Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Comprehension Questions
Document related concepts
SNP genotyping wikipedia , lookup
Genetic engineering wikipedia , lookup
Chemical biology wikipedia , lookup
Developmental biology wikipedia , lookup
Cell theory wikipedia , lookup
Monoclonal antibody wikipedia , lookup
Molecular cloning wikipedia , lookup
Cell (biology) wikipedia , lookup
Evolution of metal ions in biological systems wikipedia , lookup
Cell-penetrating peptide wikipedia , lookup
Biochemistry wikipedia , lookup
DNA vaccination wikipedia , lookup
DNA-encoded chemical library wikipedia , lookup
Nucleic acid analogue wikipedia , lookup
Biomolecular engineering wikipedia , lookup
Molecular paleontology wikipedia , lookup
Cre-Lox recombination wikipedia , lookup
Transcript
Comprehension Questions Name………………………………….….. Target Grade………………….. Grade achieved………………. __/ __% Woodford High School 1 1. Read the following passage. In a human, there are over 200 different types of cell clearly distinguishable from each other. What is more, many of these types include a number of different varieties. White blood cells, for example, include lymphocytes and granulocytes. 5 10 Although different animal cells have many features in common, each type has adaptations. associated with its function in the organism. As an example, most cells contain the same organelles, but the number may differ from one type of cell to another. Muscle cells contain many mitochondria, while enzyme-secreting cells from salivary glands have particularly large amounts of rough endoplasmic reticulum. The number of a particular kind of organelle may change during the life of the cell. An example of this change is provided by cells in the tail of a tadpole. As a tadpole matures into a frog, its tail is gradually absorbed until it disappears completely. Absorption is associated with an increase in the number of lysosomes in the cells of the tail. Use information from the passage and your own knowledge to answer the following questions. (a) Explain the link between. (i) mitochondria and muscle cells (lines 6 - 7); ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (3) (ii) rough endoplasmic reticulum and enzyme-secreting cells from salivary glands (lines 7 - 8). ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) Woodford High School 2 (b) Use information in the passage to explain how a tadpole’s tail is absorbed as a tadpole changes into a frog. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. (2) (c) Starting with some lettuce leaves, describe how you would obtain a sample of undamaged chloroplasts. Use your knowledge of cell fractionation and ultracentrifugation to answer this question. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. (6) (Total 13 marks) Woodford High School 3 2. Read the following passage. Herpes viruses cause cold sores and, in some cases, genital warts. Scientists are well on the way to producing an antibody which will counteract herpes infection. This antibody works by sticking to the virus and blocking its entry into cells. It has proved very effective in animal tests. 5 One drawback with this approach, however, is that antibodies are at present produced using hamster ovary cells. This method is expensive and only produces limited amounts. A new technique is being developed to produce antibodies from plants. It involves introducing the DNA which codes for the required antibody into crop plants such as maize. Use information from the passage and your own knowledge to answer the questions. (a) (i) What is an antibody? ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (ii) Describe how antibodies are produced in the body following a viral infection. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (6) (b) Describe how the antibody gene could be isolated from an animal cell and introduced into a crop plant such as maize (lines 7-8). Woodford High School 4 ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (4) (c) Taking a course of these antibodies from plants to treat a herpes infection would not produce long-term protection against disease. Explain why. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (d) Explain one advantage of using antibodies from plants to treat a disease, rather than antibodies produced in an experimental animal (lines 5-6). ..................................................................................................................................... ..................................................................................................................................... (1) (Total 15 marks) Woodford High School 5 3. 5 10 Read the following passage. The sequence of bases in a molecule of DNA codes for proteins. Different sequences of bases code for different proteins. The genetic code, however, is degenerate. Although the base sequence AGT codes for serine, other sequences may also code for this same amino acid. There are four base sequences which code for the amino acid glycine. These are CCA, CCC, CCG and CCT. There are also four base sequences coding for the amino acid proline. These are GGA, GGC, GGG and GGT. Pieces of DNA which have a sequence where the same base is repeated many times are called “slippery”. When “slippery” DNA is copied during replication, errors may occur in copying. Individual bases may be copied more than once. This may give rise to differences in the protein which is produced by the piece of DNA containing the errors. Use information in the passage and your own knowledge to answer the following questions. (a) Different sequences of bases code for different proteins (lines 1 – 2). Explain how. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. ............……….............................................................................................................. (2) (b) The base sequence AGT codes for serine (lines 2 – 3). Give the mRNA codon transcribed from this base sequence. ............……….............................................................................................................. (2) (c) Glycine-proline-proline is a series of amino acids found in a particular protein. Give the sequence of DNA bases for these three amino acids which contains the longest “slippery” sequence. ............……….............................................................................................................. (2) Woodford High School 6 (d) (i) Explain how copying bases more than once may give rise to a difference in the protein (lines 9 – 10). ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (ii) At what stage in the cell cycle would these errors in copying DNA bases occur? ........................................................................................................................... (1) Woodford High School 7 (e) Starting with mRNA in the nucleus of a cell, describe how a molecule of protein is synthesised. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6) (Total 15 marks) Woodford High School 8 4. 5 10 Read the following passage. Straw consists of three main organic substances – cellulose, hemicellulose and lignin. Cellulose molecules form chains which pack together into fibres. Hemicellulose is a small molecule formed mainly from five-carbon (pentose) sugar monomers. It acts as a cement holding cellulose fibres together. Like hemicellulose, lignin is a polymer, but it is not a carbohydrate. It covers the cellulose in the cell wall and supplies additional strength. In addition to these three substances, there are small amounts of other biologically important polymers present. The other main component of straw is water. Water content is variable but may be determined by heating a known mass of straw at between 80 and 90°C until it reaches a constant mass. The loss in mass is the water content. Since straw is plentiful, it is possible that it could be used for the production of a range of organic substances. The first step is the conversion of cellulose to glucose. It has been suggested that an enzyme could be used for this process. There is a difficulty here, however. The lignin which covers the cellulose protects the cellulose from enzyme attack. Use information from the passage and your own knowledge to answer the following questions. (a) (i) Give one way in which the structure of a hemicellulose molecule is similar to the structure of a cellulose molecule. ........................................................................................................................... ........................................................................................................................... (1) (ii) Complete the table to show two ways in which the structure of a hemicellulose molecule differs from the structure of a cellulose molecule. Hemicellulose Cellulose .................................................................. .................................................................. .................................................................... .................................................................... .................................................................. .................................................................. .................................................................... .................................................................... (2) Woodford High School 9 (b) Name one biologically important polymer, other than those mentioned in the passage, which would be found in straw. ..................................................................................................................................... (1) (c) Explain why the following steps were necessary in finding the water content of straw: (i) heating the straw until it reaches constant mass (line 9); ........................................................................................................................... ........................................................................................................................... (1) (ii) not heating the straw above 90°C (line 9). ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (d) A covering of lignin protects cellulose from enzyme attack (line 14). Use your knowledge of the way in which enzymes work to explain why cellulose-digesting enzymes do not digest lignin. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) Woodford High School 10 (e) Describe the structure of a cellulose molecule and explain how cellulose is adapted for its function in cells. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6) (Total 15 marks) Woodford High School 11 5. Answers should be written in continuous prose. Credit will be given for biological accuracy, the organisation and presentation of the information and the way in which the answer is expressed. The following extract has been taken from a dictionary of biological terms. cell membrane: a membrane found either on the outside of a cell or within it. Cell membranes are extremely thin. They are only about 7 nm thick and so cannot be seen with a light microscope. A transmission electron microscope however shows a cell membrane consists of three lines forming a sandwich. The two outer lines are dark in colour while there is a lighter one in between. As it is impossible, even with an electron microscope, to see how the actual molecules are arranged in a cell membrane, it is necessary to produce a model to explain the membrane’s properties. The most accurate model of membrane structure that has been developed is the fluid mosaic model and this can be used to describe most of the properties of a cell membrane. Cell membranes play a very important part in the biology of cells and they are particularly important in regulating the movement of substances into and out of cells. Source: W.J.E. INDGE, The Complete A-Z Biology Handbook (Hodder & Stoughton) 1997 (a) (i) Describe the structure of a cell membrane. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (5) (ii) Describe two ways in which the appearance of a plant cell wall would differ from a cell membrane when viewed with an electron microscope. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) Woodford High School 12 (b) Describe the part played by cell surface membranes in regulating the movement of substances into and out of cells. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... (6) (c) Describe how the distribution of cell membranes in a prokaryotic cell such as a bacterium differs from that in a cell from a plant leaf. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... (4) (QWC 3) (Total 20 marks) Woodford High School 13 6. Read the following passage. DNA tests were used to confirm the identity of deposed Iraqi leader Saddam Hussein, after his capture in December 2003. DNA tests were carried out to prove the suspect was not one of the many alleged “look alikes” of the former leader. 