Chemical Elements & Water3.1.1State that the most frequently
occurring chemical elements in living things are carbon, hydrogen,
oxygen and nitrogen.
3.1.2State that a variety of other elements are needed by living
organisms, including sulphur, calcium, phosphorus, iron and
3.1.3State one role for each of the elements mentioned in
3.1.2.Sulphur Required for synthesis of two amino acids.Calcium -
Messenger by binding to calmodulin and other proteins which
regulate transcription and other processes in the cell.Phosphorus:
Part of DNA molecules and of the phosphate groups in ATP.Iron:
Required for synthesis of cytochromes which are proteins used
during electron transport for aerobic cell respiration.Sodium:
Raises the solute concentration in the cytoplasm which causes water
to enter by osmosis.
3.1.4Draw and label a diagram showing the structure of water
molecules to show their polarity and hydrogen bond formation.
3.1.5Outline the thermal, cohesive and solvent properties of
Thermal Strong H bondsLarge heat capacity High boiling &
freezing point EvaporationCohesive H bonds attract molecules to
each other strong cohesionSolvent Polarity Dissolves most
3.1.6Explain the relationship between the properties of water
and its uses in living organisms as a coolant, medium for metabolic
reactions and transport medium. Cooling properties evaporation
removes heat energy from organisms Solvent properties allows
dissolved substances to be transported & a medium for metabolic
Carbohydrates, Lipids & Proteins3.2.1Distinguish between
organic and inorganic compounds.Organic contains carbon, found in
living organismsInorganic no carbon (some exceptions e.g. CO2, HCO3
3.2.2Identify amino acids, glucose, ribose and fatty acids from
diagrams showing their structure.
7.5.1Explain the four levels of protein structure, indicating
the significance of each level.1. Primary structure Amino acid
sequence Held together by covalent bonds2. Secondary structure
Either an -helix or a -pleated sheet Formed by hydrogen bonds
between the main chain and the peptide groups.3. Tertiary structure
Secondary structure is folded to form a 3D structure Stabilised by
hydrogen bonds, ionic bonds, disulphide bridges and hydrophobic
& hydrophilic interactions between R groups.4. Quaternary
structure Tertiary structures interact to form a protein May have a
prosthetic group (inorganic compound), forming a conjugated
protein.7.5.2Outline the difference between fibrous and globular
proteins, with reference to two examples of each protein
type.Fibrous ProteinGlobular Protein
Long and narrowGlobular/ rounded
Provides strength & supportFunctional (catalyses reactions,
Repetitive amino acid sequenceIrregular amino acid sequence
Less sensitive to environment changesMore sensitive to
environment changes (temperature, pH, etc.)
Examples: collagen & keratinExamples: catalase &
7.5.3Explain the significance of polar and non-polar amino
acids.Polar amino acids have polar R groups while non-polar amino
acids have non-polar R groups, resulting in hydrophobic and
hydrophilic amino acids. For soluble proteins, polar amino acids
tend to be found on the outside, while non-polar amino acids tend
to be found on the inside. The position of proteins in membranes
can be controlled by the distribution of polar and non-polar amino
acids. Non-polar amino acids are embedded in the membrane, while
polar amino acids are outside, causing parts of the protein to
protrude out from the membrane.Polar amino acids form hydrophilic
channels in membrane proteins.The positioning of polar/ non-polar
amino acids in enzymes allows the substrate to bind to the active
site more easily.7.5.4State four functions of proteins, giving a
named example of each.Structure support for body tissueCollagen,
Hormones regulate blood/sugar levelsInsulin, glucagon
Immunity bind to antigensAntibodies, immunoglobulins
Transport transport oxygenHaemoglobin
Movement contract to move musclesMyosin, troponin
Enzymes catalyse reactionsCatalase, lipase, pepsin
3.2.3List three examples each of monosaccharides, disaccharides
and polysaccharides.Monosaccharide glucose, galactose,
fructoseDisaccharide maltose, lactose, sucrosePolysaccharide
starch, glycogen, cellulose3.2.4State one function of glucose,
lactose and glycogen in animals, and of fructose, sucrose and
cellulose in plants.Glucose production of ATPs during
respirationLactose milkGlycogen temporary energy storage
Fructose sweet taste attracts other animalsSucrose energy
storage for plantsCellulose constitutes cell walls3.2.5Outline the
role of condensation and hydrolysis in the relationships between
monosaccharides, disaccharides and polysaccharides; between fatty
acids, glycerol and triglycerides; and between amino acids and
polypeptides.Condensation is when molecules join together to form a
chain. Hydrolysis is the reverse.-Glucose +-Glucose Maltose +
H2O-Glucose +-Fructose Saccharose + H2O-Galactose +-Glucose Lactose
3.2.6State three functions of lipids. Buoyancy Warmth Energy
3.2.7Compare the use of carbohydrates and lipids in energy
storage.Carbohydrates are used as a temporary energy source because
it can be dissolved in water and transported around the body.Lipids
are used as a more permanent energy source. They contain more
energy/g than carbohydrates.
