1. Alleles that are neither dominant nor recessive to one another, so both alleles are always expressed in the phenotype. Codominant 2. The
... [Show More] inheritance of a single gene Monohybrid inheri- tance 3. Physical, behavioural, biochemical expression of an organisms genotype Phenotype 4. The type of genes an individual has Genotype 5. Alles that is always expressed in the phenotype Dominant 6. Only expressed in the phenotype when homozy- gous Recessive 7. both alleles are the same Homozygous 8. Both alleles for a specific gene are different Heterozygous 9. Position of a gene on a chromosome Loci 10. A set of instructions for a specific polypeptide Gene 11. Different forms of a gene Allele 12. 1. Expected ratios are probability 2. Sexual reproduction is random due to random fusion of gametes and random assortment homol- ogous chromosomes. 13. 1. Homologous chromosomes pair up 2. Crossing over / chiasmata form; 3. Produces new combination of alleles 4. Chromosomes separate at random 5. This produces varying combinations of genes 6. Chromatids separated at meiosis II Suggest one reason why observed ratios are not the same as expected ratios (1). Meiosis results in cells that have the haploid number of chromosomes and show genetic varia- tion. Explain how. (6) 14. 1. Refer to the specific individuals (using their number) 2. Explain what happened with the genes (passed on recessive/dominant) 3. Describe the genotype of your examples and mention their phenotype (homozygous/heterozy- gous etc) 15. Group of organisms of the same species occupy- ing a particular space at a particular time that can potentially interbreed to produce fertile offspring. 16. The total number of genes of every individual in an interbreeding population. 17. How often an allele appears in a population. desired allele/total alleles = allele frequency (deci- mal form) 18. No emigration or immigration No mutations Mating is random No natural selection 19. 1. Selection pressure exists in an environment (name it) 2. Variation exists in stated phenotype of organ- ism/ mutation occurs 3. Some individuals have the selective advantage (describe it) 4. Produces differential survival/ organisms with successful alleles more likely to survive 5. Natural selection occurs 6. Survivors breed and pass on alleles to offspring 7. Over time, there is a switch in allele frequency 20. Favours the mean phenotype. (Normal distribution becomes narrower) Pedigree Questions Mark Scheme (3) Populations Gene pool Allele frequency Hardy-Weingberg Assumptions (4) Natural selection MS (7) Stabilising selection 21. Favours one extreme end. (Normal distribution translates that way) 22. Favours both extreme phenotypes. Mean is at dis- advantage and dies. Can create two new species 23. 1. Natural disaster 2. Few survive 3. New population develops with different allele 4. Frequency to the original 24. Genetic drift that occurs after a small number of individuals colonize a new area. Allele frequency is different to the original 25. 1. Populations geographically separated (forma- tion of a river) 2. Separated populations now unable to reproduce 3. Different environments have different selective pressures so each population will accumulate dif- ferent beneficial mutations over time to help them survive so change in allele frequency 4. Two populations become so genetically differ- ent that they cannot reproduce to form fertile off- spring, so are now classed as different species 26. Populations live in the same region but occupy different habitats 27. Same area but are sexually mature at different times 28. Different species ensure successful mating by Directional Selection Disruptive Selection Genetic Drift: Bottle- neck Effect (4) Founder effect Allopatric speciation (4) Reproductive isola- tion: Ecological Reproductive isola- tion: Seasonal Reproductive Isola- specific courtship (bird dances). As this is genetic, tion: Behavioural mutation can change this 29. Plants exhibit pheromones to any mate with their own species 30. Lack of fit between sexual organs in insects Reproductive isola- tion: Incompatibility 31. 