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 sodium.

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 water.

Thermal Strong H bondsLarge heat capacity High boiling & freezing point EvaporationCohesive H bonds attract molecules to each other strong cohesionSolvent Polarity Dissolves most substances

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 reactions

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 & Carbonates)

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, transportation)

Repetitive amino acid sequenceIrregular amino acid sequence

InsolubleSoluble

Less sensitive to environment changesMore sensitive to environment changes (temperature, pH, etc.)

Examples: collagen & keratinExamples: catalase & haemoglobin

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, keratin, elastin

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 + H2O

3.2.6State three functions of lipids. Buoyancy Warmth Energy storage

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 introns.

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 semi-conservative.

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 ligase.

7.2.3State that DNA replication is initiated at many points in eukaryotic chromosomes.

Protein Synthesis3.5.1Compare the structure of RNA and DNA.DNARNA

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 to

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 alkali.

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 concentration.

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.