Latvia, 2002
Ministry of Education and Science
University of Latvia
Association of Biology Teachers
phone: +371-7034860, +371-7334125, Fax: +371-7223801, email: ulkoro@latnet.lv
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13th International Biology Olympiad
Latvia, 2002 Ministry of Education and Science
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| 4 Kronvalda boulv., Riga LV-1586, LATVIA
phone: +371-7034860, +371-7334125, Fax: +371-7223801, email: ulkoro@latnet.lv | |
B 1. The dependence of the initial reaction rate on substrate concentration for 3 different enzymes (X, Y and Z ) is given in the table:
|
Substrate concentration (arbitrary units) |
Initial rate (arbitrary units) |
||
| X | Y |
Z |
|
| 1 | 0.92 | 0.91 |
0.032 |
| 2 | 1.67 | 1.67 |
0.176 |
| 4 | 2. 85 | 2.86 |
0.919 |
| 6 | 3.75 | 3.75 |
2.180 |
| 8 | 4.40 | 4.44 |
3.640 |
| 10 | 4.90 | 5.00 |
5.000 |
| 15 | 5.80 | 6.00 |
7.337 |
| 20 | 6.23 | 6.67 |
8.498 |
| 30 | 6.80 | 7.50 |
9.397 |
| 50 | 6.00 | 8.33 |
9.824 |
| 100 | 4.20 | 9.09 |
9.968 |
1. Plot the initial rates versus substrate concentrations on the answer sheet! (1 point)
(1 point)
3. Which of the enzymes (X, Y or Z) is inhibited by its own substrate? (1 point)
B 2. For an exponentially growing culture of microorganisms the specific growth rate (µ
) is a parameter, that gives the cell biomass (g) synthesized per gram of existing cell biomass per unit of time (usually, per hour). This rate (µ
) is inversely related to the doubling time of the culture, td: µ
= ln2/td
0.7/td . Hence, the shorter the doubling time of cells, the higher is the specific growth rate of the culture.
Two microorganisms, A and B, were inoculated each in a fresh growth medium with an initial optical density (OD) of 0.1. A lag phase of 1 hr duration was observed for both cultures. Three hours after inoculation, the OD of culture A was 0.4, while that of the culture B was 1.6.
(2 points)
B 3. Calculate the intracellular millimolar (mM) concentration of potassium in Escherichia coli, if the measured potassium content is 7.8 micrograms per milligram of dry cell mass. Assume all potassium ions are free in the cytosol (not bound to macromolecules), and that the intracellular volume is 2 microlitres per milligram of dry cell mass. The atomic weight of potassium is 39 Daltons. (1 point)
B 4. A species of fungus can dissimilate glucose and produce ATP in two ways.
Aerobically: C6H12O6+ 6O2= 6 CO2+ 6 H2O,
Anaerobically: C6H12O6= 2 C2H5OH + 2 CO2
This fungus is cultivated in a glucose-containing medium. Half of the total ATP production is anaerobic.
For calculations, assume that glucose is fermented via the usual Embden-Meyerhof-Parnas glycolytic pathway, and that oxidative phosphorylation proceeds with maximum efficiency.(3 points)
B 5. For the bacteria Bacillus subtilis, several auxotrophic mutants have been obtained which need addition of aspartate, threonine or methionine to the growth medium.
Mutant Amino acid precursors Amino acid Metabolite,
that are not synthesized needed for accumulating in
by the mutant growth the medium
aspA4. Aspartate 7. Fumarate
metA1. Homocystein5. Methionine3. Homoserine
metH5. Methionine1. Homocystein
thrC6. Threonine 2. Homoserinephosphate
thrB2. Homoserinephosphate6. Threonine 3. Homoserine
thrA3. Homoserine6. Threonine 4. Aspartate
2. Homoserinephosphate5. Methionine
1. Homocystein
_________________________________________________________________________
1. What is the biosynthetic pathway for methionine biosynthesis?
2. What is the biosynthetic pathway for aspartate biosynthesis?
3. What is the biosynthetic pathway for threonine biosynthesis?
Indicate the pathway with appropriate numbers from the table (1-7) and arrows in the answer sheet!(3 points)
B6. Before a lecture, an assistant noticed that comments on an important diagram are lost. He found many of terms in a textbook, including some which were unrelated to this diagram.
1. Please help the assistant to locate the correct terms for this diagram and to place the appropriate numeric labels in the table in the answer sheet.
