Biológia | Középiskola » Biológia angol nyelven emelt szintű írásbeli érettségi vizsga megoldással, 2012

2012. május 15 Azonosító jel: BIOLÓGIA ANGOL NYELVEN EMELT SZINTŰ ÍRÁSBELI VIZSGA ÉRETTSÉGI VIZSGA 2012. május 15 8:00 Az írásbeli vizsga időtartama: 240 perc Biológia angol nyelven Pótlapok száma Tisztázati Piszkozati NEMZETI ERŐFORRÁS MINISZTÉRIUM emelt szint írásbeli vizsga 1211 Biológia angol nyelven emelt szint Azonosító jel: Important information Read the following instructions carefully before you start your work. You have 240 minutes to complete your advanced level examination. Your written paper consists of two parts. For answering correctly the first part (tasks I-IX), which is compulsory for everyone you are awarded 80 points. The second part (task X) contains two versions (A and B). YOU ONLY HAVE TO COMPLETE ONE OF THEM. You can be awarded the total of 20 points FOR ANSWERING ONLY ONE OF THE TWO OPTIONAL TASKS. It means that by answering both you cannot get more points If you do start answering both optional tasks CROSS OUT THE ONE

WITH A PEN WHICH YOU DON NOT WANT TO BE INCLUDED. If you fail to do this the examiners will automatically evaluate the ‘A’ version. Your questions are either multiple choice or open-ended ones. When answering multiple choice questions you need to write one or two CAPITAL LETTERS into the empty boxes. These are the letters of the correct answer or answers Make sure that your answers are unambiguous, otherwise they will not be accepted. If you want to correct your answer CROSS OUT THE WRONG ONE AND WRITE THE CORRECT LETTER NEXT TO IT. A D correct A D C acceptable BD wrong When answering open-ended questions, you need to write technical terms, a few words, a whole sentence or several sentences. Mind your GRAMMAR because if your answer is not understandable because of bad grammar, or because it is ambiguous (e.g it is not clear what the subject of the sentence is), it will not be acceptable even if it contains the correct terms. Each correct answer is awarded 1 point, unless

otherwise indicated. Use black or blue pen. Don’t write anything into the grey boxes. Good luck for your work. írásbeli vizsga 1211 2 / 20 2012. május 15 Azonosító jel: Biológia angol nyelven emelt szint I. Ferns and Cycads 9 points The picture on the left shows a maidenhair spleenwort (aranyos fodorka), which is a fern, while the one on the right a cycad (sago palm) which belongs to the gymnosperms. Describe the characteristics of the two plants by writing the appropriate letters into the boxes. A) true for the maidenhair spleenwort B) true for the sago palm C) true for both D) true for neither 1. Its shoots can be seen in the picture 2. It reproduces by dispersed spores 3. Its sex cells are formed by meiosis 4. It has no seeds 5. It has fruit 6. It has compound leaves 7. It has flowers 8. Its plant body is made up of different tissues 9. Water solution is conducted in xylem vessels in its stem towards the leaves 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 3

/ 20 7. 8. 9. total 2012. május 15 Azonosító jel: Biológia angol nyelven emelt szint II. Life in the Mediterranean Sea 9 points The graph shows the seasonal variation of two environmental factors influencing marine life forms in the Mediterranean Sea. The data refer to the surface waters Along the horizontal axis the initials of the months are shown, while along the vertical axis the values relative to the maximum values are placed. The distribution of plant-like and animal-like planktonic organisms (minute, floating living things) is also shown. After studying the graph answer the following questions. relative value mass of növényi phytoplankton plankton tömege j f m á m mass of állati zooplankton plankton tömege j j a hőmérséklet temperature 1. Figure 1 s o n d concentration of plant nutrients növényi tápanyagtartalom The circular arrows refer to the mixing process of surface and deep waters. Explain the physical causes of this

