19.1. Principles of genetic technology
A subsection of Biology, 9700, through 19. Genetic technology
Listing 10 of 43 questions
People with Alzheimer’s disease (AD) lose their ability to form new memories. One form of Alzheimer’s disease, called familial Alzheimer’s disease, is caused by an autosomal dominant allele of the APP gene. To study Alzheimer’s disease, identical genetically modified mice containing the dominant human APP allele have been produced. These mice are known as AD mice and are used as mouse models of Alzheimer’s disease. When these AD mice are trained to swim through a water maze, they perform poorly and cannot learn as well as normal mice. Suggest what steps will be needed to make identical genetically modified AD mice. Suggest why it is useful to have an animal model of a human disease. Researchers wanted to know if changes in gene expression were important in the inability of the AD mice to learn. Groups of normal mice and AD mice either received training to allow them to learn how to swim a water maze, or they received no training. The mice in the four groups then had mRNA extracted from the memory-forming areas of their brains. Reverse transcription of the mRNA of individuals in each group was carried out and the resulting cDNA was labelled with fluorescent nucleotides. This was then used for DNA microarray analysis using slides containing DNA sequences from 33 696 mouse genes. Explain the principles of this type of DNA microarray analysis. Table 5.1 summarises the microarray analysis of differences in gene expression for: • an untrained AD mouse compared to an untrained normal mouse • a trained AD mouse compared to a trained normal mouse. Table 5.1 training received number of genes expressed in AD mouse but not in normal mouse number of genes expressed in normal mouse but not in AD mouse total number of genes showing a difference in expression between the normal mouse and AD mouse no yes Calculate the percentage of mouse genes whose expression has been shown to be affected by training. Show your working. percentage = % State what the results in Table 5.1 show about the effect of training and learning on gene expression in brain cells. The genes which are expressed in the brains of normal mice undergoing training and learning code for proteins important in synapse and memory formation. A large number of these genes are under the control of one transcription factor, a protein called Crtc1. To try to improve learning in AD mice, researchers caused over-expression of the Crtc1 gene in the brains of AD mice, by delivering the gene to mouse brain cells using a virus vector. State the name given to this type of treatment. Suggest the effects of over-expression in the brain of the Crtc1 gene on AD mice.
9700_s18_qp_43
THEORY
2018
Paper 4, Variant 3
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Yeast cells are unicellular eukaryotes. Yeast cells respond to changes in the concentrations of the sugars galactose and glucose by switching on genes using transcription factors. shows the events that occur when the sugar galactose is present and glucose is absent in the external environment. Msn2 transcription factor proteins bind to promoters of genes A and B gene A switched on enzyme A synthesised gene B switched on protein B synthesised galactose converted to glucose Explain why enzyme A is described as an inducible enzyme. Protein B has a similar role to the protein coded for by gene Y of the lac operon. Suggest a possible role for protein B that would allow yeast cells to make use of galactose. Scientists produced genetically engineered yeast cells. The gene coding for Msn2 transcription factor proteins and the marker gene coding for green fluorescent protein (GFP) are transcribed together to produce a single mRNA molecule. The resulting Msn2 transcription factor proteins contain a GFP region as part of their structure and are called tagged Msn2 molecules. These tagged Msn2 molecules show up as green fluorescent spots when viewed using a microscope with a very high resolution. An investigation was carried out to compare the distribution of tagged Msn2 molecules in yeast cells, when glucose is absent and when glucose is present. The results are shown in Table 5.1. Table 5.1 glucose availability mean number of tagged Msn2 molecules present cytoplasm nucleus total glucose absent 3 263 2 012 5 275 glucose present 1 755 2 387 Calculate the percentage of Msn2 molecules in the cell that are located inside the nucleus, when glucose is absent. Show your working and write your answer to two significant figures. % When glucose is present, 26% of Msn2 molecules in the cell are located inside the nucleus. Suggest why this figure is different from your answer to . Another method of tagging molecules with a fluorescent colour is to use monoclonal antibodies that have GFP attached. These monoclonal antibodies bind specifically to the molecule of interest. This is called immunofluorescent tagging. Suggest reasons why immunofluorescent tagging is not a suitable choice for this investigation.
9700_s20_qp_41
THEORY
2020
Paper 4, Variant 1
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Yeast cells are unicellular eukaryotes that respond to the presence and absence of different sugars by switching genes on or off. One example of this is summarised in . If glucose is present, a sequence of events occurs. • Yeast cells metabolise glucose using constitutively expressed enzymes. • Mig1 transcription factor (A) binds to promoter B. • This stops transcription of gene C. • Production of enzyme D stops. If galactose is present and glucose is absent, a different sequence of events occurs. • The Msn2 transcription factor (E) binds to promoter B. • This activates transcription of gene C. • Enzyme D is produced and helps convert galactose to glucose. Gene F codes for the Mig1 transcription factor, A. Gene G codes for the Msn2 transcription factor, E. With reference to , identify one letter corresponding to: a structural gene a control sequence a repressor molecule Explain why enzyme D is described as inducible. Scientists have produced genetically engineered yeast cells. The gene coding for Mig1 transcription factor and the marker gene coding for green fluorescent protein (GFP) are transcribed together to produce a single mRNA molecule. The resulting Mig1 transcription factor proteins contain a GFP region as part of their structure and are called tagged Mig1 molecules. These tagged Mig1 molecules show up as green fluorescent spots when viewed using a microscope with a very high resolution. An investigation was carried out to compare the distribution of tagged Mig1 molecules in yeast cells, when glucose is absent and when glucose is present. The results are shown in Table 5.1. Table 5.1 glucose availability mean number of tagged Mig1 molecules present cytoplasm nucleus total glucose absent glucose present Calculate the percentage of Mig1 molecules in the nucleus when glucose is present. Show your working and write your answer to two significant figures. % When glucose is absent, 13% of the available Mig1 molecules are present inside the nucleus. Explain why this figure is different from your answer to . Yeast cells are unable to take in and metabolise the disaccharide sugar lactose. Some strains of yeast have been genetically engineered to overcome this, by inserting two genes from the bacterium Escherichia coli into yeast cells. Name the two bacterial genes that have been inserted into the yeast cells.
