Questions on amino acids, proteins, enzymes and DNA

Question 1

Evaluate the use of cisplatin, cis-[Pt(NH₃)₂Cl₂], as an anticancer drug, explaining how it stops cancer cells dividing.

Accepted Answer

Indicative content

inside the cell the chloride concentration is low, so the two chloride ligands are replaced, first by water and then by DNA
the platinum forms coordinate bonds to a nitrogen atom on each of two neighbouring guanine bases on one strand of DNA
this kinks the DNA so the strands cannot be separated and copied, replication is prevented and the cell cannot divide
cancer cells divide rapidly, so they are affected most, which is the benefit that makes cisplatin an effective treatment
it also acts on healthy cells that divide quickly, causing hair loss and sickness, and it damages the kidneys
the harm is limited by the lowest effective dose given in courses with breaks, and is judged worth the risk against an untreated cancer

Levels

Level 3 (5–6 marks) both the mechanism and the evaluation are complete: ligand replacement of chloride by water and then by DNA, coordinate bonds to the nitrogen atoms of two guanine bases, the prevention of replication and division, the benefit to rapidly dividing cancer cells, the harm to healthy dividing cells, and a judgement on balance
Level 2 (3–4 marks) platinum binding to DNA bases and the halt to replication are described with a benefit and a side effect; the ligand substitution, water and then guanine replacing chloride, or the final judgement is not given
Level 1 (1–2 marks) cisplatin is said to kill cancer cells, often by 'attacking' them or their membranes, with no binding to DNA and no harm to healthy cells weighed

Question 2

Give the type of reaction and the other product when glycine, H₂NCH₂COOH, and serine, H₂NCH(CH₂OH)COOH, join to form a dipeptide, and the condensed formulae of the two dipeptides that contain one of each.

Accepted Answer

B1 a condensation reaction, forming a peptide link, –CONH– B1 the other product is water B1 H₂NCH₂CONHCH(CH₂OH)COOH, with glycine's carbonyl joined to serine's nitrogen B1 H₂NCH(CH₂OH)CONHCH₂COOH, with serine's carbonyl joined to glycine's nitrogen

Question 3

Describe the primary, secondary and tertiary structure of a protein, and the bonds that maintain the secondary and tertiary structure.

Accepted Answer

B1 the primary structure is the sequence of amino acids in the polypeptide chain B1 the secondary structure is an α-helix or a β-pleated sheet B1 the secondary structure is held by hydrogen bonds between the N–H of one peptide link and the C=O of another B1 the tertiary structure is the folding of the whole chain, held by hydrogen bonds, ionic bonds and disulfide bridges between side chains

Question 4

Describe how the amino acids in a protein can be released and then identified by thin-layer chromatography.

Accepted Answer

B1 the protein is heated under reflux with 6 mol dm⁻³ hydrochloric acid, which hydrolyses the peptide links B1 the mixture is spotted on a TLC plate beside known amino acids, and the plate is stood in a solvent below the spots B1 the dried plate is sprayed with ninhydrin or viewed under ultraviolet light to show the colourless spots B1 Rf values are calculated and each spot is matched with a known amino acid on the same plate

Question 5

Describe the structure of a DNA molecule, from the nucleotide to the double helix.

Accepted Answer

B1 DNA is a polymer of nucleotides, each made of a phosphate group, 2-deoxyribose and a base B1 each strand has a backbone of alternating sugar and phosphate groups, with a base on each sugar B1 two strands are held together by hydrogen bonds between bases: adenine with thymine and cytosine with guanine B1 the two strands are twisted into a double helix

Question 6

Explain how computer modelling helps a research team that has the X-ray structure of an enzyme a bacterium needs to design a drug that inhibits it, and why the drug may have to be made as a single enantiomer.

Accepted Answer

B1 the shape of the active site and the groups lining it are known, so candidate molecules can be fitted into it on screen B1 each candidate is scored on how well it fits and the hydrogen bonds or ionic attractions it can make, so only the best are synthesised and tested B1 a molecule that binds the active site but does not react blocks the substrate, so the enzyme is inhibited B1 the active site is chiral, so only one enantiomer of a chiral drug fits; the other may be inactive or harmful

Question 7

Explain why glycine, H₂NCH(R)COOH with R = H, is not optically active, whereas alanine, with R = CH₃, is.

Accepted Answer

B1 optical activity needs a chiral centre, a carbon atom bonded to four different groups B1 the central carbon of glycine carries two hydrogen atoms, so it has no chiral centre and no enantiomers B1 the central carbon of alanine carries H, CH₃, NH₂ and COOH, four different groups, so alanine has two enantiomers

Question 8

Give the formula of the species serine, H₂NCH(CH₂OH)COOH, forms in a solution at pH 2 and in a solution at pH 12, and give the meaning of the term isoelectric point.

Accepted Answer

B1 at pH 2, ⁺H₃NCH(CH₂OH)COOH, both groups protonated B1 at pH 12, H₂NCH(CH₂OH)COO⁻; ignore ionisation of the side-chain OH B1 the pH at which the amino acid exists as a zwitterion with no overall charge

Question 9

Explain why alanine, Mr 89, is a solid that decomposes near 300 °C, whereas butanoic acid, Mr 88, is a liquid that freezes at about −5 °C.