phenomenon (apart from the storms which are more frequent in winter time). 2. The mixing process brings up water rich in phosphorous, nitrogen and heavy metal ions from the depths for the phytoplankton. In what form can plants primarily take up phosphorous and nitrogen? Write down their chemical symbols. (2 points) phosphorous: . nitrogen: 3. Of the metal ions, iron is also required in small quantities by plants and animals alike because iron is one of the components of the enzymes (cytochromes) found in the inner mitochondrial membrane. In which stage of the oxidation process are these enzymes indispensable? írásbeli vizsga 1211 4 / 20 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: 4. By contrasting Fig 1 and Fig 2 give an explanation for the fact that the ions discussed in the above questions accumulate in the non-mixing (stratified) deep water layers far away from the coast rather than on the surface (the sediment carrying effect of rivers can

be ignored) Figure 2 The rate of photosynthesis of phytoplankton is also influenced by the temperature and light intensity. Fig 2 shows the rate of photosynthesis of a marine species as a function of light intensity. (The horizontal axis shows the measure of photosynthetically active incoming radiation while the vertical one the activity of the electron transport chain operating in the chloroplasts.) 5. Explain the fact that the phytoplankton in the Mediterranean Sea is almost equally distributed across all the layers illuminated by sunlight in the winter. However, in the summer it is at a depth of 50-100 m that it reproduces the fastest. (2 points) In winter: . In summer: . . 6. Give your reasons why the mass of phytoplankton is changing in the surface waters in spring time. (2 points) Reasons for growth: . . Reasons for decline: . . . 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 5 / 20 total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: III.

Breaking the Genetic Code 11 points Robert Holley, Har Ghobind Khorana and Marshal Nierenberg received the Nobel Prize for breaking the genetic code and compiling the codon dictionary in 1968. With meticulously designed experiments they established a link between the triplets of the mRNA and the order of amino acids in proteins. 1. At the time the experiments were carried out the existence of triplets was not yet sure and there were several hypotheses put forward about the way amino acids could be coded for. On the basis of what theoretical consideration did the researchers look for base triplets rather than units consisting of two bases? Give your reason to support your answer. 2. One condition of the successful experiment was that artificial protein synthesis could be carried out in a test tube. Which cellular organelle, possible to isolate from bacteria as well, was required for the synthesis? . In the first phase of Nierenberg’s series of experiments an artificial mRNA

consisting only of uracil bases (poli-U) was synthesised. With the help of this a polypeptide chain consisting only of the amino acid phenylalanine was created. This was the first word of the codon dictionary deciphered: UUU > Phe (phenylalanine). Subsequently, mixtures of nucleotides containing the nucleotides in different proportions were applied by the researchers, and the proportion of amino acids in the polypeptides obtained was determined. If for example in a solution there are nucleotides containing only uracil (U) and cytosine (C) in a 2 :1 ratio, their frequency is p(U) = 2/3 and q(C) = 1/3. The probability of a particular triplet being formed, supposing that the bases combine randomly, equals the product of their separate/individual probabilities. Accordingly, in the mixture the probability of the base triplet UUC being formed is x (UUC) = p(U)2 q(C) = (2/3)2 (1/3) = 4/27. Since the relative frequency of phenylalanine in the polypeptide chain being synthesised was exactly

that much, it can be concluded that this base triplet coded for phenylalanine, too. U C A G U UUU phe UUC phe UUA leu UUG leu CUU leu CUC leu CUA leu CUG leu AUU ile AUC ile AUA ile AUG met GUU val GUC val GUA val GUG val írásbeli vizsga 1211 C UCU ser UCC ser UCA ser UCG ser CCU pro CCC pro CCA pro AAG pro ACU thr ACC thr ACA thr ACG thr GCU ala GCC ala GCA ala GCG ala A UAU tyr UAC tyr UAA STOP UAG STOP CAU his CAC his CAA gln CAG gln AAU asn AAC asn AAA lys AAG lys GAU asp GAC asp GAA glu GAG glu 6 / 20 G UGU cys UGC cys UGA STOP UGG trp CGU arg CGC arg CGA arg CGG arg AGU ser AGC ser AGA arg AGG arg GGU gly GGC gly GGA gly GGG gly U C A G U C A G U C A G U C A G 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: Let us suppose that, applying Nierenberg’s method in our experiment, the nucleotides containing adenine and uracil dissolved in the mixture occur in a 1/2:1/2 ratio. 3. What else do these nucleotides contain apart from adenine and