1. Occurs in the same habitat 2. Mutation causes different flowering times 3. Seasonal Reproductive separation 4. No gene flow between the organisms so differ- ent alleles passed on. 5. Disruptive natural selection 6. Eventually, different species cannot interbreed to produce fertile offspring; 32. 1. Colonisation by pioneer species 2. Change in environment by organisms present 3. Enables other species to survive as the environ- ment becomes less hostile 4. Increase in biodiversity 5. Stability increases 6. Leading to a climax community 33. 1. Collect sample, mark and release 2. Ensure marking does not affect survival of the fish 3. Allow time for fish to randomly distribute before collecting the second sample (decide time based on question) Reproductive Isola- tion: Incompatibility in Arthropods Lord Howe Island in the Tasman Sea pos- sesses two species of palm tree which have arisen via sym- patric speciation. The two species diverged from each other after the island was formed 6.5 million years ago. The flowering times of the two species are different. Using this informa- tion, suggest how these two species of palm tree arose by sympatric speciation. (6) Succession occurs in natural ecosystems. Describe and explain how succession oc- curs. (6) The mark-release-re- capture method can be used to estimate the size of a fish pop- ulation (4) 4. Population = (first sample x second sample )/ total recaptured in 2nd sample 34. 1. Same species present over long time 2. Populations stable (around carrying capacity) 35. Maximum population size that a particular environ- ment can support due to abiotic and biotic factors. 36. All the different organisms interacting in a partic- ular place at a particular time 37. The biotic and abiotic environment of a specific area and their interactions Give two features of a climax community (2) Carrying capacity (k) Community Ecosystem 38. Position an organism occupies in an ecosystem Ecological Niche 39. The niche species could potentially occupy. Fundamental Niche 40. Due to interactions with other organisms where they are outcompeted, species occupy a smaller area than the fundamental niche 41. 1. Use tapes placed at right angles to produce a grid 2. Use a random number table to generate pairs of numbers to be used as coordinates for areas on the grid. 3. Sample in those areas using a quadrat 4. Use at least 30 quadrats so that sample is rep- resentative 5. Count all squares occupied by each species being sampled (percentage cover) 42. 1. Use tape laid across the ecosystem- a transect 2. Sample through all the areas that show differ- ences 3. Take samples at regular intervals along the tran- sect (e.g quadrants every 5 metres) 4. Count all squares occupied by each species Realised Niche Random Sampling MS (same abiotic fac- tors, like the sand dunes in Wales) (5) [Make your answer specific to the ques- tion] Systematic sampling MS (Different abiotic factors, Rocky shore at Wales) (5) [Make answer specific to the question] being sampled (percentage cover) 5. Kite diagrams 43. 1. Reduce sample distance in half 2. If sample pattern of zonation remains, then sam- ple interval is okay 44. 1. Useful for areas with dense vegetation 2. Vertical zonation How do you know if your sample distance is okay ? (2) Point quadrats (2) 45. No soil primary succession 46. Succession following a disturbance that destroys a community without destroying the soil (contains seeds often) 47. A community that remains stable only because human activity prevents succession. 48. 1. Chlorophyll absorbs light which excites elec- trons 2. And causes them to be lost 49. 1. Reduced transfer of protons across thylakoid membrane 2. Less ATP produced 3. Less NADPH produced 4. Light-independent reactions slows 50. 1. Phosphorylation of glucose using ATP; 2. Oxidation of triose phosphate to pyruvate 3. Net gain of ATP 4. NAD reduced Secondary succes- sion Deflected succession Describe what hap- pens during pho- toionisation in the LDR. (2) Atrazine binds to pro- teins in the elec- tron transfer chain in chloroplasts of weeds, reducing the transfer of electrons down the chain. Explain how this re- duces the rate of pho- tosynthesis in weeds. (4) Describe the process of Glycolysis (4) 51. 