Biological Molecules - Worksheets, Questions and Revision

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A-Level · Biology

AB1 Biological Molecules

AQA 7402 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Water is essential to living organisms because of properties that arise from its molecular structure.

(a)State what is meant by describing a water molecule as polar.(1)

(b)Explain, using the polarity of water, why hydrogen bonds form between neighbouring water molecules.(2)

(c)Water has a high specific heat capacity. Explain why this property is significant for organisms living in aquatic habitats.(2)

(d)Explain why water is described as a good solvent for ions such as sodium (Na+) and chloride (Cl-).(2)

(Total for Question 1 is 7 marks)

This question is based on the required practical involving the use of qualitative reagents to identify biological molecules. A student is given four unlabelled solutions, P, Q, R and S, one of which contains glucose, one starch, one egg albumen (protein) and one vegetable oil, in no particular order.

(a)Describe how the student could use the iodine test to identify which tube contains starch, including the result that would confirm this.(2)

(b)Solution P gave a positive Biuret test. State what this indicates about the contents of tube P, and state the colour change that would have been observed.(2)

(c)The student then heated tube Q with Benedict's reagent and it remained blue (no colour change). Explain what this result indicates about tube Q, and describe the further test needed to check whether tube Q contains a non-reducing sugar.(3)

(d)Solution S was identified using the emulsion test. Describe the procedure for this test and the result that indicates a positive result for lipid.(2)

(Total for Question 2 is 9 marks)

Monosaccharides can be joined together to form larger carbohydrate molecules.

(a)Name the type of bond formed when two monosaccharides join, and name the type of reaction by which it forms.(2)

(b)Glucose exists as two isomers, α-glucose and β-glucose. State the structural difference between them.(1)

(c)State the monomer used to build (i) amylose (a component of starch) and (ii) cellulose, then describe one difference between the glycosidic bonds/chain shape formed in amylose compared with cellulose.(3)

(d)Explain how the structure of cellulose molecules relates to their function in plant cell walls.(3)

(Total for Question 3 is 9 marks)

Triglycerides are a major class of lipid found in living organisms.

(a)Name the two types of molecule that combine to form a triglyceride, and name the type of bond formed between them.(3)

(b)Explain, in terms of molecular structure, why lipids containing unsaturated fatty acids tend to have a lower melting point than lipids containing only saturated fatty acids.(3)

(c)Calculate the number of water molecules released when one molecule of triglyceride is formed from one glycerol molecule and three fatty acid molecules. Show your working.(2)

(d)A triglyceride has a relative molecular mass of 806. Its fatty acid components each contribute a relative mass of 256, and glycerol has a relative molecular mass of 92. Calculate the total relative mass lost as water when this triglyceride forms, and give your answer to an appropriate number of significant figures.(3)

(Total for Question 4 is 11 marks)

This question is based on the required practical investigating the effect of pH on the rate of an enzyme-controlled reaction. A student investigated the effect of pH on the rate of hydrolysis of starch by amylase. Amylase solution was mixed with starch solution buffered at pH 4, 5, 6, 7 and 8, all at 30 degC. Every 30 seconds a sample was removed and tested with iodine solution on a spotting tile, until the iodine no longer turned blue-black. The times taken (until a negative iodine result was first obtained) were: pH 4 = 600 s, pH 5 = 240 s, pH 6 = 90 s, pH 7 = 150 s, pH 8 = 480 s.

(a)Identify the independent variable and the dependent variable in this investigation.(2)

(b)State two variables, other than pH, that should be controlled in this investigation, and explain why controlling temperature is particularly important.(3)

(c)Using rate = 1000 / time (s) as an arbitrary rate unit, calculate the rate of reaction at pH 6. Give your answer to 2 significant figures.(2)

(d)Describe the shape of the graph obtained when rate of reaction (1/time) is plotted against pH for this data, and explain this shape in terms of enzyme structure.(4)

(e)The student only tested each pH once. Suggest one limitation of the method used to determine the end-point (the iodine test), describe an alternative method that would give continuous quantitative data, and explain why it would improve the investigation.(3)

(Total for Question 5 is 14 marks)

The initial rate of an enzyme-catalysed reaction was measured at increasing substrate concentrations, with enzyme concentration and all other conditions kept constant. Results:
Substrate concentration (mmol dm^-3): 1, 2, 4, 8, 16, 32
Initial rate (arbitrary units): 5, 9, 15, 19, 20, 20

(a)Describe the relationship shown by the data between substrate concentration and initial rate of reaction.(2)

(b)Explain, in terms of enzyme action, why the rate of reaction reaches a maximum despite further increases in substrate concentration.(3)

(c)Calculate the percentage increase in rate of reaction between substrate concentration 2 mmol dm^-3 and 4 mmol dm^-3. Show your working.(3)

(d)Predict, giving a reason, what would happen to the maximum (plateau) rate of reaction if the experiment were repeated with double the enzyme concentration, all other conditions unchanged.(2)

(Total for Question 6 is 10 marks)

Proteins are polymers built from amino acid monomers.

