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2011年国际生物奥林匹克竞赛实验试题


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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Student Code:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY
Total Points: 100 Duration: 90 minutes

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Dear Participants,
? In this test, you have been given the following 3 tasks: Task I: Protein electrophoresis (35 points) Task II: Protein quantification (35 points) Task III: Protein purification (30 points) ? ? Check your Student Code on the Answer Sheet before starting the test. Write down your results and answers in the Answer Sheet. Answers written in the Question Paper will not be evaluated. ? Make sure that you have received all the materials listed for each task. If any of the listed items is missing, raise your sign. ? ? Use pen only. You should organize your work efficiently but ensure that you complete Task II early enough to obtain the spectrophotometer readings to answer the questions that follow. ? ? Stop answering immediately after the end bell rings. After test, enclose the Answer sheets, Question paper, and Data printout in the provided envelope. Our lab assistants will collect it promptly. ? NO paper or materials should be taken out of the laboratory.

Good Luck!!

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Shared instruments:
Camera, spectrophotometer, printer

Equipment and Materials:
Equipment: 1 Power supply 2 Electrophoresis tank (with gel and buffer) 3 Micropipettes P20 and P200 4 80-well microcentrifuge tube rack 5 Wire test tube rack with 15-mL centrifuge tubes (×6) (yellow cap) 6 4-way test tube rack 7 Plastic droppers in 15-mL centrifuge tubes 8 Micropipette tips (for P20 and P200) 9 Timer 10 96-well microplate 11 Marker pen & paper label 12 600-mL beaker for waste disposal 13 Scissors and a ruler 14 Double-sticker to attach the results 15 Student Code sticker 16 Tissue paper 17 Mini centrifuge (if you need to spin down the samples in the microcentrifuge tubes) Quantity 1 1 1 each 1 1 1 2 1 each 1 1 1 each 1 1 each 1 1 1 1

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Materials: 1 Loading dye (microcentrifuge tube-L) (pink tube with orange label)

Quantity 1

2 Pre-stained protein molecular weight marker (microcentrifuge tube-M) (pink 1 tube with orange label) 3 Unknown pre-stained protein samples (microcentrifuge tubes-U1 and U2) (pink tube with orange label) 4 CBG reagent in 50-mL centrifuge tube 5 Bovine serum albumin (BSA) concentration standard (0.5 mg/mL) in microcentrifuge tube (green tube with yellow label) 6 Enzyme E in two microcentrifuge tubes: concentrations X and Y (green tube with yellow label) 7 Distilled water (microcentrifuge tube-ddH2O) (green tube with yellow label) 8 Protein sample (microcentrifuge tube-C) (blue tube with blue label) 9 Anion exchange chromatography column on 15-mL centrifuge tube 10 Anionic buffers A and B (5 mL each in two separated 15-mL centrifuge tubes) (green cap) 11 Coomassie brilliant blue G-250 (CBG) reagent 1 mL in each of six 15-mL centrifuge tubes (A1 to A3 & B1 to B3, red cap) 1 1 1 1 1 1 1 1 1

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Task I (35 points) Protein electrophoresis
Introduction: Polyacrylamide gel electrophoresis (PAGE) is a common technique for protein study. It can be used to separate different proteins based on their charges or sizes. A type of PAGE is termed SDS-PAGE, in which the negatively charged chemical, SDS, is added before protein electrophoresis. The amount of SDS that binds to proteins is proportional to the size of the protein which gives each protein a similar charge-to-mass ratio and renders the intrinsic charge of the protein insignificant, at least for this experiment. Thus, the major factor that affects the migration of protein is the molecular weight (MW) of the protein during SDS-PAGE. The relative mobility (Rf) of the protein can be calculated as the ratio of the distance migrated by the protein to that migrated by the dye-front. The value of Rf is negatively proportional to the log of its molecular weight. In the problem set, you will perform the following experiment: 1. An electrophoresis tank has been set up for SDS-PAGE, in which a polyacrylamide gel has been secured on an electrode assembly and electrophoresis buffer bath has been filled. There are 10 wells for sample loading on the top of the gel. To load the sample, use the P20 micropipette with tip to withdraw a protein sample, and carefully place the tip on the top of the well. By

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

injecting slowly the sample will sink to the bottom of the well by gravity (Figure 1). 2. If you need to practice, use the P20 micropipette with tip to withdraw 10 ?L of loading dye from microcentrifuge tube L (pink tube with orange label) on rack. Load the dye into wells 1 to 3 or 7 to 10. 3. Each of the microcentrifuge tubes M, U1 and U2 (pink tube with orange label) contains 15 ?L of protein molecular weight marker, unknown protein U1 and unknown protein U2, respectively. Use micropipette P20 to withdraw 10 ?L solution from each tube and load the samples into wells 4 to 6 as shown in Figure 1. 4. As soon as you finish sample loading, Lift the sign , lab assistants will connect the power cord to power supply and set the voltage to 200 V for you. The gel will run for 25 minutes. The timer will be set up by an assistant to countdown. 5. After finishing electrophoresis, Lift the sign , lab assistants will disassemble the electrophoresis set-up and give back your gel. Wipe clean the surface of the gel with tissue papers and label the gel with your Student Code sticker. Lab assistants will take the photo of your gel. Put the photo on the answer sheet using double-sticker (5 points).

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Figure 1 Answer the following questions: Q.1.1. (2 points) Figure 2 shows a photograph of a SDS-PAGE gel. The electrophoresis start point and dye-front are indicated. Which side of the gel should be connected to the anode (+ charge) of the power supply? Mark your answer (X) on the answer sheet.

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Figure 2

Q.1.2. (8 points) Based on the information provided in Figure 2, plot the molecular weight of the

five marker proteins versus their relative migration-Rf values on the graph paper provided (4

points). Use the graph to estimate the molecular weights of unknown proteins on lanes A and B (4

points). Write down your answers on the answer sheet.

Q.1.3. (5 points) A protein complex of molecular weight 246 kDa is composed of multiple

subunits bound by non-covalent interaction. Two protein bands of 57 and 33 kDa were identified

after SDS-PAGE. How many 57-kDa and 33-kDa subunits, respectively, are included in the
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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

protein complex? Write down your answers on the answer sheet.

Q.1.4. (5 points) The average molecular weight of amino acid residues is about 110 daltons. How

many amino acids are there in the 33-kDa protein subunit? How many nucleotides of RNA are

translated into the protein? Write down your answers on the answer sheet. Q.1.5. (5 points) Suppose the average molecular weight of nucleotides is 330 daltons. Excluding introns and the stop codon, what is the mass ratio of dsDNA that encodes the 33-kDa protein, to the 33-kDa protein? Write down your answer on the answer sheet. Q.1.6. (5 points) Suppose a protein P can bind to a protein Q (MW = 1000 daltons). The binding can be revealed by gel-mobility shift assay. 200 pmol of protein P were mixed with various amounts (0 to 500 ng) of protein Q. These mixtures were resolved by 10% (w/v) polyacrylamide gel. The gel was stained by Coomassie blue and is shown in Figure 3. Calculate the binding molar ratio of proteins P and Q. Write down your answer on the answer sheet.

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Figure 3

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Task II (30 Points) Protein quantification
Introduction: Coomassie Brilliant Blue G-250 (CBG) is a protein staining reagent. It appears a different color under different pH conditions. It looks reddish brown in acidic solution, whereas it turns blue under neutral or alkaline conditions. Since proteins can provide a relatively neutral environment, CBG will turn blue with the maximum absorbance at a wavelength of 595 nm when binding to protein. The more protein there is in a sample, the more CBG will bind to it, and thus, the higher intensity the blue color will be. In other words, the absorbance at 595 nm is proportional to the amount of protein in a sample. Based on this, one can determine the concentration of a protein by measuring the blue intensity of a sample. In the problem set, you will perform the following experiment: 1. To make BSA concentration standards (Table 1), add 0, 2, 4, 6, 8 and 10 ?L of 0.5 mg/mL BSA (green color) in A1 to A6 wells of a microplate (Figure 4). Make duplicated BSA concentration standards in B1 to B6 wells. If you make a mistake, you can repeat the procedure in wells A7 to A12 and/or B7 to B12. Adjust the total volume of each BSA solution to 10 ?L by adding an appropriate volume of H2O (Table 1).

2. Add 200 ?L of CBG reagent per well in A1 to A6 and B1 to B6. Mix and observe the color

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

change.

3. To determine the two concentrations X and Y of enzyme E, add various amounts (2, 4, 6, 8 and 10 ?L) of enzyme E (green color) in duplicate to empty wells and bring up the volume to 10 ?L with H2O.

