Notes · BiologyUK · A-Levels

Biological molecules

Life is built from a small set of molecular families - carbohydrates, lipids, proteins and nucleic acids - assembled from monomers by condensation and broken down by hydrolysis. This chapter develops the structure of each family and links it to function, then treats enzymes as biological catalysts, the storage and transfer of information in DNA and ATP, and the properties of water and inorganic ions.

7 sections~24 min reading time3 competenciesLevel Foundation 1 · Standard 4 · Advanced 2

T·0111 / 8

Exam profile

AO1 · Describe the structure of carbohydrates, lipids, proteins and nucleic acids and the reactions that build and break themAO2 · Relate molecular structure to biological function and apply the enzyme model to unfamiliar dataAO3 · Analyse and evaluate results from biochemical tests and enzyme-rate investigations

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basic level

AS-Level requires the structures of the molecular families, the condensation/hydrolysis reactions, the biochemical tests and the enzyme model with its four factors.

higher level

The full A-Level expects the same content applied more quantitatively and synoptically - enzyme inhibition, DNA replication as evidence, and the properties of water and ions read across into later ecology and physiology topics.

Depth

Text

Contents · 7 sections

Biological molecules

Monomers, polymers and the two universal reactions#

FoundationAQA 7402 3.1.1DfE GCE Biology - biological molecules

A condensation reaction

Fig. 1A glycosidic bond forms by condensation and is broken by hydrolysis; one water molecule is released per bond formed.

A monomer is a small molecule that acts as a building block, and a polymer is a large molecule built from many monomers joined in a chain. The three polymer families you meet at A-Level are carbohydrates (monomers = monosaccharides), proteins (monomers = amino acids) and nucleic acids (monomers = nucleotides). Recognising the monomer-polymer relationship is the single organising idea of the whole chapter: once you know how one condensation reaction works, you understand how every polymer in the cell is assembled.

Polymers are built by condensation reactions. A condensation reaction joins two molecules together, forms a new chemical bond and releases a molecule of water. The same logic applies whichever family you are considering: a glycosidic bond forms between two monosaccharides, a peptide bond between two amino acids, and a phosphodiester bond between two nucleotides, and in every case one water molecule is released for each bond made. This is why a polymer of

n

monomers is joined by

n-1

bonds and releases

n-1

molecules of water.

Polymers are broken down by hydrolysis, the exact reverse of condensation. A hydrolysis reaction uses a molecule of water to break a chemical bond between two molecules (the word means splitting - lysis - by water - hydro). Digestion in the gut, the mobilisation of stored starch or glycogen, and the recycling of cellular molecules are all hydrolysis reactions; each bond broken consumes one water molecule. Enzymes such as amylase, protease and lipase catalyse these hydrolyses.

Because condensation and hydrolysis are exact opposites, the cell can build and dismantle the same molecule as its needs change, and the water balance of the two directions is symmetric. A common examination move is to give an unfamiliar polymer and ask you to predict the number of water molecules released on synthesis or required for complete hydrolysis - always

n-1

for a linear chain of

n

residues, or one per bond for a branched molecule where you count the bonds directly.

n\ \text{monomers} \rightarrow \text{polymer} + (n-1)\,\text{H}_2\text{O}

Condensation of a linear polymer

Joining $n$ monomers forms $n-1$ bonds and releases $n-1$ water molecules; complete hydrolysis reverses this and consumes $n-1$ water molecules.

Worked example

Counting water molecules in polymer synthesis

A glycogen branch is assembled from 200 alpha-glucose monomers joined into a single chain. How many water molecules are released, and how many would be needed to hydrolyse the chain completely back to glucose?

Count the bonds
A linear chain of $n$ monomers contains $n-1$ glycosidic bonds, so $200 - 1 = 199$ bonds are formed.
Water released on synthesis
Each condensation releases one water molecule, so 199 water molecules are released.
Water used on hydrolysis
Hydrolysis reverses each condensation, so breaking all 199 bonds consumes 199 water molecules.
$n - 1 = 200 - 1 = 199$

Result: 199 water molecules are released on synthesis and 199 are required for complete hydrolysis.

Typical mistakes

Writing that condensation 'removes' water without saying a bond is formed, or that hydrolysis 'adds' water without saying a bond is broken.
Giving $n$ water molecules for a chain of $n$ monomers instead of $n-1$ .