1. Pyruvate —> Lactate whilst oxidising NADH so it can enter glycolysis. Pyruvate —> ethanol in plants Anaerobic respiration [Animals] [Plants] (2) 52. 1. Excessive algae growth due to leached nitrates. Eutrophication (4) 2. Plants below cannot photosynthesise and die. 3. Aerobic bacteria feed on fish in water. They respire and use up oxygen in water. 4. Fish and other aquatic organisms die due to lack of oxygen in water so they cannot respire. 53. I - (F+R) Net production 54. Nitrogen fixing bacteria found in soil or in legumi- nous plants convert N2(g) —> NH4+ (in soil) 55. Organic N compounds decomposed to NH4+ by saprobionts. Nitrogen Fixation (1) Ammonification (2) 56. NH4+ --> NO2- --> NO3- by nitrifying bacteria. Nitrification (3) 57. Anaerobic denitrifying bacteria converts nitrogen compounds back into N2(g). Prevent this by aerat- ing soil. 58. 1. CO2 reacts with Ribulose bisphosphate 2. Produces 2 Glycerate 3-phosphate molecules using rubisco. 3. Glycerate 3 phosphate reduced to triose phos- phate 4. Using reduced NADP 5. Using energy from ATP 6. Out of 6 Triose phosphate molecules, 5 are recy- cled into the calvin cycle (light independent reac- tion) and 1 is saved to form glucose. 59. 1. Energy is lost between trophic levels 2. Energy is lost via excretion and respiration Denitrification (4) Describe the light-in- dependent reaction of photosynthesis. (6) Farming cattle for hu- mans to eat is less efficient than farming 60. 1. Constant volume of oxygen 2. Constant concentration of respiratory substrate 3. Constant concentration of bacteria 4. Time must be maintained as constant for each dish to incubate 61. 1. Cells that desired protein is expressed in are selected and should contain a large amount of mRNA for the protein 2. Identify and extract mRNA for desired proteins 3. Use reverse transcriptase to convert mRNA —> cDNA (complementary to the mRNA template and single stranded) 4. DNA polymerase used to make cDNA double stranded 5. Insert gene into plasmid using restriction and ligase enzymes 6. Insert plasmid into bacterium and clone it. 62. 1. Order of animo acids in protein sequenced al- lowing mRNA to be determined, therefore cDNA to be sequenced 2. Feed cDNA into computer 3. Check sequence passes international standard biosafety 4. Computer designs series of small overlapping strands of single stranded DNA called oligonu- cleotides which are assembled into desired gene 5. Oligonucleotides joined together to make gene and gene replicated using PCR 6. PCR also makes copied genes double stranded 7. Gene inserted into bacterial cell or other organ- ism (in vivo cloning) crops because of en- ergy transfer. Explain why. (2) Apart from tempera- ture and pH, give two variables the scien- tist would have con- trolled when prepar- ing the liquid medium cultures. (2) Using reverse tran- scriptase to make a specifically selected protein (6) Outline the method used to manufacture genes (Gene ma- chine) (8) 8. Genes in transformed organisms checked using genetic sequencing. Those containing errors are rejected 63. 1. DNA fragments - to be copied 2. DNA polymerase - obtained from bacteria Ther- mus Aquaticus that lives in hot springs making it thermostable 3. Primers - Short sequences of nucleotides called oligonucleotides that have a set of bases that are complementary to those at each end of the DNA fragments 4. Nucleotides - Containing all four bases 64. A short strand of DNA that is complementary to a known DNA sequence (e.g a mutated allele) la- belled with a fluorescent or radioactive tag that makes it identifiable. 65. 1. Thermal cycled heats to 95 degrees celsius 2. 2 separated strands now used as templates for building complementary strandd 66. 1. Cycler cools the mixture to 55 degrees celsius, allowing hydrogen bonds to reform. 2. Primers attach to ends of desired sequence in DNA fragments. Primers allow the attachment of DNA polymerase 67. 1. Cycler raises the temperature of DNA to 72 de- grees celsius (optimum temp for Thermus Aquati- cus DNA polymerase) 2. DNA polymerase attaches to primer and syn- thesises new strands that are complementary to template DNA 3. Two new double stranded DNA fragments are formed. 