1. The site of cellular respiration in the eukaryotic cells is called:
Mitochondria are membrane-bound cell organelles that generate most of the chemical
energy needed to power the cell's biochemical reactions.
C is the correct answer.
2. Eukaryotic organisms that play an important role in recycling nutrients are
Fungi are the major decomposers of nature; they break down organic matter which would
otherwise not be recycled, and make it possible for members of the other kingdoms to
be supplied with nutrients.
NB: bacteria are prokaryotic organisms.
B is the correct answer.
3. The following figure shows the structure of Ribonuclease enzyme before and after addition of urea.
Urea causes the denaturation of Ribonuclease enzyme. Thus, the latter completely
loses its activity.
B is the correct answer.
4. The germ cell of drosophila has 8 chromosomes, at the end of meiosis I each of the daughter cells obtained has
During anaphase I, homologous chromosomes are pulled apart and move to opposite ends
of the cell. At the end of meiosis I, each daughter cell has one copy of homologous
chromosomes (4 chromosomes) and sister chromatids are joined together at the level
of the centromere (each chromosome has two chromatids)
D is the correct answer.
5. Birds on the sides of the road take flight when a car approaches. After many cars pass without harming the birds, the birds do not fly anymore when a car approaches. This is an example of
Habituation occurs when animals are repeatedly exposed to the same stimuli, and eventually stop responding to that stimulus.
D is the correct answer.
6. The wings of the bird and wings of the bee are ________________ while the arm of the monkey and the flipper of the whale are _____________.
Structures that have no apparent function and appear to be residual parts from a past
ancestor are called vestigial structures. Thus B, D and E are incorrect.
Structures that are similar in unrelated organisms are called analogous structures
because they have separate evolutionary origins.
Structures that are similar in related organisms (similar embryonic origin) are
called homologous structures. They are inherited from a common ancestor.
So, C is the correct answer.
7. If an energy pyramid has 4 step trophic levels and the first step has 700 Kcal energy, then the amount of energy available at the highest trophic level is
10% of the energy at any trophic level is transferred to the next level. In a 4 step
trophic level, the energy is transferred 3 times from the first trophic level to the
fourth. That means that 10% of the energy is lost 3 times.
The amount of energy transferred from the lowest to the highest trophic level in a 4
step trophic level pyramid is 700 x (10%)3 = 0.7 kcal.
D is the correct answer.
8. Which of the following are intermolecular attractions?
Ionic and covalent bonds are intramolecular forces that hold atoms together within a
molecule (I and III). Hydrogen bonds and the Van der Waals force are intermolecular
forces are forces that exist between molecules (II and IV).
Thus, D is the correct answer.
9. In gel electrophoresis, DNA bands are separated according to
Gel electrophoresis is a technique used to separate DNA fragments according to their
size.
The solubility of fragments, the sequence of nucleotides and the composition of
nitrogenous bases don’t affect the separation in an electrophoresis technique.
D is the correct answer.
10. When a red blood cell is placed in a hypotonic solution
A hypotonic solution has a lower solute concentration than the cytosol.
When a red blood cell is placed in a hypotonic solution, water flows from the
solution into the cell due to osmosis. Therefore, the cell swells and bursts.
A is the correct answer.
11. Which of the following statements is NOT true about plant mitosis?
Centrioles do not exist in plant cells. In plants, mitosis takes place with
microtubules forming spindle fibers but without the help of centrioles.
In an animal cell, a cleavage furrow forms at the equatorial cortex after anaphase.
In plants, a rigid “cell plate” is formed in the middle of the cell to form two
cells.
C is the correct answer.
12. It occurs during strenuous exercise when the body cannot keep up with the demand of oxygen.
During strenuous exercise, your respiratory and cardiovascular systems cannot
transport enough oxygen to your muscle cells. Thus the alternative pathway is lactic
acid fermentation because it’s a type of anaerobic respiration that occurs when
muscles have too little oxygen for aerobic respiration to continue.
