9700_s20_qp_43
A paper of Biology, 9700
Questions:
10
Year:
2020
Paper:
4
Variant:
3
Login to start this paper & get access to powerful tools
1
The courgette plant, Cucurbita pepo, produces edible fruits that vary in colour and shape. Fruit colour in courgettes is controlled by the gene A/a. Fruit shape in courgettes is controlled by the gene B/b. • A yellow fruit is produced when the dominant allele A is present. • A round fruit is produced when the dominant allele B is present. Genes A/a and B/b occur on different chromosomes. Table 1.1 shows the genotypes and phenotypes of four different varieties of courgette with respect to their fruit colour and shape. Table 1.1 name of variety genotype fruit colour fruit shape Defender aabb green long Floridor AABB yellow round Golden Dawn AAbb yellow long Tondo di Piacenza aaBB green round The varieties Golden Dawn and Tondo di Piacenza were grown in the same garden and cross-pollination occurred between them. The gardener grew these cross-pollinated F1 seeds into plants that formed fruits. The gardener did not know the genotypes of the parent plants and did not know that cross-pollination had occurred. State the phenotype of the fruits of the F1 plants and explain why it was unexpected for the gardener. The gardener crossed two of these F1 plants. Complete with the F1 gametes, F2 genotypes and F2 phenotypes. State the ratio of fruit phenotypes in the F2 offspring. ratio of fruit phenotypes Watermelons, Citrullus lanatus, are plants in the same family as courgettes. They produce large round edible fruits that usually contain many hard seeds. Seeds are the structures formed when the male and female gametes fuse at fertilisation. In the 1990s a triploid (3n) watermelon plant was developed. To produce the triploid watermelon plant, a normal diploid parent plant (2n = 22) was crossed with an artificially created tetraploid plant (4n = 44). Triploid watermelon plants develop edible fruits but these are sterile and do not contain seeds, making them more enjoyable to eat. Explain why the fruits of the triploid plants are sterile and do not contain seeds. Watermelons are attacked by watermelon mosaic virus (WM. In 1965, a WMV-resistant plant in the same family, Cucurbita ecuadorensis (2n = 40), was found growing wild in South America. State why a WMV-resistant variety of watermelon cannot be obtained by breeding Cucurbita ecuadorensis with a normal diploid watermelon. Outline how the WMV-resistant trait of Cucurbita ecuadorensis could be transferred to watermelon plants.
16. Inheritance  →  16.2. The roles of genes in determining the phenotype
16. Inheritance  →  16.1. Passage of information from parents to offspring
Open
2
Diabetes insipidus (Dis a condition in humans that causes a person to have an excessive thirst, which leads to increased drinking. One form of DI is caused by a tumour in the region of the hypothalamus concerned with osmoregulation. Explain how a tumour develops in the hypothalamus. Suggest how a tumour in the hypothalamus can lead to a person producing a large volume of dilute urine. The nervous and endocrine systems of a mammal co-ordinate responses to changes in the internal and external environment. Table 2.1 compares some features of the nervous and endocrine systems. Complete Table 2.1. Table 2.1 feature nervous system endocrine system signal impulse method of transmission of signal in blood type of communication chemical duration of effect
15. Control and coordination  →  15.1. Control and coordination in mammals
14. Homeostasis  →  14.1. Homeostasis in mammals
Open
3
The collared flycatcher, Ficedula albicollis, and the pied flycatcher, F. hypoleuca are two closely related species of bird. DNA analysis has shown that speciation from a common ancestor occurred approximately 1 million years ago. A study was carried out on the island of Öland, Sweden. In Öland, the breeding areas of the two bird species overlap and small numbers of hybrid flycatchers are produced. • Birds were captured and their DNA was analysed to identify whether each bird was F. albicollis, F. hypoleuca or a hybrid. • Sperm samples were taken from the male birds. Table 3.1 shows the percentage of males of each bird type with normal sperm. Table 3.1 bird type percentage of males with normal sperm F. albicollis F. hypoleuca male hybrid • The researchers observed that female birds mostly choose mates of their own species based on plumage and song. • Hybrid flycatchers are produced when female F. albicollis mate with male F. hypoleuca that have a song that is similar to F. albicollis. • Analysis showed that all female hybrids were sterile. The group of eggs a female bird lays at a single time in its nest is called a clutch. The offspring in the nest are looked after by a male-female pair. Sometimes the male in the male-female pair does not provide the sperm that fertilise the eggs of the female. Table 3.2 shows: • the percentage of clutches with eggs that hatched • the percentage of extra-pair nestlings (offspring in the nest fathered by a male that was different from the male of the male-female pair). Table 3.2 parents of nest percentage of clutches with eggs that hatched percentage of extra-pair nestlings male female F. albicollis F. albicollis 94.5 17.2 F. hypoleuca F. hypoleuca 89.3 22.4 hybrid F. albicollis or F. hypoleuca 38.0 100.0 Discuss the pre-zygotic and post-zygotic isolating mechanisms that maintain F. albicollis and F. hypoleuca as separate species. Explain how the two species F. albicollis and F. hypoleuca could have evolved from one original ancestral population. A single-nucleotide polymorphism (SNP) is caused by a base pair substitution mutation in a specific region of DNA. One method of identifying whether two individuals have the same SNP is to: • use a specific primer and polymerase chain reaction (PCR) • add a restriction enzyme • carry out gel electrophoresis • stain with a dye to compare banding patterns. Explain why: a specific primer is used a restriction enzyme is added gel electrophoresis is carried out. The method of identifying whether two individuals have the same SNP method was carried out to compare species A and species B. shows the banding patterns that were observed. A species B well well Describe and suggest an explanation for the results obtained in .
