higher human biology essay questions

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Higher Human Biology: Extended Response Questions/Markschemes

Description.

• human biology
• extended response
• physiology and health
• neurobiology and communication
• immunology and public health
• human cells

Resource summary

Human cells.

Compare the location and functions of embryonic and tissue (adult) stem cells. (7) 1. Stem cells are undifferentiated cells capable of repeated division to both more stem cells and cells that will later differentiate to form specialised cells. 2. Embryonic stem cells (ESCs) are found in the inner cell mass of the blastocyst. 3. ESCs are capable of repeated division… 4. …to form more ESCs and other cell types. 5. ESCs are: capable of forming all the other cell types of the body / pluripotent 6. Tissue stem cells (TSCs) are found throughout the juvenile and adult body. 7. TSCs are capable of repeated division to form more TSCs and other cell types. 8. TSCs: can only form cells of the organ to which they belong / are multipotent. 9. e.g. bone marrow tissue cells can only give rise to bone marrow cells, red blood cells, platelets, phagocytes and lymphocytes. Give an account, with examples, of the different body tissue types and their functions. (8) 1. Body tissue cells derive from somatic stem cells … 2. … by repeated mitosis. 3. Epithelial tissue - covers the organ surfaces. 4. Protection - skin / secretion - intestinal glands / absorption - villi. 5. Connective tissue - gives shape to organs and supports them. 6. Protection - skull bones / structural framework - ribs / storage of energy - adipose tissue / connecting body organs - blood / connecting epithelial to muscle tissue - cartilage (in tendons) 7. Muscle tissue - which causes locomotion or movement within organs. 8. Skeletal muscle - locomotion / smooth muscle - in arterioles control of access to capillary bed / cardiac muscle - contraction of the heart. 9. Nervous tissue - which transmits messages between the central nervous system and the rest of the body (and within the central nervous system). 10. Neurons - conduct impulses / glial cells - maintain a constant environment for neurons. Describe the way in which the use of stem cells has contributed to: A. corneal transplants; (3) B. skin grafts. (3) A. Corneal transplants: 1. Traditionally, the person would have to wait for donated corneal tissues to become available. 2. With the use of stem cells isolated from the patient's healthy eye, new tissue can be cultured… 3. …which can be transplanted back into the damaged eye, 4. and will repair the damaged corneal epithelium. B. Skin grafts: 1. Traditionally, a section of the patient's healthy skin is cultured over 3 weeks. 2. During this time, dehydration and infection are risks for the patient. 3. Skin stem cells from the patient can be cultured to produce new epidermal cells much more quickly. 4. This reduces the time during which the patient is at risk. A. Describe the location and structure of DNA. (8) B. Give an account of the replication of DNA. (7) A. Describe the location and structure of DNA.(maximum of 8 marks): 1. DNA is located on chromosomes in the nucleus. 2. A DNA molecule consists of two strands wound in a double helix. 3. Each strand consists of subunits called nucleotides. 4. A nucleotide consists of a deoxyribose sugar molecule, a phosphate group and an organic / nitrogenous base. 5. The deoxyribose and phosphate are linked to their neighbours to form a sugar-phosphate backbone. 6. There are four organic bases: adenine (A), thymine (T), guanine (G) and cytosine (C). 7. Bases are linked in complementary pairs A-T and G-C. 8. Bases link the two DNA strands by hydrogen bonds. 9. The two DNA strands have an antiparallel structure / explanation. 10. Deoxyribose is found at the 3′ end of each strand and phosphate is found at the 5′ end. B. Give an account of the replication of DNA.