Year 12 Biology Knowledge Organiser PDF

A level biology Year 1 Knowledge organiser NAME: CLASS TEACHER: Study tip: Retrieval practice using knowledge organisers will help you embed fundamental concepts/facts/vocabulary, leaving you with more ‘brain space’ to focus on exam technique and further developing your understanding of the subject. Effective use of knowledge organisers involves active testing:  Look, cover, check key concepts and vocabulary  Teach someone else  Test yourself or ask someone to test you  Make a mind map from memory and check progress 1 TABLE OF CONTENTS 1.1 Topic 1: Biological Molecules................................................................................................................. 3 1.1.1 B1.1 Carbohydrates....................................................................................................................................................... 3 1.1.2 B1.2 Lipids..................................................................................................................................................................... 7 1.1.3 B1.3 Proteins................................................................................................................................................................. 9 1.1.4 B1.4 DNA and protein synthesis.................................................................................................................................. 11 1.1.5 B1.5 Enzymes.............................................................................................................................................................. 17 1.1.6 B1.6 Inorganic ions...................................................................................................................................................... 23 1.1.7 B1.7 Water.................................................................................................................................................................. 23 1.2 B2.1 Eukaryotic and prokaryotic cell structure and function................................................................. 25 1.3 B2.2 Viruses........................................................................................................................................ 30 1.4 B2.3 Eukaryotic cell cycle and division.................................................................................................. 34 1.5 B2.4 Sexual reproduction in mammals................................................................................................. 38 1.6 B2.5 Sexual reproduction in plants....................................................................................................... 42 1.7 B3.1 Classification............................................................................................................................... 45 1.8 B3.2 Natural selection......................................................................................................................... 48 1.9 B3.3 Biodiversity................................................................................................................................. 50 B4.1 Surface area to volume ratio and B4.2 Cell transport mechanisms........................................................... 52 1.10 B4.3 Gas exchange.............................................................................................................................. 56 1.11 B4.4 Circulation................................................................................................................................... 62 1.12 B4.5 Transport of gases in the blood.................................................................................................... 69 1.13 B4.6 Transfer of materials between the circulatory system and cells..................................................... 69 1.14 B4.7 Transport in plants...................................................................................................................... 72 2 Knowledge Organiser Edexcel AS Biology B 1.1 TOPIC 1: BIOLOGICAL MOLECULES 1.1.1 B1.1 Carbohydrates Recall the difference between monosaccharides, disaccharides and polysaccharides. Recall the structure of the hexose glucose (alpha and beta) and the pentose ribose. Describe how monosaccharides join to form disaccharides through condensation reactions forming glycosidic bonds, and how these can be split through hydrolysis reactions. Describe how polysaccharides are made through condensation reactions forming glycosidic bonds, and how these can be split through hydrolysis reactions. Explain how the structure of glucose, starch, glycogen and cellulose relates to their function. Key vocabulary You MUST know the meaning of the terms below. Monomer Repeating subunits e.g. monosaccharides Polymer Long chains of repeating subunits e.g. polysaccharides Condensation reaction Two monomers combine, producing a covalent bond, and a molecule of water is eliminated. Hydrolysis reaction A monomer is released from a polymer by the addition of a molecule of water. The covalent bond between the monomers is broken. Isomer Molecules with the same molecular formula e.g. C6H12O6 but a different structural formula. Covalent bond Strong chemical link formed between two atoms when they share electrons. Glycosidic bond The covalent bond between sugar monomers. 3 Carbohydrates (CH2O)n Type of carbohydrate Examples you need to know Monosaccharides Pentoses (C5H10O5) e.g. Ribose and deoxyribose Hexoses (C6H12O6) e.g. Alpha glucose, fructose, and galactose Disaccharides 1. Maltose (or malt sugar) is glucose–glucose. It is formed 1 on digestion of starch by amylase, because this enzyme breaks starch down into two- 2 glucose units. 2. Sucrose (or cane sugar) is glucose–fructose. It is common in plants because it 3 is less reactive than glucose, and it is their main transport sugar. 3. Lactose (or milk sugar) is galactose–glucose. It is found only in mammalian milk. Polysaccharides Starch – Energy store in plants Glycogen – Energy store in humans Cellulose – Major component of plant cell walls 4 Monosaccharides Alpha glucose showing numbered carbon atoms in Beta glucose: Note the position of the hydroxyl the carbon ring structure. LEARN group on carbon 1. LEARN DIAGRAM DIAGRAM Exam tip: Remember H is above C- 1 in alpha glucose but below C-1 in beta glucose. EXAM TIP: If you are asked to draw a condensation reaction, don’t forget to include the water molecule. LEARN DIAGRAM 1.4 Glycosidic bond Test yourself 1. Look, cover; check the key words to make sure you know them off by heart. 2. Make a mind map from memory of the points covered and make sure you can draw the diagrams included in the table above. 3. Consolidate your notes by writing up any additional points from lesson and filing the information in a logical order in your subject folder. 4. Exam question practice followed by self-assessment and evaluation of performance. 5 Polysaccharides Starch  Made from Amylose and Amylopectin.  Both polysaccharides are made from alpha glucose monomers.  Amylose: unbranched chain consisting of 1,4 glycosidic bonds  Amylopectin: 1,4 glycosidic bonds hold the monomers together, but it has branches along its length caused by 1,6 glycosidic bonds. Starch is used for storage in plants because:  During hydrolysis, glucose can be released and used in respiration to produce ATP.  Compact as it forms a helical structure held in shape by hydrogen bonds  Insoluble so it does not affect the water potential of the storage cell’s cytoplasm. Glycogen Highly branched molecules of amylopectin Ideal for storage in animal cells because:  During hydrolysis, glucose can be released and used in respiration to produce ATP.  Highly branched and can be hydrolysed faster than starch.  Insoluble so it does not affect the water potential of the storage cell’s cytoplasm. Cellulose  Formed from beta-glucose monomers with 1,4 Glycosidic bonding.  Straight chains of cellulose become bound together by hydrogen bonds, to form cellulose fibres called microfibrils.  These fibres are arranged in different directions in the cell wall, so the cell wall is strong but flexible. 