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1. Outline a method for carrying out gene therapy, using a named example. (Total 8 marks) 2. Discuss the potential benefits and possible harmful effects of genetic modification. (Total 7 marks) 3. Potatoes with more starch have a lower percentage water content. This has an advantage in the transport, cooking and processing of potatoes. In a strain of Escherichia coli scientists found an enzyme which increases the production of starch. Using biotechnology, the gene for this enzyme was transferred to potatoes, increasing their starch content (transgenic potatoes). The gene was transferred to three potato varieties to create three transgenic lines. The table shows the mean amount of starch and sugar contained in three lines of transgenic potatoes and normal potatoes (control), after storage for four months at 4C. Carbohydrate / % of fresh weight Potato Transgenic Control Line Sugar Starch I 0.60 11.07 II 1.56 11.61 III 1.46 12.74 Mean 1.21 11.81 I 5.14 5.88 II 5.61 3.70 III 4.32 6.35 Mean 5.02 5.31 [Source: Stark et al, (1999), Annals of the New York Academy of Sciences, 792, pages 26–36] (a) State which line of transgenic potato has the greatest amount of starch. .................................................................................................................................... (1) 1 (b) (i) Compare the levels of carbohydrate between the transgenic lines and the control potatoes. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (2) (ii) Suggest reasons for these differences. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (2) 2 Potato tubers were harvested from the field and stored in high humidity at 4C for three months. After this period, the tubers were stored at 16C, and samples were removed after 0, 3, 6 or 10 days, cut into strips, and fried. The colour of the fried potatoes was then measured and values reported using a 04 rating (light to dark), where a score of 2 or lower indicates acceptable colour. The results are shown in the table. [Source: Stark et al, (1999), Annals of the New York Academy of Sciences, 792, pages 26–36] (c) Evaluate the effect of transferring the E. coli gene on the suitability of the potatoes for frying. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (2) 3 An important part of storage management is to delay sprouting of potatoes. A second sample of potatoes was harvested from the field and stored at high humidity for three months at 4C. Storage temperature was then raised to 16C and a sample of potatoes were examined daily and scored for the number of sprouts longer than 0.5 cm. The number of days it took for 50% to sprout is indicated in the graph for control varieties (C1, C2 and C3) and three transgenic varieties (T1, T2 and T3) of potatoes. [Source: Stark et al, (1999), Annals of the New York Academy of Sciences, 792, pages 26–36] (d) Deduce how the E. coli gene affects the storage of potatoes. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (2) 4 (e) Discuss three possible harmful effects of genetically modified potatoes. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (3) (Total 12 marks) 4. Why is it possible for a gene from one organism to be introduced and function in a different organism? A. All organisms are made of cells. B. All organisms have nuclei. C. The genetic code is universal. D. All organisms have ribosomes. (Total 1 mark) 5 5. Weeds growing with crop plants can reduce yields because they compete for nutrients, water and sunlight. Synthetic chemical herbicides are often used to control these weeds. Herbicides are classified by the kinds of plants they kill and their mechanism of action. Broad-spectrum herbicides kill many different kinds of plants, but often kill the crop plant as well. Genetic engineering can create resistance to specific broad-spectrum herbicides which may solve the problem in crop plants. Different genes from bacterial sources known to protect against the effects of individual herbicides were engineered into corn plants (Zea mais). The resistance of normal and genetically engineered corn plants was measured and compared by calculating the percentage of plants that survived for 200 days with regular herbicide treatments. Graph 1 Exposure of Normal and Resistant Plants to Different Herbicides 100 80 60 Percentage of plants surviving 200 days 40 20 0 GP-R+BR-R BR-R+GU-R BR-R+SU-R GU-R+SU-R GP+BR BR+GU BR+SU GU+SU Herbicide treatment on resistant plants (a) (i) Herbicide Resistant Genes GP Glyphosate GP-R BR Bromozymil BR-R GU Glufosinate GU-R SU Sulfonylurea SU-R Calculate the difference between the survival of engineered plants and normal plants treated with Glyphosate (GP). .......................................................................................................................... .......................................................................................................................... (1) 6 (ii) Identify the engineered plant which shows the greatest difference in resistance to herbicide treatment. .......................................................................................................................... .......................................................................................................................... (1) (iii) Suggest a reason for the difference in survival of the normal plants treated with Glyphosate (GP) and Bromozymil (BR). .......................................................................................................................... .......................................................................................................................... (1) (b) (i) Define the term genetically modified crop. .......................................................................................................................... .......................................................................................................................... (1) (ii) State an example of a genetically modified plant other than corn. .......................................................................................................................... .......................................................................................................................... (1) 7 The graph below represents data from experiments in which plants were genetically engineered with more than one resistance gene. Graph 2 Exposure of resistant plants to combinations of herbicides 100 80 60 Percentage of plants surviving 200 days 40 20 0 GP+BR BR+GU BR+SU GU+SU Herbicide treatment on resistant plants (c) (i) Using both graphs, compare the data for BR-R with the data for SU-R, and for BR-R + SU-R in the same plant. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (2) (ii) Suggest a possible reason for these results. .......................................................................................................................... .......................................................................................................................... (1) 8 (d) Evaluate the effects on survival when combining two herbicide resistance genes in the same plant. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (3) (Total 11 marks) 6. Up to two additional marks are available for the construction of your answers. (2) (a) Outline how the process of meiosis can lead to Down’s Syndrome. (4) (b) Discuss the advantages and disadvantages of genetic screening for chromosomal and genetic disorders. (8) (c) Describe the technique for the transfer of the insulin gene using E. coli. (6) (Total 20 marks) 7. Up to two additional marks are available for the construction of your answers. (2) (a) Explain why enzymes are substrate specific and why their activity is affected by substrate concentration. (8) (b) Outline the use of restriction enzymes (endonucleases) and DNA ligase in gene technology. (6) (c) Outline the role of two enzymes found in the digestive system of humans. (4) (Total 20 marks) 9 8. (a) State two procedures used for the preparation of a DNA profile. ..................................................................................................................................... ..................................................................................................................................... (1) The following part of a DNA profile was used as evidence in a criminal investigation. DNA profiles of two suspects labelled S1 and S2 were compared to the DNA profile taken from the scene of the crime labeled E. [Source: Solomon and Berg, (1995), The World of Biology, Saunders Harcourt Brace College, Publishers Orlando, page 238] (b) Analyse the profiles to determine which suspect was present at the crime scene. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (Total 3 marks) 9. cystic fibrosis / SCID / other named example of gene therapy; details of a symptom of the condition; normal form of gene must be inserted into affected cells; 10 eg SCID caused by lack of normal gene for ADA; genetic screening before birth to test for SCID; stem cells removed from umbilical cord / bone marrow; (ADA) gene inserted into a retrovirus / adenovirus; virus inserts the gene into the stem cells; virus inserts gene into chromosome; recombinant stem cells injected into blood system; ADA synthesized successfully in these cells; eg cystic fibrosis caused by lack of normal gene for a chloride channel; test for cystic fibrosis using DNA from a mouthwash; gene for chloride channel / CFTP gene inserted into plasmid; recombinant plasmid cloned in bacteria; (recombinant) plasmids inserted into liposomes; liposomes sprayed through nose into lungs; liposomes fuse with lung cells; normal allele expressed in lung cells; relief of symptoms temporary; (Plus up to [2] for quality) 8 max [8] 10. named example of desired outcome eg herbicide resistance; Award [6 max] if no named example given. Award [5 max] if both possible benefits and possible harmful effects are not addressed. Possible benefits: [4 max] benefits include more specific (less random) breeding than with traditional methods; faster than traditional methods; some characteristics from other species are unlikely in the gene pool / selective breeding cannot produce desired phenotype; increased productivity of food production / less land required for production; less use of chemicals (eg pesticides); food production possible in extreme conditions; less expensive drug preparation; eg pharmaceuticals in milk; human insulin engineered so no allergic reactions; may cure genetic diseases; 11 Possible harmful effects: [4 max] some gene transfers are regarded as potentially harmful to organism (especially animals); release of genetically engineered organisms in the environment; can spread and compete with the naturally occurring varieties; some of the engineered genes could also cross species barriers; technological solution when less invasive methods may bring similar benefits; reduces genetic variation / biodiversity; [7] 11. (a) line 3 / line III (b) (i) (ii) transgenic have more (%) starch than control; transgenic have less sugar than control; transgenic greater total amounts (% fresh weight) of carbohydrates than control; greater difference between (%) starch and (%) sugar in transgenic than control; transgenic contain (gene which produces) enzyme which is active and (produces more starch); stored sugar is used to produce starch so lower in transgenic; 1 2 max 2 (c) all transgenic have acceptable fry colour; control potatoes have acceptable fry colour only after 10 days storage; transgenic have acceptable fry colour after any length of storage / all values below 2; 2 max (d) gene / it delays sprouting; delay in sprouting increases the length of time the potatoes can be