BIOL10008 2024 Sample Exam

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University of Melbourne *

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MISC

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Biology

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May 31, 2024

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18

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Practice Exam Page 1 of 18 Student ID number: Semester 1, 2024 BIOL10008 – Foundational Biology: Life’s Machinery Reading time: 15 minutes Writing time: 80 minutes Permitted Materials Writing items only. Instructions to Students This exam has 18 pages and is printed double sided. Please ensure that you have the correct number of pages before proceeding. Write your student ID number at the top of this page. This exam paper contains both multiple-choice questions and short answer questions. There are 44 questions, worth 64 marks total. For each multiple-choice question, you should select the CORRECT or MOST APPROPRIATE answer by fully shading in the circle bubble next to your answer of choice. If you make a mistake, erase your first answer completely before fully shading in the correct answer, or put a neat cross mark through the incorrect option, and then fully shade in the correct answer. NOTE: If more than one answer is recorded, no credit will be gained for that question. All multiple-choice questions are worth 1 mark, unless otherwise stated next to the question. For the short-answer questions, you must write your answers in the printed boxes provided below each question on the exam paper. Your answers should be in full sentences in legible English. Extra room in the form of blank pages has been provided at the end of the exam paper in case you run out of space for answering questions. If you do use these pages, please indicate at the bottom of the relevant question box that you have done this by clearly writing 'Extra Page’. If you require blank paper for rough working, ask an invigilator. You should attempt every question. The marks allocated to each question serve as a reflection of time to spend on each. This paper MUST NOT be removed from the exam hall.
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Practice Exam Page 3 of 18 QUESTION 1: Complex polysaccharides form many structural components of biological systems, the following are examples: Chitin found in insect exoskeletons and starch found in plant cells. Chitin found in insect exoskeletons and cellulose in plant cell walls. Glycogen found in fungi and cellulose in plant cell walls. Lactose found in milk and cellulose in plant cell walls. QUESTION 2: The endomembrane system is comprised of the following components: Nuclear envelope, endoplasmic reticulum, Golgi apparatus, lysosome / vacuole and plasma membrane. All internal membrane structures within the eukaryotic cell. Nuclear envelope, endoplasmic reticulum, Golgi apparatus, mitochondria, vacuole and plasma membrane. Endoplasmic reticulum, Golgi apparatus, lysosome / vacuole and plasma membrane. QUESTION 3: A glycosidic bond is formed between two sugar molecules. Which of the following statements describes this process? (2 marks) A hydrogen bond formed between two sugar molecules to form a disaccharide. A dehydration reaction that occurs between the same two carbons on each sugar molecule. A reaction between two sugar molecules that always results in a branch-like structure. A dehydration reaction that can occur between different carbons on each sugar molecule. QUESTION 4: Active transport across a lipid bilayer describes: (2 marks) A process involving a protein embedded in the lipid bilayer that restricts the movement of a molecule in an energy dependent manner. Energy dependent movement of a molecule across a lipid bilayer by a transport protein, often against a concentration gradient. The energy dependent movement of molecules across a semi-permeable membrane, facilitated by polysaccharides embedded in a lipid bilayer. The process in which a protein embedded in a lipid bilayer enables the movement of water across a semi-permeable membrane.
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Practice Exam Page 4 of 18 QUESTION 5: The four major biomolecules of biological systems are: Water, carbohydrates, lipids, and protein. Protein, nucleic acids, carbohydrates and carbon. Protein, nucleic acids, carbohydrates and lipids. Nucleic acids, water, carbohydrates and protein. QUESTION 6: The diagram below represents a DNA replication bubble in a prokaryote. Which of the following statements is true for this diagram? Q is the lagging strand. The enzyme at R is helicase. The arrow at T points to the 5’ end of the DNA strand. The structures at M are RNA primers. QUESTION 7: How many genomes are present in a plant cell? Two One Four Three
Practice Exam Page 5 of 18 QUESTION 8: The rate of diffusion of a gas across a membrane is: the same for all types of membranes. inversely proportional to the thickness of the membrane. independent on a partial pressure gradient. independent of membrane thickness. QUESTION 9: Proteins are digested into small peptides and amino acids in the mouth stomach ileum duodenum QUESTION 10: Lipids and carbohydrates share a common function in cells, namely: they are only found in cell membranes. they can both be used as an energy store. they form similar unbranched linear structures. they are biosynthesised by the same biochemical pathway. QUESTION 11: The secretion of hormones by the pancreas is an example of autocrine signalling. direct signalling across gap junctions. paracrine signalling. endocrine signalling.