(maximum of 7 marks): 1. DNA is unwound and unzipped… 2. …by the enzyme helicase… 3. …to form two template strands. 4. DNA polymerase is the enzyme which adds nucleotides to the new DNA strand. 5. DNA polymerase needs a primer to start replication. 6. DNA polymerase can only add complementary nucleotides to the deoxyribose / 3′ end of the DNA strand. 7. This results in one strand / the leading strand being continuously replicated, 8. and the other strand / the lagging strand being replicated in fragments, 9. which are joined together (by the enzyme ligase). Give an account of the structure and function of proteins. (8) A. Protein structure (maximum of 4 marks): 1. Proteins consist of amino acids joined together (in chains). 2. The amino acids are joined by strong peptide bonds… 3. …to produce the primary structure of the polypeptide. 4. Further bonding between amino acids… 5. …produces the secondary and tertiary structures. 6. Folding of the polypeptide chains forms the three-dimensional shape of the protein. B. Protein function (maximum of 4 marks): 1. Some proteins are enzymes + named example (e.g. the digestive enzyme amylase). 2. Some proteins are hormones + named example (e.g. insulin). 3. Some proteins are antibodies which help the body fight infections. 4. Some proteins transport substances + named example (e.g. haemoglobin). 5. Some proteins provide structure + named example (e.g. collagen). Give an account of gene expression under the following headings. A. Transcription (6) B. Post-translational modification (PTM) (4) A. Transcription (maximum of 6 marks): 1. Transcription is the formation of a mRNA molecule on a DNA template. 2. DNA unwinds, and strands separate by the action of helicase. 3. RNA nucleotides attach to exposed bases of DNA. 4. DNA A pairs with RNA U, DNA T - RNA A, DNA G - RNA C and DNAC - RNA G. 5. RNA polymerase joins nucleotides (to each other). 6. This produces the primary transcript. 7. Which contains introns and exons. 8. Exons are protein coding sections of mRNA and introns are non-coding sections. B. Post-translational modification (PTM) (maximum of 4 marks): 1. It takes place after the polypeptide has been completed on the ribosome. 2. It allows several proteins to be formed from one gene. 3. It may involve enzyme cutting and combining of polypeptide chains. 4. e.g. insulin from pro-insulin. 5. It may involve adding of phosphate or carbohydrate groups. 6. Addition of phosphate enables enzymes / receptors to be switched 'on' and 'off'. Give an account of gene mutation. (9) 1. A gene mutation is a change in the base type or sequence in a gene. 2. In a substitution (mutation) one base is replaced by another. 3. If substitution produces a new stop codon it is a nonsense mutation. 4. If substitution affects introns and exons it is a splice-site mutation. 5. If substitution changes one mRNA codon it is a missense mutation. 6. In a deletion (mutation) a base is removed. 7. In an insertion (mutation) a base is added. 8. A change to a single nucleotide / base is a point mutation. 9. Insertion and deletion (mutations) can potentially cause frameshift mutation. 10. A frameshift mutation alters all the triplets following it. 11. If any of these mutations occurs in a protein-coding gene, then the protein produced may be altered (or not produced at all). Give an account of the amplification of DNA sequences by the Polymerase Chain Reaction (PCR). (8) 1. The components of the PCR (template DNA, primers, nucleotides and the Taq DNA polymerase) are mixed together. 2. Heating to 90 - 95°C… 3. …separates (denatures) the double-stranded DNA (containing the target DNA sequence). 4. Annealing at 54°C… 5. …allows the primers to bind (by forming hydrogen bonds) to their complementary sequences on the separated strands 6. Primers are short DNA sequences (each about 20 bases long), 7. that bind at either side of the target sequence (one on each of the complementary strands of the target DNA). 