6 Knowledge Organiser Edexcel AS Biology B Topic 1: Biological Molecules 1.1.2 B1.2 Lipids Describe how a triglyceride is synthesised, including the formation of ester bonds during condensation reactions between glycerol and three fatty acids. Recall the differences between saturated and unsaturated lipids. Explain how the structure of lipids relates to their role in energy storage, waterproofing and insulation. Explain how the structure and properties of phospholipids relate to their function in cell membranes. Key vocabulary You MUST know the meaning of the terms below. Glycerol A small, 3-carbon molecule with three alcohol (OH) groups. Fatty acid Fatty acids are long molecules made of a non-polar hydrocarbon chain with a polar carboxyl acid group at one end. LEARN DIAGRAM Ester bond Covalent bond between a fatty acid and a glycerol molecule. Triglyceride LEARN DIAGRAM Triglycerides are found in fatty (or adipose) tissue. They are used for: 1. Energy storage. 2. Insulation. 7 3. Waterproofing. Hydrophobic Will not interact with water. Hydrophilic Will interact with, or dissolve in, water. Unsaturated fatty acid LEARN DIAGRAM Fatty acid chains in a triglyceride with C=C double bonds are called unsaturated fatty acids Saturated fatty All carbon atoms held together with single covalent bonds. acid No C=C double bonds present. LEARN DIAGRAM Phospholipid  Phospholipids have a similar LEARN structure to triglycerides, but with a DIAGRAM phosphate group in place of one fatty acid chain.  Phospholipids have a polar hydrophilic "head" (the negatively- charged phosphate group) and two non-polar hydrophobic "tails" (the fatty acid chains). Test yourself 5. Look, cover; check the key words to make sure you know them off by heart. 6. Make a mind map from memory of the points covered and make sure you can draw the diagrams included in the table above. 7. Consolidate your notes by writing up any additional points from lesson and filing the information in a logical order in your subject folder. 8. EXAM QUESTIONS A12-A15 on Teams. Exam question practice followed by self-assessment and evaluation of performance. Ask yourself why you did not get all the marks? Why was that the answer on the mark scheme? 8 Knowledge Organiser Edexcel AS Biology B 1.1.3 B1.3 Proteins Recall the general structure of an amino acid. Describe the formation of polypeptides and proteins from amino acid monomers linked by peptide bonds in condensation reactions. Explain the role of ionic, hydrogen and disulphide bonding in the structure of proteins. Describe how the primary, secondary, tertiary and quaternary structure of a protein determines the properties of proteins, including collagen and haemoglobin. Explain how the structure of collagen and haemoglobin are related to their function. Key terms and structures you MUST know Amino acid LEARN THIS DIAGRAM Zwitterion A neutral molecule with a positive and negative electrical charge. Peptide bond Two amino acids undergo a condensation reaction to form a dipeptide (2 amino acids). A peptide bond is formed between the amino acids and a molecule of water is also produced. Dipeptide Two amino acids covalently joined together by a peptide bond. Polypeptide Polymer of 3 or more amino acids held together by peptide bonds. Primary The sequence of amino acids in a polypeptide. Determined by the DNA base sequence of a gene. structure 9 Secondary Folding of the polypeptide chain into an alpha helix or beta pleated sheet. This folding is held structure together by hydrogen bonds between the hydrogen of one amino acid and the oxygen of an adjacent carboxyl group. Tertiary The folding of the polypeptide into a complex 3D shape. This tertiary structure is held together by structure hydrogen bonds, disulphide bridges between cysteine residues and ionic bonds between charged R groups. Quaternary Two or more polypeptides combine together to form a functional protein. structure NOT ALL PROTEINS HAVE A QUATERNARY STRUCTURE. Fibrous Tertiary structure that is long and straight e.g. collagen protein  Found in bones and tendons Provides strength to tissues because:  Its molecule is a helix made of 3 polypeptide chains  The chains are held together by a large number of covalent bonds and hydrogen bonds between the collagen fibres.  The chains bind tightly because every third amino acid is a glycine (small structure) that fits inside the small space created by the triple helix.  Prolines are either side of the glycines, these have big R groups the repel each other. This maintains the strong, insoluble, fibrous structure.  Collagen molecules cross link with each other providing further strength. Globular Tertiary structure that is spherical e.g. Haemoglobin protein It is able to transport oxygen because:  Its molecule has four polypeptide chains; 2 alpha – globin chains and 2 Beta – globin chains  Each polypeptide chain has an iron-containing haem group – a prosthetic group  Each haem group combines with one oxygen molecule in the lungs and releases this oxygen molecule in the body tissues. 10 Knowledge Organiser Edexcel AS Biology B 1.1.4 B1.4 DNA and protein synthesis Recall the structure of DNA, including the structure of the nucleotides (purines and pyrimidines), base pairing and the two sugar-phosphate backbones. Recall the role structure and role of phosphodiester bonds and hydrogen bonds in the structure of DNA. Describe how DNA is replicated semi-conservatively, including the role of DNA helicase, polymerase and ligase. Recall that a gene is a sequence of bases on a DNA molecule coding for a sequence of amino acids in a polypeptide chain. Recall the structure of mRNA including nucleotides, the sugar phosphate backbone and the role of hydrogen bonds. Recall the structure of mRNA including nucleotides, the sugar phosphate backbone and the role of hydrogen bonds. Recall the structure of tRNA, including nucleotides, the role of hydrogen bonds and the anticodon. Describe transcription and translation, including where they occur and the role of sense and anti-sense DNA, mRNA, tRNA and the ribosomes. Describe the genetic code, including triplets coding, start and stop codons, degenerate and non- overlapping nature, and that not all the genome codes for proteins. Explain the term gene mutation referring to base deletions, insertions and substitutions. Explain the effect of point mutations on amino acid sequences, as illustrated by sickle cell anaemia in humans. Key terms and structures you MUST know Nucleotide Monomer from which all nucleic acids (DNA and RNA) are made. Nucleotide structure 11 Pentose sugar Ribose in RNA or deoxyribose in DNA Organic base Purines always pair with pyrimidines. G with C and A with T(U). Phosphate group Negatively charged group, and gives nucleic acids their acidic properties. The structure of DNA Double helix 2 polynucleotide chains held together by hydrogen bonds between the bases. 12 Each polymer strand consists of nucleotides held together by phosphodiester bonds between carbon-3 of one deoxyribose and carbon-5 of the next deoxyribose in the chain. The two strands are antiparallel, i.e. run in opposite directions. The two strands are held together by hydrogen bonds between the organic bases:  Adenine and thymine (2 hydrogen bonds between each base)  Cytosine pairs with guanine (3 hydrogen bonds between each base) Semi-conservative replication of DNA Before the cell divides the DNA is replicated. Daughter cells produced are clones (genetically identical to parent cell). 1. DNA helicase Breaks hydrogen bonds holding the two polynucleotide strands together and separates the two strands. 2. Template strands Each strand acts as a template for the new strand. 3. Complementary The unpaired bases on each template strand attract free nucleotides with base pairing bases that are complementary to those on the template strand. 4. DNA polymerase Phosphodiester bonds are form between adjacent nucleotides by the enzyme DNA polymerase. 5. DNA ligase One strand is continuously synthesised whilst the other is made in short fragments. These short fragments are joined by the enzyme DNA ligase. 13 RNA Single strand of nucleotides Messenger RNA (mRNA) Transfer RNA (tRNA)  Straight polynucleotide  75 nucleotides in length  Shorter than DNA  Folded into a clover leaf shape, held together by hydrogen bonds and complementary base pairing.  One end has an amino acid binding site  The other has an anticodon. The genetic code The sequence of organic bases in a molecule of DNA contains the code for making polypeptides. Each amino acid is encoded by a sequence of 3 bases i.e. base triplet in DNA. The code is:  Universal – present in all organisms, same triplet = same amino acid.  Non-overlapping – each base forms part of one triplet only.  Degenerate – some amino acids are encoded by more than one base triplet. 