(Continuation see on the next page)
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Term | Number | Term | Number | |
| A-1 | Amino acid | A-2 | Growing polypeptide | ||
| B-1 | Pentose | B-2 | Growing DNA strand | ||
| C-1 | Fatty acid | C-2 | Growing RNA strand | ||
| D-1 | Small ribosomal subunit | D-2 | Alpha subunit of RNA polymerase | ||
| E-1 | tRNA | E-2 | Nuclear pore | ||
| F-1 | IgG | F-2 | P site | ||
| G-1 | Receptor | G-2 | Centriole | ||
| H-1 | Aminoacyl-tRNA synthetase | H-2 | Large ribosomal subunit | ||
| I-1 | Protein kinase | I-2 | A-site | ||
| J-1 | Glucokinase | J-2 | Z-site | ||
| K-1 | Aminoacyl-tRNA | K-2 | Peptidyl-tRNA | ||
| L-1 | Inductor | L-2 | DNA polymerase | ||
| M-1 | Operator | M-2 | Spliceosome | ||
| N-1 | N - end | N-2 | Adenylate cyclase | ||
| O-1 | C - end | O-2 | Capsomer | ||
| P-1 | 5` - end | P-2 | Single stranded DNA | ||
| R-1 | 3` - end | R-2 | Codon | ||
| S-1 | Nucleotide | S-2 | Initiation codon | ||
| T-1 | Lysosome | T-2 | Gene | ||
| U-1 | Sigma subunit of RNA polymerase | U-2 | Terminal transferase |
2. Which component of this diagram has (give the number) peptidyl transferase activity? (5 points)
B 7. The growth of bacteria is studied. For a period of exactly one duplication, the sample is moved from an environment with a light nitrogen isotope (14N) to an environment with heavy nitrogen isotope (15N). After this the sample is again transferred to the environment with light nitrogen for a period of two duplications.
1. What is the composition of double-stranded DNA (in %) of light and heavy nitrogen isotopes after the experiment?
| A. Only light | B. In between | C. Only heavy |
From these cells two types of mRNA {mRNA (A) and mRNA (B), respectively, expressed from two different genes} were isolated. Both mRNAs were found to contain an identical number of nucleotides. The nucleotide composition of each mRNA was estimated as (see the table).
| mRNA | A % | C % | G % | T % | U % |
| A | 17 | 28 | 32 | 0 | 23 |
| B | 27 | 13 | 27 | 0 | 33 |
2. What is the nucleotide composition of double-stranded genomic DNA in the coding part of the genes A and B, respectively.
| dsDNA | A % | C % | G % | T % | U % |
| A | |||||
| B |
3. What curve in the plot below represents the DNA melting profile of the coding part of genes A and B, respectively?
% of melting
Temperature °C
Curve 1 2 3 4 5
(3 points)
B 8.
The pie charts show the relative amounts of types of membrane found in two types of cells.
Suggest, why liver cells (answer 1 ) possess significantly more smooth ER, while pancreatic cells (answer 2 ) have more rough ER. Chose the correct statements (A to E ) from the left column and pair them with the appropriate numbers (1 to 5) from the right column.
| Process | Structure | Number | |
| A. | Higher synthesis of lipids | In nuclear membrane of pancreatic cells | 1 |
| B. | Higher proteolytic activity | In glycogen particles of liver cells | 2 |
| C. | Higher lipolytic activity | In endoplasmic reticulum of pancreatic cells | 3 |
| D. | Higher protein-secretory activity | In mitochondria of liver cells | 4 |
| E. | Hihger ATP-synthesizing activity | In endoplasmic reticulum of liver cells | 5 |
(2 points)
 |
B9. The diagram shows an apparatus made by a student to investigate the effect of temperature on the activity of ethanol fermentation of yeast. The conical flask contains 2.5 g yeast suspended in 2% sucrose solution. The meniscus moves down the glass tube (5ml micropippete) during fermentation. |
The table shows the amount of suspension (ml) pushed in the glass tube due to CO2 accumulation at regular time intervals.
| Time
(min.) |
40C | 100C | 200C | 350C | 550C |
| 1 | 0 | 0.2 | 0.4 | 0.7 | 0 |
| 2 | 0 | 1.0 | 1.3 | 1.2 | 0.1 |
| 3 | 0.1 | 1.9 | 2.2 | 2.8 | 0.2 |
| 4 | 0.2 | 3.1 | 3.3 | 4.4 | 0.3 |
| 5 | 0.3 | 4.0 | NO RESULT | NO RESULT | 0.4 |
1. Plot the data on CO2 accumulation at different temperatures.
2. Estimate the average rate of CO2 production (ml CO2/min) for the yeast suspension at 200C using the values obtained in the period between 2 and 4 minutes.
3. Estimate the specific rate of CO2generation (millimoles CO2/(min g)) at 200C.
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