AS Revision summary
Unit 1 Topic 1: Lifestyle, health and risk Topic 2: Genes and health Unit 2 Topic 3: Topic 4:
SNAB Biology AS Revision Summary
Understanding the specification
The following summary tries to explain what is meant by the specification statements for the SNAB AS course. SNAB includes a lot of material because it helps you to understand ideas or because it is interesting, but it can be a little confusing when it comes to knowing what to revise. This brief summary will not help you to understand the biology and it is not intended to replace a revision programme, but it might help to sort the wood from the trees a little. Remember that examination questions must be set so that you could do them with no reference to the SNAB course materials. Only the statements in bold below can be examined. You will not, for example be asked questions that expect you to recall the ‘Nice to Know’ information in the book, but it might help if you understand it. About half of the examination questions will expect you to recall facts, with the other half asking you to show that you understand ideas. You will often be given unfamiliar material and asked to suggest what might be going on by using ideas that you have studied. Numbers in brackets are page references to the SNAB text books so that you can read more fully if one of the statements isn’t completely clear.
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SNAB Biology AS Revision Summary
SNAB Biology – Unit 1 revision outline Topic 1: Lifestyle, health and risk
1.1.1 Explain why many animals have a heart and circulation (mass transport to overcome limitations of diffusion) Large and multicellular means small SA:vol ratio (54) Cells can be specialised (6) Explain how the structures of blood capillaries, arteries and veins relate to their functions Capillaries – small, wall single cell thick. Rapid diffusion, leaky for tissue fluid formation (9) Arteries. – narrow lumen, thick walls (collagen, elastic fibres, smooth muscle), no valves (8). Stretch during systole, maintain high pressure (9) Veins – wide lumen, thin walls (less collagen, few elastic fibres, little muscle), valves (8). Low pressure, valves prevent back flow, flow due to skeletal muscle contraction and low pressure in thorax (9). Relate the structure and operation of the mammalian heart to its function (the cardiac cycle including diastole, atrial systole and ventricular systole) Diastole – muscle at rest, semi lunar valves prevent back flow, blood entering from veins into atria and ventricles due to elastic recoil of heart, atrio ventricular valves open. Atrial systole – atrial muscle contracts, blood fills ventricles. Ventricular systole – ventricular muscle contracts, pressure closes atrio ventricular valves and opens semi lunar valves. Thickness of chamber walls related to function (10-11). Explain the course of events that leads to atherosclerosis (endothelial damage, inflammatory response, plaque formation, raised blood pressure) Fatty deposits (plaques) narrowing and hardening arteries. Increases risk of blockage by blood clots, restricts blood flow. Damaged endothelial cells due to high BP or toxins (eg from cigarettes). Inflammatory response – WBC accumulate and collect cholesterol to form atheroma. Calcium salts and fibrous tissue added to form plaque. BP increases – positive feedback (11).
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SNAB Biology AS Revision Summary
1.1.5
Describe the blood clotting process (thromboplastin release, conversion of prothrombin to thrombin and fibrinogen to fibrin) and its role in CVD Thromboplastin released from platelets and damaged tissue. Converts prothrombin to thrombin. Converts fibrinogen to fibrin, which makes clots form (13). Describe the symptoms of CVD, ie coronary heart disease (CHD) and stroke, and the factors which increase the risk of CVD (genetic, diet, age, gender, high blood pressure, smoking and inactivity) Stroke, (12). Angina, heart attack and arrhythmia (14-15). Risk increases with age, higher in men (25). Some genetic element, but many genes involved (26) High blood pressure and endothelial damage (26). Diet and obesity, cholesterol (30, 39-41). Smoking and inactivity (42). Describe what is meant by blood pressure and explain the role of high blood pressure in CVD Hypertension (26), systolic and diastolic pressure, normal values. Describe the normal electrical activity of the heart, including the roles of the sino-atrial node (SAN), the atrio-ventricular node (AVN) and the bundle of His, and how the use of electrocardiograms (ECGs) can aid the diagnosis of CVD and other heart conditions SAN (in right atrium) – pacemaker – starts wave of depolarisation (16). AVN (between atria and ventricles) passes wave to ventricles after short delay. Uses Purkyne fibres in bundle of His to reach apex first (16). ECG measures activity from outside chest. Know what the trace shows (P is atrial wave, QRS is ventricular, T is repolarisation in diastole). Know abnormal patterns (18-19). Analyse quantitative data on illness and mortality rates to determine health risks and recognise that it is important to distinguish between correlation and causation Calculating probabilities, estimating risk (20). Correlation and causation – know the difference (24). Explain why people’s perceptions of risks are often different from the actual risks Reasons for overestimating risk (22).
