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Problema 1

each question

INSTRUCTIONS SHEET – INBO 2020

The question paper is divided into Sections A and B. All answers should be written in the answer sheet booklet only which will be collected at the end of the examination. The question paper need not be submitted to the examiner. Before starting, ensure that you have received a copy of the question paper containing a total of 39 numbered pages.

Section A Section A consists of 28 questions carrying 1 point each. All 28 questions are of multiple choice type, with only one correct answer for each question. Mark the correct answer with ‘✔’ in the answer sheet provided. The correct way of marking is shown below. Use a pen to mark your answer.

Q. No. a b c d

Each wrong answer will have negative marking as indicated in the scoring key. Section B Section B consists of 27 questions with a total of 72 points. The points for the questions in Section B vary depending on the number of answers and the complexity of the question. These points have been indicated along with the question. Contradictory answers will not be considered for marking.

SCORING KEY NO. OF CORRECT ANSWERS: X NO. OF INCORRECT ANSWERS: Y

SCORE INBO (THEORY): SECTION A: 3X – Y

SECTION B: 3X


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INDIAN NATIONAL BIOLOGY OLYMPIAD – 2020 SECTION A CELL BIOLOGY (7 points)

  1. (1 point) Lipid rafts are cholesterol-rich and glycosphingolipid-rich microdomains in the plasma membrane. Integral membrane proteins required for immune signaling and cell-cell communication are found in rafts. A few statements regarding lipid rafts are given below. i. The raft microdomains are more fluid than rest of the membrane. ii. Integral membrane proteins present in lipid rafts require a specific modification. iii. The raft microdomains do not freely mix with rest of the membrane and can induce protein clustering. iv. Lipid rafts help in preferential endocytosis of clustered signalling proteins. Pick the combination of correct statements. a. i, ii and iii b. i, ii and iv c. ii and iii only d. i and iii only

  2. (1 point) Kinesin-5 motors are tetrameric motors that bind to anti-parallel microtubules and slide them apart. The microtubules in a mitotic spindle can be divided into three groups (1) kinetochore microtubules (ii) astral microtubules and (iii) interpolar microtubules as shown below.

During mitosis, Kinesin-5 is most likely to be present on __________ microtubules and is expected to play a crucial role during _________. Choose the correct option to complete the given statement. a. kinetochore and prophase b. interpolar and anaphase c. astral and anaphase d. interpolar and metaphase

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  1. (1 point) A few cell types and action potentials are shown below.

Cell Types A1. Pacemaker Cells A2. Nerve Cells A3. Cardiac Myocytes

Match the action potentials to the respective cell type and choose the combination of correct pairs. a. A1-B2, A2-B1, A3-B3 b. A1-B3, A2-B1, A3-B2 c. A1-B1, A2-B2, A3-B3 d. A1-B3, A2-B2, A3-B1

  1. (1 point) Which of the following statements regarding surface area to volume ratio is/are true? i. Large surface area to volume ratio eliminates the need for a circulatory system. ii. Large surface area to volume ratio helps in faster exchange of nutrients. iii. Large surface area to volume ratio helps in faster rate of diffusion of nutrients inside cells.

Choose the correct option. a. (i) only b. (i) and (ii) only c. (ii) and (iii) only d. (i), (ii) and (iii)

Action potentials

B1. B2 B3

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  1. (1 point) Insulin peptides and the bonds between them are shown in the figure below. The release of ‘C’ peptide is important for the production of mature functional insulin and the signaling mediated by it.

As shown below, different Single Nucleotide Polymorphism (SNP) event/s introduced an additional site for restriction enzyme RE1 in the insulin gene. These SNPs did not change the C- peptide release. However, the mature insulin produced was non-functional.

Which SNP event/s in the insulin gene can be the most likely explanation of the above observation? a. Only SNP1 b. SNP1 & 5 c. SNP 3 & 4 d. Only SNP6

  1. (1 point) A growth factor, promoting cell growth activates a distinct cascade of phosphorylation events mediated by protein kinase enzymes. The ATP dependant phosphorylation of the target by a kinase can regulate the function of target proteins. Several statements made in this regard are given below:
  1. Energy released from ATP hydrolysis by the kinase helps in activation of the target protein.
  2. ATP hydrolysis-mediated phosphorylation often brings structural changes in the target protein.
  3. Phosphorylation can facilitate new protein-protein interactions to target proteins.
  4. Phosphorylation of the target protein must activate it. Which of the statements is/are true? a. 1 and 4. B- chain C- peptide A- chain SNP1 SNP2 SNP3 SNP4 SNP5 SNP6

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b. 2 and 3. c. 3 only. d. 2 and 4.

  1. (1 point) A circular plasmid DNA with ampicillin resistance gene, having a unique site for the restriction enzyme ‘X’, was digested completely with ‘X’ to generate cytosine residues at the 5’end of the overhangs. After treatment with a second enzyme, which is not a restriction enzyme, the digested DNA was used in a self-ligation reaction. When the ligation mixture was transformed in bacteria, no colonies were obtained on ampicillin containing selection media. Following statements are made about the second enzyme and its function.
  2. It is a kinase which phosphorylates the 5’end of the cytosine residues.
  3. It is a phosphatase which removes 5’phosphate from the cytosine residue.
  4. It is a methylase which attaches –CH3 group exclusively to the 5’ cytosine residue.
  5. It is a deaminase which removes the –NH2 group from the 5’cytosine residue. Pick the option that correctly explains the failure of colony formation. a. 1 & 3 b. 2 & 4 c. 1 & 4 d. 3 & 2

PLANT SCIENCES (5 points)

  1. (1 point) Toluidine is a metachromatic stain that can impart different colors to different plant tissues. The primary cell walls turn pink while secondary cell walls turn blue. Which of the following is likely to result?

a. Collenchyma cell walls will turn pink. b. Xylem vessels will turn pink. c. Phloem walls will turn blue. d. Chlorenchyma wall will turn blue.

  1. (1 point) The graph below shows the change in the relative water content of different tissues with change in the total water content of the leaf of Peperomia trichocarpa. The cross sections A and B next to it represent water saturated leaves and the leaves on drying respectively. Based on this information indicate which of the following statements are true.

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(i) Total water retained in the chlorenchyma is always more than that in water storage tissue at different levels of desiccation. (ii) At about 50% water content, the loss of water from the water storage tissue is about 75%. (iii) Major change in leaf thickness occurs due to shrinkage in the water storage tissue. (iv) 75% water loss in the chlorenchyma is seen when the total water content falls by 75%. Choose from the options given below. a. (i) and (ii) only b. (ii) and (iii) only c. (i), (ii) and (iii) only d. (i), (ii), (iii) and (iv)

  1. (1 point) Active biological life is substantially dependent on functional membranes. The lipids of these membranes must be present in a viscous-fluid state for the proteins and protein complexes anchored within them to fulfil their functions. The melting point and viscosity of the lipids depend on the length of the component fatty acids and their degree of unsaturation. The following table provides information on the melting points of major fatty acids found in plant membranes and also compares the composition of mitochondrial membrane of two plant specie

Topic: Conservation of Energy, Oscillations & Waves, Special Relativity Metodi: Energy Conservation Method, Conservation Laws, Simple Harmonic Motion Analysis, Wave Equation Competenze: Physical Reasoning, Mathematical Modeling Fonte: Testo (PDF) — p.1