Related topics: Amines · Chromatography
Where this sits
Nothing here is new chemistry. It is your Year 2 organic chemistry applied to the molecules of life. An amino acid is a carboxylic acid and an amine on the same carbon, so it does both jobs at once. Joining them is the condensation that makes a polyamide. Holding proteins in shape is hydrogen bonding. Choosing between enantiomers is optical isomerism. And cisplatin is ligand substitution on a Pt(II) complex.
Amino acids & the zwitterion
An α-amino acid has an amine group and a carboxylic acid group on the same carbon: H2N–CHR–COOH. Because the –NH2 sits on carbon 2 of the acid chain, the systematic names are all 2-amino…oic acids: alanine is 2-aminopropanoic acid, glycine is 2-aminoethanoic acid. The R group (the side chain) is what changes from one amino acid to the next; AQA gives you the structures in the Chemistry Data Booklet, so there is nothing to memorise.
What you are given in the exam
The last page of the Data Booklet carries six amino acids (alanine, aspartic acid, cysteine, lysine, phenylalanine and serine) drawn in the layout below, plus the four DNA bases, the phosphate ion and 2-deoxyribose. You are never asked to recall a side chain. The marks are for using those structures: joining them into a peptide, charging them up at a given pH, or finding where a hydrogen bond can form. Get used to reading them in this exact form.
The zwitterion
Carrying an acid and a base on the same molecule has a consequence you can predict: the acid group hands its proton to the base group. The –COOH loses H+, the –NH2 gains it, and the result is an ion with a positive end and a negative end but no overall charge: a zwitterion.
Key definition
A zwitterion is a molecule that contains both a positive and a negative charge but is neutral overall.
Which ion at which pH
Which species you actually get depends on the pH, and the reasoning is ordinary acid–base reasoning: in acid there is a large excess of H+, so every group that can be protonated is; in alkali there is a large excess of OH−, so every acidic proton is stripped off.
Worked example: the species at a given pH
Serine has the side chain –CH2OH. Draw the species present in a solution of serine at pH = 1 and at pH = 13.
Step 1. List the groups that can gain or lose H+: the –NH2 (can gain H+), the –COOH (can lose H+). The side-chain –OH of serine is a neutral alcohol and does neither.
Step 2. pH = 1 is a large excess of H+: the amine is protonated and the acid keeps its proton.
H3N+–CH(CH2OH)–COOH
Step 3. pH = 13 is a large excess of OH−: both acidic protons are removed: the one on the –COOH and the one that would sit on the nitrogen.
H2N–CH(CH2OH)–COO−
Check the charge each time: +1 at pH 1, −1 at pH 13. A structure whose charges do not add up to the charge you have claimed cannot be right.
Why amino acids behave like salts
In the solid, an amino acid exists as its zwitterion (an ion). So the solid is held together by strong electrostatic attractions between ions, not by the weak intermolecular forces you would expect of a small organic molecule. That is why amino acids are crystalline solids with high melting points that dissolve in water but not in non-polar solvents.
Precision points
- A zwitterion needs both charges. One charge on its own is a cation or an anion, not a zwitterion.
- Put the + on the nitrogen and the − on the oxygen, not floating at the end of the formula.
- In acid the amine is protonated to –NH3+; the –COOH stays as –COOH. In alkali the acid is deprotonated to –COO−; the amine stays as –NH2. Half-changing both is the classic wrong answer.
- If the side chain itself carries a –COOH (aspartic acid) or an extra –NH2 (lysine), it changes too. Count every group before you write the charge.
- An alcohol –OH side chain is not acidic (serine, threonine). Even in strong alkali it keeps its proton. Only the –COOH and protonated –NH3+ lose theirs. Deprotonating an alcohol –OH to –O− is a common and costly error.
Exam questions
Q1
[1 mark]
Which compound is not a 2-aminocarboxylic acid?
Number the chain from the –COOH carbon. In B the –NH2 is on carbon 3, so it is a 3-aminocarboxylic acid; in the other three it sits on carbon 2.
Q2
[1 mark]
Which structure is formed by phenylalanine in solution at pH = 3?
pH 3 is acidic, so the amine is protonated to –NH3+ and the carboxylic acid keeps its proton: the cation, overall +1. B is the zwitterion (neutral), C the anion, D the un-ionised molecule. The option structures are redrawn here; the chemistry and the answer are AQA’s.
