3.3.13 Amino Acids, Proteins and DNA

If you want to see why an amino acid is amphoteric, look at both functional groups. The –COOH group can donate H⁺ and the –NH₂ group can accept H⁺; internal proton transfer then gives the zwitterion with –NH₃⁺ and –COO⁻.

Amphoteric means it can react as both an acid and a base. The zwitterion is one ionic form that follows from having both functional groups in the same molecule.

Draw the zwitterion with –NH₃⁺ and –COO⁻. In acidic solution, protonation favours the –NH₃⁺/–COOH form; in alkaline solution, deprotonation favours –NH₂/–COO⁻.

Track the movement of H⁺ between the forms and then check the overall charge. Do not change the carbon skeleton or side chain when you change pH.

When you join amino acids into a polypeptide, the link you make is the peptide or amide bond, –CO–NH–. It forms by condensation between a –COOH group and an –NH₂ group with loss of H₂O.

For a chain of residues, keep the order fixed. Swapping the residue order gives a different peptide even when the same amino acids are present.

Hydrogen bonds between suitable polar groups help stabilise folded protein structures. Disulfide links are stronger covalent S–S bonds formed between appropriate sulfur-containing side chains and can lock distant parts of a chain together.

Keep the interaction types distinct: a disulfide bridge is a covalent bond, not a particularly strong hydrogen bond.

Primary protein structure is the amino-acid sequence. Secondary structure is local regular folding such as α-helices and β-sheets, stabilised mainly by hydrogen bonding; tertiary structure is the overall three-dimensional folding of one polypeptide chain produced by several types of interaction between side chains.

Classify what the diagram is actually showing before naming the level: sequence means primary structure; a local α-helix or β-sheet is secondary; the complete three-dimensional fold of one chain is tertiary. That avoids calling every visible fold “tertiary”.

Hydrolyse a peptide completely and you add the elements of water across the –CO–NH– links and ultimately recover the constituent amino acids, in the ionic forms appropriate to the conditions.

When reconstructing products, restore carboxyl and amino functionality at each broken peptide link and keep every side chain attached to its original α-carbon.

Spot amino-acid samples near the baseline, develop in a solvent, visualise the spots and compare Rf values with standards run under the same conditions.

Keep the practical details straight: put the solvent level below the spots, mark the origin in pencil, let the solvent rise, then locate the amino-acid spots before calculating or comparing Rᶠ values.

If the amino-acid spots are colourless on a TLC plate, you need a way to reveal them after chromatography. A locating agent such as ninhydrin or suitable UV visualisation makes the spots visible; you then use their positions or Rf values for comparison with standards.

Develop the chromatogram before using the locating method. Once the spots are visible, their positions can be measured and compared with standards; the locating reagent itself is not the identification.

To draw a dipeptide or tripeptide, join the amino acids in the stated sequence through –CO–NH– links. Each new peptide bond eliminates one H₂O, so a tripeptide contains two peptide bonds.

Write the residue order before drawing if necessary. The same three amino acids can form different sequences, so correct composition alone is not enough.

To show complete peptide hydrolysis, break every –CO–NH– link and restore the amino/carboxyl groups on the resulting residues. Keep all side chains unchanged.

If acidic or alkaline conditions are specified, use the corresponding ionic forms after you have identified the amino-acid skeletons. Do not lose track of residue identity while adjusting protonation.

When you read a protein-structure diagram, decide what scale it is showing. The amino-acid sequence is primary structure, a local α-helix or β-sheet is secondary structure, and the overall three-dimensional fold of one polypeptide is tertiary structure.

A visible helix is evidence of secondary structure; it is not by itself the whole tertiary structure. Use the scale of organisation the diagram is actually showing.

Hydrogen bonds can stabilise both regular secondary structures and parts of the tertiary fold. Covalent S–S links connect suitable sulfur-containing side chains and provide stronger cross-links within the tertiary structure.

When a question asks what holds a particular feature together, identify the interaction shown in that feature; listing every force you know does not answer the question.

Rf = distance travelled by substance ÷ distance travelled by solvent front, measured from the same origin. Rf has no units and lies between 0 and 1 for a normal chromatogram.

A useful sanity check is Rᶠ ≤ 1 because the spot cannot travel farther than the solvent front in a normal TLC run. If you obtain a value above 1, you have almost certainly reversed the distances or measured from different origins.

For an enzyme, connect function to folding. The amino-acid sequence folds into a particular three-dimensional structure, and that shape creates the active site that determines which substrates bind and how catalysis occurs.

Changing the protein structure can change the active site even though many peptide bonds remain intact. Function depends on the folded shape and the active site that shape creates.

