9700_w18_qp_42
A paper of Biology, 9700
Questions:
10
Year:
2018
Paper:
4
Variant:
2

Login to start this paper & get access to powerful tools

1
2
Researchers have found evidence of natural selection in humans. • Originally, in human populations it was only babies and children that needed to digest the milk sugar, lactose. The gene coding for the enzyme lactase (LCT gene) was switched off before adulthood. • Today, in many populations, some adult individuals have lactose intolerance, which means they cannot digest lactose. Lactose intolerance leads to side-effects such as abdominal pain after eating food containing lactose. • A mutation has been identified that keeps the LCT gene switched on. An adult who has this mutation is able to digest lactose. This is called lactose persistence. • Lactose persistence increased in populations in Europe several thousand years ago. • The increase in lactose persistence in Europe coincided with an increase in farming of cows for milk. Natural selection has caused this increase in lactose persistence. State the type of selection that has caused this increase. Explain why there was selection for lactose persistence in humans several thousand years ago. Lactose intolerance and lactose persistence were investigated in a test population in Europe. The mutation which causes lactose persistence is in a regulatory gene (T/t). • People with lactose intolerance have the genotype tt. • People with lactose persistence have the genotypes TT and Tt. • 166 people were tested for their genotype. • 58 people were found to have lactose intolerance. The Hardy–Weinberg principle can be used to calculate allele, genotype and phenotype frequencies in populations. The Hardy–Weinberg equations are shown in : p + q = 1 p2 + 2pq + q2 = 1 Calculate the frequency of allele T. Show your working. frequency of allele T = When the calculated phenotype frequencies were compared to those in the general population in Europe, it was found that the percentage of people with lactose intolerance in this test population was much higher than in the general population. Suggest two reasons why the percentage of people with lactose intolerance was much higher in the test population than in the general population. In eukaryotes, gene expression is controlled by transcription factors, coded for by regulatory genes. Outline ways in which transcription factors carry out their role. It is estimated that 2% of human DNA consists of genes coding for proteins (structural genes). Of the remaining 98%, some of the DNA consists of regulatory genes and control sequences that together control gene expression. State one type of control sequence found in human DNA. A study of human evolution identified the location of mutations that result in a change in human phenotype. The study found most examples of mutations had occurred in regulatory genes, not structural genes. Suggest and explain why most changes in human phenotype are due to mutations in regulatory genes.
3
Mammals such as sheep, Ovis aries, and goats, Capra hircus, are important agricultural animals that are sometimes kept together in mixed flocks. Very occasionally, live offspring are born from a mating between a male sheep and a female goat. In sheep 2n = 54 and in goats 2n = 60. Calculate the diploid chromosome number of the hybrid offspring of a sheep and a goat. Outline why the classification of sheep and goats suggests that hybridisation between them should not be likely to occur. Normal (wild-type) goats have a gold and black coat colour pattern, known as bezoar, and are also horned (have horns). Domestic goats may have a white coat and may be hornless (do not have horns). These variations are coded for by two unlinked genes: • white coat colour, coded for by the dominant allele of the gene A/a • hornless, coded by the dominant allele of the gene H/h. A cross between a white hornless goat and a bezoar horned goat produced offspring of four different phenotypes. Draw a genetic diagram to show the genotypes of the two parents, their gametes and the offspring, and the phenotypes of the offspring. Horns on agricultural animals such as goats and cattle can be dangerous to the farmer and to other animals. Horns are often prevented from growing in 5-day-old animals by a stressful procedure called disbudding. Genetic modification can cause a deletion in the allele h coding for horns in cattle embryos, so that the allele no longer codes for a functional protein and the embryos grow into cattle that are hornless. State an ethical advantage of this example of genetic modification. Suggest why genetic modification that causes a deletion in the horned allele, in established breeds of dairy cattle, is preferable to selective breeding for hornless animals.
5
Traditional techniques for genetically modifying organisms use three enzymes: • restriction endonuclease • reverse transcriptase • DNA ligase. For example, these enzymes have been used to produce genetically modified pigs containing the GFP gene coding for green fluorescent protein, originally sourced from jellyfish. Outline how these three enzymes could be used in genetically engineering a transgenic pig containing the GFP gene. A new technique that aims to cause a deletion in a gene uses an enzyme called Cas9 nuclease. It is injected into zygotes along with an RNA sequence (the guide RNA) that is complementary to a target gene. The Cas9 nuclease causes a deletion in the target gene in the zygotes, preventing the expression of that gene. The toxicity and efficiency of the new technique was tested on four groups of pig zygotes. These pig zygotes were produced by IVF using: • ova from a female non-transgenic pig. • sperm from a male transgenic pig whose somatic cells contained one copy of the GFP gene per cell. The pig zygotes in three groups were injected with different concentrations of Cas9 nuclease and guide RNA targeted at the GFP gene. The fourth group of pig zygotes (control group) was not injected with Cas9 nuclease and guide RNA. Explain why the GFP gene was chosen for testing the new technique. Some of the zygotes in each group survived and after six days each had developed into a group of cells called a blastocyst. The blastocysts were counted using a light microscope. A filter was then added to the microscope, so that only blastocysts expressing the green fluorescent protein showed up. These were counted and the results are summarised in Table 5.1. Table 5.1 concentration of Cas9 nuclease and guide RNA / ng mm–3 number of blastocysts seen under white light number of blastocysts seen under filter 0 68 Calculate the percentage of zygotes in the control group that were transgenic. Show your working. % Explain whether the percentage you calculated for is higher or lower than expected. Name a statistical test that would allow you to test the significance of the difference between the percentage you calculated in and the expected percentage. State the best concentration of Cas9 nuclease and guide RNA to use to cause a deletion in the GFP gene and give reasons for your choice. shows the results from a second trial of the new technique, analysed by electrophoresis. • Lanes 1–4 show DNA from four pigs born after Cas9 nuclease was used to cause a deletion in a target gene coding for a cell surface protein. • Lane 5 shows DNA from their surrogate mother. • Lane 6 shows DNA from another normal pig for comparison. The size of the DNA fragments is given in kilobase pairs as shown in . 1 kbp is 1000 base pairs of DNA. The target gene measures 6 kbp and codes for a cell surface protein that is essential for the disease virus PRRSV to infect cells in the pig’s body. 6 kbp 4 kbp Explain what indicates about the success of the new technique in causing a deletion in a gene in pigs so that they show resistance to PRRSV.
6
7
8