64 base triplets or mRNA codons Some mRNA codons act as START and STOP codons during transcription. Using the DNA code to make a polypeptide Gene DNA base sequence that encodes the amino acid sequence of a polypeptide. Sense strand of DNA A gene is a region in the sense strand of a DNA molecule that codes for a particular polypeptide. Antisense strand of Used as the template for mRNA formation. By transcribing the antisense strand, DNA mRNA has a base sequence complementary to the antisense strand of DNA and, therefore, it is the same as the sense strand of DNA (except uracil replaces thymine). Transcription DNA base sequence of antisense strand is used to make a molecule of mRNA. Translation Ribosomes ‘read’ the mRNA code and assemble amino acids, coordinated by tRNA, into a polypeptide. 14 Transcription 1. RNA polymerase attaches to the start of the gene at the promoter site. RNA polymerase is a huge protein complex, consisting of 17 separate proteins, each with many polypeptide chains. One of these is a helicase enzyme. 2. DNA unwinds as the hydrogen bonds are broken between the complementary base pairs. 3. Free RNA nucleotides pair with complementary DNA nucleotides on the antisense strand. RNA polymerase moves along the gene, catalysing the formation of phosphodiester bonds between the RNA nucleotides. 4. Pre - mRNA is formed. Many copies of mRNA can be made from one gene. 5. In EUKARYOTIC cells, genes contain coding regions called EXONS and non-coding regions called INTRONS. Pre-mRNA is formed after transcription. 6. Splicing occurs to remove the introns and join the exons together. 7. Mature mRNA produced. 8. mRNA leaves the nucleus via a nuclear pore in the nuclear envelope and attaches to a ribosome in the cytoplasm. 15 Translation 1. Ribosomes have 2 binding sites. The first binding site binds to the beginning of the mRNA molecule. The first codon on the mRNA is always AUG. 2. A molecule of tRNA with the complementary anticodon, UAC, arrives carrying its specific amino acid. 3. In the figure above the ribosome has moved along the molecule of mRNA and now has both binding sites filled by tRNA. 4. Accompanied by the hydrolysis of ATP, enzymes in the ribosome catalyse the formation of a peptide bond between two amino acids. 5. The ribosome continues to move along the mRNA one codon at a time and the peptide grows one amino acid at a time. 6. The final codon is a ‘stop’ codon. Once reached, the ribosome detaches from the mRNA and releases the polypeptide. Gene mutations Mutations may affect the polypeptide produced unless they occur in non-coding regions of DNA, the altered base triplet still codes for the same amino acid, or the new amino acid in the sequence has no effect on key parts of the tertiary structure of the protein. A change in the DNA base sequence can occur in different ways: Base deletion One nucleotide lost from the sequence Both result in a frameshift. This is where the Base insertion One nucleotide added to the sequence encoded amino acid sequence is altered following the mutation. Base substitution One nucleotide used in place of Affects one base triplet and the single amino acid it another. encodes. This is called a point mutation. Point mutation and sickle-cell anaemia A point mutation in the gene encoding the beta-globulin chains of haemoglobin. This results in GTC instead of GAG; The amino acid valine added instead of glutamic acid. Effect: RBCs become sickle shaped. Carry oxygen less effectively. Mutation: Recessive allele. Carriers display sickle cell trait and have increased resistance to malaria. 16 Knowledge Organiser Edexcel AS Biology B 1.1.5 B1.5 Enzymes Recall the structure of enzymes as globular proteins and the describe concepts of specificity and the induced fit hypothesis. Recall that enzymes are catalysts that reduce activation energy. Explain how temperature, pH, substrate and enzyme concentration affect the rate of enzyme activity. Describe how the initial rate of enzyme activity can be measured and why this is important. Explain how enzymes can be affected by competitive, non-competitive and end-product inhibition. Recall that enzymes catalyse a wide range of intracellular reactions as well as extracellular ones. Key terms you MUST know Activation energy Energy barrier that must be overcome for a chemical reaction to proceed. Catalyst Enzymes are biological catalysts. They are needed in smaller concentrations than the reactants and are not used up or chemically changed by the reaction. Active site Part of the enzyme the substrate binds to when forming an enzyme-substrate complex. The shape of the substrate is complementary to the active site. Tertiary structure All molecules have a 3D shape, so examiners expect you to use the term ‘tertiary structure’ when referring to the active site of an enzyme. Substrate The reactant(s) in the enzyme - catalysed reaction. Induced-fit hypothesis The tertiary structure of an enzyme’s active site is not normally an exact complement to the shape of the substrate. Binding of the substrate molecule(s) changes the tertiary structure of the active site, causing it to become an exact complementary fit. Denatured Permanent change to the tertiary structure of an enzyme. Hydrogen bonds are broken initially and the active site shape is altered. Optimum The most favourable condition that results in the fastest rate of reaction e.g. optimum temperature. Inhibitor A substance that combines with an enzyme molecule and reduces the rate of the reaction it catalyses. Competitive inhibitor Chemicals that have a similar shape tot the substrate. Temporarily bind to the active site of the enzyme, preventing enzyme – substrate complexes from forming. Increasing substrate concentration reduces effect of inhibitor. Non-competitive inhibitor Chemicals that do not have a similar shape to the true substrate. They bind to an allosteric site. This changes the tertiary structure of the enzyme, so that the active shape is no longer complementary to the substrate. Allosteric site A binding site other than the active site. Intracellular enzymes Catalyse all reactions inside cells Extracellular enzymes Synthesised inside cells but secreted outside by exocytosis. Catalyse reactions outside the cell. 17 Enzyme and substrate  An enzyme is a globular protein which acts as a biological catalyst by speeding up the rate of a chemical reaction  Enzymes are not changed or consumed by the reactions they catalyse and thus can be reused  Enzymes are typically named after the molecules they react with (called the substrate) and end with the suffix ‘-ase’ For example, lipids are broken down by the enzyme lipase Active Site 1. The active site is the region on the surface of the enzyme which binds to the substrate molecule 2. The active site and the substrate complement each other in terms of both shape and chemical properties 3. Hence only a specific substrate is capable of binding to a particular enzyme’s active site The lock and key model of enzyme action 18 The induced fit hypothesis According to the induced fit model, the enzyme’s active site is not a completely rigid fit for the substrate. Instead, the active site will undergo a conformational change when exposed to a substrate to improve binding. This theory of enzyme-substrate interactions has two advantages compared to the lock and key model:  It explains how enzymes may exhibit broad specificity (e.g. lipase can bind to a variety of lipids)  It explains how catalysis may occur (the conformational change stresses bonds in the substrate, increasing reactivity) Denaturing All enzymes possess an indentation or cavity to which the substrate can bind with high specificity – this is the active site  The shape and chemical properties of the active site are highly dependent on the tertiary structure of the enzyme  Like all proteins, enzyme structure can be modified by external factors such as high temperatures and extreme pH  These factors disrupt the chemical bonds which are necessary to maintain the tertiary structure of the enzyme  Any change to the structure of the active site (denaturation) will negatively affect the enzyme’s capacity to bind the substrate 19 Types of inhibition Competitive inhibition involves a molecule, other than the substrate, binding to the enzyme’s active site  The molecule (inhibitor) is structurally and chemically similar to the substrate (hence able to bind to the active site)  The competitive inhibitor blocks the active site and thus prevents substrate binding  As the inhibitor is in competition with the substrate, its effects can be reduced by increasing substrate concentration Non-competitive inhibition involves a molecule binding to a site other than the active site (an allosteric site)  The binding of the inhibitor to the allosteric site causes a conformational change to the enzyme’s active site  As