8. Extension at 72°C. 9. Taq polymerase now adds nucleotides to the 3′ ends of the primers and extends them into new complementary strands. 10. Repeated cycles multiply the target DNA exponentially… 11. …since each new double strand separates to become two templates for further synthesis. Describe the control of the action of enzymes which are continuously present in the cell. (10) 1. Enzymes that are in constant use are continuously present. 2. The presence of the substrate may initiate a reaction. 3. The build-up of the product may reverse the direction of a reaction. 4. The active site is the part of the enzyme molecule to which the substrate (temporarily) binds. 5. Competitive inhibitors resemble the substrate and compete with it for the active site. 6. Increased substrate concentration reduces the effect of competitive inhibition. 7. Competitive inhibition is reversible. 8. Non-competitive inhibitors bind to allosteric sites… 9. …and alter the shape of the sites, preventing the substrate from binding to them. 10. Enzyme activators bind to allosteric sites and alter the shape of the active site to fit the substrate. 11. Feedback inhibition involves non-competitive inhibition. 12. Products from reactions late in metabolic pathway act as non-competitive inhibitors to enzymes earlier in the pathway. State what is meant by the term 'feedback inhibition', give an example of it, and explain its advantage to the cell. (7) 1. Feedback inhibition in a metabolic pathway occurs when an enzyme's activity is reduced… 2. …by an increase in the concentration of a product from a later step in the pathway. 3. In the energy investment stage of glycolysis… 4. …phosphofructokinase is inhibited by ATP… 5. …which is released by glycolysis and the cytochrome system. 6. If the ATP concentration rises in the cell, phosphofructokinase will become less active… 7. …and the production of ATP will be reduced, or vice versa. 8. The production of ATP will be matched to the cell's need for energy… 9. …and resources will be conserved. Give an account of the enzymes and coenzymes involved in the citric acid cycle. (6) 1. Each step in the citric acid cycle is mediated by a different enzyme. 2. Dehydrogenase enzymes remove hydrogen and electrons. 3. Coenzymes NAD / NAD+ and FAD accept hydrogen and electrons. 4. NAD / NAD+ accepts hydrogen to form NADH. 5. FAD accepts hydrogen to form FADH2. 6. NADH / FADH2 / reduced coenzymes carry hydrogen and electrons to the electron transport chain… 7. …which results in the generation of ATP. Give an account of the electron transport chain. (8) 1. NADH and FADH2 carry H+ ions and high energy electrons… 2. …from glycolysis and the citric acid cycle. 3. At the inner mitochondrial membrane, NADH and FADH2 pass the hydrogen ions and high energy electrons… 4. …to carriers in the electron transport chain. 5. The electron transport chain is a collection of proteins attached to the inner mitochondrial membrane. 6. As the high energy electrons released from NADH and FADH2 pass along the chain, … 7. …they release their energy in stages. 8. This energy is used to pump H+ ions into the intermembrane space. 9. The final electron acceptor is oxygen. 10. The (negatively charged) oxygen combines with two H+ ions to form water. Give an account of the role of creatine phosphate in muscles. (8) 1. Muscle cells do not contain any reserves of ATP which will fuel more than 2s of activity. 2. Creatine phosphate is an energy reserve found in cells with high and variable energy demands… 3. …such as muscle / brain. 4. Creatine phosphate is used by fast twitch muscle fibres. 5. Creatine phosphate breaks down anaerobically… 6. …to release energy and phosphate… 7. …which is used to convert ADP and Pi to ATP at a fast rate. 8. There is only sufficient creatine phosphate to fuel strenuous muscle activity for a maximum of 10s. 9. When energy demands are low, creatine phosphate reserves are restored… 10. …using ATP provided by aerobic respiration.