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SNAB Biology AS Revision Summary
1.1.11
Analyse data on energy budgets and diet so as to be able to discuss the consequences of energy imbalance BMR and energy requirements (38). BMI and obesity (39). Distinguish between monosaccharides, disaccharides and polysaccharides (glycogen and starch – amylose and amylopectin) in terms of their structure and their role in providing and storing energy (β-glucose and cellulose are not required at this stage). Students should recognise the structural formulae for α-glucose and maltose and the monomers which make up sucrose and lactose Monosaccharides (alpha glucose 32) immediate energy supply in respiration, disaccharides (maltose – glucose amd glucose, 33 , sucrose – glucose and fructose, lactose – glucose and galactose). Glycogen in animals and bacteria. Storage, polymer of glucose, many side branches make it easy to break down 1-4 and 1-6 bonds (35). Starch in plants. Amylose straight chain polymer of glucose with only 1-4 bonds, amylopectin branched with 1-4 and 1-6 (35).
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Describe how monosaccharides join to form polysaccharides through condensation reactions forming glycosidic bonds, and how these can be split through hydrolysis reactions
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SNAB Biology AS Revision Summary
1.1.14
Recognise that glycerol with three fatty acids attached is a lipid and specifically a triglyceride, describe the formation of ester bonds in condensation reactions and recognise differences between saturated and unsaturated lipids (36-37)
Saturated 1.1.15
unsaturated
Calculate body mass indices (BMIs) using the formula BMI = body mass (kg)/height 2 (m) and explain their significance BMI calculations and importance (39). 20-24.9 normal. Discuss the possible significance for health of blood cholesterol levels and levels of high-density lipoproteins and low-density lipoproteins (HDLs and LDLs) Lipoprotein used to transport cholesterol. High levels linked to CHD. LDL include saturated fats, stay in circulation when in excess and end up in plaques (31). HDL include unsaturated fats – transport cholesterol to liver for breakdown and lowers blood cholesterol (41). Describe how the effect of caffeine on heart rate in Daphnia can be investigated practically Activity 1.21 – know the procedure. Discuss how individuals, by changing their diet, taking exercise and not smoking, can reduce their risk of coronary heart disease. Sensible advice pages 45-49.
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SNAB Biology AS Revision Summary
Topic 2: Genes and health
1.2.1 Describe the properties of gas exchange surfaces (large surface to volume ratio, thickness of surface, difference in concentration) and explain how the structure of the lung provides a large surface area to volume ratio Alveoli provide large SA, circulation of blood and breathing maintain diffusion gradients, really thin layer of cells in alveoli presents minimum barrier. Fick’s law not specified, but it has been given in a question. (55-56) Describe the structure of the unit membrane (fluid mosaic model) and how its structure depends on the properties of the phospholipids Phospholipid bilayer, hydrophobic tails and hydrophilic heads cause stability but fluidity in plane of membrane. Mosaic provided by glycoproteins (in one layer or both). Cholesterol to limit fluidity. Functions of proteins – signaling and recognition, enzymes, transport. (60-62) Describe how the effect of temperature on membrane structure can be investigated practically Activity 2.7 – know the technique. Explain what is meant by osmosis in terms of the diffusion of free water molecules through a partially permeable membrane (consideration of water potential is not required) Osmosis is movement of water across a partially permeable membrane. Depends on concentration of free water molecules, so since more solute molecules mean fewer free water molecules the gradient will be towards a more dilute solution. (62-63)
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SNAB Biology AS Revision Summary
1.2.5
Explain what is meant by passive transport (diffusion, facilitated diffusion), active transport (including the role of ATP), endocytosis and exocytosis and describe the involvement of carrier and channel proteins in membrane transport Diffusion – movement of particles from higher to lower concentration. Passive – no energy provided by cell – small and uncharged molecules can cross the membrane (eg oxygen, carbon dioxide) Follows diffusion gradient (62). Facilitated diffusion – larger charged particles need water-filled channels provided by membrane channel proteins. Still passive, can allow control when only certain particles allowed through. Can also be gated to open and close. Carrier proteins are a bit more complicated since they