a result of this change, the active site and substrate no longer share specificity, meaning the substrate cannot bind  As the inhibitor is not in direct competition with the substrate, increasing substrate levels cannot mitigate the inhibitor’s effect Examples of Enzyme Inhibition Enzyme inhibitors can serve a variety of purposes, including in medicine (to treat disease) and agriculture (as pesticides)  An example of a use for a competitive inhibitor is in the treatment of influenza via the neuraminidase inhibitor, RelenzaTM  An example of a use for a non-competitive inhibitor is in the use of cyanide as a poison (prevents aerobic respiration) 20 End – product inhibition End-product inhibition (or feedback inhibition) is a form of negative feedback by which metabolic pathways can be controlled  In end-product inhibition, the final product in a series of reactions inhibits an enzyme from an earlier step in the sequence  The product binds to an allosteric site and temporarily inactivates the enzyme (via non-competitive inhibition)  As the enzyme can no longer function, the reaction sequence is halted and the rate of product formation is decreased End-product inhibition functions to ensure levels of an essential product are always tightly regulated  If product levels build up, the product inhibits the reaction pathway and hence decreases the rate of further product formation  If product levels drop, the reaction pathway will proceed unhindered and the rate of product formation will increase Example Isoleucine is an essential amino acid, meaning it is not synthesised by the body in humans (and hence must be ingested). Food sources rich in isoleucine include eggs, seaweed, fish, cheese, chicken and lamb In plants and bacteria, isoleucine may be synthesised from threonine in a five-step reaction pathway In the first step of this process, threonine is converted into an intermediate compound by an enzyme (threonine deaminase). Isoleucine can bind to an allosteric site on this enzyme and function as a non-competitive inhibitor. As excess production of isoleucine inhibits further synthesis, it functions as an example of end-product inhibition This feedback inhibition ensures that isoleucine production does not cannibalise available stocks of threonine. 21 Factors affecting the rate of enzyme activity Increasing temperature causes: Increases rate of collisions between enzyme and substrate molecules. It denatures the enzyme molecules, changing the tertiary structure of their active sites. The peak on the graph represents the optimum temperature where the rate of enzyme activity is highest. Amino acids can ionise in different ways, depending on the pH of the solution. The ionic bonds within the globular protein are altered which alters the tertiary structure of the protein. At optimum pH the tertiary structure of the enzyme’s active site is complementary to that of its substrate. At low enzyme concentrations, there are fewer enzyme-substrate collisions and the rate of reaction is slow. The enzyme concentration is said to be a limiting factor. At low substrate concentrations, there are fewer enzyme-substrate complexes and the rate of reaction is slow. The substrate concentration is the limiting factor. Increase substrate concentration, increase rate until something else becomes a limiting factor. The two types of inhibitor can be distinguished experimentally by carrying out a substrate vs. rate experiment in the presence and absence of the inhibitor. If the inhibition is reduced at high substrate concentration then the inhibitor is a competitive one. 22 Knowledge Organiser Edexcel AS Biology B 1.1.6 B1.6 Inorganic ions 1.1.7 B1.7 Water Describe and explain the role of nitrate ions, calcium ions, magnesium ions and phosphate ions in plants. Explain the importance of the dipole nature of water leading to hydrogen bonding. Explain the significance of the following properties of water high specific heat capacity, polar solvent, surface tension surface tension, incompressibility and maximum density at 4 °C to organisms. Key terms you MUST know. Anion Negatively charged ion Cation Positively charged ion Dipole Water molecules are dipolar. The larger oxygen nucleus draws electrons away from the hydrogen atoms resulting in: a small positive charge on each hydrogen atom and a negative charge on the oxygen atom. Hydrogen A relatively weak chemical link formed by the attraction between a weakly positive bond atom and a weakly negative atom. Inorganic ions Calcium ions Form calcium pectate in plant cell walls Magnesium ions Used to produce chlorophyll Phosphate ions Used to make ADP and ATP Nitrate ions Used to make nucleic acids and amino acids 23 Properties of water The hydrogen bonding of water molecules gives water a number of properties that are important in biology: Cohesion Molecules of water hydrogen bond together. This is important in plant transport of water. High specific heat capacity A large amount of heat is required to break the hydrogen bonds holding water molecules together. Seas and lakes therefore tend to have temperatures that are more stable than air temperatures. High latent heat of A large amount of heat is required to turn liquid water into water vaporisation vapour. The evaporation of water from the surface of organisms has a strong cooling effect. Polar solvent Because water molecules are polar, it is an important solvent in which ions and charged compounds can all dissolve. The reactions inside cells all occur in aqueous environments. High surface tension At a water-air interface, the cohesion between water molecules gives the water surface high surface tension. Creates habitats for organisms such as insects. Incompressibility Water cannot be easily compressed, making it ideal for supporting plant tissues and the bodies of soft-bodied animals. Maximum density at 4C Even though a lake or a pond might be covered in ice, there will be a layer of warmer water beneath it. 24 Knowledge Organiser Edexcel AS Biology B Topic 2: Cells, Viruses and Reproduction of Living Things 1.2 B2.1 EUKARYOTIC AND PROKARYOTIC CELL STRUCTURE AND FUNCTION Recall that cell theory states that cells are a fundamental unit of structure, function and organisation in all living organisms. Recall that in complex organisms, cells are organised into tissues, organs, and organ systems. Recall and describe ultrastructure of prokaryotic cells and the structure of organelles, including nucleoid, plasmids, 70S ribosomes and cell wall. Distinguish between Gram positive and Gram-negative bacterial cell walls and understand why each type reacts differently to some antibiotics. Recall and describe the ultrastructure of eukaryotic cells. Describe the functions of the nucleus, nucleolus, 80S ribosomes, rough and smooth ER, mitochondria, centrioles, lysosomes, Golgi apparatus, cell wall, chloroplasts, vacuole and tonoplast. Recall how magnification and resolution can be achieved using light and electron microscopy. Describe the importance of staining specimens in microscopy. Key terms you MUST learn Tissue A group of similar cells working together to carry out a particular function Organ Different tissues working together to perform a particular function Organ system Different organs working together to perform a particular function Prokaryote Cells without a nucleus Eukaryote Cells with a nucleus Ultrastructure Features of a cell that can be seen with an electron microscope but not with a light microscope e.g. membrane - bound organelles Magnification Measure of how much larger an image of an object is than the actual object itself. Resolution A measure of the ability to distinguish between two very close objects. E.g. 0.2um is a better resolution than 0.9um. 25 Comparison between prokaryotic cells and eukaryotic cells Feature Prokaryotic cells Eukaryotic cells Size 1 – 10 um 10 – 800um DNA molecules Short, circular, and not associated Long, linear, and wrapped around with proteins forming a nucleoid. proteins forming chromosomes. Plasmids may also be present Chloroplasts and mitochondria have short, circular DNA molecules not associated with proteins. Ribosomes Smaller 70S Larger 80S (70S ribosomes in chloroplasts and mitochondria) Membrane-bound Absent Present organelles Cell wall Present Fungi – Chitin cell wall Made of peptidoglycan – a polymer Plants and algae – cellulose cell wall. of amino acids and monosaccharides Calcium goes into pectate to hold plant cells together. Animals – no cell wall Cell surface Present Present membrane Cytoplasm Present Present Nucleus Absent Present Ultrastructure of prokaryotic cells Gram-positive and Gram-negative bacteria Gram-positive bacteria Gram-negative Thick wall of peptidoglycan with no outer layer of Thin wall of peptidoglycan plus an outer layer of lipopolysaccharide lipopolysaccharide. Peptidoglycan stains purple with crystal violet stain Outer layer prevents staining of peptidoglycan with crystal violet. Stained pink with counter-stain safranin. 