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Contents
Matura 2020 in biology. On June 16 at 9:00 a.m. Matura exam in biology (advanced level) began. Upon completion of the exam, we publish a biology question sheet with all the answers compiled by our expert. Check what the 2020 exam in biology looked like or verify your answers.
Matura exam 2020 in biology: worksheet and answers
Below, we publish a complete sheet with high school graduation questions in biology (advanced level) and answers prepared by our expert. There are also helpful comments for some questions.
Task 1.1
A
Task 1.2
When the temperature of the environment rises, the fluidity of the cell membrane of a unicellular organism increases. This phenomenon is counteracted by the change in the composition of the cell membrane, which consists in increase the share of molecules with longer hydrocarbon chains with a smaller number of double bonds.
Task 1.3
Cholesterol
Exercise 2.
This statement is not true as the number of fat cells may increase due to the ability of fibroblasts to differentiate into fat cells, thereby leading to an increase in the number of adipocytes.
Comment: Many the research have shownthat in an adult human the number of fat cells does not change. In other words – by sticking to the information from the task, it should be assumed that the number of fat cells may increase, but it is not necessarily consistent with the actual state (or at least reported in the research).
Task 3.1
1. P
2. P
3. F
Task 3.2
Cork cells have walls that undergo suberinization – this causes the death of the protoplast and, consequently, its disappearance. Thus, the inside of the cork cells is filled with air – replacing the protoplast containing mainly water with air causes the tissue to become a heat-insulating layer, protecting the plant against both low (e.g. in winter) and high (e.g. during a fire) temperatures.
Task 3.3
Sklerenchyma – this tissue fulfills a protective function (e.g. as a part of the seed coat, where it protects the soft interior of the semen, endosperm and embryo against mechanical damage), located in the stem (e.g. in the form of wood or bast fibers), it stiffens it, and the form of stone cells in some fruits limits the activity of herbivores.
Task 4.1
Unlike other types of parenchyma, aerenchyma is characterized by highly developed intercellular spaces that allow the circulation of gases.
Task 4.2
Due to the presence of numerous air-filled voids, it reduces the specific weight, increasing the buoyancy of the plant and allowing it to float on the surface (e.g. water leaves).
Task 5.1
FADHMore2 i NADH+H+: terminal oxidation on the respiratory chain
Acetylo-CoA: Krebs cycle (= citric acid cycle = tricarboxylic acid cycle)
Task 5.2
Lauric acid contains 12 carbon atoms, the acetyl group in acetyl-CoA contains two carbon atoms, and therefore 12/2 = 6. As a result of beta-oxidation of this acid, 6 acetyl-CoA molecules are formed.
18/6 = 3, so 3 lauric acid molecules are needed to produce one molecule of cholesterol.
Task 5.3
Liver
Task 5.4
A (glycogen), because 3 (they breathe anaerobically and obtain energy only through glycolysis).
Comment: Information on the type of reserve substance in intestinal parasites (glycogen) is given in the introduction, so the task is limited to the choice of justification. The first answer is clearly incorrect – glucose, not glycogen, is the immediate substrate of cellular respiration. The second answer is also not correct – intestinal parasites do not have an active lifestyle that would require a large stockpile. Moreover, if this were the case, fats give you a better volume-to-energy ratio. The third answer is not semantically correct – cellular respiration is a process that involves glycolysis, the Krebs cycle, and terminal oxidation along the respiratory chain. Depending on the type of the final electron acceptor (whether it is oxygen or, for example, nitrogen compounds), aerobic and anaerobic respiration can be distinguished. Animals do not perform anaerobic respiration. Correctly formulated answer should read “they live in anaerobic conditions and obtain energy only in the process of glycolysis (fermentation)”. Replacing fermentation (which only involves glycolysis and reduction of pyruvate) with anaerobic respiration is a common mistake.
Exercise 6.
