Student Biology Olympiad. University Stage

On March 18, the first stage of the student biology olympiad took place. Here is a summary table of its results:

Code

Surname

Course

Raw scores

Recalculated (worst-best) results

Place

FBI

BIR

Genetics

BH/MB

FBI

BIR

Genetics

BH/MB

Sum

Shlakhtar M.

III

18,5

21,4

25,0

89,9

Vasylyenko V.

10,9

0,0

10,9

Sheiko V.

III

22,8

10,7

25,0

6,3

64,8

III

Kovalev V.

III

16,3

10,7

12,5

0,0

39,5

Ostras D.

0,0

25,0

12,5

0,0

37,5

Drogvalenko N.

III

25,0

21,4

0,0

12,5

58,9

Fedorova A.

III

12,0

14,3

25,0

76,2

{"translated_text":"On March 18, the first stage of the student olympiad in biology took place. Here is the summary table of its results:\n\nCode\nSurname\nCourse\n\"Raw\" scores\nRecalculated (worst-best) results\nPlace\n\n\nFBR\nBIR\nGenetics\nBC/MB\nFBR\nBIR\nGenetics\nBC/MB\nTotal\n\n\n10\nShlakhter M.\nIII\n41\n8\n3\n4\n18.5\n21.4\n25.0\n25.0\n89.9\nI\n\n\n12\nVasylenko V.\nI\n34\n2\n1\n0\n10.9\n0.0\n0.0\n0.0\n10.9\n\n\n\n\n13\nSheiko V.\nIII\n45\n5\n3\n1\n22.8\n10.7\n25.0\n6.3\n64.8\nIII\n\n\n18\nKovalev V.\nIII\n39\n5\n2\n0\n16.3\n10.7\n12.5\n0.0\n39.5\n\n\n\n\n21\nOstras D.\nV\n24\n9\n2\n0\n0.0\n25.0\n12.5\n0.0\n37.5\n\n\n\n\n37\nDrogvalenko N.\nIII\n47\n8\n1\n2\n25.0\n21.4\n0.0\n12.5\n58.9\n\n\n\n\n42\nFedorova A.\nIII\n35\n6\n3\n4\n12.0\n14.3\n25.0\n25.0\n76.2\nII\n\nThe first part of the results presents the raw scores as they were assigned by the teachers who checked the tasks. Obviously, simply summing them would not make it possible to evaluate each participant's result: the \"weight\" of the task in plant physiology and biochemistry turns out to be much greater than the genetics problem. Therefore, the scores for all tasks are recalculated so that 0 points corresponds to the lowest result for that particular task, and 25 points corresponds to the highest. The resulting sum is quite easy to interpret. A score of 100 would be earned by a student who achieved the highest score on all tasks; 0, respectively, would be earned by someone who answered all tasks worse than everyone else.\nAccordingly, I place was taken by Mykhailo Shlakhter, II — Anna Fedorova, III — Valeriia Sheiko. The next stage of the olympiad (which will take place in Zhytomyr) will be attended by Shlakhter and Fedorova.\nTasks\nTest in Plant Anatomy, Physiology, and Biochemistry\nVasyl Vasylovych Zhmurko provided a test consisting of 60 questions. We will not reproduce this test here. The test score is the number of correct answers out of 60. Mykola Drogvalenko performed best on this test.\nBiology of Individual Development\nQuestion author — Yevhen Oleksandrovych Kyosia.\nThe silkworm (Bombyx mori) has a very peculiar method of sex determination in ontogenesis. Sex chromosomes play a leading role in this process. External factors, such as temperature increases, can stimulate different types of parthenogenetic development of eggs, but they do not have a significant effect on sex determination in silkworms developing from fertilized eggs.\nThe heterogametic sex in silkworms is the female, meaning that males have two identical sex chromosomes (ZZ), and females have two different ones (ZW). Unlike the fruit fly (Drosophila melanogaster), in silkworms the decisive factor in sex determination is not the balance of genes on sex chromosomes and autosomes, but the presence or absence of the W chromosome. In this respect, the silkworm resembles humans and other mammals, in which sex is determined by the presence of the Y chromosome. However, in mammals, a specific region of the Y chromosome encodes a protein (TDF, testis determining factor) that \"switches on\" the male type of embryonic development, whereas in silkworms, the W chromosome does not encode any proteins at all.