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Selasa, 21 Juni 2011

16th International Biology Olympiad





THEORY EXAMINATION
Part 1

Total time available:  2.5 hours (150 minutes)

Total points available:  ~80



GENERAL INSTRUCTIONS

Please check that you have the appropriate examination papers and answer sheets.

It is recommended that you manage your time in proportion to the points allotted for each question.


IMPORTANT

Use the answer sheets provided to record your answers.

Ensure your name and three digit code number is written on the top of each page of the answer sheets. 

Use the 2B pencil provided to fill in the mark sense icons on the answer sheet.

.



1.        Various forces are important to the interactions contributing to the tertiary structure of a protein.  The figure below is a diagram showing several possible interactions.  Please match the numbered interactions with correct names. (1 point)


    1. hydrogen bond
    2. Hydrophobic interaction
    3. Peptide bond
    4. Disulphide bond
    5. Ionic bond
Interactions
Answer: A-E
1

2

        3

        4

2. Which of the following is/are NOT correct about cytokinesis of plant cells? (1 point)
  (1) Plant cells form cell plates
  (2) Cytokinesis can start during mitosis
  (3) Plant cells have contracting ring
(4) Membrane fusion connects cell plate and cytoplasmic membrane of mother cell.
(5) Golgi apparatus does not participate cytokinesis of a plant cell until two daughter cells are formed.

  1. 1, 2, 4,
  2. 3
  3. 3, 5
  4. 4, 5
  5. 4

3.  DNA ligase is an important enzyme that connects DNA fragments.  Which of the following is/are true about DNA ligase? (1 point)

1) It is important to DNA replication process
2) It is important in molecular cloning
3) It requires DNA fragments have sticky ends
4) It could cut DNA molecules in the presence of ATP and Mg2+.
5) It requires ATP for its function because the 3’-hydroxyl group of a DNA fragment needs to be phosphorylated before the DNA molecules could be ligated.

A.    1, 2, 3
B.     2, 3, 5.
C.     1, 2
D.    1, 5
E.     1, 2, 4

Questions 4-6: Checkpoints in the cell cycle are very important in regulation of cell cycle.  The following three questions are about the cells cycles and checkpoints.

4. Two animal cells at different phase in cell cycle can be induced to fuse to form a single cell with two nuclei.  This system provides a very useful tool for studying cell cycle.
Which of the following is correct? (1 point)
  A. When a cell in M phase is fused with a cell in G1 phase, the nucleus in M phase stops the process of mitosis.
  B. When a cell in M phase is fused with a cell in G2 phase, the nucleus in G2 phase starts mitosis process.
  C. When a cell in G2 phase is fused with a cell in G1 phase, both nuclei start mitosis process.
  D. When a cell in M phase is fused with a cell in G1 phase, the nucleus in G1 phase starts DNA synthesis.
  E. When a cell in M phase is fused with a cell in G1 phase, the nucleus in M phase stops mitosis process.


5. Which of the following are true about checkpoints in cell cycle? (1 point)
(1) If a cell in G1 phase does not receive a signal at the G1 checkpoint, the cell usually goes into G0 phase.
      (2) A cell must receive a signal at G2 checkpoint to go into mitosis.
      (3) A cell must receive a signal at M checkpoint to go into mitosis.
      (4) The protein factors that control checkpoints in cell cycle are mostly present in nuclei.
      (5) Cell cycle in unicellular organisms does not have checkpoints.

A.    1, 2
B.     1, 3,
C.     1, 3, 4
D.    2, 3, 4
E.     1, 5

 6. In cloning the first mammal, researchers used mammary cell as a donor of nucleus and fused it with a nucleus-less egg.  Which of the following is correct? (1 point)
       A. The mammary cell was in G1 phase
       B. The mammary cell was in G2 phase
       C. The mammary cell was in S phase
       D. The mammary cell was in M phase
       E. The mammary cell was in G0 phase



7.        Cyanobacteria (blue-green algae) are a group of very important bacteria that perform photosynthesis.  Which of the following is/are true about cyanobacteria. (1 point)

(1)They are gram-negative bacteria
(2) They produce oxygen in photosynthesis
(3) All cyanobacteria can fix nitrogen
(4) Some cyanobacteria can live with fungi symbiotically
 (5) The blue-green color of cyanobacteria comes from chlorophyll

  1. All are correct
  2. 1, 2, 3, 4
  3. 1, 2, 3
  4. 1, 2, 4      
  5. 1, 2

Questions 8-9 are about biotechnology of transgenic organisms or genetically modified organisms (GMO).

8.        In creating “golden rice” that produces betacarotenes in rice kernels, the genes responsible for betacarotene synthesis are transformed.  Which of the following is/are true? (1 point)
(1)   The researcher used normal rice for transformation
(2)   The researcher used Ti plasmid for transformation
(3)   The researcher used a dicot plant for transformation first followed by crossing between the dicot plant and rice plant
(4)   Golden rice has a higher nutritional value than normal rice
(5)   Beside Agrobacterium, the researcher also used another bacterium, Escherichia coli in construction of transforming vectors.

A.    All are correct
B.     1, 2, 4, 5               
C.     1, 2, 3
D.    1, 2
E.     1, 3, 4, 5

9.      When a DNA fragment under control of a promoter was transformed into tobacco plants with Ti plasmid, the transgenic plants showed a lower activity of CO2 fixation.  Biochemical examination found that the transgenic plant had a lower amount of Rubisco, a key enzyme for Calvin cycle.  Which of the following is/are likely to be the reason(s) for the phenotype? (1 point)

(1)   The DNA fragment was transformed into chloroplasts and resulted in interference with chloroplast transcription.
(2)   Genetic exchange between the transformed DNA fragment and host chromosomal DNA resulted in insertion of Ti plasmid into chromosome, leading to a lower expression of Rubisco genes
(3)   The transformed DNA fragment interfered normal transcription of the gene encoding large subunit of Rubisco
(4)   The transformed DNA fragment interfered normal transcription of the gene encoding small subunit of Rubisco
(5)   The transformed DNA fragment encoded a cytoplasmic protein that prevents Rubisco activation by bicarbonate.

A.    1, 2, 5
B.     1, 3, 4
C.     1, 4,
D.    4,
E.     3    

10.  Which of the following is/are true about endosymbiosis? (1 point)
(1)   Both plastid and lysozome are products of endosymbiosis
(2)   A eukaryotic cells could engulf another eukaryotic cells to establish symbiotic relationship
(3)   Cyanobacteria are ancestors of plastids and mitochondria
(4)   Cyanobacteria lost their chlorophyll b gene in endosymbiosis.
(5)   Flagella of some eukaryotic cells are derived from cyanobacteria

A.    1, 3, 5
B.     1, 2
C.     2, 4
D.    2     
E.     4
11. Which of the following graphs correctly displays the relationship of blood flow velocity in humans as the blood flows from the aorta à arteries à arterioles à capillaries à venules à veins à venae cavae: (1 point)     B

A

B

C
D

E


12. Lowering the level of a hedge with a hedge trimmer stimulates the hedge to become bushy because: (1 point)
A.  It stimulates the production of ethylene gas.
B.  Removing the apical meristems makes more auxin, which stimulates lateral branch buds to grow.
C.  Removing the apical meristems makes less ethylene, which stimulates lateral branches to grow
D.  Removing the apical meristems results in less auxin, which then allows lateral branches to grow.
E.  Removing the lateral buds results in apical dominance under the influence of cytokinins
13.    Which of the following is/are true about telomeres? (1 point)

(1) Telomeres are present in all eukaryotic DNA
(2) Telomeres are present in bacterial plasmids
(3) Telomeres are required for replication fork formation
(4) Telomeres are unique sequences of eukaryotic chromomes
(5) Telomeres are required for maintaining chromosomal length

A.    1, 3, 5,
B.     3, 4, 5
C.     4, 5
D.    2
E.     3

14. For terrestrial and most aquatic environments, neither animal nor plant life could exist without the metabolic "services" provided by: (1 point)
A.  chemoheterotrophs
B.  extremophile archaeans
C.  Fungi      
D.  Homo sapiens
E.   Fertilizer



15. The inner ear of humans, and most other mammals, is sensitive to body position and balance.  What organ(s) is/are responsible for this? (1 point)
A.  cochlea
B.  cochlea and basilar membrane
C.  semicircular canals
D.  semicircular canals and cochlea
E.   semicircular canals, utricle, and saccule       

16. Flukes are often parasites in or on another animals.  They could cause diseases in human beings.  Blood fluke (Schistosoma mansoni) is a parasitic trematode that infects men.  Which of the followings is NOT true about its life cycles. (1 point)

A.      There are two types of larvae in the fluke
B.       It reproduce asexually in the human host     
C.       The larvae need water to swim
D.      Its infection of human being is through skin
E.       An intermediate host is often required for completion of their life cycle.