Make the active site three-dimensional and specific in an enzyme-catalysis explanation. A suitable substrate binds through complementary shape and intermolecular interactions, forming an enzyme–substrate complex and giving a lower-energy route.

“Fits the active site” is a useful start, but the chemistry is the specific interactions and orientation produced by the stereospecific site.

For competitive-style inhibition, picture the inhibitor occupying the same active site the substrate needs. If it can make enough of the right interactions there, it blocks substrate binding and reduces the rate.

The inhibitor does not need to look identical to the substrate; it needs enough complementary features to compete successfully for the same binding region.

Computer modelling can predict how candidate molecules fit and interact with an enzyme active site, helping chemists prioritise structures for synthesis and testing. It supports experiments; it does not replace biological testing.

Think of modelling as a screening step. You can compare predicted shape, charge and intermolecular contacts for many candidates, then send the most promising structures forward for synthesis and experimental testing.

If the active site is chiral, the two enantiomers do not present their groups in exactly the same three-dimensional arrangement. One may line up the required interactions much better than the mirror image, so you can get very different binding strengths.

The same functional groups occupy different positions in the two mirror images, so one enantiomer may make the required set of contacts in the active site while the other cannot line them up simultaneously. That changes binding strength.

Use the data-booklet structures of phosphate, 2-deoxyribose and the four bases as supplied. The skill is recognising the correct atoms/sites used to assemble nucleotides and DNA, not memorising every atom in those structures.

Read the supplied structures carefully before joining anything. This avoids drawing a plausible-looking bond at the wrong atom.

A nucleotide contains one phosphate group, one 2-deoxyribose unit and one base. The base attaches to the sugar, and phosphate also connects through the sugar framework.

Build the unit from the supplied structures and preserve the correct attachment atoms. The base is not bonded directly to the phosphate group.

Trace one DNA strand along the repeating sugar–phosphate backbone. Covalent links join phosphate groups to 2-deoxyribose units, while the bases project from the sugars.

When identifying the backbone, follow the repeating phosphate–sugar sequence. The bases carry the sequence information but are not part of that repeating backbone.

When you draw the DNA backbone, keep the bases attached to the 2-deoxyribose units and out of the repeating phosphate–sugar chain. The backbone itself is built from covalently linked phosphate and 2-deoxyribose units.

Place each base on the deoxyribose unit at the correct attachment site shown by the supplied structures. The base does not form part of the repeating phosphate–sugar backbone.

Picture DNA at the next level up: you have two complementary polynucleotide strands arranged as a double helix. Each strand has a phosphate–2-deoxyribose backbone with the bases projecting inward for pairing.

Complementary base pairing joins the two strands: A pairs with T and C with G. That means the base sequence on one strand determines the matching sequence on the other.

For complementarity, I’d learn the specific pairs first: A with T and C with G. Their hydrogen-bond donor and acceptor patterns fit one another, so the sequence on one strand determines the matching sequence on the other.

Complementary pairing means the sequence on one strand constrains the sequence on the other: wherever one strand has A the partner has T, and wherever it has C the partner has G. The strands are held together by those specific non-covalent interactions.

For cisplatin, keep the geometry attached to the name: it is a cis, square-planar Pt(II) complex. The adjacent chloride positions matter because that arrangement allows the platinum centre to form the DNA cross-links associated with its anticancer action.

In cis-[Pt(NH₃)₂Cl₂], the two chloride ligands occupy adjacent positions around Pt(II). That cis geometry is part of the identity of the drug; changing the relative ligand positions gives a different complex with different biological behaviour.

Ligand substitution lets platinum bind to nitrogen atoms on guanine bases, producing links that distort DNA and interfere with replication.

Follow the causal chain all the way through: Pt–N bonding can make cross-links within DNA, the helix is distorted, and the replication machinery can no longer copy the DNA normally. That is how the coordination chemistry reaches the cell-division effect.

For cisplatin as a drug, link both sides of the judgement to the same chemistry. DNA reactivity can inhibit replication in cancer cells, but it can also damage healthy dividing cells, causing toxicity and limiting the dose.

A balanced answer links both the benefit and the harm to the drug’s chemistry. A generic “pros and cons” statement does not do that.

If platinum forms cross-links in DNA, follow the consequence: the DNA structure is distorted, normal replication is disrupted and cell division can be inhibited. That is the molecular-to-biological chain I’d want in the explanation.

Connect the molecular event to the biological consequence: Pt–DNA bonding → structural distortion/cross-linking → disrupted replication.

Remember that cisplatin’s DNA reactivity is not exclusive to tumour cells when you explain its side effects. Healthy cells that divide rapidly can also be damaged, so the same mechanism that gives anticancer activity can produce toxicity.

That lack of complete selectivity helps explain the adverse effects and limits the dose that can be used safely.