Source: AQA A-Level Chemistry past papers.
The peptide link
Put two amino acids next to each other and the –COOH of one reacts with the –NH2 of the other. The acid loses its –OH, the amine loses an H, the two leave together as water, and the carbon and nitrogen left behind are joined. That new –CONH– group is an amide link. In a protein it is called a peptide link, and the reaction that makes it is a condensation.
Key definition
A peptide link is the –CONH– group formed when the carboxylic acid group of one amino acid and the amine group of another react together, losing a molecule of water.
Two amino acids, two dipeptides
Two different amino acids can join in two different orders, and the two products are different compounds: cysteine–serine has the cysteine amine group free, serine–cysteine has the serine amine group free. They have the same molecular formula and the same relative molecular mass, so a mass spectrum cannot tell them apart, but they have different retention times, which is why chromatography is the tool of choice for sequencing.
Worked example: drawing a named dipeptide
Draw the dipeptide formed between cysteine and serine, cysteine first.
Step 1. Read the two side chains off the Data Booklet: cysteine is –CH2SH, serine is –CH2OH. Both sit on the CH of an H2N–CH–COOH backbone.
Step 2. Decide which end reacts: “cysteine first” means the cysteine’s –COOH joins the serine’s –NH2. So cysteine keeps its free –NH2 on the left, serine keeps its free –COOH on the right.
Step 3. Write the chain, link drawn in full:
H2N–CH(CH2SH)–CONH–CH(CH2OH)–COOH
Step 4. Check before you stop: one peptide link with its N–H; the correct side chain on the correct carbon; a free amine at one end and a free acid at the other. AQA awards one mark for the link and one for the two side chains. A dipeptide drawn with the right link but the wrong R groups scores only the first.
Drawing a peptide: the checklist
- Count the residues the question names: dipeptide is two, tripeptide is three.
- Read each side chain off the Data Booklet and hang it below the correct CH.
- Draw each link in full: C=O as a double bond and the H on the nitrogen.
- Leave a free –NH2 at the left end and a free –COOH at the right, unless the question shows a section inside a chain, when both ends carry trailing bonds instead.
Longer chains work the same way. Three amino acids give a tripeptide with two peptide links and two water molecules lost; a protein is the same reaction run hundreds of times, giving a chain called a polypeptide.
Hydrolysis: running it backwards
Heating a protein with hot aqueous acid (or alkali) breaks every peptide link and puts the water back: hydrolysis. It is the same reaction that breaks a polyamide, because a peptide link is an amide link. What you get depends on which reagent you used, and marks are lost here for ignoring that.
Precision points
- Draw the peptide link in full: C=O as a double bond and the H on the nitrogen. AQA accepts a minimum of –CONH–, but a nitrogen drawn without its hydrogen is not creditworthy.
- Asked for a dipeptide when the question said tripeptide? You lose the structure mark even if every link is right. Count the residues named in the question.
- Reagent and conditions for hydrolysis: aqueous HCl (any strong acid or alkali) and reflux/heat. Adding “concentrated” is ignored; a stated temperature above 200 °C is not accepted.
- Trailing bonds are not needed on a peptide you have been asked to draw, but they are needed on a polymer repeat unit. Know which one you are drawing.
Exam questions
Q3a
[2 marks]
Part of the structure of a protein is shown. Each amino acid is shown using the first three letters of its name.
–Cys–Ser–Asp–Phe–
Draw a structure for the –Cys–Ser– section of the protein. Use the Data Booklet to help you answer this question.
Show answer
Correct peptide link: AQA allows a minimum of –CONH–. 1 mark
The correct amino acid R groups: –CH2SH for cysteine and –CH2OH for serine. 1 mark
A dipeptide can only score the first mark: the question shows a section inside a chain, so the ends carry trailing bonds rather than a free –NH2 and –COOH. Trailing bonds themselves are not needed.
Q3b
[1 mark]
Name the other substance formed when two amino acids react together to form part of a protein chain.
Show answer
Water. 1 mark
H2O is allowed.
Source: AQA A-Level Chemistry past papers.
Primary, secondary & tertiary structure
A protein is not a floppy string. The chain folds into one precise shape, and that shape is what does the job, so AQA describes it at three levels, each held together by different bonding.