Alcoholic fermentation doesn’t occur in human cells.
Calvin cycle occurs in plant cells.
Glycolysis is the first step in the breakdown of glucose to extract energy for
cellular metabolism and it’s not an alternative pathway for aerobic respiration.
Krebs cycle (aerobic) is a central driver of cellular respiration.
Therefore B is the correct answer.
13. It occurs in the cytoplasm where one glucose molecule is broken apart into two molecules of pyruvate.
Glycolysis is a series of reactions that extract energy from glucose by splitting it
into two molecules of pyruvates.
During lactic acid fermentation pyruvate is transformed into lactate.
During alcoholic fermentation pyruvate is transformed into ethanol.
Krebs cycle occurs in the mitochondria.
Calvin cycle is a part of photosynthesis.
Therefore C is the correct answer.
14. A piece of DNA is amplified by
Polymerase Chain Reaction (PCR) is a fast technique used to "amplify" small segments
of DNA.
C is the correct answer.
15. Speciation is the result of
Speciation occurs when a group within a species separates from other members of its
species and develops its own unique characteristics.
Temporal isolation occurs when two populations differ in their periods of activity
or reproductive cycles. Geographic isolation occurs when two populations occupy
different habitats or separate niches within a common region. These two isolation
lead to speciation.
D is the correct answer.
16. Which of the following is NOT true about blood circulation?
The right part of the heart has deoxygenated blood while the left part of the heart
has oxygenated blood. The left atrium receives the oxygenated blood by the pulmonary
vein. The left ventricle receives the blood from the atrium and pumps it to all the
cells of the organism. While the right ventricle pumps the deoxygenate blood to the
lungs by the pulmonary artery.
E is the correct answer.
17. The gene for color blindness is carried on the non-homologous segment of sex chromosome X. The allele for color blindness is recessive. We can conclude that
The gene for color blindness is carried on the non-homologous segment of X
chromosome. Since males (XY) always only inherit the Y from their fathers, sons
cannot inherit color blindness from their fathers. Additionally, they only have one
X chromosome inherited from their mothers, and will then only require one copy of
the allele on X for the disease to affect them, even if the gene is recessive.
Assuming the allele for colorblindness is ‘d’, and the allele for ‘normal vision’ is
‘N’, we can conclude that all boys of a color blind mother
(Xd Xd) are affected
(Xd Y).
The chance for a carrier mother (XN Xd) and a normal father
(XNY) to have color blind
boys is 50% of boys (or 25% of children).
The chance for a carrier mother (XN Xd) and a color blind
father (Xd Y) to have color
blind girls is 50% of girls or 25% of children.
E is the correct answer.
18. Which statement is true about evidence of evolution?
Evidence for evolution comes from many different areas of biology: anatomy, molecular biology, biogeography, fossils and direct observation. Since I, II and III belong to those areas, E would be the correct answer.
19. A phase of cellular respiration that occurs in the cristae of mitochondria is
C is the correct answer
20- The amount of DNA in a cell obtained at the end of mitosis is 3.3 a.u.
At the end of mitosis (at the end of telophase) the amount of DNA is 3.3 a.u in each
daughter cell obtained. Each daughter cell will undergo interphase to be prepared
for another mitosis. Following one of these daughter cells, it will then start G1
with 3.3 a.u. After G1, during S phase, the amount of DNA will be duplicated to 6.6
a.u. (at the end of S phase). The cell will start phase G2 with 6.6 a.u..
A is the correct answer.
21. A couple with normal phenotype, who each have a family history of albinism, is expecting a baby. They had DNA fingerprinting in order to determine their genotypes and to know if the fetus is going to be albino or not. The following document shows the fingerprinting for this couple and their fetus.
