9700_m23_qp_42
A paper of Biology, 9700
Questions:
10
Year:
2023
Paper:
4
Variant:
2
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1
is a drawing of a longitudinal section (LS) of a human kidney. D A B C Use the letters A, B, C and D in to complete Table 1.1. Each letter may be used once, more than once or not at all. For each description, list all the letters that are correct. Table 1.1 description region of kidney location of loops of Henle location of Bowman’s capsules location of glomeruli contains urine at final concentration The volume and water potential of the urine produced by the kidney vary according to the water potential of the blood. This is a result of osmoregulation. Describe the role of aquaporins in osmoregulation. Describe the role of the brain in osmoregulation when the water potential of the blood increases above the set point.
14. Homeostasis  →  14.1. Homeostasis in mammals
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Interferon-alpha (IFN-α) can be produced as a recombinant human protein to treat some types of cancer. The gene IFNA2 codes for IFN-α. One method of producing recombinant IFN-α uses genetically engineered Escherichia coli bacteria that contain recombinant plasmids. Each recombinant plasmid contains: • the gene IFNA2 • three regulatory sequences of the lac operon (promoter, operator and lac• a gene for antibiotic resistance, AMPR. Each of the sequences for the lacI gene and AMPR gene contains its own promoter. As a result, these genes are always expressed in E. coli bacteria that contain this recombinant plasmid. is a diagram of the recombinant plasmid. The promoter regions of the lacI gene and AMPR gene are not shown. IFNA2 AMPR lacI o p e r a t o r p r o m o t e r The start of transcription of the gene IFNA2 by E. coli with the recombinant plasmid shown in needs to be controlled to obtain an optimum yield of IFN-α. Scientists investigated the effect of two inducers of transcription on the production of recombinant IFN-α: • lactose, which is converted to allolactose in E. coli • IPTG, which is a synthetic molecule with a very similar structure to allolactose. IPTG cannot be broken down by E. coli. The scientists grew three cultures of E. coli containing the recombinant plasmid in the same growth medium. The growth medium contained glucose, amino acids, essential vitamins and minerals. The growth medium did not contain lactose. After four hours, either lactose or IPTG at the same concentration was added to two of the cultures of E. coli. As a control, the third culture of E. coli was grown without adding lactose or IPTG. The concentration of recombinant IFN-α in the cultures was measured at different times over a period of 28 hours. The results are shown in . time / hours concentration of IFN-α / μg dm–3 key culture to which IPTG added culture to which lactose added control culture The regulatory sequences of the lac operon contained in the recombinant plasmid are involved in the control of transcription of the gene IFNA2. Explain the role of the gene lacI in the control of transcription of the IFNA2 gene between 0 hours and 4 hours. With reference to , describe the changes in the concentration of recombinant IFN-α in the culture containing IPTG from when IPTG was added at 4 hours to the end of the experiment at 28 hours. Suggest one reason for the difference between the concentration of recombinant IFN-α in the culture at 8 hours in the presence of lactose and the concentration of recombinant IFN-α in the culture at 8 hours in the presence of IPTG. Suggest one reason for the change in the concentration of recombinant IFN-α in the culture containing IPTG from 12 hours to 16 hours. The gene AMPR in the plasmid shown in codes for a protein that provides resistance to the antibiotic ampicillin. Suggest how AMPR allows genetically engineered E. coli containing the recombinant plasmid to be identified. Bacteria can evolve antibiotic resistance through natural processes. Outline how bacteria can evolve to become resistant to antibiotics.