4. Add 200 ?L of CBG reagent per well to the diluted enzyme E. Mix and observe the color change.

5. Lift the sign , lab assistants will accompany you to measure the absorbance values of your samples at 595 nm using spectrophotometer. Put your Student Code on the print-out data with marker pen.

6. Return to your work bench, and put the result on the answer sheet using double-sticker. Table 1 Well of a microplate Materials 0.5 mg/mL BSA (?L) H2O (?L) Diluted BSA concentration (mg/mL) A1 & B1 0 10 0 A2 & B2 2 8 A3 & B3 4 6 A4 & B4 6 4 A5 & B5 8 2 A6 & B6 10 0

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

A1 B1

A2 B2

A3 B3

A4 B4

A5 B5

A6 B6

Figure 4

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Answer the following questions: Q.2.1. (10 points) Calculate the concentrations of BSA in each sample (10 ?L) and fill in the blanks in the table on the answer sheet (Q.2.1.1. 5 points). Use these values to plot a standard curve of BSA concentrations (X-axis) versus mean absorbance values of duplicated standards (Y-axis) on the answer sheet (Q.2.1.2. 5points). Q.2.2. (12 points) For both X and Y choose the best absorbance value within the range of the BSA standard curve to determine concentration and fill in the table on the answer sheet. Q.2.3. (8 points) Based on the values you chose, calculate the original concentrations (X and Y) of enzyme E from the standard curve of BSA concentration. The concentrations should be expressed in units of mg/mL. Write down your answers on the answer sheet.

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Task III (35 points) Protein purification
Introduction: Column chromatography is commonly used for purification of proteins. The column is made by packing solid porous material (stationary phase) in a column filled with buffer solution (mobile phase). The protein solution to be separated is loaded on top of the column and allowed to percolate into the solid matrix (stationary phase). A reservoir at the top supplies elution buffer constantly which flows through the matrix and passes out of the column at the bottom (the eluent). Since proteins interact with the solid matrix to different degrees, individual proteins migrate faster or more slowly through the column depending on their properties. Therefore, one can obtain purified proteins by collecting eluent at different times ( Figure 5). Ion-exchange chromatography can be used to separate proteins with different electric charge at a given pH. In anion exchange chromatography, negatively charged proteins bind to the positively charged stationary phase. The bound proteins will be eluted using a solution containing anions to compete with proteins for the adsorption of solid matrix. In the practical, proteins are eluted first with buffer containing a lower concentration of anion, than with buffer containing a higher concentration of anions. Since differently-charged proteins interact with the stationary phase with different strengths, they can be separately eluted by different

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

concentrations of anionic buffers.

Figure 5

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

In the problem set, you will perform the following experiment (5 points): 1. Label six 15-mL centrifuge tubes (yellow cap) a1 to a3 and b1 to b3 accordingly, with a marker pen. 2. Take the anion chromatography column (Figure 6A), un-plug the tube and allow the solution to be drained by gravity. Plug the tube intermediately when the liquid surface reaches the top of the disc (Figure 6A, white arrow). Do not over-dry the gel as it may affect protein purification. 3. Withdraw 200 ?L of protein solution from microcentrifuge tube C (blue tube with blue label) using a P200 micropipette, and apply the sample to the chromatography column slowly by touching the filled pipette tip lightly against the inside wall of the tube (Figure 6B). 4. Un-plug the column and allow the protein sample buffer to drain out, then transfer the column to centrifuge tube a1 (yellow cap). Withdraw 3 mL of anion buffer A (blue cap) with a plastic dropper and apply the solution to the gel by touching the pipette tip against the wall of the tube (Figure 6C). 5. Collect ~1 mL eluent in centrifuge tubes a1 to a3 (yellow cap) sequentially. It takes about 2 to 3 minutes for each tube. 6. Allow the contents of the column to drain entirely out then transfer the column to centrifuge tube b1 (yellow cap). Withdraw 3 mL of anion buffer B (blue cap) with a plastic dropper and

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

apply the solution to gel by touching pipette tip against the wall of the tube (Figure 6C). 7. Collect ~1 mL eluent in centrifuge tubes b1 to b3 (yellow cap) sequentially. It takes about 2 to 3 minutes for each tube. 8. Withdraw 50 ?L of eluent from tubes a1 to a3 & b1 to b3 (yellow cap) and transfer to centrifuge tubes A1 to A3 & B1 to B3 (red cap), respectively. Mix and observe color change. The CBG reagent (see introduction in Task II) in tubes A1 to A3 & B1 to B3 will turn blue when it reacts with the eluted protein. 9. After finishing all the experiments, Lift the sign , lab assistants will take a photo of your experiment results and put a stamp mark on your answer sheet. Without a stamp mark you will not be evaluated for Q3.1.1 AND Q3.1.2

Figure 6

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Q.3.1. (7 points) Mark the deepest color change (X) on the answer sheet (Q.3.1.1. 5 points). Which of the following buffers (buffer A or buffer B) can be used to elute the protein? Mark your answer (X) on the answer sheet (Q.3.1.2. 2 points). Q.3.2. (5 points) Enzyme A is a protein whose surface is evenly distributed with electric charges. If enzyme A can be eluted from anionic exchange chromatography by a high concentration of anionic buffer, what is the property of enzyme A with respect to electric charge? Mark (X) the answer on the answer sheet. (A) High negative net charges (B) Low negative net charges (C) Zero net charge (D) Low positive net charges (E) High positive net charges

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Q.3.3. (4 points) Amino acids differ in the chemical nature of the R group (side chain). Figure 7 shows four amino acids A, B, C, and D in their prevailing ionic forms at pH 7.2, with the side chain marked in a white box. Which of the following amino acids in Figure 7 would be present more frequently on enzyme A? Write down your answer (X) on the answer sheet.

Figure 7 Q.3.4. (5 points) Hydrophobic interaction chromatography can be used to separate proteins based on their hydrophobicity. To perform the chromatography, protein samples were first treated with buffer containing a high concentration of salts such as ammonium sulfate (NH4)2SO4, which will remove water molecules from the protein surface. This causes the hydrophobic area on the surface of the protein to be exposed. When the salt-treated proteins are subjected to chromatography, they will be adsorbed on the stationary phase through hydrophobic interactions. The higher the hydrophobicity of the protein, the stronger the adsorption. As salt concentration can affect the hydrophobic interaction between the protein and the stationary phase, different proteins can be separately eluted by using different concentrations of salt-containing buffers. If enzyme A is highly hydrophobic, which of the following buffers should be used to separate

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

enzyme A from other proteins by chromatography? Mark (X) the answer on the answer sheet. (A) Low-salt buffer (B) High-salt buffer (C) Buffer without salt (D) Low-salt buffer first then high-salt buffer (E) High-salt buffer first then low-salt buffer Q.3.5. (4 points) If enzyme A is highly hydrophobic, which of the amino acids in Figure 7 would be present more frequently on enzyme A? Mark (X) the answer on the answer sheet. Q.3.6. (5 points) Gel filtration chromatography separates proteins based on their sizes. The gel, or stationary phase, consists of cross-linked polymer beads with engineered pores of a particular size. Small proteins enter the pores and move slowly through a complex path. Large proteins cannot enter the pores and so take a short path through the column, around the beads. Table 2 is a list of gels and their fractionation ranges. Suppose both enzyme A (22 kDa) and protein B (44 kDa) are single-subunit proteins. Which gel is best suited for the task of purifying enzyme A from a mixture containing enzyme A and protein B, using gel filtration chromatography. Mark your answer (X) on the answer sheet.

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IBO – 2011 TAIWAN PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY

Table 2 Types of stationary phase G-10 G-15 G-25 G-50 G-75 G-100 G-150 G-200 Fractionation range (MW, Da) <700 <1500 1,000-6,000 1,500-30,000 3,000-70,000 4,000-150,000 5,000-400,000 5,000-800,000

Q.3.7. (5 points) Assume that the concentration of total proteins in the original solution is 1 mg/mL and the activity of enzyme A is 0.5 units in 1 mL protein sample. The concentration of total proteins after purification is 0.1 mg/mL and the activity of enzyme A is 1 unit in 1 mL protein sample. Calculate the purification factor (number of times purified) of enzyme A. Write down your answer on the answer sheet.