17. In animal behavior a sing stimulus could trigger a fixed action pattern (FAP).  Which of the following is Not an example of sign stimulus-FAP? (1 point)

A.     Some moths fold their wings and drop to the ground when they detect an ultrasonic signal from bats.
B.     A wasp finds its nest according to the surrounding objects.
C.     A newly hatched bird cheeping loudly in begging for food when its parent returns to nest.     
D.     Breeding mayflies lay eggs when they detect water.

18. Some crows feed on mollusks.  The crows grasp the prey and fly upwards to certain height before they drop the preys onto a rock to break the shells. If the shell wasis not broken by the first drop, the crows will pick it up and drop it again until it is broken.  In one experiment, researchers found the following relationship between the heights of drop and the number of dropping times to break the shells.  (1 point)

            Height of drop (m)      Number of drops required to break shell
1                                                                                                                                            67
2                                                                                                                                            46
3                                                                                                                                            18
4                                                                                                                                            6
5                                                                                                                                            5
6                                                                                                                                            4
12                                                                                                                                        3



According to optimal foraging theory, which of the following is the most likely height that the crows would fly to drop the shells?
A.    6.5 m
B.     4.5 m    
C.     2.5 m
D.    3.5 m
E.     12.5 m

19. The figure below shows cytological and biochemical changes of a human infected by HIV.  There are three curves in the figure labeled as 1 through 3. Which of the followings is correct? (1 point)

A.    Curve 1 represents viral numbers
Curve 2 represents concentration of antibodies against HIV
Curve 3 represents humoral and cell-mediated immunity      

B.     Curve 1 represents humoral and cell-mediated immunity
Curve 2 represents concentration of antibodies against HIV
Curve 3 represents viral numbers

C.     Curve 1 represents humoral and cell-mediated immunity
Curve 2 represents viral numbers
Curve 3 represents concentration of antibodies against HIV


D.    Curve 1 represents concentration of antibodies against HIV
Curve 2 represents humoral and cell-mediated immunity
Curve 3 represents viral numbers

E.     Curve 1 represents viral numbers
Curve 2 represents humoral and cell-mediated immunity
Curve 3 represents concentration of antibodies against HIV



20. The figure below shows a generalized life cycle of fungi.  Which of the followings is/are true? (1 point)

           


(1)   Spores are generally haploid
(2)   Cycle I is sexual life cycle and cycle II is asexual life cycle
(3)   Diploid fungi are formed after plasmogamy
(4)   There are two types of mycelia that mate even thought they may look alike.

A.    1, 2,
B.     1, 3
C.     1, 4             
D.    1, 2, 4
E.     1, 3, 4

Questions 21-24. The hyperthermophilic archeon, Pyrococcus furiosus, has an unusual phosphofructokinase. It catalyzes the following reaction:
Fructose-6-phosphate + ADP --> Fructose-1,6-bisphosphate + AMP
It was found that the addition of glucose, pyruvate, phosphoenolpyruvate, citrate and fructose-2,6-bisphosphate did not show any effect on the reaction rate. The effects of ATP and AMP addition were shown as Lineweaver-Burk plots:

Answer the following questions:
21. Which of the following statements is TRUE ? (1 point)
            A. The reaction is ATP-dependent.
            B. The reaction is ADP-dependent.  
            C. The reaction is AMP-dependent
            D. Neither of the above answers is true.

22. What is the effect of ATP or AMP on the reaction rate? (1 point)
            A. Allosteric Stimulation
            B. Allosteric inhibition
            C. Competitive inhibition     
            D. Uncompetitive inhibition
            E. Mixed inhibition

23. Does this phosphofructokinase play an important role in the regulation of glycolysis in Pyrococcus furiosus? (1 point)
            A. Yes
            B. No     
            C. The conclusion cannot be drawn.

24. Pyrococcus furiosus phosphofructokinase was purified and gave a single band at 52 kDa on SDS-polyacrylamide gel electrophoresis. Its native molecular mass determined by gel filtration chromatography was approximately 190 kDa. The protein is: (1 point)
            A. monomer
            B. dimer
            C. trimer
            D. tetramer   
            E. hexamer

25. Match the following names or descriptions to the right biochemical compounds listed. (2 points)


          Answer
[A-G]
1. 
2. 
3.  
4.  
5. 


1. Nucleoside found in DNA
2. Phospholipid
3.  A yeast fermentation product
4. Monosaccharide
5. Iron-sulfur center



26. Antibiotics are antimicrobial substances produced by some organisms to prevent growth of other organisms. Match the following antibiotics as inhibitors to their cellular targets: (1 point)

A. Cell wall synthesis
B. Plasma membrane formation
C. DNA replication
D. RNA transcription
E. Protein translation
                       

Answer
(A-E)
1. Polymyxins

2. Tetracycline

3. Rifampin

4. Penicillin

5. Mitomycin








27. Glucose labeled with 14C at C-1 is incubated with the glycolytic enzymes and necessary cofactors. What is the distribution of 14C in the pyruvate that is formed? (1 point)
A.    The label is in the methyl carbon atom of pyruvate.     
B.     The label is in the carboxyl carbon atom of pyruvate.
C.     The label is in both the methyl and carboxyl carbon atoms of pyruvate.
D.    The label is in the middle carbonyl carbon atom of pyruvate.

28. A common moiety for NADP, NAD, FMN, FAD, and coenzyme A is: (1 point)
A.    A pyrimidine ring
B.     A three ring structure
C.     An ADP          
D.    A pyranose ring
E.     A triphosphate group

29. Which of the following statements is/are correct? (1 point)
(2)   The citric acid cycle oxidizes the acetyl CoA derived from fatty acid degradation.
(3)   The citric acid cycle produces most of the CO2 in anaerobic organisms.
(4)   The citric acid cycle provides succinyl CoA for the synthesis of carbohydrates.
(5)   The citric acid cycle provides carbon skeletons for amino acid synthesis.
A.           1, 2, 5,
B.            3, 5,
C.            2, 4
D.           2, 3,
E.            2, 5             

30. Key enzymatic differences between liver, kidney, muscle and brain account for their differences in the utilization of metabolic fuels. Which of the following does NOT represent such a biochemical difference? (1 point)

A.    The liver contains glucose 6-phosphatase, whereas muscle and the brain do not. Hence muscle and the brain, in contrast with the liver, do not release glucose into the blood.
B.     The liver has little of the transferase needed to activate acetoacetate to acetoacetyl CoA. Consequently, acetoacetate and 3-hydroxybutyrate are exported by the liver and be used by heart muscle, skeletal muscle and the brain.
C.     Under conditions of prolonged starvation, the fatty acids stored in the adipose tissues will be converted into ketone bodies there before being transported to the brain and muscle for complete oxidation.        
D.    Lactate dehydrogenase does not appear to exist in the heart muscle. As a result, the heart depends on aerobic oxidation to obtain the energy for its continuous pumping.
31. An organelle in eukaryotic cell is spherical or ovoid with a diameter of 0.1 to 1.5 mm and consists of a single membrane. It participates in a variety of metabolic processes, including H2O2-based respiration and lipid metabolism. This organelle is most likely? (1 point)
A.    Mitochondrion
B.     Peroxisome      
C.    Endoplasmic reticulum
D.    Lysosome
E.     Endosome
32. A red alga has two major kinds of photosynthetic pigments: phycobilisomes that absorb green light and chlorophylls that absorb red and blue light.  A student performed an experiment and obtained measurement data as shown in table.  Note, all light is used with light intensity at 100 mol.m-2.s-1  
Light quality
Photosynthetic oxygen evolution rate
Blue light only
          28
Green light only         
          65
Red light only
47
Blue and green
150
Blue and red  
73
Green and red
146

Which of the following is/are NOT correct? (2 points)
(1) Blue light absorbed was less efficient for photosynthetic electron transfer because the blue light is mostly absorbed by chlorophyll b.
            (2) Red light is more efficiently absorbed by chlorophyll than blue light.
            (3) Emerson enhancement effect is observed in this experiment.
(4) It is predicted that more overlapping is present in long wavelength region and than the short wavelength region between phycobilisome absorption spectrum and chlorophyll absorption spectrum.
           
A.    1, 2, 4
B.     1, 3, 4
C.     3, 4   
D.    1, 2
E.     1

33.    The figure below shows the nitrogen cycle on earth.  Finish the table below according to the information provided. (1 point)













Bacteria:
Answer:
A through E.  Note, there could be more than one answer

(1) Able to form nodule with plants



(2) Able to denitrify



(3) Able to nitrify



(4) Able to use ammonium as energy source



(5) Able to fix nitrogen from air





34.    A researcher found that seeds from a plant could inhibit growth of some fungi.  He isolated some substances from the seeds and performed analyses.  The figure below is the result.  He also ran a regular SDS-gel electrophoresis that separated molecular standard proteins from 14 kDa to 100 kDa. 

Treatment 1: no addition of the substance.
            Treatment 2: addition of the substance.
Treatment 3: addition of substance that was treated with -mercaptoethanol (BME)
Treatment 4: same as treatment 3 except that BME was removed before addition of the substance to the fungal culture.
Treatment 5: the substance was treated at 80°C for 20 min before the addition to the fungal culture.
Treatment 6: the substance was treated at 80°C for 20 min in the presence of BME before the addition to the fungal culture.
Treatment 7: the substance was treated with trypsin.
Treatment 8: only trypsin was added to the fungal culture.