Practice Exam Page 6 of 18 The following information is relevant for Questions 12 – 18 Seeds contain storage reserves to enable the initial germination and emergence of a seedling from the soil prior to photosynthesis. The following table outlines the approximate content of each of the major biomolecules and water found in both wheat and canola seeds. Lipid (%) Carbohydrate (%) Protein (%) Nucleic Acids (%) Water (%) wheat seed 4 70 15 1 10 canola seed 50 20 25 1 4 QUESTION 12: Referring to the data, identify which macromolecule is the main storage reserve in the wheat seed? (1 mark) Carbohydrate Lipid Protein QUESTION 13: Explain why you made this choice in Q13. (1 mark) QUESTION 14: Select one other role or structure of this macromolecule in the wheat seed? (1 mark) Signalling between cells Control of cell function Carrier of complex molecules across membranes Component of cell walls
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Practice Exam Page 7 of 18 QUESTION 15: A feature of the eukaryotic cell is that many metabolic processes can be partitioned from each other. State one disadvantage of this. (1 mark) QUESTION 16: Select where in the plant the storage reserve found in wheat seeds would be converted into usable energy? (1 mark) Chloroplast Golgi body Peroxisome Mitochondria Nucleus Cytosol Endoplasmic reticulum QUESTION 17: Describe two processes where there is conversion of the storage reserve to usable energy. (2 marks)
Practice Exam Page 8 of 18 QUESTION 18: Both wheat and canola seeds contain only a very small proportion of water compared to typical plant cells. Describe the role of water in the functioning of a cell and explain what the effect of low water would be on the metabolism and the structure of the cells in either the wheat or canola seeds. (2 marks) QUESTION 19: Below is a diagram showing 4 stages of the cell cycle, M, G1, S and G2. From your knowledge of the cell cycle, what can be concluded about the DNA content of the cell? (2 marks) At W is double that at X At Z is the same as at W At X is half that at Z At W is twice that at Z
Practice Exam Page 9 of 18 QUESTION 20: If a molecule of DNA consists of 22 percent cytosine, what percentage of the molecule is adenine? 22 28 44 56 QUESTION 21: Choose the area on the image below that indicates the tissue responsible for the majority of photosynthetic output. Region A Region B Region C Region D
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Practice Exam Page 10 of 18 The following information is relevant for Questions 22 - 26 Scientists discovered that a parasitic, jellyfish-like eukaryotic organism, Henneguya salminicola , does not have a mitochondrial genome and has lost almost all the nuclear genes involved in transcribing and replicating mitochondria. It has mitochondria-like organelles, with folds in the inner membrane not usually seen in normal mitochondria. H. salminicola lives in cysts in the flesh of the host salmon, with almost no oxygen present. In the experiments investigating H. salminicola , a closely related fish parasite, Myxobolus squamalis , was used as a control. This parasite clearly shows a mitochondrial genome and lives in an environment that has some oxygen available. QUESTION 22: Explain what the role of oxygen in cellular respiration is. (1 mark) QUESTION 23: Choose the correct sequence for the major steps involved in typical eukaryotic cellular respiration. (1 mark) Oxidation of pyruvate into Acetyl Co-A – Citric Acid cycle – Conversion of glucose into pyruvate - Electron Transport Chain - Chemiosmosis Conversion of glucose into pyruvate– Citric Acid cycle – Oxidation of pyruvate into Acetyl Co-A - Electron Transport Chain - Chemiosmosis Oxidation of pyruvate into Acetyl Co-A – Conversion of glucose into pyruvate - Citric Acid cycle - Chemiosmosis – Electron Transport Chain Chemiosmosis – Electron Transport Chain - Conversion of glucose into pyruvate - Oxidation of pyruvate into Acetyl Co-A – Citric Acid cycle Conversion of glucose into pyruvate - Oxidation of pyruvate into Acetyl Co-A – Citric Acid cycle – Electron Transport Chain - Chemiosmosis Conversion of glucose into pyruvate– Chemiosmosis - Citric Acid cycle – Oxidation of pyruvate into Acetyl Co-A - Electron Transport Chain