The three levels
- Primary structure: the sequence of amino acids in the chain, joined by peptide links.
- Secondary structure: the α-helix or β-pleated sheet the chain takes up, held by hydrogen bonds between the C=O and the N–H groups of the backbone. The C=O bond is polar, so its oxygen is δ− and has lone pairs: it can hydrogen-bond to the δ+ hydrogen of an N–H (or O–H) elsewhere in the chain.
- Tertiary structure: the overall three-dimensional shape of the folded chain, held by interactions between the R groups: hydrogen bonds, S–S (disulfide) bridges (AQA call these sulfur–sulfur bonds) and ionic interactions.
The sequence –Cys–Ser–Asp–Phe– is a primary structure: it says which amino acids, in which order, and nothing about shape. Everything after that follows from it.
Secondary structure: the helix and the sheet
Tertiary structure: the R groups meet
Fold that coiled or pleated chain up again and the R groups come into contact with each other. What happens next depends entirely on which side chains meet, and the six in the Data Booklet cover all three possibilities.
| Side chains that meet | Interaction | Strength |
|---|---|---|
| –CH2SH and –CH2SH (two cysteines) | disulfide (S–S) bridge, covalent | strongest |
| an –NH3+ and a –COO− (lysine and aspartic acid) | ionic interaction | middle |
| an –OH and a C=O (serine and aspartic acid) | hydrogen bond | weakest of the three |
The covalent S–S bridge is the strongest, then the ionic interaction, then the hydrogen bond. The bars show the order only, not measured values.
Worked example: naming the interaction and ranking it
A folded protein brings a lysine R group close to an aspartic acid R group, and elsewhere two cysteine R groups meet. Name each interaction and say which is stronger.
Step 1. Read the side chains: lysine ends in –NH2, aspartic acid ends in –COOH, cysteine ends in –SH.
Step 2. Lysine + aspartic acid: in the cell the amine is protonated to –NH3+ and the acid is deprotonated to –COO−, so oppositely charged groups attract: an ionic interaction.
Step 3. Cysteine + cysteine: the two –SH groups are oxidised together to form an S–S disulfide bridge.
Step 4. Rank them: the disulfide bridge is a covalent bond; the ionic interaction (and any hydrogen bond) is not. Covalent wins, so the disulfide bridge is stronger.
That last sentence is the one that earns the mark. “Disulfide bridges are stronger” on its own is an assertion; “because a disulfide bridge is a covalent bond” is the reason.
Precision points
- Secondary structure is held by hydrogen bonds between the backbone C=O and N–H. Tertiary structure is held by interactions between the R groups. Putting S–S bridges in the secondary structure is the standard error.
- “Stronger” is not an explanation. Say why: a disulfide bridge is a covalent bond, a hydrogen bond is not.
- Primary structure is the sequence. A list of amino acids in order says nothing about shape.
Exam questions
Q4a
[1 mark]
Proteins are polymers made from amino acids. Part of the structure of a protein is shown, each amino acid given as the first three letters of its name.
–Cys–Ser–Asp–Phe–
Identify the type of protein structure shown.
A sequence of amino acids and nothing else is the primary structure. Secondary would describe an α-helix or β-pleated sheet; tertiary would describe the folded 3-D shape.
Q4b
[4 marks]
R groups can interact and contribute to protein structure. Explain why the strength of the interaction between two cysteine R groups differs from the strength of the interaction between a serine R group and an aspartic acid R group. Use the Data Booklet to help you answer this question.
Show answer
Two cysteine R groups form a disulfide bridge (stated or described). 1 mark
Serine and aspartic acid R groups form hydrogen bonds. 1 mark
Disulfide bridges are stronger than hydrogen bonds. 1 mark
Because disulfide bridges are covalent bonds (while hydrogen bonds are not). 1 mark
The first mark can also be scored by a correct diagram showing at least –S–S–.
Q4c
[1 mark]
Deduce the type of interaction that occurs between a lysine R group and an aspartic acid R group.
Show answer
Ionic bond. 1 mark
Source: AQA A-Level Chemistry past papers.
Separating & identifying amino acids
Hydrolyse a protein and you have a mixture of amino acids in one tube. To find out which ones, separate them by thin-layer chromatography. The plate is coated with a thin layer of silica or alumina (the stationary phase) and the solvent that creeps up it is the mobile phase.