He found no protein could be detected in this molecular mass range with coomassie stain even though the substance(s) showed Coomassie binding in solution.  The substance(s) is colorless, but had a strong absorption in UV region.  Which of the following is/are correct? (2 point)

(1)   The substance(s) contains protein
(2)   The substance(s) has disulphide bond that is important to the function
(3)   The substance(s) is stained poorly with Coomassie blue
(4)   The substance(s) is a protein with molecular mass smaller than 14 kDa.
(5)   The substance(s) is resistant to trypsin treatment.

A.    1, 2, 3, 4, 5     
B.     1, 2, 4
C.     1, 3, 4
D.    1, 4
E.     1, 5

35.         Calculate the pI value of aspartic acid.  Its pK1 is 2.09, pK2 is 3.86, pK3 is 9.82. (1 point)
A.    pI = 5.26
B.     pI = 2.98         
C.     5.96
D.    6.84

Question 36-40. To complete a life cycle, e.g. from zygote to gametes, plants need to constantly incorporate environmental information to ensure all the organs required for the life cycle to be properly initiated from growth tip. Flowering is a process with the most sophisticated morphological changes, in which, specific environmental signals are often required.

36. When we call a plant as a “short-day plant”, it exactly means: (1 point)

A.       The plant flowers in winter
B.       The plant flowers when day is shorter than 12 hours
C.       The plant flowers only in the equator area
D.       The plant flowers when night is longer than its own critical night length   
E.        A and D

37. Which of the following is the photoreceptor that responses to day-length? (1 point)
A.           Chlorophyll
B.            Carotenoids
C.            Cytochrome
D.           Phytochrome             
E.            Retinal
38. Which of the following statements is correct? (1 point)

A.           A flower is a reproductive organ
B.            A flower lacking any of sepal, petal, stamen or carpel is an imperfect flower
C.            Most grasses have imperfect flowers
D.           Floral parts in all angiosperm are arranged as four whorls
E.            Floral parts are sequentially initiated at the floral meristem    
39. One the means to prevent self-fertilization in plants is self-incompatibility. Which of the following statements is/are true about self-incompatibility? (1 point)
(1)          The plants that show self-incompatibility have a unique stigma structure.
(2)                                      The flowers of the plants that show self-incompatibility only produce pollens when stigmas fail to develop.
(3)                                      Self-incompatibility is analogous to animal immune response in that both have the ability to distinguish the cells of “self” from those of “nonself”.
(4)                                      A pollen from one plant will only develop pollen tube on its own stigma if a pollen from another plant is present on the stigma.
(5)                                      A pollen from one plant will develop pollen tube on its own stigma, but will not be able to fertilize the egg.

A.    1, 2
B.     3, 4, 5
C.     4, 5
D.    3
E.     3, 5        
40. Where do you find the cells undergo meiosis in plants? (1 point)

A.       From shoot apical meristem
B.       From pollens
C.       From embryo sacs
D.       From corolla
E.        From ovule    
41. Which of the following structures of plants consists of haploid cells? (1 point)

A.       Sporophytes
B.       Sporocytes
C.       Spores
D.       Tapetum
E.        Gemetophytes     
         
Questions 42-45. Algae play very important roles in ecosystems.  They are also diverse in pigmentation.

42. Red algae differ from green algae and brown algae in that (1 point)

A.      Red algae produce agar
B.       Red algae do not produce chlorophyll a
C.       Red algae do not have sexual reproduction
D.      No unicellular red alga has been found
E.       Red algae do not produce flagelatted cells in their life cycle.  

43. Dinoflagellates are a group of algae.  Their pigments are similar to brown algae.  Therefore, the pigments of a typical dinoflagellate are similar to: (1 point)

            A. Pigments of Chlamydomonas
            B. Pigments of Volvox
            C. Pigments of a diatom            
            D. Pigments of a red alga
            E. Pigments of blue-green algae


44. According to their pigmentation, which algal group would be most likely to perform photosynthesis in deepest water? (1 point)
A.                 Red algae          
B.                 Green Algae
C.                 Brown algae
D.                 Golden algae






45.         Seaweeds are large marine algae and they play very important role in marine ecosystems.  Which of the following is/are NOT true about seaweeds? (1 point)

(1)   Most seaweeds are brown algae.
(2)   Diatoms can sometimes be large enough to be included as seaweeds.
(3)   Seaweeds have complicated structures such as leaves.
(4)   Seaweeds live in deep water
(5)   They use holdfast to absorb nutrients
A.    1, 2, 3, 4,    
B.     2, 3, 4, 5,
C.     1, 3, 4, 5,
D.    1, 2, 4, 5
E.     1, 2, 3, 4, 5

46.         Apoptosis was first described in nematodes and later was found to be present in many organisms.  Which of the following is NOT true about apoptosis? (1 point)
A.    It was discovered by cell lineage analysis of nematodes
B.     It is a critical process in animal development.
C.     It is controlled by single gene     
D.    It is not found in insects
E.     Proteases and nucleases participate in apoptosis.
47.         After synthesis, proteins are translocated either by nonvesicular transport or by vesicular transport.  Answer A for vesicular or B for Non-vesicular to indicate how the protein indicated is transported. (0.2x9, 1.8 points)


Proteins:
Answer A or B
1. cytoskeletal proteins

2. Mitchondrial proteins

3. Lysosomal proteins

4. Nuclear proteins

5. Cytosolic enzymes

6. integral plasma membrane proteins

7. secreted proteins

8. Chloroplast proteins

9. Peroxisomal protein



48.         An action potential in neurons is characterized by all the following except that (1 point)
A.           It is initiated by opening of voltage-gated potassium channels  
B.           It is regarded as a regenerative response
C.           It is regarded as a all-or-none response
D.           It does not degrade in magnitude with space or time
E.            It is characteristic of transmembrane potential changes that occur in most axons.

49. The resting potential in most neurons is primarily due to the permeability of (1 point)
F.      Calcium
G.    Chloride
H.    Sodium
I.       Potassium      
J.       Magnesium

50. Which of the following cell cycle phases is usually the shortest in duration? (1 point)
A. G1
B. G0
C. G2
D. S    
E. M

51. Which of the following is/are often used for protein purification? (1 point)
(1)   Gel filtration chromatography
(2)   Ion exchange chromatography
(3)   Salt precipitation
(4)   SDS-electrophoresis
(5)   Substrate affinity chromatography

A.    all of the above
B.     1, 2, 3, 4,
C.     1, 2, 4, 5
D.    1, 2, 3, 5    
E.     2, 3, 4, 5



52. Which of the following is/are important in ATP synthesis? (1 point)
(1)   P700
(2)   P680
(3)   P450
A.    1
B.     2
C.     3       
D.    1, 2           
E.     1, 2, 3

53. Which of the following statements about mRNA is correct? (1 point)
(1)   All mRNA has a cap at its 5’ end
(2)   All mRNA has a poly A tail at its 3’ end
(3)   Its synthesis is performed by RNA polymerase
(4)   The stability of mRNA regulates abundance of its coding protein.
(5)   The codons on mRNA pair with anti-codons of tRNA through A-T, G-C hydrogen bonds

  1. 1, 2, 3, 4,
  2. 3, 4, 5,
  3. 1, 2,
  4. 3, 4         
  5. 3

54. Which of the following about tRNA is/are correct? (1 point)
(1)   There are stem-loop structures
(2)   It consumes ATP in synthesis of aminoacyl tRNA
(3)   tRNA is synthesized by RNA polymerase III
(4)   tRNA is synthesized as a precursor and was processed before it is functional.
(5)   Although theoretical number of tRNA molecules is 61, the actually number of tRNA molecules in most of the cell is smaller, partially because some anticodons can recognize more than one codon.

A.    1, 2, 3
B.     1, 2, 4
C.     1, 2, 5       
D.    1, 2, 3, 4, 5,
E.     2, 3, 4, 5,
55. Which of the following is/are NOT true about Freeze-fracture method in electron microscopy? (1 point)
(1)   Low temperature is used to weaken hydrogen bonding.
(2)   It is often used to observe structures within membrane.
(3)   Particles observed on fractured faces are often liposomes
(4)   Both eukaryotic and prokaryotic cells can be observed with this method
(5)   This method actually observes a replica of specimen.