Practice Exam Page 11 of 18 QUESTION 24: This lack of a complete mitochondrial genome in H. salminicola suggests: (1 mark) The entire mitochondrial genome of H. salminicola has moved to the nucleus. All salmon parasites lack mitochondrial genomes. H. salminicola never had mitochondria. The ancestors of H. salminicola once had mitochondria but have lost this over time. There was no endosymbiotic event in the evolution of H. salminicola . Mitochondrial genomes are not necessary in functioning mitochondria QUESTION 25: Compare the expected differences in ATP production between H. salminicola and M. squamalis in cellular respiration. Use your understanding of the process of cellular respiration to explain your choice. (3 marks) QUESTION 26: In eukaryotes, a proton gradient is created in an important cell structure involved in energy production. Explain whether H. salminicola would be able to create a proton gradient in this structure and why . (2 marks)
Practice Exam Page 12 of 18 QUESTION 27: Fossil evidence indicates that the earliest forms of life on Earth resembled: Lokiarchaea Stromatolites Slime mould Seaweed QUESTION 28: By convention, a protein is presented in the following way: from the N-terminus to the C-terminus as a series of beads with R-groups facing down with peptide bonds highlighted QUESTION 29: Which of the following statements is true about the physical properties of gas exchange? The solubility of a gas is independent of the salinity of the solution. The solubility of CO 2 is much lower than for O 2 . An increase in temperature decreases the solubility of CO 2 and O 2 in water. CO 2 is insoluble in water. QUESTION 30: Phospholipid bilayers have the following properties: (2 marks) They are barriers to the movement of all molecules and require transporters to facilitate movement. They are semi-permeable barriers that allow water, some gases and small molecules to cross. They are semi-permeable barriers that prevent water movement but allow uncharged molecules to cross. They form semi-permeable barriers that only allow water to cross.
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Practice Exam Page 13 of 18 QUESTION 31: Amino acids are joined to form proteins through peptide bonds. Which of the following describes the process involved? (2 marks) A dehydration reaction between the R-group between the first amino acid and the amino group of the second amino acid. A hydrogen bond forming between each amino acid as they assemble and form a protein. A dehydration reaction between the carboxyl group of the first amino acid and the amino group of the second amino acid. The R-groups of each amino acid forming covalent bonds through dehydration reactions. QUESTION 32: What feature of the bases Cytosine (C) and Guanine (G) found in nucleic acids distinguishes their interaction? (2 marks) The interaction between the bases G and C occurs via a covalent bond and creates the conditions for the formation of the DNA double helix. The presence of bases G and C would indicate a DNA molecule, since RNA does not contain both G and C. Regions of the double helix rich in G and C require more energy to separate due to presence of three hydrogen bonds between these two bases. The bases G and C are both purine bases and interact through two hydrogen bonds. QUESTION 33: In a cricket species, Schistocerca gregaria , males have 23 chromosomes and females 24. Females have 2 sex chromosomes, XX, and males only one sex chromosome, XO (where ‘O’ indicates the absence of a chromosome). Gametes are produced in both sexes by meiosis. At which stage of meiosis would some of the gametes of the male cricket contain 12 double stranded chromosomes? (2 marks) Anaphase 2 Metaphase 1 Anaphase 1 Telophase 2
Practice Exam Page 14 of 18 The following information is relevant for Questions 34 – 38 Syringa vulgaris flowers can be of three colours: white, cream, pink, lilac or rose. There are two pathways involved in petal colour production shown below. Each step of the pathway is controlled by a gene with two alleles. Genes W, X, Y result in an enzyme which catalyses that step of the pathway. The loci assort independently. The mutations at each locus are recessive and mean no enzyme is produced. When lilac and pink pigments are produced the flower is rose colour. QUESTION 34: What phenotype would a flower of genotype WW; XX; yy have? (1 mark) Lilac Rose Cream Pink White QUESTION 35: Explain what it means to say these three loci “assort independently”? (1 mark) QUESTION 36: Choose from below, the genotype for a pink flower: (1 mark) ww; xx; yy ww; XX; YY Ww; Xx; Yy ww; xx; YY WW; xx; YY