(a)Name the type of bond formed between adjacent amino acids during protein synthesis, and name the reaction type by which it is formed.(2)

(b)Describe what is meant by the primary, secondary and tertiary structure of a protein.(3)

(c)Explain how disulfide bridges, together with hydrogen bonds and ionic bonds, contribute to the tertiary structure of a globular protein such as an enzyme.(3)

(d)Compare the structure of a globular protein (e.g. haemoglobin) with that of a fibrous protein (e.g. collagen), relating structure to function.(2)

(Total for Question 7 is 10 marks)

Enzyme inhibitors reduce the rate of enzyme-catalysed reactions by different mechanisms.

(a)Define what is meant by a competitive inhibitor.(2)

(b)Explain why increasing substrate concentration can reduce the effect of a competitive inhibitor but has no effect on the action of a non-competitive inhibitor.(2)

(c)A non-competitive inhibitor binds to an allosteric site. Explain how this affects enzyme activity.(3)

(d)A biochemist adds increasing concentrations of a non-competitive inhibitor to a reaction mixture containing a fixed, saturating substrate concentration. Predict and explain the effect on the rate of reaction as inhibitor concentration increases.(3)

(Total for Question 8 is 10 marks)

Nucleic acids store and transmit genetic information.

(a)Describe the structure of a DNA nucleotide.(3)

(b)State two differences between the structure of DNA and RNA.(2)

(c)Explain how complementary base pairing and hydrogen bonding allow DNA to act as a stable store of genetic information.(3)

(Total for Question 9 is 8 marks)

Adenosine triphosphate (ATP) is described as the universal energy currency of cells.

(a)Describe the structure of an ATP molecule.(2)

(b)Explain why ATP is described as a useful, immediate source of energy for cells, referring to the hydrolysis of ATP.(2)

(c)State two cellular processes that require energy released from ATP hydrolysis.(2)

(Total for Question 10 is 6 marks)

Starch, glycogen and cellulose are all polysaccharides built from glucose monomers, but each has a structure suited to a different biological function. Discuss the relationship between the structure of starch (as amylose and amylopectin), glycogen and cellulose, and their functions in living organisms.

(Total for Question 11 is 6 marks)

A student used a colorimeter to determine an unknown concentration of glucose using a calibration curve.

(a)A stock glucose solution has a concentration of 200 mmol dm^-3. The student needs to prepare 10 cm³ of a 1 in 4 dilution of this stock. Calculate the volume of stock solution and the volume of distilled water required, and state the resulting glucose concentration.(3)

(b)The student's calibration line had the equation: absorbance = 0.004 x concentration (mmol dm^-3). Before measuring, the unknown sample was diluted 1 in 5 (since its absorbance was too high to read directly), and the diluted sample gave an absorbance reading of 0.52. Calculate the glucose concentration of the original (undiluted) unknown sample.(3)

(c)Explain why the student diluted the unknown sample before measuring its absorbance, and explain why extrapolating (reading a value beyond the range of) the calibration curve would reduce the validity of a result.(3)

(Total for Question 12 is 9 marks)

The rate of an enzyme-catalysed reaction was measured at 20 degC and found to be 4.0 arbitrary units. When repeated at 30 degC (all else unchanged), the rate was found to be 8.6 arbitrary units. The Q10 value describes how much the rate of reaction changes for every 10 degC rise in temperature, where Q10 = rate at (T + 10 degC) / rate at T.