Materials: aquatic slipper culture, microscope, slides and coverslips, pipette, graphite powder
Procedure: Using a pipette, draw about 5 ml of the water slipper culture and sprinkle 5 drops on the glass slide. A small volume of powdered graphite should be poured into the water on the glass slide, covered with the glass slide and observed using a microscope (magnification 50x). The movement of graphite particles in the immediate vicinity of the shoes should be observed – if the particles rotate, it means that the shoes are moving using cilia that generate a whirling motion.
Comment: The volume of the solution, the number of drops, or the volume of powdered graphite are of course arbitrary.
Exercise 7.
When opening the stomatal apparatus, the vacuole swells (as a result of the osmotic influx of water). This organelle puts pressure on the cytoplasm, which in turn puts pressure on the cell wall (increasing its turgor). The cell is therefore being pushed from the inside; because its wall is unevenly thickened, the cell bends – the outer, thin walls become more curved than the inner, thickened ones. Thus, the stomatal apparatus is opened.
Task 8.1
A, E
Task 8.2
It is a dioecious plant, as a given individual produces only male or female flowers, but does not produce both types (male and female) of flowers.
Task 8.3
1. A post with a large, brush-like mark – the large surface of the post and its specific shape increase the chance of depositing pollen carried by the wind
2. The flowers are gathered in inflorescences growing on long and slender axes – the long axes move the inflorescences away from the stem and leaves, which could prevent the accumulation of pollen carried by the wind, and their laxity causes the inflorescences to move under the influence of the wind, which in the case of flowers can lead to efficient pollen discharge (carried by the wind).
Task 9.1
Does high temperature accelerate the flowering of the radish plant? / Effect of elevated temperature on the flowering time of a radish flower.
Task 9.2
B (dicots)
1. Flowers with perianth differentiated into calyx sepals and crown petals
2. Pile root system
Comment: Other possibilities include pinnate innervation of the leaves; leaves having a petiole.
Task 9.3
B, D, E
Task 10.1
C (sexual dimorphism)
Task 10.2
They are nestbirds – parents lead the chicks to the water after hatching (sometimes several kilometers), which proves that the young are able to move effectively shortly after hatching; in contrast to the nesters, which after hatching spend a long time in the nest, unable to exist independently.
Task 10.3
Limiting the availability of breeding sites.
Exercise 11.
1. P
2. P
3. P
Task 12.1
Iodine deficiency leads to lowering the rate of metabolism in the body, as a result reduction hormone secretion thyroid gland.
Task 12.2
People on such a diet are at risk of iodine deficiency because: (1) they do not eat iodine-rich animal products (eg fish, eggs, milk) – thus reducing iodine intake; while (2) consuming large amounts of plant foods — particularly legumes often considered a meat substitute — they provide the body with glycosides that reduce the utilization (eg, absorption) of iodine. Thus, such a restrictive diet leads to a low intake and use of iodine, and thus problems with maintaining the proper level of this element in the body.
Comment: It is not clear what the use of dietary iodine means – the answer was that use meant e.g. absorption, thus reducing absorption would lead to problems with maintaining normal levels. At the same time, it is possible that glycosides restrict iodine utilization after it has been absorbed in the gut – so iodine levels could theoretically be normal, but its utilization would not be possible.
Exercise 13.
B (resorption) in the 2nd (XNUMXst order nephron spiral tubule).
Task 14.1
This disease consists, inter alia, in on the deficiency of surfactant – a substance that reduces the surface tension of the alveoli. Its deficiency leads to collapse of the alveoli (atelectasis), limiting the efficiency of gas exchange. This in turn leads to an increase in the concentration of carbon dioxide in the blood, stimulating the respiratory center to initiate more frequent breaths (hence increased breathing) – this is the body’s response to compensate for the low efficiency of gas exchange.
Tasks 14.2
Strongly developed rough endoplasmic reticulum – the surfactant is a compound composed of phospholipids, proteins and carbohydrates. Proteins to be then linked to lipids and / or sugars (glycosylated) are synthesized by ribosomes on the surface of the rough endoplasmic reticulum (RER). The presence of an extensive RER therefore allows for the intensive production of protein-lipid-sugar complexes, i.e. also a surfactant.