Active revision

A polypeptide is formed from 148 amino acids. State the number of peptide bonds present and the number of water molecules released during its synthesis.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Carbohydrates: sugars and polysaccharides#

StandardAQA 7402 3.1.2DfE GCE Biology - carbohydrates

Comparing the three glucose polysaccharides

Fig. 2Alpha-glucose gives compact stores (starch, glycogen); beta-glucose gives strong structural cellulose.

Monosaccharides are the simplest carbohydrates and the monomers of the family; the most important is glucose, a hexose with the formula

\text{C}_6\text{H}_{12}\text{O}_6

. Glucose exists as two isomers that differ only in the arrangement of the hydroxyl group on carbon 1: in alpha-glucose it points below the ring, in beta-glucose above it. This tiny difference has enormous consequences for the polysaccharides each can form, and is a favourite starting point for structure-function questions.

Two monosaccharides join by a condensation reaction to form a disaccharide linked by a glycosidic bond. Maltose is two alpha-glucose units (a 1,4-glycosidic bond); sucrose is glucose joined to fructose; lactose is glucose joined to galactose. Glucose, maltose, fructose and galactose are reducing sugars and give a positive Benedict's test (a brick-red precipitate on heating); sucrose is a non-reducing sugar and only tests positive after it is first hydrolysed with acid, then neutralised, then heated with Benedict's.

Storage polysaccharides are built from alpha-glucose. Starch, the plant store, is a mixture of amylose (an unbranched, helical 1,4-linked chain that is compact and good for storage) and amylopectin (a 1,4- and 1,6-linked branched chain whose many ends allow rapid hydrolysis). Glycogen, the animal store, is like amylopectin but even more highly branched, so it is very compact and can be mobilised quickly to meet the higher metabolic demands of animals. Both are insoluble, so they do not affect the water potential of the cell and do not diffuse out.

Cellulose, in contrast, is built from beta-glucose. To form a straight chain, alternate beta-glucose molecules must be flipped through 180 degrees, giving long, straight, unbranched chains. Many parallel chains are held together by large numbers of hydrogen bonds between them, forming microfibrils of great tensile strength that make the plant cell wall rigid and resist the turgor pressure of the cell. The same monomer polymerised through a different bond geometry therefore gives a structural material rather than a store - the clearest single illustration of structure determining function in the whole subject.

\text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O}

Formation of a disaccharide

Two hexoses (total $\text{C}_{12}\text{H}_{24}\text{O}_{12}$ ) condense to a disaccharide plus one water; note the loss of $\text{H}_2\text{O}$ .

Worked example

Identifying reducing and non-reducing sugars

A solution turns Benedict's reagent orange when first tested. A second portion of the same solution, boiled with hydrochloric acid, neutralised, then tested with Benedict's, turns brick-red. What can you conclude?

Interpret the first test
The orange colour is an intermediate positive result, so a reducing sugar (such as glucose or maltose) is present before any hydrolysis.
Interpret the second test
The stronger brick-red after acid hydrolysis shows that even more reducing sugar is now present; hydrolysis has broken a non-reducing sugar (such as sucrose) into its reducing monosaccharides.
Combine
The solution therefore contains both a reducing sugar and a non-reducing sugar. The deeper colour after hydrolysis is the diagnostic evidence for the non-reducing sugar.

Result: Both a reducing sugar and a non-reducing sugar (revealed by hydrolysis) are present.

Typical mistakes

Saying cellulose is made of alpha-glucose, or forgetting that alternate beta-glucose units are inverted.
Claiming sucrose gives a positive Benedict's test directly, without the acid hydrolysis step first.

Active revision

A student obtains a green colour with Benedict's solution before hydrolysis and a brick-red colour after acid hydrolysis and neutralisation. Explain what these results show about the sugars present.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Lipids: triglycerides and phospholipids#

StandardAQA 7402 3.1.3DfE GCE Biology - lipids

A triglyceride

Fig. 3Glycerol condenses with three fatty acids, forming three ester bonds and releasing three water molecules.

Lipids are a diverse group united by being non-polar and insoluble in water but soluble in organic solvents; they are not polymers because they are not built from a repeating monomer. A triglyceride is formed by condensation between one molecule of glycerol and three fatty acids, forming three ester bonds and releasing three water molecules. Each fatty acid has a carboxyl group and a long hydrocarbon tail, and it is those tails that make the molecule hydrophobic.

