olympiads.hbcse.tifr.res.in Indian National Biology Olympiad INBO 2016 : Homi Bhabha Centre For Science Education

Name of the Centre : Homi Bhabha Centre For Science Education
Name Of The Exam : Indian National Biology Olympiad INBO 2016
Name Of The Subject : Biology
Document type : Previous Question Papers
Year : 2017
Website : http://olympiads.hbcse.tifr.res.in/how-to-prepare/past-papers/
Download Model/Sample Question Paper :
INBO 2016 : https://www.pdfquestion.in/uploads/13129-inbo2016.pdf
INBO 2015 : https://www.pdfquestion.in/uploads/13129-inbo2015.pdf
INBO 2014 : https://www.pdfquestion.in/uploads/13129-inbo2014.pdf
INBO 2013 : https://www.pdfquestion.in/uploads/13129-inbo2013.pdf

Indian National Biology Olympiad INBO Question Paper :

Cell Biology : (7 points)
1. (1 point) There are several types of enzyme catalyzed reactions. In one type of enzyme catalyzed reaction, in addition to the catalytic site to which the substrate (X) binds, the enzyme also has a site to which some other substance (Y) can bind. When Y binds to such an enzyme, the enzyme can still bind to the substrate but cannot convert it to the product. Which of the following will occur in such a case?

Related : Homi Bhabha Centre For Science Education Indian National Mathematical Olympiad Question Paper INMO 2017 : www.pdfquestion.in/13123.html

i. The affinity of the enzyme for the substrate will reduce.
ii. Vmax of the reaction will decrease.
iii. Y will alter the conformation of X.
iv. The conformation of the catalytic site will be altered by binding of Y.
v. The effect of Y can be overcome by increasing the concentration of X.
a. Only i, iii and v
b. Only ii, iv and v
c. Only i, ii and v
d. Only ii and iv

2. (1 point) In an experiment, E. coli cells growing at 37°C were shifted to 20°C and grown for a few generations. Which of the following changes in the membrane would help the E. coli cells adapt to the new environment?
a. Increase in the unsaturated fatty acid content.
b. Increase in the number of integral membrane proteins.
c. Increase in the phospholipid content.
d. Increase in the length of the hydrophobic tail.

3. (1 point) The shaker (sh) gene in Drosophila, when mutated, shows a typical behavior of continuous leg shaking. When the action potential of axons of such a shaker mutant of Drosophila was studied, it showed the following graph.

The graph indicates that the defect lies in the:
a. functioning of activation gate of voltage-gated Na+ channel.
b. functioning of inactivation gate of voltage-gated Na+ channel.
c. functioning of voltage-gated K+ channel.
d. strength of the electrical stimulus.

4. (1 point) The nuclear membrane disappears during cell division. After completion of cell division, it re-appears during the interphase. Which of the following contributes towards formation of the nuclear membrane?
a. Spindle fibre proteins
b. Cytoskeletal elements
c. Endoplasmic reticulum
d. Golgi bodies

5. (1 point) Eosin is an acidic stain that is widely used for staining cytoplasm of eukaryotic cells. Which of the following is responsible for this specificity?
a. Ability of eosin to bind with water molecules in cytoplasm.
b. Ability of eosin to bind with amino acids in cytoplasm.
c. Eosin can cross the plasma membrane but cannot cross the nuclear membrane and hence accumulates in cytoplasm.
d. Ability of eosin

6. (1 point) In an experiment of mitotically dividing animal cells, nuclei of cells in G1 and G2 phases were removed. In the subsequent step, the G2 phase nuclei were introduced in enucleated cells of G1 phase. If these cells are cultured, what will be the consequence?
a. Cells will abort cell cycle and enter G0 phase.
b. Cells will shift directly from G1 to G2 phase. )
c. Cells will continue to stay in G1 phase.
d. Cells will proceed from G1 to S phase.

7. (1 point) Propidium iodide (PI) is a dye that can intercalate and stain cellular genome upon cell fixation. The amount of DNA indicated by the fluorescence intensity of PI in a population of cells which are in different stages of the cell cycle is depicted in the following graph. P, Q and R respectively indicate:
a. (G2), (S & G1) and (M).
b. (G1), (G2&M) and (S).
c. (G1), (G2) and (S&M).
d. (S & G2), (G1) and (M).

Plant Sciences : (3 points)
8. (1 point) The process of photosynthesis needs carbon dioxide to diffuse from the atmosphere into the leaf and into the carboxylation site of RUBISCO. This happens through a series of steps. Each step in this diffusion pathway imposes resistance to CO2 diffusion. This is depicted in the diagram. Various resistances are listed below.
i. Boundary layer resistance
ii. Stomatal resistance
iii. Liquid phase resistance
iv. Intercellular air space resistance
Which of the following shows the correct combination?
a. P – i; Q – iii; R – ii; S – iv
b. P – i; Q – ii; R – iii; S – iv
c. P – iii; Q – iv; R – ii; S – i
d. P – iv; Q – iii; R – ii; S – i

9. (1 point) In Angiosperms, double fertilization provides the stimulus to the formation of seeds. Growth (measured as change in volume) of ovule, embryo and endosperm in Pisum is plotted in the graph.
I, II and III respectively represent:
a. endosperm, ovule and embryo.
b. ovule, endosperm and embryo.
c. embryo, ovule and endosperm.
d. endosperm, embryo and ovule.

10. (1 point) The diploid chromosome number in the leaf cell of Pisum is 14. The diagram given below depicts the ovule structure immediately after fertilization. (Note: PEN refers to primary endosperm nucleus)
a. nucellus:7; integument:14; PEN:21; egg cell:7.
b. nucellus:14; integument:14; PEN:21; zygote:14.
c. embryo sac:7; integument:14; PEN:14; zygote:14.
d. nucellus:14; integument:7; PEN:14; zygote:14.