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

22nd INTERNATIONAL BIOLOGY OLYMPIAD 10th — 17th, 2011 Taipei, Taiwan

PRACTICAL TEST 1 BIOCHEMISTRY AND CELL BIOLOGY
Total Points: 100 Duration: 90 minutes

ANSWER KEY

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Q.1.1. (2 points)

Start point Anode (+ charge)

Dye-front

X
Q.1.2. (4+4 points)

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Protein Molecular weight

Lane A

Lane B

56 (53-59)
(kDa) Q.1.3. (5 points)

72 (68-76)

57 kDa Number Q.1.4. (5 points)

33 kDa

2

4

Amino acid Number

Nucleotide

300

900

Q.1.5. (5 points) DNA

:

Protein

18 :

1

Q.1.6. (5 points) P

:

Q

=

1

:

2

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Task I photo - protein electrophoresis (5 points)

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Task II result sheet - Protein quantification

BSA X Y

Q.2.1.1. (5 points)

Well of a microplate Materials 0.5 mg/mL BSA (?L) H2O (?L) Diluted BSA 0 concentration (mg/mL) A1 & B1 0 10 A2 & B2 2 8 A3 & B3 4 6 A4 & B4 6 4 A5 & B5 8 2 A6 & B6 10 0

0.1

0.2

0.3

0.4

0.5

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Q.2.1.2. (5 points) Standard curve for BSA

Q.2.2. (12 points) Solution X Sample volume (?L) H2O (?L) Optical density - OD595 nm 2 8 0.859 4 6 6 4 8 2 10 0 2 8
0.452

Solution Y 4 6 6 4 8 2 10 0

0.521 0.586 0.631 0.691

out of range

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Q.2.3. (8 points)

Concentration of X (mg/mL)

Concentration of Y (mg/mL)

2.0 (1.6-2.4)

0.26 (0.21-0.31)

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Q.3.1.1. (5 points)

Tube Color change

A1

A2

A3

B1

B2

B3

X

Q.3.1.2. (2 points)

A Buffer

B

X

Q.3.2. (5 points)

A

B

C

D

E

X

Q.3.3. (4 points)

A

B

C

D

X

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IBO – 2011 TAIWAN PRACTICAL TEST 1 – ANSWER KEY BIOCHEMISTRY AND CELL BIOLOGY

STUDENT CODE:

Q.3.4. (5 points)

A

B

C

D

E

X

Q.3.5. (4 points)

A

B

C

D

X

Q.3.6. (5 points)

G-10

G-15

G-25

G-50

G-75

G-100

G-150

G-200

X

Q.3.7. (5 points)

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Student Code:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY
Total Points: 100 Duration: 90 minutes

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Dear Participants,
? In this test, you have been given the following 2 tasks: Task I: The observation of the sciatic nerve of American bullfrog (58 points) Task II: The observation of tissue morphology and the match of their functionality (42 points) Check your Student Code on the Answer Sheet before starting the test. Write down your results and answers in the Answer Sheet. Answers written in the Question Paper will not be evaluated. Make sure that you have received all the materials listed for each task. If any of the listed items is missing, raise your sign. Use pen ONLY. You must complete Task I first. Stop answering and put down your pen immediately after the end bell rings. After test, enclose both the Answer sheets and Question paper test sheets in the provided envelope. Our Lab assistants will collect it promptly. No paper or materials should be taken out from the laboratory.

? ? ? ? ? ? ? ?

Good Luck!!

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Equipment and Materials:
For task I: The observation of the sciatic nerve of American bullfrog. Instruments/materials Bullfrog specimen Dissecting tray Round plastic petri dish Ringer ’s solution (in wash bottle) Pin (in a glass bottle) Cotton thread (in a glass bottle) Electric stimulating device Wet paper Gloves Dissecting equipment : scissors (large) scissors (small) fine forceps Quantity 1 1 1 500 10 2 1 1 1 1 1 2 unit piece piece piece mL pieces pieces set set pair pair pair pairs

For task II: The observation of tissue morphology and the match of their functionality Instruments/materials Microscope Prepared tissue sections (marked A to J) Coloured pictures of tissue sections (numbered 1 to 9) on 3 sheets of A4 paper Quantity 1 10 1 unit set slides set

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Task I (58 points) The observation of the sciatic nerve of the American bullfrog.
Introduction: The sciatic nerve is a branch of the sacral plexus. It is the thickest and longest nerve tract in the body, extending from the vertebral column to the foot. The sciatic nerve includes the distributed sensory and motor nerves that control most sensory and motor activities of the lower extremities. Mediated by the sciatic nerve, sensory signals from the lower limbs are transmitted to the brain. Similarly, muscle contraction of the lower extremities can be stimulated by nerve impulses from the brain. The aim of this experiment is to observe and isolate the sciatic nerve from the bullfrog. Experiment Procedure: Step 1 to 5: (Do not allow the tissues to dry out. To keep the tissues wet, a small amount of Ringer’s solution may be added onto the tissue at anytime.) 1. Carefully check if all the experiment instruments/materials are fully provided. Raise your sign if you have any problem. 2. Put the bullfrog specimen on the dissecting tray. 3. First, carefully observe the 10 pairs of spinal nerves extending from the vertebrae of the bullfrog. Next, locate the sciatic nerve that is formed by pairs of spinal nerves VII, VIII and IX (as shown in Fig. 1). 4. Turn on the switch (on/off) on the electric stimulating device. The red light will be lit up immediately, indicating that the device is functioning. 5. Simultaneously stimulate the sciatic nerve with the two electric wires that are separately connected to the (+) and (-) electrodes of the electric stimulating device. Fig. 3 indicates the position of the sciatic nerve emerging from the spinal cord. Observe the contracting response of the hind limb.
Electrodes (+) (-)
Spinal Nerve I Spinal Nerve II Spinal Nerve III Spinal Nerve IV Spinal Nerve V Spinal Nerve VI Spinal Nerve VII Spinal Nerve VIII Spinal Nerve IX Spinal Nerve X Sciatic Nerve
OFF ON

Switch

Region of stimulation

Red LED

Fig. 1. Spinal nerve

Fig. 2. Electric stimulating Fig. 3. Sciatic nerve device

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Q.1.1. (9 points) When you have finished the above five steps, raise the sign to notify the Lab assistant to videotape the contraction. Step 6 to 10: (To keep the tissues wet, a small amount of Ringer’s solution may be added onto the tissues at any time) 6. Use a pair of scissors to make a circular cut through the skin spanning the circumference of the upper part of one thigh of the bullfrog. Starting from the cutting point, completely peel off the skin by hand to remove it from the hind limb (Fig. 4) It may be necessary to cut some connections between the skin and underlying tissue. 7. Lay the bullfrog on the dissecting tray with its back facing up. 8. Push two pins separately into both ends of the gastrocnemius and separate it from the tibiofibular (shinbone) (Fig. 5). 9. The sciatic nerve is located in a trough surrounded by thigh muscles. Carefully separate the muscles on both sides of the trough and to expose the pale yellow colored sciatic nerve. Pass a cotton thread underneath the sciatic nerve to label it. 10. Stimulate the cotton thread-labeled sciatic nerve with the provided electric stimulating device and observe the contracting response of the gastrocnemius. Cotton thread

Skin

Pin Sciatic nerve

Gastrocnemius Fig. 4. Demonstration of skin peeling Fig. 5 Sciatic nerve and gastrocnemius

Q.1.2. (8 points) When you have finished steps 6 to 10, raise the sign to notify the Lab assistant to videotape the contraction.

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Step 11 to 12: (To keep the tissues wet, a small amount of Ringer’s solution may be added onto the tissues at anytime) 11. Completely separate and isolate the INTACT sciatic nerve – gastrocnemius muscle preparation from the bullfrog specimen and place it in a petri dish, as shown in Fig. 6. (The sciatic nerve must be at least 2cm long). 12. Stimulate the sciatic nerve with the electric stimulating device and observe the contraction response of the gastrocnemius.

Tendon (A) Gastrocnemius Sciatic nerve-gastrocnemius connection (B) Sciatic nerve

Student code and scale bar Fig. 6 Isolation of 「sciatic nerve-gastrocnemius」tissue Q.1.3. (40 points) When you have finished steps 11-12, raise the sign to notify the lab assistant to check your specimen integrity and to videotape the contraction.