Fatty acids are described as saturated when every carbon-carbon bond in the tail is a single bond (the chain is saturated with hydrogen), and unsaturated when one or more carbon-carbon double bonds are present. Double bonds put kinks in the tails so the molecules cannot pack closely; unsaturated triglycerides therefore have lower melting points and tend to be liquid oils at room temperature, whereas saturated triglycerides are solid fats. This is the structural reason animal fats are solid and plant oils are liquid.

Triglycerides are excellent energy stores: they are highly reduced, so their oxidation releases about twice as much energy per gram as carbohydrate, and being insoluble they can be stored in large amounts without affecting cell water potential. Stored under the skin and around organs they also provide thermal insulation and physical protection, and their metabolic oxidation yields metabolic water - valuable to organisms such as desert mammals.

A phospholipid is a modified triglyceride in which one fatty acid is replaced by a phosphate group. This makes the molecule amphipathic: the phosphate head is polar and hydrophilic while the two fatty-acid tails are non-polar and hydrophobic. In water, phospholipids spontaneously arrange into a bilayer, heads facing the water on both surfaces and tails shielded in the centre - the structural basis of every cell membrane, developed fully in the Cells chapter.

Lipids are detected by the emulsion test: the sample is dissolved in ethanol, then the solution is added to water. A lipid, being insoluble in water, comes out of solution as tiny droplets that scatter light and produce a cloudy white emulsion - a positive result. A control with no lipid stays clear.

\text{glycerol} + 3\ \text{fatty acids} \rightarrow \text{triglyceride} + 3\,\text{H}_2\text{O}

Triglyceride synthesis

Three ester bonds form, releasing three water molecules; hydrolysis reverses this.

Worked example

Counting bonds and water in a triglyceride

A cell hydrolyses 5 triglyceride molecules completely to glycerol and fatty acids. How many ester bonds are broken and how many water molecules are used?

Bonds per molecule
Each triglyceride contains three ester bonds (one per fatty acid).
Total bonds
$5 \times 3 = 15$ ester bonds are broken.
Water used
Hydrolysis uses one water molecule per bond, so 15 water molecules are consumed.
$5 \times 3 = 15$

Result: 15 ester bonds are broken and 15 water molecules are used.

Typical mistakes

Calling triglycerides polymers, or saying they contain glycosidic or peptide bonds rather than ester bonds.
Confusing the roles of the hydrophilic phosphate head and hydrophobic tails in a phospholipid bilayer.

Active revision

Explain, in terms of bonding and energy, why a triglyceride releases about twice as much energy per gram on oxidation as a carbohydrate such as glycogen.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Proteins: from amino acids to four levels of structure#

StandardAQA 7402 3.1.4DfE GCE Biology - proteins

Four levels of protein structure

Fig. 4Each level of folding is stabilised by particular bonds; the primary sequence dictates all the others.

The monomer of a protein is the amino acid. Every amino acid has the same general structure: a central carbon bonded to an amino group (

\text{-NH}_2

), a carboxyl group (

\text{-COOH}

), a hydrogen atom and a variable R group (side chain). There are twenty amino acids that differ only in their R group, and it is the sequence of R groups that ultimately determines a protein's shape and function.

Amino acids join by condensation to form a peptide bond between the amino group of one and the carboxyl group of the next, releasing one water molecule. A chain of many amino acids is a polypeptide. This sequence of amino acids is the primary structure, and because it is coded directly by a gene, it determines every higher level of folding that follows - change one amino acid and you may change the whole protein.

The secondary structure arises when the polypeptide coils or folds through hydrogen bonds between the carboxyl and amino groups of the backbone (not the R groups). Two patterns recur: the alpha-helix, a right-handed coil, and the beta-pleated sheet. These local structures are stabilised entirely by weak but numerous hydrogen bonds, which is why heat and extremes of pH can disrupt them.

The tertiary structure is the overall three-dimensional shape of a single polypeptide, produced by interactions between the R groups: hydrogen bonds, ionic bonds, disulfide bridges (covalent, between cysteine residues) and hydrophobic interactions. This precise folding creates the specific shape - for example an enzyme's active site - on which function depends. The quaternary structure applies to proteins made of more than one polypeptide chain, held together by the same kinds of R-group interactions; haemoglobin, with four chains and four haem groups, is the standard example.

