1. The site of cellular respiration in the eukaryotic cells is called:

A) Smooth endoplasmic reticulum
B) Golgi bodies
C) Mitochondria
D) Ribosomes
E) Chloroplasts

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

I.   Protists
II.  Fungi
III. Plants
IV.  Bacteria
A) I only
B) II only
C) I and III
D) II and IV
E) I, II and IV

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.


Which statement is true about Ribonuclease enzyme?

    A) Urea causes the denaturation of the Ribonuclease enzyme but the enzyme maintains its activity.
    B) Urea causes the denaturation of the Ribonuclease enzyme and the enzyme loses its activity.
    C) Urea causes the denaturation of the Ribonuclease enzyme and the enzyme maintains 50 % of its activity.
    D) Urea causes a modification in the sequence of amino acids of the Ribonuclease enzyme to make it inactive.
    E) Urea causes a modification in the number of amino acids of the Ribonuclease enzyme without affecting its activity.

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

    A) 8 chromosomes made up of 2 chromatids.
    B) 8 chromosomes made up of 1 chromatid.
    C) 16 chromosomes made up of 2 chromatids.
    D) 4 chromosomes made up of 2 chromatids.
    E) 4 chromosomes made up of 1 chromatid.

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

    A) Classical conditioning
    B) Operant conditioning
    C) Imprinting
    D) Habituation
    E) Insight learning

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 _____________.

A) Homologous structures, analogous structures
B) Analogous structures, vestigial structures
C) Analogous structures, homologous structures
D) Homologous structures, vestigial structures
E) Vestigial structures, homologous structures

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

    A) 700 kcal
    B) 70 kcal
    C) 7 kcal
    D) 0.7 kcal
    E) 0.07 kcal

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?

I.   Ionic bonds
II.  Hydrogen bonds
III. Covalent bonds
IV.  Van der Waals force
A) I only
B) I and III
C) II and III
D) II and IV
E) II, III and IV

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

A) Their solubility in agarose gel
B) The sequence of nucleotides
C) Composition of nitrogenous bases
D) Size of fragment
E) Size of fragment and sequence of nucleotides

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) The cell swells and bursts since water flows into the cell.
B) The cell shrinks since the water flows outside the cell.
C) The cell maintains its shape since water diffuses in and out of the cell.
D) The cell swells and bursts since water flows outside the cell.
E) The cell shrinks since water flows into the cell.

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?

I.   A cleavage furrow is formed in the middle of the cell during cytokinesis.
II.  A cell plate is formed in the middle of the cell during cytokinesis.
III. The centromere divides and sister chromatids separate to opposite poles of the cell.
IV.  Centrioles change into asters and migrate to opposite poles of the cell during prophase.
A) I only
B) II only
C) I and IV
D) II and IV
E) II, III and IV

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.

A) Alcoholic fermentation
B) Lactic acid fermentation
C) Glycolysis
D) Krebs cycle
E) Calvin cycle

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.

A) Alcoholic fermentation
B) Lactic acid fermentation
C) Glycolysis
D) Krebs cycle
E) Calvin cycle

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

    A) Gel electrophoresis
    B) DNA recombination
    C) PCR
    D) Addition of restriction enzymes
    E) Transcription

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

I.   Random mating
II.  Temporal isolation
III. Geographic isolation
IV.  No mutation in the population
    A) I only
    B) II only
    C) I and IV
    D) II and III
    E) I, II and III

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?

    A) Deoxygenated blood is pumped by the right ventricle to the lungs through pulmonary artery.
    B) Oxygenated blood reaches the left atrium by pulmonary vein.
    C) All arteries carry oxygenated blood except the pulmonary artery.
    D) All veins carry deoxygenated blood except the pulmonary vein.
    E) The right part of the heart has oxygenated blood while the left part has deoxygenated blood.

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

I.   Sons cannot inherit color blindness from their fathers.
II.  The chance for a carrier mother and a normal father to have color blind boys is 50 % of children.
III. A color blind mother and a normal father have all their boys having color blindness.
IV.  The chance for a carrier mother and a color blind father to have color blind girls is 25 %.
    A) I only
    B) III only
    C) I, II and III
    D) II, III, and IV
    E) I, III, and IV

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?