9700_s20_qp_43
THEORY
2020
Paper 4, Variant 3
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Variable number tandem repeats (VNTRs) are repetitive, non-coding sections of DNA. A particular VNTR is located at the same locus in different individuals, but the number of repeats in that VNTR varies between individuals. Explain how, in the process of genetic fingerprinting, gel electrophoresis is able to distinguish between the VNTRs that occur at the same loci of different individuals. Gel electrophoresis is also used in genetic screening. The mutation of the β-globin gene which gives rise to sickle cell anaemia removes a recognition site of a restriction enzyme, R, as shown in . R cuts DNA at the sites indicated by arrows ( ). The lengths of the resulting fragments are shown in kilobases . 1.1 kb normal allele of ȕ-globin gene, HbA sickle cell allele, HbS 0.2 kb 1.3 kb non-coding DNA shows an electrophoresis gel with a stained band of DNA from an individual who was homozygous for the normal allele for β-globin, HbA HbA. This band is the 1.1 kb fragment shown in . The 0.2 kb fragment is not shown. Complete by drawing the stained DNA that would result from an individual who is heterozygous for the sickle cell allele, HbA HbS. Put your answer on to . stained 1.1 kb DNA fragment HbA HbA individual HbA HbS individual direction of movement of DNA fragments electrophoresis gel Describe the different circumstances in which this genetic screening for the sickle cell allele, HbS, might be used.
9700_s12_qp_41
THEORY
2012
Paper 4, Variant 1
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Variable number tandem repeats (VNTRs) are repetitive, non-coding sections of DNA. A particular VNTR is located at the same locus in different individuals, but the number of repeats in that VNTR varies between individuals. Explain how, in the process of genetic fingerprinting, gel electrophoresis is able to distinguish between the VNTRs that occur at the same loci of different individuals. Gel electrophoresis is also used in genetic screening. The mutation of the β-globin gene which gives rise to sickle cell anaemia removes a recognition site of a restriction enzyme, R, as shown in . R cuts DNA at the sites indicated by arrows ( ). The lengths of the resulting fragments are shown in kilobases . 1.1 kb normal allele of ȕ-globin gene, HbA sickle cell allele, HbS 0.2 kb 1.3 kb non-coding DNA shows an electrophoresis gel with a stained band of DNA from an individual who was homozygous for the normal allele for β-globin, HbA HbA. This band is the 1.1 kb fragment shown in . The 0.2 kb fragment is not shown. Complete by drawing the stained DNA that would result from an individual who is heterozygous for the sickle cell allele, HbA HbS. Put your answer on to . stained 1.1 kb DNA fragment HbA HbA individual HbA HbS individual direction of movement of DNA fragments electrophoresis gel Describe the different circumstances in which this genetic screening for the sickle cell allele, HbS, might be used.
9700_s12_qp_43
THEORY
2012
Paper 4, Variant 3
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Bacteria were grown in a medium containing 15N. After several generations, all of the DNA contained 15N. Some of these bacteria were transferred to a medium containing the common isotope of nitrogen, 14N. The bacteria were allowed to divide once. The DNA of some of these bacteria was extracted and analysed. This DNA was all hybrid DNA containing equal amounts of 14N and 15N. Some bacteria from the medium with 15N were transferred into a medium of 14N. The bacteria were allowed to divide twice. The graph shows the percentages of 14N and 15N in the DNA of these bacteria. 14N 14N / 15N percentage of DNA Some bacteria from the medium with 15N were transferred into a medium of 14N. The bacteria were allowed to divide three times. What would be the percentages of 14N and 15N in the DNA extracted from these bacteria?
9700_w11_qp_12
MCQ
2011
Paper 1, Variant 2
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In 1984, the geneticist Alec Jeffreys invented a DNA testing technique, known as DNA profiling, that produces a DNA banding pattern on a gel. The DNA banding pattern is unique to each individual. DNA profiling can be used in police forensic work to catch criminals. Since 1987, police in many countries have collected and stored DNA from crime scenes to create DNA profiles, which they try to match with the DNA profiles of criminal suspects. DNA at a crime scene may be obtained from hairs and traces of blood, semen and saliva. Explain why PCR may be needed before DNA from a crime scene can be profiled. Explain why electrophoresis produces a DNA banding pattern on a gel. GEDmatch is described as ‘an open data personal genomics website’. It can be used by people who want to upload their DNA data to trace their ancestors and other relatives. In 2018, police in the USA solved a large number of serious crimes. Some of these crimes had been unsolved for over thirty years. The police used GEDmatch to profile DNA taken from crime scenes and to look for matching DNA profiles. In many cases the police found partial matches to the relatives of criminals. This allowed the criminals to be identified and then charged on the basis of a complete DNA profile match. Suggest why the police strategy of comparing crime scene DNA with the GEDmatch database was so successful. Explain why GEDmatch is an example of bioinformatics. The first successful conviction resulting from the use of GEDmatch by the police was widely reported. Some journalists and broadcasters thought that the GEDmatch website should not have been used by the police in this way. In the days following the news, the number of citizens choosing to upload their DNA data to GEDmatch increased from 1500 to 5000 a day. Comment on the social and ethical issues raised by this first successful conviction.
9700_w21_qp_41
THEORY
2021
Paper 4, Variant 1
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Questions Discovered
43