change shape to pass particles across the membrane. Follows diffusion gradient (62). Active transport – active, required energy from ATP provided by cell. Uses carrier proteins in membrane, against diffusion gradient (64). Exocytosis and endocytosis – bulk transport out and in. Vesicles form inside cell and fuse with membrane to transport out, bits of extracellular fluid engulfed to bring stuff in. Requires energy so active (65). Nice summary of transport methods on 65. Describe the basic structure of mononucleotides as a phosphate group, deoxyribose or ribose and a base eg thymine, uracil, cytosine, adenine and guanine. Describe the structures of DNA and RNA as polynucleotides composed of mononucleotides linked in condensation reactions. Describe complementary base pairing and the hydrogen bonding involved in the formation of the DNA double helix
1.2.6
Polymer is produced by making the sugar-phosphate backbone, bases are not involved. Complementary bases pair A-T, (A-U in RNA), C-G. Due to shape and number of hydrogen bonds possible. H-bonds between bases links two polymer strands together to form double helix in DNA. (74-75)
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SNAB Biology AS Revision Summary
1.2.7
Explain the process of protein synthesis (transcription, translation, transfer RNA, messenger RNA, ribosomes, the role of start and stop codons) explain the roles of the template (antisense) DNA strand in transcription, codons on messenger RNA and anticodons on transfer RNAs Some triplets code for start or stop (beginning and end of a gene) (76). Transcription – making a copy of the base sequence on the template strand of DNA. Result is mRNA which can move to cytoplasm, leaving DNA in place for another day (77). Transcription because language is the same – bases. Translation – converting base sequence to sequence of amino acids, so changing language. tRNA is the translator – it know swhich amino acid to carry for a given triplet (anticodon when on tRNA). Matches anticodon with codon on mRNA to supply correct amino acid. Occurs on a ribosome in the cytoplasm. Peptide bonds link amino acids in order specified by mRNA (and therefore DNA) (77- 80). Explain the nature of the genetic code (triplet code, non-overlapping and degenerate) and describe a gene as being a sequence of bases on a DNA molecule coding for a sequence of amino acids in a polypeptide chain Triplet code – three bases on a strand code for one amino acid 22 amino acids and 64 codes, so degenerate code – some amino acids have more than one code. Learn this definition of a gene – sequence bounded by start codon and stop codon (76). Describe the basic structure of an amino acid (structure of specific amino acids are not required). Describe the formation of polypeptides and proteins as amino acid monomers linked by peptide bonds in condensation reactions. Explain the significance of a protein’s primary structure in determining its three-dimensional structure and properties as a globular or fibrous protein
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Primary structure is sequence of amino acids. Different properties of R groups determine properties of polypeptide, causing folding etc (57). Globular proteins tightly folded and soluble in water. 3D (tertiary structure) shape crucial, eg enzymes are globular proteins. Fibrous proteins don’t fold – stay as chains and often cross linked together into strands. Insoluble, eg keratin in hair, collagen in tendons, bones, skin and blood vessels (59).
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SNAB Biology AS Revision Summary
1.2.10
Explain the mechanism of action and specificity of enzymes in terms of their three dimensional structure and explain that enzymes are biological catalysts that reduce activation energy Enzymes have properties of catalysts (make reaction more likely to occur without being changed by it), but are also protein. Active site shape and charge distribution important (tertiary protein structure, 3D shape). Physical contact required between active site and substrate (enzyme + substrate, enzyme-substrate complex, enzyme + products) (70). Induced fit theory – substrate fits active site a bit, then induces it to fit better as it binds (70). Lock and key hypothesis – links shapes of substrate and active site (70). Specificity of enzymes explained by shape of active site. Activation energy – energy needed to start a reaction. Enzymes reduce activation energy needed to normal body conditions (71). Describe how enzyme concentrations and substrate concentrations can affect the rates of reactions and how the effect of enzyme concentration on reaction rate can be investigated practically Activity 2.10, core practical. All about number of enzyme-substrate complexes that can be formed. Note that the enzyme concentration graph is about initial rate, ie when plenty of substrate around (72).