Proteins are detected by the biuret test: adding sodium hydroxide then dilute copper(II) sulfate gives a lilac/purple colour if peptide bonds are present, and stays blue if they are not. The test detects the peptide bond itself, so it responds to any polypeptide regardless of the amino acids involved.

n\ \text{amino acids} \rightarrow \text{polypeptide} + (n-1)\,\text{H}_2\text{O}

Polypeptide formation

A chain of $n$ amino acids contains $n-1$ peptide bonds and releases $n-1$ water molecules.

Worked example

Peptide bonds and water in a polypeptide

A polypeptide is 305 amino acids long. State the number of peptide bonds it contains and the number of water molecules released when it was synthesised.

Peptide bonds
A chain of $n$ residues contains $n-1$ peptide bonds, so $305 - 1 = 304$ bonds.
Water released
Each peptide bond forms by condensation, releasing one water molecule, so 304 water molecules are released.
$305 - 1 = 304$

Result: 304 peptide bonds and 304 water molecules.

Typical mistakes

Attributing the alpha-helix to R-group bonds instead of backbone hydrogen bonds.
Forgetting that the biuret test detects the peptide bond, so it works for any protein.

Active revision

Explain why boiling a solution of an enzyme abolishes its catalytic activity, referring to the levels of protein structure and the bonds involved.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Enzymes: mechanism and the factors affecting rate#

AdvancedAQA 7402 3.1.4.2DfE GCE Biology - enzymes

An enzyme lowers activation energy

Fig. 5The enzyme provides a route with a lower activation energy; the overall energy change is unaffected.

Enzymes are biological catalysts - proteins with a specific tertiary structure that speed up metabolic reactions without being used up. They work by lowering the activation energy of a reaction: the minimum energy that colliding molecules must have to react. By binding the substrate in the active site and straining or orienting its bonds, the enzyme provides an alternative route with a lower energy barrier, so reactions that would be negligibly slow at body temperature proceed rapidly.

The relationship between enzyme and substrate is described by the induced-fit model, which has superseded the older lock-and-key idea. The active site is not a rigid, pre-formed complement to the substrate; instead, as the substrate binds, the active site changes shape slightly to mould around it, straining the substrate's bonds and catalysing the reaction. Specificity still arises because only a substrate with a complementary shape can induce the correct change, so each enzyme catalyses only one type of reaction.

Temperature raises the rate at first because molecules have more kinetic energy, collide more often and more collisions exceed the activation energy - roughly doubling the rate for each 10 degree rise up to the optimum. Above the optimum, the increasing vibration breaks the weak hydrogen and ionic bonds holding the tertiary structure together; the active site changes shape and the enzyme denatures, so the rate falls sharply. Denaturation is usually irreversible.

pH affects rate because hydrogen ions interfere with the ionic and hydrogen bonds that maintain the active site's shape. Each enzyme has an optimum pH at which its active site is the ideal shape; on either side the active site distorts and the rate falls, and extreme pH denatures the enzyme. Substrate concentration raises the rate until every active site is occupied as substrate arrives - the enzyme concentration then becomes limiting and the rate plateaus at a maximum (

V_{max}

). Raising enzyme concentration raises the rate provided substrate is in excess.

Inhibitors reduce enzyme activity. A competitive inhibitor has a shape similar to the substrate and binds to the active site, blocking the substrate; its effect can be overcome by adding more substrate. A non-competitive inhibitor binds elsewhere (an allosteric site), changing the shape of the active site so the substrate no longer fits; adding more substrate does not relieve it. Distinguishing the two from rate-versus-substrate graphs is a classic AO3 task.

\text{rate} = \dfrac{\Delta\ \text{amount of product (or substrate)}}{\Delta\ \text{time}}

Rate of an enzyme reaction

Read the initial (steepest) part of a progress curve, where substrate is not yet limiting, to compare conditions fairly.

Rate against substrate concentration

Fig. 6Rate rises with substrate concentration then plateaus at Vmax when all active sites are occupied.

Rate against temperature

Fig. 7Rate rises to an optimum, then falls sharply above it as the enzyme denatures.

Worked example

Calculating an initial rate

In a catalase investigation, 18 cm3 of oxygen was collected in the first 20 seconds. Calculate the mean rate of reaction over this interval in cm3 s^-1.