5 10 Firstly, the DNA was extracted from the mouth of the captured man using a swab. Great care was taken to check that the swab did not become contaminated with any other DNA. DNA extracted from the swab was then subjected to a standard technique called the polymerase chain reaction (PCR), which takes a couple of hours. Lastly, the sample was “typed” to give the genetic fingerprint. This was produced within 24 hours of Saddam Hussein’s capture. Tests for use in criminal cases often take much longer because samples are very small or contaminated. It appears that Hussein’s genetic fingerprint was already stored away for comparison. This was obtained from personal items such as his toothbrush. DNA from the toothbrush would have been subjected to PCR before a DNA fingerprint could have been obtained. Source: adapted from SHAONI BHATTACHARYA, New Scientist 15 December, 2003 Use information from the passage and your own knowledge to answer the questions. (a) Describe how the technique of genetic fingerprinting is carried out and explain how it can be used to identify a person, such as Saddam Hussein. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6) Woodford High School 14 (b) Explain how DNA could be present on a toothbrush (line 12). ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (c) (i) Explain why the polymerase chain reaction was used on the sample of DNA from the toothbrush (lines 12-13). ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (ii) Explain one way in which the polymerase chain reaction differs from DNA replication in a cell. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) Woodford High School 15 (d) Tests for use in criminal cases often take much longer because samples are very small or contaminated (lines 8-10). Explain why it takes longer to obtain a genetic fingerprint if the sample is (i) very small; ........................................................................................................................... ........................................................................................................................... (1) (ii) contaminated. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (Total 15 marks) 7. Answers should be written in continuous prose. Credit will be given for biological accuracy; the organisation and presentation of the information and the way in which the answer is expressed. Read the following passage. Photosynthesis takes place in the chloroplasts. These are disc-shaped organelles surrounded by an outer envelope consisting of two layers of membrane. Inside, there are further membranes which are arranged in stacks called grana. Surrounding these is the stroma. Chlorophyll and other lightcapturing pigments are found on the membranes of the grana and it is here that the light-dependent reaction takes place. This generates the ATP and reduced NADP which are used in the light-independent reaction in the stroma. Woodford High School 16 (a) (i) Describe the way in which ATP and reduced NADP are produced in the light-dependent reaction of photosynthesis. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (5) (ii) Explain how ATP and reduced NADP are used in the light-independent reaction of photosynthesis. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (4) Woodford High School 17 (b) Using the information in the passage, describe how the structure of a chloroplast is adapted to its function in photosynthesis. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3) (Total 12 marks) 8. Answers should be written in continuous prose. Credit will be given for biological accuracy, the organisation and presentation of the information and the way in which the answer is expressed. Read the following passage Bt is a toxin made by the soil bacterium, Bacillus thuringiensis. It is very toxic to insects so it is an effective insecticide. Unfortunately, resistant strains have developed in the diamond-back moth, whose caterpillars are important pests of crops of the cabbage family. To try to overcome some of the problems of resistance, a programme of crop management was developed. Studies showed that the allele which confers resistance on diamond-back moths was recessive. Farmers were encouraged to leave a few untreated fields in which moths that were susceptible to Bt would survive. If the rest of the fields were then sprayed with Bt, a few resistant individuals would remain. However, because they were likely to mate with susceptible individuals from the untreated fields, the number of resistant moths in the population would not increase. Woodford High School 18 (a) Many populations of insect pest have become resistant to insecticides. Explain how selection can result in an insect population which is resistant to a particular insecticide. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (5) (b) (i) Explain how the programme of crop management described in the second paragraph prevents the number of resistant moths from increasing. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (ii) Explain why this programme would not work if the resistance allele were dominant. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (6) (Total 11 marks) Woodford High School 19 9. Read the following passage. 5 Shark-fin soup is an expensive delicacy. To provide the basic ingredient, fishermen catch the sharks, hack the fins off and throw the dead bodies back into the ocean. But sharks are slow to mature and produce only a few offspring at a time, so they are vulnerable to overfishing. Monitoring the shark-fin trade is difficult, as once a fin has been cut off, it can be extremely difficult to work out precisely from which species it was taken. The DNA from different species of sharks shows some differences in base sequence. This has enabled a new genetic fingerprinting technique to be developed. This technique would allow conservationists and fisheries managers to assess which of the 400 shark species are most threatened by the trade in shark fins. 10 15 An identification process has been developed using a range of “primers”. These are short pieces of single-stranded DNA that are complementary to a particular sequence of DNA. Each primer is specific to the DNA of one shark species. The primers are added to DNA taken from a shark’s fin and the polymerase chain reaction is carried out. Only two primers, one at each end of a certain piece of DNA, will bind. The piece of DNA between the primers is replicated by the polymerase chain reaction. The primers that bind are specific to a particular species of shark and the length of the DNA fragment replicated differs for each species. When this DNA is run in an electrophoresis gel it produces a single band, enabling the researchers to identify which species of shark is involved. Use information from the passage and your own knowledge to answer the questions. (a) (i) Explain why the DNA for each species of shark shows differences in base sequence (line 6). ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (2) (ii) Each primer is specific to the DNA of one shark species (line 12). Explain why a particular primer will only bind to the DNA of one species. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (2) (iii) The length of the replicated DNA fragment is different for each species. Explain why this is important in identifying the shark species involved. Woodford High School 20 ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3) (b) In conventional DNA fingerprinting, a series of bands is produced on the electrophoresis gel, resembling the rungs of a ladder. When the DNA in this new genetic fingerprinting technique is run in an electrophoresis gel it produces just one of these ‘rungs’. Explain the reason for the difference in the number of ‘rungs’ produced. ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... (2) Woodford High School 21 (c) Describe the polymerase chain reaction. ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... ..............................................................................................................…..................... (6) (Total 15 marks) 10. Read the following passage. Despite its bad press, cholesterol is essential. We cannot manage without it. It is an important component of plasma membranes. Myelin, the substance surrounding many nerve fibres, is involved in the rapid conduction of nerve impulses and myelin is rich in cholesterol. 5 Being insoluble in water, cholesterol cannot be transported in solution in the blood plasma. Instead, it is packaged in lipoprotein particles. The main carrier of cholesterol is low-density lipoprotein (LDL). Each LDL particle has a cholesterol core protected by an outer coat and topped by a special protein molecule. Woodford High School 22 10 15 20 How is LDL-packaged cholesterol taken up by cells? Plasma membranes are studded with binding sites for this "topping" protein. These LDL receptors are made of protein with some sugar chains attached, and their numbers increase or decrease according to the cell's needs for cholesterol. After latching on to LDL receptors, LDL particles are pulled into the cytoplasm and processed in various ways. This regulatory mechanism, however, cannot control cholesterol concentrations outside cells when large amounts of cholesterol are present in the blood. The excess cholesterol is eventually deposited in artery walls. This leads to an increased risk of thrombosis. In the 1980s, researchers purified the LDL receptor molecule and determined the sequence of its 839 amino acids. They also isolated the LDL receptor gene. Mutation of this gene gives rise to a condition known as familial hypercholesterolemia (FH). All the evidence we have, such as that all affected individuals have at least one affected parent, and that male to male transmission is possible, indicates that the mutant allele is dominant and located on one of the autosomes. The FH allele is found in a high frequency in some populations. Among South African Afrikaners, for example, 1 in 100 are FH heterozygotes while 1 in 30 000 are homozygous for the FH allele. These people are all at risk of premature death from coronary heart disease. Source: adapted from MANGE and MANGE, Basic Human Genetics (Sinaeur Associates Inc.) 1994 Use information from the passage and your own knowledge to answer the following questions. (a) Explain how myelin is involved in the rapid conduction of nerve impulses (lines 2-3). ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (b) Describe how negative feedback is involved in controlling the concentration of cholesterol in the cytoplasm of a cell. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3) (c) (i) Explain how a mutation of the LDL receptor gene can lead to a high concentration of Woodford High School 23 cholesterol in the blood. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (2) (ii) Describe how excess cholesterol deposited in artery walls can lead to an increased risk of thrombosis (lines 14-15). .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (2) (d) What is the minimum number of nucleotides in the mRNA molecule that codes for the LDL receptor? Explain how you arrived at your answer. ..................................................................................................................................... ..................................................................................................................................... (1) (e) All individuals affected with FH have at least one affected parent (line 19). Explain how this shows that the FH allele is not recessive to the normal allele. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) Woodford High School 24 (f) Use the data in the last paragraph to estimate the number of people in the South African Afrikaner population at risk from premature death from coronary heart disease because of FH. Give your answer per 100 000 of the population. Show your working. Answer .................................... per 100 000 (3) (Total 15 marks) 11. Answers should be written in continuous prose. Credit will be given for biological accuracy, the organisation and presentation of the information and the way in which the answer is expressed. Read the following passage. The living state requires a constant input of energy, and the most fundamental difference between animals and plants is the way they obtain their energy. Animals take in food - organic compounds - and release chemical energy during respiration; green plants absorb light energy from the sun, converting it to chemical energy in the process of photosynthesis. Source. adapted from RIDGE.I.(ED), Plant Physiology (Hodder and Stoughton, 1991) (a) Describe how plants absorb light energy from the sun and use this energy to produce useful substances in the light-dependent stage of photosynthesis. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (5) Woodford High School 25 (b) Describe how the products of the light-dependent stage of photosynthesis are used in the Calvin cycle and how carbohydrate is synthesised as a result of the cycle. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6) (c) Describe the similarities between photosynthesis and respiration. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6) (Total 17 marks) 12. Answers should be written in continuous prose. Credit will be given for biological accuracy, the organisation and presentation of the information and the way in which the answer is expressed. Read the following passage. DNA acts as a template for the synthesis of RNA molecules. Only in this way does the genetic information stored in the DNA molecule become directly useful to the cell. RNA synthesis is a highly selective process. In most mammalian cells only about 1% of the genetic information is transcribed into functional RNA sequences. Genetic information must be passed on from cell to cell and from generation to generation. Before a cell can divide it must produce a new copy of each of its chromosomes. Thus cell division is preceded by a special “DNA-synthesis phase” during which the DNA is replicated. In mammals, the process of meiosis forms the gametes through which DNA is passed on to the next generation. Meiosis produces haploid cells which have half the number of chromosomes of the original parent cell and are genetically different from each other. (a) Describe the similarities and differences between: (i) the structures of RNA and DNA; Woodford High School 26 ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (5) (ii) the processes of replication and transcription. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (6) (b) Suggest why only about 1% of the genetic information is transcribed into functional RNA sequences in most mammalian cells. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (Total 13 marks) Woodford High School 27 13. Read the following passage. All mammals can dive. They are able to hold their breath and swim below the surface. For most of them, this is a very limited facility involving brief dives to depths of no more than a few metres. Seals, however, have many adaptations which allow them to spend far longer under water and to dive to much greater depths. 5 A diving human breathes in deeply and enters the water with the lungs fully inflated. Seals do not do this. They exhale before they dive. Full lungs would make it energetically expensive to swim down through the water. As they cannot take down an oxygen supply in the lungs, they must take it in other ways. Their main oxygen store is the blood. 10 Seals have greater blood volumes than terrestrial mammals. A Weddell seal, for example, has about 150 cm3 of blood per kg of body mass, twice the corresponding value for humans. In addition, the seal’s blood contains more haemoglobin. The combined result is that the seal’s oxygen store is over three times that of a human of comparable mass. Not only do seals have more haemoglobin in their blood, they also have a higher concentration of another oxygen-binding pigment, myoglobin. Myoglobin is what makes meat red. The darker the meat, the greater the concentration of myoglobin. Weddell seal muscles are almost black, so great is the concentration of myoglobin. Crabeater seals forage for krill near the surface and their muscles are no darker than uncooked beef. 15 20 The combined store of oxygen in a 450 kg Weddell seal is about 30 litres. The average rate of oxygen consumption in tissue is about 250 cm3 kg–1 hour–1, so we can estimate how long the oxygen store should allow this seal to remain under water. We know, however, that Weddell seals can remain submerged for much longer than this. How do they do this? 25 When a seal dives, changes occur in its blood system. The brain is very sensitive to oxygen deprivation so its oxygen supply must be maintained. On the other hand, most other systems, such as the gut and muscles, are able to function without oxygen. When a seal dives, the heart rate slows right down. At the same time, a ring of muscle, the caval sphincter, contracts round the main vein bringing blood back to the heart from the abdomen. This prevents any more blood returning from the liver, gut and muscles of the back. Blood flow to the brain is unimpeded although that to the rest of the body is reduced by about 90%. 30 Woodford High School 28 35 40 Once their oxygen stores are exhausted, the organs outside the heart-brain-lung system continue to derive energy anaerobically from glycolysis and to accumulate lactate. It is the accumulation of lactate in the muscles which gives rise to fatigue. During rest and recovery, this lactate is processed in the liver. Blood samples from Weddell seals have shown that there is no significant increase in blood lactate concentrations until the dive time exceeds 25 minutes. After this, lactate accumulates and reaches a concentration of about 230 mg per 100 cm3 in dives of 60 minutes duration. Source: adapted from BONNER, Seals and sea lions of the world (Blandford) 1994 Use information from the passage and your own knowledge to answer the following questions. (a) Explain why full lungs would make it ‘energetically expensive to swim down through the water’ (lines 6-7). ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (b) (i) The graph shows the dissociation curve for myoglobin. 100 myglobin human haemoglobin 80 Percentage 60 saturation with oxygen 40 20 0 0 1 2 3 4 5 Partial pressure of oxygen / kPa 6 Use this graph to explain how the presence of myoglobin in its muscles can be of benefit to a seal. Woodford High School 29 .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (2) (ii) Weddell seals get their food by diving to great depths. Explain the link between the colour of a Weddell seal’s muscles and the animal’s diving habits. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (3) (c) (i) Use the figures in paragraph 4 to calculate the time you would expect a 450 kg Weddell seal to be able to remain under water, respiring aerobically. Explain your working. Answer ............................................... (2) Woodford High School 30 (ii) Weddell seals can remain under water for longer than this. Describe two adaptations of the blood system which allow them to remain under water longer. 1 ....................................................................................................................... .......................................................................................................................... 2 ....................................................................................................................... .......................................................................................................................... (2) (d) Describe one way in which the change in blood flow to the organs of the body of a diving seal differs from that in a human undergoing moderate exercise. ..................................................................................................................................... ..................................................................................................................................... (1) (Total 12 marks) 14. Answers should be written in continuous prose. Credit will be given for biological accuracy, the organisation and presentation of the information and the way in which the answer is expressed. Read the following passage. A potential new cancer treatment involves what are known as magic bullets. Cancer cells have cell-surface antigens which are not found on normal cells. Antibodies are produced to one of these antigens by the monoclonal antibody technique. Once the antibody molecules have been made, an enzyme is attached to them and the antibody-enzyme complex is injected into the patient. A drug, which causes cell lysis, is then injected in an inactive form. Woodford High School 31 (a) Use the information in the passage to suggest how the drug kills only cancer cells. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3) (b) Explain the role of B-lymphocytes and T-lymphocytes in the defence of the body against a virus infection. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6) (c) Immunisation programmes may use either attenuated or dead microorganisms. Suggest why there might be problems for the patient when using these vaccines. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3) (Total 12 marks) Woodford High School 32 15. Read the following passage. In Uzbekistan, lorries bring them vast distances to improve the desert soil. In the Scottish hills, they helped a sheep farmer to increase the size of his flocks. The vital ingredient in both cases is an annelid, the earthworm. Scientists and farmers are only just beginning to appreciate the full value of this animal in maintaining the fertility of the soil. When many of the larger species of earthworm burrow through the soil, they literally eat their way through it. The faeces they produce, known as worm casts, contain soil particles and undigested plant material. Fresh worm casts also contain enzymes such as protease, amylase, lipase and cellulase which continue to break down organic material after they have been lost from the body of the worm. 5 Earthworms play an important part in the cycling of organic matter. In European woodlands, a few common species such as Lumbricus terrestris feed on dead leaves. They are responsible for breaking up a large amount of the leaf litter into smaller fragments. They greatly increase soil fertility. At least part of this must be due to increases in the mineralised nitrogen that they make available for plant growth. Earthworm corpses decay rapidly. In one investigation, it was found that they had completely disappeared from the soil after three weeks at 12 °C. Of the nitrogen added to the soil as a result, 25 % was in the form of nitrates and 45% as ammonia. The average lifespan of an earthworm is about a year so this represents a substantial contribution to the cycling of nitrogen. 10 Earthworms also add nitrogen to the soil in their excretory products. The exact amount that they add depends on the activity of the worm. It seems likely that earthworms produce more excretory nitrogen when they are living on food reserves stored in their tissues than when they are actively feeding. The ratio of carbon to nitrogen (the C : N ratio) in organic matter is of considerable importance because plants cannot absorb mineralised nitrogen unless the C : N ratio of the organic matter in the soil is in the order of 20 : 1 or less. The C : N ratio of freshly fallen leaves is much higher than this, being, for example 43:1 for oak. Earthworms feeding on leaf litter gradually lower its C : N ratio as they break the material down and use the products for respiration. 15 20 25 Source: adapted from C.A. EDWARDS and J.R. LOFTY, Biology of Earthworms (Chapman and Hall) 1977 Woodford High School 33 Answers 1. (a) (i) Mitochondria site of respiration; Production of ATP / release of energy; For contraction; Do not award credit for making or producing energy. (ii) (b) (c) Enzymes are proteins; Proteins synthesised/made on ribosomes; Lysosomes produce/contain enzymes; Which break down/hydrolyse proteins/substances/cells of tail; 1. Chop up (accept any reference to crude breaking up); 2. Cold; 3. Buffer solution; 4. Isotonic / same water potential; 5. Filter and centrifuge filtrate; 6. Centrifuge supernatant; 7. At higher speed; 8. Chloroplasts in (second) pellet; 3 2 2 max 6 [13] 2. (a) (i) protein/immunoglobulin; specific to antigen; idea of “fit’/complementary shape; (ii) (b) (c) (d) 1. 2. 3. 4. 5. 6. 7. 1. 2. 3. 4. 5. 6. 7. 8. 9. virus contains antigen; virus engulfed by phagocyte/macrophage; presents antigen to B-cell; memory cells/B-cell becomes activated; (divides to) form clones; by mitosis; plasma cells produce antibodies; antibodies specific to antigen; correct reference to T-cells/ cytokines; antibody gene located using gene probe; cut using restriction enzyme; at specific base pairs; leaving sticky ends/unpaired bases; cut maize/DNA /vector using same restriction enzyme; join using DNA ligase; introduce vector into maize/crop/recombinant DNA into maize; passive; person is not making own antibodies/antibodies not replaced; memory cells not produced; fewer ethical difficulties/less risk of infection; 2 max 6 max 4 max 2 max 1 [15] 3. Woodford High School 34 (a) Protein made of (chain of) amino acids; Each amino acid has its own base code/code; Triplet codes; (b) (c) (d) UCA = 2 marks TCA – 1 mark; 2 CCG; GGG GGG; 2 (i) (ii) (e) max 2 Changes base sequence; Of later triplets/amino acid codes; 2 S-phase/interphase; 1 1. mRNA leaves (nucleus) through nuclear pore; 2. To ribosome; 3. tRNA molecules bring amino acids (to ribosome); 4. Specific tRNA molecule for specific amino acid; 5. Anticodon of tRNA corresponds / complementary to codon on mRNA; 6. Peptide bonds form between amino acids; 7. tRNA detaches and collects another amino acid; 8. Ribosome moves along mRNA; max 6 [15] 4. (a) (i) both are polymers/polysaccharides/built up from many sugar units/ both contain glycosidic bonds/ contain (C)arbon, (H)ydrogen and (O)xygen; (ii) (b) (c) (d) (e) hemicellulose shorter/smaller than cellulose/fewer carbons; hemicellulose from pentose/five-carbon sugars and cellulose from hexose/glucose/six-carbon sugars; (only credit answers which compare like with like.) 1 2 protein/nucleic acid/enzyme/RNA/DNA/starch/amylose/amylopectin polypeptide; 1 (i) to make sure that all the water has been lost; 1 (ii) only water given off below 90°C; (above 90°C) other substances straw burnt/oxidised/broken down; and lost as gas/produce loss in mass; enzymes are specific; shape of lignin molecules; will not fit active site (of enzyme); OR shape of active site (of enzyme); will not fit molecule; 1. made from β-glucose; 2. joined by condensation/removing molecule of water/glycosidic bond; 3. 