Why the spots separate
Each amino acid sits somewhere on a balance between dissolving in the mobile phase and being adsorbed onto (retained by) the stationary phase. The more time it spends dissolved, the further it travels. Different amino acids have a different balance, so they travel different distances up the plate.
Locating the spots
Amino acids are colourless, so at the end you cannot see anything. Spray the dried plate with a developing (locating) agent (ninhydrin is the standard one, iodine vapour also works) or view it under ultraviolet light. Then measure each spot and work out its Rf.
Rf values
Worked example: calculating an Rf
On a developed plate the solvent front is 8.0 cm above the start line and the centre of a spot is 3.2 cm above it. Calculate the Rf value.
Step 1. Write the definition:
Rf = distance moved by the spotdistance moved by the solvent
Step 2. Substitute, keeping both distances in the same unit:
Rf = 3.2 cm8.0 cm = 0.40
Rf has no units (it is a ratio of two lengths) and it can never exceed 1, because a spot cannot outrun the solvent that carries it. Show the division: the mark scheme needs working that lands inside the accepted range, and there is no error carried forward to the identification that follows.
Precision points
- Measure both distances from the start line, and to the centre of the spot, not to its leading edge.
- The start line is drawn in pencil: ink would dissolve in the solvent and run up the plate with the sample.
- The solvent must start below the start line. If the spot is submerged it dissolves into the solvent instead of travelling up the plate.
- Rf has no units and is always less than 1. Show the division. The identification mark is lost if the working does not land in the accepted range.
Two-way chromatography
Sometimes two amino acids have almost the same Rf in one solvent and their spots overlap. Running the plate a second time in a different solvent, with the plate turned through 90°, spreads those out: a two-way chromatogram. This is Required practical 12; the full technique, along with column and gas chromatography, is on chromatography.
Exam questions
Q5a
[2 marks]
The protein fibroin can be broken down into amino acids using an enzyme. A student uses thin-layer chromatography (TLC) to identify these amino acids, and identifies two of them as alanine and serine. Use the figure below to calculate the Rf value of the unknown amino acid. Show your working. Use your Rf value and the table to identify the unknown amino acid.
| Amino acid | Rf value |
|---|---|
| tyrosine | 0.25 |
| glycine | 0.34 |
| valine | 0.64 |
| leucine | 0.73 |
Show answer
The unknown spot has moved 3.0 cm; the solvent front has moved 8.8 cm.
Rf = 3.08.8 = 0.34
Rf = 0.34 1 mark
Identity: glycine 1 mark
Relevant working is needed to arrive at 0.325–0.35, and there is no error carried forward: a wrong Rf cannot earn the identification mark.
Q5b
[1 mark]
The amino acids cannot be seen as they move during the experiment. State how the amino acids can be made visible at the end of the experiment.
Show answer
Use a UV lamp or ninhydrin. 1 mark
“Developing agent”, “locating agent” and iodine are also allowed.
Q5c
[1 mark]
State why each amino acid has a different Rf value.
Show answer
Each amino acid has a different (relative) affinity for / solubility in the stationary and mobile phases. 1 mark
Reference to different solubility in the solvent, or different affinity for the stationary phase, is allowed.
Q6
[4 marks]
A protein was hydrolysed to form a mixture of amino acids. A spot of the mixture was added to a TLC plate and the plate placed vertically in solvent 1. When the solvent front reached nearly to the top, the plate was removed and dried. The plate was then turned anticlockwise through 90° and placed vertically in solvent 2, run again, and dried. The diagram shows the final plate.
(a) Suggest a suitable reagent for the hydrolysis of a protein. (b) Suggest how the positions of the amino acids on the plate were located. (c) Deduce the minimum number of amino acids present in the original mixture. (d) Suggest why it was necessary to use two different solvents.
Show answer
(a) Concentrated HCl. 1 mark
(b) Using ninhydrin or ultraviolet light. 1 mark
(c) 7 1 mark
(d) Some of the amino acids did not separate with the first solvent (they had the same Rf value, or the same affinity, in that solvent). 1 mark
For (a), concentrations of 5 M or higher, concentrated sulfuric acid and concentrated strong alkalis are all allowed. For (b), iodine vapour is allowed. For (d), “amino acids have different Rf values in different solvents” on its own is not accepted. The point is that some did not separate first time.
