HBCSE INBO Sample Question Paper 2018 : Indian National Biology Olympiad
Name of the Centre : Homi Bhabha Centre For Science Education
Name Of The Exam : INBO – Indian National Biology Olympiad – 2018
Name Of The Subject : Biology
Document type : Sample Questions/Past Papers
Year : 2018
Website : http://olympiads.hbcse.tifr.res.in/how-to-prepare/past-papers/
HBCSE INBO Sample Question Paper
Question Paper of Indian National Biology Olympiad – 2018 Sample Questions and model Solutions is now available in the official website of Homi Bhabha Centre For Science Education.
Related : Homi Bhabha Centre For Science Education INAO 2018 Question Paper : www.pdfquestion.in/29874.html
Instructions
** 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.
Section A :
** Section A consists of 20 questions carrying 1 point each.
** All 20 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.
** Each wrong answer will have negative marking as indicated in the scoring key.
Section B :
** Section B consists of 33 questions with a total of 80 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.
Download Question Paper :
INBO QP : https://www.pdfquestion.in/uploads/INBO2018.pdf
Solutions : https://www.pdfquestion.in/uploads/INBO2018Sol.pdf
Model Questions
Section – A :
Cell Biology : (5 points)
1. (1 point) The beaded string in the diagram represents a protein backbone. The protein is divided into regions (i) – (iv). Within each region, some side chain interactions are shown. In which region of the protein, both „Van der Waals interactions? and „hydrophobic interactions? are depicted?
a. i
b. ii
c. iii
d. iv
2. (1 point) Bacteriophages take control of the host transcription machinery and sequentially express different groups of phage genes. A, B, C and D are four groups of phage genes expressed in the order A, B, C, D. Each group of genes has a specific type of promoter, each of which requires a specific sigma factor to initiate transcription. The genes encoding sigma factors to transcribe group A and group D genes are located in:
a. bacterial genome and phage genome respectively. b. bacterial genome. c. phage genome. d. phage genome and bacterial genome respectively.
3. (1 point) Following is a DNA profile of five individuals of which two are parents. Which of the remaining profiles belong to children of these parents?
a. A and D
b. C and E
c. B and C
d. C and D
4. (1 point) Which one of the following molecules is least likely to pass through the plasma membrane?
a. CO2
b. CH3-CH2-OH
c. NaHCO3
d. CH3-CO-CH3
5. (1 point) Bacterial core RNA Polymerase (RNAP) can bind to DNA. However, in order to initiate transcription from specific promoter sequences, it needs to associate with a sigma factor to form the holo-enzyme. Study the table below and choose the correct explanation for this phenomenon. (The numbers in the table are arbitrary.)
a. Sigma factor provides catalytic site essential for transcription.
b. Holo RNAP is able to bind to a longer stretch of DNA sequence thus increasing the probability of binding to a promoter sequence.
c. Binding of sigma factor increases the catalytic efficiency of core RNAP.
d. Sigma factor destabilizes the non-specific RNAP-DNA complex and strengthens the RNAP-promoter complex.
Plant Sciences : (4 points)
6. (1 point) Which of the following adaptations help plants grow under a canopy?
I. Increasing the specific leaf area (ratio of leaf area to dry mass).
II. Increase in the chlorophyll a/b ratio in the light harvesting complex.
III. Reduced tillering (production of lateral shoots) if the plants belong to grass species.
IV. Elongation of petioles to enhance access to sunlight.
a. II and IV only
b. I, II, and III only
c. I, III and IV only
d. I, II, III and IV
7. (1 point) The following characters are found in many trees that grow in temperate forests.
i. Pollen are shed at the beginning of growing season before the leaves develop.
ii. Shedding of pollen is timed to avoid high humidity and rain.
In such trees, the type of pollination is most likely to be:
a. entomophilic.
b. anemophilic.
c. ornithophilic.
d. chiropterophilic.
8. (1 point) In an experiment, water potentials of soil and a plant growing in the soil were measured for 8 days. The results are shown below. Mark the correct interpretation. (White and black boxes on the X-axis indicate day and night respectively.)
a. The plant is growing in a soil that is watered optimally throughout the experimental period.
b. Graph P indicates the transient wilting of the plant upto 6 days.
c. The plant is most likely to be a xerophyte.
d. Later days represent the wilting of younger leaves of the plant.
9. (1 point) Cross sections of two leaves (P and Q) are shown.
P and Q respectively most likely belong to:
a. a xerophyte and a hydrophyte.
b. a mesophyte and a halophyte.
c. a mesophyte and a hydrophyte.
d. a hydrophyte and a mesophyte.
Animal Sciences : (2 points)
10. (1 point) Oxygen content of tissues/organs of Australian sea lion at two developmental stages is shown in the bar graph.
Three tissues represented by X, Y and Z are respectively:
a. Brain, heart and lungs.
b. Lungs, blood and muscles.
c. Lungs, heart and brain.
d. Blood, muscles and lungs.
11. (1 point) During the growth of marine fish, initially, yolk sac and skin are the principal sites where transport protein “X” is located. Later, skin and gills become principal sites and finally „X? becomes localized to gills. The protein “X” is:
a. glycogen synthase.
b. glucokinase.
c. Na+ K+ ATPase.
d. glucose permease.