From the above DNA fingerprinting, we can conclude that
Parents are normal but they carry the allele for albinism (recessive). The fetus
carries two alleles for albinism (double band). Therefore, he/ she will have the
condition.
C is the correct answer.
Questions 22 and 23
The following graph studies the activity of 2 different enzymes as a function of pH
22. From the above graph we can conclude that
According to the graph, enzyme A is active at an acidic pH (0 < pH < 4.5), and enzyme
B is active at an alkaline pH (6 < pH < 11).
The pH is highly acidic in the
stomach and becomes alkaline in the small intestine.
Therefore, enzyme A is
found in the stomach and enzyme B in the small intestine.
C is the correct
answer.
23. At pH 7,
According to the graph, the activity of enzyme A becomes 0% above pH = 4.5, thus its
activity is 0% at pH = 7. While enzyme B is active at 6 < pH < 11, therefore, at
pH=7 its activity is 50%.
C is the correct answer.
24. Chlorophyll b has _________ color and fucoxanthin has _________ color.
Each pigment has a specific color. Chlorophyll b has a green color and fucoxanthin
has a brown color.
B is the correct answer.
25. Nitrogen fixation is
There are five stages in the nitrogen cycle:
During the first stage, nitrogen moves from the atmosphere into the soil by nitrogen
fixing bacteria.
A is the correct answer.
26. Young birds follow the first moving object they see after they are born. This is
Imprinting is a simple and highly specific type of learning that occurs at a
particular life stage. For example, when ducklings hatch, they imprint on the first
adult animal they see, typically their mother.
C is the correct answer.
27. The clownfish live among the sea anemone tentacles. The clownfish protects the sea anemone by chasing away the attackers, in turn, the sea anemone protects the clownfish from predators. This type of symbiosis is
A, B and C are three types of symbiotic relationships. Among these relationships,
mutualism represents the relationship described in the given. Mutualism is defined
as: two organisms of different species "work together," each benefiting from the
relationship.
C is the correct answer.
Questions 28 and 29
The following figure shows one of the immune reactions.
28. Which is true concerning the above immune reaction?
Physical and chemical barriers form the first line of defense (innate immunity). Non-specific immunity forms the second line of defense and involves in general phagocytes. Specific immunity is the third line of defense and involves in general lymphocytes. The figure represents the phagocytosis of bacteria entering the skin due to an injury. B is the correct answer.
29. The rush of immune cells to the area of injury is the cause behind
Redness: due to the additional number of erythrocytes passing through the area.
Heat: caused by the increased movement of blood through dilated vessels into the
environmentally cooled extremities.
Swelling is the result of increased passage of fluid into the surrounding tissues,
the infiltration of cells and the deposition of connective tissues.
C is the correct answer.
30- The first carnivore in the food chain is
Every food chain starts with a primary producer (autotrophs like plants…). The second
trophic level, one step away from the primary producer, consists of organisms that
eat the producer. They are herbivores. The third trophic level, two steps away from
the primary producer, consists of organisms that eat herbivores. These are the first
carnivores.
B is the correct answer.
31. RNA processing involves
Transcription involves the formation of mRNA. Translation involves the formation of
peptide. Replication involves the formation of DNA. During RNA processing, introns
are removed and exons are joined together.
A is the correct answer.
Questions 32, 33, 34
The following pedigree shows the transmission of an autosomal recessive disease in a
certain family.
32. The genotype of individual I 1 is
I1 has affected children. The disease is recessive which means his children both have
two copies of the diseased allele, and got one copy from each parent.
Therefore, I1 has at least one copy of the disease allele ‘d’.
Since he is normal,
he also carries the normal allele ‘N’.
Therefore, his genotype is Nd.
B is the correct answer.
33. The genotype of individual II 2 is
II2 is affected and since the disease is recessive, II2 has both copies of the allele
and the genotype is definitely dd.
C is the correct answer.