(a)Calculate the Q10 value for this reaction between 20 degC and 30 degC.(2)

(b)Assuming the same Q10 value continues to apply, calculate the predicted rate of reaction at 40 degC.(2)

(c)Evaluate the assumption made in part (b) that the same Q10 value continues to apply at 40 degC, given that this is an enzyme-catalysed reaction.(3)

(Total for Question 13 is 7 marks)

Mark scheme · AB1 Biological Molecules

Question 1

(a) B1 a molecule with an uneven/unequal distribution of charge, having partially positive and partially negative regions oe
(a) Answer: A molecule with an unequal distribution of charge across it.
(b) M1 oxygen is more electronegative than hydrogen, so shared electrons are pulled towards oxygen, making oxygen slightly negative (delta-) and hydrogen slightly positive (delta+) oe
(b) A1 the delta+ hydrogen of one water molecule is attracted to the delta- oxygen of a neighbouring water molecule, forming a hydrogen bond oe
(b) Answer: Electronegativity difference creates delta+/delta- charges, allowing attraction (hydrogen bonding) between neighbouring molecules.
(c) M1 a high specific heat capacity means a large amount of energy must be absorbed or released before the temperature of water changes noticeably, so temperature changes slowly oe
(c) A1 this buffers aquatic habitats against rapid temperature fluctuations, providing a relatively stable thermal environment for organisms/enzymes oe
(c) Answer: Water resists rapid temperature change, giving aquatic organisms a thermally stable habitat.
(d) M1 the delta+ hydrogens of water molecules are attracted to (surround) negative ions such as Cl-, and the delta- oxygens are attracted to (surround) positive ions such as Na+ oe
(d) A1 this hydration shell separates the ions from one another, allowing them to dissolve/move in solution oe
(d) Answer: Polar water molecules surround and separate ions, dissolving them.

Question 2

(a) M1 add iodine dissolved in potassium iodide solution to a sample of each solution oe
(a) A1 a positive result (starch present) is shown by a colour change from orange/brown to blue-black oe
(a) Answer: Add iodine/potassium iodide solution; blue-black colour confirms starch.
(b) B1 tube P contains protein (peptide bonds present)
(b) B1 colour change from blue to purple/lilac/violet oe
(b) Answer: Tube P contains protein; colour changes blue to purple.
(c) M1 tube Q does not contain a reducing sugar, since no brick-red/orange precipitate formed oe
(c) M1 to test for a non-reducing sugar, a fresh sample must first be boiled with dilute hydrochloric acid (to hydrolyse any glycosidic bonds), then neutralised with sodium hydrogencarbonate solution oe
(c) dM1 the sample is then retested with Benedict's reagent and heated; if a brick-red precipitate now forms, a non-reducing sugar (e.g. sucrose) is confirmed present oe (dependent on M1 hydrolysis step being described)
(c) Answer: No reducing sugar in Q; must hydrolyse with acid, neutralise, then retest with Benedict's for a non-reducing sugar.
(d) M1 add ethanol to the sample, shake/mix thoroughly, then pour the resulting solution into a separate tube of water oe
(d) A1 a positive result is shown by a white/cloudy emulsion forming (in the water) oe
(d) Answer: Mix with ethanol, then add to water; white emulsion confirms lipid.

Question 3

(a) B1 glycosidic bond
(a) B1 condensation reaction
(a) Answer: Glycosidic bond; condensation reaction.
(b) B1 the -OH group on carbon 1 is below the plane of the ring in α-glucose and above the plane of the ring in β-glucose oe
(b) Answer: Position of the -OH group on carbon 1 differs (below the ring in α, above in β).
(c) B1 (i) α-glucose
(c) B1 (ii) β-glucose
(c) B1 amylose monomers are joined by 1,4-glycosidic bonds forming a coiled/helical chain, whereas cellulose monomers are joined by 1,4-glycosidic bonds (with alternate glucose rotated 180 degrees) forming long straight/unbranched chains oe
(c) Answer: Amylose: α-glucose, coiled chain. Cellulose: β-glucose, straight unbranched chain.
(d) M1 β-glucose monomers joined by 1,4-glycosidic bonds, with alternate glucose molecules rotated 180 degrees, produce long straight, unbranched chains oe
(d) M1 many hydrogen bonds cross-link adjacent parallel cellulose chains, bundling them into microfibrils oe
(d) A1 the large number of cross-links gives microfibrils high tensile strength, allowing cellulose to provide structural support/rigidity to the plant cell wall oe
(d) Answer: Straight chains cross-linked by hydrogen bonds into strong microfibrils give the cell wall tensile strength.

