Vision is dependent on retinol (vitamin A) present in retina cells. Retinol is oxidized to the photosensitive chemical 11-cis-retinal and isomerizes to 11-trans-retinal on absorption of light.

Outline how the formation of 11-trans-retinal results in the generation of nerve signals to the brain.

11-trans retinal no longer fits into the rhodopsin/protein
OR
11-trans retinal is ejected from the rhodopsin/protein

leads to conformational change in rhodopsin/protein «to opsin generating signals»

[2 marks]

Explain the shape of the curve at low oxygen partial pressure up to about 5 kPa.

Sketch a graph on the axes above to show the effect of decreasing pH on the binding of oxygen to hemoglobin (the Bohr Effect).

Outline the effect of decreasing pH on the oxygen saturation of hemoglobin.

binding of O₂ «to one active site» affects shape of Hb/other active sites
OR
binding of one O₂ «molecule» affects binding of other O₂ «molecules»

increasing affinity of Hb to O₂
OR
enhanced binding of «further» O₂ «molecules»
OR
cooperative binding

[2 marks]

sketching right shift of curve on graph

[1 mark]

decreases «oxygen saturation»

Accept “hemoglobin binds to O₂ with less affinity".

[1 mark]

Draw the structures of the main form of glycine in buffer solutions of pH 1.0 and 6.0.

The pK_a of glycine is 2.34.

Calculate the pH of a buffer system with a concentration of 1.25 × 10⁻³ mol dm⁻³ carbonic acid and 2.50 × 10⁻² mol dm⁻³ sodium hydrogen carbonate. Use section 1 of the data booklet.

pK_a (carbonic acid) = 6.36

Sketch the wedge and dash (3-D) representations of alanine enantiomers.

UV-Vis spectroscopy can be used to determine the unknown concentration of a substance in a solution.

Calculate the concentration of an unknown sample of pepsin with an absorbance of 0.725 using section 1 of the data booklet.

Cell length = 1.00 cm

Molar absorptivity (extinction coefficient) of the sample = 49650 dm³ cm⁻¹ mol⁻¹

A different series of pepsin samples is used to develop a calibration curve.

Estimate the concentration of an unknown sample of pepsin with an absorbance of 0.30 from the graph.

Penalize charge on incorrect atom once only.

Penalize missing hydrogens or incorrect bond connectivities once only in Option B.

Accept condensed structural formulas.

Accept skeletal structures.

[2 marks]

ALTERNATIVE 1

«pH = 6.36 + log $\left(\frac{2.50\times {10}^{-2}}{1.25\times {10}^{-3}}\right)$ =»

7.66

ALTERNATIVE 2

«K_a = 4.4 × 10^–7 = [H⁺] $\left(\frac{2.50\times {10}^{-2}}{1.25\times {10}^{-3}}\right)$, [H⁺] = 2.2 × 10^–8 mol dm^–3»

«pH =» 7.66

Do not accept “«pH =» 8”.

[1 mark]

Penalize missing hydrogens or incorrect bond connectivities once only in Option B.

Wedges AND dashes must be used.

[1 mark]

«$\frac{0.725}{49650\text{ d}{\text{m}}^3\text{c}{\text{m}}^{-1}\text{mo}{\text{l}}^{-1}\times 1.00\text{cm}}=$» 1.46 × 10⁻⁵ «mol dm⁻³»

[1 mark]

0.65 «μg cm^–3»

Accept any value in the range 0.60–0.70 «μg cm^–3»

[1 mark]

State the name of the component of DNA responsible for the migration of its fragments to the positive electrode in gel electrophoresis.

In 2010, scientists claimed that they had discovered a species of bacteria capable of incorporating arsenic in place of phosphorus into the bacterial DNA. This claim has since proved controversial. Suggest one technique or evidence that might help support the claim.

phosphato/phosphate «group»  

Do not accept “phosphoric acid”, “phosphorus” or any formula.

mass spectrometry / X ray diffraction/crystallography / nuclear magnetic resonance «spectroscopy»
OR
bacteria able to grow in absence of phosphorus
OR
reproducible data

Accept abbreviations (eg, MS, NMR).
Accept “elemental analysis” or “atomic absorption spectroscopy/AA(S)”.

Amino acids act as buffers in solution. In aspartic acid, the side chain (R group) carboxyl has pK_a = 4.0. Determine the percentage of the side chain carboxyl that will be ionized (–COO^–) in a solution of aspartic acid with pH = 3.0. Use section 1 of the data booklet.

«–COOH $\rightleftharpoons$ –COO^– + H⁺    (–COOH = HA ; –COO^– = A^–)»
pH = pK_a + log $\frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{H}\mathrm{A}\right]}$ / 3.0 = 4.0 + log $\frac{\left[-\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\right]}{\left[-\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$ / –1.0 = log $\frac{\left[-\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\right]}{\left[-\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$

10^–1=$\frac{\left[-\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\right]}{\left[-\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$

«percentage ionized/–COO^– = $\frac{1}{1+10}$ × 100 =» 9.1 «%»
Award [3] for correct final answer.

Outline why this molecule absorbs visible light.

With reference to its chemical structure, outline whether this pigment is found in aqueous solution in the cells or in the lipid-based membranes.

A student investigated the ability of anthocyanins to act as pH indicators. He extracted juice from blackberries and used a UV-vis spectrophotometer to produce absorption spectra at different pH values. His results are shown below.

Deduce the colour of the juice at each pH, giving your reasoning. Use section 17 of the data booklet.

«extensive» conjugation «of double bonds»/delocalization «of electrons»
OR
«many» alternating single/C–C AND double/multiple/C=C bonds

in aqueous solution AND hydroxyl/OH/ionic/oxonium/O⁺ «groups»

Accept “polar/hydroxy” for “hydroxyl”.
Do not accept “OH^- /hydroxide/ oxygen”.

pH 2: «absorption peak 520 nm» red AND pH 11: «absorption peak 620 nm» blue

complementary/opposite colour observed «to wavelength absorbed»
OR
pH 2: «absorption peak 520 nm» green absorbed AND pH 11: «absorption peak 620 nm» orange absorbed

Award [1 max] if colour absorbed and colour observed are correct for either at pH 2 or pH 11.

(i) Estimate the K_m values of the two enzymes.

(ii) Suggest, with a reason, which enzyme will be more responsive to changes in the concentration of glucose in the blood.

(i) Outline what is meant by product inhibition as it applies to hexokinase.

(ii) Product inhibition of hexokinase does not affect its K_m value. Using this information, deduce the type of binding site that the inhibitor attaches to.

i

K_m hexokinase: approx. 1.7 «mmol dm^–3»
AND
K_m glucokinase: approx. 8.5 «mmol dm^–3»

Accept answers in the range 1.0-2.0 for hexokinase and 7.0-9.0 for glucokinase.

ii

glucokinase as it is not saturated «with substrate at normal concentration of blood glucose»
OR
glucokinase as its saturation increases with increased glucose concentration in the blood

Accept “at the normal levels of blood glucose concentration, relative velocity of glucokinase still dependent on concentration of glucose”

i

glucose-6-phosphate lowers enzyme activity/acts as enzyme inhibitor

ii

«inhibitor binds at» allosteric site

Accept “outside/away from active site”.