\nSo how does the silkworm's W chromosome influence development?\nThe answer is clear even from the title of the article (published in 2014) that described the mechanism under discussion: \"A single female-specific piRNA is the primary determiner of sex in the silkworm\".\nGenetics\nQuestion author — Nataliia Yevhenivna Volkova.\nThree white-flowered delphinium cultivars were crossed with each other and the following results were obtained: \"Magic White Fountain\" × \"Percival\" — all offspring are white; \"Magic White Fountain\" × \"Galahad\" — all offspring are red; \"Percival\" × \"Galahad\" — all offspring are white. In crosses of red first-generation plants with plants of each cultivar, the following progeny were obtained: with \"Magic White Fountain\" — ¼ red: ¾ white; with \"Percival\" — 1/8 red: 7/8 white;with \"Galahad\" — ½ red: ½ white.\nWhat is the genetic basis of white flowering in these delphinium cultivars? Determine the genotypes of all plants mentioned in the problem. Provide explanations to accompany your answer as fully as possible.\n[IMG_1]\nGrading criteria: 5 — problem solved and explained; 4 — problem solved, no explanations; 3 — reasoning went in the right direction but did not reach completion; 2 — reasoning went off track; 1 — thoughts were about the wrong things; 0 — no thoughts.\nBiochemistry / Molecular Biology\nQuestion author — Kateryna Vasylivna Kot.\nIn DNA molecules isolated from two unidentified bacteria X and Y, the number of adenine nucleotides is 61,827 and 85,550 nucleotides, respectively. The length of the DNA molecule of bacterium X is 5.68×10⁻³ cm, and of bacterium Y is 4.93×10⁻² mm. Determine the percentage of adenine, thymine, cytosine, and guanine nucleotides out of the total number for each of the bacteria, given that the isolated DNA molecules are in the B-form.\nOne of these bacteria lives in hot springs (64°C). Which do you think — X or Y — is it? Justify your answer.\nFor reference: the length of one turn of B-DNA is 3.4 nm; one turn contains 10 nucleotide pairs.\nThe point distribution was as follows:\n— to indicate the percentage proportions (A+T)/(G+C) for bacterium X — 0.5 points for each value, total — 2 points;\n— to indicate the percentage proportions (A+T)/(G+C) for bacterium Y — 0.5 points for each value, total — 2 points;\n— to indicate which bacterium — X or Y — is the thermophile — 1 point;\n— to justify your choice — 1 point.\nThus, the maximum possible number of points for the task is 6.\nResults:\n\nParticipant code\nScore\nExplanations\n\n\n10\n4\nCalculations are written out in very detail and competently. With explanations.\nThe participant gave correct nucleotide percentage ratios for both bacteria. However, they chose the wrong bacterium as the thermophile. And the justification was somewhat questionable. The issue is not so much the length of the DNA, but its composition.\n\n\n13\n1\nErrors were made in the calculations. Instead of the total number of nucleotides, the participant decided to use the number of nucleotide pairs, which led to incorrect answers.\nThe thermophile bacterium was chosen correctly. Perhaps guessed? But still, for the correct answer — 1 point.\nHowever, the justification for the reasons is completely wrong. The participant confused AT and GC pairs and the number of bonds between them.\n\n\n12\n0\nNot a single answer was given to any question.\n\n\n18\n0\nSomething got mixed up in the calculations, resulting in incorrect percentage ratios.\nThe participant did not answer the question about who is the thermophile and why at all.\n\n\n21\n0\nNo percentage ratios. The wrong bacterium was chosen. And the explanation has errors (AT and GC pairs and the number of bonds between them were confused)\n\n\n37\n2\nNo percentage ratios, although initial calculations are present. For some reason, the participant did not finish the calculations.