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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Task II (42 points)
Identify tissues based on their morphology and match their functionality
Introduction: The vertebrate physiological system is established by the functional coordination of 11 organ systems, which include the skin, skeletal, muscle, nervous, endocrine, cardiovascular, lymphatic, respiratory, digestive, urinary, and reproductive systems. Identify the specimens on the slides (30 points) Slides A to J are tissue sections from vertebrates. Identify the tissues or cell types, based on their characteristic features, using the microscope.
1. Vein 6. Blood (human) 11. Smooth muscle 16. Pancreas 2. Artery 7. Ovary 12. Cardiac muscle 17. Intestine 3. Ganglion 8. Testis 13. Kidney 18. Gastric tissue 4. Neuron 9. Lung 14. Cartilage 19. Skin 5. Blood ( frog ) 10. Skeletal muscle 15. Bone 20. Rectum

Q.2.1. (30 points) Match each slide specimen (A to J) with its correct name from the 20 different tissue/organ names listed in table above. (Note: there is ONLY one correct answer for each specimen). Fill in the correct number in the answer sheets. Identify the photographed tissue sections and match their correct functions (12 points) Fig. 1-9 are enlarged pictures of tissue sections of different mammalian tissues. Based on their structural features, identify the tissue and answer the questions below. The functions of 11 organs are listed in the following table. Each specific function is assigned an alphabetic letter (A to K). Symbol Functional description A B C D E F G H I J K Producing vitamin D3 Producing erythropoietin Producing urea Producing surfactant to reduce surface tension Regulating the homeostasis of the pH of body fluid Helping vein compression and promoting blood flow back to the heart Digesting proteins Secreting secretin Producing inhibin Major organ for the storage of calcium and phosphate Producing progesterone
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IBO – 2011 TAIWAN PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY

Q.2.2. (13 points) Correctly write down the functional symbols, i.e. the alphabetic letters (A to K), on the answer sheets. Match the organ with their functionalities. Note: some organs may have more than one function (1 point will be deducted for each incorrect answer and the minimum score will not be less than zero).

8

Fig 1.

H&E stain (400 X)

Fig 2.

H&E stain (100 X)

Fig 3.

H&E stain (200 X)

Fig 4.

H&E stain (200 X)

Fig 5.

H&E stain (200 X)

Fig 6.

H&E stain (40 X)

Fig 7.

(200 x)

Fig 8.

H&E stain (100 X)

Fig 9.

H&E stain (200 X)

IBO – 2011 TAIWAN PRACTICAL TEST 2 – ANSWER KEY ANIMAL PHYSIOLOGY AND ANATOMY

STUDENT CODE:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 2 ANIMAL PHYSIOLOGY AND ANATOMY
Total Points: 100 Duration: 90 minutes

ANSWER KEY

1

IBO – 2011 TAIWAN PRACTICAL TEST 2 – ANSWER KEY ANIMAL PHYSIOLOGY AND ANATOMY

STUDENT CODE:

Q.1.1. (9 points) When you have finished the above five steps, raise the sign to notify the Lab assistant to videotape the contraction. Lab assistant signature

The 1st checkpoint

Points

Leg limb muscle contraction

0 or 9

Q.1.2. (8 points) When you have finished the steps 6 to 10, raise the sign to notify the Lab assistant to videotape the contraction. Lab assistant signature

The 2nd checkpoint Indicate the scietic nerve correctly

Points

0 or 4 (nerve must be intact)

Gastrocnemius contraction

0 or 4

2

IBO – 2011 TAIWAN PRACTICAL TEST 2 – ANSWER KEY ANIMAL PHYSIOLOGY AND ANATOMY

STUDENT CODE:

Q.1.3. (40 points) When you have finished the steps 11-12, write down the results of your observation on the answer sheets. And, raise the sign to notify the Lab assistant for checking the results and videotape the contraction.

The 3rd checkpoint Points

Lab assistant signature

Isolation of “sciatic nerve” 0 or 15 (must be more than 2 cm) Isolation of “gastrocnemius” (must be intact; including A and B parts) Contraction of the isolated sciatic nerve-gastrocnemius tissue. 0 or 10 0 or 15

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IBO – 2011 TAIWAN PRACTICAL TEST 2 – ANSWER KEY ANIMAL PHYSIOLOGY AND ANATOMY

STUDENT CODE:

Q.2.1. (30 points) Match each slide specimen (A to J) with its correct name from 20 different tissue/organ names listed in above table. (Note: only one correct answer for each specimen). Fill in the correct number in the answer sheets. Slide specimen A. B. C. D. E. F. G. H. I. J. Answers(3 points each)

4 9 15 10 14 11 18 5 12 17

4

IBO – 2011 TAIWAN PRACTICAL TEST 2 – ANSWER KEY ANIMAL PHYSIOLOGY AND ANATOMY

STUDENT CODE:

Q.2.2. (13 points) Correctly write down the functional symbols, i.e. the alphabetic letters (A to K), on the answer sheets. Match the organ with their functionalities.

Note: some organs may have more than one function (1 point will be deducted for each incorrect answer, minimum score will not be less than zero).

Figure 1. 2. 3. 4. 5. 6. 7. 8. 9.

Functional symbol

B,E A C D,E F I, K J G,H I

5

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Student Code:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS
Total Points: 100 Duration: 90 minutes

Dear Participants,
? In this test, you have been given the following 2 tasks:
1

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Task I: Reconstruct the phylogenetic tree for the given spiders (60 points) Task II: Test of species association in a community (40 points) ? ? ? ? ? ? ? Check your Student Code on the Answer Sheet before starting the test. Write down your results and answers in the Answer Sheet. Answers written in the Question Paper will not be evaluated. Please make sure that you have received all the materials listed for each task. If any of the listed items is missing, please raise your sign. Use pen only. You can use a ruler and a calculator provided. Check the condition of the spiders in the first 5 minutes. If any of the legs is missing, please raise your sign. No replacement of the spiders is possible after 5 minutes. Stop answering and put down your pen immediately after the end bell rings. After test, our lab assistants will check the condition of the spiders and fill out the spider checklist at the end of your answer sheet. Each undamaged spider in the original vial will get you one bonus point. Please put down student code and sign after the check is done. Enclose both the Answer Sheets and Question Paper in the provided envelope after the spider check is finished. Our Lab Assistant will collect it promptly.

?

Good Luck!!

2

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Equipments and Materials:
Equipment: 1 Dissecting microscope 2 Four sheets of colored pictures and one sheet of black and white pictures: Figures (figure 1-3 to 1-12) Figure (figure 2-1) 3 Forceps 4 5 6 7 Petri dish 70% ethanol Plastic dropper 1-m quadrat cardboard (represented by a small cardboard in a zip lock bag) 4 1 2 2 1 1 1

1

Materials: 1 Four spider samples in glass vials (W, X, Y, Z)

1

3

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

TASK I: (60 points) Reconstruct the phylogenetic tree for the given spiders

cervical

A

B
Figure 1-1 External morphology of spider. A. Dorsal view. B. Ventral view.
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IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

A: two transverse eye rows, with 4 eyes on each eye row

B: two rows, with 6 eyes on the first eye row

C: four rows, with 2 first row eyes smallest and 6 posterior eyes arranged in a hexagonal configuration

D: three rows, with the 4 first E: three rows, with the 2 row eyes much smaller than anterior median eyes much the others larger than the others

F: three rows, with 4 eyes on the second eye row

G: eight eyes with one diad (a cluster of 2 eyes) and two
triads (a cluster of 3 eyes)

H: six eyes in two triads

I: six eyes in three diads

Figure 1-2 Eye arrangements (A key is provided on pages 7-9) Legends and Abbreviations of figures 1-3 to 1-12 Figure 1-3 Book lungs. A. Two pairs. B. One pair. Figure 1-4 Spinnerets. A. Three pairs. B. Two pairs. Figure 1-5 Cribellum. A. Absent. B. Present. Figure 1-6 Calamistrum on metatarsus IV. A. Absent. B. Present. Figure 1-7 Figure 1-8 Figure 1-9 Figure 1-10 Tarsi claw. A. Three claws. B. Two claws. Claw tufts. A. Absent. B. Present. Base of anterior spinnerets (AS). A. Widely separated. Grades of legs. A. Prograde. B. Laterigrade.
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B. Close or in contact.

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Figure 1-11 Tibia and metatarsus of legs I and II have a series of long spines interspersed with much shorter setae. A. Absent. B. Present. Figure 1-12 Double-rowed trichobothria on femora IV. A. Absent. B. Present.

1-1 Key to species of some common spiders There are many living creatures in the world. For unfamiliar creatures, scientists usually choose a suitable key, the most commonly used tool, to find out its name. A key uses dichotomous statements (a or b) of diagnosed characters to divide a larger group of taxa into two smaller subgroups (indicated by numbers or taxon names). Beginning with the number 1, choose a more likely statement (a or b) for the specimen and then go to the number shown at the end of the statement, and so on. Go through the key, until a taxon name is shown. A key for some common spiders of the world is given below.