Identify the change and time
Volume of oxygen = 18 cm3; time = 20 s.
Apply the rate equation
$\text{rate} = \dfrac{18}{20}$ .
$\text{rate} = \frac{18\ \text{cm}^3}{20\ \text{s}} = 0.90\ \text{cm}^3\,\text{s}^{-1}$
State with units
The mean initial rate is 0.90 cm3 s^-1.

Result: 0.90 cm3 s^-1.

Typical mistakes

Saying enzymes are 'killed' by heat - proteins denature, they are not alive.
Reading a rate from the plateau of a progress curve rather than its initial linear gradient, so conditions are compared unfairly.

Active revision

An enzyme-controlled reaction produced 24 cm3 of oxygen in the first 30 seconds and 30 cm3 in total after 90 seconds. Calculate the mean initial rate over the first 30 seconds and explain why it is greater than the mean rate over the full 90 seconds.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Nucleic acids, DNA replication and ATP#

AdvancedAQA 7402 3.1.5AQA 7402 3.1.6DfE GCE Biology - nucleic acids and ATP

Comparing DNA and RNA

Fig. 8DNA and RNA share the nucleotide plan but differ in sugar, one base and strandedness.

The monomer of a nucleic acid is the nucleotide, made of a pentose sugar, a phosphate group and a nitrogen-containing organic base. In DNA the sugar is deoxyribose and the bases are adenine, thymine, cytosine and guanine; in RNA the sugar is ribose and thymine is replaced by uracil. Nucleotides join by condensation, forming phosphodiester bonds between the phosphate of one and the sugar of the next, building a sugar-phosphate backbone.

DNA is a double helix of two antiparallel polynucleotide strands held together by hydrogen bonds between complementary base pairs: adenine pairs with thymine (two hydrogen bonds) and cytosine with guanine (three). This complementary base pairing means the two strands carry the same information in mirror form, which is the key to accurate replication, and it forces the equalities of Chargaff's rule - the amount of adenine equals thymine and cytosine equals guanine.

DNA is copied by semi-conservative replication. DNA helicase breaks the hydrogen bonds and unwinds the helix; each old strand acts as a template; free DNA nucleotides align opposite their complementary bases and are joined by DNA polymerase to form a new strand. Each daughter molecule therefore contains one original (conserved) strand and one new strand - hence semi-conservative. The classic Meselson-Stahl experiment, using heavy nitrogen (

^{15}\text{N}

) then light nitrogen (

^{14}\text{N}

), produced a single band of intermediate density after one generation, providing the definitive evidence for this mechanism over the conservative and dispersive alternatives.

ATP (adenosine triphosphate) is the universal immediate energy currency of the cell: a nucleotide made of adenine, ribose and three phosphate groups. It is hydrolysed by ATP hydrolase to ADP and an inorganic phosphate ion (

\text{P}_i

), releasing a small, usable amount of energy for cellular work such as active transport, muscle contraction and synthesis. Being a small, single, immediate release makes ATP more manageable than glucose - it is a rechargeable carrier, not a store.

ATP is resynthesised from ADP and

\text{P}_i

by ATP synthase during respiration and photosynthesis, so the ADP-ATP cycle turns constantly. The released inorganic phosphate can also phosphorylate other molecules, making them more reactive - for example the phosphorylation of glucose at the start of glycolysis. Because the cell holds only a small pool of ATP, it must be regenerated continuously, which is why respiration never stops.

\text{ATP} + \text{H}_2\text{O} \rightleftharpoons \text{ADP} + \text{P}_i + \text{energy}

ATP hydrolysis and resynthesis

ATP hydrolase releases usable energy; ATP synthase reverses the reaction during respiration and photosynthesis.

Semi-conservative replication

Fig. 9Each daughter molecule keeps one parental strand and gains one new strand.

Worked example

Using complementary base pairing

In a double-stranded DNA molecule, 32% of the bases are adenine. Calculate the percentage of cytosine.

Pair adenine with thymine
By complementary base pairing, %thymine = %adenine = 32%.
Find the remaining fraction
A + T = 32 + 32 = 64%, so C + G = 100 - 64 = 36%.
Split C and G equally
%cytosine = %guanine, so each is $36 \div 2 = 18\%$ .
$\%C = \frac{100 - 2(32)}{2} = 18\%$

Result: Cytosine is 18%.