A.    1, 3
B.     2
C.     3, 4,
D.    4,5
E.     3      

Questions 56-57. Yeast is one of the ideal organisms for the study of cellular, developmental and genetic processes. It can grow either on fermentable or non-fermentable carbon sources. With this property, people can isolate and analyze different yeast mutants associated with certain functions of subcellular organelles.
56. When the yeast mutant cannot grow on oleate, the mutant has defect in which organelle? (1 point)
A.      Mitochondria
B.       Lysosome
C.       Peroxisome            
D.      Nucleus
E.       Endoplamic reticulum 
57. When a yeast mutant cannot grow on glycerol, the mutant has defect in which organelle? (1 point)
A.      Mitochondria     
B.       Lysosome
C.       Peroxisome
D.      Nucleus
E.       Endoplamic reticulum
58. Which of the following is NOT a factor influencing membrane fluidity? (1 point)
A.    Number of double bonds in lipids
B.     Temperature
C.     Flip-flop move of lipids  
D.    Cholesterol
Questions 59-61 are about food digestion in mammalian digestive system.
59. Which of the following is NOT involved directly in protein digestion? (1 point)
A.    Trypsin
B.     Dipeptidase
C.     Aminopeptidase
D.    Carboxypeptidase
E.     Enteropeptidase        
60. Which of the following enzymes is NOT functionally present in small intestine? (1 point)
A.    Nucleases
B.     Lipase
C.     Chymotrypsin
D.    Pancreatic amylases
E.     Pepsin  
61. Many hormones are involved in food digestion and absorption.  Please match the functions with appropriate hormones. (1 point)
A.    Regulation of blood sugar
B.     Stimulation of bicarbonate release
C.     Stimulation of gallbladder to contract and release bile
D.    Stimulation of secretion of gastric juice.
Hormones
Fill your answer
1. Cholecystokinin

2. Gastrin

3. Secretin

4. Insulin


Questions 62-63.  Flowering is one of the most sophisticated processes in plants.  By analysis of flowering mutants and through other studies, researchers proposed an ABC model (hypothesis) to explain gene regulation of flower structures.  Three classes of genes are involved: class A, class B and class C.
In this model, sepal is produced where gene A is active, petal is produced where genes A and B are active; Stamen is produced where genes B and C are active and Carpel is produced where gene C is active.  When gene A is missing, gene C takes its place and when gene C is missing, gene A takes gene C’s place.
62. According to the ABC model, which of the following mutants will produce a phenotype shown below? (1 point)


A.    A mutant lacking gene A
B.     A mutant lacking gene B      
C.     A mutant lacking gene C
D.    A mutant lacking genes A and B
E.     A mutant lacking genes B and C
63. It has been demonstrated that Genes A, B and C encode transcription factors. Which of the following is NOT a property of transcription factors? (1 point)
A.    DNA-binding
B.     Interaction with other proteins
C.     Degradation by protease
D.    RNA binding          
E.     Participation of other gene regulation.


64. PCR (polymerase chain reactions) is one of the most powerful methods in molecular biology.  Which of the following is/are NOT true about PCR? (1 point)
(1)   Primers are needed in PCR
(2)   A DNA polymerase that can tolerate high temperature is needed in PCR
(3)   ATP is needed in PCR
(4)   A DNA template is needed in PCR
A.    1, 2
B.     2, 3          
C.     3
D.    1, 3,
E.     2, 4
65. Nitrogenous wastes of animals are released to their environments in different forms.  Which of the following statements is/are true about animal nitrogenous wastes? (1 point)
(1) Urea is excreted by many marine fishes.
(2) Ammonia is so toxic that it is rarely used by any animals
(3) The animals in dry environments could excrete uric acid
(4) The form of nitrogenous waste is often an adaptation to animal habitats.
A.    1, 2, 3, 4
B.     1, 4,
C.     1, 2, 4
D.    3, 4
E.     1, 3, 4      
66. Among the nitrogenous wastes, urea, uric acid and ammonia have the toxicity in following order: (1 point)
A.    Ammonia > uric acid > urea  
B.     Urea > ammonia > uric acid
C.     Uric acid > urea > ammonia
D.    Ammonia > urea > uric acid
E.     Urea > ammonia > uric acid
F.      Uric acid > urea > ammonia

Question 67-69. Equilibrium dialysis is a method often used to determine dissociation constant KD for a ligand-binding protein.  In this method, a protein at a know concentration is put into several dialysis tubes and each dialysis tube containing the protein is dialyzed against solutions containing the ligand at various ligand concentrations.  Because the protein can not move cross the dialysis tube membrane while the ligand can, the ligand is “trapped” by the protein inside the dialysis tube and it creates a higher concentration of the ligand in dialysis tube than that outside the dialysis tube.  The dissociation constant of the ligand can thus be determined according to the following formula:
                        
Where [M] is the concentration of free protein (no bound ligand) in dialysis tube, [L] is the concentration of the ligand and [ML] is the concentration of the protein with bound ligand.  Therefore, KD is the ligand concentration when [M] equals [ML].  [MT] = [M] + [ML].               
Where [MT] is the total concentration of the protein
The table below shows the measurement result of a calcium-binding protein.  The protein has a molecular mass of 20 kDa and the concentration of the protein in equilibrium dialysis is 1 mg.ml-1. 


Calcium concentration in dialysis solution (M)
Calcium concentration in dialysis tube (M)
[M]/[MT]
20
30

50
68

100
129

200
237

400
442

600
647

1000
1050

1500
1548

2000
2049

Please calculate the values of [M]/[MT] at each concentration and plot the data (Calcium concentration in solution vs [M]/[MT]) with the plotting paper shown below.

67. How many calcium molecules does one protein molecule bind? (1 point)
A.    1          
B.   2
C.  3
D.    4
E.  It can not be determined

68. What is the KD of the protein? (3 point)
A.    30 M
B.     78 M
C.     95 M        
D.    104 M
E.     200 M
69. There are two calcium-binding proteins, protein A and protein B.  If protein A has a KD of 250 nM and protein B has a KD of 400 nM, which of the following is/are NOT correct? (2 point)
(1)   Protein A binds calcium more tightly than protein B.
(2)   Half of protein B will have bound calcium at the concentration of 400 nM.
(3)   It is more difficult to release the bound calcium from protein B.  
(4)   When protein A and protein B are mixed at equal molar concentration, more protein A will have bound calcium than protein B at a calcium concentration of 250 nM.
(5)   When protein A and protein B are mixed at equal molar concentration, equal amount of protein A and protein B will have bound calcium at a calcium concentration of 400 nM.
A.    1, 2, 5
B.     2, 4,
C.     3, 4
D.    4, 5
E.     3, 5            
70. Which of the following are amniotes? (1 point)
(1)   Bony fishes (Osteichthyes)
(2)   Reptiles
(3)   Chondrichthyes
(4)   Cyclostomata
(5)   Mammalia
(6)   Amphibia
(7)   Aves

A. 1, 4, 6, 7
B. 2, 3, 5
C. 2, 5, 7               
D. 2, 4, 5, 6
E. 2, 5, 6, 7
F. 4, 5, 6, 7
G. 5, 6, 7


71. The figure below shows schematic structures of an amniotic egg.  Please name the structures labeled by numbers 1 through 7. (1 point)
A. amnion   B. embryo   C. allantois  D. chorion. E. yolk sac.  F. gut   G. allantois cavity








 


Answer
A-G









 



1
2
3
4
5
6
7

 





















72. Fill appropriate answers based on the functions of the structure shown the figure of question 71. (1 point)




Main Function
Answer:
A-G
(1) It protects the embryo in a fluid-dilled cavity that prevents dehydration.

(2) It provides nutrients for embryo


(3) It functions as a disposal sac for metabolic wastes

(4) It is rich in blood vessels and it forms sac for collecting waste



 












Questions73-74. Compare 4 different invertebrates:
(1) Spider.
(2) Grasshopper,
(3) Millipedes
(4) Shrimp
73. Fill in appropriate answers according to the following descriptions. (1 point)
A.    1 pair of antennae, 3 pairs of legs
B.    
Answer: A-F
(1)


(2)


(3)


(4)



 
1 pair of antennae, more than 4 pairs of legs
C.     2 pairs of antennae, 4 pairs of legs
D.    2 pairs of antennae, more than 4 pairs of legs
E.     No antennae, 3 pairs of legs
F.      No antennae, more than 3 pairs of legs
74. Fill in appropriate answers according to organs of excretion and gas exchange. (1 point)
A.      Excretion with Malpighian tubules and gas exchange with tracheal system.
B.       Excretion with Malpighian tubules and coxal gland, gas exchange with tracheal system.
C.       Excretion with maxillary gland and gas exchange with gill and tracheal system.
D.      Excretion with Malpighian tubules and coxal gland, gas exchange with book lung
E.      
Answer
A-F
(1)

(2)

(3)

(4)





 
Excretion with Malpighian tubules and coxal gland, gas exchange with book lung and tracheal system
F.        Excretion with maxillary gland and gas exchange with gill.


Questions 75-84. The figure below is a diagram for ultra-structure of a cell.. 


75. If you are provided with two electron microscopic pictures, one from of pancreas gland cells and the other from endothelial cells of proximal tubule of a kidney’s nephron.  Which of the structure shown in the figure will be more developed in pancreas gland cell? (1 point)

Answer: choose one from A through G.          

76. As the cells grow, the surface area of each cell increases.  Which structure is the location where the lipids are synthesized for plasma membrane synthesis? (1 point)

Answer: choose one from A through G.    

77. If you treat the cells for a short period of time with 3H-Uracil followed by detecting the labeled cellular structure with autoradiography, which structure will have the highest silver grains (strongest labeling)? (1 point)

Answer: choose one from A through G.