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IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Key to species of some common spiders 1a Two pairs of book lungs (Fig. 1-3A) ……………………………………………2 1b One pair of book lungs (Fig. 1-3B) …………………………………………… 3 2a Three pairs of spinnerets (Fig. 1-4A)……………………………………… A. aus 2b Two pairs of spinnerets (Fig. 1-4B) ……………………………………… M. bus 3a With a cribellum in front of the spinnerets (Fig. 1-5B), and a calamistrum on metatarsus IV (Fig. 1-6B) …………………………………………………Z. cus 3b Without the cribellum and calamistrum (Figs. 1-5A, 1-6A) …………………… 4 4a With six eyes …………………………………………………………………… 5 4b With eight eyes ………………………………………………………………… 6 5a six eyes in three diads (Fig. 1-2I) ………………………………………… S. dus 5b six eyes in two triads (Figs. 1-2H) ……………………………………… P. eus 6a Tarsi with two claws (Fig. 1-7B), with or without claw tufts ………………… 7 6b Tarsi with three claws (Fig. 1-7A), never with claw tufts (Fig. 1-8A) ……… 10 7a Eyes in three or four rows (Figs. 1-2C, D, E, F) ……………………………… 8 7b Eyes in two rows (Figs. 1-2A, B) ……………………………………………… 9 8a Eyes arranged in three rows in 2-4-2 conformation; with a pair of remarkably large anterior median eyes (AMEs) (Fig. 1-2E)……………………………………………T. fus 8b Eyes arranged in 2-4-2 three rows (Figs. 1-2F); AMEs not as above …… C. gus 9a Base of both anterior spinnerets separated from each other or wide apart (Fig. 1-9A); Legs prograde (Fig. 1-10A) ……………………………………… Z. hus 9b Bases of both anterior spinnerets in contact (Fig. 1-9B); Legs laterigrade (Fig. 1-10B) …………………………………………………………………… T. kus 10a Eye group hexagonal, eyes arranged in 4 rows, in a 2-2-2-2 pattern (Fig. 1-2C)……………………………………………………………………… O. lus 10b Eye group not hexagonal……………………………………………………… 11 11a Eyes in two rows (Figs. 1-2A, B)……………………………………………… 12 11b Eyes in three rows (Figs. 1-2D, E, F)………………………………………P. mus 12a Tibia and metatarsus of legs I and II armed with series of long spines interspersed with much shorter setae (Fig. 1-11B) …………………………………… M. nus 12b Legs I and II without such spine arrangement ………………………………… 13 13a Femora IV with a proximal cluster of double-rowed trichobothria (Fig. 1-12B) …………………………………………………………………… L. ous 13b Femora IV without such trichobothria (Fig. 1-12A)……………………… N. pus You have four spider specimens coded W to Z, respectively.
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Identify some of their (Caution!

characters with the aid of figures 1-1 to 1-12 and identify all spiders using the key.

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

You may take out the specimen from the vials for identification. When you do so, you should place a spider in the petri dish with some 70% alcohol to examine its characters under the stereomicroscope. Because the spider ’s body is very fragile, the best way to handle the specimen is to gently grasp its legs with a pair of forceps to move it in or out from the vial. Don’t break the spider’s body or its legs. Undamaged spiders in their original vials will get extra bonus points. Please handle everything with care! alcohol at all times to prevent desiccation). Spiders should be kept in 70%

Q1.1.1 (4 points for each correct spider; 16 points total) Match each spider code with the correct taxon name respectively in your Answer Sheet Note: each spider code can only be used once; repeated taxa cells will not be given any points. Q1.1.2 (0.65 points for each cell; 13 points total) If a spider has the characters listed in the left column of the table in your Answer Sheet, indicate with a “+” and if the character is absent, indicate with a “-”. (Penalty of 0.2 points for each wrong answer, minimum 0 point) 1-2 Reconstruct a phylogenetic tree for eight spiders Data matrix 1-1 represents character entries (a to t) for a group of hypothetical organisms A to H. Based on Data Matrix 1-1, Taxon A serves as the outgroup and the other 7 organisms (Taxa B to H) are ingroups. Character state 0 represents the pleisiomorphy (ancestral character) and states 1-6 are apomorphies (derived characters). “-“ represents missing character. We may reconstruct a cladogram (cladistic tree) by using synapomorphies (shared derived characters). Each change represents one step of the evolutionary events (indicated by the character and its state, e.g., e-5, t-4). The following tree (Figure 1-13) is the only resulting most parsimonious cladogram that shows all the character changes on the tree. Numbers 1 to 15 represent 15 steps of the tree. Data Matrix 1-1
Taxa A B C D E F G H

Character a 0 1 1 1 1 1 1 1 b 0 1 1 1 1 1 1 1 c 0 0 0 0 0 0 0 0 d 0 1 1 1 0 0 1 0 e 0 5 6 3 1 1 4 1 f 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 0 1 h 0 1 0 0 0 0 1 1 m 0 1 0 0 0 0 1 1 n 0 1 0 0 0 0 1 1 o 0 0 0 0 0 1 0 0 p 0 0 0 0 0 0 0 0 s 0 2 2 2 1 1 2 2 t 3 4 -

8

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Figure 1-13

The most parsimonious cladogram reconstructed from data matrix 1-1.

Q1.2. (1.2 points for each cell; 18 points total) Fill in the character code and state (e.g., e-1) in the Answer Sheet for each of the 15 steps. 1-3 Based on the cladogram (figure 1-13), answer the following questions: Q1.3.1. (2 points) How many steps of the cladogram are there in total? Q1.3.2. (2 points) Besides character e-1, which character is homoplasious (i.e., not an homologous character)? Q1.3.3. (2 points) Which of the following taxon is the sister group of taxon {C, D}? (A) {E, F} (B) {H, B, G} (C) {F} (D) {H} (E) {B, G} Q1.3.4. (Each correct answer will get 0.4 points, 2 points total) Mark with an “X” in the “True” cell in the Answer Sheet if the characters given below appeared prior to the evolution of character m-1 in the cladogram; otherwise, mark the “False” cell. Character s-1 s-2 a-1 g-1 d-1

9

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Q1.3.5. (1 point for each cell; 5 points total) To what kind of grouping do the following taxa belong? Use code “I” for polyphyletic, “II” for paraphyletic, or “III” for monophyletic grouping. Taxon {H} {B, C, G, H} {C, D, E, F} {B, G, H} {B, E, G}

10

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

TASK II: (40 points) Test of species association in a community
The basic idea of community organization is that species tend to be associated in a non-random manner. One way to understand their association conditions is to use a 2 × 2 contingency table (Table 2-1-0). If a sample contains both species x and y, it is defined as type “a”. If a sample contains only species y, species x, or no species, then it is defined as type “b”, “c”, or “d” respectively. Table 2-1-0 Species x Species y Present Absent Total Present a c a+c Absent b d b+d Total a+b c+d n

n=a+b+c+d Probability of obtaining species x, P(x) = (a+c)/n Probability of obtaining species y, P(y) = (a+b)/n Joint probability (JP): the probability of both species x and y being present, JP = P(x) × P(y) Expected joint occurrences = n x JP Significance level for Chi-squared statistical test (χ2) Significance level (α) 0.05 df 1 2 3 3.841 5.991 7.815 6.635 9.210 11.345 0.01

Figure 2-1 (the figure on a separate paper) is a distribution map of two plant species, Plant-A (○) and Plant-B (●), and a sympatric spider species, Spider (?), in a hypothetical community. Each square is 0.5 × 0.5 m2.

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IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

2-1 Association between Plant-A (○) and Spider (?): analyzed by quadrat method. Put a 1-m square quadrat on Figure 2-1 using the following 40 randomly assigned coordinates as the center (i.e., 2 × 2 complete squares) and determine the type of each quadrat. N-11, S-8, F-10, Q-18, O-16, K-2, L-4, M-17, M-4, H-17 X-2, K-11, T-19, M-8, P-10, G-8, B-19, M-19, S-10, O-12 J-18, D-7, B-17, I-11, B-10, G-13, V-16, C-3, F-5, R-15 L-2, Q-11, R-5, G-11, K-10, T-10, X-9, R-3, O-3, F-16 For example, a coordinate of C-3 would look like this:

Q2.1.1. (1 point each; 9 points total) Write down your results in Table 2-1-1 and complete all the blank cells. Answer the following in your Answer Sheet: Q2.1.2a. (0.6 points) Calculate P (Plant-A). Q2.1.2b. (0.6 points) Calculate P (Spider). Q2.1.2c. (0.6 points) Calculate JP (Plant-A and Spider) Q2.1.2d. (0.6 points) Calculate the expected joint occurrences. Q2.1.2e. (0.6 points) Two species are more likely to be positively associated if the actual observation of the joint occurrence is greater than the expected one, and negatively associated if the actual observation is smaller than the expected one. What kind of association exists between Plant-A and Spider? [Answer Code: P for positive association, N for negative association .] 2-1-3 A simple Chi-squared statistical test (χ2) with one degree of freedom (df = 1) is calculated as follows: n=a+b+c+d n (ad-bc)2 2= χ (a+b)(c+d)(a+c)(b+d) Q2.1.3. (2 points) Based on Table 2-1-1, calculate χ2 (to the fourth decimal place).
12

