INASMO - 2022
IBO TEAM SELECTION 2nd ROUND EXAM REGULATIONS
1. Exam will be held on Saturday, February 19th, Saturday at 9:30 AM in Stirling Colleges, which is
found in each city.
2. Students who show at least 60% success in the second round will be eligible for the third round.
3. There are 50 questions. Students will have 150 minutes working time for the question paper.
4. Do not open this booklet until told to do so by the invigilator.
5. This is a multiple-choice question paper. Each question is followed by answers marked A, B, C and D
6. In these tests there is only one correct answer to each question. If more than one alternative is marked,
that answer will automatically be counted wrong.
7. SCORING RULES:
Correct answer : + 2 points
Wrong answer : − 0.25 point
No answer
: no point
8. The exam results will be announced by the school administration at the school where you had the
second-round exam.
9. 3rd Round Exam will be held in April 30, 2022. (Could be rearranged according to the Educational
Syllabus).
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
10. Exam topics are the following topics:
The Chemistry of Life (12 points)
The Chemical Context of Life, Water and Life, Carbon and the Molecular Diversity of Life, the
Structure and Function of Large Biological Molecules
The Cell (20 points)
A Tour of the Cell, Membrane Structure and Function, An Introduction to Metabolism, Cellular
Respiration and Fermentation, Photosynthesis, Cell Communication, The Cell Cycle
Genetics (40 points)
Meiosis and Sexual Life Cycles, Mendel and the Gene Idea, The Chromosomal Basis of Inheritance,
The Molecular Basis of Inheritance, Gene Expression: From Gene to Protein, DNA Tools and
Biotechnology
Plant Form and Function (28 points)
Vascular Plant Structure, Growth, and Development, Resource Acquisition and Transport in Vascular
Plants, Soil and Plant Nutrition, Angiosperm Reproduction and Biotechnology, Plant Responses to
Internal and External Signals
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
1. The ionic bond of sodium chloride is formed when
5. Which of the following take place as an ice cube
cools a drink?
a. Chlorine gains a proton from sodium
b. Sodium gains an electron from chlorine
c. Sodium and chlorine share a pair of electrons
d. Chlorine gains an electron from sodium
a. Molecular collusions in the drink increases
b. Kinetic energy in the drink increases as energy flow
to the water molecules from ice cube
c. Evaporation of the water in the drink decreases
d. The electronegativity of water molecules increases
2. An element “X” has 7 electrons in its outer shell.
Which is true about element “X”?
6. And a human genome, the first of which took over
10 years to sequence, could be completed at today’s
pace in day or less. The number of genomes that have
been fully sequenced has exploded. All these
progresses are result of a new field which use of
computer software and other computational tools that
can handle and analyze these large data sets. What is
that biology related interdisplinary field called?
a. It is stable and chemically non-reactive
b. Its atom contains 7 protons
c. It can accept an electron to acquire noble gas
configuration
d. It can lose an electron to become stable
3. One of the buffers that contribute to pH stability in
human blood is carbonic acid (H2CO3). Carbonic acid
is a weak acid that dissociates into a bicarbonate ion
(HCO3-) and a hydrogen ion (H+).
Thus, H2CO3  HCO3- + H+ .
If the pH of blood drops, one would expect
a. Genomics
b. Bioinformatics
c. Molecular computing science
d. Genetics
a. A decrease in the concentration of H2CO3 and an
increase in the concentration of HCO3b. The concentration hydroxide ion (OH-) to increase
c. The concentration of bicarbonate ion to increase
d. The HCO3- to act as a base and remove excess H+
with the formation of H2CO3
7. A cell cycle consists of:
4. Which of the following contains nitrogen in addition
to carbon, oxygen and hydrogen?
8. How do cells at the completion of meiosis compare
with the diploid cell from which they were derived?
a. A steroid such as testosterone
b. A polysaccharide such as cellulose
c. An amino acid such as glycine
d. An alcohol such as ethanol
a. They have twice the amount of cytoplasm and half
the amount of DNA.
b. They have half the number of chromosomes and
half the amount of DNA.
c. They have the same number of chromosomes and
half the amount of DNA.
d. They have the same number of chromosomes and
the same amount of DNA.
a. mitosis and meiosis.
b. G1, the S phase, and G2.
c. prophase, metaphase, anaphase, and telophase.
d. interphase and mitosis.
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
9. During the light-dependent reactions of
photosynthesis (photophosphorylation), light energy is
converted to chemical potential energy through the
process of chemiosmosis in the chloroplasts.