Question 4

(a) B1 glycerol
(a) B1 (three) fatty acids
(a) B1 ester bond
(a) Answer: Glycerol and three fatty acids joined by ester bonds.
(b) M1 unsaturated fatty acids contain one or more C=C double bond(s), which create a kink/bend in the hydrocarbon tail oe
(b) A1 this prevents the fatty acid chains from packing closely together
(b) A1 resulting in weaker intermolecular forces between chains, so less energy is needed to separate them, giving a lower melting point (liquid/oil at room temperature) oe
(b) Answer: Kinks from C=C bonds stop close packing, weakening intermolecular forces and lowering melting point.
(c) M1 3 ester bonds are formed (one between glycerol and each fatty acid), each by a condensation reaction that releases one water molecule oe
(c) A1 3 (water molecules) cao
(c) Answer: 3 water molecules.
(d) M1 total mass of reactants = 92 + (3 x 256) = 860
(d) M1 mass lost as water = 860 - 806
(d) A1 = 54 (g mol^-1) awrt 54
(d) Answer: 54 g mol^-1 lost as water.

Question 5

(a) B1 independent variable = pH (of the buffer solution)
(a) B1 dependent variable = time taken for the iodine test to become negative/for starch to be completely hydrolysed oe
(a) Answer: IV: pH. DV: time for starch to disappear (iodine test negative).
(b) B1 volume/concentration of amylase solution
(b) B1 volume/concentration of starch solution (allow total reaction volume)
(b) A1 temperature must be controlled because it independently affects the kinetic energy of molecules and therefore enzyme activity/rate of reaction, which would confound (mask) the effect of pH being investigated oe
(b) Answer: Control amylase concentration and starch concentration; temperature must be fixed as it independently affects rate and would confound results.
(c) M1 rate = 1000 / 90
(c) A1 = 11 (arbitrary units) awrt 11
(c) Answer: 11 (arbitrary rate units, 2 s.f.).
(d) M1 rate increases from pH 4 to a maximum then decreases (a bell-shaped curve), with the highest rate at pH 6 (ft from the candidate's own calculated rates)
(d) M1 at the optimum pH, the ionic and hydrogen bonds maintaining the enzyme's tertiary structure are at their most stable, so the active site is complementary in shape to the substrate (starch) oe
(d) M1 at pH values above or below the optimum, excess H+ or OH- ions disrupt the ionic/hydrogen bonds holding the tertiary structure together, altering the shape of the active site oe
(d) A1 so fewer/no enzyme-substrate complexes can form and the rate of reaction decreases oe
(d) Answer: Bell-shaped curve peaking at pH 6; deviation from optimum pH disrupts bonds maintaining active site shape, reducing enzyme-substrate complex formation.
(e) B1 the iodine end-point is judged qualitatively by eye (subjective) and only gives one data point per pH rather than showing how rate changes continuously oe
(e) B1 use a colorimeter to measure the absorbance/percentage transmission of the iodine-starch complex at regular time intervals oe
(e) B1 this gives continuous, objective, quantitative data, allowing the rate at any time point to be calculated (e.g. from the gradient of a graph) oe
(e) Answer: Iodine end-point is subjective and only one data point; using a colorimeter gives continuous, objective, quantitative absorbance data.

Question 6

(a) M1 as substrate concentration increases from 1 to 16 mmol dm^-3, the initial rate of reaction increases (rapidly at first) oe
(a) A1 above 16 mmol dm^-3 (up to 32) the rate levels off/plateaus, becoming independent of further increases in substrate concentration oe
(a) Answer: Rate rises with substrate concentration then plateaus above about 16 mmol dm^-3.
(b) M1 at low substrate concentration there are more free/unoccupied active sites than substrate molecules, so rate is limited by substrate availability oe
(b) M1 as substrate concentration rises, a greater proportion of active sites become occupied by substrate at any moment, forming more enzyme-substrate complexes oe
(b) A1 at high substrate concentration nearly all active sites are occupied (saturated) at any one time, so enzyme (not substrate) concentration becomes the limiting factor and rate cannot increase further oe
(b) Answer: Active sites become saturated with substrate, so enzyme concentration becomes limiting.
(c) M1 increase in rate = 15 - 9 = 6
(c) M1 percentage increase = (6 / 9) x 100
(c) A1 = 66.7% awrt 66.7 (allow 67%)
(c) Answer: 66.7% (awrt).
(d) B1 the maximum (plateau) rate would be higher than 20 (arbitrary units)
(d) B1 because doubling enzyme concentration provides more active sites available at any one time, so a higher substrate concentration (and higher rate) is needed before all active sites become saturated oe
(d) Answer: Plateau rate would increase, since more active sites are available before saturation occurs.

