India 2017 — INBO2017-Question.pdf

Fonte: gare di altri paesi/India/biologia/INBO2017-Question.pdf · Apri PDF · apri PDF p.2 Cluster: Fluidi Soluzioni (stessa cartella): · · · · · · · · · · ·

Problema 1

1. (1 point) In a lung cancer cell line, it is observed that a single point mutation (cysteine to phenylalanine) makes a protein defective in phosphorylation function. The loss of function is most likely due to: a. increase in solubility of the protein in aqueous environment. b. alteration of protein conformation essential for protein-protein interaction. c. increase in the net charge of the protein leading to loss of active site. d. decrease in hydrophobic interactions of the protein with other proteins.

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.2

Problema 2

2. (1 point) Which of the following proteins are likely to be active during mitosis? i. Telomerase ii. Primase iii. Topoisomerase iv. Kinesin a. ii, iii & iv only b. iii & iv only c. i, ii & iii only d. only iv

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.2

Problema 3

3. (1 point) A tuberculosis patient died within a few minutes of second intravenous administration of streptomycin and the post-mortem report gave anaphylactic shock as the cause of death. This is explained as: a. a sudden surge in IgM synthesis leading to complement activation causing fatal drop in blood pressure. b. a vigorous precipitation of soluble antigens in body fluids triggered by the IgA insurgence leading to blockage of capillaries to vital organs. c. IgE induced mast cell degranulation triggering abrupt dilation of peripheral blood vessels resulting in a precipitous drop in blood pressure. d. IgD induced agglutination of antigen-bearing cells in vital organs leading to multi-organ failure. 2

Topic: Fluid Mechanics Metodi: Bernoulli’s Equation, Hydrostatic Equilibrium, Continuity Equation Competenze: Physical Reasoning Fonte: Testo (PDF) — p.2

Problema 4

4. (1 point) Apoptosis is programmed cell death characterized by the formation of apoptotic bodies whereas necrosis is characterized by bursting of cells. Which of the following condition/s lead/s to necrosis? i. Gaps and nicks formed in DNA during replication cannot be repaired. ii. Developing neurons fail to make synapses with neighbouring cells. iii. Heart muscle cells damaged by oxygen depletion followed by cardiac infarction.

a. i and ii b. ii and iii c. i only d. iii only

PLANT SCIENCES (3 points)

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.3

Problema 5

5. (1 point) Diagrams A – D show the external or internal structures of leaves from different groups of plants.

A

B

C D

Match these diagrams with the plants described by statements I - IV.

I. The sexual reproductive structures are formed on an independent non-dominant plant body. 3

II. Plants in this group have leaves with a haploid number of chromosomes. III. Plants in this group are seed-bearing and amongst the earliest evolved extant trees. IV. The mega gametophytes develop within enclosed ovules.

a. I-B, II-A, III-D, IV-C b. I-B, II-D, III- C, IV-A c. I-C, II-A, III-B, IV-D d. I-D, II-B, III-A, IV-C

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.3

Problema 6

plants are given

6. (1 point) A few statements regarding the sexual and asexual modes of reproduction in plants are given.

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.4

Problema 7

1. In sexual reproduction, progeny are genetically different from each other.

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.4

Problema 8

2. In asexual mode of reproduction, progeny are genetically identical to each other but different from the parent.

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.4

Problema 9

3. Sexual reproduction is more conducive for driving evolution.

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.4

Problema 10

reproduction

4. A minor change in the habitat may adversely affect all offspring derived by asexual reproduction.

Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.4

Problema 11

5. A bisexual plant grown in isolation is always incapable of sexual reproduction.

Which of these statements are correct? a. 1, 3 and 4 b. 1, 2 and 5 c. 3, 4 and 5 d. 2, 3 and 4

7. (1 point) Which one of the following plant traits is NOT associated with colonization on land from aquatic habitat?

a. Presence of apical meristems. b. Absence of alternation of generations. c. Formation of multicellular, dependent embryos. d. Production of gametes within multicellular organs. 4

ANIMAL SCIENCES (6 points)

8. (1 point) Fluid filtration across capillaries depends on several factors. Which of the following causes net filtration at the arterial end and net reabsorption at the venous end? a. Greater osmotic pressure of plasma at the arterial end of capillary than that at the venous end. b. Greater interstitial fluid osmotic pressure at venous end of capillary than that at arterial end. c. Greater capillary hydrostatic pressure than osmotic pressure of plasma at arterial end and the same being lower than plasma colloidal osmotic pressure at venous end. d. Sum total of plasma colloidal osmotic pressure & interstitial fluid osmotic pressure at venous end being greater than the sum total at the arterial end capillary.

9. (1 point) Concentrations of cations & anions present in intracellular & extracellular fluids of human body are depicted in the graph.

P, Q, R and S respectively are:

a. K+, Na+, Cl- and organic anions b. Proteins, bicarbonates, K+ and Na+ c. Mg++, proteins, bicarbonates and Na+ d. Ca++, lactate, K+ and Cl- 5

10. (1 point) Typically a skeletal muscle has origin in one bone and insertion in the other across a joint. Consider the insertion of X and Y. Which one of the following statements is correct?

a. Muscle Y is responsible for rotation across the joint. b. Muscle X is responsible for extension and Y for flexion. c. Muscle X is for stronger flexion, while Y brings about weaker flexion. d. Muscle X is for quick flexion while Y produces strong flexion.