26 Antibiotics that prevent cell wall formation are effective Antibiotics that prevent cell wall formation are not e.g. Penicillin effective due to presence of outer layer. Ultrastructure of eukaryotic cells Cell component Function Cell surface membrane Controls the movement of ions and molecules into and out of the cytoplasm. Contains proteins that act as cell receptors or self-antigens. Cell wall NOT IN ANIMAL CELLS. Provides support for plant cell. Centrioles In animal cells, produce spindle fibres. Chloroplasts Site of photosynthesis. Contains its own circular DNA. Golgi apparatus Modifies proteins and packages polymers in vesicles for transport Lysosomes Small vesicles that release hydrolytic enzymes Nucleolus Site of ribosome synthesis Nucleus Contains the cell’s DNA in the form of chromosomes Mitochondria Produce ATP during aerobic respiration. Contains circular DNA in the matrix. Permanent vacuole Contains cell sap. Maintains turgor of plant cells. Stabilizes pH of cytosol. Rough endoplasmic reticulum Synthesises proteins and transports them to the golgi. Smooth endoplasmic reticulum Synthesis and transport of steroids and lipids Tonoplast Controls the movement of water and ions between the cytoplasm and vacuole 80S ribosomes Synthesise polypeptides 27 Studying cells. A comparison of light and electron microscopes Feature LIGHT microscope ELECTRON microscope Radiation source Light Electron beams How magnification is achieved Glass lenses – condenser, Electromagnets – condenser and eyepiece and objective lenses objective. Viewing image Using eyepiece Observing image on a fluorescent Natural colour of sample screen preserved. Black and white image Types of stain Chemical dyes that bind to Heavy metal ions that absorb specific material and provide electrons contrast. Resolution Lower resolution Very high resolution Type of specimen Living and dead Dead only Artefacts can be formed through Artefacts can be formed during staining procedure. specimen preparation. Much more likely with this type of microscope. Cost Cheap to buy (£100-600) Expensive (£1000000+) Sample preparation Simple Complex (requires trained staff) Types of electron microscope Transmission electron microscope Scanning electron microscope Pass a beam of electrons through the specimen. Pass a beam of electrons over the surface of the The electrons that pass through the specimen are specimen in the form of a ‘scanning’ beam. detected on a fluorescent screen on which the Electrons are reflected off the surface of the image is displayed. specimen as it has been previously coated in heavy 28 metals. It is these reflected electron beams that are focussed on the fluorescent screen in order to Higher resolution than SEM. make up the image. Sections of specimens must be very thin, as electrons must pass through. Lower resolution than TEM 2D images produced. Thicker specimens can be used. 3D images produced. Cell ultrastructure Hair cells Magnification and measurements Bigger units to smaller units = multiply Smaller units to bigger units = divide 29 Rules of biological drawing Sharp pencil Lines clear, no gaps or crossing lines Draw only what you see No colour Label in pencil Use a ruler for label line No arrow heads, only lines Add magnification and scale bar Knowledge Organiser Edexcel AS Biology B 1.3 B2.2 VIRUSES Recall what the classification of viruses is based on [inc λ (lambda) phage (DNA), tobacco mosaic virus and Ebola (RNA) and human immunodeficiency virus (RNA retrovirus)]. Recall the lytic cycle of a virus and latency. Recall that viruses are not living cells and how antivirals work by inhibiting virus replication. Describe how viruses can be difficult to treat and how the focus of disease control should be on preventing the spread (example: 2014 Ebola outbreak in West Africa). Evaluate the ethical implications of using untested drugs during epidemics. Key terms you MUST know Capsid Outer protein coat of a virus 30 Lytic cycle Virus genetic material replicates independently of host DNA. New viruses are released by lysis (cell bursting) Latency A period of inactivity Lysogenic cycle Viral genetic material incorporated into host cell DNA. Spread through host cell replication. Viral genetic material remains latent. Environmental trigger causes activation of viral genetic material, which then enters the lytic cycle. Metabolism All the chemical reactions occurring in living cells Nucleic acid DNA or RNA Reverse transcriptase Enzyme found in retro viruses. Used to convert RNA to DNA. LEARN THIS TABLE Phage lambda TMV 31 Ebola HIV 32 CASE STUDY: EBOLA 2014 Ebola timeline Prevention of spread Isolate affected areas Quarantine patients Protective equipment Safety protocols used used when handling blood Hand washing with alcohol- Sterilisation of medical Careful disposal of Safety protocols based sanitising gel equipment medical equipment regarding disposal of the dead. Ebola drug development Drug development can take many years Vaccines Clinical trials fast-tracked for use against Ebola. July 2015 – reported that one vaccine was 100% effective in clinical trials involving 400 patients. Ethics Treatment Issues to consider when deciding to use a drug not ZMAPP – experimental drug fully tested are:  Comprises of 3 antibodies associated with  Severity of disease surviving the disease, produced by GM  Availability of other treatments tobacco plants  Effectiveness of disease control  Effective in treating monkeys, not trialled  Informed consent in humans  Assessment of safety and ethics  Used to treat 7 people – some died. 33 Knowledge Organiser Edexcel AS Biology B 1.4 B2.3 EUKARYOTIC CELL CYCLE AND DIVISION Recall that the cell cycle is where cells divide into two identical daughter cells, and this process consists of three main stages: interphase, mitosis and cytokinesis. Explain what happens to genetic material during the cell cycle, including the stages of mitosis. Describe how mitosis contributes to growth, repair and asexual reproduction. Explain how meiosis results in haploid gametes, including the stages of meiosis. Explain that meiosis results in genetic variation through recombination of alleles, including independent assortment and crossing over. Describe what chromosome mutations are, as illustrated by translocations. Explain how non-disjunction can lead to polysomy, including Down’s syndrome, and monosomy, including Turner’s syndrome. Key terms Differentiation The process of cell Meiosis Reduction division resulting in 4 genetically specialisation different, haploid, daughter cells. Cell cycle A sequence of events Haploid One set of chromosomes in the nucleus of consisting of interphase, a cell mitosis, and cytokinesis Interphase A time of cellular activity Gamete Haploid sex cell consisting of G1 phase, S phase and G2 phase. Mitosis Division of the nucleus Zygote Fertilised egg cell resulting in 2 genetically identical, diploid, daughter cells. Diploid 2 sets of chromosomes in Homologous Two chromosomes of the same size and the nucleus of a cell shape, one originating from each parent. They contain the same genes, but different alleles. Cytokinesis Cytoplasmic division at the Mutation Change in the DNA base sequence. end of mitosis Alleles Different versions of one Chiasmata Points at which non-sister chromatids gene. cross over during prophase 1 of meiosis Genes Short section of DNA Trisomy 3 copies of the same chromosome in one coding for a specific diploid cell. polypeptide. Chromatid One arm of a replicated Chromosome Compact X-shaped form of chromatin chromosome formed (and visible) during mitosis 34 The cell cycle Mitosis Key events  Chromosomes shorten and thicken  Become visible as replicated chromosomes consisting of 2 sister chromatids held together at the centromere.  Nuclear envelope breaks down  Centrioles move to poles of cell  Spindle fibres form.  Spindle fibres attach to centromere  Chromosomes line up at the equator of the cell  Centromeres divide  Contraction of spindle fibres separates sister chromatids.  