I.   Closely related organisms go through similar stages in their embryonic development.
II.  Closely related aerobic organisms have cytochrome c with an identical chain of amino acids.
III. Organisms that have common ancestors have common biochemical pathways.
    A) I only
    B) II only
    C) I and II
    D) II and III
    E) I, II and III

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

    A) Glycolysis
    B) Krebs cycle
    C) Electron transport chain
    D) Calvin cycle
    E) Fermentation

  • Glycolysis occurs in the cytoplasm.
  • Krebs cycle occurs in the mitochondrial matrix.
  • Electron transport chain occurs in the inner membrane of the mitochondria.
  • Calvin cycle occurs in the chloroplast.
  • Fermentation occurs in the cytoplasm.

C is the correct answer


20- The amount of DNA in a cell obtained at the end of mitosis is 3.3 a.u.

    A) The amount of DNA in this cell is 6.6 a.u during G2 of Interphase and 3.3 a.u during G1
    B) The amount of DNA in this cell is 6.6 a.u during G1 of Interphase and 3.3 a.u during G2 .
    C) The amount of DNA in this cell is 6.6 a.u during G1 and S – phase of Interphase.
    D) The amount of DNA in this cell is 3.3 a.u during G1 and S – phase of Interphase.
    E) The amount of DNA in this cell is 6.6 a.u during G2 and beginning of S – phase of Interphase.

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

    A) The allele for albinism is dominant and the fetus will have the condition
    B) The allele for albinism is dominant and the fetus will not have the condition
    C) The allele for albinism is recessive and the fetus will have the condition
    D) The allele for albinism is recessive and the fetus will not have the condition
    E) The allele for albinism is recessive and masks the normal allele

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

    A) Enzyme A is found in the intestine and enzyme B is found in the stomach
    B) Enzyme A is found in the mouth and enzyme B is found in the intestine
    C) Enzyme A is found in the stomach and enzyme B is found in the intestine
    D) Enzyme A is found in the stomach and enzyme B is found in the mouth
    E) Both enzymes A and B are found in the stomach

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,

    A) The amount of substrate obtained by enzymes A and B is 50 % and 0% respectively
    B) The amount of substrate obtained by enzymes A and B is 10 % and 0% respectively
    C) The amount of substrate obtained by enzymes A and B is 0 % and 50% respectively
    D) The amount of substrate obtained by enzymes A and B is 0 %
    E) The amount of substrate obtained by enzymes A and B is 100%

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.

    A) Yellow, green
    B) Green, brown
    C) Orange, red
    D) Green, yellow
    E) Green, orange

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

    A) The process by which nitrogen fixing bacteria remove nitrogen from the atmosphere
    B) The process by which consumers convert nitrogen gas to amino acids
    C) The process by which primary producers release nitrogen to the atmosphere
    D) The process by which bacteria convert nitrogen gas to amino acids
    E) The process by which primary producers convert ammonia to nitrogen gas

There are five stages in the nitrogen cycle:

  1. Nitrogen fixation
  2. Mineralization
  3. Nitrification
  4. Immobilization
  5. Denitrification

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

    A) Classical conditioning
    B) Operant conditioning
    C) Imprinting
    D) Insight learning
    E) Habituation

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) Parasitism
    B) Commensalism
    C) Mutualism
    D) Predation
    E) Speciation

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?

    A) It is the 1st line of defense and the cells involved are phagocytes.
    B) It is the 2nd line of defense and the cells involved are phagocytes.
    C) It is the 3rd line of defense and the cells involved are phagocytes.
    D) It is the 1st line of defense and the cells involved are B-lymphocytes.
    E) It is the 2nd line of defense and the cells involved are B- lymphocytes.