34. The genotype of individual II 4 is
II4 is normal but with affected siblings so she definitely has one allele N inherited
from a parent. Her parents (I1 and I2) have affected children while being normal
themselves, which means they are both carriers with a Nd genotype. Doing the Punnett
square, II4 has 50% chance of being Nd and 25% chance of being NN.
Since we don’t know the phenotypes of her children, her genotype cannot be known
definitively.
E is the correct answer.
35. The following table shows different experiments applied on mice to determine the mode of action of the pituitary gland.
Lot of mice | Experiment | Result |
Lot 1 | Mice with normal pituitary gland | Normal development of the ovaries and the uterus and normal sexual cycle |
Lot 2 | Ablation (removal) of the pituitary gland | Decrease in the size of ovaries and uterus and stop of the sexual cycle |
Lot 3 | Ablation (removal) of the pituitary gland and injection of pituitary extracts (FSH and LH hormones) | Normal development of the ovaries and uterus and normal sexual cycle |
Lot 4 | Ablation (removal) of the pituitary gland and grafting it in the abdomen | Decrease in the size of ovaries and uterus and stop of the sexual cycle |
Lot 5 | Ablation (removal) of the pituitary gland and grafting it within its area | Normal development of the ovaries and uterus and normal sexual cycle |
The ablation of the pituitary gland causes a decrease in the size of the ovaries and
uterus and puts a stop to the sexual cycle. However, LH and FSH injections reverse
the effects of the ablation, return the sexual cycle to normal and both the ovaries
and uterus to normal development.
This means that the pituitary gland controls the development of the uterus, ovaries
and sexual cycle through secretion of LH and FSH in the bloodstream (I and III).
Since grafting the pituitary gland was only effective on site (not in the abdomen),
this gland only works in its site (IV).
Answer D is correct.
36. A woman visits the gynecologist who asks her to take some blood tests to monitor the level of some female hormones. The results showed that she has a constant low level of LH and FSH as well as low level of estrogen and progesterone. How can you explain the low level of progesterone and estrogen?
FSH hormone is a follicle-stimulating hormone that triggers, at a specific level, the
development of follicles. While an acute rise of LH hormone level triggers
ovulation.
C is the correct answer.
37. 35% of the nitrogenous bases that form a DNA segment is Guanine, then the percentages of Thymine is
%G = 35%; G is complementary to C therefore there are the same amount of G and C
bases: %G = %C → %C= 35%.
70% of the bases are G and C, which leaves 30% of the bases to be A and T. ( %A + %T
=
100 - 70 = 30%)
A is complementary to T, there is an equal amount of A and T bases => %A = %T = 30/2
=
15%.
A is the correct answer.
38. Sacs of hydrolytic enzymes that play a role in apoptosis of the cell are
A lysosome is a membrane-bound cell organelle that contains digestive enzymes. It’s
involved with various cell processes. If the cell is damaged beyond repair,
lysosomes can help it to self-destruct in a process called programmed cell death or
apoptosis.
E is the correct answer.
Questions 39, 40, 41
The following table shows different crosses made between pea plants of long and
short stems.
Cross | Result | |
I | Long stem plants x short stem plants | 100% long stem plants |
II | Short stem plants x short stem plants | 100% short stem plants |
III | Long stem plants x long stem plants | 60 long stem plants and 20 short stem plants |
IV | Long stem plants x short stem plant | 80 long stem plants and 79 short stem plants |
39. The genotype of parents in cross III
In cross I, 100% of the progeny have long stem, despite one parent having short stem;
we can conclude that long stem is dominant over short stem.
In cross III, parents are both long stem, so they each have at least one L allele.
However, 20 out of 80 of F1 in cross III have short stem (equivalent of 25% of F1).
The short stem allele being recessive, this means they have two copies of the s
allele and inherited one s allele from a parent and another s allele from the other
parent. Therefore, both parents are Ls.
B is the correct answer.