The diverse functions of biological molecules depend on their structure and shape.

Classify vitamins A, C and D as either mainly fat- or water-soluble, using section 35 of the data booklet.

The diverse functions of biological molecules depend on their structure and shape.

Deduce the straight chain structure of deoxyribose from its ring structure drawn in section 34 of the data booklet.

The diverse functions of biological molecules depend on their structure and shape.

Draw the nitrogenous base that is paired with guanine in DNA, showing the hydrogen bonds between the bases. Use section 34 of the data booklet.

The diverse functions of biological molecules depend on their structure and shape.

Retinal is the key molecule involved in vision. Explain the roles of cis- and trans-retinal in vision and how the isomers are formed in the visual cycle.

all three correct ✔

–CH₂– must be placed next to $\mathrm{CHO}$ AND 2OHs on central carbons must be on same side (LHS or RHS) ✔

Accept crosses in place of $C$ on three middle carbons.

cytosine drawn ✔

appropriate representation of three hydrogen bonds AND between correct atoms ✔

Structure of cytosine must be given for M1.

Ignore missing hydrogens on carbon atoms in cytosine.

Dashed lines (horizontal or vertical) OR dots can be used to represent hydrogen
bonds.

Only award M2 if M1 correct.

Any three of:
cis-retinal binds to «the protein» opsin
OR
cis-retinal «binds to opsin and» forms rhodopsin ✔

opsin extends conjugation in retinal
OR
conjugation in rhodopsin is larger/more extended/involves more atoms than that in retinal
OR
rhodopsin allows absorption of visible/blue/green light ✔

when visible light is absorbed cis-retinal changes to trans-retinal ✔

change «to trans-retinal» triggers an electrical/nerve signal ✔

trans-retinal detaches from opsin AND is converted back to cis-retinal
OR
trans-retinal is converted back to cis-retinal through enzyme activity ✔

Classification of the three vitamins A, C and D as fat-soluble, water-soluble and fat-soluble was very well answered and nearly all scored the single mark here.

The straight chain structure of deoxyribose was poorly represented. All sorts of errors were seen, such as not having –CHO at one end, not having the methylene group adjacent to –CHO etc.

This question proved to be very challenging and although many were able to identify and draw cytosine correctly, few were able to score M2 which required the representation of the multiple hydrogen bonds between guanine and cytosine. It was very disappointing that such a high percentage of candidates did not understand the nature of a hydrogen bond and equally tried to represent a hydrogen bond using a full line notation (typical of a covalent bond), instead of the more customary dashed lines (either horizontal or vertical) or dots type representation, used to represent hydrogen bonds.

This question which centred on retinal as the key molecule in vision was well answered and many scored all three marks. The most common incomplete answer was "cis-retinal changing to trans-retinal" without stating that this occurs when visible light is absorbed. Some also were unclear of the difference between opsin and rhodopsin.

Proteins are polymers of amino acids.

A paper chromatogram of two amino acids, A1 and A2, is obtained using a non-polar solvent.

© International Baccalaureate Organization 2020.

Determine the ${\mathrm{R}}_{\mathrm{f}}$ value of A1.

Proteins are polymers of amino acids.

The mixture is composed of glycine, $\mathrm{Gly}$, and isoleucine, $\mathrm{Ile}$. Their structures can be found in section 33 of the data booklet.

Deduce, referring to relative affinities and ${\mathrm{R}}_{\mathrm{f}}$, the identity of A1.

Proteins are polymers of amino acids.

Glycine is one of the amino acids in the primary structure of hemoglobin.

State the type of bonding responsible for the α-helix in the secondary structure.

Proteins are polymers of amino acids.

Sketch and label two oxygen dissociation curves, one for adult hemoglobin and one for foetal hemoglobin.

Proteins are polymers of amino acids.

Explain why the affinity for oxygen of foetal hemoglobin differs from that of adult hemoglobin.

$0.70$ ✔

Accept any value within the range “$\mathit{0}\mathit{.}\mathit{67}\mathit{-}\mathit{0}\mathit{.}\mathit{73}$”.

Ile AND larger R_f ✔

more soluble in non-polar solvent «mobile phase»
OR
not as attracted to polar «stationary» phase ✔

Only award M2 if Ile is identified in M1.

hydrogen/$\mathrm{H}$ bonding «between amido hydrogen and carboxyl oxygen atoms» ✔

both curves sigmoidal shape AND starting at zero ✔

foetal hemoglobin showing greater affinity/steeper/higher gradient ✔

Do not penalize if convergence is not approached for M1.

Both curves must be labelled to score M2.

Any two of:

contains two gamma/$\gamma$ units «instead of two beta/$\beta$ units found in adults»
OR
differs in amino acid sequence «from the two beta//$\beta$ units found in adults» ✔

less sensitive to inhibitors/2,3-BPG/DPG ✔

receives ${\mathrm{O}}_2$ from «partly deoxygenated» blood so can work at low $p{\mathrm{O}}_2$ ✔

low $p{\mathrm{CO}}_2$ in foetal blood increases affinity for ${\mathrm{O}}_2$ ✔

hemoglobin concentration in foetal blood greater than in the mother ✔

It was surprising that more did not manage to determine the Rf value of A1 within the acceptable range of "0.67 to 0.73". An out of range value of 0.75 was frequently seen.

Few candidates recognised that A1 was Ile so no marks were scored. Even of those candidates that did identify A1 as IIe, many did not mention larger Rf and therefore M1 was lost. There appeared to be in general, poor understanding of the basic principles of paper chromatography.

A majority stated that hydrogen bonding is the bonding responsible for the alpha-helix in the secondary structure.

The better candidates scored both marks in this question on oxygen dissociation curves. Many scored M2 for showing foetal hemoglobin having a higher gradient. Frequently sigmoidal curves were not drawn which lost M1 and often both curves were not explicitly labelled which was necessary to score M2.

This question required an explanation as to why the affinity for oxygen of foetal hemoglobin differs from that of adult hemoglobin. This question was poorly answered and only the stronger candidates managed to score both marks, usually by stating that foetal hemoglobin contains two gamma units instead of the two beta units found in adult hemoglobin, required for M1 and often by stating that there is less sensitivity to inhibitors for M2.

Deduce the pH range in which glycine is an effective buffer in basic solution.

Enzymes are biological catalysts.

The data shows the effect of substrate concentration, [S], on the rate, v, of an enzyme-catalysed reaction.

Determine the value of the Michaelis constant (K_m) from the data. A graph is not required.

Outline the action of a non-competitive inhibitor on the enzyme-catalysed reaction.

The sequence of nitrogenous bases in DNA determines hereditary characteristics.

Calculate the mole percentages of cytosine, guanine and thymine in a double helical DNA structure if it contains 17% adenine by mole.

«pH range» 8.6–10.6

Accept any value between 8.2 and 11.0.

[1 mark]

«K_m =» 0.67 «mmol dm^–3»

Do not penalize if a graph is drawn to determine the value.

