwitterions and the Isoelectric Point

2.1 The zwitterion

2.2 Behaviour in acid and in base

2.3 Calculating the isoelectric point

2.4 Acidic and basic amino acids

2.5 Electrophoresis and solubility

3. Preparation of α-Amino Acids

3.1 From proteins

3.2 Amination of α-halo acids

3.3 Gabriel phthalimide synthesis

3.4 Strecker synthesis

4. Chemical Reactions of Amino Acids

4.1 Reactions of the carboxyl group

4.2 Reactions of the amino group

4.3 Reaction of both groups: effect of heat

5. Peptides

5.1 The peptide bond

5.2 Geometry of the peptide bond

6. Proteins: Roles and Classification

7. Structure of Proteins

7.1 Primary structure

7.2 Secondary structure

7.3 Tertiary structure

7.4 Quaternary structure

8. Denaturation of Proteins

9. Finding the Primary Structure

9.1 Partial hydrolysis

9.2 End-group analysis

10. Tests for Proteins

11. Enzymes

11.1 How enzymes work

12. Uses of Proteins

13. The Whole Topic on One Page: Mind Map

14. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is a peptide bond?

What is a zwitterion?

What is the isoelectric point of an amino acid?

What are the four levels of protein structure?

What is denaturation of proteins?

How do enzymes work?

Which protein topics are most important for NEET?

How are amino acids and proteins tested in JEE?

Previous year questions on Proteins

Ready to master Biomolecules?

3. Preparation of $α$ -Amino Acids

3.2 Amination of $α$ -halo acids