40- Cross IV is
A test cross can be used to determine whether a dominant phenotype (long stem) is
homozygous or heterozygous by crossing it with a known phenotype (short stem, ss,
recessive homozygous).
In cross IV, we obtained offsprings with 50% long stem plants (80) and 50% short
stem plants (79). Working with Punnett squares done where the long stem plants are
homozygous in one and hybrid in another, this phenotype percentage is obtained when
the long stem parent is Ls.
Therefore, the long stem plant is hybrid.
D is the correct answer.
41. If one of the plants obtained in cross II is crossed with a plant obtained from cross I, the result would be
Plants obtained from cross I have Ls genotype. Plants obtained from cross II have ss
genotype. The results of a cross between Ls and ss are as follow:
50% Ls and 50% ss
s (50%) | s (50%) | |
L (50%) | Ls (25%) | Ls (25%) |
s (50%) | ss (25%) | ss (25%) |
42. Raises blood calcium
Parathyroid hormone helps prevent low calcium levels by acting on the bones,
intestine and kidneys. It stimulates the release of calcium by bones into the
bloodstream, the absorption of calcium by the intestine and the conservation of calcium by the kidneys.
C is the correct answer.
43. Promotes water retention
Antidiuretic hormone stimulates water reabsorption at the level of kidneys.
B is the correct answer.
44. Lowers blood glucose
Glucagon and insulin are two hormones secreted by the pancreas to regulate blood
glucose level. Glucagon increases blood glucose level, while insulin decreases blood
glucose level.
E is the correct answer.
45. The dorsal root of the spinal nerve reaching the leg of a frog is destroyed, then a thermal stimulation is done on the frog’s leg.
Dorsal nerve roots carry sensory neural signals to the central nervous system. If
this root is destroyed, the frog will not feel the stimulation and the message can’t
reach the central nervous system. The frog wouldn’t be able to respond to the
stimulus via a reflex. Therefore, the frog will not feel the stimulation and won’t
react to it.
B is the correct answer.
46. When the plant is found in a dry area, the guard cells of the stomata
Guard cells use osmotic pressure to open and close stomata allowing plants to
regulate the amount of water and solutes within them. To open, the guard cell
becomes turgid, and to close, it becomes flaccid to stop transpiration. In a dry
area the plant conserves water and stops the transpiration.
Therefore E is the
correct answer.
47. Yellow grasshoppers are easily caught by predators, whereas green grasshoppers hide from predators by camouflage. This is called
The tendency of individuals that are better adapted to a particular environment to be more successful at surviving and producing offspring is called the survival of the fittest. Therefore the given example corresponds to the survival of the fittest (answer C).
48. Which is NOT true about Tundra?
Extremely cold climate, low biotic diversity, simple vegetation structure, short
season of growth and reproduction, energy and nutrients in the form of dead organic
material form some of tundra characteristics.
Answer C is correct
Questions 49 and 50
A water plant placed under bright light gives off bubbles of oxygen. The table below
shows the bubbles of oxygen produced per minute with respect to the distance of
plant from light (cm).
Distance from light (cm) | Bubbles of oxygen produced per minute |
40 | 5 |
30 | 8 |
20 | 22 |
10 | 39 |
49. What conclusion can we draw out of the given table?
In reference to the given table, when the amount of light decreases (the distance
from light increases), the amount of oxygen produced decreases. III is correct.
At a distance of 10 cm from the light, 39 bubbles of oxygen are produced per minute,
therefore at a distance of 15 cm from the light (distance increases), there would be
less bubbles of oxygen produced than at 10 cm (amount of oxygen produced decreases).
IV is correct.
Answer E is correct.
50- The reason behind the results obtained in the above table is
In the presence of light, plants undergo photosynthesis. During photosynthesis,
plants release oxygen. Thus, if the rate of photosynthesis increases, the amount of
oxygen produced increases. As the plant gets closer to light, the rate of
photosynthesis increases.