Q7a
[2 marks]
A student hydrolyses a sample of the peptide endomorphin-2 to break it down into its constituent amino acids, then analyses the mixture by TLC. State a reagent and the conditions needed for the hydrolysis.
Show answer
Reagent: (aqueous) HCl: the name or formula of any strong acid or alkali. 1 mark
Conditions: reflux / heat. 1 mark
Warm, hot and high temperature are allowed for heat, but not a temperature above 200 °C. Alternatively a protease (peptidase) at warm temperature, and with an enzyme, “hot” or above 50 °C is not allowed. “Concentrated” and pressure are ignored.
Q7b
[3 marks]
The apparatus used for the TLC is a beaker containing the solvent, with the plate standing in it. There is a piece of the apparatus missing, and this omission will result in an inaccurate chromatogram. Identify the missing piece of apparatus, then state and explain why it is needed.
Show answer
Missing piece: a lid / cover on the beaker. 1 mark
Then any two of: it prevents the escape of vapour / evaporation of solvent from the beaker; so the atmosphere in the beaker is saturated with solvent vapour; to reduce evaporation from the plate. 2 marks
For the third point, “so the solvent can rise up the plate” and “to avoid the plate drying out” are both allowed.
Q7c
[2 marks]
After the solvent has risen up the plate, the student removes it and sprays it with a developing agent. Name a suitable developing agent and state why it is needed.
Show answer
Name: ninhydrin. 1 mark
Why: the amino acids are colourless / to make them visible. 1 mark
Iodine is allowed; UV is ignored here because the question asks for an agent to spray. Naming the final colour (“it turns them purple”) is not enough on its own.
Q8
[3 marks]
A tripeptide, L, is partially hydrolysed with concentrated hydrochloric acid to produce two dipeptides and the amino acids alanine (ala), lysine (lys) and serine (ser). The two dipeptides are separated by chromatography, giving spots M and N on the chromatogram below. Use the chromatogram and the Rf values in the table to identify the two dipeptides present in spots M and N, then deduce the order of the amino acids in tripeptide L.
| Dipeptide | ala-lys | ala-ser | lys-ser | lys-ala | ser-ala | ser-lys |
|---|---|---|---|---|---|---|
| Rf value | 0.55 | 0.85 | 0.10 | 0.20 | 0.15 | 0.45 |
Show answer
Spot M: ser-ala 1 mark
Spot N: ala-lys 1 mark
Order in tripeptide L: ser-ala-lys 1 mark (this order only)
Measuring gives Rf ≈ 0.15 for M and ≈ 0.55 for N. The two dipeptides overlap at alanine (ser-ala and ala-lys), so the shared residue sits in the middle and the tripeptide reads ser-ala-lys.
Source: AQA A-Level Chemistry past papers.
Enzymes & drug action
Enzymes are proteins, and they work as catalysts. The folded tertiary structure leaves a cavity (the active site) whose shape is complementary to one particular substrate molecule. The substrate binds there, reacts, and leaves; the enzyme is unchanged.
Because that cavity is a definite three-dimensional shape, it is stereospecific. A chiral substrate has two enantiomers that are non-superimposable mirror images, and only one of them is complementary to the site. The other cannot fit (a left hand in a right glove), so the enzyme catalyses the reaction of one enantiomer and leaves the other untouched.
Key definitions
The active site of an enzyme is a region of its surface with a shape complementary to the substrate, where the reaction is catalysed.
A stereospecific active site can bond to only one enantiomeric form of a substrate or drug, because only that enantiomer is complementary to its three-dimensional shape.
Drugs as enzyme inhibitors
That is also how a whole class of drugs works. A molecule shaped to fit the active site of an enzyme can bind there and block it, stopping the enzyme doing its job. The drug acts as an enzyme inhibitor. Because the fit has to be so precise, computers are used to design molecules with the right complementary shape before any of them are made.
Exam questions
Prilocaine is used as an anaesthetic in dentistry. Figure 1 shows the structure of prilocaine.
Q9a
[1 mark]
Draw a circle around any chiral centre(s) in Figure 1.
Show answer
One circled carbon atom only: the carbon attached to CH3, to the C=O, to H and to NH. 1 mark
It is the only carbon in the molecule with four different groups on it.