Which of the following statements about this process
is FALSE?
a. During photophosphorylation, the chloroplast
stroma becomes more acidic than the interior of the
thylakoid membranes.
b. Protons diffuse through protein channels which are
ATP synthetase molecules.
c. ATP is synthesized from ADP and Pi on the stroma
side of the thylakoid membranes in the chloroplast.
d. During photophosphorylation, water ionizes to form
H+ and OH- ions, yielding an electron to Photosystem
II.
10. Which of the following statements about the
Calvin cycle are FALSE?
a. The first stable product of the cycle is 3phosphoglycerate.
b. The reducing power of NADPH is used to reduce 3phosphoglycerate to glyceraldehyde 3-phosphate
(PGAL).
c. The carboxylation of the 5-carbon sugar ribulose 1,5
bisphosphate (RuBP) is catalyzed by the rubisco
enzyme (ribulose bisphosphate
carboxylase/oxygenase).
d. Two molecules of ATP are synthesized for each
"turn" of the cycle.
12. Which of the following evaluations is correct about
regulation processes given in the diagram given
above?
a. Regulation process A is about post-translational
modification of enzymes but regulation process of B is
about regulation of gene expression
b. Regulation process A is about regulation of enzyme
activity but regulation process of B is about regulation
of enzyme production
c. Regulation process A is about pre-translational
modification of enzymes but regulation process of B is
about regulation of translation
d. Regulation process B is about regulation of enzyme
activity but regulation process of A is about regulation
of enzyme production
11. A soluble red dye is injected into the phloem of a
young plant about half-way up the stem. At the same
time a yellow dye is injected into the xylem at the
same height along the stem. Which statement best
explains where the two dyes will be found one day
later?
a. The tip of the plant (farthest from the soil) will
contain only red dye; the tip of the root (deepest into
the soil) only yellow dye.
b. The tip of the plant will contain only yellow dye; the
tip of the root only red dye.
c. The tip of the plant will contain both red and yellow
dye; the tip of the root only red dye.
d. The tip of the plant will contain only red dye; the tip
of the root both red and yellow dye.
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
13. A …..I….. operon is one that is usually on;
binding of a …..II….. to the operator shuts off
transcription. An example for this operon is …..III…..
operon. But an …..IV….. operon is one that is usually
off; a molecule called an …..V….. inactivates the
repressor and turns on transcription. An example for
this operon is …..VI….. operon.
Which of the followings is correct for information
about operons given above?
I
A Repre
ssible
B Repre
ssible
C Induci
ble
D Repre
ssible
II
Repr
essor
Indu
cible
Indu
cer
Repr
essor
III
Lacto
se
Repre
ssor
Lacto
se
Trypt
ophan
IV
Induci
ble
Induc
er
Repre
ssible
Induci
ble
V
Induc
er
Trypt
ophan
Repre
ssor
Induc
er
Some genetic materials are given below.
I. Plasmids
II. Transposons
III. Viruses
IV. Insertion sequences (IS)
16. Which of the following evaluations is correct?
a. All of them are mobile genetic elements and all can
be used as vector for gene transfer
b. All of them are mobile genetic elements and all can
be used as vector for gene transfer except IS because
there is no ORI region in IS for replication after
transfer of gene
c. All of them are mobile genetic elements but only
plasmids and transposons can be used as vector for
gene transfer
d. All of them are mobile genetic elements but any of
them can be used as vector for gene transfer because of
their low capacity of gene
VI
Trypt
ophan
Lacto
se
Trypt
ophan
Lacto
se
17. The polymerase chain reaction, PCR, can produce
many copies of a specific target segment of DNA.
PCR is a three-step cycle brings about a chain reaction
that produces an exponentially growing population of
identical DNA molecules.
Which of the following is correct about PCR?
14. Chemical modifications to histones and DNA of
chromatin influence both chromatin structure and gene
expression. Which of the following modifications
results in loosening of chromatin and initiation of
transcription accordingly?
a. First step is cooling for denaturation, second step is
heating for annealing and the last step is replication by
using heat resistant polymerase enzyme
b. First step is cooling for denaturation, second step is
replication by using heat resistant polymerase enzyme
and the last step is heating for annealing
c. First step is replication by using heat resistant
polymerase enzyme, second step is cooling for
denaturation and the last step is heating for annealing
d. First step is heating for denaturation, second step is
cooling for annealing and the last step is replication by
using heat resistant polymerase enzyme
a. DNA acetylation
b. DNA methylation
c. Histone acetylation
d. Removal of phosphate groups (dephosphorylation)
next to a methylated amino acid in histone
15. Prions are slow-acting, virtually indestructible
infectious proteins that cause brain diseases in
mammals. How can prions propagate within infected
cells or organisms?
a. By using cellular mechanism, they can propagate
within cells or organisms
b. Ribosomes of infected cells produce prions so they
propagate by this way
c. Prions propagate by converting normal proteins into
the prion version.
d. Rough endoplasmic reticulum in infected cells
produces prions so they propagate by this way
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
18. In most nuclear transplantation studies, only a
small percentage of cloned embryos have developed
normally to birth. Which of the following options
is/are appropriate solution for this problem?