Practice Exam Page 15 of 18 QUESTION 37: When the W/w locus is homozygous recessive, explain what occurs to the phenotype of the flower overall? (1 mark) QUESTION 38: If you performed the cross between Ww; Xx; YY × Ww; Xx; YY what outcome in terms of phenotypic ratios would you expect in the offspring? Explain how you determined this. (3 marks) QUESTION 39: In guinea pigs there are 2 autosomal genes. One determines coat colour with alleles B for black and b for brown. A second locus determines whether the guinea pig has patches of white or solid colour with alleles S for patches of white and s for solid. When pure-breeding guinea pigs with black fur with white patches are crossed to pure-breeding guinea pigs with solid brown fur, all the F 1 were black and had patches of white. If the F 1 were allowed to interbreed to produce the F 2 , then which of the following statements is true about the F 2 generation? (2 marks) 2/16 would be genotype BB ; ss . 3/16 would have fur that was brown and have white patches. 1/16 would have fur that was solid and black. 9/16 would be genotype Bb ; Ss .
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Practice Exam Page 16 of 18 QUESTION 40: Plant and animal cells differ in the following: Animal cells have a nucleolus that controls DNA replication. Plant cells have a cytoskeleton for cytosolic streaming. Animal cells contain an endomembrane for secreting macromolecules. Plant cells have plastids for photosynthesis and cell walls for structure. QUESTION 41: Which statement best describes water flow through the tissues of plant roots? (2 marks) Water moves through the root via the symplast until it reaches the Casparian strip and is then forced to flow via the apoplast to the central vasculature. Water moves through the root via the apoplast and symplast until it reaches the Casparian strip and is then forced to flow via the symplast to the central vasculature. Water moves freely through the root via the apoplast and symplast until it reaches the central vasculature. Water moves through the root via the apoplast until it reaches the Casparian strip and is then forced to flow via the symplast to the central vasculature. QUESTION 42: Consider these diagrams of fish and mammal circulatory systems. Which of the following statements is correct? (2 marks) T and W are both pulmonary circuits. Ventricles only ever pump blood to the gas exchange organ. Blood pressure at S would be higher than at T. Blood at position X would be more highly oxygenated than blood at Y.
Practice Exam Page 17 of 18 QUESTION 43: The microtubule-associated proteins (MAPs) kinesin and dynein can: (2 marks) Form a complex between microtubules and actin filaments to interact with chromosomes during cell division. Interact with the plasma membrane and microtubules to define the shape of animal cells. Bind actin filaments at the negative end and prevent disassembly of the polymer. Specifically interact with microtubules and vesicles and actively move them around the cell. QUESTION 44: Frogs can have spots of 3 different colours; black, brown and white. Crosses were performed using frogs of different genotypes and the ratios of their offspring recorded in the table below. Cross Parents Offspring 1 black × brown all black 2 black × brown ½ black; ½ brown 3 brown × brown ¾ brown; ¼ white 4 white × white all white 5 black × black ¾ black; ¼ white 6 black ×brown ½ black; ¼ brown; ¼ white 7 black × black ¾ black; ¼ white From these data, it can be concluded that: (3 marks) cross 2 on its own indicates which phenotype is dominant. there is one gene locus and 3 alleles involved in these crosses. the brown parents are always heterozygous. white can be homozygous or heterozygous. END of EXAMINATION Turn the page for extra writing space and the appended material.
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