[1 mark]

does not compete for active site
OR
binds to allosteric site/away from «enzyme» active site
OR
alters shape of enzyme

reduces rate/V_max

[2 marks]

«% cytosine + % guanine = 100% – 17% – 17% = 66%»

Cytosine: 33 «%» AND Guanine: 33 «%»

Thymine: 17 «%»

[2 marks]

Identify the structural feature which enables rhodopsin to absorb visible light.

Outline the change that occurs in the retinal residue during the absorption of visible light.

«extensive system of» conjugation/alternating single and double «carbon to carbon» bonds
OR
delocalized electrons «over much of the molecule»

Accept “delocalization”.

[1 mark]

cis«-retinal» converts to trans«-retinal»
OR
one of the C=C «fragments in retinal» changes «its configuration» from cis to trans

[1 mark]

An aqueous buffer solution contains both the zwitterion and the anionic forms of alanine. Draw the zwitterion of alanine.

Calculate the pH of a buffer solution which contains 0.700 mol dm^–3 of the zwitterion and 0.500 mol dm^–3 of the anionic form of alanine. 

Alanine pK_a = 9.87.

Penalize incorrect bond linkages or missing hydrogens once only in 8 (a) and 8 (c) (i).

[1 mark]

«pH = 9.87 + log$\left(\frac{0.500}{0.700}\right)$»

«= 9.87 – 0.146»

= 9.72

pH can be deduced by an alternative method.

[1 mark]

Outline how its structure allows it to be negatively charged in the body.

Deduce the nucleotide sequence of a complementary strand of a fragment of DNA with the nucleotide sequence –GACGGATCA–.

phosphate groups «in nucleotides fragments are almost completely» ionized

Do not accept just “phosphate «groups»”.

[1 mark]

–CTGCCTAGT–

[1 mark]

Explain the shape of the curve from 0 to X kPa.

Explain why carbon monoxide is toxic to humans.

oxygen binds to first active site «of deoxygenated heme» AND alters shape of other active sites
OR
cooperative binding

affinity of partially oxygenated hemoglobin for oxygen increases

[2 marks]

CO is a competitive inhibitor «of oxygen binding to hemoglobin»
OR
CO has greater affinity for hemoglobin «than oxygen»

less oxygen is transported
OR
uptake of oxygen decreases
OR
causes hypoxia

Do not penalize “CO binds irreversibly” if included in answer.

[2 marks]

Determine the value of the Michaelis constant, K_m, including units, from the graph.

Sketch a second graph on the same axes to show how the reaction rate varies when a competitive inhibitor is present.

Outline the significance of the value of K_m.

«K_m = [substrate] at $\frac{1}{2}$ V_max»

4.2 x 10^–3

mol dm^–3

Accept answers in the range of 3.5 x 10^–3 to 5.0 x 10^–3 mol dm^–3.

M2 can be scored independently.

[2 marks]

graph to right of curve AND finish at same V_max

Do not penalize if curve does not finish exactly at same V_max as long as it is close to it (since drawn curve does not
flatten out completely at V_max = 0.50).

[1 mark]

K_m is inverse measure of affinity of enzyme for a substrate / K_m is inversely proportional to enzyme activity
OR
high value of K_m indicates higher substrate concentration needed for enzyme saturation
OR
low value of K_m means reaction is fast at low substrate concentration

Idea of inverse relationship must be conveyed.

Accept “high value of K_m indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.

Accept “low value of K_m indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.

[1 mark]

State the half-equation for the reduction of molecular oxygen to water in acidic conditions.

Outline the change in oxidation state of the iron ions in heme groups that occurs when molecular oxygen is converted to water.

O₂ + 4H⁺ (aq) + 4e^– → 2H₂O (l)

Accept any balanced equation with any integer or fractional coefficients.

+2 to +3

OR

+1

OR

increases «by 1»

The stability of DNA is due to interactions of its hydrophilic and hydrophobic components.

Outline the interactions of the phosphate groups in DNA with water and with surrounding proteins (histones).

Water:
hydrogen/H-bonds

OR

ion–dipole interactions

Proteins:
ionic bonds/interactions

OR

hydrogen/H-bonds

OR

ion–dipole interactions

Ignore “London/dispersion/vdW/dipole–dipole interactions” stated for water and/or proteins.

Describe how DNA determines the primary structure of a protein such as insulin.

triplet/genetic code

OR

sequence of three bases/nucleotides

instruction for «particular» amino acid

[2 marks]

Determine the value of the Michaelis constant, K_m, by annotating the graph.

The malonate ion acts as an inhibitor for the enzyme.

Suggest, on the molecular level, how the malonate ion is able to inhibit the enzyme.

Draw a curve on the graph above showing the effect of the presence of the malonate ion inhibitor on the rate of reaction.

K_m labelled on x-axis as the [succinate ion] at ½V_max

OR

horizontal line at ½V_max AND vertical line down to x-axis

«K_m ＝» 6.5 x 10^–3 «mol dm^–3»

Annotation of graph required for M1.

Accept any specific value in the range 6.0 x 10^–3 to 7.5 x 10^–3 «mol dm^–3».

similar shape/size/structure «as succinate ion/substrate»

competes for the active site «with the succinate ion/substrate»

Accept “competitive inhibitor” for M2.

Award [1 max] if non-competitive inhibition is correctly described.

same V_max reached at higher [substrate]

Explain with reference to the binding site on the enzyme how a non-competitive inhibitor lowers the value of V_max.

Outline the significance of the value of the Michaelis constant, K_m.

binds at allosteric site

OR

binds away from active site

changes shape of active site

OR

renders active sites ineffective

[2 marks]

K_m is inverse measure of affinity of enzyme for a substrate

OR

K_m is inversely proportional to enzyme activity

OR

high value of K_m indicates higher substrate concentration needed for enzyme saturation

OR

low value of K_m means reaction is fast at low substrate concentration

Idea of inverse relationship must be conveyed.

Accept “high value of Km indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.

Accept “low value of Km indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.

[1 mark]

Hemoglobin’s oxygen dissociation curve is shown at a given temperature. Sketch the curve on the graph at a higher temperature.

Outline two differences between normal hemoglobin and foetal hemoglobin.

curve below original curve «showing lower affinity for oxygen» beginning at 0

Award mark if end of student curve does not finish at same location as original curve.

[1 mark]

Any two of:

foetal hemoglobin has higher affinity for oxygen «than normal hemoglobin»

foetal hemoglobin is less sensitive to inhibitors/2,3-bisphosphoglycerate/2,3-BPG/DPG «than normal hemoglobin»

foetal hemoglobin contains two gamma units instead of the two beta units found in adult hemoglobin

[2 marks]

Name the type of link between the two monosaccharide residues.

Outline how the two monomer structures, galactose and glucose, differ.

Outline the difference between their structures.

Outline why cellulose is an essential part of human diet.

«1,4-»glycosidic ✔

Do not accept “glucosidic”.

H and OH are reversed/in different positions on C-4 ✔

C-4 must be specified.

Do not penalize if reference is made to H and OH above and below ring/in alpha and beta positions on C-4 incorrectly.

Starch: α«-glucose/links»

AND

Cellulose: β«-glucose/links» ✔

Accept “Starch: coiled/spiral structure OR cross-links AND Cellulose: uncoiled OR straight chains/linear polymer OR no/few cross-links”.