Physiology and Health

Describe:A. the development of ova in the ovary; (6)B. and their possible fates after ovulation. (2)A. Development of ova in the ovary (maximum of 6 marks):1. The ovary contains many immature ova.2. Each ovum develops in a follicle.3. Follicles protect the ova.4. Follicles secrete hormones / oestrogen.5. Every 28 days, a (mature) follicle moves to the surface of the ovary.6. An ovum is released into the oviduct.7. The follicle develops into the corpus luteum…8. …which releases hormones / progesterone.B. After ovulation (maximum of 2 marks):1. The ovum may be fertilised to form a zygote.2. The fertilised ovum / zygote divides many times to form the blastocyst.3. The blastocyst implants in the endometrium.4. Unfertilised ova pass out of the body.Give an account of negative feedback control under the headings:A. testosterone production; (3)B. the luteal phase of the menstrual cycle. (3)A. Testosterone production (maximum of 3 marks):1. The pituitary gland releases LH/ICSH.2. LH stimulates the interstitial cells of the testes to release testosterone.3. Increasing level of testosterone inhibits/reduces the production/release of LH.4. Pituitary gland releases less LH…5. …so testes/interstitial cells release less testosterone.B. The luteal phase of the menstrual cycle (maximum of 7 marks):1. After ovulation, the high level of LH causes the follicle to develop into the corpus luteum.2. The corpus luteum secretes progesterone and oestrogen.3. During this phase the secretion of oestrogen and progesterone rise to a maximum and then decline.4. The high levels of oestrogen inhibit the pituitary from secreting FSH.5. The high levels of progesterone inhibit the pituitary from secreting LH.6. The resulting low level of FSH suppresses the development of further follicles.7. The low level of LH causes the corpus luteum to degenerate…8. …and progesterone secretion to fall to a minimum.9. The falling level of progesterone at the end of the cycle triggers the start of menstruation.10. The low level of oestrogen at the end of the cycle causes the pituitary to increase secretion of FSH.Give an account of fertile periods under the headings:A. men; (4)B. women. (6)A. Men (maximum of 4 marks):1. Men are fertile from puberty to death.2. Fertility does decrease with age.3. Men are continuously fertile / produce sperm constantly.4. Testosterone stimulates sperm production.5. Testosterone production is stimulated by ICSH from pituitary.6. Testosterone and ICSH interact in a negative feedback loop.B. Women (maximum of 6 marks):1. Women are fertile from puberty to the menopause / age 45 - 55.2. Women are cyclically fertile.3. Only fertile for a few days each month around time of ovulation.4. Ovulation is stimulated by a surge in LH secretion.5. LH is secreted by the pituitary gland.6. Oestrogen and progesterone from ovary interact in a negative feedback loop…7. …with FSH and LH from the pituitary.8. Time of ovulation can be determined by recording body temperature and cervical mucus.9. Fertile period lasts from a few days before to a few days after ovulation.Give an account of the tests which can be carried out once a woman has been confirmed as pregnant, under the headings:A. screening tests; (6)B. diagnostic tests. (4)A. Screening tests (maximum of 6 marks):1. Screening indicates possibility of a condition.2. Ultrasound dating scan at 8-14 weeks…3. …indicates age of fetus / likely due date.4. Ultrasound anomaly scan at 18-20 weeks…5. …indicates possible unusual development, associated with e.g. Down's syndrome.6. Biochemical tests on blood samples.7. Levels of marker chemicals inappropriate to stage of pregnancy, e.g. α-feto protein…8. …indicate possible presence of Down's syndrome.B. Diagnostic tests (maximum of 4 marks):1. Diagnostic tests confirm the presence of condition.2. Amniocentesis removes cells from the amniotic fluid…3. …and are carried out at about 18 weeks4. Chorionic villus sampling removes cells from the placenta…5. …and are carried out at 10-12 weeks.6. Karyotyping - examination of fetal chromosomes.Give an account of the exchange of materials between the blood and the body tissues under the headings:A. movement of fluid in and out of the blood; (6)B. substances transferred. (4)A. Movement of fluid in and out of the blood (maximum of 6 marks):1. Exchange only occurs in the capillaries / capillary beds.2. Fluid moves because of pressure differences.3. Fluid leaves blood because hydrostatic pressure exceeds osmotic pressure.4. Fluid re-enters blood because osmotic pressure exceeds hydrostatic pressure.5. Fluid leaves capillaries