Comment: Other possibilities are the presence of the Golgi apparatus (similar explanation as above – some types of glycosylation take place in the RER, some in the Golgi apparatus; moreover, proteins synthesized in the RER often end up in the Golgi apparatus where they segregate), the presence of numerous mitochondria (which provide they don’t produce, energy for surfactant synthesis), the presence of microvilli (i.e. folds of the cell membrane that increase the surface area and thus the intensity of the secretion of substances outside the cell).
Exercise 15.
1. Change the ambient temperature from 20 degrees C to 10 degrees C
Increase in metabolic rate – this change leads to the release of more energy in the form of heat energy, thus preventing the body from cooling down.
2. Change the ambient temperature from 40 degrees C to 50 degrees C
Increased blood supply to the skin – this change causes that a greater volume of blood flows through the skin, which in the body acts as a heat carrier. Thus, thermal energy is more easily released into the environment, and therefore it is dispersed, which prevents the body from overheating.
Comment: If the temperature is raised from 40 to 50 degrees C, sweating can also be indicated (water evaporation, not just its appearance on the skin) leads to the dissipation of thermal energy from the body. Care should be taken that the answer does not contain statements that could suggest the production or destruction of energy (energy is not produced, but changes form), nor can it be destroyed.
Task 16.1
Child: l1l1 the l1l2
Mother: Ll1 the Ll2
Father: l1l1 the l1l2
Task 16.2
Father: Ll2 Mother: Ll2
Gamety: L, l2 L, l2
Possible combinations in descendant: LL, Ll2, Ll2,2l2 => hereditary lactose intolerance only occurs in genotype l2l2so the probability is 25%.
Task 16.3
Lactose, present in human milk, is a substance B. (nutritional)and as a result of its hydrolysis are formed 2. (galactose and glucose).
Task 16.4
Small intestine
Task 16.5
Determining the type of intolerance is necessary to select the correct medical procedure – in the case of genetic intolerance, it will be necessary to adjust the diet (e.g. replacing dairy products containing lactose with products from which lactose has been removed, e.g. by fermentation); however, if the intolerance is acquired, it may indicate the presence of other diseases (e.g. bacterial infections), so additional tests should be performed to identify the actual cause of lactose intolerance.
Task 16.6
1. Congenital lactose intolerance: this variant requires the elimination of lactose from the diet immediately after birth; it is considered a disease because it prevents the nourishment typical of humans (and mammals in general), ie the ingestion of milk (containing lactose) by newborns.
2. Primary lactose intolerance: this variant requires the elimination of lactose from the diet only at a later stage of development; it is not considered a disease because the proportion of milk (and therefore lactose) in the diet decreases with age until it is completely eliminated. Thus, this type of intolerance does not require the use of a diet different from that typical for humans.
Exercise 17.
Based on the analysis of the presented pedigree, it can be excluded that the polydactyly present in this family is the result of a mutation: coupled with the dominant sex.
Rationale: Assuming sex-linked inheritance, the dominant ill male from generation I would have genotype XAY and his partner XaXa. Thus, all sons would be healthy (inheriting Xa after mother, Y after father), and all daughters would be sick (inheriting XA after the father, and Xa after mother). Since this is not the case, this type of mutation can be ruled out.
Task 18.1
BBDD, BBDd, BbDD, BbDd
Task 18.2
Female genotype: BbDd
genotype samca: bbdd
Crossword:
Offspring phenotypes and their numerical ratio:
black: white: brown – 1: 2: 1
Task 19.1
1 – Thymine dimer causes deformation of a fragment of the DNA molecule.
2 – The enzyme cuts the damaged DNA strand.
3 – The enzyme synthesizes the missing strand section according to the principle of complementarity.
4 – The enzyme joins the free ends of the newly added fragment and the old DNA.
