The shape of the lesson
By the end of the lesson, learners can
- Zwitterions and the isoelectric point: Slide the pH in the simulator and watch the amino acid change form.
- Peptides and electrophoresis: Two amino acids join head to tail.
- Zwitterion: H3N+CH(R)COO−, high m.p.
- Acid: +1, to the cathode. Alkali: −1, to the anode.
- Peptide bond: –CONH–, condensation, lose H2O.
- Electrophoresis: charge sets direction; charge and size set speed.
One picture every idea returns to
An α-amino acid has the general formula RCH(NH2)COOH. In the solid and in neutral solution the COOH group donates H+ to the NH2 group, forming a zwitterion, H3N+CH(R)COO−: the strong ionic attractions between zwitterions give amino acids high melting points and solubility in water. The pH at which the amino acid has no overall charge is its isoelectric point. In acid the COO− is protonated, giving the positive ion H3N+CH(R)COOH; in alkali the NH3+ loses H+, giving the negative ion H2NCH(R)COO−. Amino acids join by condensation, losing water, to form peptide (amide) bonds, –CONH–; peptides are hydrolysed back to amino acids by heating with acid or alkali. In electrophoresis, ions move in an electric field at a set pH: the direction depends on the sign of the charge, and the speed on the size of the charge and of the ion.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, An acid and a base in one molecule.. Every amino acid carries an acidic COOH and a basic NH2 on the same carbon. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Zwitterions and the isoelectric point | Slide the pH in the simulator and watch the amino acid change form. Project tab 5 of the lab (Nitrogen compounds (A Level) lab, zwitterions) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Peptides and electrophoresis | Two amino acids join head to tail. Project tab 6 of the lab (Nitrogen compounds (A Level) lab, peptides and electrophoresis) and let a student drive it while the class predicts. | Connect-Extend-Challenge |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Peptides and electrophoresis, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| In acid, an amino acid is positive (NH 3 +, COOH) and moves to the cathode; in alkali it is negative (NH 2, COO −) and moves to the anode. | The peptide bond is an amide link made by condensation, not addition. Two different amino acids give two dipeptides, not one, because the order matters. |
Amino acids and peptides: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Draw (as condensed formulae) the forms of alanine, CH3CH(NH2)COOH, present (a) at low pH, (b) at its isoelectric point, (c) at high pH.
Glycine melts with decomposition above 230 °C, but ethanoic acid melts at 17 °C. Explain.
Write the condensed formulae of the two dipeptides formed from glycine and alanine, and name the bond formed.
Calculate the Mr of the tripeptide gly-gly-gly (glycine Mr 75.0), and state how many water molecules are needed to hydrolyse it completely.
A mixture of lysine (isoelectric point 9.7), glycine (6.0) and aspartic acid (2.8) is separated by electrophoresis in a buffer at pH 6.0. Predict and explain where each spot ends up.
State the conditions used to hydrolyse a protein to its amino acids, and explain why the products are present as ions.
Answer key and marking notes
- (a) CH3CH(NH3+)COOH.
- (b) CH3CH(NH3+)COO− (zwitterion).
- (c) CH3CH(NH2)COO−.
- Glycine exists as zwitterions, H3N+CH2COO−, held by strong ionic attractions in a lattice.
- Ethanoic acid molecules are held only by hydrogen bonds and weaker intermolecular forces, which need far less energy to overcome.
- Gly-ala: H2NCH2CONHCH(CH3)COOH.
- Ala-gly: H2NCH(CH3)CONHCH2COOH.
- The –CONH– group is the peptide (amide) bond, formed by condensation with loss of H2O.
- Two peptide bonds form, so two H2O are lost: 3 × 75.0 − 2 × 18.0 = 189.0.
- Complete hydrolysis breaks 2 peptide bonds, using 2 H2O.
- Lysine: the pH is below its isoelectric point, so it is protonated and positive; it moves to the cathode.
- Glycine: at its isoelectric point it is a neutral zwitterion and stays on the start line.
- Aspartic acid: the pH is above its isoelectric point, so it is negative and moves to the anode.
- Heat under reflux with 6 mol dm−3 HCl(aq) for several hours (or with NaOH(aq)).
- In acid the amino acids are protonated (H3N+ groups); in alkali they are present as carboxylate ions (COO−).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A mixture of aspartic acid, glycine and lysine is separated by electrophoresis at pH 6.0. Which row identifies spots P, Q and R?
- AP lysine, Q glycine, R aspartic acid
- BP aspartic acid, Q glycine, R lysine
- CP glycine, Q lysine, R aspartic acid
- DP lysine, Q aspartic acid, R glycine
Which formula shows the zwitterion of glycine?