19. Genetic technology  →  19.1. Principles of genetic technology
19. Genetic technology  →  19.2. Genetic technology applied to medicine
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Salmon can be genetically modified (GM) to produce increased quantities of growth hormone, which is a protein. GM salmon modified in this way have a faster growth rate and reach their maximum body mass at a younger age than non-GM salmon. Within any population of salmon there is variation in body mass. This is an example of continuous variation. Explain what is meant by continuous variation and how it can be caused. Scientists investigated whether injection of very young non-GM salmon with recombinant growth hormone could cause an increase in the growth rate of the salmon. The scientists used two groups of non-GM salmon: • a control group of salmon that were not injected with recombinant growth hormone • an experimental group of salmon that were injected with 1.0 µg of recombinant growth hormone at the start of the experiment and once a week for the next six weeks. The mean body mass of the salmon in the two groups at the start of the experiment was the same (5.3 g). After six weeks, the body mass of every salmon was measured again. The results are summarised in Table 3.1. Table 3.1 no injection with recombinant growth hormone injected with recombinant growth hormone number of non-GM salmon 28 body mass / g range 6.5–8.6 7.2–12.7 mean (xr ) 7.7 9.4 standard deviation 0.4 1.1 A student decided that a t-test should be performed on the results shown in Table 3.1. Calculate the value of t for the results shown in Table 3.1 using the formula for the t-test: t = n s n s x x + - r r f p Give your answer to two decimal places. Show your working. t = The critical value at p = 0.05 for these data is 2.01. The student used the results in Table 3.1 and the t-test to conclude that the injections of recombinant growth hormone cause an increase in the growth rate of the non-GM salmon. Comment on the extent to which the conclusion made by the student can be supported. Suggest one advantage, other than cost, of farming GM salmon that produce increased quantities of growth hormone instead of farming non-GM salmon that are injected with recombinant growth hormone each week.
17. Selection and evolution  →  17.1. Variation
19. Genetic technology  →  19.3. Genetically modified organisms in agriculture
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Array comparative genome hybridisation (aCGH) is a technique involving the use of a microarray to analyse a genome or sections of a genome. Outline the steps required to prepare the genome of an individual so that the genome is ready for analysis using a microarray chip. DiGeorge syndrome is a dominant inherited disease in humans. DiGeorge syndrome is caused by deletion of a large number of nucleotides from chromosome 22. The number of nucleotides deleted varies between individuals in a range from 800 000 to 3 100 000. The largest deletions can cause the removal of up to 46 protein-coding genes from the chromosome. shows the results of aCGH using a microarray specific for the section of chromosome 22 within which the DiGeorge syndrome deletion occurs. The microarray analysed DNA from two individuals: • one with DiGeorge syndrome • one who did not have DiGeorge syndrome (control DNA for comparison). In the aCGH results shown in : • Each small circle represents the results from a single probe on the microarray. • The x-axis shows the position of each probe on chromosome 22. The position is shown as distance along the chromosome in millions of nucleotides. • A result close to 100% fluorescence on the y-axis means that the DNA from the individual with DiGeorge syndrome fluoresces at the same intensity as the control DNA for that probe. • A result close to 50% fluorescence on the y-axis means that the DNA from the individual with DiGeorge syndrome fluoresces half as much as the control DNA for that probe. 16.0 17.0 18.0 position of probe on chromosome 22 / millions of nucleotides 19.0 20.0 21.0 fluorescence of DNA from an individual with DiGeorge syndrome as a percentage of the fluorescence of control DNA With reference to , estimate the number of nucleotides deleted from the affected chromosome 22 in the individual with DiGeorge syndrome. Give your answer to the nearest 100 000 nucleotides. Explain how the microarray technique works to give the results shown in . Suggest why the phenotypes of two individuals with DiGeorge syndrome can be different.
16. Inheritance  →  16.2. The roles of genes in determining the phenotype
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Meiosis is described as a reduction division because the number of chromosomes in the daughter cells is reduced by half. Table 5.1 describes some of the events that take place during four of the different stages of meiosis in an animal cell. Table 5.1 stage of meiosis spindle fibres diagram metaphase I attach to centromeres and arrange homologous pairs of chromosomes at the equator of the cell anaphase I re-form spindle in daughter cells telophase II disassemble Complete Table 5.1 by: • outlining the behaviour of the spindle fibres during anaphase I • identifying the stage of meiosis in which spindle fibres re-form the spindle in daughter cells • drawing a diagram to show telophase II. You do not need to add labels to your diagram showing telophase II. Explain the need for a reduction division during meiosis.