Task 19.2
A – nuclease
B – DNA polymerase
C – ligase
Task 19.3
The disease is in most cases caused by a defect in one of the enzymes involved in the DNA repair system associated with the removal of thymine dimers. Such dimers are often produced by the action of ultraviolet (UV) rays. Since sunlight includes ultraviolet light, they expose skin cells to damage in the form of thymine dimers. Since the repair mechanism of the described damage in people suffering from parchment skin does not work properly, exposure to the sun’s rays may lead to the accumulation of thymine dimers in their DNA and, for example, the transformation of neoplastic cells.
Task 20.1
This bacterium is a mutualistic species for humans – thanks to the presence of appropriate genes, it takes part in the synthesis of carbohydrates and vitamins that are necessary for humans. Moreover, it produces signaling molecules that stimulate the formation of blood vessels necessary for the functioning of the intestines. Same relationship B. thetaiotaomicron with man it is very strong, i.e. typical of mutualistic systems.
Task 20.2
Vitamin K
Task 21.1
Way of spreading: hydrochoria
Features of diasporas: 1. Presence of air crumb allowing floating in water 2. Covering the seed coat with wax to prevent soaking
Task 21.2
1. P
2. P
3. F
Task 21.3
Birds that used the island as a resting place, and later also a nesting place, contributed to increasing the species diversity of the island’s flora. It was caused by birds bringing diasporas from the mainland – such diasporas show adaptations to zoochoria. (1) exozoochory is possible, then diasporas are characterized by, for example, the presence of hooks, a rough surface, various protrusions that allow them to be caught on skin or feathers, and thus transferred over the open sea from the mainland to an island; or (2) endozoochoria, where diasporas (eg fruit) have fleshy parts (eg pericarp) that attract birds to eat the fruit and disperse it as it passes through the digestive tract with the faeces.
Task 21.4
The rise of the gull colony contributed to the wave of colonization for two reasons: (1) the greater number of birds nesting on the island meant that diasporas were transferred from the mainland to the island more frequently and in greater numbers; (2) gull droppings may have contributed to local soil fertilization, thus allowing new species to more effectively inhabit the island.
Comment: It must be remembered that the number of species on the newly created island is the result of two processes – dispersion and effective settlement. The dispersion itself will not lead to an increase in the richness of the flora.
Task 22.1
1. T
2. T
3. N
Task 22.2
This is because the crossing of these two subspecies leads to the mixing of their genetic information, thus catfish are born, whose genetic information (and thus phenotype, unique ecological function, ecological niche, etc.) does not fully coincide with that of a wildcat.
Comment: In other words, the offspring of a wildcat and a cat are neither a domestic cat nor a wildcat.
Task 22.3
1. Threat to their numbers – an extensive road network means that wildcats are forced to cross them from time to time, which may lead to hitting these animals; fragmentation of habitats may lead to a reduction in the number of the wildcat’s food base (e.g. frogs, young roe deer) and thus reduce the size of the predator
2. Threat to their genetic diversity – habitat fragmentation as a result of road development; Wildcats can avoid crossing roads, thus the flow of genes between populations is stopped, which reduces their genetic diversity.
Matura 2020 – change of date
Originally, the final exam session was to start on May 4. In connection with the coronavirus epidemic, classroom classes in schools are suspended from March 12 (from March 25, schools are obliged to distance education). The dates of external examinations, including matura exams, were also postponed. Some of the provisions relating to the conduct of the secondary school-leaving examination have also been changed.
Matriculation sheets and answers:
- Polish language: basic level
- Polish language: advanced level
- Mathematics: elementary level
- English language: basic level
The high school graduate – as before – has to take three obligatory written exams: in Polish, in a foreign language and in mathematics at the basic level. He must also take at least one written exam in optional subjects; up to six.
- Also check: Matriculation exams schedule
Matura 2020. Which subjects to choose from?
The subjects to choose from are: biology, chemistry, philosophy, physics, geography, history, history of art, history of music, computer science, Latin and ancient culture, social studies, languages of national and ethnic minorities, regional language, as well as mathematics, Polish and modern foreign languages. Exams in the subject of your choice are taken at the advanced level. Therefore, this group also includes – at this level – mathematics, Polish and foreign languages, which are compulsory at the basic level.