Answer A is correct.
51. The uptake of large dissolved molecules by the cell which requires energy is
The five alternatives are cellular processes. Among these processes, pinocytosis is
defined as a biological process of taking in fluid together with its contents into
the cell.
B is the correct answer.
52. The following bar graph shows the amount of food X in different organs of the digestive system.
We can conclude that the type of food X is
Digestive enzymes include:
In reference to the bar graph, we can conclude that the digestion of food X starts in the stomach (by pepsin enzyme) and continues in the small intestine (by trypsin enzyme). Therefore, the correct answer is D.
Questions 53, 54, 55, 56
A genetic condition is controlled by 2 alleles dominant “S” and recessive “s”. This
genetic condition affects only homozygous recessive individuals. In a population of
size 10,000, there are 36 affected individuals.
53. The frequency of q is
q corresponds to the frequency of the recessive allele ‘s’. The frequency of affected
individuals (ss) will then correspond to q2.
In a population of 10000, 36 individuals are affected, therefore:
q2 = 36/ 10000 = 3.6x10-3
q = √(q2) = 0.06= 6%
54. The frequency of p is
p + q = 1
q= 0.06 => p= 0.94 = 94%
55. The frequency of SS (homozygous dominant)
p corresponds to the frequency of the dominant allele ‘S’.
The frequency of
homozygous dominant individuals (SS) will then correspond to p2.
p2 = 0.94 = 0.8836= 88.36%
56. The frequency of ss (homozygous recessive)
q2 = 36/ 10000= 3.6 x 10-3 = 0.36%
57. The frequency of Ss (hybrid)
The frequency of hybrid individuals (Ss) corresponds to 2pq.
Applying Hardy-Weinberg equation: p2 + 2pq + q2 = 1
2pq = 1- 0.8836- 3.6 x 10-3 = 0.1128 = 11.28%
Questions 58 and 59
The following figure shows a cell during one of the phases of mitosis.
58. The phase of mitosis shown in the above figure is
During metaphase, chromosomes line up at the equatorial plate and sister chromatids
of each chromosome are captured by microtubules from opposite spindle poles.
The
correct answer is C.
59. By the end of mitosis of the above cell,
Each cell starts its cycle with interphase during which DNA is duplicated to obtain
chromosomes with 2 chromatids each ready to be separated during mitosis.
A cell that undergoes mitosis gives at the end of the division two identical
daughter cells. The cell shown in the figure started with 2n= 4 (4 chromosomes of 2
chromatids each) and will end with 2 cells each contains 2n= 4 (4 chromosomes of 1
chromatid each).
B is the correct answer.
60- The heterotroph hypothesis states
Bacteria are prokaryotes and have been the very first organisms to live on Earth. So
the first cells were prokaryotes. The primordial atmosphere was oxygen-free so there
were only anaerobic heterotrophic bacteria.
Answer A is correct.
61- Which of the following m-RNA strands is complementary to the given transcribed DNA strand?
CCGATAGATCAAA transcribed DNA strand
m-RNA is complementary to the transcribed DNA with the substitution of T by U.
Answer B is correct.
Questions 62 and 63
The cockroach is an insect known for its rapid escape. In order to identify the
stimulus that causes its escape and the sense organ that receives the stimulus, the
following experiments were performed.
Experiment | Result | |
Experiment 1 | Cockroach was placed in a glass box with a speaker producing the voice of the toad | Cockroach doesn’t run away |
Experiment 2 | Cockroach was placed alone in the glass box with air blowing on the cockroach | Cockroach runs away |
Experiment 3 | The antenna of the cockroach are covered with petroleum based jelly which acts as an insulator, and then blowing air on the cockroach | Cockroach runs away |
Experiment 4 | The legs of the cockroach are covered with petroleum based jelly then blowing air on the cockroach | Cockroach doesn’t run away |
62. Based on the above experiments, the stimulus that causes the cockroach to escape is
Experiment 1 shows that the voice of the predator didn’t cause the cockroach to run
away. Experiments 2, 3 and 4 show that only the feeling of air produced by the
predator causes the cockroach to run away.