by pressure filtration.6. The capillary walls determine what may pass through them.7. Fluid entering the tissues is called tissue fluid.8. Tissue fluid is similar to blood plasma but lacks plasma proteins.B. Substances transferred (maximum of 4 marks):i. All substances in the blood except cells and large protein molecules pass out into tissue fluid.ii. Tissue fluid supplies cells with oxygen, glucose, amino acids, fatty acids, hormones, mineral ions. (any 2 for 1 mark, any 4 for 2 marks)iii. Waste materials from cells are passed back into the blood + any example, e.g. CO2, lactic acid. (full mark requires an example)iv. Products of cells are passed back into the blood + any example, e.g. hormones. (full mark requires an example)v. Excess / 10% tissue fluid passes into the lymphatic system.Give an account of the cardiac cycle under the following headings:A. atrial systole; (3)B. ventricular systole; (5)C. diastole. (2)A. Atrial systole (maximum of 3 marks):1. During atrial systole, the sino-atrial node / pacemaker, sends waves of nerve impulses across the walls of the atria…2. …which makes them contract simultaneously.3. The pacemaker is situated in the wall of the right atrium.4. Blood is forced through the atrio-ventricular/AV valves into the ventricles.B. Ventricular systole (maximum of 5 marks):i. The SA node also stimulates the atrio-ventricular (AV) node…ii. …which is situated between the atria and the ventricles.iii. The AV node sends nerve impulses down the septum and into the walls of the ventricles…iv. …making them contract simultaneously…v. …from the bottom up, which is the most efficient way of emptying them.vi. As the ventricles contract the AV valves close…vii. …preventing backflow of blood (into the atria).C. Diastole. (maximum of 2 marks):I. During diastole the atria and ventricles relax.II. The semi-lunar valves close.III. The AV valves open to allow blood to pass into the ventricles (again).Give an account of cholesterol in the body under the headings:A. sources and removal of cholesterol; (3)B. high density lipoproteins and low density lipoproteins. (7)A. Sources and removal of cholesterol (maximum of 3 marks):1. Cholesterol is formed in all body cells.2. The liver is the greatest producer.3. Cholesterol is present in animal foods e.g dairy products / meat / poultry / fish.4. Cholesterol is eliminated from the body in the bile.B. High density lipoproteins and low density lipoproteins(maximum of 7 marks):i. Lipoproteins are made in the liver.ii. They are responsible for the transport of cholesterol.iii. High density lipoproteins (HDLs) carry cholesterol from the body cells to the liver.iv. Low density lipoproteins (LDLs) carry cholesterol from the liver to body cells.v. LDLs attach to LDL-receptors on the cell membrane of most cells.vi. The production of LDL-receptors is controlled by negative feedback.vii. LDLs are carried into the cells to release their cholesterol.viii. Excess LDLs circulate in the blood and can become absorbed into the atheromas in plaques on artery walls.ix. A high ratio of HDLs to LDLs lowers the level of cholesterol in the blood…x. …reducing the development of atherosclerosis.Give an account of Type 2 diabetes under the headings:A. causes; (2)B. effects; (6)C. diagnosis and treatment (2)A. Causes (maximum of 2 marks):i. Type 2 diabetes results from reduced cellular sensitivity to insulin……ii. …caused by a reduction in the number of insulin-receptors on liver cells.iii. It is associated mainly with increasing obesity in adulthood.B. Effects (maximum of 6 marks): Diabetes of all forms means that the body is unable to control the rise in blood glucose levels after meals. Diabetics have chronically high blood glucose levels… …which damage blood vessels walls. This leads to the development of vascular diseases in large blood vessels… …e.g. atherosclerosis / stroke / myocardial infarction. In small blood vessels of vulnerable tissues it can cause haemorrhaging in the retina, resulting in reduced vision / damage to the glomeruli, resulting in renal failure / slowing down of impulses in the peripheral nervous system, resulting in loss of control and sensation. (1 mark for naming a cause and 1 additional mark for naming the resultant effect) C. Diagnosis and treatment (maximum of 2 marks) Diabetes is diagnosed by a glucose tolerance test… …which involves blood tests taken before, and 2 hours after taking a drink containing 75g of glucose. The patient should have fasted for 8 hours before the test. Treatment is usually administered by adjusting diet… …reducing energy intake and increasing exercise levels.