Sister chromatids pulled to opposite poles of the cell.  2 nuclear envelopes form around each group of chromosomes  Chromosomes begin to unravel to form chromatin. Cytokinesis Cell surface membrane pinches in to the centre to separate the 2 cells completely. Plant cells do this by forming a new cell wall from the centre to the outside of the cell. 35 Meiosis - Crossing over occurs Independent assortment occurs Crossing over in prophase 1  Non – sister chromatids in a homologous pair become entangled and form chiasma.  Can result in breakage of DNA strands.  Broken fragments can re-join to the wrong chromosome resulting in a new combination of alleles from where the original break occurred. 36 Chromosome mutations: Non-disjunction If an abnormal gamete with too many or too few chromosomes is fertilised then the resulting zygote will show polysomy in that chromosome, since there will be the wrong number of chromosomes. Some examples of polysomy are shown below.  Trisomy 21  Egg cell with 2 copies of chromosome 21 fetilised by a sperm cell with 1 copy, a zygote is formed with a total of 3 copies.  Symptoms of Down syndrome include particular recognisable facial features, short height, heart defects, poor vision, severe learning difficulties and a shorter life expectancy  Non-disjunction of the 2 X chromosomes during egg production.  If an egg cell containing 0 X chromosomes is fertilised by a sperm cell containing 1 X chromosome, the resulting zygote has only 1 X chromosome. (X0).  This condition is called monosomy.  X0 individuals appear female but their sex organs do not mature at adolescence, and they are sterile. Chromosome translocation A change in the combination of genes on a chromosome due to broken fragments of chromosome attaching to a non-homologous chromosome. 37 Knowledge Organiser Edexcel AS Biology B 1.5 B2.4 SEXUAL REPRODUCTION IN MAMMALS Describe the processes of oogenesis and spermatogenesis. Describe the events of fertilisation from the first contact between the gametes to the fusion of nuclei. Describe the early development of the embryo to blastocyst stage. Key terms you MUST know Gametogenesis Gamete formation Spermatogenesis Sperm cell formation Oogenesis Egg cell formation Oocyte Produced by rapid mitosis of germline epithelium cell in the ovary. Ovum Formed when the secondary oocyte completes meiosis II after fertilisation. Polar body Formed after unequal cell division of the primary oocyte. Spermatocyte Produced by rapid mitosis of germline epithelium cell of seminiferous tubule (testes) Spermatid Immature sperm cells Sperm Mature male gamete Acrosome reaction Occurs when hydrolytic enzymes in the acrosome of the sperm are released and digest a way to the cell surface membrane of the secondary oocyte. Cortical reaction Occurs to prevent polyspermy. Granules from the cortex of the secondary oocyte form a barrier to prevent entry of another sperm. Zygote Fertilised egg cell (2n) Blastocyst Hollow ball of 128 cells produced by mitosis in the first 7 days after fertilisation. Blastomere Inner cell mass of the blastocyst. Will develop into the embryo. Trophoblasts Outer layer of the blastocyst forming extra-embryonic tissue. Gametes 38 Germline cell divides by mitosis Primary spermatocyte formed (2n) Primary oocyte formed (2n) These primary cells complete meiosis I 2 Secondary spermatocytes formed (n) 1 Secondary oocyte and 1 polar body formed (n) Meiosis II completed immediately Meiosis II completed only after fertilisation Secondary oocyte divides to form the ovum and 1 polar body. The first polar body also divides giving a total of 3 polar bodies formed. 4 spermatids formed that mature into sperm cells Ovum and 3 polar body produced 39 Fertilisation 1. Numerous sperm use their flagella to swim through the follicle cells. 2. The sperm release hydrolytic enzymes from their acrosomes by exocytosis (the acrosome reaction). These enzymes partially digest the zona pellucida, allowing many sperm to swim through. The enzymes from many sperm are required to digest a path through the zona pellucida. 3. When the first sperm reaches the ovum cell membrane the two membranes fuse and the sperm nucleus enters the cytoplasm of the ovum. The rest of the sperm cell degrades, so the sperm only contributes a nucleus to the zygote. All the cytoplasm and organelles (Including the DNA in the mitochondria) are contributed by the ovum. 4. The entry of the sperm nucleus triggers a series of reactions in the ovum (called the cortical reaction). Cortical granules fuse with the cell membrane releasing enzymes that cause the elements of the zona pellucida to cross-link, preventing any other sperm from entering the ovum and preventing polyspermy. 5. The entry of the sperm nucleus also stimulates the secondary oocyte nucleus to complete meiosis II, forming a third polar body and the haploid female nucleus. 6. The male and the female nuclei do not fuse immediately, but instead the DNA in both nuclei replicate and condense to form chromosomes. Then both nuclear envelopes disappear leaving the two sets of chromosomes free in the cytoplasm. The cell is now a diploid cell – the zygote, which immediately starts its first mitosis. 40 Embryonic development Day Fertilisation takes place near the top of the oviduct and the zygote continues to move down the oviduct 0 driven by the beating of the cilia that line the oviduct. Day As it moves the zygote divides repeatedly by cleavage – a special form of mitosis with short interphase 1-3 and no growth phase. The total mass of the cells is the same as the original zygote (which is why the ovum had to be such a large cell). Day After four days and three divisions the ball of 8 cells (the morula) arrives in the uterus. 4 Day After seven divisions there are 128 cells, arranged in a hollow ball called a blastocyst. 7 Day This blastocyst embeds itself in the lining of the uterus (the endometrium) in a process called 8-9 implantation. This is the start of pregnancy and the embryo can now receive nutrients from the mother’s cells, allowing it to grow. Test yourself 1. Look, cover; check the key words to make sure you know them off by heart. 2. Make a mind map from memory of the points covered and make sure you can draw the diagrams included in the table above. 3. Consolidate your notes by writing up any additional points from lesson and filing the information in a logical order in your subject folder. 4. EXAM QUESTIONS on Moodle. Exam question practice followed by self-assessment and evaluation of performance. Ask yourself why you did not get all the marks? Why was that the answer on the mark scheme? 41 Knowledge Organiser Edexcel AS Biology B 1.6 B2.5 SEXUAL REPRODUCTION IN PLANTS Describe how a pollen grain forms in the anther and the embryo sac forms in the ovule. Describe how the male nuclei are formed by division of the nucleus in the pollen grain reach the embryo sac, inc the roles of tube nucleus, pollen tube and enzymes. Describe the process of double fertilisation inside the embryo sac to form a triploid endosperm and a zygote. Key terms you MUST know Pollen grain A pollen grain is a microscopic body that contains the male reproductive cell of a plant Megaspore mother cell A diploid cell in plants that undergoes meiosis to create four haploid megaspores Megaspores The megaspore, or large spore, germinates into a female gametophyte, which produces egg cells. Embryo sac The female gametophyte of a seed plant, within which the embryo develops Female gamete One nucleus in the embryo sac which will form the embryo plant after fertilisation Polar nuclei 2 nuclei in the embryo sac which will form the primary endosperm cell after fertilisation. Microspore mother cell A diploid cell in plants that undergoes meiosis to create four haploid microspores. Microspores The microspore, or small spore, germinates into a male gametophyte, which produces haploid nuclei (in a pollen grain). Generative nucleus One of the two nuclei in an angiosperm pollen grain (compare tube nucleus). It divides to produce two male gamete nuclei. Pollen tube nucleus One of the two nuclei formed by mitotic division of a microspore during the formation of a pollen grain that is held to control subsequent growth of the pollen tube and that does not divide again Pollination Transfer of pollen from a male part of a plant to a female part of a plant, enabling later fertilisation and the production of seeds Double fertilisation This process involves the joining of the embryo sac with two male gametes. Zygote Fertilised egg cell (2n). Primary endosperm cell Product of the male nucleus fusing with two polar nuclei to ultimately form the endosperm tissue. Seed Produced from the embryo sac. Fruit Produced from the ovary wall. Gametophyte The gametophyte is the sexual phase in the life cycle of plants 42 Sexual reproduction in plants The gametes of a flowering plant are not cells; they are haploid nuclei within a spore. Formation of female gametes Megasporogenesis Megaspore mother cell divides by meiosis. 