4. PCR is repeated around 25 times creating over 50 million copies of template DNA Requirements for PCR (4) DNA Probe In Vitro Step 1 (Tem- plate DNA double stranded —> single stranded) (2) In Vivo Step 2 (An- nealing the Primer) (2) In Vivo Step 3: Syn- thesis (4) 68. 2^x where x = number of cycles Number of DNA frag- ments produced = 69. 1. Easily manipulated 2. Restriction sites well known 3. Easily taken up by cells in culture as bacteria absorb plasmids 70. 1. Target DNA (spliced prior so introns removed) is cut using restriction endonucleases to create fragments with sticky ends. 2. Promoter and terminator regions (STOP codon) are added allowing transcription. 3. Same restriction endonuclease is used to cut open plasmid (vector) to allow complementary sticky ends. 4. DNA ligase is used to incorporate the DNA frag- ment into the plasmid, forming recombinant DNA 5. Recombinant DNA is taken up by bacteria in Ca2+ solution to increase membrane permeability, forming a transformed cell. 71. 1. Genes for ampicillin and tetracyline resistance added before human gene is added. 2. Human gene inserted in the middle of tetracy- cline gene, omitting tetracycline resistance. 3. Bacteria that can grow on ampicillin but not tetracycline have recombinant plasmid. 72. 1. Human gene is inserted in the middle of GFP gene (isolated from jellyfish) 2. GFP no longer made in transformed cells but have ampicillin resistance so is grown on agar plate 3. So transformed cells are not fluorescent 73. 1. Human gene inserted between B-galactosidase gene. Why are bacterial plasmids used in In Vivo cloning ? (3) Preparation and In- sertion of DNA frag- ments (5) Identifying Trans- formed Bacteria Cells (Replica plating) (3) Identifying trans- formed Bacteria Cells (Green Fluorescent Protein) (3) Identifying trans- formed Bacteria Cells 2. B-galactosidase + lactose analogue —> blue 3. Transformed colonies are not blue 74. 1. Nucleotides with radioactive phosphate. 2. Probe identified using photographic plate. (B-galactosidase and lactose analogue) (3) Radioactively La- belled Probes (2) 75. 1. Emit light under specified conditions Fluorescent Labelled Probes 76. 1. Gel plate is prepared to hold gel matrix. Gel comb is used to create pores in the agarose gel. 2. Gel tray placed in electrophoresis chamber and is filled with a buffer (allowing electric current to flow through). 3. This separates the DNA fragments, as DNA is negatively charged. 4. DNA fragments migrate towards positive elec- trode. 5. Gel has fine pores, so small fragments move faster than large ones. 6. Results in separated fragments in the form of bands, with the smallest fragments nearest to the positive electrode. Gel Electrophoresis (6) 77. 1. After gel electrophoresis, bands are transferred Southern Blots (6) to nylon membrane as gel is fragile. 2. DNA probe is made complementary to tar- get gene by DNA sequencing techniques (known genotypes) 3. Quantity of DNA probe increased by PCR 4. DNA tested is turned into single stranded DNA 5. Strands mix with copies of DNA probe causing DNA probes to bind to complementary bases on strands causing DNA hybridisation. 6. DNA is then washed clean of any unattached probes, and marked DNA is identified. 78. Fluorescent insitu hybridisation probes can be used to identify mutations of nearly any human gene. FISH Probes 79. 1. Heritable changes in gene function 2. Without changes to the base sequence of DNA 80. 1. Methyl groups added to a tumour suppressor gene; 2. The transcription of tumour suppressor genes is inhibited 3. Leading to uncontrolled cell division. 81. 1. EGCG binds to active site of DNMT 2. DNMT cannot methylate promoter region of tu- mour suppressor gene 3. Transcription factors can bind to promoter re- gion 4. RNA polymerase stimulated Define what is meant by epigenetics (2) Explain how in- creased methylation could lead to cancer. (3) Increased methyla- tion of the pro- moter region of a tumour suppressor gene causes one type of human throat cancer. In this type of throat cancer, can- cer cells are able to pass on the in- creased methylation to daughter cells. The methylation is caused by an en- zyme called DNMT. Scientists have found that a chemical in green tea, called EGCG, is a com- petitive inhibitor of DNMT. EGCG en- ables daughter cells to produce messen- ger RNA (mRNA) from the tumour sup- pressor gene. Suggest how EGCG allows the production 82. 