21.I. Mads-box genes in animals are the regulatory
equivalent of Hox genes in plants.
II. Homeotic genes in animals are called Hox genes.
III. Homeobox genes code for a domain that allows
a protein to bind to DNA and to function as a
transcription regulator.
IV. Modifications within humans are due to single
nucleotide polymorphisms, inversions, deletions,
and duplications.
V. Small changes in regulatory sequences of certain
genes can lead to major changes in body form.
a. Epigenetic change like acetylation of histones must
be reversed in the nucleus from a donor animal
b. Epigenetic change like methylation of DNA must be
reversed in the nucleus from a donor animal
c. Epigenetic change like phosphorylation of amino
acids next to a methylated amino acid in histone must
be reversed in the nucleus from a donor animal
d. All are correct because all kinds of epigenetic
changes must be reversed in the nucleus from a donor
animal
Five information sentences are given above. Choose
the correct options in the table below about these
sentences?
19. Vertebrate genomes can produce more than one
polypeptide per gene because of
A
B
C
D
a. an alternative splicing of RNA transcripts
b. an alternative starting points while transcription of
RNAs
c. an alternative starting points while replication of
DNA
d. usage of alternative different enzymes while
transcription of RNAs
I
False
Correct
False
False
II
Correct
Correct
Correct
False
III
Correct
False
Correct
Correct
IV
Correct
Correct
False
Correct
V
False
False
Correct
Correct
22. Arrange the following five events in an order that
explains the bulk flow of substances in the phloem.
20. Which of the followings contribute to genome
evolution?
I. Water diffuses into the sieve tube elements
II. Leaf cells produce sugar by photosynthesis
III. Solutes are actively transported into sieve elements
IV. Sugar is transported from cell to cell via the
apoplast and/or symplast
V. Sugar moves down the stem
a. Duplication
b. Rearrangement
c. mutations of DNA
d. All of them
a. II, I, IV, III, V
b. I, II, III, IV, V
c. II, IV, III, I, V
d. IV, II, I, III, V
23. During winter, tree sap can sometimes freeze and
"cavitation" (the formation of an air pocket) may
occur. Which one of the following mechanisms of sap
transport would you expect to be most immediately
affected by cavitation?
a. symport
b. pressure flow (mass flow)
c. cohesion transpiration
d. root pressure
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
24. The pressure-flow model of phloem sap transport
involves _____.
28. You are growing plants hydroponically, measuring
CO2 usage. The plants initially grow robustly, but
soon CO2 assimilation drops. Young leaves continue
to grow normally, but older leaves begin to yellow.
Which of the following is the MOST likely cause for
the observed changes?
a. decreased hydrostatic pressure at the sugar source
end as a result of reduced water potential
b. increased hydrostatic pressure at the sugar source
end as a result of reduced water potential
c. decreased hydrostatic pressure at the sugar source
end as a result of increased water potential
d. increased hydrostatic pressure at the sugar source
end as a result of increased water potential
a. The hydroponic medium lacks magnesium.
b. The plants are preparing to flower.
c. The plants are mutants which overproduce
gibberellins.
d. The hydroponic medium lacks oxygen, causing the
roots to suffocate.
25. Plants can modify their water potential by opening
and closing their stomata to modulate the rate of
respiration according to environmental conditions.
Which environmental condition would cause the
stomata to close?
29. Students are sketching diagrams of the shoot
system of angiosperms for a plant anatomy class.
These lists describe diagrams made by four students.
Which diagram represents the shoot system
incorrectly?
a. Increased temperature
b. High oxygen concentration
c. High relative humidity
d. High light levels
a. Leaves, stem, fruit, flowers
b. Stem, fruit, leaves, branches
c. Flowers, leaves, branches, stem
d. Stem, hair roots, leaves, flowers, branches
26. On a field trip, students collect a few samples to
analyze back in their classroom. One student picks a
blade of grass in the field and identifies it as a dicot
leaf, but his partner thinks it is a monocot. Which
explanation supports his partner's opinion?
