Results of the biology olympiad for students of the biology faculty of Kharkiv National University named after V.N. Karazin
On February 28, a mini-olympiad in biology was held among biology department students. The results are presented here.
On February 28, a mini-biology Olympiad was held at the Faculty of Biology. The winner of previous Olympiads, Valeriia Sapozhnikova, could not participate. Nevertheless, by the decision of the jury (Kateryna Vasylivna Kot, Yevheniia Oleksandrivna Kiosia, and myself, Dmytro Andriiovych Shabanov, as the head), Valeriia Sapozhnikova will still be sent from our faculty to the All-Ukrainian Olympiad. Thus, in the competition we held, one winning place was contested, which ensures participation in the Olympiad in Zhytomyr.
At the beginning, I want to announce the tasks and discuss the criteria by which they were evaluated.
Task 1.
Task 1a.
Freshly harvested corn kernels have a sweet taste due to their high sugar content. Corn sold a few days after harvest is less sweet.
To preserve the sweet taste of freshly harvested corn, the husked cobs are immersed in boiling water for 2–3 minutes ("blanched") and then cooled in cold water. Corn treated in this way retains its sweet taste when frozen.
What biochemical processes underlie such treatment, and why does the "loss" of sweet taste occur if this treatment is not carried out? Explain.
Task 1b.
The enzyme glucose oxidase catalyzes the oxidation of β-D-glucose to D-glucono-δ-lactone, and then to gluconic acid, with the formation of hydrogen peroxide. This enzyme exhibits high specificity for β-glucose and does not act on α-glucose.
However, the glucose oxidase reaction is widely used for the clinical determination of the total glucose content in blood—that is, in a solution containing a mixture of β- and α-D-glucose. Why do you think this is possible?
What are the advantages of using glucose oxidase compared to Fehling's reagent?
For reference: Fehling's test is a qualitative reaction that occurs due to the conversion of free hemiacetal (terminal) hydroxyls of a carbohydrate into an aldehyde group capable of oxidation, which gives the substance reducing properties. Thus, when interacting with such carbohydrates, red Cu2O is formed from the complex compound of copper with Rochelle salt (Fehling's reagent).
Answers and comments on results
No. 1. High temperature is a denaturing agent for enzymes that convert sweet mono- and small oligosaccharides into tasteless starch. Cold inhibits the action of those enzymes that have not had time to denature. And various explanations on the topic of enzymes, carbohydrates, and denaturation are also welcome.
No. 2. As a result of mutarotation, all the alpha form will convert to the beta form as the beta form is oxidized. And it will also be oxidized. The result is that all the glucose is oxidized, converted into gluconic acid and peroxide. This is what is used in the laboratory test. Fehling's reaction detects not only glucose but also many other reducing carbohydrates (students can learn about this directly from the reference on the reaction if they read carefully and recall chemistry, even school-level chemistry), so its use is not preferred, and the results for glucose are not accurate.
Since there were 2 questions, and each contained 2 sub-questions, points were awarded on the following basis:
0 — incorrect answer; 1 — almost correct answer, but not fully covered; 2 — correct answer.
Summary:
For question 1a, one could get 2+2=4 points; for question 1b, one could get 2+2=4 points.
The maximum possible number of points for both tasks is 8 points.
Unfortunately, no one got 8 points. Participants numbered 1, 2, and 3 scored the same maximum number of points. All of them have 6 points. I wanted to single out someone among them, but I couldn't. One wrote one thing well, another wrote another. One wrote nonsense about one thing, another about another. Some works were amusing. One work was upsetting (0 points scored).
Task 2.
The figure shows the chromosomal sets of a female and a male fruit fly (Drosophila melanogaster). Fruit flies are amphimictic (i.e., they reproduce sexually, with fertilization). Gametes are formed in both males and females through meiosis, but in males, meiosis occurs without crossing over.
Estimate the probability that a fruit fly will not inherit any genes from its grandfather? And from its grandmother?
Answer
Surprisingly, not a single participant correctly solved the rather simple problem about the fly and its grandfather. Participant No. 6 gave a final score close to the correct one, but their solution is practically impossible to understand – it is formatted in such a specific way.
The key to the correct solution lies in the fact that crossing over does not occur during meiosis in male fruit flies. In scientific literature, this situation is denoted by the term "achiasmatic meiosis." This phenomenon is most fully studied in fruit flies, although it is also known in a number of other invertebrates (including some mollusks, planarians, copepods, mites, spiders, scorpions, and many groups of insects), as well as in some flowering plants. In most cases, achiasmatic meiosis is confined to gametogenesis and, as a rule, occurs only in the heterogametic sex.
So, what is the probability that our fruit fly (let's call it the proband) will not inherit any genes from its grandfather or grandmother?
It is quite natural that the proband fly has two grandfathers (and as many grandmothers), so the probabilities for them must be assessed independently.
Let's start with the maternal grandmother. For her, the probability of not passing any genes to the proband is practically zero, as she will definitely pass on her mitochondrial DNA genes to her daughter (the proband's mother) and subsequently to the proband.
For the maternal grandfather, this probability is also vanishingly small, as in the meiosis of the proband's mother, his genes will mix with the genes of his "wife" as a result of crossing over, and almost certainly some of them will end up in each egg cell of the proband's mother, and thus in the proband's genome.
The situation with the paternal grandfather is fundamentally different, as crossing over does not occur in the proband's father. This has two interesting consequences:
Firstly, the only source of recombination in the gametogenesis of the proband's father (and thus, the source of gamete diversity) remains the independent assortment of chromosomes in anaphase I. This means that the genetic diversity of gametes that this male fruit fly can produce is essentially very small.