Any two of:

helps food pass through intestine

OR

adds bulk/dietary fibre ✔

reduces appetite

OR

helps prevent obesity ✔

prevents constipation

OR

reduces risk of hemorrhoids/diverticulosis/Crohn’s disease/irritable bowel syndrome/bowel cancer ✔

A graph showing saturation of oxygen against partial pressure of oxygen is shown.

Explain the shape of the graph from 0 to 50 % saturation.

Explain why carbon monoxide is very toxic and how it may be possible to treat carbon monoxide poisoning.

binding of oxygen/O₂ «to one active site» affects shape of Hb/other active sites

OR

binding of one oxygen/O₂ «molecule» affects binding of other oxygen/O₂ «molecules» ✔

increasing affinity of Hb to oxygen/O₂

OR

enhanced binding of «further» oxygen/O₂ «molecules»

OR

cooperative binding ✔

Toxicity:

carboxyhemoglobin/Hb–CO does not readily dissociate

OR

CO + Hb $\rightleftharpoons$ Hb–CO AND forward reaction favoured

OR

affinity of carbon monoxide/CO for hemoglobin is «200 times/much» higher than that of oxygen/O₂

OR

competitive inhibitor of oxygen/O₂ binding ✔

Treatment:

moving patient to fresh air

OR

«in severe cases» inhaling pure oxygen/O₂

OR

high pressure oxygen/O₂ chamber ✔

Accept “move away from carbon monoxide/CO source” OR “remove carbon monoxide/CO source”.

Identify the type of inhibition shown in the graph.

Determine the value of $V_{\max }$ and $K_{\mathrm{m}}$ in the absence and presence of the inhibitor.

Outline the significance of the value of the Michaelis constant, $K_{\mathrm{m}}$.

non-competitive «inhibition» ✔

✔✔✔

Award [3] for four values correct.
Award [2] for three values correct.
Award [1] for two values correct.
Ignore units.
Accept $\mathit{\pm }\mathit{0}\mathit{.}\mathit{1}$ for $K_{\mathit{m}}$ and $V_{max}$. No ECF applied.

$K_{\mathrm{m}}$ is an inverse measure of affinity of substrate for enzyme
OR
higher $K_{\mathrm{m}}$ indicates higher substrate concentration is needed for enzyme saturation
OR
low value of $K_{\mathrm{m}}$ means reaction is fast at low substrate concentration ✔

Idea of “inverse relationship” must be conveyed.

Most scored the one mark here for identifying the correct type of inhibition, namely, non-competitive inhibition.

The stronger candidates scored all three marks here for the four correct values. Most scored at least one mark for the V_max values in the absence and presence of the inhibitor. The Km values sometimes were outside the permitted ±0.1 range.

The stronger candidates scored all three marks here for the four correct values. Most scored at least one mark for the V_max values in the absence and presence of the inhibitor. The Km values sometimes were outside the permitted ±0.1 range.

State the feature of DNA that determines the primary structure of proteins synthesised by a cell.

Suggest one concern about the use of genetically modified, GM, food.

«triplet» sequence/«specific» order of «nitrogenous» bases

OR

codon ✔

Any one of:

long-term «health» effects unknown ✔

can cause allergic reactions ✔

possible transfer of genetic material to other/wild species ✔

concern that power over farming is concentrated in hands of multinationals

OR

dependent on multinationals ✔

labelling differences between countries «means informed choice not possible» ✔

Accept “outcrossing”.

Draw the structure of the dipeptide Asp–Phe using section 33 of the data booklet.

Describe, using another method, how a mixture of four amino acids, alanine, arginine, glutamic acid and glycine, could be separated when placed in a buffer solution of pH 6.0.

Suggest why alanine and glycine separate slightly at pH 6.5.

Calculate the ratio of [A⁻] : [HA] in a buffer of pH 6.0 given that pK_a for the acid is 4.83, using section 1 of the data booklet.

amide link (eg, CONH) ✔

correct order and structures of amino acids ✔

NOTE: Accept a skeletal formula or a full or condensed structural formula.
Accept zwitterion form of dipeptide.
Accept CO–NH but not CO–HN for amide link.

Any three of:
«gel» electrophoresis «technique»
OR
mixture «in buffer solution» placed on gel/paper ✔

voltage/potential «difference» applied ✔

amino acids move differently «depending on pH/isoelectric point» ✔

compare/measure distances travelled/R_f values ✔

NOTE: Accept “mixture placed on plate covered with polyacrylamide «gel» OR “mixture put in a gel «medium»”.

different sizes/molar masses/chain lengths «so move with different speeds» ✔

NOTE: Do not accept “different side-chains/R-groups/number of carbons”.

«6.0 = 4.83 + log $\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$»

«log$\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$= 1.17»

«[A⁻] : [HA] =» 14.8 : 1 ✔

NOTE: Accept “15:1”.
Do not accept 1:14.8.

The graph shows the change in oxygen partial pressure in blood, measured at different pH values.

Explain the effect of changing pH on the percentage saturation of hemoglobin at a given partial pressure of oxygen.

Explain the biomagnification of the pesticide DDT.

Vitamins are organic compounds essential in small amounts.

State the name of one functional group common to all three vitamins shown in section 35 of the data booklet.

as pH decreases, protons/CO₂ bind to allosteric sites
OR
as pH decreases, protons/CO₂ act as non-competitive inhibitor
OR
active/binding site changes shape ✔

saturation decreases
OR
more oxygen released
OR
affinity to oxygen decreases ✔

accumulates in fat/tissues/living organisms
OR
cannot be metabolized/does not break down «in living organisms»
OR
not excreted / excreted «very» slowly ✔

passes «unchanged» up the food chain
OR
increased concentration as one species feeds on another «up the food chain» ✔

NOTE: Accept “lipids” for “fat”.

hydroxyl ✔

NOTE: Accept “hydroxy” but not “hydroxide”.
Accept “alkenyl”.

Do not accept formula.

Draw a circle around the functional group formed between the amino acids and state its name.

Name:

A mixture of phenylalanine and aspartic acid is separated by gel electrophoresis with a buffer of pH = 5.5.

Deduce their relative positions after electrophoresis, annotating them on the diagram. Use section 33 of the data booklet.

Aspartic acid is obtained synthetically as a racemic mixture. Draw the three‑dimensional shape of each isomer showing their spatial relationship to each other. Use section 33 of the data booklet.

Name:
amide/amido/carboxamide    [✔]

Note: Accept “peptide bond/linkage”.

Phe: must be on the origin    [✔]

Asp: any position on the left/anode/+ side    [✔]

    [✔]

Candidates were challenged to draw a circle around the functional group formed between the two amino acids in a dipeptide. Commonly the central C=O or the N-H groups were circled instead of the amide. Some candidates then named the functional group as a ketone or amine, however most candidates stated “peptide” here and scored a mark.

Most candidates correctly deduced the positions of the two amino acids after electrophoresis.

Was very poorly answered. Many candidates had difficulty drawing a 3D structure and showing the spatial relationship of two enantiomers. An incorrect amino acid residue was often used, or incorrect bond connections drawn.

The iodine number is the maximum mass of iodine that reacts with 100 g of an unsaturated compound.

Determine the iodine number of stearidonic acid, C₁₇H₂₇COOH.