Q9b
[1 mark]
Identify the functional group(s) in the prilocaine molecule. Tick the box(es) corresponding to the functional group(s): amide, amine, ester, ketone.
Show answer
Two ticks only: amine and amide. 1 mark
The C=O with an N attached is an amide; the second nitrogen, bonded only to carbon and hydrogen, is a secondary amine. There is no C–O single bond to a second carbon, so no ester, and the C=O is not flanked by two carbons, so no ketone.
Q9c
[3 marks]
Prilocaine is completely hydrolysed in the human body to give a mixture of products. Draw the structures of the two organic products formed in the complete hydrolysis of prilocaine in acidic conditions.
Show answer
Choosing the correct bond to hydrolyse: the amide C–N bond, not the amine. 1 mark
The carboxylic acid product, HOOC–CH(CH3)–NH–CH2CH2CH3. 1 mark
The amine product, as the phenylammonium ion C6H5NH3+. 1 mark
The protonated form of the acid product is also allowed, since the conditions are acidic; C6H5NH3+ may be written with the charge outside a square bracket.
Q9d
[2 marks]
Isomers F and G are optical isomers. Isomer F is the active compound in the medicine ibuprofen. In the manufacture of ibuprofen both isomers F and G are formed. An enzyme is then used to bind to isomer G and catalyse its hydrolysis; after the products of hydrolysis of G are removed, a pure sample of isomer F is collected. Explain how a structural feature of this enzyme enables it to catalyse the hydrolysis of isomer G but not the hydrolysis of isomer F.
Show answer
The enzyme has an active site. 1 mark
The active site is stereospecific: it has the correct stereochemistry for, or a shape complementary to, the G enantiomer. 1 mark
The opposite argument (explaining that the site is not complementary to F) scores equally.
Source: AQA A-Level Chemistry past papers.
DNA: nucleotides & base pairing
DNA (deoxyribonucleic acid) is a polymer too, built from monomers called nucleotides. One nucleotide is three pieces bonded together: a phosphate ion, the pentose sugar 2-deoxyribose, and one of four bases (adenine, cytosine, guanine or thymine). All of those structures are printed in the Data Booklet.
Base pairing
The two strands are complementary and twisted around each other into a double helix. They are held together by hydrogen bonds between the base pairs, and the pairing is not a matter of choice: it is fixed by where the hydrogen-bond donors (N–H) and acceptors (a lone pair on N or O) sit on each base.
The two base pairs
- Adenine pairs with thymine: two hydrogen bonds.
- Cytosine pairs with guanine: three hydrogen bonds.
Worked example: reading a strand
One strand of a DNA fragment reads A–G–G–C–T. Give the sequence of the complementary strand, and calculate the total number of hydrogen bonds holding the fragment together.
Step 1. Pair each base: A with T, G with C, G with C, C with G, T with A. The complementary strand is T–C–C–G–A.
Step 2. Count the bonds pair by pair: A–T uses 2; each of the three G–C or C–G pairs uses 3; the final T–A uses 2.
2 + 3 + 3 + 3 + 2 = 13 hydrogen bonds
Notice the consequence: a stretch of DNA rich in C and G takes more energy to separate than one rich in A and T, because there are more hydrogen bonds per base pair to break.
Precision points
- The bonds along each strand are covalent, between the phosphate of one nucleotide and the 2-deoxyribose of the next. Only the bonds across the middle, between base pairs, are hydrogen bonds.
- A–T is two hydrogen bonds, C–G is three. Swapping them, or pairing A with G, is the standard error.
- A nucleotide is phosphate + 2-deoxyribose + base, all three, and the sugar is the one in the middle.
Exam question
Q10
[1 mark]
Which row shows a pair of bases that can link two strands of DNA with three hydrogen bonds? Use the Data Booklet to help you answer this question.
Cytosine–guanine is the three-hydrogen-bond pair. Adenine–thymine is a real pair but uses only two. Adenine–guanine (two purines) and cytosine–thymine (two pyrimidines) are not complementary at all.
Source: AQA A-Level Chemistry past papers.