1 : 4 link specified or described; 4. “flipping over” of alternate molecules; 5. hydrogen bonds linking chains/long straight chains; 6. cellulose makes cell walls strong/cellulose fibres are strong; 7. can resist turgor pressure/osmotic pressure/pulling forces; 8. bond difficult to break; 9. resists digestion/action of microorganisms/enzymes; (allow maximum of 4 marks for structural features) Woodford High School 2 max 2 max 6 max 35 [15] 5. (a) (i) Double layer of phospholipid molecules; Detail of arrangement of phospholipids; Intrinsic proteins/protein molecules passing right through; Some with channels/pores; Extrinsic proteins/proteins only in one layer/on surface; Molecules can move in membrane/dynamic/membrane contains cholesterol; Glycocalyx/carbohydrates attached to lipids/proteins; (ii) Thicker; Single layer/presence of fibrils in cell wall; plasmodesmata; max 5 2 (b) Non-polar/lipid soluble molecules move through phospholipid layer/bilayer; Small molecules/water/gases move through phospholipid layer/bilayer; Ions/water soluble substances move through channels in proteins; Some proteins are gated; Reference to diffusion; Carriers identified as proteins; Carriers associated with facilitated diffusion; Carriers associated with active transport/transport with ATP/pumps; Different cells have different proteins; Correct reference to cytosis; max 6 (c) Absence of nuclear envelope/membrane; Membrane bounded organelles; Such as mitochondria/chloroplast/vacuole/lysosome; and membrane systems/endoplasmic reticulum/Golgi; Mesosomes in prokaryotes; max 4 Quality of language Aspect of work Grammar, punctuation and spelling of an acceptable standard 1 Material presented in an appropriate scientific style with due regard to correct use of technical terms 1 Argument clearly and logically presented 3 [20] 6. (a) 1. DNA is cut; 2. using restriction enzyme; 3. electrophoresis; 4. separates according to length/mass/size; 5. DNA made single-stranded; 6. transfer to membrane/ Southern blotting; 7. apply probe; 8. radioactive/ single stranded/ detected on film/ fluorescent; 9. reference to tandem repeats/VNTRs/minisatellites; 10. pattern unique to every individual; Woodford High School 6 max 36 (b) (c) cells on toothbrush; DNA present in cell; (i) (ii) (d) toothbrush gives small sample of DNA/ need more DNA for analysis; PCR gives many copies; uses heat; to separate strands; OR PCR replicates pieces of DNA; because DNA has been cut; OR primer added in PCR; to initiate replication 2 2 2 max (i) PCR/amplification needed; 1 (ii) other DNA present; need to identify ‘required’ DNA from rest; 2 [15] 7. (a) (i) In context of ATP formation light raises energy level of / excites electrons; pass through carriers; energy released; ATP formed from ADP + P; In context of producing reduced NADP protons / H+ ions; from photolysis / water; electrons; (ii) (b) GP converted to triose phosphate / GALP; this involves reduction; reduced NADP provides reducing power / hydrogens; ATP supplies energy for this reaction; phosphate from ATP; for production of RuBP; Membranes / (disc) shape provides large surface for light absorption; layering of membrane allows a lot of pigment; (permeable) membrane allows diffusion of gases / carbon dioxide; membranes provide surface for attachment of electron / hydrogen acceptors; stroma / matrix containing enzymes for Calvin cycle / light–independent reactions; max 5 max 4 max 3 [12] 8. (a) Insecticide resistance already in population; (resulting) from mutation; resistant insects are not killed (by insecticide)/survive; (And are able to) reproduce/breed; passing on the relevant allele/gene to the next generation/offspring; resulting in increasing frequency of resistance allele in population. Woodford High School max. 5 37 (b) (i) Surviving/resistant moths are homozygous recessive/rr; moths from untreated fields/non-resistant will be RR/Rr; crossing these produces heterozygotes/Rr; non resistant are susceptible and die; Allow annotated diagrams (ii) (If the allele were dominant) all heterozygotes would survive (and be able to breed); pass on dominant allele/gene; increasing frequency/number/of resistant individuals/moths; max. 6 [11] 9. (a) (i) Different genes/characteristics/features; Base sequence determines protein; Different species have different protein sequences; Reference to mutations; (ii) (iii) (b) (c) max 2 Primer has different DNA sequence; DNA specific / complementary base-pairing; 2 Electrophoresis separates DNA; (So they can be) identified by position on gel; Smaller/shortest fragments travel furthest/quicker / or reverse argument; 3 (conventional) Many lengths/all DNA / (new) one length; Each rung is DNA of one/specific length; 1 Heat DNA; 2 Breaks hydrogen bonds/separates strands; 3 Add primers; 4 Add nucleotides; 5 Cool; 6 (to allow) binding of nucleotides/primers; 7 DNA polymerase; 8 Role of (DNA) polymerase; 9 Repeat cycle many times; 2 max 6 [15] 10. (a) Insulates nerve fibre / axon / does not allow passage of ions / charge; Ions only pass at non-myelinated points / nodes / action potential only occurs at node; Saltatory conduction (is faster) / description of ‘jumping’; (b) (c) Rise / fall in cholesterol concentration in cytoplasm / cell; Reject references only to plasma concentrations; Fall / rise in cholesterol receptors (in plasma membrane); Leads to fall / rise in cholesterol / cholesterol returns to norm; (i) Mutation produces receptor with different shape / tertiary structure / not specific to LDL; So LDL will not bind to it and be absorbed / removed from the blood; Do not penalise ‘active site’. Woodford High School 2 max 3 2 38 (ii) (d) (e) (f) Endothelium / lining of artery torn / damaged; Atheroma / plaque / underlying cells come into contact with blood; Triggers blood clotting mechanism; OR Artery narrowed by plaque / atheroma; May be blocked by clot from elsewhere; 2 max 2517; Accept 2514 or 2511 if explanation refers to start / and stop codons. 1 If recessive would inherit one allele from each parent; reject ‘gene’ Parents could be heterozygotes / carriers; Parents / heterozygotes / carriers would not show the condition; NB points 2 and 3 may appear in one linked sentence. 