State two functions of lipids in the body.

Outline one effect of increased levels of low-density lipoproteins in the blood.

ALTERNATIVE 1:

4 C=C bonds/4 carbon to carbon double bonds ✔

mass of iodine per mole of acid = «4 × 253.80 g mol^–1 =» 1015.2 «g mol^–1» ✔

iodine number «$=\frac{1015.2\text{g}\text{mo}{\text{l}}^{-1}}{276.46\text{g}\text{mo}{\text{l}}^{-1}}\times 100$» = 367 ✔

ALTERNATIVE 2:

4 C=C bonds/4 carbon to carbon double bonds ✔

«$\frac{100\text{g}}{276.46\text{g}\text{mo}{\text{l}}^{-1}}\times 4$ =» 1.447 mol of I₂ «reacts with 100 g»  ✔

iodine number «= 1.447 mol × 253.80 g mol^–1» = 367 ✔

Award [3] for correct final answer.

Any two of:

«structural» components of cell membranes ✔

energy storage/utilization ✔

«thermal/electrical» insulation ✔

transport «of lipid-soluble molecules» ✔

hormones/chemical messengers ✔

Accept other specific functions, such as “prostaglandin/cytokine/bile acid synthesis”, “cell differentiation/growth”, “myelination”, “storage of vitamins/biomolecules”, “signal transmission”, “protection/padding of organs”, “precursors/starting materials for the biosynthesis of other lipid”.

Any one of:

atherosclerosis/cholesterol deposition «in artery walls» ✔

heart/cardiovascular disease ✔

stroke ✔

Accept “arteries become blocked/walls become thicker”.

Describe the interaction responsible for the secondary structure of a protein.

Explain the action of an enzyme and state one of its limitations.

Contrast the actions of non-competitive and competitive inhibitors of an enzyme and state their effects on the maximum rate of reaction, V_max, and the Michaelis–Menten constant, K_m.

hydrogen bonding ✔

between C=O and H–N «groups» ✔

Accept a diagram which shows hydrogen bonding for M1 and which shows the interaction between O of C=O and H of NH groups for M2.

Accept “between amido/amide/carboxamide” but not “between amino/amine” for M2.

Enzyme action:

Any two of:

substrate binds to active site ✔

weakens bonds in substrate ✔

lowers activation energy

OR

provides alternate pathway ✔

increases rate of reaction

OR

acts as catalyst ✔

substrate specific ✔

Limitation:

Any one of:

temperature dependent ✔

pH dependent ✔

can be sensitive to heavy metal ions ✔

sensitive to denaturation ✔

can be inhibited ✔

substrate specific ✔

Accept “favourable orientation/conformation of the substrate «enforced by enzyme»” for M1.

Do not accept “substrate specific” as both an enzyme action and a limitation.

Award [1] for each action.

Award [1] for any two effects stated correctly.

Award [2 max] if both actions and effects are switched to incorrect inhibitor types.

The absorption spectrum of β-carotene is shown below.

Explain its colour in terms of its absorption bands. Use section 17 of the data booklet.

The absorption spectrum of chlorophyll a is shown below.

Suggest how the combination of chlorophyll a and carotenoids is beneficial for photosynthesis.

400–424 «nm» absorption band/violet AND 424–490 «nm» absorption band/blue    [✔]

complementary/opposite colour observed
OR
yellow/orange observed    [✔]

Note: Accept “400–500 «nm» absorption band” for M1.

extends energy absorption spectrum «for photosynthesis»    [✔]

Very few candidates scored the first mark. Two absorption bands were required but most candidates only mentioned one. However, most candidates scored the second mark for stating the colour of β-carotene.

Most candidates gave good answers to suggest how a combination of chlorophyll a and carotenoids is beneficial for photosynthesis.

A Michaelis–Menten plot for an enzyme-catalysed reaction is shown.

Sketch a curve to show the effect of a competitive inhibitor.

Suggest, based on the Michaelis–Menten plot, how a competitive inhibitor such as ethanol reduces the toxicity of methanol.

Enzymatic activity is studied in buffered aqueous solutions.

Calculate the ratio in which 0.1 mol dm⁻³ NaH₂PO₄ (aq) and 0.1 mol dm⁻³ Na₂HPO₄ (aq) should be mixed to obtain a buffer with pH = 6.10. Use section 1 of the data booklet.

pK_a (NaH₂PO₄) = 7.20

   [✔]

Note: Line must start at origin, to the right of original line and bending toward the same V_max.

Km is higher /same Vmax reached at higher [substrate]    [✔]

slower reaction rate
OR
gives time to excrete/eliminate methanol    [✔]

«pH = pK_a + log $\frac{[{\text{HPO}}_4^{2-}]}{\text{[}{\text{H}}_2{\text{PO}}_4^{-}]}$ / 6.10 = 7.20 + log $\frac{[{\text{HPO}}_4^{2-}]}{\text{[}{\text{H}}_2{\text{PO}}_4^{-}]}$ »

log $\frac{[{\text{HPO}}_4^{2-}]}{\text{[}{\text{H}}_2{\text{PO}}_4^{-}]}$ = «6.10 – 7.20 =» –1.10
OR
$\frac{[{\text{HPO}}_4^{2-}]}{\text{[}{\text{H}}_2{\text{PO}}_4^{-}]}$ = «10^–1.10 =» 0.079    [✔]

NaH₂PO₄ : Na₂HPO₄ = 12.6 : 1    [✔]

Note: Award [2] for correct final answer.

Most candidates correctly sketched a curve to show the effect of a competitive inhibitor on a Michaelis-Menten plot.

Many scored 1 out of 2 for stating that the inhibitor causes a slower reaction rate.

The calculation of the ratio of a conjugate acid-base pair to create a buffer with a specific pH was poorly done. Some candidates wrote a ratio without indicating which compound each value referred to and thus could not score. Many candidates used the concentrations of 0.1 mol dm⁻³ for both compounds and could not proceed.

The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.

Explain this in terms of their saturated fatty acid composition.

Fats contain triglycerides that are esters of glycerol and fatty acids. Deduce an equation for the acid hydrolysis of the following triglyceride.

The addition of partially hydrogenated cocoa butter to chocolate increases its melting point and the content of trans-fatty acids (trans-fats).

Outline one effect of trans-fatty acids on health.

coconut oil has higher content of lauric/short-chain «saturated» fatty acids
OR
cocoa butter has higher content of stearic/palmitic/longer chain «saturated» fatty acids    [✔]

longer chain fatty acids have greater surface area/larger electron cloud    [✔]

stronger London/dispersion/instantaneous dipole-induced dipole forces «between triglycerides of longer chain saturated fatty acids»    [✔]

Note: Do not accept arguments that relate to melting points of saturated and unsaturated fats.

correct products    [✔]

correctly balanced     [✔]

Any one of:
«increased risk of» coronary/heart disease    [✔]

«increased risk of» stroke    [✔]

«increased risk of» atherosclerosis    [✔]

«increased risk of type-2» diabetes    [✔]

increase in LDL cholesterol    [✔]

decrease in HDL cholesterol    [✔]

«increased risk of» obesity    [✔]

Candidates had difficulty explaining the melting points of fats in terms of length of carbon chain, and referred instead to an explanation of saturated and unsaturated fat structures.