1. Chemoreceptors in aorta and carotid artery de- tect decrease in pH. 2. Send impulses to medulla oblongata 3. More impulses to SAN 4. By sympathetic nerve 5. Resulting in higher heart rate 83. 1. Electrical activity only through Bundle of His 2. Wave of electrical activity passes through both ventricles at the same time 84. 1. Saltatory conduction (depolarisation occurs along whole length of axon) 2. Nerve impulses slowed 3. Less impulses converted into neurotransmitters at neuromuscular junction 85. 1. SAN releases wave of depolarisation across atria, causing atria to contract 2. Wave of depolarisation reaches AVN 3. AVN releases another wave of depolarisation 4. Bundle of His transmits waves down the septum to the purkyne fibres in the ventricle walls 5. Before ventricles contract, delay occurs to make sure ventricles fill with blood. 86. 1. Detected by baroreceptors in the wall of the aorta and the carotid artery. 2. Impulses sent to the medulla oblongata, then back to the SAN via parasympathetic nerve. 3. Decrease in frequency of electrical signals, so of mRNA in daughter cells. (3) Exercise causes an increase in heart rate. Describe the role of receptors and of the nervous system in this process. (4) When the heart beats, both ventricles contract at the same time. Explain how this is coordinated in the heart after initiation of the heartbeat by the SAN. (2) Damage to the myelin sheath can cause muscular paralysis. Explain how. (3) SAN —> AVN (5) Heart rate control: In- crease in Pressure (3) heart rate decreases to ensure blood vessels do not burst 87. 1. Detected by baroreceptors in the aorta and Heart rate control: carotid artery. Decrease in Pressure 2. Impulses sent to the medulla oblongata and (3) back to the SAN via sympathetic nerve 3. Increase in frequency of electrical signals, so heart rate increases. 88. 1. High visual acuity The fovea of the eye 2. Each cone is connected to a single neurone of an eagle has a high 3. Cones send separate sets of impulses to brain density of cones. Ex- plain how the fovea enables an eagle to see its prey in detail. (3) 89. 1. High visual sensitivity The retina of an owl 2. Several rods connected to a single neurone has a high density of 3. Spatial summation to overcome threshold rod cells. Explain how this enables an owl to hunt its prey at night. (3) 90. 1. High visual sensitivity Rod Characteristics 2. Can detect light in low light intensity as many (The Where) rod cells are connected to a single bipolar cell (spatial summation) 3. Low visual acuity (quality of vision) 4. Rhodopsin 91. 1. Most concentrated in fovea Cone Characteristics 2. High visual acuity (The What) 3. Can only respond to high light intensity due to each cone being connected to one bipolar cel 4. Different types of iodopsin (red, green and blue) to absorb different wave lengths of light 92. 1. Membrane more permeable to K+ and less per- meable to Na+. 2. 2 K+ pumped in and 3 Na+ actively transported out. 93. 1. Pressure causes membrane to become stretched 2. Sodium ion channels in membrane open and sodium ions move in 3. Greater pressure means more channels open. so more sodium ions enter. 94. 1. Threshold reached 2. Which causes maximal response/ All or nothing principle 95. 1. IAA diffuses to the shaded side of the shoot promoting growth and cell elongation 2. Causes the shoot to bend towards the light source (positive phototropism) 96. 1. IAA diffuses towards the shaded side of the root and inhibits growth and cell elongation. 2. Root bends away from the light source (negative phototropism) 97. 1. IAA diffuses towards the bottom due to gravity 2. Increase in IAA concentration promotes growth and cell elongation, so the shoot tends and growths against gravity (negative geotropism) 98. 