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Describe the process of DNA replication (semi-conservative, including the role of DNA polymerase) Replication is copying of DNA. Compare it with transcription. Semi conservative – each new DNA molecule has one old strand and one new strand. Controlled by DNA polymerase – splits original pair of strands and promotes pairing of new nucleotides with exposed bases. (81)
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SNAB Biology AS Revision Summary
1.2.13
Explain how errors in DNA replication can give rise to mutations and explain how CF results from one of a number of possible gene mutations Gene mutation is change in the base sequence of a gene. Changing a triplet can change the amino acid code of mRNA, so polypeptide is different. This might affect 3D structure, which is crucial to function (82). Lots of different mutations of the same gene cause CF. The effect of the mutation on the shape (and therefore function) of the CFTR protein determines the severity of the CF (82). Explain the terms genotype, phenotype, recessive, dominant, homozygote and heterozygote and use a knowledge of genetics (including the interpretation of pedigree diagrams) to answer questions about monohybrid inheritance including CF, albinism, thalassaemia and garden pea height and seed morphology Genotype – alleles present whether expressed or not. Phenotype – observable effect of the genotype (what it looks like) (84-85). Allele – possible form of a gene. Recessive – allele expressed only when homozygous. Dominant – allele expressed whenever present. Homozygote – individual possessing only one type of allele for a gene (eg FF, ff). Heterozygote – individual having different alleles for a gene (eg Ff) (85). Monohybrid – inheritance pattern of a single gene. Remember – genotypes of individuals have two letters (alleles), gametes are haploid and only have one. CF – recessive, normal is dominant. Thalassaemia – book says it’s recessive, then says the heterozygote is different so it could be codominant. Albinism is recessive. Know about yellow and green, tall and short peas. (86) Explain how the expression of the CF gene impairs the functioning of the gaseous exchange, digestive and reproductive systems Gas exchange – sticky mucus can’t be removed by cilia. Pathogens grow in mucus (especially anaerobes where oxygen can’t penetrate). DNA from dead white cells attacking pathogens makes mucus thicker (53-54). Mucus blocks bronchioles so some alveoli not ventilated. Can be harder to get air out, so parts of lung overinflate and lose elasticity (56). Digestion – mucus blocks pancreatic duct so enzymes not released into duodenum. Food not properly digested and can’t be absorbed. Enzymes trapped in pancreas can cause damage, possibly diabetes if islets of Langerhans affected. Reproduction – mucus blocks cervix to prevent sperm reaching egg, males can lack vas deferens or have it blocked by mucus, so sperm can’t reach the urethra (73).
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SNAB Biology AS Revision Summary
1.2.16
Describe the principles of gene therapy and distinguish between somatic and germ line therapy Gene therapy – normal copies of the gene inserted into cells that need to express it (eg lining of lung). Inserted gene is transcribed and translated, and functioning protein is produced (88). Can use virus particles or liposomes to get gene into cells (88). Not very successful – hard to get gene into enough cells, and cells are constantly replaced so treatment needs to be repeated regularly. Somatic gene therapy treats cells that never make gametes. Inserted genes are never passed to next generation. Germ line therapy modifies eggs or sperm so that every cell will be changed, including those in testes and ovaries, so the genes will be passed on and become a part of the species genotype. Currently illegal (89). Describe how gel electrophoresis can be used to separate DNA fragments of different length Technique is summarised on page 91. Chop up DNA with restriction enzymes. Load on agarose gel, apply current to draw DNA towards positive electrode, speed depends on size of fragment. Transfer to nylon sheet (southern blotting), apply probe or stain to see the fragments. Describe how the genetic profiles produced by gel electrophoresis can be used in genetic screening using gene probes Gene probe -single strand of DNA with complementary base sequence to known gene sequence. Apply to gel, wash off anything that hasn’t bound to DNA. Usually radioactive, so can detect whether they’ve stuck to the DNA. (92) Explain the uses of genetic screening in the identification of carriers, prenatal testing (amniocentesis and chorionic villus sampling) and embryo testing Test potential parents to see if they have a recessive mutation. Allows assessment of risk. Testing embryo or foetus – amniocentesis samples fluid around foetus containing some foetal cells. Done at 15-17 weeks, 1% risk of abortion. CVS sample placenta which includes foetal cells. Done at 8-12 weeks, 1-2% risk. PIGD – preimplantation genetic diagnosis. Done using IVF, cell from embryo taken at 8-16 cell stage, tested and only implanted in uterus if found to be normal. Expensive and unreliable. (92-93) Discuss the social, ethical, moral and cultural issues related to genetic screening. Rights of embryo – some cultures deem them to have full rights at conception, so any tinkering is wrong (no consent, abortion not allowed). Duties – eg of parent to prevent harm at any stage of development. Maybe to avoid misery of particularly bad disorder. Four frameworks on pages 94-95 – Rights and duties, maximising good, decision making, virtuous lives.