Genetics & Evolution : (5 points)
12. (1 point) Consider a case where a mutation „R? occurred in some prairie moles influencing their behavior/physiology. In which of the following cases, it can be asserted that the mutant allele frequency will increase in the population?
i. The mutation results in enhanced monogamy.
ii. The mutation results in greater average number of surviving offspring as compared to the wild type population.
iii. The mutation results in longevity in the mole population.
iv. The mutation results in greater average time spent together by paired male and female moles compared to the wild type population.
a. i and ii only
b. iii and iv only
c. ii only
d. iii only
13. (1 point) The wild type lac operon is inducible. Mutations in the operator region (OC) can make the operon constitutive, i.e. the operon is active even in the absence of the inducer such as lactose. Constitutive expression can also be observed if the lac repressor is not synthesized. Partial diploids (merodiploids) can be developed in E. coli by transforming E. coli with a plasmid carrying a genomic DNA fragment. Of the following genotypes of merodiploids which one of them would show constitutive expression of the lac operon resulting in synthesis of functional proteins?
[?I : deletion of lac I gene, I+ : Wild type lac I gene, lac Z-, Y- & A- : mutated lac Z, Y& A genes, lac Z+Y+A+ : wild type lac Z, Y & A genes, OC: operator constitutive mutant, O+: wild type operator]
a. ?I O+ Z+Y+A+/ I+ O+ Z+Y+A-
b. ?I O+ Z-Y-A-/ I+ O+ Z+Y+A+
c. I+ OC Z+Y+A+ / I+ O+ Z+Y+A+
d. I+ O+ Z+Y+A-/ I+ OC Z-Y-A-
14. (1 point) A set of genes that are linked to one another is called as a linkage group. The minimum number of linkage groups in an organism is equal to its haploid number of chromosomes. Which of the following events can change the linkage group of a given gene?
a. Paracentric inversion
b. Chromosomal translocation
c. Meiotic recombination
d. Pericentric inversion
15. (1 point) Graph below depicts the relationship of fitness with the body size for a population. The type of selection occurring and the most probable future trend for average body size for this population will be: (Choose from the graphs P-S and I –IV):
a. P and IV
b. Q and II
c. R and III
d. S and I
16. (1 point) One can observe three phenotypic variants in the 4 o?clock plant, Mirabilis jalapa, namely, variegated leaves, green leaves and yellow leaves. If pollen from a branch that bore only variegated leaves fertilize ovules from a branch that had only green leaves, the leaves of the plantlets will be:
a. only green.
b. only yellow.
c. either green or yellow.
d. only variegated.
Ecology : (1 point)
17. (1 point) Predator-prey relationships observed in nature can also be studied in the laboratory using controlled experimental conditions. The results often vary depending on the choice of experimental conditions, namely homogeneous or heterogeneous environment (microcosm). All the parameters in homogenous microcosm are uniform while in heterogeneous system, they can vary and can have varied impact on the populations studied. Following graphs (A, B and C) depict three predator-prey interactions.
Match the correct interpretation to each of the graphs and choose from the option.
I. The microcosm is homogeneous.
II. After introduction of predator, both the predator and prey eventually die out.
III. The microcosm is heterogenous.
IV. Prey can hide in some kind of refuge where predator cannot enter.
V. There is intermittent immigration of prey and predator.
a. A: I, II B: III, IV C: I, V
b. A: II, III B: I, II C: III, V
c. A: I. II B: IV C: I, II
d. A: II, IV B: I, II C: I, IV
Ethology : (2 points)
18. (1 point) Great tits are passerine birds found in Europe as well as Asia. Along with forests, they also live in densely inhabited urban regions. The following graphs depict the frequencies and durations of songs of birds adapted to forest and urban environments.
Which of the following statements is correct?
a. As compared to urban great tits, forest tits are required to spend more energy towards foraging and protecting the territory and hence have evolved low frequency songs which are energetically less demanding.
b. Noise made by vehicles bustling through cities occupy lower frequency channels hence only great tits that sing high-pitched songs can be clearly heard.
c. Short-note and high-pitched songs by urban tits require more energy and thus urban tits declare their greater relative fitness to other members of the species.
d. Songs by forest tits have longer notes with lower frequencies and can reach much longer distances in forest ecosystem.
19. (1 point) Migratory hummingbirds often guard winter territories to maintain exclusive access to nectar produced by certain flowers during their migration. This behavior depends on a number of factors including flower density, nectar yields of individual flowers, hummingbird density and ambient environmental conditions, which, in turn, would influence the metabolic costs incurred by the bird during foraging, territorial defence and other activities. Suggest the parameters that may have been plotted on the X-axis and Y-axis respectively in the following graph.
a. X-axis: Hummingbird density; Y-axis: Flower density
b. X-axis: Flower density; Y-axis: Hummingbird density
c. X-axis: Territory size; Y-axis: Flower density
d. X-axis: Flower density; Y-axis: Territory size
Biosystematics : (1 point)
20. (1 point) Which of the following support/s common ancestry of Humans, Cheetah, Whale and Bat?
i. Homologous organs
ii. Analogous organs
iii. Vestigial organs
a. ii and iii only
b. i and ii only
c. i and iii only
d. i only