Quite a few candidates had no idea how to write an equation for the acid hydrolysis of a given triglyceride. Many struggled with the structure or molecular formula of glycerol. Some created products of (R-O-CO-OH)₃ and alkanes, even though the question told them that triglycerides are esters of glycerol and fatty acids. Some got the products almost correct but wrote O-H-CO-R for the fatty acid so failed to score the first mark. Some forgot to balance the equation.

Outlining an effect of trans-fatty acids on health was generally answered well, with only a few vague answers.

Unreacted salicylic acid may be present as an impurity in aspirin and can be detected in the infrared (IR) spectrum.

Name the functional group and identify the absorption band that diff erentiates salicylic acid from aspirin. Use section 26 of the data booklet.

Name:

Absorption band:

Deduce the protons responsible for signals X and Y by marking them on the structure of aspirin in (a). Use section 27 of the data booklet.

Identify the splitting pattern of signals X and Y.

X:

Y:

Name:
hydroxyl   [✔]

Absorption band:
3200–3600 «cm^–1»  [✔]

Note: Accept “phenol” OR “alcohol” but not “hydroxide”. 

correct X   [✔]
correct Y   [✔]

Note: X and Y must be near the Hs.

X: singlet AND Y: singlet  [✔]

Many candidates scored a mark for naming a functional group that differentiates salicylic acid from aspirin. Some incorrectly said ether or carboxylic acid. Many candidates also scored for identifying the absorption band although 1700-1750 was a popular incorrect answer.

There was considerable confusion with indicating protons responsible for ¹H NMR signals. Often entire functional groups were circled or carbon atoms and not hydrogen atoms were circled.

There was also great difficulty in identifying the splitting pattern of the signals. It was rare to see both signals identified as singlets.

Retinal is the key molecule involved in vision. Explain the roles of cis and trans-retinal in vision and how the isomers are formed in the visual cycle.

Any three of:

cis-retinal binds to «the protein» opsin

OR

cis-retinal «binds to opsin and» forms rhodopsin

rhodopsin extends conjugation in retinal

OR

rhodopsin allows absorption of visible/blue/green light

when visible light is absorbed cis-retinal changes to trans-retinal

change «to trans-retinal» triggers an electrical/nerve signal

trans-retinal detaches from opsin AND is converted back to cis-retinal

OR

trans-retinal is converted back to cis-retinal through enzyme activity

Do not accept “cis-retinal to trans-retinal” alone without reference to absorption of visible light.

[Max 3 Marks]

Draw the structure of the repeating unit of starch and state the type of linkage formed between these units.

Type of linkage:

Formulate the equation for the complete hydrolysis of a starch molecule, (C₆H₁₀O₅)_n.

Calculate the energy released, in kJ g⁻¹, when 3.49 g of starch are completely combusted in a calorimeter, increasing the temperature of 975 g of water from 21.0 °C to 36.0 °C. Use section 1 of the data booklet.

Explain how the inclusion of starch in plastics makes them biodegradable.

continuation bonds AND −O attached to just one end AND both H atoms on end carbons must be on the same side    [✔]

Type of linkage:
glycosidic    [✔]

Note: Square brackets not required.

Ignore “n” if given.

Mark may be awarded if a polymer is shown but with the repeating unit clearly identified.

Accept “ether”.

(C₆H₁₀O₅)_n (s) + nH₂O (l) → nC₆H₁₂O₆ (aq)     [✔]

Note: Accept “(n-1)H₂O”.

Do not award mark if “n” not included.

q = «mcΔT = 975 g × 4.18 J g^–1 K^–1 × 15.0 K =» 61 100 «J» / 61.1 «kJ»    [✔]

«heat per gram = $\frac{61.1\text{ kJ}}{3.49\text{ g}}$ =» 17.5 «kJ g^–1»    [✔]

Note: Award [2] for correct final answer.

Any two of:
carbohydrate grains swell/break plastic into smaller pieces    [✔]

inclusion of carbohydrate makes the plastic more hydrophilic/water soluble     [✔]

carbohydrates are broken down/hydrolysed/digested by bacteria/micro-organisms    [✔]

plastic becomes more accessible to bacteria as holes/channels are created in it     [✔]

«presence of» carbohydrate weakens intermolecular/London/dispersion forces between polymer chains in the plastic    [✔]

Note: Accept “starch” for “carbohydrate” throughout.

Do not accept “carbohydrates are broken down/hydrolyzed”.

Candidates were required to draw the structure of the repeating unit of starch given the ring structure as a starting point. This proved extremely difficult with very few candidates scoring a mark. Commonly, the structure of $a$-glucose was given, or an attempt was made to draw a polymer. Naming the type of linkage formed was answered well.

Also proved challenging, with many candidates unable to write an equation for the hydrolysis of a starch molecule (C₆H₁₀O₅)_n. The n was often omitted from otherwise correct equations or the product was incorrectly given as (C₆H₁₂O₆)_n.

The incorrect mass was frequently used when calculating energy released from combustion of starch in a calorimeter. Those who used the mass of water correctly frequently stopped when energy in kJ or J was calculated, and did not seem to notice that the question asked for the energy to be calculated in kJg⁻¹ so a further calculation was required.

Responses to explain how including starch in plastics makes them biodegradable were sometimes lacking in detail. Some candidates simply said that “starch is soluble in water”. Some said that “starch can be broken down/hydrolyzed” but omitted the key words “by bacteria or microorganisms”.

Outline why cellulose fibres are strong.

Any two of:

long straight/unbranched chains

multiple hydrogen bonds «between chains»

microfibrils

OR

rigid/cable structure

[2 marks]

Some proteins form an α-helix. State the name of another secondary protein structure.

Compare and contrast the bonding responsible for the two secondary structures.

One similarity:

One difference:

Explain why an increase in temperature reduces the rate of an enzyme-catalyzed reaction.

State and explain how a competitive inhibitor affects the maximum rate, V_max, of an enzyme-catalyzed reaction.

Suggest two reasons why oil decomposes faster at the surface of the ocean than at greater depth.

Oil spills can be treated with an enzyme mixture to speed up decomposition.

Outline one factor to be considered when assessing the greenness of an enzyme mixture.

β/beta pleated/sheet  [✔]

One similarity:
hydrogen bonding
OR
attractions between C=O and N–H «on main chain»  [✔]

One difference:
α-helix has hydrogen bonds between amino acid residues that are closer than β-pleated sheet
OR
H-bonds in α-helix parallel to helix axis AND perpendicular to sheet in β-pleated sheet
OR
α-helix has one strand AND β-pleated sheet has two «or more» strands
OR
α-helix is more elastic «since H-bonds can be broken easily» AND β-pleated sheet is less elastic «since H-bonds are difficult to break»  [✔]

Note: Accept a diagram which shows hydrogen bonding between O of C=O and H of NH groups for M1.

Accept “between carbonyl/amido/amide/carboxamide” but not “between amino/amine” for M1.

enzyme denatured/ loss of 3-D structure/conformational change
OR
«interactions responsible for» for tertiary/quaternary structures altered  [✔]

shape of active site changes
OR
fewer substrate molecules fit into active sites  [✔]

V_max unchanged  [✔]

at high substrate concentration substrate outcompetes inhibitor/need a higher
substrate concentration to reach V_max  [✔]

Note: Accept suitable labelled diagram.