DNA Structure3.3.1Outline DNA nucleotide structure in terms of
sugar (deoxyribose), base and phosphate.
3.3.2State the names of the four bases in DNA.Cytosine, Guanine,
Adenine & Thymine3.3.3Outline how DNA nucleotides are linked
together by covalent bonds into a single strand.A covalent bond is
formed between the phosphate group of one nucleotide and the sugar
of another nucleotide during a condensation reaction.
3.3.4Explain how a DNA double helix is formed using
complementary base pairing and hydrogen bonds.The two strands of
DNA are held together by weak hydrogen bonds which bond the base
pairs. G-C is held together by a three hydrogen bonds, while A-T is
held together by two hydrogen bonds. The complementary base pairing
system means that the strands must be antiparallel as the bases
must face each other.
3.3.5Draw and label a simple diagram of the molecular structure
of DNA.7.1.1Describe the structure of DNA, including the
antiparallel strands, 35 linkages and hydrogen bonding between
purines and pyrimidines. Pyrimidines one ringed structure (T &
C) Purines double ringed structure (A & G) Purines link to
pyrimidines via hydrogen bonds (AT 2, GC 3)
7.1.2Outline the structure of nucleosomes.
7.1.3State that nucleosomes help to supercoil chromosomes and
help to regulate transcription.
7.1.4Distinguish between unique or single-copy genes and highly
repetitive sequences in nuclear DNA.Single copy genes have one
locatable region on a DNA molecule and are used during protein
synthesisHighly repetitive sequences can repeat up to 10 000 times
on a DNA molecule. They are not involved in protein synthesis.
7.1.5State that eukaryotic genes can contain exons and
DNA Replication3.4.2Explain the significance of complementary
base pairing in the conservation of the base sequence of
DNA.Adenine always pairs up with guanine, while cytosine pairs up
with thymine. Because DNA replication is semi-conservative, when
the DNA strands are separated, the base sequence is conserved when
the free base attaches to the corresponding base, resulting in two
new identical strands.3.4.3State that DNA replication is
7.2.1State that DNA replication occurs in a 5 3direction.
3.4.1Explain DNA replication in terms of unwinding the double
helix and separation of the strands by helicase, followed by
formation of the new complementary strands by DNA
polymerase7.2.2Explain the process of DNA replication in
prokaryotes, including the role of enzymes (helicase, DNA
polymerase, RNA primase and DNA ligase), Okazaki fragments and
deoxynucleoside triphosphates.1. Helicase unwinds and separates the
DNA strands by breaking the hydrogen bonds between the base pairs2.
SSB proteins prevent the two strands from sticking together again3.
RNA primase adds RNA primer - short sequences of RNA to the
strands, allowing DNA Polymerase III to begin adding nucleotides in
a 5 3 direction.4. The nucleotides are originally
deoxyribonucleoside triphosphates but they lose two phosphate
groups to release energy.5. The antiparallel nature of the DNA
strands means that one strand is synthesised discontinuously
(lagging strand), while the other is synthesised continuously
(leading strand). 6. The lagging strand is replicated in fragments
and RNA primer is added as the original DNA strand is constantly
unwound. These fragments are called Okazaki fragments.7. DNA
Polymerase I removes the RNA primer and replaces them with DNA8.