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

2-1-4 The strength of the association between the two species can be estimated from a coefficient (V), defined as follows:

V?

ad ? bc (a ? b)(c ? d )(a ? c)(b ? d )

The V value varies from -1 (strongly negative association) to +1 (strongly positive association) and it is zero when there is no association. Q2.1.4a. (2 points) Calculate the V value from Table 2-1-1 (to the fourth decimal place). Q2.1.4b. (2 points) According to the V value, what can be hypothesized about the strength of the association exists between the two species? (Mark your answer with an “X” in the cell) 2-2 The following table shows data derived from 40 randomly placed 2-m square quadrats. Table 2-2-1 Spider (?) Plant-A(○) Present Absent Total Present 14 8 22 Absent 16 2 18 Total 30 10 40

The expected joint occurrence is 16.5. The Chi-squared statistical test (χ2) with one degree of freedom (df = 1) is calculated as χ2 = 3.3670. V = -0.2901 Based on Table 2-2-1 answer the following questions: Q2.2.1a. (2 points) According to the expected joint occurrence, what can be hypothesized about the kind of association between Plant-A and Spider? [Answer Code: P for positive association, N for negative association] Q2.2.1b. (2 points) According to the V value, what can be hypothesized about the strength of the association between the two species? (Mark your answer with “X” in the cell on the Answer Sheet) 2.2.2. (6 points total) Answer the questions: Mark with an “X” on the Answer Sheet whether each of the following statements are true or false. Q2.2.2a. (2 points) Both tests of association using 1-m and 2-m square quadrats (sections 2-1 and 2-2), allowed us to reject the null hypothesis of random distribution.
13

IBO – 2011 TAIWAN PRACTICAL TEST 3 ECOLOGY AND SYSTEMATICS

Q2.2.2b. (2 points) The larger the quadrat size used, the more accurate the results. Q2.2.2c. (2 points) Increasing the sampling efforts in the quadrat method should improve the accuracy for the results for species association. 2-3 Association between Plant-A (○) and Plant -B (●): analyzed by the nearest neighbor method. Tally up the frequencies of the nearest neighbor of each plant systematically for all individuals. Fill in the totals in the table printed in the Answer Sheet. Q2.3.1. (0.5 points for each cell; 3 points total) Write down your results in Table 2-3-1 and complete all the blank cells. Q2.3.2a. (2 points) Based on Table 2-3-1, with one degree of freedom (df = 1), calculate χ2 (to the fourth decimal place). Q2.3.2b. (3 points) Are these two plant species randomly distributed, associated or segregated? (Mark your answer with an “X” in the cell) 2-4 Mark with an “X” on the Answer Sheet whether each of the following are true or false. (4 points total) Q2.4.1. (2 points) The null hypothesis of the χ2 test for the nearest neighbor method is that both Plant-A and Plant-B are randomly distributed. Q2.4.2. (2 points) Using the nearest neighbor method to test species association can avoid the quadrat-size effect.

14

IBO – 2011 TAIWAN PRACTICAL TEST 3 – FIGURE SYSTEMATICS AND ECOLOGY

A: two pairs

B: one pair

Figure 1-3 Book lungs A: three pairs B: two pairs

Figure 1-4 Spinnerets A: absent B: present

Figure 1-5

Cribellum

1

IBO – 2011 TAIWAN PRACTICAL TEST 3 – FIGURE SYSTEMATICS AND ECOLOGY

A: absent

B: present

Figure 1-6

Calamistrum on metatarsus IV

A: three claws

B: two claws

Figure 1-7 Tarsi claws A: absent B: present

Figure 1-8 Claw tufts

2

IBO – 2011 TAIWAN PRACTICAL TEST 3 – FIGURE SYSTEMATICS AND ECOLOGY

A: widely separated

B: close or in contact

Figure 1-9 Base of anterior spinnerets (AS) A : prograde B: laterigrade

Figure 1-10 Grades of Legs A: absent B: present

Figure 1-11 Tibia and metatarsus of legs I and II with series of long spines interspersed with much shorter setae.

3

IBO – 2011 TAIWAN PRACTICAL TEST 3 – FIGURE SYSTEMATICS AND ECOLOGY

A: absent

B: present

Figure 1-12 Double-rowed trichobothria on femora IV

4

IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 3
ECOLOGY AND SYSTEMATICS Total Points: 100 Duration: 90 minutes

ANSWER KEY

1

IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

Q.1.1.1. (4 points for each correct spider; 16 points total) Note: each spider code can only be used once, or the grades of these cells will not be counted. Taxon name A. aus C. gus L. ous M. bus N. pus O. lus P. eus Spider code Taxon name P. mus P. nus Spider code

X

Z Y W

S. dus T. fus T. kus Z. cus Z. hus

Q.1.1.2. (0.65 points for each right answer cell; 13 points total)
(Penalty of 0.2 point for each wrong answer, minimum 0 point)

Taxon Eyes in two rows

W

X

Y

Z

- + + - -

- + + - -

+ + + - -

+ + + - +

Tarsi with three claws

Bases of both anterior spinnerets in contact Calamistrum present on metatarsus IV Present a cluster of double-rowed trichobothria on femora IV

Please put all spider specimens back to the original vials.

2

IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

Up to 4 points bonus will be granted to students who keep the intact specimens.

Q1.2. (1.2 points for each cell; 18 points total) 1 a-1 or b-1 2 b-1 or a-1 3 s-2 4 h-1 or m-1 or n-1 5 h-1 or m-1 or n-1

6 h-1 or m-1 or n-1

7 d-1

8 e-4

9 s-1

10 t-3

11 d-1

12 e-6

13 e-3

14 o-1

15 g-1

Q1.3.1.

Q1.3.2.

Q1.3.3.

19

d-1

B

(2 points for each cell)

Q1.3.4. (Each correct answer will get 0.4 points, 2 points total)

Character s-1 s-2 a-1 g-1

True X X

False X

X

3

IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

d-1

X

Q1.3.5. (1 point for each cell; 5 points total) Taxon {H} {B, C, G, H} {C, D, E, F} {B, G, H} {B, E, G} Kind of grouping

III II II III I

Q2.1.1. (1 point each; 9 points total) Table 2-1-1 Spider(?) Plant-A(○) Present absent Total Present Absent Total

2 4 6

10 24 34

12 28 40

Q2.1.2a.

Q2.1.2b.

Q2.1.2c.

Q2.1.2d.

Q2.1.2e.

0.3

0.15

0.045

1.8

P

(0.6 points for each cell)

Q2.1.3.(2 points)

0.0373

4

IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

Q2.1.4a. (2 points)

0.0306

Q2.1.4b. (2 points) Association Strong -
V value -1=V≦-0.6

Moderate -
-0.6<V≦-0.2

None
-0.2<V<0.2

Moderate +
0.2≦V<0.6

Strong +
0.6≦V=1

X

Q2.2.1a. (2 points)

N

Q2.2.1b. (2 points)
Association V value

Strong -
-1=V≦-0.6<

Moderate -
-0.6<V≦-0.2

None
-0.2<V<0.2

Moderate +
0.2≦V<0.6

Strong +
0.6≦V=1

X

5

IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

Q2.2.2a. (2 points) True False X

Q2.2.2b. (2 points)

Q2.2.2c. (2 points) X

X

Q2.3.1. (0.5 points for each cell; 3 points total) Table 2-3-1 Species Plant-A (○) Plant-B (●) Total Species of nearest neighbor Plant-A (○) Plant-B (●) Total

24 21 45

16 19 35

40 40 80

Q2.3.2a. (2 points)

0.4571

Q2.3.2b. (3 points) randomly distributed associated segregated X

Q2.4.1 (2 points) True False X

Q2.4.2 (2 points) X

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IBO – 2011 TAIWAN PRACTICAL TEST 3 – ANSWER KEY ECOLOGY AND SYSTEMATICS

STUDENT CODE:

Check list of the spider condition in their original vials (Filled out by the LAB ASSISTANTS after test) Taxon Damaged Undamaged W X Y Z

Signed by Inspector:

Student Code:

(Without Student Code written here, the 4 bonus points will not be awarded)

7

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Student Code:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS
Total Points: 100 Duration: 90 minutes

1

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Dear Participants,
? In this test, you have been given the following 2 tasks: Task I: Plant anatomy (60 points) Task II: Plant physiology and genetics (40 points) Check your Student Code on the Answer Sheet and Template paper before starting the test. Write down your results and answers in the Answer Sheet. Answers written in the Question Paper will not be evaluated. Make sure that you have received all the materials listed for each task. If any of the listed items is missing, lift the sign. Ensure that you organize the sequence of your tasks efficiently. Stop answering immediately when the end bell rings. After test, enclose the Answer sheets, Question paper, Data printout, and the stamped Template paper (without slides) in the provided envelop. Our lab assistant will collect it promptly. No paper or materials should be taken out of the laboratory.