17. Selection and evolution  →  17.3. Evolution
18. Classification, biodiversity and conservation  →  18.1. Classification
Open
4
Lung epithelial cells have a thin layer of watery mucus on their surface. The normal allele of the CFTR gene codes for a transport protein that transports chloride ions out of epithelial cells. is a diagram of part of the cell surface membrane and the mucus layer of an epithelial cell with normal CFTR proteins. outside cell thin mucus layer cell surface membrane inside cell Cl – water Cystic fibrosis (CF) is a genetic disorder caused by having two recessive alleles of CFTR. In severe cases of CF, the transport proteins are not added to the cell surface membrane. This causes the mucus layer to be thick and sticky. Explain why the absence of CFTR proteins will cause the mucus layer to be thick and sticky. The probability of a baby having CF when both parents are heterozygous carriers for CF is 25%. It is possible to carry out prenatal screening to check for CF by using one of these tests: • amniocentesis, using cells from the amniotic fluid • chorionic villus sampling, using cells from the placenta. Both tests slightly increase the probability of the pregnancy failing . Outline the advantages of carrying out prenatal screening for CF. Embryos produced by IVF may be screened for genetic abnormalities: • to test for a specific genetic disease, such as cystic fibrosis • to check whether there is an abnormal number of chromosomes present. To improve the success of implantation and pregnancy, only embryos without any form of genetic abnormality are transferred to the woman’s uterus. A new double screening method was trialled where a single embryo biopsy was taken and used to test for a specific genetic disease and to check the number of chromosomes. In the trial, 1122 embryos were tested using this double screening method. In the trial, of the 1122 embryos tested: • 50.6% did not have a genetic disease • 27.5% did not have a genetic disease and did not have an abnormal number of chromosomes (normal embryos). Only normal embryos were transferred into the women. The percentage of embryo transfers that resulted in pregnancy was calculated. The results of the trial using double screening of a single biopsy were compared to the results of IVF procedures that used standard screening methods, as shown in Table 4.1. Table 4.1 IVF method percentage of embryo transfers that resulted in pregnancy IVF with standard screening IVF with double screening Using the data in Table 4.1, discuss the social and ethical considerations of double screening for cystic fibrosis and chromosomal abnormalities in a single biopsy.
19. Genetic technology  →  19.2. Genetic technology applied to medicine
4. Cell membranes and transport  →  4.1. Fluid mosaic membranes
Open
5
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.
19. Genetic technology  →  19.1. Principles of genetic technology
14. Homeostasis  →  14.1. Homeostasis in mammals
16. Inheritance  →  16.3. Gene control
Open
6
Structures and compounds involved in the light dependent stage of photosynthesis are listed, A to J. A – thylakoid membrane F – NADP B – starch grain G – electron C – chlorophyll a H – proton D – chlorophyll b I – ATP E – water J – chloroplast envelope Complete Table 6.1 by matching each description with one letter chosen from A to J to show the correct structure or compound. You may use each letter once, more than once or not at all. Table 6.1 description letter accessory pigment location of ATP synthase acts as reaction centre transports hydrogen atoms diffuses through ATP synthase broken down in photolysis GP (PGA) and TP (triose phosphate) are intermediates of the Calvin cycle of the light independent stage. Some GP and TP are used for the synthesis of organic compounds. Name a polysaccharide synthesised as a result of TP production. Name the additional element that is required for the production of amino acids from Calvin cycle intermediates.
13. Photosynthesis  →  13.1. Photosynthesis as an energy transfer process
Open
7
shows part of the process of respiration in a mitochondrion. ETC A D C B NAD reduced NAD Krebs cycle water ATP ADP + Pi e– e– H+ H+ H+ H+ With reference to , name: area A process B substance C State the type of chemical compound that is represented by D. State the process by which ATP can be synthesised directly during glycolysis or the Krebs cycle. The elephant seal, Mirounga angustirostris, spends most of its life in the ocean. is an elephant seal. Elephant seals can stay underwater for up to two hours. During this time, respiration continues. shows the mass of blood in the body, as a percentage of total body mass, for the elephant seal and for three other mammals. elephant seal mass of blood as a percentage of total body mass dog mammal human cow shows that elephant seals have a higher mass of blood as a percentage of total body mass than humans. Calculate how many times greater this figure is for elephant seals compared to humans. Show your working and write your answer to two decimal places. answer Suggest why an elephant seal needs such a large mass of blood as a percentage of total body mass. Elephant seals have a very thick layer of adipose tissue under their skin. Adipose cells are rich in fat molecules. Suggest why 1 g of fat will produce more ATP than 1 g of carbohydrate as a respiratory substrate in aerobic respiration. Adipose tissue plays an important role in thermoregulation in elephant seals. State the role of adipose tissue in thermoregulation. The hypothalamus in the brain is the control centre for thermoregulation. Outline how a change in temperature of the external environment results in an impulse arriving at the hypothalamus. When the blood temperature of a mammal decreases, one response is that its body secretes more adrenaline. Suggest how an increase in adrenaline results in an increase in blood temperature.
12. Energy and respiration  →  12.2. Respiration
8. Transport in mammals  →  8.1. The circulatory system
Open
8
Sampling can be used to find the distribution and abundance of species in an area. Students sampled a rocky shore from a high tide area to a low tide area. They decided to use a belt transect. Describe how you would carry out a belt transect to assess the distribution and abundance of organisms in an area. The results of an ecological investigation may suggest that there is a relationship between two variables, for example the abundance of a plant species and light intensity. Scattergraphs can be drawn between the two variables. shows three scattergraphs, A, B and C. y x A y x B y x C State which scattergraphs show a relationship between the two variables, x and y. Name a statistical test you would carry out to assess the strength of any relationship between two variables.
18. Classification, biodiversity and conservation  →  18.2. Biodiversity
Open