Neurobiology and Communication

Give an account of the nervous system under the headings:A. divisions of the nervous system; (5)B. homeostatic control. (5)A. Divisions of the nervous system(maximum of 5 marks):1. The nervous system is divided into the central nervous system and the peripheral nervous system.2. The central nervous system consists of the brain and spinal cord.3. The peripheral nervous system comprises all the sensory and motor neurons which connect it to the rest of the body.4. The peripheral nervous system is sub-divided into the somatic and the autonomic nervous systems.5. The somatic nervous system controls the voluntary activity of the skeletal muscles (and thus all movement) via its sensory and motor neurons.6. The autonomic nervous system is further divided into the sympathetic and the parasympathetic nervous systems, which act antagonistically.B. Homeostatic control (maximum of 5 marks):i. The autonomic nervous system plays an important part in many involuntary homeostatic processes…ii. …by conducting impulses through its sensory and motor neurons to the smooth muscle of artery walls, the cardiac muscle of the heart, and to glands.iii. The sympathetic nervous system prepares the body for action, 'fight or flight'…iv. …by speeding up heart and breathing rates, and slowing down peristalsis and intestinal secretions.v. The parasympathetic nervous system calms the body down, 'rest and digest'…vi. …by slowing down the heart and breathing rates, and increasing peristalsis and intestinal secretions.Give an account of Short-Term Memory under the headings:A. increasing memory span; (3)B. serial position effect; (5)C. transfer from STM to LTM. (2)A. Increasing memory span (maximum of 3 marks):1. Memory span is the number of items that can be retained in STM.2. The normal short-term memory span is 7± 2 items.3. Information is retained in STM for 15-30s.4. 'Chunking' helps short-term memory in particular as several items are grouped as one.B. Serial position effect (maximum of 5 marks):i. A large number of items is shown briefly to the subjects so that they cannot all be retained in short-term memory.ii. Items recalled by the subjects are recorded.iii. Subjects usually recall items presented early and late in the series.iv. Items at the end of the series are still retained in STM (the recency effect).v. Items from the start of the series have been transferred to long-term memory (primacy effect).vi. Items from the middle of the list have been displaced from STM…vii. …and have not been transferred to LTM.C. Transfer from STM to LTM (maximum of 2 marks):1. Rehearsal - repetition of information.2. Elaboration of meaning by linking to other memories/emotions.3. Organisation - linking to other similar memories.Compare and contrast sensory and motor neurons and describe events that occur at a synapse. (10)1. Sensory neurons pass messages from sense organs to the central nervous system whereas motor neurons transfer messages from the CNS to muscles and glands.2. Both neurons consist of cell body, axon and dendrites.3. The cell body is found part way along the axon of a sensory neuron, whereas the axon grows out from one side of the cell body in the motor neuron.4. In each case, the axon is wrapped in a myelin sheath with nodes every few millimetres.5. At a synapse, neurotransmitters cross from the pre-synaptic neuron to the post-synaptic neuron.6. Neurotransmitters include acetylcholine and noradrenaline.7. The type of receptor on the post-synaptic dendrite, to which the transmitter chemical binds, determines whether the next neuron is inhibited or excited.8. Acetylcholine is immediately degraded by an enzyme.9. Noradrenaline is reabsorbed by active uptake.10. Synapses filter out single weak impulses, but can sum several weak impulses.Give an account of the mode of action of recreational drugs (with examples), under the headings:A. effects on the brain; (4)B. modes of action. (6)A. Effects on the brain (maximum of 4 marks):1. Recreational drugs mainly affect the reward circuit of the brain…2. …by stimulating increased secretion of dopamine…3. …which causes feelings of euphoria and relaxation…4. …and leads quickly to addiction.5. The changed neurochemistry leads to changes in mood / cognition / perception / behaviour. (mention of all four examples gains this mark)B. Modes of action (maximum of 6 marks): Stimulation of release of neurotransmitter, … …e.g. MDMA stimulates release of serotonin. Agonists , imitating the action of a neurotransmitter, … …e.g. cannabis binds to cannabinoid receptors and suppresses GABA secretion. Antagonists bind to receptors and prevent neurotransmitter from doing so, … …e.g. ethanol binds to GABA receptors and depresses impulse generation. Inhibiting re-uptake of neurotransmitter, …v …e.g. cocaine blocks the re-uptake of serotonin / norepinephrine / dopamine. Inhibiting neurotransmitter degradation, … …e.g. tobacco contains two MAO inhibitors which suppress dopamine breakdown. Describe ways in which humans communicate under the headings:A. non-verbal communication; (4)B. verbal communication. (6)A. Non-verbal communication (maximum of 4 marks):1. Non-verbal communication involves gestures, signs, facial movements and posture.2. Non-verbal communication aids verbal communication.3. Attitudes and emotions are signalled by non-verbal communication.4. Mirroring of non-verbal communication strengthens bonds.5. Examples of non-verbal communication include winking, folding arms and smiling. (three listed for one mark)6. Any two examples explained in terms of meaning conveyed.B. Verbal communication (maximum of 6 marks): Language uses symbols to represent information Language enables information to be organised into categories and hierarchies. This organisation of information accelerates learning. Organisation of information aids intellectual development. The ability of humans to communicate verbally has resulted in the transmission of knowledge. (plus explanation) The ability of humans to communicate verbally has resulted in the development of culture. (plus explanation) The ability of humans to communicate verbally has resulted in social evolution. (plus explanation) The ability of humans to communicate verbally has resulted in the transmission of knowledge, development of culture and social evolution. Give an account of group behaviour and social influence under the headings:A. social facilitation; (3)B. deindividuation; (4)C. influences that change beliefs. (4)A. Social facilitation (maximum of 3 marks):1. Performance of a task may be improved in the presence of others.2. The presence of others in a competitive situation may enhance performance3. Detailed example of the point above4. The presence of an audience may improve performance.5. Detailed example of the point aboveB. Deindividuation (maximum of 4 marks): Deindividuation is responsible for the loss of identity in a crowd. Detailed example of the point above Deindividuation leads to diminished restraints on behaviour. Deindividuation leads to behaviour which would not be shown by individuals on their own. Detailed example of the point above C. Influences that change beliefs (maximum of 4 marks): Internalisation is the changing of one's beliefs as a result of persuasion. Detailed example of the point above Identification is the changing of one's beliefs to those of an admired influencing source. Detailed example of the point above