Question 7

(a) B1 peptide bond
(a) B1 condensation reaction
(a) Answer: Peptide bond; condensation reaction.
(b) B1 primary structure is the sequence of amino acids in the polypeptide chain, joined by peptide bonds oe
(b) B1 secondary structure is the folding/coiling of the polypeptide into an α-helix or β-pleated sheet, held by hydrogen bonds between (backbone) -NH and -C=O groups oe
(b) B1 tertiary structure is the further folding of the polypeptide into a specific 3D shape, held by ionic bonds, hydrogen bonds and disulfide bridges between R groups oe
(b) Answer: Primary = amino acid sequence; secondary = helix/sheet held by H bonds; tertiary = 3D folding held by ionic, hydrogen and disulfide bonds.
(c) M1 disulfide bridges are strong covalent bonds that form between the R groups of two cysteine residues, holding parts of the polypeptide chain firmly together oe
(c) M1 hydrogen bonds form in large numbers between polar R groups, and ionic bonds form between oppositely charged R groups (e.g. -COO- and -NH3+) oe
(c) A1 together these bonds fold the polypeptide into a precise, stable 3D tertiary structure, which for an enzyme determines the specific shape of the active site oe
(c) Answer: These bonds collectively fold and stabilise the polypeptide into its precise 3D shape, forming the active site.
(d) B1 globular proteins fold into a compact, roughly spherical, soluble shape suited to transport/catalytic roles, whereas fibrous proteins form long, insoluble strands suited to structural/mechanical roles oe
(d) B1 globular proteins (e.g. haemoglobin) have varied, irregular amino acid sequences allowing specific binding, whereas fibrous proteins (e.g. collagen) have repetitive amino acid sequences giving a regular structure suited to strength/support oe
(d) Answer: Globular = compact, soluble, varied sequence, functional roles; fibrous = long strands, insoluble, repetitive sequence, structural roles.

Question 8

(a) B1 a molecule with a similar shape to the substrate
(a) B1 that binds (reversibly) to the active site of the enzyme, blocking the substrate from binding oe
(a) Answer: A substrate-shaped molecule that binds to the active site, blocking the substrate.
(b) M1 increasing substrate concentration increases the frequency of substrate molecules binding to the active site relative to the inhibitor, out-competing it, so the competitive inhibitor's effect can be overcome oe
(b) A1 a non-competitive inhibitor binds away from the active site (at an allosteric site), so increasing substrate concentration cannot displace it/prevent its binding oe
(b) Answer: Substrate out-competes a competitive inhibitor for the active site, but cannot displace a non-competitive inhibitor bound elsewhere.
(c) M1 the inhibitor binding at the allosteric site causes the tertiary structure of the enzyme to change shape oe
(c) M1 this changes the shape of the active site so it is no longer complementary to the substrate oe
(c) A1 the substrate can no longer bind (no enzyme-substrate complexes form), so the rate of reaction decreases oe
(c) Answer: Binding at the allosteric site distorts the active site so the substrate cannot bind, lowering the rate.
(d) M1 the rate of reaction decreases as inhibitor concentration increases
(d) M1 because more enzyme molecules have an inhibitor bound at the allosteric site (active site distorted) at any one time oe
(d) A1 since substrate is fixed/saturating, but fewer functional enzyme molecules are available, the rate falls even though substrate is in excess oe
(d) Answer: Rate decreases progressively, as an increasing proportion of enzyme molecules are inhibited despite excess substrate.

Question 9

(a) B1 a pentose sugar (deoxyribose)
(a) B1 a phosphate group
(a) B1 a nitrogen-containing (organic) base, one of adenine, thymine, cytosine or guanine
(a) Answer: Deoxyribose sugar, phosphate group and a nitrogenous base.
(b) B1 DNA contains deoxyribose sugar; RNA contains ribose sugar
(b) B1 DNA is usually double-stranded; RNA is usually single-stranded (allow: DNA contains thymine, RNA contains uracil instead)
(b) Answer: DNA has deoxyribose and is double-stranded; RNA has ribose and is single-stranded.
(c) M1 adenine always pairs with thymine (by 2 hydrogen bonds) and cytosine always pairs with guanine (by 3 hydrogen bonds) oe
(c) M1 the large number of hydrogen bonds between the two strands, together with the double helix structure, makes the DNA molecule chemically stable oe
(c) A1 so the genetic information (base sequence) is protected from damage and can be copied/replicated accurately (semi-conservatively) oe
(c) Answer: Fixed base pairing and many hydrogen bonds give DNA a stable structure that protects and allows accurate copying of genetic information.