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SNAB Biology AS Revision Summary
Topic 3: Voice of the genome
2.3.1 Describe the ultrastructure of a typical eukaryotic cell (nucleus, ribosomes, rough and smooth,endoplasmic reticulum, mitochondria, centrioles, Iysosomes, nucleolus) (100) Nucleus: nuclear envelope, pores, heterochromatin, euchromatin, nucleolus, nucleoplasm, DNA & protein (histones) Ribosomes: rRNA & protein, site of protein synthesis RER: membranes (lamellae) and ribosomes, protein synthesis SER: no ribosomes Mitochondria: aerobic respiration, matrix, cristae, elementary particles, DNA & small ribosomes Centrioles: microtubules, spindle formation Lysosomes: membrane bound bag of enzymes Nucleolus: rRNA synthesis, zero to several in a nucleus Explain the role of the rough endoplasmic reticulum (rER) and the Golgi apparatus in protein trafficking within cells Production in RER, transport to Golgi, packaging and activation in Golgi, formation of secretory vesicles (101) Distinguish between the ultrastructures of eukaryotic and prokaryotic cells Prokaryotes: no membrane bound organelles (no nucleus, mitochondria, ER, Golgi, vesicles), smaller ribosomes, cell wall not cellulose (98) Prokaryotes o Bacteria and blue-green bacteria are prokaryotes. o Cell does not have a nucleus. o DNA is not associated with any proteins and lies free in the cytoplasm. o Do not contain membrane-bound cell organelles. o o Most prokaryote cells are extremely small with diameters between 0.5 and 5 µm. A cell wall is always present. Eukaryotes o Animals, plants, fungi and protoctists are eukaryotes. o Cell contains a nucleus. o DNA is associated with proteins and is in the nucleus. o o o Contain discrete membrane-bound organelles such as nuclei, mitochondria and chloroplasts (plants only). Eukaryotic cells are larger with diameters of 20 µm or more. Not all eukaryotes have a cell wall.
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SNAB Biology AS Revision Summary
2.3.4
Explain the role of DNA replication and mitosis in the cell cycle (107) Replication - identical copy of DNA Mitosis - two identical nuclei (not cells - cell division usually follows) Explain the significance of mitosis for growth and asexual reproduction Cells produced have complete genome - can give rise to new organism or tissue (113) Describe the stages of mitosis and how they can be observed practically (110) Practical method (root tip squash) Activity 3.8 & 3.9
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SNAB Biology AS Revision Summary
2.3.7
Explain how mammalian gametes are specialised for their functions including the acrosome reaction The ovum is much larger than the sperm - the ovum cytoplasm contains protein and lipid food reserves for the early development of the embryo; the small sperm can move easily. The ovum has jelly-like coating; the sperm has no such coating - the ovum coating thickens after the first sperm penetrates, preventing entry of other sperm. The sperm has a tail; the ovum has no tail - this allows the sperm to move through the cervix and uterus to reach the ovum in the Fallopian tube. Sperm have an acrosome (a specialised lysosome); ova do not - this allows sperm to release digestive enzymes which break down the jelly-like coating of the ovum enabling the sperm to penetrate it. Explain the importance of fertilisation in sexual reproduction Restores diploid number of chromosomes Combines genome from two cells - important for genetic variation Explain how meiosis results in the halving of chromosome numbers and the introduction of variation through random assortment (the stages of meiosis, crossing over chiasmata are not required) Meiosis as one replication and two divisions Cells produced are haploid Separation of homologous chromosomes at first division (105) Random assortment - mixing of maternal and paternal chromosomes - random assortment of homologues at metaphase I (106) Explain what is meant by stem cells, pluripotency and totipotency Stem cells - multipotent (can become a limited range of cell types) eg blood stem cells in bone marrow form red, white cells and platelets Pluripotent - cells able to produce a wide range of cell types, but not all (found in embryo) Totipotent - able to form a complete organism (early embryo cells, cause twinning), common in plants etc Discuss the moral, ethical and spiritual implications of stem cell research Know the potential uses of stem cells, and pros and cons of use (115-117)
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SNAB Biology AS Revision Summary
2.3.12