78. Which structure is assembled in nucleus and then transported to cytoplasm? (1 point)

Answer: choose one from A through G.    

79. Erythropoietin (EPO) is hormone that stimulates production of erythrocytes.  EPO is a highly glycosylated and secretive protein.  Which structure would be responsible for the synthesis of EPO? (1 point)
Answer: choose one from A through G.          

80. Which structure would be the site for initial glycosylation of EPO? (1 point)
Answer: choose one from A through G.          

81. Which structure would be the site for final glycosylation of EPO? (1 point)

Answer: choose one from A through G.              

82. Which structure is essential for the transport of EPO inside the cell? (1 point)

Answer: choose one from A through G.        

83. The receptor for EPO is a membrane protein.  Which structure is responsible for EPO’s receptor synthesis? (1 point)

Answer: choose one from A through G.      

84. Which structure has ability to synthesize some proteins that are not encoded by nucleus. (1 point)
Answer: choose one from A through G.             

END of PART I





 














16th International Biology Olympiad

Beijing
July, 2005




THEORY EXAMINATION
Part 2

Total time available:  2.5 hours (150 minutes)

Total points available:  ~80

Questions 85-92. Sex determination in fruit flies and mammals are both XY type, that is, XX leads to female and XY leads to male.

85. Some organisms have abnormal sex chromosomes such as XO (only have one X chromosome) or XXY (extra X chromosome).  The cause of the abnormal sex chromosome is: (1 point)
A.    Error occurred in mitosis of fertilized egg.
B.     Gene mutation
C.     Error occurred in first division in meiosis in gamete formation.      
D.    Sex chromosomes in gametes are either lost or doubled in fertilization.

86. In organisms with XXY chromosome type, there is an extra X chromosome. How do you determine if this X chromosome is from sperm or egg? (1 point)

A.    Karyotype
B.     In situ hybridization
C.     RFLP  
D.    DNA sequencing


87. In mammals, XO leads to female and XXY leads to male.  In fruit flies, XO leads to male and XXY leads to female.  Which of the following is NOT correct? (1 point)

A.          Y chromosome in mammals is necessary for formation of a male organism.
B.           The Y chromosome in mammals is required for sex organ to develop.
C.           The Y chromosome in fruit flies is not functional.     
D.          Number of X chromosome in fruit flies has impact on sex determination.

88. In mammals with abnormal sex chromosomes, the number of individuals with XO chromosome type is far fewer than the number of individuals with XXY chromosome type.  It is therefore predicted that: (1 point)

A.    The individuals with XO chromosome are less capable of surviving than that with XXY chromosome.   
B.     The individuals with XO chromosome are less capable of reproducing than that with XXY chromosome.
C.     The difference is related to sex of the individuals (XO leads to female and XXY leads to male).
D.    None of the above.

89. In both fruit flies and mammals, XX leads to female and XY leads to male. The gene products encoded by two X chromosomes of female individuals are nearly identical to those encoded by one X chromosome of male individuals. This is accomplished by gene dosage compensation.  In mammals, it is accomplished by converting one X chromosome into Barr body (X inactivation). Which of the following about Barr body is/are correct? (1 point)
(1)                               Only normal female individuals have Barr bodies.
(2)                               Only normal male individuals don’t have Barr bodies.
(3)                               Barr body can be used to determine sex of human beings.
(4)                               The maximum number of Barr body is one
(5)                               The number of Barr bodies equals the number of X chromosomes minus one.
A. 1, 3, 5
B. 2, 5
C. 4
D. 5                 
E. 1, 4, 5


90. No Barr body can be observed in normal female fruit flies because (1 point)
A.    X chromosome of fruit flies is too small
B.     There is no mechanism of dosage compensation in fruit flies
C.     There is no X inactivation in fruit flies       
D.    Heterochromatin is difficult to detect in fruit flies.

91. The fur color of cats is determined by genes on X chromosome.  XY is the dominant allele for orange fur, while Xy is the recessive allele for black fur. Which of the following is true about the fur color of the offsprings from a XYXy female cat and XYYmale cat? (1 point)
A.       They are all orange
B.       All the female are orange and half the male are orange
C.       Regardless of sex, half are orange, the other half have furs that are mosaic of orange and black.
D.       Those with mosaic furs are all female.   
92. One of the genes controlling sweat gland in human is located on X chromosome.  Two twin sisters show different phenotypes of the sweat gland.  One has no sweat gland on her left arm while the other has on her left arm. Which of the following statements is/are true? (1 point)
(1)     The twin cannot be true twin. 
(2)     They both are heterozygotic of the gene.
(3)     The reason for the different phenotype is random X inactivation.
(4)     X inactivation must occur after first division of the zygote.
A.    1, 2, 3
B.     1
C.     2, 3
D.    3
E.     2, 3, 4    

93. Mycorrhizae are symbiotic associations of fungi and plant roots.  Which of the following is/are true about mycorrhizae? (1point)
(1) They are often harmful to plant roots while beneficial to fungi.
(2) They are often beneficial to plants but harmful to fungi
(3) They are helpful for plants to absorb water and minerals.
(4) They could even help the older root region above the root hair area to supply minerals to plants. 
A. 1, 3, 4
B. 2, 3, 4
C. 3, 4     
D. 3

94. Stomata of a plant open when guard cells (1point)
A. accumulate water by active transport.
B. sense an increase in CO2 in the air spaces of the leaf.
C. become more turgid because of an influx of K+, followed by the osmotic entry of water.    
D. sense that water content of whole plant is low.

95. Which of the following processes of plants could be regulated by phytochrome? (1point)
(1) seed germination
(2) flowering
(3) shoot elongation
(4) open and closure of stomata
A. 1, 2, 3, 4
B. 1, 2, 3                  
C. 1, 2
D. 1

96. If N represents population size, r represents the difference in per capita birth rates and death rates, K represents the maximum sustainable population size, t represents time, which of the following equations best describes logistic growth of the population? (1point)

           
97. Which of the following is usually the limiting process of phosphorous cycles? (1point)
A. Decomposition          
B. Utilization in primary production
C. Release from soil
D. Sedimentation

98. Which of the following ecosystems has the lowest primary production per square meter? (1point)
A. a salt marsh
B. an open ocean             
C. a grassland
D. a tropical rain forest

99. Which of the following is/are true about Archaea and Eubacteria? (1point)
(1) They don’t have nuclear envelope
(2) They both have branched chain in membrane lipids
(3) They have one kind of RNA polymerase
(4) They have circular chromosomes.
A. 1, 2, 4
B. 1, 4                   
C. 2, 3
D. 1, 2, 3

100. Four major groups of fungi are recognized.  They are chytrids, zygote fungi, sac fungi and club fungi.  Chytrids differ from other three groups in that (1point)
A. Chytrids don’t have sexual reproduction
B. They are all aquatic.
C. They have cell walls made of cellulose
D. They have flagellated cells in their life cycles.           

101. Chlorophyll a is involved in both light energy absorption and primary charge separation of photosynthesis.  Which of the following are true about the chlorophyll a? (1point)
(1) The position of chlorophyll a in photosystems has a strong influence on the function of chlorophyll a.
(2) Chlorophyll a in photosynthetic reaction center is chemically modified so that it performs charge separation.
(3) Part of chlorophyll a is structurally related to heam group of hemoglobin.
(4) Part of chlorophyll a is structurally related to carotenoids.
A. 1, 2, 3, 4
B. 1, 3, 4                 
C. 3, 4
D. 1, 2

102. In measurement of photosynthetic electron transfer, intact chloroplasts are isolated and used for electron transfer rates under different conditions.  Which of the following is correct? (1point)
A. Addition of an uncoupler leads to an increased rate of electron transfer. 
B. Cyclic electron transfer starts only when linear electron transfer is inhibited.
C. ATP synthesis could only be observed with continuous light illumination.
D. Oxygen evolution by chloroplast suspension is absolutely dependent upon the presence of CO2.



103. The figure shown below is a diagram of evolutionary tree. Which of the following statements about evolution are true and deductible from the figure? (2 points)

(1) All eucaryotic cells contain mitochondria.
(2) Symbiosis of the eucaryotic ancestor with autotrophic cell precede the symbiosis with the cell taking advantage of the oxidative metabolism.
(3) There is common ancestor of eubacteria and eukaryota, archaebacteria are a group with unique and independent origin.
(4) Ancestral eukaryote was anaerobic.
(5) None of the recent photosynthetic bacteria are related to the chloroplasts.
(6) Mitochondria and chloroplasts has similar genomes.
(7) Mitochondria are present in the cells of the plants, animals and fungi.
(8) Fungi lost chloroplasts during evolution.
(9) Bacteria are highly homogenous group of organisms which quickly diversified genomes and metabolisms during the last billion years.
(10) Chloroplasts and mitochondria are results of independent endosymbiotic events.