1. IAA diffuses to the bottom due to gravity 2. IAA inhibits cell elongation in the roots so the roots bend and grow downward (positive geotro- pism) Explain how the rest- ing potential of -70 mV is maintained in the sensory neurone when no pressure is applied. Explain how applying pressure to the Pacin- ian corpuscle pro- duces the changes in membrane poten- tial recorded by mi- croelectrode P. (3) Why is membrane po- tential never higher than it is ? (2) Phototropism in Shoots (2) Phototropism in Roots (2) Geotropism in Shoots (2) Geotropsim in Roots (2) 99. Taxes Directional response in movement due to stimuli Positive = towards stimuli Negative = Against stimuli 100. Non directional response in movement towards a stimulus. (Humans moving around in the cold to stay warm) Kineses 101. Rapid, unconscious responses to stimuli Reflexes 102. 1. Action potential arrives at the synaptic knob, de- polarising it so voltage gated Ca²+ channels open and Ca²+ diffuse into the synaptic knob 2. Vesicles containing Acetylcholine fuse to the presynaptic membrane 3. Acetylcholine is released into the synaptic cleft via exocytosis 4. Acetylcholine diffuses across the synaptic cleft and binds to cholinergic receptors. 5. Na+ channels on the post synaptic membrane open (when neurotransmitters bind to receptors on Channel protein) and Na+ diffuses into the post synaptic neurone causing depolarisation. 6. If threshold is reached, action potential is formed 103. 1. Acetylcholine is removed from synaptic cleft and degraded by acetylcholine esterase to prevent continuous impulses 2. Products of the degradation of acetylcholine are transferred back into the presynaptic neurone and Na+ channels close, allowing post synaptic neurone to repolarise and reach resting potential 104. 1. Mimics the shape of neurotransmitters so binds to post synaptic membrane receptors and causes active potential (agonist) 2. Bind to receptors, but block new complexes from forming, preventing action potential (antago- nist) Transmission of a cholinergic synapse. (6) Reabsorption of Acetylcholine after transmission of neu- rotransmitters (2) How can drugs af- fect synaptic trans- mission? (4) 3. Bind to Acetylcholine esterase so fewer ESCs formed, causing a continuous impulse 4. Can cause the release or blocks the release of neurotransmitters. 105. 1. Myelination and saltatory conduction (provides electrical insulation) 2. Axon diameter (wider the faster 3. Temperature (higher the faster) 106. 1. Myelination provides electrical insulation 2. Saltatory conduction (depolarisation at nodes of Ranvier) (depolarisation jumps from node to node) 3. In non-myelinated depolarisation occurs along length of axon 107. 1. Surround the axon and provide electrical insu- lation via myelin sheath. 2. Carry out phagocytosis to remove cell debris 108. 1. Inside of postsynaptic neurone becomes more negative (hyperpolarisation at -80mV) by diffusion of Cl-. 2. More Na+ required to reach threshold for action potential 3. So it is unlikely causing an inhibitory effect 109. One or more presynaptic neurons transmit impuls- es in rapid-fire order in a short period of time 110. Many neurons collectively stimulate an action po- tential by combining the neurotransmitters they release. Factors affecting speed of conduc- tance Explain why the speed of transmis- sion of impulses is faster along a myeli- nated axon than along a non-myelinat- ed axon. (3) Schwann Cells How can cholinergic synapses be inhibito- ry? (3) Temporal summation Spatial summation 111. 1. Unidirectional as neurotransmitter receptors are How are neuromus- only on the post synaptic membrane 2. Only excitatory 3. Connects motor neurones and muscles cular junctions differ- ent ? (5) 4. End point of action potential 5. Acetylcholine binds to receptors on muscle fibre membranes. 112. 1. Impulse is received from receptors, Na+ chan- nels open, so more Na+ enter the neurone, causing depolarisation. 2. If depolarisation reaches threshold, voltage gat- ed Na+ channels are activated, leading to a higher influx of Na+, creating an action potential (+40mV). 3. When +40mV reached, voltage gated Na+ chan- nels close, whilst voltage-gated K+ channels open, so repolarisation occurs as K+ leave the neurone. 4. When most of the K+ have left the neurone, hyperpolarisation occurs which close the voltage gated K+ channels. 5. Sodium-potassium pump returns cell back to its resting potential (2 K+ in for 3 Na+ out) 113. 