30. Modified organs are part of survival strategies of
plants. Which plant has a flattened, photosynthetic
stem that could be mistaken for a leaf?
a. The leaf displays a thin lamina.
b. There is no petiole.
c. The margins are serrated.
d. The venation is parallel.
a. Fern
b. Cactus
c. Potato
d. Iris
27. Scientists are cataloguing slides of plant crosssections. They are interested in finding examples of
secondary growth. Which example contributes to
secondary growth?
31. Which one of the followings is a photosynthetic
plant parasite?
a. Mistletoe
b. Rafflesia
c. Indian pipe
c. Staghorn fern
a. Apical meristem, which contributes to increase in
length
b. Vascular cambium, which contributes to increase in
thickness or girth
c. Root region, which shows an increase in root hairs
d. Stems, which show an increase in number of leaves
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
32. ………………. ensures that endosperm develops
only in ovules where the egg has been fertilized:
a. Fertilization
c. Pollination
36. de-etiolation means:
a. When a stem reaches light, the plant undergoes
profound change.
b. When a shoot reaches light, the plant undergoes
profound change.
c. When a leaf reaches light, the plant undergoes
profound change.
d. None of them
b. Double fertilization
d. Seed Development
33. The plant that provides the roots is called:
a. Scion
c. Stock
b. Callus
d. Shoot
34. Which one of the following plants is not belonged
to short day plants:
a. Chrysanthemums
c. Soybean
37. An individual of the genotype AaBbCcDd was
crossed with the one aabbccdd and the following
results were obtained.
b. Poinsettias
d. Dandelions
aBCD 42
Abcd 43
ABCd 140
abcD 145
aBcD 6
AbCd 9
ABcd 305
abCD 310
35. In cattle, the polled (hornless) condition is
dominant to the horned condition. Coat colour can be
red, white or roan (red with White patches). Both
genes are carried on autosomes and they are not
linked. A cross was carried out between a cow and a
bull, both of which had the roan coat colour and both
were heterozygous for the polled condition. Which of
the following statements are true about the offspring
from the cross, assuming that the cross was carried out
several times to produce a lot of offspring?
Which of the following shows the arrangement of the
genes and their distance (in centimorgans)?
1. The chance of producing white polled and white
horned offspring is the same.
2. The chance of producing roan polled offspring is
three times that of producing roan horned.
3. There is an equal chance of producing red polled
and White polled offspring.
4. Statistically there should be more roan horned
offspring than any other type.
5. The chance of producing roan polled offspring is
twice that of producing white polled.
a. 1 & 2
b. 2 & 3
c. 3 & 4
d. 2, 3 & 5
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
41. Hemoglobin in the erythrocytes of adults is composed
of a combination of two a-globin molecules and two bglobin molecules. Sickle-cell anemia is caused by the
substitution of a single amino acid in the b-globin subunit.
In 1957, Vernon M. Ingram and his colleagues investigated
the amino acid sequences of normal and sickle-cell anemia
hemoglobins in several short peptide chains obtained by
trypsin digestion. A difference in the “fourth peptide”
between both types of b-globin was found and further
hydrolytic digestion of the “fourth peptides” revealed six
hydrolyzed products.
38. When a dominant allele (A) is alone, it causes a
brown fur color but when it is with another dominant
allele which has an epistatic effect, the fur color is
white.
According to this, which of the following shows the
true genotypes of the individuals in the family tree
given above?
1
a. Aabb
b. AaBb
c. AaBb
d. Aabb
2
AaBb
aabb
aaBb
aabb
3
Aabb
Aabb
AaBb
Aabb
4
AaBb
aaBb
Aabb
aaBb
# the “fourth peptide” products of normal b-globin were
(amino acid residues are abbreviated by the following
letters: V=valine, H= histidine, L= leucine, T= threonine,
P= proline, E= glutamic acid and K= lysine):
V—H
V—H—L
V—H—L—T
T—P—E
T—P—E—E—K
E—K
5
aabb
aabb
Aabb
aabb
# the “fourth peptide” products of b-globin of sickle cell
anemia were
V—H
V—H—L
V—H—L—T
T—P—V
T—P—V—E—K
E—K
39. Which statement regarding the amount of genomic
DNA per cell (M) during the cell cycle is correct?
a. M [G1] = M [meiosis prophase II]
b. M [meiosis prophase II] = 2 x M [meiosis prophase
I]
c. M [G1] = M [G2]
d. M [G2 after mitosis] < M [G2 after meiosis]
From these results, how many amino acids is the “fourth
peptide” composed of and what was the substituted position
of amino acid residue counting from the N-terminus? From
the following, choose the one statement which is most
appropriate. Assume that this fourth peptide contains only
one molecule of T (threonine).