Cisplatin: stopping DNA replicating
Cisplatin, [Pt(NH3)2Cl2], is a Pt(II) complex used as an anticancer drug. Platinum(II) has a co-ordination number of 4 and the complex is square planar: the four ligands sit at the corners of a square around the platinum, at 90° to each other. Two geometric arrangements are possible, and only the cis isomer, with the two chloride ligands next to each other, is the drug.
How it stops replication
Once cisplatin is inside a cell, chemistry you already know takes over: ligand substitution. First one chloride ligand is replaced by a water molecule. Then that water is replaced as a bond forms between the platinum and a nitrogen atom on a guanine base, and the remaining chloride is replaced the same way by a nitrogen on another guanine. The result is a cross-link: the platinum is bonded to two adjacent guanine bases on the same strand at once.
Why that stops a cancer cell
To divide, a cell must replicate its DNA. Locking two neighbouring guanines to the same platinum distorts the double helix, so the machinery that copies DNA cannot work along it: DNA replication is prevented and the rapidly dividing cancer cell dies.
Worked example: the first substitution
After cisplatin enters a cell, one chloride ligand is replaced by a water molecule to form a complex ion, B. Give the equation for this reaction.
Step 1. Work out what B is: take one Cl− out of [Pt(NH3)2Cl2] and put a neutral H2O in. Removing a 1− ligand and replacing it with a neutral one raises the charge by one, so B is [Pt(NH3)2Cl(H2O)]+.
Step 2. Balance, and account for the chloride: the Cl− that left is a product in its own right.
[Pt(NH3)2Cl2] + H2O → [Pt(NH3)2Cl(H2O)]+ + Cl−
Step 3. Check the charges balance: 0 on the left; (+1) + (−1) = 0 on the right. If the charges do not balance, the formula of the complex is wrong.
Benefits weighed against harm
Cisplatin cannot tell a cancer cell from a healthy one. Any rapidly dividing cell (hair follicles, bone marrow, the lining of the gut) is attacked in the same way, which is why the drug has serious adverse effects. Society has to weigh those against the benefit of treating the cancer, and that judgement is part of what you are expected to be able to discuss.
Precision points
- The process cisplatin prevents is DNA replication. “Mitosis”, “DNA synthesis”, “cell replication” and “damages DNA” are all rejected.
- It is the two chloride ligands that are replaced, never the ammonia ligands.
- Drawing the cross-link: the platinum must keep both NH3 ligands and must be drawn cis. A trans arrangement scores zero, though a charge on the Pt and any wedges or dashes are ignored.
- Show either the lone pairs on both nitrogen atoms or two arrows for the co-ordinate bonds. That is a mark in its own right.
Exam questions
Q11a
[1 mark]
Cisplatin, [Pt(NH3)2Cl2], is used as an anti-cancer drug. Cisplatin works by causing the death of rapidly dividing cells. Name the process that is prevented by cisplatin during cell division.
Show answer
DNA replication. 1 mark
Mitosis, DNA synthesis, DNA transcription, “cell replication” and “damages DNA” are not accepted.
Q11b
[2 marks]
After cisplatin enters a cell, one of the chloride ligands is replaced by a water molecule to form a complex ion, B. Give the equation for this reaction.
Show answer
[Pt(NH3)2Cl2] + H2O → [Pt(NH3)2Cl(H2O)]+ + Cl−
Correct formula and charge of B. 1 mark
Correct balancing and charges in the equation. 1 mark
The second mark can still be earned if the only error in B is its charge, provided Cl− is the other product. Complexes written without square brackets, or without brackets around H2O, are allowed; any additional different species loses the second mark.
Q11c
[2 marks]
When the complex ion B reacts with DNA, the water molecule is replaced as a bond forms between platinum and a nitrogen atom in a guanine nucleotide. The remaining chloride ligand is also replaced as a bond forms between platinum and a nitrogen atom in another guanine nucleotide. Complete a diagram of two adjacent guanine nucleotides to show how the platinum complex forms a cross-link between them.
Show answer
Pt drawn as a cis-diammine complex bonded to the correct nitrogen atoms. The platinum must show its two ammonia ligands. 1 mark
Both lone pairs shown, or two arrows indicating the co-ordinate bonds. 1 mark
Drawn as trans scores nothing for the first mark. Any charge on the Pt and any wedges or dashes are ignored. The second mark is still available even if the bonds to platinum are drawn from the wrong nitrogen atoms.
The figure above this exam block is the completed cross-link.