Any two of:
surface water is warmer «so faster reaction rate»/more light/energy from the sun  [✔]

more oxygen «for aerobic bacteria/oxidation of oil» [✔]

greater surface area [✔]

Any one of:
non-hazardous/toxic to the environment/living organisms  [✔]

energy requirements «during production» [✔]

quantity/type of waste produced «during production»
OR
atom economy [✔]

safety of process [✔]

Note: Accept “use of solvents/toxic materials «during production»”.

Do not accept “more steps involved”.

This question was well answered with many scoring the mark although there were quite a few incorrect responses that answered “beta-helix” rather than “beta-pleated sheet”.

Almost all the candidate’s stated hydrogen bonding as the similarity between the 2 types of secondary structures but lost marks on the difference between them.

This question was well answered where most candidates received one mark for identifying that the enzyme will denature with an increase in temperature. However, many candidates did not continue with the explanation that the shape of the active site changes.

Many candidates stated correctly that Vmax remains unchanged but only some mentioned that a higher substrate concentration was required to reach Vmax for the second mark.

Many candidates received two marks for this part while some candidates only suggested one reason or repeated the same reason (for example - heat and energy from the sun) even though the question clearly asked for two reasons.

The candidates struggled with this part and gave journalistic or vague answers that cannot be awarded marks. Atom economy was mentioned correctly by a few candidates.

DNA is a biopolymer made up of nucleotides. List two components of a nucleotide.

Any two of:

pentose «sugar»

OR

deoxyribose

phosphate «group»

«organic» nitrogenous base

OR

nucleobase

OR

nucleic base

OR

purine

OR

pyrimidine

Accept names or formulas.

Accept “ribose” for M1.

Do not accept “phosphoric acid”.

Accept the four bases together: “adenine, cytosine, thymine, guanine”.

[2 marks]

Outline which pK_a value should be used when calculating the pH of the solution, giving your reason.

Calculate the pH of the glutamine solution using section 1 of the data booklet.

Describe what is meant by the genetic code and how it relates to protein synthesis.

pK_a2 AND pH of solution > pH of isoelectric point «as anion present»
OR
pK_a2 AND zwitterion has lost H⁺ to become anion «so in basic solution»
OR
pK_a2 AND «only» anion «and zwitterion» present   [✔]

«pH = 9.1 + log $\left[\frac{0.30}{0.60}\right]$ »
«= 9.1 + (–0.3)» = 8.8  [✔]

sequence of bases in DNA  [✔]

codon/triplet code/each set of three bases codes for an amino acid  [✔]

Several candidates stated correctly that pK_a2 should be used with a reason whereas others wrote pKa1, which was incorrect.

Majority of the candidates calculated correctly the pH of the glutamine solution while other candidates managed the ECF mark from part a (i).

This part was poorly answered by many candidates and was unable to state that genetic code is a sequence of bases in DNA and each codon codes for an amino acid. Often complex answers were written which were incorrect.

Outline why the complex formed between Fe²⁺ and oxygen is red. Refer to the diagram above and section 17 of the data booklet.

Explain the shape of the curve.

Sketch another line to show the effect of an increase in body temperature on the oxygen saturation of hemoglobin.

extensive conjugated system
OR
extensive delocalized bonding system
OR
extended system of alternating double and single bonds  [✔]

absorbs green
OR
complementary to red light  [✔]

sigmoid/S-shaped
OR
as oxygen binds to one active site, shape of other active sites change  [✔]

affinity of other sites for oxygen increases/ability to bind oxygen is increased by initial binding of oxygen
OR
cooperative binding   [✔]

Note: Accept description of sigmoid/S-shaped curve if not stated for M1.

This part was fairly well answered with the majority of the candidates managed one mark. The second mark was missed because the candidates did not write extensive conjugated system or extensive delocalised bonding system.

Many candidates managed one mark by describing the graph and stating that the affinity of other sites for oxygen increases/cooperative bonding. Several candidates missed that it was a sigmoid/S-shaped curve.

Many candidates were able to sketch another line to show the effect of an increase in body temperature on the oxygen saturation of haemoglobin.

Deduce the structural formula of phosphatidylcholine.

Identify the type of reaction in (a).

Lecithin is a major component of cell membranes. Describe the structure of a cell membrane.

Lecithin aids the body’s absorption of vitamin E.

Suggest why vitamin E is fat-soluble.

Phospholipids are also found in lipoprotein structures.

Describe one effect of increased levels of low-density lipoprotein (LDL) on health.

phosphodiester correctly drawn  [✔]

both ester groups correctly drawn  [✔]

Note: Accept protonated phosphate.

Accept phosphodiester in centre position.

condensation  [✔]

Note: Accept “esterification”.

Accept “nucleophilic substitution/S_N”.

phospholipid bilayer/double layer
OR
two layers of phospholipids  [✔]

polar/hydrophilic heads facing aqueous environment AND non-polar/hydrophobic tails facing away from aqueous environment  [✔]

Note: Award [2] for a suitably labelled diagram.

Award [1] for a correct but unlabelled diagram.

Accept “polar/hydrophilic heads on outside AND non-polar/hydrophobic tails on inside” for M2.

long non-polar/hydrocarbon chain «and only one hydroxyl group»
OR
forms London/dispersion/van der Waals/vdW interactions with fat  [✔]

Any one of:
atherosclerosis/cholesterol deposition «in artery walls»  [✔]

increases risk of heart attack/stroke/cardiovascular/heart disease/CHD  [✔]

Note: Accept “arteries become blocked/walls become thicker”, “increases blood pressure”, or “blood clots”.

Do not accept “high cholesterol”.

Almost all the candidates struggled with this part. Although the phosphodiester was a challenging mark it could be awarded in both the protonated and deprotonated form. The two ester groups were required for the second mark. Candidates were not able to draw correctly for both marks, and many left this question blank.

This part was very well answered.

This question was another one where the first part was fairly well answered, but the explanation or second mark was often not correct or incomplete.

Many candidates missed the idea of a long or large non-polar chain when describing the structure of vitamin E. Simply stating non-polar chain was not sufficient for the mark.

Candidates were required to state one effect of increased LDL. Majority of the candidates scored well on this part. High cholesterol is not an acceptable answer but was frequently seen.

State the name of the functional group forming part of the ring structure of each monosaccharide unit.

Classify, giving your reason, the hexose (six-membered) ring of sucrose as an α or β isomer.

Sketch the cyclic structures of the two monosaccharides which combine to form sucrose.

acetal
OR
ether  [✔]

Note: Accept “glycosidic bond/linkage” but not “glucosidic”.

Do not accept “hemiacetal”.

α-isomer AND hydroxyl group on carbon 1 and –CH₂OH are trans
OR
α-isomer AND hydroxyl group on carbon 1 is below plane of ring
OR
α-isomer AND glycosidic linkage between rings is below plane of ring  [✔]

Note: Accept “ether linkage” for M3.

  [✔]

[✔]

This was reasonably answered although there were some candidates who stated ester or hemiacetal, which is incorrect.