4 haploid megaspores produced. 3 megaspores degenerate (break down). 1 enlarges to become the single-celled embryo sac. The haploid nucleus within the sac divides by mitosis to form 8 haploid nuclei. The cytoplasm of the embryo sac does not divide. One of these 8 nuclei is the female gamete. Two of these nuclei are the polar nuclei 43 Formation of male gametes Microsporogenesis Within the lobes of the anther are the microspore mother cells. Each microspore mother cell divides by meiosis 4 haploid microspores produced. The haploid nucleus within each microspore divides once by mitosis. 2 haploid nuclei produced. One is the generative nucleus One is the pollen tube nucleus. These nuclei remain in a single-celled pollen grain. Its cytoplasm does not divide. EXAM TIP: Don’t refer to pollen grains as gametes; the gametes are haploid nuclei within the pollen grains. Double fertilisation Fertilisation of a flowering plant involves:  Generative nucleus dividing by mitosis to produced 2 haploid male gametes.  Fusion of one male gamete with the female gamete inside the embryo sac.  Fusion of one male gamete with 2 polar nuclei inside the embryo sac. How do the male gametes reach the embryo sac? POLLINATION 1. Pollen grain lands on stigma. 2. Pollen grain absorbs water from the stigma, swells and bursts. 3. A pollen tube is produced. 4. The pollen tube nucleus controls the growth of the tube, through the tissue, to the embryo sac. 5. The generative nucleus follows the pollen tube nucleus. 6. The generative nucleus divides once by mitosis. 7. The two nuclei are the male gametes. 8. In the embryo sac, the pollen tube tip and nucleus break down. 9. One male gamete fuses with the female gamete and an embryo forms. 10. One male gamete fuses with the two polar nuclei. 11. A triploid primary endosperm cell is formed. A fuel store for the developing embryo. 12. The embryo sac becomes the seed. 13. Ovary tissue forms the fruit. 44 Knowledge Organiser Edexcel AS Biology B Topic 3: Classification and Biodiversity 1.7 B3.1 CLASSIFICATION Recall that the classification system consists of a hierarchy of domain, kingdom, phylum, class, order, family, genus and species. Explain the limitations of the definition of a species as a group of organisms with similar characteristics that interbreed to produce fertile offspring. Describe why it is difficult to assign organisms to any one species or to identify new species. Explain how gel electrophoresis can be used to distinguish between species and determine evolutionary relationships. Recall that DNA sequencing and bioinformatics can be used to distinguish between species and determine evolutionary relationships. Explain the role of scientific journals, the peer review process and scientific conferences in validating new evidence supporting the accepted scientific theory of evolution. Explain the evidence for the three-domain model of classification as an alternative to the five-kingdom model and the role of the scientific community in validating this evidence. Key terms you MUST know Taxa (taxon) Groups of organisms within the hierarchy of the biological classification system. Species A group of organisms with similar characteristics that interbreed to produce fertile offspring. Binomial Two names given to all organisms; Genus and species. Genus is a noun; the species name nomenclature is an adjective. Names are usually in Latin – a dead language so less likely to be corrupted. Written in italics or underlined e.g. Homo sapien. Hierarchy Groups within groups with no overlapping taxa. Bioinformatics The storage and indexing of electronic information for future analysis and use. Genomes Complete set of genes in an organism Phylogenetic tree A diagram that attempts to show when different species diverged from a common ancestor. Validity Suitability of the investigative procedure to answer the question being asked. Peer review The strict process through which a paper submitted for publication in a scientific journal is validated by independent experts in the same field before being published. Accurate Close to the true value Precise Little variation between repeated measurements of the same variable. Autotroph An organism that is able to synthesise complex organic compounds from simple inorganic compounds using an external energy source. Heterotroph An organism that feeds on other organisms. Saprobiont An organism that feeds by secreting digestive enzymes into its environment and absorbing the digested products. Parasite An organism that lives in or on another living organism (its host), causing it damage. 45 Classification Artificial classification The ordering of organisms into groups on the basis of non-evolutionary features Natural classification Attempts to show evolutionary relationships between organisms. Taxonomy  Organisms are put into taxa.  These taxa are organised into a hierarchy.  Largest number of organisms that are least similar to smallest number of organisms that are most similar.  Species is the smallest taxon. Limitations of the definition of a species:  How much variation can be tolerated in a species before you classify organisms into a different species?  Differences in species are often not obvious.  Hybrid animals have been shown to produce fertile offspring on rare occasions. Definition of a species Morphology definition – members of the same species have similar characteristics. Problem: can be difficult to distinguish between structures that have evolved from a common ancestor and those which arose independently to do a similar job e.g. wings. Ecological definition – members of the same species have the same ecological niche. Reproduction definition - members of the same species can breed together in their natural environment to produce fertile offspring but cannot breed with members of other species. Evolutionary definition - members of the same species share a unique common ancestor with each other but not with members of any other species. This is the most modern definition of a species and is the basis of phylogenetics – placing species in their correct evolutionary family tree. DNA technology and classification The relatively new field of DNA technology has made a significant contribution to establishing evolutionary relationships. How? 46 DNA hybridisation DNA sequencing Investigate the similarity between DNA 1. DNA is isolated from the test organism. molecules of different organisms. 2. The isolated DNA is hydrolysed into single-stranded fragments of different lengths. 1. Heat DNA samples from two 3. The fragments are mixed with DNA nucleotides and DNA different organisms in two different polymerase so that new strands of DNA are produced. tubes. 4. Some of the nucleotides are structurally slightly different from 2. The DNA strands then separate as normal and carry a fluorescent label. hydrogen bonds are broken. 5. When one of the labelled nucleotides is included in a developing 3. The two samples are mixed and DNA strand, it stops any more nucleotides from being added. cooled. 6. The result is a mixture of strands of different length, each with a 4. The separated strands hybridise as fluorescent label that shows the last base that was added to the complementary bases form sequence. hydrogen bonds between the 7. The strands are separated by electrophoresis. different strands. 8. In automated DNA sequencing machines, the process is carried 5. The hybridised strands are reheated out using gel within a capillary tube and laser scanners. until they separate again. The fewer 9. When placed in an electric field the negatively charge DNA the number of hydrogen bonds fragments move towards the positive electrode. The separate by between them: size with the smallest fragments travelling faster. 6. The lower the temperature at which 10. Lasers detect the fluorescence as they pass through the gel they separate capillary tube, allowing identification of the DNA base sequence. 7. The less similar their DNA base 11. Overlaps in the sequences of fragments allow scientists to sequences deduce the base sequence of the entire genome. 8. The less closely related the organisms. Bioinformatics and DNA sequencing enable comparisons to be made of the entire genomes of different organisms. Scientists can then deduce how many gene mutations would be required to cause the observed differences in these genomes. Knowing the frequency of mutations, scientists can put a timeline to these mutations. In this way the divergence of two species can be estimated and a phylogenetic tree created. 