Accepted Answer

B1 in alanine the COOH group donates a proton to the NH₂ group, forming the zwitterion ⁺H₃NCH(CH₃)COO⁻ B1 the zwitterions form an ionic lattice with strong attractions between opposite charges, which need a lot of energy to overcome B1 butanoic acid molecules are held only by hydrogen bonds and van der Waals forces, which are much weaker

Question 10

Explain how a disulfide bridge forms in a protein, what it does to the protein's structure, and why it survives warming that breaks hydrogen bonds.

Accepted Answer

B1 the SH groups in the side chains of two cysteine residues form a covalent S–S bond B1 the bond holds two parts of the chain together, fixing the folded tertiary structure B1 a covalent bond is far stronger than a hydrogen bond, so much more energy is needed to break it

Question 11

Deduce the two amino acids in a dipeptide that is hydrolysed and run on a TLC plate beside standards. The distances and the standards' Rf values are given in the table.

Accepted Answer

M1 Rf = distance moved by the spot ÷ distance moved by the solvent front, for each spot A1 spot 1: 1.9 ÷ 7.5 = 0.25, so glycine A1 spot 2: 4.4 ÷ 7.5 = 0.59, so valine

Question 12

Calculate the Mr of the tripeptide gly-ala-gly. Mr of glycine is 75.0, of alanine 89.0 and of water 18.0.

Accepted Answer

M1 two peptide links form, so two molecules of water are lost M1 Mr = 75.0 + 89.0 + 75.0 − 2 × 18.0 A1 Mr of the tripeptide = 203.0, with no unit

Question 13

Explain why an enzyme catalyses the reaction of only one enantiomer of a chiral substrate.

Accepted Answer

B1 the enzyme is built from one enantiomer of each amino acid, so its active site is chiral B1 only one enantiomer of the substrate has its groups in the positions to bind at the site's binding points B1 the other enantiomer does not fit, so it is not converted: the enzyme is stereospecific

Question 14

Suggest how a drug could be designed to inhibit an enzyme whose active-site shape is known.

Accepted Answer

B1 a molecule is designed with a shape similar to the substrate's, so that it binds in the active site B1 the molecule does not react, so it blocks the active site and stops the substrate binding B1 computer modelling is used to test how candidate molecules fit and bond in the site before they are made

Question 15

Explain why adenine pairs only with thymine and guanine only with cytosine in DNA, and state the number of hydrogen bonds in each pair.

Accepted Answer

B1 only these pairs line up hydrogen bond donors, N–H, opposite acceptors, N or O, at the right distance B1 adenine and thymine are joined by two hydrogen bonds B1 cytosine and guanine are joined by three hydrogen bonds

Question 16

Suggest the formula and the overall charge of the species that lysine, H₂N(CH₂)₄CH(NH₂)COOH, isoelectric point about 9.7, forms at pH 7.

Accepted Answer

B1 pH 7 is below the isoelectric point, so lysine holds more protons than in its neutral form B1 ⁺H₃N(CH₂)₄CH(NH₃⁺)COO⁻, both amine groups protonated and the acid group ionised B1 the overall charge is +1

Question 17

Explain why, in electrophoresis with a buffer at pH 6.0, alanine (isoelectric point 6.0) stays at the starting line while aspartic acid (isoelectric point 2.8) moves towards the positive electrode.

Accepted Answer

B1 at pH 6.0 alanine is at its isoelectric point, so it exists as the zwitterion with no overall charge B1 a species with no overall charge is not attracted to either electrode, so alanine does not move B1 pH 6.0 is above aspartic acid's isoelectric point, so both COOH groups have lost protons, the ion is negative and it is attracted to the positive electrode

Question 18

Give the equation for the esterification of glycine with ethanol, the conditions used, and the form the ester takes in the reaction mixture.

Accepted Answer

B1 H₂NCH₂COOH + CH₃CH₂OH ⇌ H₂NCH₂COOCH₂CH₃ + H₂O B1 heated with a concentrated sulfuric acid catalyst B1 in the acid the amine group is protonated, so the ester is present as its salt, ⁺H₃NCH₂COOCH₂CH₃

Question 19

Suggest a buffer pH that would make glycine (isoelectric point 6.0) and lysine (isoelectric point 9.7) move in opposite directions in electrophoresis.

Accepted Answer

B1 any pH between 6.0 and 9.7, such as pH 8 B1 at that pH glycine is above its isoelectric point and is negative, moving to the positive electrode, while lysine is below its isoelectric point and is positive, moving to the negative electrode

Question 20

Explain why cisplatin is supplied as the cis isomer alone rather than as a mixture with the trans isomer.

Accepted Answer

B1 the trans isomer cannot bind to two neighbouring guanines, so it does not stop cancer cells dividing B1 it would still be taken up by cells and add to the side effects, and the dose of the active isomer in a mixture would be uncertain

Measurement	Value
distance moved by solvent front	7.5 cm
distance moved by spot 1	1.9 cm
distance moved by spot 2	4.4 cm
Rf of glycine standard	0.25
Rf of alanine standard	0.38
Rf of valine standard	0.59
Rf of leucine standard	0.72