The gaps between the Okazaki fragments are filled in by DNA
7.2.3State that DNA replication is initiated at many points in
Protein Synthesis3.5.1Compare the structure of RNA and
Deoxyribose sugarRibose sugar
Thymine baseUracil base
Double strandedSingle stranded
Double helixSingle spiral
7.3.1State that transcription is carried out in a direction.
3.5.2Outline DNA transcription in terms of the formation of an
RNA strand complementary to the DNA strand by RNA
polymerase.7.3.3Explain the process of transcription in
prokaryotes, including the role of the promoter region, RNA
polymerase, nucleoside triphosphates and the terminator.
1. The promoter region is identified and RNA polymerase binds to
it.2. RNA polymerase unzips & separates the DNA strands3. One
of the exposed strands, or the template strand, is used to
synthesise a complementary RNA sequence by RNA polymerase in a 5 3
direction.4. As the ribonucleoside triphosphates are covalently
bonded they form nucleosides by releasing two phosphate groups to
provide energy for the bonding.5. Thymine is replaced by uracil6.
Transcription continues until the RNA polymerase encounters the
terminator.7. RNA polymerase is released and the mRNA separates
from the DNA which recoils.3.5.3Describe the genetic code in terms
of codons composed of triplets of bases.A codon consists of three
bases. Each codon codes for an amino acid, which link together to
form proteins. DNA therefore regulates protein
synthesis.7.3.2Distinguish between the sense and antisense strands
of DNA.The antisense strand is the template strand upon which the
RNA is synthesised. The sense strand is the one which corresponds
to the RNA which is synthesised.
7.3.4State that eukaryotic RNA needs the removal of introns to
form mature mRNA.
3.5.5Discuss the relationship between one gene and one
polypeptide.Genes store the information used to make polypeptides.
Three genes make up a codon. The order of the codons determines the
order of the amino acids, which in turn determines the type of
polypeptide. 7.4.1Explain that each tRNA molecule is recognized by
a tRNA-activating enzyme that binds a specific amino acid to the
tRNA, using ATP for energy. 1. Each tRNA-activating enzyme
recognises one tRNA molecule. The 3D structure and chemical
properties allow the enzymes to distinguish between the tRNA
molecules.2. The enzyme binds an ATP molecule to the amino acid to
provide energy.3. At the 3 end of the tRNA molecule is a CCA
nucleotide sequence which the amino acid attaches to.4. The tRNA
molecule is able to participate in translation with the energy
provided by the bond.
7.4.2Outline the structure of ribosomes, including protein and
RNA composition, large and small subunits, three tRNA binding sites
and mRNA binding sites.Ribosomes are made of protein, which
provides stability, and ribosomal RNA (rRNA), which provides
catalytic activity. A ribosome consists of a large subunit and a
small subunit. The large subunit has three tRNA binding sites
(aminacyl site, peptidyl site and an exit site), while the small
subunit has an mRNA binding site. However, the exit site only
allows the tRNA to exit, and two sites (A and P) are therefore used
by the tRNA to synthesise the peptide. 3.5.4Explain the process of
translation, leading to polypeptide formation.7.4.6Explain the
process of translation, including ribosomes, polysomes, start
codons and stop codons.1. The small subunit of the ribosome binds
to the 5 end of the mRNA and moves it along until the start codon
(AUG) is reached.2. The tRNA with the matching anticodon binds to
the start codon and the large subunit is aligned with the tRNA so
the tRNA is at the P site.3. The next tRNA bonds with the mRNA at
the A site.4. The amino acid in the P site forms a peptide bond
with the amino acid in the A site. 5. The large subunit moves
forward over the smaller subunit, and the smaller subunit rejoins
the larger subunit, allowing the first tRNA to be released, and the
next mRNA codon to be exposed.6. The second tRNA with the two amino
acids is now at the P site, while a new tRNA with the matching
anticodon to the mRNA starts to bond at the A site.7. The process
is repeated, allowing a long polypeptide to be synthesised.8. When
the ribosome reaches the stop codon on the mRNA, the polypeptide is
released.9. Several ribosomes can translate the mRNA at the same
time. This cluster of ribosomes is called a polysome.