C is the correct answer.
63. Based on the above experiments, the sense organs that receive the stimulus are
When the antennae are insulated with a petroleum based jelly, the cockroach runs away because it can still feel the air. Therefore, antenna are not the sense organs that receive the stimulus (air). But when the legs are covered with a petroleum based jelly, the cockroach does not run away because it cannot feel the air anymore. We can conclude that the legs are the sense organs that receive the stimulus (air) and the correct answer is B.
Questions 64 and 65
The following graph shows the changes in the total volume of ice in Antarctica since
1992.
64. Based on the data in the above graph, the greatest volume of ice was lost
Between 1994 and 2000 (6 years) the volume lost was 300 gigatons: 300/6= 50
Between 2002 and 2006 (4 years) the volume lost was 250 gigatons: 250/4= 62.5
Between 2008 and 2016 (8 years) the volume lost was 1500 gigatons: 1500/8= 187.5
Between 2002 and 2016 (14 years) the volume lost was 2000 gigatons: 2000/14= 142.85
Between 1994 and 2016 (22 years) the volume lost was 2500 gigatons: 2500/22= 113.63
The highest ratio corresponds to alternative C. Therefore, answer C is correct.
65. The most reasonable explanation for the loss of ice since 1994 is
Glaciers provide a scientific record of how climate has changed over time. Through studying glaciers, we gain valuable information about the extent to which the planet is rapidly warming. They provide scientists a record of how climate has changed over time. The rise of earth’s temperature causes the ice melting. Therefore, B is the correct answer.
Questions 66 and 67
The following graph shows the percentage composition of mitochondria in different
cells of the mouse.
66. Based on the data given in the above bar graph, the approximate percentage of mitochondria in the liver is
In reference to the bar graph E is the correct answer.
67. The amount of mitochondria in the left ventricle is more than that in the left atrium. How can you explain this variation in the amount of mitochondria?
The blood circulatory system delivers nutrients and oxygen to all cells in the body. Oxygenated blood coming from lungs arrives to the left atrium. The latter pumps blood to the left ventricle, and this action requires a small amount of energy, thus, a small amount of mitochondria. The left ventricle pumps blood to all body cells, and this action requires a high amount of energy, thus, a high amount of mitochondria. Therefore, D is the correct answer.
Questions 68 and 69
The following graph shows the logistic growth in the population of yeast.
68. The cause of rapid growth of phase A is probably due to
When resources are available (I), yeasts reproduce rapidly (II) and few yeasts die (IV). This leads to a rapid growth of the population. On the other hand, if the medium contains a low quantity of resources, many yeasts will die. Therefore, E is the correct answer.
69. The cause behind slow growth in phase B may be due to
Due to a high rate of reproduction in phase A, resources won’t be available for the
high number of yeasts (IV). Thus, the birth rate decreases and the death rate
increases (I). Therefore, the growth of yeast population slows down.
C is the
correct answer.
Questions 70 and 71
The following document shows different experiments done in vitro to study the
digestion of lipids.
70- Referring to the data in the above document, bile
Experiment 2 shows that bile emulsifies lipids but can’t digest them into fatty acid and glycerol. But when pancreatic juice is added (experiment 3), oil drops are digested into fatty acids and glycerol. Therefore, B is the correct answer.
71- The enzyme found in pancreatic juice that digests lipids is
D is the correct answer
Questions 72 and 73
The following document shows the Dichotomous key for leaves.
72. Referring to the above dichotomous key table, leaf A is_______ and leaf B is ________
Leaf A is magnolia (leaf edge is smooth) and leaf B is walnut (leaflets attached at
multiple points).
C is the correct answer.