A. 1, 2, 5
B. 3, 4, 7
C. 4, 7, 10             
D. 6, 8, 10
E. 4, 9, 10

104. The figure shown below is an image of a DNA molecule. Structure of the DNA molecule can undergo dramatic and highly regulated changes during the cell cycle. Which of the following statements are true about the cross-like structure on the image. (2 points)
(1)          During the replication all four DNA strands in the double helix are covalently interconnected.
(2)          In the prophase of the mitosis chromosomes highly condense and get interconnected via covalent bonds.
(3)          During the prophase of the first meiotic division recombination between sister chromatids takes place and new covalent bonds are temporarily formed which results in the formation of the cross-like conformation of the DNA.
(4)          Image is photomontage of the forbidden conformation of the DNA molecule.
(5)          During the prophase of the first meiotic division recombination between homologous chromosomes takes place and new covalent bonds are temporarily formed which results in the cross-like conformation of the DNA.
(6)          Cross-like structures of the DNA molecules could be observed in the nucleus of the B-cells and T-lymphocytes during their development.
(7)          Figure shows situation in the cytosol of the bacterial cell where translation and transcription are not separated spatially.
(8)          Some viruses use formation of the cross-like structures to integrate into the host chromosomes.
(9)          In the apoptotic cells DNA is cleaved and finally forms unusual cross-like conformation - useful marker of the final stages of the programmed cell death.
(10)      Figure shows unusual type of the replication in the Archaebacteria when three double helixes are formed from one precursor DNA double helix.

A. 5, 6, 8              
B. 1, 3, 8
C. 6, 8, 10
D. 2, 7, 9
E. 4, 6, 10
105. Siamese cat is an example of the animals with melanin synthesized in both sexes mostly at the body extremities. That makes snout, ears, tail and feet much darker than the rest of the body. Explanation of this type of the body coloration is that: (1 point)
  1. Only at the body extremities the enzyme tyrosinase (responsible for the synthesis of the melanin) is synthesized.
  2. The only places where one of the X chromosomes that have dominant allele of the tyrosinase is NOT inactivated
  3. Melanin is synthesized only in the colder parts of the body because Siamese cat bears temperature sensitive allele for enzyme producing melanin.             
  4. Melanocytes are localized only at the snout, ears, tail and feet – the rest of body is without melanocytes.
  5. The body extremities are more exposed to the UV-radiation which stimulates production of the melanin.


106. Retinoblastoma protein (Rbp) and p53 are examples of the anti-oncogens. Which of the following statements about these proteins is true? (1 point)
A. Mutation in the p53 (when p53 lost its regulatory function) can stop the cell cycle.
B. Overproduction of the Rbp in the retina can cause cancer.
C. Cells with mutated p53 are predisposed to malignancy.             
D. Cells with mutated Rbp are resistant to malignancy.
E. Various viruses incorporated homologs of the p53 and Rb into their genomes and use these proteins for the transformation of the host cell.

107. Extracellular matrix is responsible for the mechanoelastical properties of the tissues. Which of the following molecules is NOT a component of the extracellular matrix is: (1 point)
A. elastin
B. cytokeratin           
C. laminin
D. kolagen
E. chondroitin sulphate
108. Prions are unique infectious agents formed only from protein called PrP. What are the true statements about prions? (1 point)

(1)               prion protein has an exceptionaly stable conformation
(2)               mutated form of the PrP can predispose to Creutzfedt-Jacob disease in human
(3)               wt form of prion protein is expressed in the brains of the healthy animals
(4)               spongiform encephalopathy is an typical phenotype of the prion caused disease
(5)               prion disease are restricted only man, cow and sheep - because only these species express PrP
(6)               prions are small viruses with symmetrical capsid without DNA or RNA
(7)               prion disease is highly infectious and could be transmitted via body fluids
(8)               prion disease could be transmitted via transplantation or canibalism
(9)               mouse with genetic knock-out for the PrP is resistant to the prion disease
(10)           prion disease could be cured by the bone marrow transplantation

A.  1, 4, 6, 7
B.  2, 3, 4, 5
C. 2, 3, 8, 9         
D. 4, 6, 8, 9
E.  1, 3, 9, 10

109. Algae were supplied with a radioactive isotope of Carbon, 14C, and allowed to photosynthesis. After a period of time, the light was switched off and the algae were left in the dark. The graph shows the relative amount of some radioactive labelled compounds over the period of the experiment. (1 point)


Which line is representing the amount of glycerate 3-phosphate (3GP), ribulose biphosphate (RuBP) and sucrose? (1 point)
Fill out the correct number of the line in the correct box.
    
Compound
Line
(1) 3GP

(2) RuBP

(3) Sucrose


110. Methylene blue acts as a hydrogen acceptor. It is blue in oxidised state, but goes colourless when it is reduced by accepting hydrogen atoms. (1 point)

Methylene blue + hydrogen  ®  reduced methylene blue
(blue)                                      (colorless)

A student likes to investigate this reaction.
He prepares four test tubes as shown below


Tube A
Tube B
Tube C
Tube D
Distilled water
-
2 ml
2 ml
2 ml
Glucose solution
2 ml
2 ml
-
2 ml
Methylene blue solution
1 ml
1 ml
1 ml
-
Yeast solution
2 ml
-
2 ml
2 ml

All tubes were incubated at a temperature of 30 °C. The colour was recorded at the start and after intervals of 5 and 15 minutes. The results are shown in the table.



Colour of content
Tube A
Tube B
Tube C
Tube D
At start
Blue
Blue
Blue
colourless
After 5 minutes
colourless
Blue
Blue
colourless
After 15 minutes
colourless
Blue
Pale blue
colourless

Which test tube can be characterized as a control in this investigation and which test tube is useless? (1 point)
Fill out the correct letter


Tube
(1) Control

(2) Useless


111. Morgan crossed Drosophila of two known genotypes, BbVv x bbvv, where B, the wild-type (grey) body, is dominant over b (black body) and V (wild-type wing) is dominant over v (vestigial, a very small wing). Morgan expected to see four phenotypes in a ratio 1:1:1:1. But he observed:
Wild type:              965
Black vestigial:      944
Grey vestigial:        206
Black normal:         185
These results were explained in assuming linkage of alleles together with genetic recombination (crossing over).
In this particular example the recombinant frequency (defined as the ratio of recombinants in relation to the total offspring) is: (1 point)
A. 0.205
B. 0.170              
C. 0.108
D. 0.900
E. 0.080
112. 70% of the population of Beijing is able to taste phenylthiocarbamide. The ability to taste (T, taster) is dominant over the inability to taste (t, non-taster).
What percentage of the offspring of 'tasters' will be non-tasters? (2 points)
A.  25%
B.  15%
C.  13%      
D.  20%
E.   7.5%

Questions 113-116. Wild type individuals of Drosophila have red eyes and straw-coloured bodies. A recessive allele of a single gene in Drosophila causes glass eye and a recessive allele of a different gene causes ebony body.
A student crosses pure breeding wild type flies with pure breeding flies having glass eye and ebony body and the resulting F1 flies showed all the wild type phenotype for both features. On crossing the F1 flies among themselves the student expect a 9:3:3:1 ratio but the results are not like that. The actual offspring showed:

Eye
Body
Number of flies in F2
Wild
Wild
164
Wild
Ebony
37
Glass
Wild
59
Glass
Ebony
28

There are two possibilities:
-        The differences from 9:3:3:1 are coincidental (null hypothesis accepted).
-                 The differences do not occur by coincidence (null hypothesis rejected).

You are required to check this applying the c2 (chi square) test.

For this situation, e.g. degree of freedom, the following diagram with c2 values should be used:

Question 113.  The alculated c2 is? (1 point)

          A. 10.11
          B. 2.84
          C. 14.33
          D. 11.40               

Question 114. Indicate the degree of freedom (df) for this test: (1 point)

          A. 2
          B. 3              
          C. 4
         

Question 115. Determine the probability that the deviation of the observed results from expected results is due to chance alone. (1 point)
A. About 1%              
B. About 2%
C. About 5%
D. About 8%


Question 116. To explain the observed deviation of the 9:3:3:1 ratio the student suggested some possibilities.
(1) linkage of both the alleles
(2) crossing over
(3) incomplete dominance
Which combination of suggestions is the correct explanation? (1 point)
                   A. 1, 2           
            B. 1, 3
            C. 2, 3
            D. 1, 2, 3

117. Which of the following diagram offers a correct representation of the urea content in urine of a person in hunger strike, who finally died. (1 point)


                                                                                    

118. Wilhelm von Osten gave performances with his horse called smart Hans. He stated that he taught his horse to make calculations. But in fact this isn’t true at all. He had taught the horse to mind his hidden but trigging indications. As a result the horse made the desired movements: swinging the correct number of times with his foreleg. After that the horse got some reward.
What kind of learning behaviour is this? (1 point)
A. adaptation
B. conditioned reflex             
C. habituation
D. imitation
E. imprinting
F. insight
G. sign stimulus

119. A snail crawling across a board will withdraw into its shell when you drop a marble on the board. Repetition of dropping marble will lead to weaker withdraw action and in the end the snail will ignore the marble dropping. Which of the following terms do apply for the disappearance of the withdraw action? (1 point)

(1)   adaptation
(2)   conditioning
(3)   habituation
(4)   imprinting
(5)   insight
(6)   learned behaviour
(7)   ritualisation
(8)   trial and error

A.    1, 2, 3, 5
B.     3, 5, 8
C.     6, 7, 8
D.    2, 3, 4, 5
E.     3, 4, 5, 6             

120. Bonsai trees need water with very low lime content. Which types of water could be used to water them? (1 point)

(1)      Carbonated mineral water
(2)      Rain water
(3)      Tap water with high water hardness
(4)          Tap water with high water hardness treated by leaving it over night with a mix of peat and crushed stones and filtrating it before use
(5)      Molten snow
A             1, 5
B              2, 5
C              1, 3
D             4, 5
E               2, 4, 5      


121. Observe the diagrams 1 to 4 representing cross sections of the ovaries of different flowers.
Match the numbers in front of the placentation type (A-D) with the corresponding diagram.