\nHowever, the thermophile bacterium was indicated correctly and the correct explanation was given.\n\n\n42\n4\nPercentage ratios are correctly indicated for bacterium X. But not for Y.\nThe bacterium was chosen correctly. The explanation is correct.\n\nFor bacterium X:\n1) For convenience in calculations, let's convert the DNA length value from cm to nm: lDNA = 5.68×10⁻³ cm = 5.68×10⁻³×10⁷ nm = 5.68×10⁴ nm\n2) Knowing that the length of one turn of B-DNA is 3.4 nm, and one turn contains 10 nucleotide pairs, let's determine the length of 1 nucleotide pair (1 bp). It will be equal to 0.34 nm.\n3) Now we can determine the number of nucleotide pairs (bp) in the DNA molecule: Nbp = lDNA / l1 bp, Nbp = 5.68×10⁴ nm / 0.34 nm = 16.71×10⁴ = 167,100 nucleotide pairs.\n4) Let's determine the total number of nucleotides (n) in the DNA molecule. Nn = Nbp×2, Nn = 167,100×2 = 334,200 nucleotides.\n5) Let's determine the percentage of adenine nucleotides out of the total number of nucleotides: If 334,200 — 100%, then 61,827 — x%, from which x = 6,182,700 / 334,200 = 18.5%\n6) And according to Chargaff's rule A=T, G=C. Therefore T = 18.5%. G+C = 100% - (A+T) = 100 - 37 = 63%. G = C = 63 / 2 = 31.5%\nFor bacterium Y:\n1) For convenience in calculations, let's convert the DNA length value from mm to nm: lDNA = 4.93×10⁻² mm = 4.93×10⁻²×10⁶ nm = 4.93×10⁴ nm\n2) The number of nucleotide pairs (bp) in the DNA molecule: Nbp = 4.93×10⁴ nm / 0.34 nm = 14.5×10⁴ = 145,000 nucleotide pairs.\n4) The total number of nucleotides (n) in the DNA molecule: Nn = 145,000×2 = 290,000 nucleotides.\n5) The percentage of adenine nucleotides out of the total number of nucleotides: If 290,000 — 100%, then 85,550 — x%, from which x = 8,555,000 / 290,000 = 29.5%\n6) And according to Chargaff's rule A=T, G=C. Therefore T = 29.5%. G+C = 100% - (A+T) = 100 - 59 = 41%. G = C = 41 / 2 = 20.5%\nWho is the thermophile and why?\nThe thermophile is bacterium X due to its higher percentage of GC pairs: 63% (compared to 41% in the DNA of bacterium Y). And since it is specifically between G and C that 3 hydrogen bonds form (between A and T — only 2), the presence of a greater number of guanine and cytosine nucleotides in the molecule makes it stronger, more stable."}

Participant Code

Grade

Explanation

Calculations are described in great detail and substance. With explanations. The participant provided correct nucleotide percentage ratios for 2 bacteria. However, they chose the wrong bacterium as the thermophile. And the justification was somewhat debatable. It's not just about DNA length, but also its composition.

Errors were made in the calculations. Instead of the total number of nucleotides, the participant took the number of nucleotide pairs, which led to incorrect answers. The thermophilic bacterium was chosen correctly. Perhaps guessed? But, nevertheless, 1 point for the correct answer. But the justification for the reasons is completely incorrect. The participant confused AT and GC pairs and the number of bonds between them.

No answer was given to any question.

Something got mixed up in the calculations, resulting in incorrect percentage ratios. The participant did not answer the question about who the thermophile is and why at all.

No percentage ratios. The wrong bacterium was chosen. And the explanation contains errors (confused AT and GC pairs and the number of bonds between them).

There are no percentage ratios, although initial calculations are present. For some reason, the participant did not complete them. However, the thermophilic bacterium is indicated correctly and the explanation is correct.

Percentage ratios for bacterium X are indicated correctly. For Y – they are not. The bacterium is chosen correctly. The explanation is correct.