2 1003.3 / 1003; Two marks for reason from below: (As dominant,) both heterozygote and homozygote at risk; (Heterozygotes 1 in 100 so) 1000 are heterozygous; (Homozygotes 1 in 30000 so) 3 / 3.3 homozygous; 3 max [15] 11. (a) Light absorbed by/strikes,chlorophyll/photosystem/PSI/PSII; electrons excited; pass down chain of carriers; energy released/transferred; producing ATP from ADP and phosphate; reduced NADP/formed with electrons; photolysis of water /allow light splits water; (water) supplies protons/H+ ions to reduce NADP; (b) (c) RuBP converted to GP; RuBP as carbon dioxide acceptor/combines with carbon dioxide; GP converted to triose phosphate/TP/GALP; this reaction is a reduction; reduced NADP provides hydrogen; ATP provides energy; some triose phosphate/TP/GALP converted to glucose/carbohydrate; some triose phosphate used to produce RuBP ATP supplies phosphate for this reaction; Both processes involve: Transfer of energy/conversion of energy from one form to another; Use and produce ATP; chain of electron carriers; located on membranes; detail of process (eg ref to chemiosmotic theory); involve cycle of reactions; oxidation and reduction/redox reactions involved; and coenzymes; processes are controlled by enzymes; some common intermediates/GALP is common to both; max. 5 max. 6 max. 6 [17] Woodford High School 39 12. (a) (i) Alike both have phosphate/phosphoric acid/PO4; bases/named bases/accept letters; nucleotides; pentose sugar; Different DNA deoxyribose; DNA thymine; DNA double stranded; DNA larger/longer; DNA one form RNA 3 types; (ii) max. 5 Alike H bonds break/DNA unwinds/DNA unzips; between (complementary) bases; complementary nucleotides/bases added/DNA acts as template; same, correctly named, enzymes e.g. polymerase; Different uracil/thymine used; all copied or only section copied respectively; one strand used transcription, two in replication; DNA/mRNA produced; enzymes that are different, correctly named; (b) Only some genes transcribed; only one strand transcribed; different protein/enzyme required by different cells; intron ref/some DNA does not code/non sense codes/junk DNA; stutter sequences/repeat DNA; max. 6 max.2 [13] 13. (a) More work done / more energy / ATP required; To overcome greater buoyancy; Look for idea of buoyancy, not term. (b) (i) (ii) (c) 2 Retains oxygen until the partial pressure is low / myoglobin has high affinity for oxygen; Partial pressure of oxygen late in dive low; Gives up oxygen (readily) at low partial pressures; 2 max Remains under water for long time; Unable to breathe while under water; Greater amount of myoglobin can release / provide more oxygen; More myoglobin, the darker the muscle; 3 max (i) 0.267 hours or 16 minutes;; Error, but 30 00 cm3 divided by 250 = 1 mark (ii) Heart rate slows; Less blood flowing to / from muscles / liver / gut / blood flow to body other than brain reduced by 90%; Large volume of blood compared with body mass; Greater concentration of haemoglobin; Woodford High School 2 40 Less blood required to supply heart / to heart muscle; Reject imprecise answers relating to ‘organs’ / ‘the body’. (d) Less blood to muscles / heart muscle / skin; 2 max 1 [12] 14. (a) Antibody binds/eq/recognises only to cancer cells; because of antibody-antigen binding/eg; enzyme activates the drug; at cancer cells only; (b) (c) max 3 B lymphocytes produce antibodies/involved in humoral response; T lymphocytes involved in cell mediated immunity; Macrophages present antigens; (specific) B lymphocytes recognise/bind to antigen; increase in numbers by mitosis; produce plasma cells (which make antibodies); antibodies bind to and clump/ agglutinate virus; memory cells produced by 1st exposure/cloned on 2nd exposure; T lymphocytes(helpers) produce lymphokines/chemicals; which aid B lymphocyte cloning; encourages phagocytes to engulf clumped virus; killer T cells kill virus infected cells; max. 6 Process of killing organisms might not be 100% efficient; live organisms might give rise to full-blown disease; attenuated organisms are non-virulent; but might mutate to virulent forms; immunity can decline - booster injections required; named side effects, eg allergies; less effective due to changed antigens; max. 3 [12] 15. (a) One mark for each correct column 2 Animalia/animal Annelida/annelid Class Order Family; Lumbricus (L.) terrestris; (b) (i) (ii) Enzymes are proteins; large molecules so not reabsorbed; are not used up in reactions (which they catalyse); enzymes are not themselves digested; max. 2 Add starch to worm casts; test for reducing sugars with Benedict’s / test for disappearance Woodford High School 41 of starch with iodine; need for control with boiled worm casts / soil; (c) (d) (e) Fungicide also killing earthworms; earthworms break leaves down into smaller pieces; making more surface for microbial action; OR soil fungi / fungal decomposers killed; less decomposition of leaves; (g) max 2 (i) Nitrogen as inorganic ions/nitrate/ammonia / nitrite; 1 (ii) Nitrogen in waste products of metabolism/urea/uric acid /ammonia; (ignore references to egestion) 1 (i) Any TWO from: Protein/amino acid/nucleic acid/ATP / urea; 1 Decomposers/saprophytic/putrifying bacteria release ammonia; ammonia (nitrite) nitrate; (named) nitrifying bacteria / nitrification; 3 (i) Reduces surface area minimising water loss; 1 (ii) Using food stores resulting in excretory nitrogen; 1 (ii) (f) 3 Addition of nitrogen; from excretion / decay / enzymes; removal of carbon; when lost as carbon dioxide / during respiration; max. 3 [20] 16. Woodford High School 42 Using information in the passage and your own knowledge, answer the following questions. (a) Copy and complete the following table to show how the earthworm, Lumbricus terrestris, is classified. Kingdom Phylum Oligochaeta Opisthopora Lumbricidae Genus Species (2) (b) (i) Explain why digestive enzymes from the earthworm’s gut are found in its faeces. (2) (ii) Describe briefly how you could show that fresh worm casts contain the enzyme, amylase. (3) (c) In an orchard where the trees had been sprayed with a fungicide for many years researchers found that there were very few earthworms but a normal population of soil microorganisms. The soil in this orchard was covered by a thick layer of dead leaves which were only decomposing slowly. In a nearby orchard which had never been sprayed, the leaves decomposed much faster. As a result, there was only a thin layer of decomposed leaves. Suggest an explanation for the difference in the rate of decomposition of the leaves in the two orchards. (2) (d) Explain what is meant by: (i) mineralised nitrogen (line 13-14); (1) (ii) excretory nitrogen (line 21). (1) Woodford High School 43 (e) (i) Name two organic compounds in the body of a dead earthworm that would contain nitrogen. (1) (ii) Explain how bacteria in the soil make nitrogen in organic compounds in the body of a dead earthworm available to plants. (3) (f) In hot, dry weather an earthworm curls up into a ball and stops moving. (i) How does this behaviour help the earthworm to survive these conditions? (1) (ii) Explain how earthworms still contribute to soil nitrogen even though they have stopped moving. (1) (g) Explain two ways in which earthworms can lower the C : N ratio of organic matter (line 27). (3) (Total 20 marks) Woodford High School 44