? ? ? ? ? ? ?

Good Luck!!

2

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Task I (60 points) Plant Anatomy
Equipment:
1. Slides (in carrier box) 2. 3. 4. 5. 6. 7. 8. 9. 10. Cover slips (in carrier box) Compound microscope (with 4X, 10X, and 40X objective lenses) Ocular micrometer (installed within the lens) Single sided razor blade (in carrier box) Petri dish (in carrier box) Forceps (in carrier box) Marker pen Kimwipes Paper towel Quantity 20 30 1 1 5 1 1 1 1 1 1 Quantity 20 mL/vial 5-10 mL/vial 1 vial 1 slide each 2 petri dishes

11. Waste basket

Materials:
1. Double distilled water (labeled as “ddH2O” in carrier box) 2. 1 M hydrochloric acid (HCl) (in carrier box) 3. Transparent nail polish (in carrier box) 4. Section slides X, Y, and Z of the root of plant K in slide box K 5. Four-compartment plastic petri dish (Containing tissue samples from plants V, W, M, N, P, Q, R, S in each compartment) 6. Template paper (with student code) for placing the slides with sections you made and for documenting

1 sheet

3

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Part A: Structure of Plant Root (5 points total)
Introduction :

Figure 1

The structure of a typical plant root tip

There are three slides (X, Y, and Z), which are transverse sections (located within the circular label on the slides) from different regions of the root of plant K. You need to determine whether plant K is monocot or dicot to answer some questions later. Observe these sections under a compound microscope and answer the following questions. Q1.A.1.(2 points each, 6 points total) Sections X, Y, and Z each correspond to a part of the root depicted in Figure 1. Check [X] the correct answers on the answer sheet. Q.1.A.2. (single answer, 4 points) What is the direction of maturation of primary xylem in the root (tissues mature earlier ? tissues mature later)? Check [X] the correct answers on the answer sheet.

Part B: Structure of Plant Stem(6 points total)
Introduction : Carefully prepare transverse sections of proper thickness from the stem segments of plants V and W and place them on separate slides. Add a drop of water to the sections and cover with cover slips. Observe under the microscope, and answer the following questions. You need to determine whether plants V and W are monocot or dicot to answer some questions later. When you finish this part, place your slides with sections on the template paper, lift the sign and the lab assistant will stamp in the boxes with the slides on the Template Sheet. . Q.1.B.(3 points for each plant, points are given when all correct answers are selected; 6 points total) What are the distribution patterns of vascular bundles in the stems of plant V and W? Check [X] the correct answers on the answer sheet.
4

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Part C: Structure of Plant Leaf(14 points total)
Introduction : First identify the upper and lower epidermis of the leaves of plants M and N. Answer the following questions. You need to determine whether plants M and N are monocot or dicot to answer some questions later. When you finish this part, place your slides with sections on the template paper, lift the sign and the lab assistant will stamp in the boxes with the slides on the template sheet. . Q.1.C.1(8 points) Observation of stomata of plant M : Prepare upper and lower epidermis strips of the leaves, either by peeling them off with bare hands or by scraping off the unwanted tissues and leaving only the epidermis. Place these epidermal tissues on separate slides, with the epidermal side facing upward. Add a drop of water to each tissue sample and cover with a cover slip. Observe under the microscope and use the ocular micrometer for measurements. The smallest unit of scale length of the micrometer is approximately 30 μm when observing under the 4X objective lens. Answer the following questions in the answer sheet. (a) Measurement of stomatal size on upper epidermis: i) Under the 40X objective lens, what is the length of each smallest scale unit of the ocular micrometer? (1 point) ii) Measure the lengths of 3 guard cells, then average their lengths. (3 points) (b) Measurement of stomatal density on lower epidermis: i) Under the 40X objective lens, what is the approximate area of the field of view? (1 point) ii) Observe 3 fields of view, and calculate the number of stomata in each. Then work out the average stomatal density. (3 points) Q.1.C.2 (6 points, points are given when all correct answers are selected) Observation of leaf tissue of plant N: Use the provided transparent nail polish to paint the upper and lower epidermis of leaves. When the nail polish has dried, carefully peel off the layers and place on separate slides, with the epidermal side facing upward. Add a drop of water on each sample, cover with cover slips, and examine under the appropriate objective lens. Determine the stomatal distribution of the upper and lower epidermis, and deduce the habitat of plant N. Check [X] the correct answers on the answer sheet.
5

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Part D: Monocot or Dicot Plants (5 points)
Q.1.D ( 1 point each, 5 points total) Determine whether plants K, V, W, M, and N are monocot or dicot. Check [X] the correct answers on the answer sheet.

Part E: Calcium Crystals in Plant Cells(20 points)
Introduction : Some plants have idioblasts that can form polygonal calcium oxalate crystals or calcium carbonate crystals. Use the four plant materials (P, Q, R, S) to carefully prepare transverse sections of appropriate thickness with a clean razor blade and place the sections on separate slides. Add a drop of water to each section, and cover with cover slips. Observe under the microscope and check for the presence of crystals. If crystals are present, locate the region of crystal distribution in the tissue (most crystals present in or absent from the cells of vascular bundles), carefully open the cover slip, remove the excess water around the sections, and add a few drops of HCl. Add cover slips again and observe the samples under the microscope and deduce the types of crystals that are present. When you finish this part, place your slides with sections on the template paper, lift the sign and the lab assistant will stamp in the boxes with the slides on the template sheet. Q.1.E (6 points for each plant with the presence of crystals, points are given when all correct answers are selected; 2 points for the plant with crystals absent; 20 points total) Using your observations, fill in the corresponding letters in the table in the answer sheet. Plants: P, Q, R, S Location of crystals: A (most crystals present in cells of vascular bundles) B (crystals absent from the cells of vascular bundles) Crystal type: C (polygonal calcium oxalate crystal); D (calcium carbonate crystal)

Documentation (5 points total)
When you finish all the parts (A to E) of Task I, double check whether all your slides have been stamped for this task. If not, lift the sign and the lab assistant will check your Template Sheet (0.5 point for each slide present, 5 points total).

6

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Task II (40 points) Plant Physiology and Genetics
Shared Equipment
ELISA reader

Equipment:
1. 2. 3. Micropipettes P200 and P1000 Micropipette tips for P200 and P1000 96-well microplate

Quantity 1 each 1 box each

1
12 (6 extra unlabeled) 1 each

4. 1.5 mL microcentrifuge tubes
(for preparation of standard solutions, use those labeled 0 μM, 25 μM 50 μM, 100 μM, 200 μM, 400 μM ) 5. 80-well microcentrifuge tube rack / 4-way test tube rack 6. Vortex mixer

7. Marker pen Materials:
1. 2. 3. 4. Phosphate detection solution (labeled as “Solution A”) 400 μM KH2PO4 solution (labeled as “Solution B”) Double distilled water (labeled as “ddH2O”) 6 samples to be tested (allotted in microcentrifuge tubes, labeled as sample #1, #2, #3, #4, #5, & #6)

1 1
Quantity 10 mL/tube 10 mL/tube 50 mL/vial

Introduction: Phosphate is an important plant nutrient found in cell membranes, nucleic acids, and energy compounds like ATP. When lacking phosphates, plant growth and development can be dramatically affected. Plants can sense changes in phosphate concentration in the environment and accordingly regulate their gene expression, changing the activity of phosphate transport proteins on the cell membranes to maintain homeostasis of phosphate concentration within the cells. Using the model plant Arabidopsis, scientists discovered that root cells respond to phosphate-sufficient (Pi-sufficient; e.g. 1mM) or phosphate-deficient (Pi-deficient; e.g. 10μM) conditions as depicted below in Figures 2 and 3, respectively:

7

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Figure 2

Figure 3

When Arabidopsis is in Pi-sufficient conditions (Figure 2), the protein Z negatively regulates the level of the protein T on the plasma membrane, responsible for the transport of phosphate into the cell. This regulation avoids excessive phosphate absorption that leads to toxicity. On the other hand, when the plant is in Pi-deficient condition (Figure 3), the transcription factor X will enhance the expression of gene Y and thus increase the level of protein Y. Protein Y can promote the degradation of protein Z, leading to an increase of protein T level, and consequently a higher absorption of phosphate. In general, the phosphate level in the shoot of a plant is proportional to the uptake efficiency of phosphate in the root. Phosphate level in plants will be considerably affected when the expression of genes encoding T, X, Y, and Z is perturbed by a mutation or insertion of a transgene. Therefore, plant biologists can utilize such mutant or transgenic plants to determine the role and relationship of these genes in the regulatory mechanism of phosphate homeostasis. There are 6 samples in microcentrifuge tubes, which are extracts from the shoots of five Arabidopsis lines (A to E) grown under either Pi-sufficient (1 mM) or Pi-deficient (10 μM) conditions (listed in Table 1). Arabidopsis A is wild type and B to E are from either a knockout (KO; null mutant with complete loss of function of the gene) mutant line or an over-expression lines of gene T, X, Y, or Z. You will be measuring the phosphate level in each sample and determine the identities of the samples based on the principles shown in Fig. 2 and 3. Each sample is derived from 20 seedlings of fresh weight shown in Table 1 and brought to final volume of 10 mL with ddH2O.
8

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Table 1 Sample No. 1 2 3 4 5 6 Plant A A B C D E [Pi] in medium 1 mM 10 μM 1 mM 1 mM 1 mM 1 mM Fresh weight of seedlings (mg) 40.4 17.3 28.0 39.2 30.6 33.8

Use the provided equipment and solutions to measure the phosphate concentration in each sample in accordance with the experimental procedures. Experimental Procedures: 1. Use the 400-μM KH2PO4 solution (Solution B), ddH2O, and pre-labeled 1.5-mL microcentrifuge tubes to prepare the following concentrations of phosphate solutions for a standard curve: 0, 25, 50, 100, 200, 400 μM. Vortex the samples to mix solutions thoroughly. For each concentration, there should be at least 0.5 mL of solution. Use the P200 micropipette with fresh tips to transfer 0.1 mL of each standard solution into the 96-well microplate at specified positions (as in Figure 4; make 2 replicates for each standard). Figure 4 Positions of standards and samples in 96-well microplate Standards Samples Standards A B C D E F G H 1 2 3 4 5 6 7 8 9 10 11 12 2. Transfer 0.1 mL of each sample into the 96-well microplate at specified positions (as in Figure 4, make 2 replicates for each sample). 3. Add 0.1 mL of the phosphate detection solution (Solution A) into the wells that contain the standards and the samples. Mix by gently tapping the side of the plate. 0 25 50 100 200 400 #1 #2 #3 #4 #5 #6 #1 #2 #3 #4 #5 #6 0 25 50 100 200 400

9

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

4.

Lift the sign after you finish Step 3, and wait for lab assistants to guide and help you with measuring the absorbance of the reaction mixtures with the ELISA reader at 820 nm.

5. The lab assistant will print out the data for you. Put your student code on the print-out. 6. Answer the following questions: Q.2.1. (18 points total) Calculate the mean values of the absorbance for each sample and standard. Use the graph paper on the answer sheet to plot a standard curve. (0.5 point for each standard point correctly plotted) Determine the phosphate concentrations of the samples in ?M and the nmol phosphate per mg of seedling fresh weight for sample # 1 to # 6. Fill in your results in the table of answer sheet. (2 points for each phosphate concentration measured, 0.5 point for each nmol/mg of phosphate calculated) Q.2.2. (Multiple answers, 4 points. Points given when all correct answers are selected) For each of the following statements, determine whether they are true or false explanations for those plants having higher phosphate content (nmol/mg) than the wild type. Check [X] the correct answers on the answer sheet. (A) X cannot be activated in the plant, thus leading to an increase in phosphate uptake. (B) Loss of function of Gene Y in the plant causing an increase in phosphate uptake. (C) Loss of function of Gene Z in the plant causing an increase in phosphate uptake. (D) The plant harbors a transgene that over-expressed gene Y, causing the loss of inhibition of protein T, leading to higher activity in phosphate uptake. (E) Protein T of the plant has a defect and is unable to transport phosphate efficiently. (F) The transcription factor X of the plant has a mutation, and is unable to bind the promoter of gene Y. Q.2.3. (2.5 points each, 10 points total) Using the results from the experiment, assign the corresponding plant (B, C, D, or E) to the correct description on the answer sheet. Q.2.4. (Single answer, 4 points) If a wild type plant (W), a gene X knockout mutant (X), and a gene Y knockout mutant (Y) were all grown in the same Pi-deficient condition, what would be the phosphate level in their shoots (rank from the lowest to highest)? Check [X] the correct answer on the answer sheet. (A) X < W < Y (B) Y < W < X (C) W < X < Y (D) W < Y < X (E) X < Y < W (F) Y < X < W
10

IBO – 2011 TAIWAN PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

Q.2.5. (Single answer, 4 points) Western blot is a technique to detect protein levels with the use of specific antibodies. Which of the following would be the most likely result of the Western blot analysis of protein Y and Z from the total protein extracts of samples # 1 and # 2? Check [X] the correct answer on the answer sheet.

11

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

22nd INTERNATIONAL BIOLOGY OLYMPIAD July 10-17, 2011 Taipei, Taiwan

PRACTICAL TEST 4 PLANT ANATOMY, PHYSIOLOGY, AND GENETICS
Total Points: 100 Duration: 90 minutes

ANSWER KEY

1

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

Q1.A.1.(2 points each, 6 points total)Check [X] the correct answers Part A Section X Section Y Section Z Part B Part C Part D

X X X

Q.1.A.2. (Single answer, 4 points)Check [X] the correct answers

X
Q.1.B.(3 points for each plant, points are given when all correct answers are selected; 6 points total) Check [X] the correct answers Distribution pattern of vascular bundles in the stem arranged in a ring scattered in ground tissue solid vascular cylinder with star-like xylem Plant V X X Plant W

central core of parenchyma cells surrounded by rings of xylem and phloem
2

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

Q.1.C.1(8 points) Fill in the correct answers (a) i) 3 _μm. (1 point) μm. (3 points)

ii) ___40 _μm (± 10) (b) i) ___ 0.2 (± 0.05) (± 50)

___mm2. (1 point) _ __(stomatal number /mm2). (3 points)

ii) ___ 150

Q.1.C.2 (6 points, points are given when all correct answers are selected) Check [X] the correct answers True False

Few or no stomata on the upper epidermis

X

Many stomata on the lower epidermis

X

An aquatic plant

X

3

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

Q.1.D ( 1 point each, 5 points total) Check [X] the correct answers Plant K V W M N X X X Monocot Dicot X X

4

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

Q.1.E (6 points for each plant with the presence of crystals, points are given when all correct answers are selected; 2 points for the plant with the absence of crystals; 20 points total) Fill in the corresponding letters Plant (P, Q, R, or S) Presence of crystals P Q S Location of crystals (A or B) B B B Crystal type (C or D) D D C

Absence of crystals

R

Documentation of the template paper
(0.5 point for each slide present, 5 points total)

5

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

Q.2.1. (18 points total) (0.5 point for each standard point correctly plotted)

6

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

(2 points for each phosphate concentration measured, 0.5 point for each nmol/mg of phosphate calculated) Sample Plant # Fresh weight of seedlings (mg) 1 2 3 4 5 6 A A B C D E 40.4 17.3 28.0 39.2 30.6 33.8 Average phosphate concentration of extract (μM) 160 ± 10% 33 ± 10% 75 ± 10% 150 ± 10% 380 ± 10% 300 ± 10% nmol of phosphate per mg of seedling fresh weight (nmol/mg) 39.6 ± 10% 19.0 ± 10% 26.8 ± 10% 38.3 ± 10% 124.2 ± 10% 88.8 ± 10%

Q.2.2. (Multiple answers, 4 points. Points given when all correct answers are selected) Check [X] the correct answers (A) True False X X (B) (C) X (D) X X X (E) (F)

7

BO – 2011 TAIWAN PRACTICAL TEST 4 – ANSWER KEY PLANT ANATOMY, PHYSIOLOGY, AND GENETICS

STUDENT CODE:

Q.2.3. (2.5 points each, 10 points total)

Check [X] the correct answers B C D E

Knockout mutant plant of gene X

X

Knock out mutant plant of gene Z

X

Plant with defective protein T

X

Transgenic plant with gene Y over-expression

X

Q.2.4. (Single answer, 4 points) (A) (B) (C) (D)

Check [X] the correct answers (E) (F) X

Q.2.5. (Single answer, 4 points) (A) (B) (C) X (D)

Check [X] the correct answers (E)

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