7.4.3State that translation consists of initiation, elongation,
translocation and termination.Initiation P site is
occupiedElongation P & A sites are occupiedTranslocation E
& P sites are occupied, amino acids have formed
peptidesTermination no sites are occupied.7.4.4State that
translation occurs in a 5 3 direction.
7.4.5Draw and label a diagram showing the structure of a peptide
bond between two amino acids.
7.4.7State that free ribosomes synthesize proteins for use
primarily within the cell, and that bound ribosomes synthesise
proteins primarily for secretion or for lysosomes.
Enzymes3.6.1Define enzyme and active site.Enzyme a biological
catalystActive site the site on an enzyme which the substrate binds
3.6.2Explain enzymesubstrate specificity.Enzymes are specific to
their substrates as the shape and chemical properties of the enzyme
and substrate must match in order for the enzyme to bind to the
substrate. 3.6.3Explain the effects of temperature, pH and
substrate concentration on enzyme activity.As the temperature of
the enzymes environment increases, the enzyme activity increases
until the optimum temperature is reached, at which point the enzyme
activity decreases. At low temperatures, the enzyme is unable to
bind to its substrate due to an insufficient number of particles
moving fast enough. At high temperatures, the protein of the enzyme
denatures, changing the shape of the active site.
Enzymes have an optimum pH, above or below which the enzymes
will begin to denature as their environment is too acidic or too
As the substrate concentration increases, the enzyme activity
increases due to an increase in the number of random collisions.
However, the enzyme activity plateaus after a certain point as
there are not enough active sites for all the substrate particles
to bind to.
3.6.4Define denaturation.Denaturation is a change in the
structure of a protein, resulting in a change in biological
properties.3.6.5Explain the use of lactase in the production of
lactose-free milk.Lactase is an enzyme which breaks lactose down
into galactose and glucose. To produce lactose free milk, lactase
can either be added directly to the milk, or it can be immobilised
and bound to an inert porous substance e.g. alginate beads. The
milk is then run over/ through the beads, resulting in lactose free
milk.7.6.1State that metabolic pathways consist of chains and
cycles of enzyme-catalysed reactions.7.6.2Describe the induced-fit
model.The active site of the enzyme does not fit the substrate
perfectly when the substrate first attaches to it. The active site
changes shape as it binds to the substrate, and only then does it
perfectly fit. As the substrate binds to the active site, its bonds
are weakened and the activation energy of the reaction is reduced.
This allows different substrates to bind to one enzyme.
7.6.3Explain that enzymes lower the activation energy of the
chemical reactions that they catalyse.For a reaction to occur, the
system needs a specific amount of activation energy. The activation
energy is the energy required to overcome the bonds in the
reactants. Enzymes work by reducing the amount of activation energy
required for the reaction to occur, resulting in a faster reaction.
As the enzyme binds to the substrate, the bonds within the
substrate weaken, resulting in a reduction in the amount of energy
required for the reaction to begin.
7.6.4Explain the difference between competitive and
non-competitive inhibition, with reference to one example of
each.Competitive inhibitors stop the reaction from occurring by
binding to the active site of the enzyme. This prevents the
substrate from binding to the enzyme, resulting in a decrease in
the number of reactions occurring. The effect of competitive
inhibitors can be reduced by increasing substrate
Non-competitive inhibitors bind to the allosteric site instead.
This causes the enzymes active site to change, preventing the
substrate from fitting into, and therefore binding to the active
site of the enzyme. The effect of non-competitive inhibitors is not
reduced by an increase in substrate concentration.
Melonate is an example of a competitive inhibitor. It binds to
the active site of the dehydrogenase enzyme, preventing succinate
from binding to dehydrogenase.
ATP is a non-competitive inhibitor. It binds to the allosteric
site of phosphofructokinase when there is an excess amount of ATP,
resulting in a decrease in the production of ATP.
7.6.5Explain the control of metabolic pathways by end-product
inhibition, including the role of allosteric sites.Metabolic
pathways involve many reactions catalysed by enzymes. When there is
an excess of the end-product, production of it is no longer
required. The end-product can therefore act as a non-competitive
inhibitor by binding to the allosteric site of the first enzyme,
resulting in a shutdown of the entire pathway. This prevents excess
production of a product.