Question 10

(a) B1 adenine (base) attached to a ribose sugar, forming adenosine
(a) B1 with three phosphate groups attached in a chain oe
(a) Answer: Adenine + ribose (adenosine) with three phosphate groups attached.
(b) M1 hydrolysis of ATP (catalysed by ATP hydrolase) removes the terminal phosphate group, forming ADP and inorganic phosphate (Pi), releasing energy oe
(b) A1 this single-step, rapid, reversible reaction releases a relatively small, manageable quantity of energy exactly when and where it is needed in the cell oe
(b) Answer: ATP hydrolysis to ADP + Pi in one rapid step releases small, manageable amounts of energy on demand.
(c) B1 active transport (across membranes)
(c) B1 any other valid named energy-requiring process, e.g. muscle contraction, DNA replication, protein synthesis, cell division/mitosis oe
(c) Answer: Active transport; and one of muscle contraction / DNA replication / protein synthesis / cell division.

Question 11

Level 3 (5-6): A detailed and accurate account covering the structure and function of at least two of starch, glycogen and cellulose, with clear, logically linked explanations connecting structural features (e.g. branching, alpha/beta-glycosidic bonds, coiling, hydrogen bonding) to biological function (storage compactness, rapid hydrolysis, structural strength). Relevant scientific terminology is used accurately and consistently throughout.
Level 2 (3-4): A generally accurate account of the structure and function of the polysaccharides, but coverage may be less detailed, may focus mainly on one polysaccharide, or the links between structure and function are only partially explained. Scientific terminology is mostly appropriate with some minor inaccuracies.
Level 1 (1-2): Basic or isolated statements are made about the structure or the function of one or more polysaccharides, with little or no linkage made between structure and function. Limited use of scientific terminology.
Level 0 (0): No relevant content is provided.
Indicative content:
- Starch is a mixture of amylose (unbranched chain of alpha-glucose joined by 1,4-glycosidic bonds, coiled/helical shape) and amylopectin (branched, with 1,4 and 1,6-glycosidic bonds).
- Amylose's coiled, compact shape makes it well suited to storage, taking up little space.
- Amylopectin's branching gives many terminal ends, allowing enzymes to hydrolyse it rapidly, releasing glucose quickly when energy is needed.
- Starch is large and insoluble, so it does not affect the water potential of the cell (osmotically inactive), unlike free glucose.
- Glycogen has a structure similar to amylopectin but is even more highly branched, giving more ends for even faster hydrolysis, suited to the higher metabolic/respiratory demand of animal cells.
- Glycogen is the main storage polysaccharide in animals (e.g. stored as granules in liver and muscle cells).
- Cellulose is made of beta-glucose monomers joined by 1,4-glycosidic bonds, with alternate glucose molecules rotated 180 degrees, producing long, straight, unbranched chains.
- Many hydrogen bonds cross-link adjacent parallel cellulose chains into bundles called microfibrils.
- Microfibrils are arranged in bundles/fibres within the plant cell wall, giving it high tensile strength, providing structural support and preventing the cell bursting under turgor pressure.

Question 12

(a) M1 volume of stock solution = 10 / 4 = 2.5 cm³
(a) M1 volume of distilled water = 10 - 2.5 = 7.5 cm³
(a) A1 resulting concentration = 200 / 4 = 50 mmol dm^-3 cao
(a) Answer: 2.5 cm³ stock + 7.5 cm³ water; concentration = 50 mmol dm^-3.
(b) M1 concentration of diluted sample = 0.52 / 0.004
(b) M1 = 130 mmol dm^-3, then multiply by the dilution factor of 5 to find the original concentration
(b) A1 = 650 mmol dm^-3 awrt 650
(b) Answer: 650 mmol dm^-3.
(c) M1 the undiluted sample's absorbance was too high/outside the range of the calibration standards to be measured accurately, so it was diluted to bring the reading within the calibrated range oe
(c) M1 extrapolating the calibration curve assumes the same directly proportional relationship between absorbance and concentration continues beyond the highest standard tested oe
(c) A1 but this relationship may not remain linear at higher concentrations (e.g. due to saturation of the colour reaction), so a concentration read from an extrapolated curve may be inaccurate/less valid oe
(c) Answer: Dilution keeps the reading within the calibrated range; extrapolation assumes the proportional relationship holds beyond tested values, which may not be valid.