40. The strand of DNA molecule isolated from E. coli
bacteria has sequence 5’ GТАGCCТАCCCАТАGG
3’. Assume that an mRNA is transcribed from the
corresponding double-stranded DNA, the template
strand being complementary to the strand isolated.
What is the sequence of this mRNA?
a. It was composed of 8 amino acids and the 6th amino acid
was substituted.
b. It was composed of 8 amino acids and the 3rd amino acid
was substituted.
c. It was composed of 7 amino acids and the 6th amino acid
was substituted.
d. It was composed of 7 amino acids and the 3rd amino acid
was substituted.
a. 3’ – CAUCGGAUGGGUAUCC – 5’.
b. 5’ – GUAGCCUACCCAUAGG – 3’.
c. 5’ – GGAUACCCAUCCGAUG – 3’.
d. 5’ – CACAGAUACCCAGAUG – 3’.
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
42. Karyotypes represent a display of the
chromosomes present in eukaryotic cells. The
following diagram shows a mutated human female
karyotype.
44. Which of the following diagrams represents
anaphase I of meiosis?
a. I
b. II
c. IV
d. V
45. The following question refers to this figure of a
simple metabolic pathway:
Which chromosomal mutation could be observed from
the diagram?
If A, B, and C are all required for growth, a strain that
is mutant for the gene-encoding enzyme A would be
able to grow on which of the following media?
a. Deletion
b. Duplication
c. Translocation
d. Insertion
a. minimal medium
b. minimal medium supplemented with nutrient A only
c. minimal medium supplemented with nutrient B only
d. minimal medium supplemented with nutrient C only
43. A DNA molecule has 160 base pairs and 20
percent of adenine nucleotides. How many cytosine
nucleotides are present in this molecule?
46. A plot of reaction rate (velocity) against
temperature for an enzyme indicates little activity at
0°C and 45°C, with peak activity at 35°C. The most
reasonable explanation for the low velocity at 0°C is
that
a. 20 cytosine nucleotides.
b. 60 cytosine nucleotides.
c. 32 cytosine nucleotides.
d. 48 cytosine nucleotides.
a. the hydrogen bonds that define the enzyme's active
site are unstable
b. the substrate becomes an allosteric regulator
c. there is too little activation energy available
d. the co-factors required by the enzyme system lack
the thermal energy required to activate the enzyme
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND
47. A researcher made an interesting observation about
a protein made by the rough endoplasmic reticulum
and eventually used to build a cell’s plasma
membrane. The protein in the membrane was actually
slightly different from the protein made in the ER. The
protein was probably altered in the
50. Green olives may be preserved in brine, which is a
20-30% salt solution. How does this method prevent
contamination by microorganisms?
a. Bacterial cell walls are shrivelled up by salt, causing
the cell to burst.
b. High salt concentrations lower the pH, thus
inhibiting the process of glycolysis.
c. High salt concentrations raise the pH, thus inhibiting
the process of glycolysis.
d. Bacteria can't survive in a hypertonic solution
because they lose water
a. Golgi apparatus
b. smooth endoplasmic reticulum
c. mitochondrion
d. nucleus
48. Imagine two solutions separated by a selectively
permeable membrane that allows water to pass, but not
sucrose or glucose. The membrane separates a 0.2molar sucrose solution from a 0.2-molar glucose
solution. With time, how will the solutions change?
a. Nothing happens because the two solutions are
isotonic to one another.
b. Water enters the sucrose solution because the
sucrose molecule is a disaccharide and thus larger than
the monosaccharide glucose.
c. Water leaves the sucrose solution because the
sucrose molecule is a disaccharide and thus larger than
the monosaccharide glucose.
d. The sucrose solution is hypertonic and will gain
water because the total mass of sucrose is greater than
that of glucose.
49. In glycolysis, the glucose splits into two 3-carbon
compounds: glyceraldehyde-3-phosphate and 1,3diphosphoglyceric acid. Which of the following
statements is true?
a. Only one of these compounds is used, which
explains the low efficiency of energy retrieval of cell
respiration.
b. Glyceraldehyde-3-phosphate and 1,3diphosphoglyceric acid are interconvertible.
c. One of the compounds goes through the Krebs cycle
and one becomes lactate.
d. All of the carbons in glyceraldehyde-3-phosphate
are immediately converted to carbon dioxide.
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INTERNATIONAL BIOLOGY OLYMPIAD – 2nd ROUND