1. Action potential travels to muscle fibres and depolarises the sarcolemma via T-tubules, causing the release of Ca2+ from the sarcoplasmic reticu- lum. 2. Ca2+ binds to troponin, which causes the change in shape of tropomyosin, exposing the myosin binding site on the actin filament. 3. Actin-myosin cross bridge is formed upon at- tachment. 4. ATP is hydrolysed by ATPhydrolase to detach the myosin head allowing the reattachment at a further site. Sarcomere shortens causing muscle contraction 5. When impulse stops, Ca2+ is actively transport- ed back into the sarcoplasmic reticulum. 6. This allows tropomyosin to block the actin fila- ment from binding to myosin, so muscle contrac- tion stops. How is an action po- tential generated? (5) Sliding filament The- ory (6) (like a rowing boat) 114. The maintenance of a constant internal environ- ment via physiological control systems What is homeosta- sis? (1) 115. Only myosin H Zone 116. The Z discs Z line 117. Thin filament Actin 118. Myosin Line M Line 119. Length of only actin I band 120. Thick filaments Myosin 121. Length of myosin A-band 122. Can quickly identify transformed bacteria by using UV light. (The non-fluorescent cultures are recom- binant). 123. 1. Blood enters kidneys via renal artery at a high pressure 2. Renal Artery divides into the afferent arteriolar and then the glomerulus 3. Water, ions and glucose are forced out of the glomerulus down a pressure gradient through small gaps in capillary endothelium, into the Bow- mann's capsule. 4. Pressure gradient is aided by the efferent arteri- olar being narrower than afferent 5. Proteins are left behind in blood as they are too large 6. Glomerular filtrate is formed 124. hypothalamus Suggest one advan- tage of using this gene for GFP to iden- tify bacteria that have taken up plasmids. (1) Ultrafiltration(6) 125. 1. Glucose is re absorbed by co-transport from epithelial cells of the proximal convoluted tube (PCT) to blood capillaries 2. Carried out by actively transporting Na+, creat- ing a low Na+ concentration in the epithelial cells. 3. Na+ moved in from the PCT by facilitated diffu- sion which bring in glucose as they are co-trans- ported. 4. Glucose then diffuses into blood capillaries 126. 1. Many mitochondria provide ATP for active trans- port 2. Microvilli provide large surface area for absorp- tion. 127. 1. A short, single-stranded section of DNA with complementary bases to desired gene. 2. That has a label attached to make it easily iden- tifiable 128. 1. Restriction endonuclease Location of osmore- ceptors in the body of a mammal Selective Reaborp- tion (4) Reabsorption of glu- cose takes place in the proximal tubule. Explain how the cells of the proximal tubule are adapted for this function (2). What is a DNA probe ? (2) Describe how the 2. Cuts DNA at specific base sequence by breaking DNA is broken down phosphodiester bond. 129. 1. High concentrations of specific mRNA stimu- lates production of the enzyme RDR (RNA Depen- dent polymerase) 2. RDR catalyses production of dsRNA by syn- thesising complementary RNA strand from RNA template 3. Enzymes cut up dsRNA into siRNA (small in- terfering RNA) - small fragments of RNA about 23 into smaller frag- ments (2). siRNA mark scheme (6) base pairs long 4. An enzyme (RISC) takes up the siRNA fragments and separates their two target RNA strands using the hydrolysis of ATP to provide the necessary energy 5. The siRNA created pairs with complementary bases on mRNA and the enzyme cuts mRNA into small pieces. 6. The siRNA is only complementary to a small section of the mRNA. But this is sufficient for the enzyme, which it is embedded, to bind to it. 130. 1. Lipid soluble 2. Diffuse through phospholipid bilayer 131. 1. AR is a transcription factor 2. Binds to promoter region 3. Stimulates RNA polymerase 132. 1. siRNA binds to CENP-W mRNA 2. Preventing translation of CENP-W 3. Reduced CENP-W causes reduction of spindle fibre synthesis Steroid hormones are hydrophobic. Explain why steroid hormones can rapid- ly enter a cell by passing through its cell-surface mem- brane. (2) The binding of testos- terone to an AR changes the shape of the AR. This AR molecule now enters the nucleus and stim- ulates gene expres- sion. Suggest how the AR could stimulate gene expression. (3) CENP-W is involved in the formation of spindle fibres in mito- sis. Spindle fibres are made of molecules of 133. 