Source: AQA A-Level Chemistry past papers. Parts (d) to (f) of this question are a rate-equations and Arrhenius calculation — they are worked through on rate equations.
Capstone quiz
Five real past-paper multiple-choice questions across the whole topic. Read every option before you answer. Two of these turn on a single charge.
Capstone quiz: 5 past-paper questions. Pick an answer to see whether it is right and why.
Q12
[1 mark]
Which structure shows a zwitterion of an amino acid?
A zwitterion is neutral overall. Add the charges up: A is +1 (lysine, two protonated amines), B is −1 (aspartate, two carboxylates), C is −1 (a neutral amine and one carboxylate). Only D (cysteine, with –NH3+ and –COO− and a neutral side chain) carries one of each and nets to zero. The option structures are redrawn here; the chemistry and the answer are AQA’s.
Q13
[1 mark]
Which is the main species present in an aqueous solution of aspartic acid at pH = 14?
pH 14 is a huge excess of OH−, so every acidic proton comes off, and aspartic acid has two –COOH groups, one in the backbone and one in the side chain. Both become –COO− and the amine stays neutral, giving the 2− species. Forgetting the side-chain acid gives C. The option structures are redrawn here; the chemistry and the answer are AQA’s.
Q14
[1 mark]
Which type of interaction between polypeptide chains is mainly responsible for maintaining the secondary structure of a protein in the form of an α-helix?
Secondary structure (both the α-helix and the β-pleated sheet) is held by hydrogen bonds between the backbone C=O and N–H groups. Covalent S–S bridges and ionic interactions belong to tertiary structure, between R groups.
Q15
[1 mark]
Cisplatin has the formula [Pt(NH3)2Cl2] and is an anti-cancer drug that prevents replication of DNA. When cisplatin bonds to DNA, which is the correct ligand replacement reaction?
Both chloride ligands are replaced (one by a nitrogen on each of two adjacent guanine bases), while the two ammonia ligands stay on the platinum throughout.
Q16
[1 mark]
Which statement about enzymes is not correct?
The active site is stereospecific, so only one enantiomer is complementary to it. That is exactly why a single enantiomer of a drug is often the active one. The other three statements are all correct, which is what makes C the answer to a “not correct” question.
Source: AQA A-Level Chemistry past papers.
3.3.13 Amino acids, proteins & DNA: Quick-reference summary
- Amino acids are 2-aminocarboxylic acids, H2N–CHR–COOH. Six side chains are given in the Data Booklet; the α-carbon is chiral in all of them.
- Zwitterion: +H3N–CHR–COO−: both charges, neutral overall. In acid the cation +H3N–CHR–COOH; in alkali the anion H2N–CHR–COO−. Count every acidic and basic group, side chain included.
- Peptide link –CONH–, made by condensation with loss of H2O. Draw it in full, with the H on the nitrogen. Two amino acids give two different dipeptides depending on the order.
- Hydrolysis of the peptide link (reflux with aqueous HCl, or alkali) returns the amino acids: protonated in acid, as carboxylates in alkali.
- Protein structure: primary = the sequence; secondary = α-helix and β-pleated sheet, held by hydrogen bonds between backbone C=O and N–H; tertiary = the folded 3-D shape, held by S–S bridges (covalent, strongest), hydrogen bonds and ionic interactions between R groups.
- TLC: spots separate on the balance between solubility in the mobile phase and retention by the stationary phase; locate colourless amino acids with ninhydrin or UV; Rf = distance moved by the spotdistance moved by the solvent, measured from the start line.
- Enzymes are proteins with a stereospecific active site complementary to one substrate, so only one enantiomer binds. A drug can act as an inhibitor by blocking the site, and computers help design molecules that fit.
- DNA: nucleotide = phosphate + 2-deoxyribose + base; covalent bonds between the phosphate of one nucleotide and the sugar of the next give the sugar–phosphate backbone; two complementary strands form a double helix, hydrogen bonded A–T (2) and C–G (3).
- Cisplatin [Pt(NH3)2Cl2] is square planar and cis. Its two Cl− ligands are replaced as bonds form between Pt and a nitrogen atom on each of two adjacent guanine bases on the same strand, cross-linking the DNA so replication is prevented. It harms healthy dividing cells too, so the benefits must be weighed against the adverse effects.