11. (1 point) The following pattern of venous blood drainage is typical of:

a. fish b. amphibian c. bird d. mammal 6

12. (1 point) Which of the following statements about blood capillaries is/are true? i. These are found in the peripheral region of organs and body. ii. The rate of flow of blood in capillaries is the least. iii. Their total cross section area is intermediate between those of veins and arteries. iv. They have only a thin layer of involuntary muscles in the wall.

a. i, ii, iii and iv b. Only ii, iii & iv c. Only ii d. Only i & iv

13. (1 point) Following are microscopic sections of quadriceps muscles from two athletes. The cells are stained to show the activity of myosin ATPase.

Mark the most appropriate description from the following.

a. P is likely to contain more mitochondria than Q. b. Q is likely to have large stores of glycogen than P. c. P is likely to represent muscle of a high jumper and sprinter. d. Creatine phosphate is the major source of energy in the muscle depicted in Q but not in P. GENETICS & EVOLUTION (6 points)

14. (1 point) Two statements regarding evolution are made below. I. Rates of evolution are typically very slow because natural selection is usually _P selection. 7

II. The plant population growing on high-zinc-soil is able to grow at concentrations which are otherwise lethal to plants of the same species. This is Q selection.

P and Q refer to: a. directional and disruptive selection respectively. b. stabilizing and directional selection respectively. c. directional selection. d. stabilizing selection.

15. (1 point) Pedigree analysis helps determine the pattern of inheritance of a trait among related individuals. One such pedigree is given.

Squares indicate males and circles females. Filled symbols indicate presence of trait while circles with a dot indicate carriers. The inheritance pattern shown in the given pedigree is: a. X-linked recessive b. Autosomal recessive c. Sex limited trait d. Sex influenced trait

16. (1 point) A female with normal genotype (XX, 44) showed the presence of testis determining factor gene (TDF gene) on the X chromosome. This is most likely a result of : a. X chromosome inactivation b. Dosage compensation effect c. Mutation d. Meiotic recombination 8

17. (1 point) Two alleles are found in Calico cats for coat color and are located on the X chromosome. One allele is responsible for black color while the other for orange color. Which of the following is correct about the phenotype of Calico cats? a. All male cats will have the same coat color. b. Female cats can have mosaic pattern. c. All females will show intermediate coat color. d. All female cats born to mothers with black coat color will have black coat color.

18. (1 point) Consider the following pedigree.

Assume that one of the grandparents has the genotype A1/A2 and the other A3/A4. The probability that the individual R will be homozygous for any of the given alleles is: a. 1/2 b. 1/4 c. 1/8 d. 1/16

19. (1 point) Which of the following properties of plasmids are central to the evolution of bacteria? i. Plasmids are genetically dispensable for host cells in the absence of a selection pressure. ii. Some plasmids have the ability to get transferred horizontally from one cell to another. iii. Some plasmids permit interaction between bacterial cells of different species. iv. Plasmids are naked DNA molecules incapable of survival outside a living cell. a. i & ii only b. ii & iii only 9

c. i & iii only d. i, ii & iv

ECOLOGY (1 point)

20. (1 point) After initial labeling of plants in an old field community, accumulation of radioactive phosphorus in different trophic levels was monitored over the next 25 days. The results are shown in the graph. P,Q, R & S respectively most likely represent:

a. Detrivore, herbivore, primary carnivore, secondary carnivore. b. Detrivore, omnivore, herbivore, carnivore. c. Primary carnivore, detrivore, herbivore, omnivore. d. Herbivore, omnivore, primary carnivore, secondary carnivore.

BIOSYSTEMATICS (1 point)

21. (1 point) Comparison of related sequences from different species can give clues to evolutionary relationships among proteins. A phylogenetic tree representing relationship between tubulin sequences is depicted below. 10

Filled circles indicate common ancestors. Mark the correct statement about the likely evolutionary mechanism. a. Tubulin gene most likely diverged into $\alpha$ and $\beta$ forms after speciation. b. A gene duplication event led to two identical copies of the gene within each species and then the sequences diverged further. c. A gene duplication event occurred first followed by sequence divergence. Each sequence diverged further during speciation. d. Earliest eukaryotic cell most likely contained two tubulin genes which further diverged from each other during the course of speciation.

********* END OF SECTION A ******* 11

INDIAN NATIONAL BIOLOGY OLYMPIAD – 2017

SECTION B

NOTE: Write all answers in the ANSWERSHEET ONLY. Only the answer sheets will be collected at the end of the examination.

CELL BIOLOGY (13 points)

22. (2 points) The famous experiment by Avery, MacLeod and McCarty showed that neither heat killed S-type nor live R-type pneumococci could kill mice (experiment 1) but simultaneous injection of both killed the mice just as efficiently as the live S-type (experiment 2). S-type pneumococci are known to possess a capsule structure outermost to the cell wall. Structurally, capsule is a linear homopolymer of sialic acid residues. Apart from the capsule, there are other virulence factors present in these bacteria such as adhesion protein

Topic: Conservation of Energy, Oscillations & Waves, Thermodynamics Metodi: Energy Conservation Method, Conservation Laws, Simple Harmonic Motion Analysis, Wave Equation Competenze: Physical Reasoning Fonte: Testo (PDF) — p.4