A. Axile placentation.      
B. free central placentation.         
C. Marginal placentation.                                     
D. Parietal placentation.                                       
Match the number with correct placenta type. (1 point)
type
Answer
1

2

3

4


122. Which curve shows the correct time course of the production of saliva in a human after the intake of citric acid? (1 point)


 















Questions 123-125. The behavior of eight Humboldt penguins (Spheniscus humboldti) is investigated in a larger group of penguins in a zoo enclosure. The animals can be distinguished by marks or their individual pattern of black dots on their white thorax. To document the relationship of the penguins, their nearest neighbor (closest animal in the enclosure) was recorded in short time intervals during day time in a period of several weeks. The table shows the relatively stable mean values for the frequency of neighbors for the four male (M1 – M4) and four female (F1 – F4) penguins.


M1
M2
M3
M4
F1
F2
F3
F4
S
M1

2
5
1
0
3
7
77
95
M2
2

0
9
9
75
1
2
98
M3
5
0

0
0
0
78
6
89
M4
1
9
0

80
8
0
0
98
F1
0
9
0
80

7
0
0
96
F2
3
75
0
8
7

0
0
93
F3
7
1
78
0
0
0

7
93
F4
77
2
6
0
0
0
7

92
S
95
98
89
98
96
93
93
92


Several months later the same animals were observed again yielding the following values.




M1
M2
M3
M4
F1
F2
F3
F4
S
M1

4
8
2
1
4
11
60
90
M2
4

0
12
12
65
1
5
99
M3
8
0

0
0
1
62
9
80
M4
2
12
0

70
14
0
1
99
F1
1
12
0
70

10
0
1
94
F2
4
65
1
14
10

0
3
97
F3
11
1
62
0
0
0

10
84
F4
60
5
9
1
1
3
10

89
S
90
99
80
99
94
97
84
89


During the following years the tendency of these values remained the same.

123. Analyze the tables and determine the mating system of the Humbodt penguins. (1 point)
A. promiscuity   
B. polyandry   
C. polygyny   
D. monogamy      

124. Which is the most common polygamous relationship? (1 point)
A. promiscuity   
B. polyandry             
C. polygyny   
D. monogamy

125. Which group of animals do the penguins belong to? (1 point)
  1. Ratitae (birds with flat breast and weak breast muscles)
  2. Carinatae (birds with strong breast muscles)                
  3. Neither, they are not birds
126. Substrate(s) of RUBISCO is (are): (1 point)
(1)   Phosphoenolpuruvate (PEP)
(2)   Ribulose-bis-phosphate (RuBP)
(3)   Oxaloacetic acid (OAA)
(4)   Phosphoglyceric acid (PGA)
(5)   Carbon dioxide (CO2)
(6)   Phosphoglyceraldehyde (GAP)
(7)   Oxygen (O2)
A.                          1, 2, 5
B.                           1, 5
C.                           2, 5
D.                          1, 2, 6
E.                           2, 5, 7    
127. The diagram shows a section through a mammalian ovary. The numbers indicate different develop stages. (1 point)
Choose the correct sequence of numbers in which the structures develop.

                
A. 1, 2, 3, 4, 5
B. 5, 4, 3, 2, 1
C. 5, 2, 4, 1, 3
D. 5, 2, 4, 3, 1              
E. 2, 4, 1, 3, 5

.
Questions 128-131. PKU and albinism are two autosomal recessive disorders, unlinked in human. If a normal couple produced a boy with both disorder, they want to have the second child:

128. What is the chance of the second child with PKU? (1 point)
A.1/2 
B. 1/4     
C.  2/3  
D. 1/16

129. What is the chance of the second child with both traits? (1 point)
A. 1/2 
B. 1/4 
C. 1/8 
D. 1/16      

130. What is the chance of the second child with either PKU or albinism?
A. 1/2 
B  3/4 
C.  3/8 
D.  3/16
131. What is the chance for them to have a normal child? (1 point)
A. 1/16
B. 4/9
C. 9/16 
D. 6/16

Questions 132-137. There is a patient who expressed a very rare phenotype. According to the medical source, this phenotype is seen in 1 in every 100,000 people. The family history of this patient is given below:

                        

132. How is the trait inherited? (1 point)

A.      autosomal recessive
B.       autosomal dominant   
C.       sex-linked recessive
D.      sex-linked dominant
133. If D= dominant, d = recessive, what is the genotype of II-3? (1 point)

A.    DD
B.     Dd   
C.     dd
D.    XdY
E.     XDY

134. What is the genotype of II-4? (1 point)
A.    DD or XDXD
B.     Dd or XDXd
C.     dd or XdXd      

135. If IV-2 married to a man from an unrelated family, what is the chance to get a normal child? (1 point)
A.    1/2
B.     2/3
C.     100%
D.    Cannot be determined  

136.     For the alleles D and d, which individual should be homozygous? (1 point)
A.    III-1
B.     III-2
C.     III-4
D.    III-5
E.     III-7       

137. If this trait is instead quite common in the population, then what is the chance that IV-4 is heterozygous? (1 point)
A.    1/2
B.     1/4
C.     2/3      
D.    100%

138. There are several types of human blood cells such as erythrocytes and monocytes.  They all come from stem cells. Which of the following is/are correct about the stem cells of blood cells? (1 point)
            (1) B cells come from lymphoid stem cells.
            (2) T cells come from lymphoid stem cells.
            (3) Erythropoietin stimulates production of erythrocytes from myeloid stem cells.
            (4) Neutrophils and basophils have same stem cells.
            (5) Lymphoid stem cells come from myeloid stem cells.

A.    1, 2, 3, 4, 5
B.     1, 2, 3, 4               
C.     1, 3
D.    1, 2, 4
139. Which of the following role(s) does platelets play in clotting process? (1 point)
            (1) They form plug for protection against blood loss.
            (2) They release chemical signals for fibrin formation.
            (3) They release chemical signals for reducing blood pressure.
A. 1, 2             
B. 1, 2, 3
C. 2, 3
D. 1, 3
140. Which of the following is NOT involved in allergic response in human? (1 point)
A.    Histamine.
B.     Mast cell.
C.     Plasma cell
D.    B cell.             

141. There are several sensory receptors in human skin.  Which of the following is located in deepest position of the skin? (1 point)
A. Sensory receptor for pain.
B. Sensory receptor for cold.
C. Sensory receptor for heat.
D. Sensory receptor for strong pressure.             

142. One of the mutant zebra fish has a reduced number of hair cells in neuromast of its lateral line system.  Which of the following will happen? (1 point)
(1) The mutant fish will not be able to detect depth of water.
(2) The mutant fish swims slowly.
(3) The mutant fish could not detect sound of its prey.
(4) The mutant fish could be impaired in detecting water movement around its body.
A. 1, 2
B. 3, 4
C. 4           
D. 2, 4

143. Hemoglobin is responsible for transporting oxygen from lung to tissues.  Bohr shift is one of the most important properties of hemoglobin.  Which of the following is NOT true about Bohr shift? (1 point)
A.  Additional oxygen is absorbed by hemoglobin in lung when pH decreases.         
B.Additional oxygen is released from hemoglobin at a lower pH.
C.CO2 is involved in Bohr shift.
D.  Bohr shift helps tissues to obtain more oxygen in exercise.

144. Which of the following is/are NOT true about the difference in digestive tracts of carnivores and herbivores? (1 point)
(1)   Carnivores usually have a bigger stomach.
(2)   Carnivores usually have a shorter colon.
(3)   Herbivores usually have a longer cecum.
A.    1, 2
B.     1
C.     2, 3     
D.    3

Questions 145-148. Hemophilia and color blindness are X-linked recessive traits. When a color-blind woman married to a hemophiliac man,

145. What is the chance for them to have a normal son? (1 point)
A.    50%
B.     0%, all their sons will suffer from color-blind       
C.     0%, all their sons will suffer from hemophilia
D.    It depends on the recombinant frequency.