Immunology and Public Health

Give an account of the inflammatory response. (8)1. Mast cells…2. …release histamine.3. Histamine causes vasodilation…4. …and increases capillary permeability.5. Mast cells also secrete cytokines.6. Cytokines act as signalling molecules.7. The increased blood flow and the secretion of cytokines lead to…8. …the delivery of antimicrobial proteins…9. …and clotting elements to the site of infection/damage.Give an account of clonal selection theory. (6)1. Clonal selection theory explains the way in which lymphocytes are developed to respond to specific antigens which invade the body.2. Antigens are molecules on the surface of pathogens and other foreign cells or materials which activate the immune system.3. Lymphocytes have a single type of receptor on the cell membrane.4. This receptor is specific to one antigen.5. Antigen binding leads to repeated lymphocyte division…6. …which results in a clonal population of lymphocytes.Describe the role of community responsibility in the control of the spread of pathogens. (6)1. Quality of water supply - drinking water must be safe to drink.2. Not contaminated with sewage.3. Free from harmful chemicals and bacteria.4. Supervision of food chains - enforcing minimum hygiene standards.5. In abattoirs, restaurants, fast-food outlets, supermarkets, market stalls. (any two)6. Health education.7. Waste disposal - refuse collection.8. Sewage treatment.State the aim of public health immunisation programmes and explain why they may fail to protect non-immunised individuals. (6)1. Public health immunisation programmes seek to establish herd immunity to a number of diseases.2. Herd immunity occurs when a large percentage of a population are immunised.3. This percentage is called the herd immunity threshold.4. Non-immune individuals are protected because there is a lower probability they will come into contact with infected individuals.5. Difficulties arise when widespread vaccination is not effective because of malnutrition, which weakens the immune system…6. …or poverty, which reduces vaccination rates…7. …or a vaccine being rejected by a percentage of the population.

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Higher Human Biology Revision

How to answer exam questions.

Type of questions:

identify , name, give or state questions: You need only answer or present in a brief form, single word etc

Describe : You must provide a statement/sentence not a single one word

Explain: You must provide a reason for the information given - re-read your answer and make sure you have said why and not just described what

Compare : You must demonstrate knowledge and understanding of the similarities and/or differences between topics being examin ed. Use comparative words like wider not wide, or say whereas write about both parts of the question

Calculate : You must determine a number from given facts, figures or information

Predict : You must indicate what may happen based on available information - look at data and find the last change and apply this to the next predicted change

S uggest : You must apply your knowledge and understanding to a new situation. Don't panic thing about what you have been taught which connects to the unusual situation in the question and make the link clear in your answer

What is in your assessments

There will be a longer essay (8-10 marks) when you have a choice of 2 options and a shorter extended response question (approx 5 marks)

There will be a large data handling question requiring different calculations (5-9 marks)

There will be a large experimental design question requiring different calculations (5-9 marks)

Higher Biology & Human Biology Resources

Materials for the study of H SQA Biology & Human Biology.

Consolidated Unit 1 NAB Revision Questions

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