Explain how genes can be switched on and off by DNA transcription factors and how this gene switching gives rise to specialised cells Transcription factors control gene activity (121), evidence that cytoplasm is involved (Acetabularia (119), control of development (122 fig 3.33), FOP as evidence of importance of gene control (124), chemical gradients controlling development (125) Protein transcription factors and an enzyme called RNA polymerase bind to a section of the DNA adjacent to the gene to be transcribed. This section is known as the promoter region. Once all the transcription factors and RNA polymerase have successfully bound to the promoter region, the transcription initiation complex has formed and transcription will proceed. The transcription factors may be present all the time in the cell. However, some transcription factors are only synthesised when needed. They may be present in an inactive form and will only bind when activated by signal proteins. Genes are switched off (not able to be transcribed) by the cell by preventing the binding of the transcription initiation complex to the section of the DNA adjacent to the gene. Protein repressor molecules may attach to the DNA of the promoter region blocking the attachment sites for transcription factors. Alternatively, protein repressor molecules can attach to the transcription factors preventing them forming the transcription initiation complex. In some cases signal proteins acting as transcription factors may simply not be present. Describe how the expression of a gene can be demonstrated practically by induction of β galactosidase O locus, R locus and structural genes. Only in prokaryotes (123, Activity 3.13) Explain how certain characteristics may be affected by both genotype and the environment, including human height, skin colour, hair colour and cancers Genotype controls the number of MSH receptors in skin cells. If your genotype codes for more MSH receptors, your melanocyte cells will be more active, produce more melanin and make your skin darker. The environment also affects the colour of skin because ultraviolet (UV) light increases the amount of MSH and of MSH receptors, making the melanocytes more active. As a result, more melanin is produced, making the skin darker. The melanin is packaged into organelles called melanosomes. The melanosomes are transferred to nearby skin cells where they collect around the nucleus, protecting the DNA from harmful UV light. Human albinos have a gene mutation that prevents them making melanin. They have white hair, white skin and no pigment in their eyes. Their skin colour is determined solely by their genotype; the environment has no effect. Types of variation and examples (127),effects of environment eg rabbit fur (130) Explain that cancers arise from uncontrolled cell division (detailed knowledge of the checkpoint control in the cell cycle is not required) and describe genetic, environmental and lifestyle causes of cancer More cell production than cell death. Result is a tumour. Damaged DNA causes lack of normal control of cell cycle. Oncogenes stimulate cell cycle if active. Tumour suppressor genes turn off cell cycle. (131) Don't learn 131. Inherited and environmental causes (133)
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SNAB Biology AS Revision Summary 2.3.16 Discuss the principal outcomes of the Human Genome Project and the social, moral and ethical issues which arise from it. DNA sequenced, all genes identified. Outcomes - information about genome, identification of new genes, finding drug targets, preventative medicines, understanding physiology, clues about evolution (137) Ethics mostly about knowledge and controlling information eg effects on insurance (139) Genetic screening Adequate counselling is necessary to ensure that the results and the consequences of the results arefully understood. Genetic screening is rarely 100% accurate, so the possibility of incorrect test results should be considered. A positive result for the presence of a faulty allele for a genetic condition may influence whether couples choose to have children. This decision should be considered alongside present and potential future treatments for any disorder. Deciding not to have children for genetic reasons may reduce the number of abortions carried out after prenatal screening. Positive results raise issues concerning duties to inform other related family members. Individuals identified as carriers of the genes associated with inherited diseases may face discrimination in employment or when applying for insurance. Screening of embryos to be used in IVF raises issues about making choices about which embryos should live and which should not. There are arguments for using technology to try to ensure only the healthiest babies are born. Screening of embryos during pregnancy for a particular gene disorder by chorionic villus sampling between 8 and 12 weeks of pregnancy and amniocentesis at around 15-17 weeks raise issues such as: o These procedures present a risk of miscarriage to possibly healthy fetuses. o Ensuring informed choice about whether or not to abort a fetus if it is found to have a genedisorder. o The right of the unborn child to live. o The right of the parent and family to choose whether they continue and have a child with a condition that may have major social, financial, happiness and health implications for the family.