47 Models of biological classification Five - kingdom model The three - domain model Prokaryotae Protoctista Fungi Plantae Animalia Archaea Bacteria Eukarya Based on structural differences between organisms More recent theory, based on molecular similarities How scientists come to an agreement about classification of species The scientific community validates the work of scientists and the evidence they have collected. Scientific journals Scientific conferences Peer review 1.8 B3.2 NATURAL SELECTION Explain how evolution can come about through natural selection acting on variation bringing about adaptations. Describe how organisms occupy niches according to physiological, behavioural and anatomical adaptations. Explain how reproductive isolation can lead to allopatric and sympatric speciation. Explain why there is an evolutionary race between pathogens and the development of medicines to treat the diseases they cause. Key terms you MUST know Natural selection Natural selection is the differential survival and reproduction of individuals due to differences in phenotype. It is a key mechanism of evolution, the change in the heritable traits characteristic of a population over generations. Evolution The change in the frequency of alleles of a single gene within a population over many generations. Selection pressure external factors which affect an organism's ability to survive in a given environment e.g. presence of antibiotics Intraspecific competition Competition between individuals of the same species due to more offspring being produced than can be supported by their environment. Speciation The formation of new species that can result from groups from one population becoming reproductively isolated and accumulating genetic differences. Antibiotic A substance that disrupts part of the metabolism of bacterial cells. Ecological niche Where a species can successfully exist that includes all its interactions with the living and non-living environment. 48 Summary of natural selection 1. Genetic variation in a sexually reproducing 2. Some individuals have a competitive advantage and population can out compete other members of the population (intraspecific competition). 3. This leads to differential reproductive success 4. Advantageous alleles are passed on to the next (those individuals carrying the genetic generation. advantage produce more offspring). 5. Frequency of advantageous allele increases in 6. This is evolution. the population over many generations. Types of selection Selection for the mean phenotype (intermediate) and against extremes e.g. birth weight of babies Selection for one extreme phenotype e.g. Peppered moth. Speciation – formation of a new species from exisiting species Allopatric speciation Sympatric speciation 1. A single population split in to two by a 1. This does not involve physical separation. geographical barrier. 2. Two diverging groups can be prevented from 2. This prevents gene flow between the two groups. breeding in other ways: 3. The allele frequencies of critical genes become  Seasonal isolation – two groups reproduce at different in the two groups. different times of the year 4. If the differences result in reproductive isolation  Temporal isolation – the two groups reproduce then new species have been formed. at different times of the day  Behavioural isolation – different courtship patterns. Antibiotic resistance – an evolutionary race As more bacterial species become resistant to many different antibiotics, due to horizontal gene transfer, new drugs need to be developed to treat the diseases they cause. 1. A random mutation of a gene results in an allele conferring resistance to an antibiotic. 2. In the presence of the antibiotic it will give the possessor a competitive advantage over susceptible bacteria. 3. Susceptible bacteria will be killed and the resistant bacteria will survive, reproduce (binary fission) and pass on the advantageous allele. This is an example of directional selection. 49 Knowledge Organiser Edexcel AS Biology B 1.9 B3.3 BIODIVERSITY Recall that biodiversity can be assessed within a habitat at the species level using a formula to calculate an index of diversity: Recall that biodiversity can be assessed within a species at the genetic level by looking at the variety of alleles in the gene pool of a population. Explain the ethical and economic reasons (ecosystem services) for the maintenance of biodiversity. Explain the principles of ex-situ (zoos and seed banks) and in-situ conservation (protected habitats), and the issues surrounding each method. Key terms you MUST know Biodiversity Variety of organisms living in a habitat. Community All the different populations living in the same habitat at the same time. Population A group of organisms of the same species living in the same habitat at the same time. Gene pool Sum of all genes within a population, usually considered gene by gene. Conservation Management of the environment to maintain biodiversity. Biodiversity Biodiversity can be assessed at different levels of organisation: variation within a habitat – looking at the variety of different species and variation within a species – looking at the variety of alleles of each gene. Variation within a habitat Species richness Species diversity A measure of the number Takes into account the number of individuals of each species as ell as the of different species number of species within a community. An index of diversity can be present in a community. calculated using the formula: The higher the index, the higher the species diversity. The maintenance of biodiversity Human activities are reducing biodiversity:  Ecosystem diversity is reduced by deforestation, mining and building.  Species diversity is reduced by farming, hunting and habitat destruction.  Genetic diversity is reduced by selective breeding and competition with humans. Understanding our impact on biodiversity has led to conservation – the attempt to conserve biodiversity worldwide. Conservation: The aims of conservationists include: preserving and promoting existing habitats; ensuring that natural resources are used in a way that encourages a sustainable yield. 50 Why conservation is important? Ethical reasons  Responsibility to maintain biodiversity for the next generation.  Natural ecosystems are healthy areas for recreation and leisure.  Indigenous people have a right to their traditional way of life.  The needs of everyone on the planet to avoid catastrophes such as global warming or mass extinctions Economic reasons Natural ecosystems provide:  Goods, such as timber, sources of medicines and fish  Services, such as tree roots to stabilise soil; photosynthetic organisms remove carbon dioxide from the environment and release oxygen into it.  Resources, such as gene pool resource. Alleles of genes fund in wild cereals, for example, might be useful in producing new crops that could withstand effects of global warming.  Cultural reasons including art and recreation. Types of conservation In-situ Ex-situ Attempts to maintain biodiversity by designating and Attempts to preserve biodiversity by creating seed preserving representative habitats as nature reserves. banks, botanical gardens and zoological gardens. Issues with conservation methods Advantages of in-situ conservation Advantages of ex-situ conservation 1. When a habitat is conserved the whole 1. Animals and plants that are endangered in the community is conserved with it, including wild can be protected in zoos and botanic invertebrates, plants, fungi and microbes. gardens. 2. Large numbers of animals and plants can be 2. Well-managed breeding programs can maintain conserved in-situ, with a correspondingly large and even improve genetic diversity. genetic diversity. 3. Animals in zoos have a longer life expectancy 3. Young animals learn natural skills from families than in the wild. and social groups in their natural environment. 4. Since zoos are cheaper and easily-accessible, 4. Reserves are safe environments to re-introduce they attract large numbers of visitors, which animals from captive (ex-situ) breeding helps to raise programs. 5. funds and provides an excellent opportunity to 5. Reserves can be popular tourist destinations, educate many people in the importance of which helps to fund them and to educate conservation work. visitors. Disadvantages of in-situ conservation Disadvantages of ex-situ conservation 1. Many important protected areas are in poor 1. Only a sma

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