stored; (e) populations of wild plants might be changed; can cross species barrier; allergies; ethical reasons; economic reasons (positive or negative); 2 3 max [12] 12. C [1] 12 13. (a) (b) (i) 35% (1) (units required) 1 (ii) SU-R 1 (iii) BR is a broader spectrum herbicide (than GP) / chemical composition of BR is more effective against plant than GP; natural resistance of corn (Zea mais) plants to glyphosphate (GP); application / concentration of herbicide was unequal; climatic factors affected applications; 1 max (i) genome / genetic makeup / genes / DNA (of crop) has been altered; the alteration is artificial and achieved by means of recombinant DNA technology; (ii) stay fresh tomatoes / delayed ripening reduces spoiling; frost resistant strawberries; ring spot resistant papayas; golden rice; insect resistant potatoes (lectins); 1 max 1 max N.B. Other acceptable answers may appear. Ensure that the feature is implied in the name supplied. Question asks for examples “other than corn”. (c) (i) Award [1 max] for any of the following which refer to the comparison of BR-R and SU-R in graph 1. SU-R = 60% survival, BR-R = 45% / SU-R has higher survival than BR-R; SU-R offers more resistance than BR-R; Award [1 max] for any of the following which refer to the comparison of BR-R and SU-R in graph 2. BR-R performs (slightly) better when combined with SU-R; SU-R performs less well when combined with BR-R; Award [2 max] to a candidate who combines these marks into a single statement: (ii) SU-R = 60% survival, BR-R = 45%, BR-R and SU-R combined = 50%; 2 max BR-R could interfere with SU-R expression / effects cancelled each other out; the maximum benefit for each is independent, not additive; 1 max 13 (d) a combination of any two of the four should give better protection than any one individually / there is no clear pattern in the data; reason: results of individual herbicide resistant genes might suggest results different from those actually seen in graph 2; GU-R and SU-R individually give highest resistance, (but when combined, give the same resistance); in graph 1, GP-R and BR-R give least resistance, but in graph 2 give highest resistance / GP-R and BR-R are additive; data shows interference between BR-R and GU-R; 3 max Award [1] for any additional valid statement about the effects on survival when combing two herbicide resistance genes in the same plant. [11] 14. (a) (b) Accept any of the points below if clearly drawn and correctly labelled in a diagram. in metaphase homologues in centre of cell / spindles attached; homologues are separating; one pair doesn’t separate / non-disjunction; in telophase cells divide into two; cells have either one more / one less chromosome; can occur in second division of meiosis; sister chromatids fail to separate; fertilization with one gamete / sperm / egg carrying extra chromosome; Down’s syndrome is trisomy of chromosome 21; 4 max genetic screening is testing for the presence or absence of gene / chromosome; screening for chromosomes can involve karyotyping; genetic screening is controversial; advantages: [4 max] parents can choose to avoid having children with disorder; parents can prepare for a child with a disorder; parents can use IVF to select embryos that are normal; can use gene therapy to correct the problem; treatment can start to prevent symptoms; fewer children with the disorder are born; disadvantages: [4 max] frequency of abortion can increase; parents can select embryos for sex of the child; can have harmful side effects such as depression if you know you will develop a disorder later; can create a genetic underclass; health insurance / treatment can be denied if there is genetic predisposition; 8 max 14 (c) mRNA is extracted; DNA copy of RNA is made using reverse transcriptase; plasmids are cut open with endonucleases (at specific sequences); insulin gene and plasmid are mixed together; addition of “sticky ends” to the DNA copy (so that it will combine with the cut plasmid); DNA ligase will seal the plasmid; recombinant plasmid is inserted into E. coli; E. coli is cultured; E. coli begins to make insulin; 6 max (Plus up to [2] for quality) [20] 15. (a) specificity: (active site works as a) lock and (substrate as a) key; (enzyme has) a specific shape; active site; (substrate has) a specific / complementary shape; (active site) fits substrate molecule / part of molecule / enzyme-substrate complex formed; activation energy lowered; substrate concentration: lower / medium concentration activity increases; directly proportional to concentration of substrate; random collisions more frequent; activity levels off / plateau; high concentration no change in activity; as all active sites fully utilised; Award [6 max] if only specificity or substrate concentration aspects addressed. (b) 8 max restriction enzyme: bacteria / E. coli has plasmids; plasmids / DNA cleaved / cut by enzyme; at specific points; leaving sticky ends; other species DNA cleaved / cut out by enzyme at same base sequence; suitable example; ligase: DNA added to plasmid / other DNA; spliced to plasmid / other DNA by enzyme; at sticky ends; recombinant plasmids / DNA inserted into (new) host cells; (new) host cells may be cloned; Award [4 max] if only restriction enzyme or ligase aspects addressed. 6 max (c) 15 Type / Specific example Substrate Product eg amylase / salivary amylase; starch; maltose; egprotease / pepsin; proteins; polypeptides / short peptide chains; eglipase / pancreatic lipase; lipid; glycerol and fatty acids; 4 max Award [2 max] for all three parts correct for one enzyme and [1 max] for correct enzyme and substrate or enzyme and product. (Plus up to [2] for quality) [20] 16. (a) polymerase chain reaction and electrophoresis 1 Both required for [1] (b) all bands on the DNA profile of E / evidence match those of S2 / no agreement between DNA fragments and S1; suspect 2 / S2 was present at the crime scene; 2 [3] 16