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SNAB Biology AS Revision Summary
Topic 4: Climate change
2.4.1 Explain the importance of water and inorganic ions (nitrate, calcium and magnesium) to plants Turgidity, photosynthesis, transport Nitrate - amino acid and protein synthesis Calcium - cell wall structure, membrane permeability Magnesium - chlorophyll production (158) Recall the typical ultrastructure of animal cells and contrast this with the ultrastructure of typical plant cells (presence of cell wall), chloroplasts, amyloplasts (containing starch grains), vacuole, tonoplast, plasmodesmata, pits and middle lamellae Animal
Centriole Nucleus Nucleolus Mitochondrion Rough and smooth ER Golgi Ribosome Lysosome
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Plant
Cell wall Chloroplast Large central vacuole Amyloplast Tonoplast
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Compare the structure and function of the polysaccharides starch and cellulose including the role of hydrogen bonds between β glucose molecules in the formulation of cellulose microfibrils Cellulose (148) Starch Cellulose Polymer of α-glucose Polymer of β-glucose Mixture of branched amylopectin and Long and unbranched helical amylose β-1,4 glycosidic bonds α-1,4 and α-1,6 glycosidic bonds H-bonding between molecules to form Compact storage molecule microfibrils Structural material in cell walls
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SNAB Biology AS Revision Summary 2.4.4 Describe the structure of sclerenchyma fibres and xylem vessels and where they are found in the plant stem. Describe how their physical properties enable them to be used for human benefit Activity 4.3 Distribution in a stem (153) Uses of fibres (160), activity 4.6 Xylem vessels are arranged one above the other creating long columns of cells. The cell contents die, leaving tubes that extend the full height of the stem. The sclerenchyma fibre cells are also arranged in columns. Thus xylem vessels and sclerenchyma fibres result in long, strong fibres when extracted from the plant. The tough cellulose cell walls impregnated with lignin make these fibres very strong and waterproof. Explain the relationship between structure and function in sclerenchyma fibres (support) and in xylem vessels (support and transport of water and mineral ions through the stem) Xylem vessels (156), lignin, water transport, transpiration (156), mass flow carrying mineral ions. Support (158) Explain the role of adhesion, cohesion and the transpiration stream in the movement of water through the stem Cohesion and adhesion (sticky water molecules (159) (158)) Describe how to determine the strength of fibres Activity 4.6 Compare how William Withering developed his digitalis soup with drug developing and testing nowadays Activity 4.8. Drug testing (165), Dosage and digitalis (164) Describe how to investigate the antibacterial properties of plants. Activity 4.7 Relate the structure of seeds to their role in the dispersal and survival of the plant(adaptations for dispersal, protection and nutrition of the embryo) Seed structure (166), dispersal adaptations (167 - 168) animal, wind, water, self Describe uses of starch and plant oils for humans (diet, packaging, glues, absorbent materials, fuels) and explain how the use of plant products can make resource utilisation more sustainable Starch uses (1'69) Oil uses (170)
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SNAB Biology AS Revision Summary 2.4.12 Explain how the genetic modification of plants is similar to but distinct from conventional breeding Ways in which GM of plants is similar to conventional breeding Both processes rely on mutations and genetic inheritance. Plants are produced with alleles for desired characteristics. Only plants with the desired characteristics are grown on. Plants result that are homozygous for desired characteristics. Ways in which GM of plants is distinct from conventional breeding Favourable characteristics can be chosen from almost any plant species or other organism. Gene manipulation techniques are used to transfer new genes into plants. Marker genes are used to select plant cells which contain the new gene. The whole process takes less time than conventional breeding. There is more control over which new genes are introduced into the plants. There is less control over where new genetic material is inserted. Discuss the scientific arguments for and against the use of genetically engineered plants(improved plant quality, 2.4.14 enhanced yield and consequences for the environment and health) Discuss the social and ethical arguments for and against the use of genetically engineered plants (179) Arguments for the use of GM plants Improved crop plants; this may include higher yields, higher vitamin content, plants that can manufacture drugs, drought-resistant plants and trees from which paper can be made without the need for chlorine to be used. Crops that are unpalatable to insect herbivores reduce insecticide use. Arguments against the use of GM plants Antibiotic resistance genes may be transferred to bacteria in the gut which could build up resistance to certain antibiotics used in medical treatments. There are health concerns related to the formation of harmful products in the plant by new genes, and the products are then eaten. Transfer of genes to non-GM plants may create superweeds and adversely affect food chains. Chemical use is increased in GM crops which are resistant to herbicide (weedkiller). Right to choose Ownership and profit Outline the causes of global warming - including the role of green house gases in the green house effect. Describe the sources of carbon dioxide and methane CH4, their possible role in global warming and how their levels might be controlled Greenhouse effect - radiation trapped by layer of gases (196) Gases involved, changing levels (198).