16. Inheritance  →  16.1. Passage of information from parents to offspring
5. The mitotic cell cycle  →  5.2. Chromosome behaviour in mitosis
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is a diagram representing a synapse between a chemoreceptor cell from a human taste bud and a dendrite of a sensory neurone. microvilli chemoreceptor cell synapse dendrite of sensory neurone In an experiment, different concentrations of sodium chloride solution were applied to the microvilli of the chemoreceptor cell. The membrane potential of the chemoreceptor cell and the membrane potential of the dendrite of the sensory neurone were recorded for each concentration. The resting potential of this chemoreceptor cell is –50 mV and the resting potential of the dendrite of this sensory neurone is –70 mV. The results are shown in Table 7.1. Table 7.1 concentration of sodium chloride solution / g dm–3 membrane potential / mV chemoreceptor cell dendrite of sensory neurone 0.1 –50 –70 1.0 +30 +40 10.0 +30 +40 Explain the results shown in Table 7.1. Describe the differences in structure and function between sensory neurones and motor neurones.
15. Control and coordination  →  15.1. Control and coordination in mammals
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Describe the functions of the internal membranes of the chloroplast in photosynthesis. Rubisco activase (RA) is an enzyme that has an effect on the activity of rubisco. An investigation was carried out on the effect of RA on the activity of rubisco. • Solutions of rubisco and RuBP were added to two tubes, A and B. • RA was added to tube A. • Both tubes were incubated at 25 °C for 6 minutes. • The activity of rubisco was measured every 30 seconds. All conditions were kept the same, except for the addition of RA to tube A. The results are shown in . time / min rubisco activity / arbitrary units A B Describe the results shown in and suggest an explanation for the effect of RA on the activity of rubisco.
13. Photosynthesis  →  13.1. Photosynthesis as an energy transfer process
13. Photosynthesis  →  13.2. Investigation of limiting factors
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is a diagram of a relaxed sarcomere in striated muscle. I-band I-band A-band Z-line M-line Z-line On , use label lines and letters to label: • an actin filament with the letter P • a myosin filament with the letter R. State what happens to the A-band and the I-band when the sarcomere contracts. A-band I-band The plant Strychnos toxifera produces the toxin curare, which can cause muscle paralysis in mammals. The toxin acts by binding to receptors on the cell surface membranes of muscle cells at neuromuscular junctions. Suggest how binding of curare to receptors may cause muscle paralysis. Suggest why the action of curare may lead to the death of a mammal.
15. Control and coordination  →  15.1. Control and coordination in mammals
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The passage in is about biodiversity. Complete the passage by using the most appropriate scientific terms. Biodiversity within an area can be assessed at different levels, including the species diversity, genetic diversity and ecological diversity. Species diversity can be assessed by determining the number of different species and the relative of different species in a given area. From this information, species diversity can be estimated using index of diversity. Organisms of the same species can show much genetic diversity even though they share the same . This is because they can have different combinations of . The greater the genetic diversity, the greater the ability of a species to to a changing environment. Ecological diversity is a measure of the number and range of different ecosystems and within a given area. The International Union for Conservation of Nature (IUCN) Red List of Threatened Species is updated regularly. Table 10.1 shows the numbers of endangered animal species counted every three years between 2007 and 2019. Table 10.1 year number of endangered species 7 851 9 618 11 212 12 630 14 234 Calculate the rate of increase in the number of endangered species between 2007 and 2019. Show your working. Give your answer to the nearest whole number. rate of increase = per year More species of fish were listed as endangered in 2019 than species of mammals. Suggest reasons why more fish species than mammal species are endangered.
18. Classification, biodiversity and conservation  →  18.2. Biodiversity
18. Classification, biodiversity and conservation  →  18.1. Classification
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