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

    A) Redness
    B) Heat
    C) Swelling
    D) Heat and swelling
    E) Redness and heat

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

    A) One step from the primary producer
    B) Two steps from the primary producer
    C) Three steps from the primary producer
    D) Two steps from the herbivore
    E) Three steps from the herbivore

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

    A) Removal of introns
    B) Formation of m-RNA
    C) Formation of peptide
    D) Formation of DNA
    E) Removal of exons

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

    A) NN
    B) Nd
    C) dd
    D) Maybe NN or Nd
    E) Maybe Nd or dd

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

    A) NN
    B) Nd
    C) dd
    D) Maybe NN or Nd
    E) Maybe Nd or dd

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

    A) NN
    B) Nd
    C) dd
    D) Maybe Nd or dd
    E) Maybe NN or Nd

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

By referring to experiments done on mice of five lots, we can conclude
I.   The pituitary gland favors the development of ovaries and uterus by secreting FSH and LH
II.  The pituitary gland and the ovaries do not communicate by blood
III. The pituitary gland and the ovaries communicate by blood
IV.  The pituitary gland works only in its site
    A) I only
    B) I and II
    C) II and III
    D) I,III and IV
    E) I, II and IV

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?

I.   The low level of FSH didn’t cause follicles to develop so they secreted low amounts of estrogen
II.  The low level of LH didn’t trigger ovulation then corpus luteum is not formed to secrete estrogen and progesterone
III. The low level of LH didn’t cause the follicles to develop so they secreted low amounts of estrogen
    A) I only
    B) II only
    C) I and II
    D) II and III
    E) I, II, and III

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

    A) 15 %
    B) 35%
    C) 30%
    D) 70 %
    E) 50%

%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) Ribosomes
    B) Mitochondria
    C) Nucleolus
    D) Vacuole
    E) Lysosomes

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

    A) Both have LL
    B) Both have Ls
    C) Both have ss
    D) One has LL and the other has Ls
    E) One has Ls and the other has ss

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) A test cross done between pure long and pure short stem plants
    B) A test cross done between pure long and hybrid short stem plants
    C) Self cross done between hybrid long and hybrid short stem plants
    D) A test cross done between hybrid long and pure short stem plants
    E) A test cross done between hybrid long and hybrid short stem plants

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

    A) 75 % long stem plants (LL) and 25 % short stem plants (ss)
    B) 50 % long stem plants (Ls) and 50 % short stem plants (ss)
    C) 50 % long stem plants (LL) and 50 % short stem plants (ss)
    D) 100 % plants having long stem with genotype (Ls)
    E) 100 % plants having short stem with genotype (ss)

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

    A) Glucagon hormone
    B) Antidiuretic hormone
    C) Parathyroid hormone
    D) Thyroid hormone
    E) Insulin hormone

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

    A) Glucagon hormone
    B) Antidiuretic hormone
    C) Parathyroid hormone
    D) Thyroid hormone
    E) Insulin hormone

Antidiuretic hormone stimulates water reabsorption at the level of kidneys.
B is the correct answer.

44. Lowers blood glucose

    A) Glucagon hormone
    B) Antidiuretic hormone
    C) Parathyroid hormone
    D) Thyroid hormone
    E) Insulin hormone

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.

    A) The frog will feel the stimulation and move its leg accordingly.
    B) The frog will not feel the stimulation and it won’t react to it.
    C) The frog will not feel the stimulation and it will move its leg.
    D) The frog will feel the stimulation but it will be paralyzed.
    E) The frog will feel and move its other leg accordingly.

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

    A) Become turgid and open to increase transpiration
    B) Become turgid and close to stop transpiration
    C) Maintain their shape and close to stop transpiration
    D) Become flaccid and open to increase transpiration
    E) Become flaccid and close to stop transpiration

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

    A) Adaptation
    B) Temporal isolation
    C) Survival of the fittest
    D) Artificial selection
    E) Speciation

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?

I.   It includes wide variety of animal species
II.  It is called permafrost
III. It is dominated by conifers
IV.  t is located in far northern parts of North America, Europe and Asia
    A) I only
    B) III only
    C) I and III
    D) II and IV
    E) I , III and IV

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?