73. Referring to the above dichotomous key, leaf F is __________ and leaf G is________
Leaf F is white oak (leaf edge is lobed) and leaf G is chestnut (leaflets attached at
one single point).
B is the correct answer.
74. To prove that DNA replication is semi conservative, an experiment was done on Lily roots. A lily root cell containing DNA labeled with heavy nitrogen (N15) is placed in a medium containing light nitrogen (N14) and is allowed to make 2 consecutive cell cycles.
The number of cells containing only light nitrogen (N14) at the end of 2 consecutive cell cycles is
The first cell contains DNA labeled with heavy nitrogen (N15). After the first cycle, two cells are obtained containing a hybrid DNA. After the second cycle, four cells are obtained: 2 cells contain hybrid DNA (labeled with both N15 and N14) and 2 cells contain light DNA (N14). Therefore, A is the correct answer.
75. The following figure shows a beaker containing water and solute separated by a selectively permeable membrane. What would happen after a few minutes?
Water molecules are capable to flow through a selectively permeable membrane. This process is called osmosis. It takes place from a hypotonic medium (low on solute) to a hypertonic medium (high on solute), thus, from medium 2 to medium 1. Therefore, C is the correct answer
Questions 76 and 77
The graph below shows the variation of the amount of oxygen used by the cell as a
function of the intensity of exercise.
76. Referring to the data from the above graph, we can conclude that
As the exercise intensity increases, oxygen consumption increases then remains constant. The body tries to find another energetic source so it relies on lactic acid fermentation to supply the muscles with more energy. Therefore, A is the correct answer.
77. How can you explain the plateau (constant) of oxygen consumption.
Aerobic pathway supplies the body with a high amount of energy. Oxygen consumption
increases as a function of exercise intensity. At a high level of intensity, red
blood cells become saturated with oxygen and can’t bind more oxygen molecules. Thus
the body relies on other energetic pathways (lactic acid fermentation).
B is the correct answer.
78. An experiment was done with different amounts of fertilizers to determine their effects on the productivity of green plants. The results obtained are shown in the table below:
Amount of fertilizers (a.u) | 0 | 60 | 80 | 90 |
Productivity of green plants (a.u) | 3 | 4.6 | 7 | 5.2 |
What conclusion can be drawn out of this experiment?
In reference to the table, the productivity of green plants increases to reach 7 a.u. as the amount of fertilizers increases from 0 to 80 a.u. Above 80 a.u., the productivity decreases to 5.2 a.u. Thus, fertilizers are a necessary component for plant growth but in a specific quantity. Therefore, D is the correct answer.
79. An experiment was done on nerve fibers to study the affect on transmission of the nerve impulse. Two variables were examined: variable 1 is the diameter of the nerve fiber and variable 2 is myelinated vs. unmyelinated nerve fiber. The speed of conduction of the nerve impulse was recorded as shown in the document below.
Diameter of nerve fiber (µm) | 1 | 2 | 3 | 4 |
Speed of conduction of nerve impulse in myelinated nerve fiber (a.u) | 1 | 5 | 8 | 15 |
Speed of conduction of nerve impulse in non myelinated nerve fiber (a.u) | 0.2 | 1 | 3 | 5 |
Referring to the data of the above table, we can conclude that
In reference to the table, the speed of conduction increases as the diameter of the
nerve fiber increases (from 1 to 15 a.u. in myelinated nerve fibers, and from 0.2 to
5 a.u. in non-myelinated nerve fibers). With the same diameter (1 µm), the speed of
conduction of nerve impulses is faster in myelinated nerve fibers (1 a.u. in
myelinated nerve fibers and 0.2 a.u. in non-myelinated nerve fibers).
A is the correct answer.
80- The plant in the figure below shows a very low amount of transpiration.
The factor that affects transpiration in the above figure is.
The figure shows one factor that affects transpiration: humidity, in the form of
water droplets.
D is the correct answer.