146. If their son was married to a woman whose mother was color-blind, what is the chance for them to produce a normal daughter? (1 point)
A.    0%
B.     50%         
C.     75%
D.    100%

147. If their daughter was married to a normal man whose father was color-blind, and produced 1 normal son, 4 normal daughters, 2 color-blind sons, 2 hemophiliac sons and 1 color-blind, hemophiliac son, the distance between the two genes is: (1 point)
A.    0.5
B.     0.33   
C.     0.2
D.    0.1

148. If they have a color-blind daughter, (1 point)
A.    There must be a mutation in her father’s germ line.
B.     She must have abnormalities other than color-blind.      
C.     The chance is less than 1/100,000
D.    The chance is about 1/1000

Questions 149-152. Huntington disease is a rare fatal disease. People with this disease start to show symptoms in their 40’s.  Peter’s father (John) has Huntington disease. John’s father (Peter’s grandfather), who also had this disease, had 11 children (5 sons and 6 daughters).  Among them, 6 (3 sons and 3 daughters) of them had got the disease and five died from it.
149. How is the trait inherited? (1 point)
A.    autosomal recessive
B.     autosomal dominant      
C.     sex-linked recessive
D.    sex-linked dominant

150. What is the possibility that Peter will also develop the disease? (1 point)
A.    50%      
B.     25%
C.     75%
D.    67%

151. Peter is married to a normal woman, what is the possibility that their first child will eventually develop the disease? (1 point)
A.    50%
B.     25%      
C.     75%
D.    67%
E.     0

152. If Peter’s mother-in-law died from the same disease, what is the possibility that their first child will eventually develop the disease? (1 point)
A.    3/16
B.     4/16
C.     7/16       
D.    9/16
E.     12/16


153. Trophic levels are indicated below with numbered lines in the flowchart.  Write the appropriate tropic level name in the space provided next to its number.  Write ONLY the letter of the tropic characteristic. (1 point)

NOTE:  Left-hand circle in flowchart is Heat; right-hand circle in flowchart is To detritivores.




1. __________                  

2. _________              

3. __________

4. __________
5. __________

6.  __________

7.  __________


A. energy used in cellular respiration
B. secondary consumers
C. tertiary consumers
D.  energy in wastes
E. primary producers
F. primary consumers
G. production energy




154. Match the biome in the figure below with the appropriate plotted area (a, b, c, d, e, and f) in the climograph. (1 point)

1. _______ arctic and alpine tundra

2. _______ coniferous forest

3. _______ desert

4. _______ grassland

5. ________ temperate forest

6. ________ tropical forest




155. Referring to the action potential graph below, write the letter (from the graph) that corresponds with the appropriate action potential action on the right of what is occurring at that stage of the action potential. (1 point)  Note, there could be more than one choice for each question.




1._______ The membrane is unable to respond to any further stimulation regardless of intensity

2.__ ______ Sodium gates close, and potassium gates re-open

3.____ ___ Both sodium and potassium voltage-gated channels are closed

4.________ Stimulus opening of some sodium channels



156. Molting is a process observed in insects.  Which of the following statements is/are true? (1 point)
(1)   The exoskeleton of insects is largely made of protein and chitin.
(2)   The structure of chitin is similar to that of bacterial cell wall peptidoglycan.
(3)   No enzyme has been found to digest chitin.
(4)   Molting can be observed in all arthropods.
(5)   The only place that is not covered by exoskeleton is the joints between the body and walking legs.
A.    1, 2, 4, 5
B.     1, 4                 
C.     1, 4, 5
D.    1, 5


157. The mechanism of molting has largely been revealed.  The figure below is a diagram of such a process.  Boxes A, B and C represent 3 different growth hormones and molting hormones.  Fill in the answer boxes by choosing correct letter. (1 point)


    

Answer: A-C
1. brain hormone (BH)

2. juvenile hormone (JH)

3. molting hormone (MH)



158. The figure below shows 4 different circulation systems of vertebrates. From left to right, these are the circulation systems of  (1 point)
A.    mammals, reptiles, amphibians, and fish, respectively.
B.      fish, amphibians, reptiles, and mammals, respectively.
C.      mammals, amphibians, reptiles, and fish, respectively.
D.     mammals, amphibians, fish, and reptiles, respectively.                



159. Match the numbers shown below with correct structures in the figure in question above (question158). (1 point)




Answer
A-G

1 Sinus venosus

2. Right ventricle

3 Pulmonary vein

4 Pulmonary artery

5 Conus arteriosus

6 Atrium

7 Left ventricle


 
 





Questions 160-162. The structure of a mammalian kidney is shown below. 




160. Match the following terms with correct structures shown in the figure. (1 point)

collecting duct
glomerulus
distal tubule
Bowman`s capsule
proximal tubule
ureter  
afferent arteriole  


Answer
A-G










161. The substances that be reabsorbed by proximal tubule is/are: (1 point)

(1) Na+
(2) Cl-
(3) Water
(4) Glucose
(5) Amino acids
(6) Urea
A.         1, 2, 3
B.          6,
C.          1, 2, 4, 5,
D.         1, 2, 3, 4, 5      
E.          4, 5

162. In the process of urine formation, filtration occurs in which of the following structures? (1 point)

(1)   Nephrons
(2)    Bowman’s capsule
(3)    Proximal tubule
(4)    Distal tubule
(5)    Collecting duct

A.    1, 2, 3, 4, 5
B.      2, 3, 4, 5
C.      3, 4, 5             
D.     2, 3, 4
E.      1, 2, 3, 4


Questions 163-166. The sensory transduction by a taste receptor is shown the figure below.  The sequential events of the transduction is labeled by number 1 through number 7.  A portion of the cell is magnified.

163.     Structure A which is responsible for event 3 is a  (1 point)
A.    Possium channel             
B.     Calcium channel
C.     Sodium channel
D.    Neurotransmitter channel
E.     Glycine channel

164. Structure C which is responsible for event 5 is a  (1 point)
A.           Possium channel
B.            Calcium channel     
C.            Sodium channel
D.           Neurotransmitter channel
E.            Glycine channel

165. Event 4 by structure B  (1 point)
A.           depolarizes membrane potential.           
B.            increases membrane permeability
C.            transports more sugars molecules into the cell.
D.           transports signal molecules into the cell so that the cell starts to synthesize neurotransmitters.
E.            transports precursor molecules of neurotransmitters into the cell so that the cell can synthesize neurotransmitters.


166. Which of the following statements is/are true about the action potentials shown as D and E in the figure? (1 point)
(1) They were recorded after and before sugar molecules were present, respectively.
(2) They were recorded before and after sugar molecules were present, respectively.
(3) The action potential observed after sugar reception is triggered by an increase of calcium ions which stimulate neurotransmitter release.
(4) The action potential observed after sugar reception is triggered by an increase of Potassium ions which stimulate neurotransmitter release.
(5) The action potential is recorded from taste sensory receptor cells.

    1. 2, 3
    2. 1, 3
    3. 2, 4              
    4. 2, 5
    5. 2, 4, 5
167.  Which of the following is NOT a mechanism of animal migration? (1 point)
            A. Cruising          
            B. Piloting
            C. Navigation
            D. Orientation
168. Both snake and weasel have hibernation.  Which of the following is correct? (1 point)
A. They will die when temperature decreases below the insupportable temperature.
B. Weasel will die when temperature decreases below the insupportable temperature.
C. Snake will die and weasel will wake up when the temperature decreases below the insupportable temperature.        
D. Weasel keeps low body temperature and slow heart rate during the hibernation.
169. It is possible to predict bird diversity based on forest types.  Which of the following is most critical to bird diversity for a forest? (1 point)
AForest area      
BVertical stratification                
CSpecies composition of plants
DConiferous or deciduous forests

170. Four quantity pyramids are shown below. Which is representative for plant-aphid-ladybug? (1 point)














                                                      
                                       
               A                              B                                   C                            D

171. Which of the following ecosystems has the highest net primary productivity? (1 point)
ATropical rain forest  
BOpen ocean            
CNorthern coniferous forest  
DFarm lands

172. The Figure below shows vertical distribution of some parameters (Chlorophyll, Phosphate, Primary production and Temperature) of North Pacific during summer. 

From left to right, letter a through letter d represent: (1 point)
A. Temperature, phosphate, chlorophyll and primary production     
B. Chlorophyll, phosphate, temperature and primary production
C. Primary production, phosphate, temperature, chlorophyll
D. Phosphate, temperature, primary production and chlorophyll.
173. The length of a food chain in a food web is often quite short.  Usually, the length is shorter than 5 links.  Which is mostly likely reason for the shortness of the food chain? (1 point)
A.  The population of final predator is often too large.
B.  The primary producers can sometimes be indigestible.
C.            Only about 10% of energy in on link can be converted to organic matters in next trophic level.              
D.  Wintertime is too long and low temperature limits primary productivity.

174. The figure below shows a membrane potential graph detected after a rod cell of human eyes sees light.  Which of the following is the direct trigger for the hyper-polarization? (1 point)
A.    Retinal switches from cis form to trans form.
B.     Cyclic GMP is destroyed.
C.     Transducin is activated.
D.    Potassium channel is closed
E.     Sodium channel is closed.            



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