I.   The amount of oxygen produced by the plant increases as the plant moves away from light
II.  The amount of oxygen produced by the plant is not affected by the distance of the plant away from light
III. The amount of oxygen produced by the plant decreases as the plant moves away from light
IV.  The amount of oxygen bubbles produced when the plant is placed at a distance of 15 cm is less than 39 bubbles per minute
    A) II only
    B) III only
    C) I and IV
    D) II and IV
    E) III and IV

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

I.   As the plant gets closer to light, the rate of photosynthesis increases thus producing more oxygen bubbles
II.  As the plant gets closer to light, the rate of photosynthesis increases thus absorbing more oxygen
III. As the plant moves away from light, the rate of photosynthesis increases thus producing more oxygen bubbles
IV.  As the plant moves away from light, the rate of photosynthesis decreases thus producing more oxygen bubbles
    A) I only
    B) II only
    C) III only
    D) IV only
    E) II and III

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

    A) Phagocytosis
    B) Pinocytosis
    C) Exocytosis
    D) Osmosis
    E) Diffusion

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

    A) Sucrose
    B) Starch
    C) Lipids
    D) Protein
    E) Maltose

Digestive enzymes include:

  • amylase produced in the mouth to break down large starch molecules.
  • pepsin produced in the stomach to break down proteins.
  • trypsin produced in the pancreas and released in the small intestine to break down proteins.
  • lipase produced in the pancreas and released in the small intestine to break down fats.

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

    A) 94 %
    B) 6%
    C) 0.36%
    D) 88.36%
    E) 11.28 %

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

    A) 6 %
    B) 0.36 %
    C) 11.28 %
    D) 94 %
    E) 88.36 %

p + q = 1
q= 0.06 => p= 0.94 = 94%


55. The frequency of SS (homozygous dominant)

    A) 0.36%
    B) 88.36%
    C) 94 %
    D) 6%
    E) 11.28 %

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)

    A) 88.36%
    B) 11.28 %
    C) 94 %
    D) 6%
    E) 0.36%

q2 = 36/ 10000= 3.6 x 10-3 = 0.36%


57. The frequency of Ss (hybrid)

    A) 6%
    B) 0.36%
    C) 88.36%
    D) 11.28 %
    E) 94 %

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

    A) Telophase
    B) Anaphase
    C) Metaphase
    D) Prophase
    E) Cytokinesis

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,

    A) 2 daughter cells having 4 chromosomes of 2 chromatids are produced
    B) 2 daughter cells having 4 chromosomes of 1 chromatid are produced
    C) 4 daughter cells having 2 chromosomes of 2 chromatids are produced
    D) 2 daughter cells having 2 chromosomes of 1 chromatid are produced
    E) 2 daughter cells having 8 chromosomes of 1 chromatid are produced

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

    A) First cells on Earth were anaerobic heterotrophic prokaryotes
    B) First cells on Earth were aerobic heterotrophic prokaryotes
    C) First cells on Earth were anaerobic heterotrophic eukaryotes
    D) First cells on Earth were aerobic autotrophic prokaryotes
    E) First cells on Earth were aerobic autotrophic eukaryotes

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

    A) CCGAUAGAUCAAA
    B) GGCUAUCUAGUUU
    C) GGCTATCTAGTTT
    D) GGCUAUCTAGTUU
    E) GGCUAAGAAGUUU

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

    A) Seeing the predator
    B) Smelling the odor of the predator
    C) Feeling the air produced by the movement of the predator
    D) Hearing the voice of the predator
    E) Hearing the voice of the predator and feeling the air produced by the predator’s movement

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

    A) The antennae
    B) The legs
    C) The wings
    D) The eyes
    E) The ears

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

    A) Between 1994 and 2000
    B) Between 2002 and 2006
    C) Between 2008 and 2016
    D) Between 2002 and 2016
    E) Between 1994 and 2016

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

    A) An increase in the total productivity of the world’s oceans
    B) A gradual rise in the earth’s average temperature
    C) An increase in the total amount of ice at earth’s pole
    D) An increase in the sun rays
    E) An increase in the human activity at the polar areas

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

    A) around 20 %
    B) around 12%
    C) around 10 %
    D) around 15 %
    E) around 18%

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?