- AH3N+CH2COO−
- BH2NCH2COOH
- CH3N+CH2COOH
- DH2NCH2COO−
Excess NaOH(aq) is added to a solution of alanine, CH3CH(NH2)COOH. Which species is now present?
- ACH3CH(NH2)COO−
- BCH3CH(NH3+)COO−
- CCH3CH(NH3+)COOH
- DCH3CH(NH2)COONa+
Why do amino acids have much higher melting points than carboxylic acids of similar Mr?
- Athey exist as zwitterions, held together by strong ionic attractions
- Bthey form giant covalent networks
- Cthey have stronger instantaneous dipole-induced dipole forces because of the nitrogen atom
- Dtheir peptide bonds must be broken before they melt
Glycine and alanine join to form a dipeptide. Which statement is correct?
- Ait is a condensation reaction that forms an amide (peptide) bond, –CONH–, and releases water
- Bit is an addition reaction that forms an ester bond, –COO–
- Cit is a hydrolysis reaction that uses water to form an amide bond
- Dit is a condensation reaction that forms an azo bond, –N=N–, and releases hydrogen
How many different TRIPEPTIDES can be made from one molecule each of glycine, alanine and serine?
- A6
- B3
- C1
- D27
What is the Mr of the tripeptide gly-ala-gly? (glycine Mr 75.0, alanine Mr 89.0)
- A203.0
- B239.0
- C221.0
- D185.0
The dipeptide gly-ala is heated under reflux with 6 mol dm−3 HCl(aq) until hydrolysis is complete. Which species are present in the solution?
- AH3N+CH2COOH and H3N+CH(CH3)COOH
- BH2NCH2COO− and H2NCH(CH3)COO−
- CH3N+CH2COO− only
- DCH3COOH, NH4+ and alanine
Quiz key
At pH 6.0 glycine is at its isoelectric point, so it is a neutral zwitterion and stays on the start line (Q). Lysine (isoelectric point 9.7) is in a solution more acidic than its isoelectric point, so it carries a positive charge and moves to the cathode (P). Aspartic acid (2.8) is in a less acidic solution, so it is negative and moves to the anode (R). Swapping P and R reverses the charges. The other two rows move glycine, which has no overall charge at this pH.
The acidic COOH donates H+ to the basic NH2 in the same molecule, giving H3N+CH2COO−, with no overall charge. H2NCH2COOH is the un-ionised formula, which does not represent solid glycine. H3N+CH2COOH is the cation present in acid. H2NCH2COO− is the anion present in alkali.
In the zwitterion, H3N+ is the acidic group; OH− removes H+ from it: CH3CH(NH3+)COO− + OH− → CH3CH(NH2)COO− + H2O, an ion with charge −1. The zwitterion is the form at the isoelectric point, not in excess alkali. The cation is the form in acid. Sodium is present as a separate Na+ ion; COONa+ has the wrong charge.
In the solid, amino acids are zwitterions, H3N+CH(R)COO−, so the lattice is held by electrostatic attractions between opposite charges, much stronger than the hydrogen bonds between carboxylic acid molecules. They are not giant covalent structures: covalent bonds join atoms only within each molecule. A single N atom adds few electrons, so id-id forces change little. A single amino acid contains no peptide bond.
The OH from one COOH and an H from the other molecule's NH2 are lost as H2O (condensation), joining the two through -CONH-, the peptide (amide) link. An ester link needs an alcohol OH group and would also be condensation, not addition. Hydrolysis is the reverse reaction, which uses water to break peptide bonds. Azo links come from diazonium coupling, and no hydrogen gas is released.
Any of the 3 can be at the N-terminal end, then either of the remaining 2, then the last one: 3 × 2 × 1 = 6 different sequences (for example gly-ala-ser and ser-ala-gly are different molecules). 3 counts only which amino acid is in the middle. 1 ignores the order. 27 (33) allows each amino acid to be used more than once, but only one molecule of each is available.
75.0 + 89.0 + 75.0 = 239.0, minus 2 × 18.0 for the two water molecules lost = 203.0. 239.0 forgets the condensation entirely. 221.0 subtracts only one water molecule. 185.0 subtracts three, as if three bonds formed, but three amino acids are joined by only two peptide bonds.
Water splits the -CONH- link, giving glycine and alanine. In excess strong acid both the NH2 and COO− groups are protonated, so each amino acid is present as its cation, H3N+CH(R)COOH. The anions would be present after alkaline hydrolysis. Glycine alone ignores the alanine released, and the zwitterion is the form at the isoelectric point, not in 6 mol dm−3 acid. Ethanoic acid and NH4+ would require C-C bond cleavage, which does not happen.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.