Typical mistakes

Muddling the base pairs or their hydrogen-bond numbers (A-T two, C-G three).
Describing ATP as a long-term energy store rather than an immediate, rechargeable energy carrier.

Active revision

A sample of double-stranded DNA contains 18% thymine. Calculate the percentage of guanine, showing your reasoning using complementary base pairing.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Water and inorganic ions#

StandardAQA 7402 3.1.7DfE GCE Biology - water and ions

Properties of water and their importance

Fig. 10Hydrogen bonding underlies the solvent, thermal and cohesive properties of water.

Water is a polar molecule: the oxygen atom is slightly negative and the hydrogen atoms slightly positive because oxygen pulls the shared electrons towards itself. This dipole allows hydrogen bonds to form between neighbouring water molecules - individually weak but collectively strong and responsible for nearly every biologically useful property of water. Almost every metabolic reaction occurs in aqueous solution, so water is the medium of life as well as a reactant in hydrolysis.

Water is an excellent solvent for polar and ionic substances, which dissolve when their charged parts are surrounded by water's dipoles; this lets metabolites be transported (in blood and phloem) and lets reactions occur in solution. Its high specific heat capacity means a large input of energy is needed to raise its temperature, so bodies of water and cell contents resist rapid temperature change - buffering organisms and habitats against fluctuation.

Water has a high latent heat of vaporisation because many hydrogen bonds must be broken to evaporate it; this makes evaporation (sweating, transpiration, panting) an efficient cooling mechanism, since a great deal of heat is removed for a small mass of water lost. Cohesion - the attraction of water molecules to one another through hydrogen bonds - produces high surface tension and, crucially, allows continuous columns of water to be drawn up the xylem under tension during transpiration.

Inorganic ions occur in solution in organisms, each in characteristic concentrations, and take part in many processes. Hydrogen ions (

\text{H}^+

) determine pH and therefore enzyme activity; iron ions (

\text{Fe}^{2+}

) are part of haemoglobin and carry oxygen; sodium ions (

\text{Na}^+

) are needed for the co-transport of glucose and amino acids across membranes; and phosphate ions (

\text{PO}_4^{3-}

) form the backbone of nucleotides (DNA, RNA and ATP) and are used to phosphorylate other compounds. Recognising these named roles is a frequent low-tariff mark that also sets up later topics.

Q = mc\,\Delta T

Heat energy and temperature change

$m$ is mass, $c$ the specific heat capacity and $\Delta T$ the temperature change; a high $c$ means a large $Q$ is needed for a given $\Delta T$ .

Worked example

Energy to warm body water

Calculate the heat energy needed to raise the temperature of 2.0 kg of water by 1.5 degrees C. Take the specific heat capacity of water as 4200 J kg^-1 degrees C^-1.

State the equation
$Q = mc\,\Delta T$ with $m = 2.0$ kg, $c = 4200$ J kg^-1 degrees C^-1, $\Delta T = 1.5$ degrees C.
Substitute
$Q = 2.0 \times 4200 \times 1.5$ .
$Q = 2.0 \times 4200 \times 1.5 = 12600\ \text{J}$
State the answer
$Q = 12\,600$ J, or 12.6 kJ - a large energy for a small temperature rise, illustrating water's high specific heat capacity.

Result: 12 600 J (12.6 kJ).

Typical mistakes

Describing properties of water without explaining them via hydrogen bonding, or without giving a biological consequence.
Confusing high specific heat capacity (resists temperature change) with high latent heat of vaporisation (cooling on evaporation).

Active revision

Explain how two named properties of water make it suitable as a transport medium in the blood and as a coolant during sweating.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Biology 7402 specification

Biological molecules

7 sections~24 min reading time3 competenciesLevel Foundation 1 · Standard 4 · Advanced 2

Biological molecules

Key points

Counting water molecules in polymer synthesis

Key points

Identifying reducing and non-reducing sugars

Key points

Counting bonds and water in a triglyceride

Key points

Peptide bonds and water in a polypeptide

Key points

Calculating an initial rate

Key points

Using complementary base pairing

Key points

Energy to warm body water

Sources

Biological molecules

Key points

Counting water molecules in polymer synthesis

Key points

Identifying reducing and non-reducing sugars

Key points

Counting bonds and water in a triglyceride

Key points

Peptide bonds and water in a polypeptide

Key points

Calculating an initial rate

Key points

Using complementary base pairing

Key points

Energy to warm body water

Sources