I.   The size of the ventricle is bigger than the size of atrium.
II.  The left ventricle pumps the blood with larger force to all parts therefore it needs a larger amount of energy produced by cellular respiration.
III. The left atrium pumps blood to the left ventricle with smaller force than that of the left ventricle therefore it needs less amount of energy produced by cellular respiration.
IV.  The left atrium pumps the blood with big force to all body parts therefore it needs a big amount of energy produced by cellular respiration.
    A) I only
    B) I and II
    C) I and IV
    D) II and III
    E) I, II and III

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

I.   Unlimited resources
II.  Yeasts reproduce rapidly
III. Many yeast cells die
IV.  Few yeast cells die
    A) II only
    B) I and II
    C) I and III
    D) II and IV
    E) I, II and IV

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

I.   Decrease in birth rate and increase in death rate
II.  Increase in birth rate and decrease in death rate
III. Increase in nutrient quantity in the medium
IV.  Depletion of nutrients in the medium
    A) I only
    B) I and III
    C) I and IV
    D) II and III
    E) II and IV

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

    A) digests lipids into fatty acids and glycerol
    B) emulsifies lipids to make it easier for the enzymes to digest lipids
    C) digests and emulsifies lipids
    D) maintains the temperature of the tube
    E) has no effect on the digestion of lipids

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

    A) Pancreatic amylase
    B) Trypsin
    C) Maltase
    D) Lipase
    E) Pepsin

  • Pancreatic amylase: breaks down starch molecules
  • Trypsin: breaks down proteins in the small intestine
  • Maltase: breaks down maltose molecules
  • Lipase: breaks down fats
  • Pepsin: breaks down proteins in the stomach

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 ________

    A) White oak, chestnut
    B) Walnut, magnolia
    C) Magnolia, walnut
    D) Magnolia, chestnut
    E) White oak, walnut

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________

    A) Magnolia, walnut
    B) White oak, chestnut
    C) Magnolia, chestnut
    D) White oak, walnut
    E) Walnut, chestnut

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

    A) 2 cells
    B) 4 cells
    C) 8 cells
    D) 16 cells
    E) 32 cells

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?

    A) Osmosis of water from medium 1 to medium 2 takes place.
    B) Diffusion of the solute from medium 1 to medium 2 takes place.
    C) Osmosis of water from medium 2 to medium 1 takes place.
    D) Diffusion of the solute from medium 2 to medium 1 takes place.
    E) Nothing changes.

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

I.   As the exercise becomes difficult, the body relies more on lactic acid fermentation.
II.  Low intensity exercises do not require increased oxygen uptake.
III. Difficult exercise requires additional oxygen in order to generate extra ATP for muscles.
IV.  During exercise the body relies only on the aerobic pathway.
    A) I only
    B) III only
    C) I and II
    D) II and IV
    E) I, II and IV

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.

I.   Very high intensity exercise does not require increased amount of oxygen.
II.  Red blood cells are saturated with oxygen and can’t bind more oxygen molecules
III. The body uses CO2 instead of O2 when doing very high intensity exercises.
    A) I only
    B) II only
    C) III only
    D) I and II
    E) II and III

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?

I.   Fertilizers are not a necessary component for the growth of green plants.
II.  Fertilizers are needed regardless of its quantity.
III. Suitable amount of fertilizers increases the productivity of green plants.
IV.  The excessive use of fertilizers has a negative effect on the productivity of green plants.
    A) I and II
    B) II and III
    C) II and IV
    D) III and IV
    E) II, III and IV

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

I.   The speed of conduction of the nerve impulse increases as the diameter of the nerve fiber increases.
II.  The speed of conduction of nerve impulse decreases as the diameter of the nerve fiber increases.
III. The speed of conduction of nerve impulse is faster in a myelinated nerve fiber.
IV.  The speed of conduction of nerve impulse is faster in a non- myelinated nerve fiber.
    A) I and III
    B) I and IV
    C) II and III
    D) II and IV
    E) I, III and IV

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.

    A) Temperature
    B) Amount of water in soil
    C) Sunlight
    D) Humidity
    E) Temperature and humidity

The figure shows one factor that affects transpiration: humidity, in the form of water droplets.
D is the correct answer.