1. Requires DNA fragment, DNA polymerase, DNA nucleotides and primers 2. Heat to 95 °C to break hydrogen bonds and separate strands 3. Reduce temperature to 55 degrees Celsius so primers bind to DNA strands 4. Increase temperature to 75°C, DNA polymerase joins nucleotides and repeat method 134. 1. Na+ actively transported out of the ascending limb, creating low water potential 2. The ascending limb is impermeable to water so water can only move out of the descending limb by osmosis 3. Water then enters blood capillaries by osmosis 4. At the hairpin of the loop, Na+ ions naturally diffuse out as water potential is lowest here. 135. 1. Osmoreceptors in hypothalamus detect blood water potential changes. 2. Osmoreceptors shrink when blood potential falls, causing the release of ADH (due to dehydra- tion often) 3. ADH travels to posterior pituitary gland where it's secreted into the blood. 4. Travels to the kidney and binds to receptors on the surface of the collecting duct and activates enzyme phosphorylase. 5. Causes vessicles containing aquaporins to in- a protein called tubu- lin. SiRNA causes inter- ference of expression of CENP-W gene. Suggest how siRNA could reduce spin- dle fibre synthesis. (3) (made up question) Describe and explain how the polymerase chain reaction (PCR) is used to amplify a DNA fragment. (4) Loop of Henle (4) Hormones can alter collecting duct per- meability (7). corporate into the cell surface membrane 6. This increases water permeability and urea per- meability. 7. Urea leaves the collecting duct, so water leaves and is reabsorbed into the blood. 136. 1. To be hydrolysed to glucose 2. For respiration to provide ATP 137. 1. Low pH changes the shape of Ca2+ receptors 2. Fewer calcium ions bind to troponin. 3. Fewer tropomyosin molecules move away. 4. Fewer binding sites on actin revealed 5. Fewer actin-myosin cross-bridges form 138. 1. Reaction with ATP breaks binding of actin-myosin bridge 2. Provides energy to move myosin head 139. 1. Produced by beta cells in the islets of Langer- hans in response to detecting high glucose levels. 2. Insulin attaches to receptors on target cells and changes the tertiary structure of the channel pro- teins, so more glucose is absorbed via facilitated diffusion. 3. More protein carriers are incorporated into the membranes so even more glucose is absorbed from the blood into the cells. 4. Glycogenesis occurs catalysed by activating en- zymes, so glycogen is stored in the cells (liver and muscles). Glycogen granules are present in skele- tal muscle. Explain their role in skeletal muscle. (2) During rigorous exer- cise, the pH of skele- tal muscle tissue falls. This fall in pH leads to a reduction in the ability of calcium ions to stimulate muscle contraction. Suggest how. (5) What is the role of ATP in myofibril con- traction ? (2) Insulin Action (4) 140. 1. Produced by alpha cells in pancreas in response to low glucose levels. 2. Glucagon attaches to receptors on the surface of target cells 3. This stimulates adenylate cyclase to con- vert ATP into cyclic adenosine monophosphate (cAMP). cAMP acts as a second messenger. 4. cAMP activates protein kinase which hydrolyses glycogen --> glucose 5. Can also activate enzymes involved in gluconeo- genesis. 141. Molecules inside a cell that transmit signals in- tercellularly from the binding of an extracellular molecule to a receptor. 142. 1. Body cannot produce insulin 2. Starts in childhood 3. Autoimmune disease 4. Treatment: Insulin injections 143. 1. Receptors on target cells lose responsiveness to insulin as cells become resistant to insulin 2. Usually develops in adulthood due to obesity and poor diet. 3. Diet contains too much sugar 4. Treatment: Increasing exercise, regulating car- bohydrate intake and sometimes insulin injec- tions. Glucagon/Adrena- line Action [adrenaline does not respond to low glucose levels] (4) Second Messenger Model Type 1 Diabetes (4) Type 2 diabetes (4) [Show Less]