This part was very poorly answered. Majority of the candidates had no idea about the reason whether the six-membered ring was an alpha or beta isomer.

This question was poorly answered. Many candidates lost marks due to sloppy drawing and incorrect bond linkages. Some candidates did not separate the two monosaccharides as instructed.

List two components of nucleotides.

Explain how the double-helical structure of DNA is stabilized once formed.

Any two correct for [1]:
pentose «sugar»
OR
deoxyribose ✔

phosphate/phosphato «group»/residue of phosphoric acid ✔

NOTE: Accept “−OPO₃²⁻/−OPO₃H⁻/−OPO₃H₂” but not “PO₄³⁻”.

«organic» nitrogenous base
OR
nucleobase
OR
nucleic base
OR
purine
OR
pyrimidine ✔

NOTE: Accept the four bases together: “adenine/A, guanine/G, cytosine/C, thymine/T”.
Accept names or formulas.

Any two of:
H-bonding between bases in each pair ✔
hydrophobic interactions/π-stacking between bases ✔
polar/charged/hydrophilic groups in sugar-phosphate backbone interactions with aqueous solution/water
OR
H-bonding AND ion-dipole interactions between phosphato «groups» andwater/histones ✔

Accept "phosphate groups are hydrophilic and form H-bonds with water".
Accept “H-bonding with histones”.

Outline the significance of the Michaelis constant K_m.

Compare the effects of competitive and non-competitive inhibitors.

K_m is inverse measure of affinity of enzyme for a substrate
OR
K_m is inversely proportional to enzyme activity
OR
high value of K_m indicates higher substrate concentration needed for enzyme saturation
OR
low value of K_m means reaction is fast at low substrate concentration ✔

NOTE: Idea of inverse relationship must be conveyed.
Accept “high value of K_m indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.
Accept “low value of K_m indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.

NOTE: Accept “outside/away from active site” for “allosteric site”.
Award [1] for any two correct effects from any of the six listed.

Describe the function of chlorophyll in photosynthesis.

Compare and contrast the structures of starch and cellulose.

One similarity:

One difference:

Explain why maltose, C₁₂H₂₂O₁₁, is soluble in water.

absorbs/traps light «energy» ✔

initiates redox reactions
OR
transfers electrons ✔

One similarity:
1−4/glycosidic linkage
OR
glucose monomers/residues ✔

NOTE: Accept “both are polysaccharides”.

One difference:
starch has α-glucose AND cellulose has β-glucose «monomers»
OR
starch can form coiled/spiral/helical chains «and straight chains» AND cellulose cannot/can only form straight chains/can only form a linear structure
OR
starch «in amylopectin» also has 1−6 glycosidic links AND cellulose does not ✔

NOTE: Accept "cellulose has alternate glucose monomers upside down with respect to each other AND starch does not".

«solubility depends on forming many» H-bonds with water ✔
maltose has many hydroxyl/OH/oxygen atom/O «and forms many H-bonds» ✔

NOTE: Reference to “with water” required.
Accept “hydroxy” for “hydroxyl” but not “hydroxide/OH^–”.
Reference to many/several OH groups/O atoms required for M2.

The iodine number is the number of grams of iodine which reacts with 100 g of fat. Calculate the iodine number of oleic acid.

The chemical change in stored fats causes rancidity characterized by an unpleasant smell or taste.

Compare hydrolytic and oxidative rancidity.

State one similarity and one difference in composition between phospholipids and triglycerides.

Similarity:

Difference:

«one C=C bond»
«1 mole iodine : 1 mole oleic acid»

« $\frac{100\times 253.80}{282.46}$ =» 89.85 «g of I₂» ✔

NOTE: Accept “90 «g of I₂»”.

NOTE: Award [1] for any two sites or conditions from any of the four listed.
Accept “high temperature” for “heat”. Accept "lipase" for "enzyme".
Do not accept just “double bond”.
Accept “air” for “oxygen” and “UV/sun” for “light”.
Ignore any reference to heat/high temperature as a condition for oxidative.

Similarity:

«derived from» propane-1,2,3-triol/glycerol/glycerin/glycerine

OR
«derived from» at least two fatty acids
OR
contains ester linkages
OR
long carbon chains ✔

NOTE: Do not accept “two fatty acids as both a similarity and a difference”.
Do not accept just “hydrocarbon/carbon chains”.

Difference:

phospholipids contain two fatty acids «condensed onto glycerol» AND triglycerides three
OR
phospholipids contain phosphate/phosphato «group»/residue of phosphoric acid AND triglycerides do not ✔

NOTE: Accept “phospholipids contain phosphorus AND triglycerides do not".
Accept “phospholipids are amphiphilic AND triglycerides are not” OR “phospholipids have hydrophobic tails and hydrophilic heads AND triglycerides do not”.

Outline what is meant by genetically modified organisms.

Outline one benefit of the use of these products.

organism whose genetic material/DNA has been altered by genetic engineering techniques «involving transferring DNA between species»   [✔]

Note: Accept “any living organism that possesses a novel combination of genetic material obtained through the use of modern biotechnology”.

Any one of:
increased resistance to pests/micro-organisms    [✔]

increased shelf-life of food    [✔]

increased nutritional value    [✔]

greater crop yield     [✔]

greater tolerance of crops to adverse climatic/soil/growing condition    [✔]

Candidates had difficulty outline the meaning of genetically modified organisms with many simply repeating the question by stating modified DNA without including technology or engineering techniques.

Outlining one benefit of using GMOs was well answered.

Outline why anthocyanins are coloured.

Explain why the blue colour of a quinoidal base changes to the red colour of a flavylium cation as pH decreases.

highly conjugated systems

OR

alternating single and double bonds

OR

many delocalized electrons

electron transitions occur when visible light is absorbed

[2 marks]

gaining protons

decreases electron density/extent of conjugation «in aromatic backbone»

increases energy of electron transitions

[3 marks]

Explain how the structure of vitamin A is important to vision using section 35 of the data booklet.

vitamin A oxidized to «11-cis-»retinal

extended conjugation

OR

extensive delocalization

cis-retinal converts to trans-retinal through absorption of light

Accept “vitamin A/hydroxyl/hydroxy/alcohol/CH₂OH group oxidized to aldehyde/CHO «group in retinal»”.

[3 marks]

Ascorbic acid and retinol are two important vitamins.

Explain why ascorbic acid is soluble in water and retinol is not. Use section 35 of the data booklet.

ascorbic acid: many hydroxyl/OH groups AND retinol: few/one hydroxyl/OH group
OR
ascorbic acid: many hydroxyl/OH groups AND retinol: long hydrocarbon chain    [✔]

ascorbic acid: «many» H-bond with water
OR
retinol: cannot «sufficiently» H-bond with water    [✔]

Note: Do not accept “OH⁻/hydroxide”.

Explanations of why ascorbic acid is soluble in water and retinol is not were poor. Very few referred to the ability to form hydrogen bonds with water. Some said “hydroxide” instead of “hydroxyl” and thus failed to score a mark. Commonly, candidates scored 1 mark for saying that ascorbic acid has many hydroxyl groups and retinol has a long hydrocarbon chain.