Secondly, each individual chromosome of the proband's father (regardless of which parent it was inherited from) is either passed on to offspring intact or not passed on at all. Thus, the proband's father can produce gametes with a fairly high probability that do not contain any chromosomes from one of his parents.
The probability that each individual chromosome from the grandfather will be passed from the father to the proband is 1/2 (this is Mendel's first law in its purest form). There are a total of 4 grandfather's chromosomes in the karyotype of the proband's father, so the probability that none of them will be passed on (and thus, none of the grandfather's genes) is (1/2)^4, which is 1/16, or 6.25%. With exactly the same probability, all 4 chromosomes from the grandfather will be passed from the father to the proband (and thus, none of the grandmother's chromosomes). In the remaining 14/16, or 87.5% of cases, the proband will inherit genes from both the grandmother and the grandfather from the father.
The sex of the proband and the specifics of sex chromosome inheritance are irrelevant to solving the problem. The probability of the paternal grandmother's mitochondrial DNA being passed to the proband through the father's sperm is non-zero, but it can probably be neglected.
Answer: the probability of not inheriting any genes from the grandfather is 6.25%, from the grandmother - the same.
Task 3.
How has the number of living organism species inhabiting the Earth changed over geological time? What factors influence it?
Answer and comments
There is no suitable illustration at hand that would describe the dynamics of both marine and terrestrial forms, so we can use an estimate of the dynamics of marine life forms. It should be emphasized that the dynamics of the number of marine taxa are better studied due to better conditions for the burial of marine organisms. It is clear that a significant number of terrestrial species appear starting from the Devonian-Carboniferous, and then the growth of the total number of species should be even more dramatic.
Here is an illustration taken from Oleksandr Markov's website. It shows the dynamics of marine animal genera throughout the Phanerozoic. The numbers on the graph indicate key events, the descriptions of which can be found on Markov's website at the link. It is clear that the number of genera is closely related to the number of species (the fluctuations of which are more dramatic). The number of species of plants, fungi, and protists is significantly smaller than the number of animal species, and its consideration will not significantly change the overall picture. Finally, there is no clarity on what should be considered a "species" in prokaryotes, the dynamics of the number of their forms are practically unknown, but they also have a weak impact on the overall picture. This picture can be described as follows. The number of species is practically always increasing, but during crises, it falls more or less sharply. The main content of the last geological epochs is extremely rapid growth. We live in a time of the greatest diversity of life forms in history. Over the past centuries, a slight (against the general background) decrease in the number of species has begun. It can be considered an extremely sharp extinction due to its speed, but it has not yet significantly changed the overall picture of diversity.
The most important factors influencing the change in the number of species: — an increase in the number of potential ecological niches as life diversity grows; — geological and cosmic perturbations; — changes in the character of atmospheric and hydrospheric circulation (which change the character and diversity of climatic zones); — the position of continents, which sometimes leads to the formation of global ranges, and sometimes fragments the biosphere into separate zones; — geologically rapid anthropogenic restructuring of the biosphere (in the most recent times).
There are constant disputes between paleontologists-geologists and paleontologists-biologists about the extent to which the evolution of the biosphere is driven by geological causes. Probably, the contribution of abiotic factors (climate change, volcanic eruptions, asteroid impacts, etc.) was quite moderate, and the evolution of the biosphere was primarily guided by its internal logic.
I must admit that when assigning grades, I revised my initial assessment. The answers were quite unsuccessful. I assigned a grade, recalculated the points, and saw that it was impossible to determine the first place by the sum of points. I reviewed the works again and raised my grade for the work that I immediately considered the best (participant #2). This work lacks much of what I would like to see in the answers, but this work turned out to be the only one that had no errors or nonsense. This determined the final distribution of places.
Results
| Surname | Course | Code | Task 1 | | | Task 2 | Task 3 | Proportion of points earned from maximum, % | | | Total, % | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | | | 1a | 1b | Total | | | Task 1 | Task 2 | Task 3 | | | Telichenko Vladyslav | II | 1 | 2+0 | 2+2 | 6 | 1.5 | 3 | 33.3 | 20.0 | 16.7 | 70.0 | | Drohvalenko Mykola | II | 2 | 2+2 | 0+2 | 6 | 0.5 | 6 | 33.3 | 6.7 | 33.3 | 73.3 | | Moskalev Vitaliy | II | 3 | 2+1 | 2+1 | 6 | 1 | 3 | 33.3 | 13.3 | 16.7 | 63.3 | | Petrova O. | II | 5 | 0+2 | 0+0 | 2 | 1.5 | 3 | 11.1 | 20.0 | 16.7 | 47.8 | | Ostras Danylo | IV | 6 | 2+0 | 1+0 | 3 | 2.5 | 4 | 16.7 | 33.3 | 22.2 | 72.2 | | Shlakhter M. | II | 7 | 2+2 | 1+0 | 5 | 1.5 | 4 | 27.8 | 20.0 | 22.2 | 70.0 | | Cherepashchuk Ivan | I | 9 | 0+0 | 0+0 | 0 | 0 | 0 | 0.0 | 0.0 | 0.0 | 0.0 | | Shulik Viktoriia | IV | 10 | 1+1 | 0+0 | 2 | 1.5 | 3 | 11.1 | 20.0 | 16.7 | 47.8 |
Thus, Drohvalenko Mykola is declared the winner of the faculty Olympiad. I ask him to find me on March 3rd to arrange his application for participation in the All-Ukrainian Olympiad in Zhytomyr, along with Sapozhnikova Valeriia.
On behalf of the jury, I congratulate Mykola and thank all the Olympiad participants.
I ask all biology faculty students to unite with all our fellow citizens in the face of external threats. Please take care of yourselves and do not fall for provocations from manipulators. Remember: it is in the interest of all citizens of Ukraine that our country remains free, peaceful, and independent!