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2.4.15
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SNAB Biology AS Revision Summary 2.4.16 Recall the role of the carbon cycle in regulating atmospheric carbon dioxide levels and discuss the methods that can help to reduce atmospheric levels of carbon dioxide(including the use of biofuels and reafforestation) Carbon cycle (206). Biofuels (209), Reafforestation (210) Discuss the possible relationship between CO2levels and global warming and how this can be investigated practically Link between carbon dioxide and temperature change (198) Activity 4.23 Describe and analyse data from different types of evidence for and against global warming including temperature records, pollen in peat bogs and dendrochronology and appreciate that scientific theories must be supported by evidence Temperature records (190, 195), dendrochronology (193), peat bogs (191) Appreciate that data can be extrapolated to make predictions, that these are used in models of future climate change, and that these models have limitations Mathematical models and predictions (202), complexity and limitations (203) Explain how climate change (rising temperature, changing rainfall patterns and changes in seasonal cycles) can affect plants and animals (distribution of species, development and life cycles) Species distribution (183), growth rate and development (186), development (188) Evidence that many species that currently reach their northern limit of distribution in, for example, southern England are shifting northwards. The changing distribution of animals may be a direct response to rising temperature, changes in rainfall patterns or the result of a shift in distribution ofthe plants or animals they feed on. Some species may become rare even though they can cope with the changing climatic conditions because new species moving into the areas prey on them or their food sources. The distribution of diseases infecting both animals and plants may also change and diseases could spread rapidly, given a suitable climate. Higher temperatures may lead to higher plant yields with increases in the rate of photosynthesis up to the point where light intensity or another factor becomes the limiting factor. But a lack of soil moisture that is likely to accompany rising temperatures would see a decline in plant yields. It is probable that crop production in cooler temperate regions will benefit from climate change whereas warmer tropical regions may suffer. Animals are likely to be affected where temperature acts as an environmental cue or trigger for development or behaviour. For many species, the hatching of eggs or the emergence of adults is synchronised with periods of maximum food availability. If a mismatch occurs between hatching and peak food abundance the survival of the species may be jeopardised. The egg incubation temperature of some reptiles determines the sex of the offspring.
2.4.17
2.4.18
2.4.19
2.4.20
21
SNAB Biology AS Revision Summary 2.4.21 Explain the effect of increased temperature on the rate of enzyme activity Exponential rise to optimum then denaturation (186) Describe how to investigate the effects of temperature on the development of organisms(eg plant growth or brine shrimp hatch rates) Activity 4.18 Discuss the way in which scientific conclusions about controversial issues can sometimes depend on who is reaching the conclusions, including their ethical and cultural perspectives. Dealing with controversy and bias (199)
2.4.22
2.4.23
22
SNAB Biology AS Revision Summary
Terms used in the exams
Advantages, disadvantages
Here there will be two (or more) sets of data, structures, functions, processes or events to be referred to and the answer must relate to both. One process, or whatever, is required to be compared with another. It is important that the answers are comparative and that the feature being referred to is clearly stated.
Compare, contrast, distinguish between, differs from
As with advantages and disadvantages, here there will be two (or more) sets of data, structures, functions, processes or events to be referred to and the answer must relate to both. It is important to select equivalent points and keep them together. Compare generally indicates that similarities as well as differences are expected; contrast, distinguish between or differs from indicate that the focus should be on the differences.
Describe
This may be related to a biological event or process, or to data presented in a table, graph or other form. The description must be concise and straightforward, using relevant biological terms rather than vague generalisations. The trend should be presented in words or translated into another form. If interpreting numerical data, it is often appropriate to refer to the figures, and these should be ‘manipulated’ in some way, for instance the trend could be quantified or the percentage difference over a period of time calculated.
Discuss
Give a considered account of a particular topic about which a degree of uncertainty exists.
Distinguish
Identify appropriate differences in a given context.
Explain, give explanations, give reasons
The answer would be expected to draw on biological knowledge to give reasons or explanations for the information or data given. Usually 2- or 3-mark answers are required and the answer should go beyond just repetition or reorganisation of the information or data presented. Students should check that their response answers the question, ‘Why …?’.
Name, state, give
Indicate that short, factual answers are needed, possibly with precise use of biological terms or the name of a structure. Often one-word answers are sufficient.
Suggest/suggestion
Implies that the answer may include material or ideas which have not been learnt directly from the specification. A reasonable suggestion, using biological knowledge and understanding of related topics, is required.
Using the information in the diagram/on the graph/in the table/features visible in the diagram
Refer only to the information presented in the question and not other examples or features, which may be perfectly correct but are not shown and are, therefore, not what the examiners require.
Steve Hall March 2008
23
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