uracil? (2 points) . and 4. In what type of chemical reaction are nucleotides bound to one another? . 5. How many different triplets can be formed if the bases randomly combine in the experiment? Determine the number of different types of amino acids in the proteins synthesised on the basis of these mRNA chains. Explain the difference between the two numerical values.! (4 points) How many different triplets are there? How many different types of amino acids are there? . Reasons for the differences: . . . . 6. Determine the expected ratio of the amino acids isoleucine (ile) and leucine (leu) in the polypeptide chain synthesised in the mixture if it is assumed that there is an equal chance of the two types of the nucleotides being linked to each other in the synthetic mRNA. Ratio of leucine to isoleucine: . 7. Let us suppose that a polypeptide chain of 4 amino acids were produced in the above experiment. Give the number of different combinations in which the amino acids can be linked

to one another in the chain 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 7 / 20 7. total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: IV. Microscopic Investigation 9 points We made a cross section of one of the organs of a dicotyledonous plant and the specimen was viewed under a light microscope. 1. Let us start the investigation under low power of magnification Several tissues can be seen. Which of the following tissues can be seen? Write the letter of the correct answer into the boxes. (2 points) A) paranchyma D) conducting tissue B) epithelium E) connective tissue C) epidermis 2. Now let us move on to higher magnification Explain how exactly the magnification of the microscope can be calculated. Under this magnification it can be seen that at the top and the bottom parts of the specimen the cells are arranged into a single layer where they fit closely together. At the top part they look more or less identical, while at the bottom

different, bean-shaped cells can be observed as well. 3. Which tissue is being investigated? 4. In the mid-section of our specimen one of the tissues observed has two, differently shaped cell types with chloroplasts visible inside them. What are the characteristics of this tissue? (2 points) A) Some of its cells are loosely packed in a sponge-like arrangement. B) Some of its cells preserved their ability to divide. C) Some of its cells have a columnar arrangement. D) Its cells can be covered with a layer of cuticle. E) Its cells can be covered with hair like structures. 5. In the course of our further investigations in one particular part of a specimen a lot of small, undifferentiated cells with thin cell walls can be seen. Which type of tissue is investigated? . 6. Could the specimen be prepared from the leaf of a monocotyledonous plant? Give your reason to support your answer. . 7. Could the specimen be the leaflet of a moss plant? Give your reason to support your answer. . 1. 2.

írásbeli vizsga 1211 3. 4. 5. 6. 8 / 20 7. total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: V. Levels of Organisation and Nutritional Types 7 points Write the number of the species or group of living organisms from 1 to 9 into the appropriate spaces of the figure according to their levels of organisation and nutritional types. Each correct answer is 1 point Eukaryotes Heterotrophs Animals with pseudotissue Chemotrophs 1. 2. 3. 4. giant amoeba Chlamydomonas (single-celled green algae) baker’s yeast nitrifying bacteria, which rely on the energy deriving from oxidising nitrogen containing inorganic compounds for fixing/reducing atmospheric carbon dioxide. 5. fresh water sponge 6. bacterium causing pneumonia 7. ferns and their relatives (Pteridophytes) 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 9 / 20 7. total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: VI. Enforced Diet 8 points Life in Cuba

took a sudden turn for the worse twenty years ago when the Soviet Union, whose main priority (.) up until then was to keep the regime in Cuba alive, collapsed() The graveness of the collapse is well reflected in the fact that famine became widespread in the country. () Due to the shortage of food the average calorie intake of Cubans dropped from 2899 kilocalories, which was regarded as too much anyway, to 1863 kilocalories. In the meantime, due to severe restrictions on fuel, masses of people were compelled to walk and cycle, which doubled the number of physically active inhabitants compared to earlier data. As a result of all this the body mass index of the Cubans fell by 1.5 points between 1990 and 1995 (The body mass index is a number indicating the extent of obesity and is obtained by dividing the body mass expressed in kilograms by the square of height in meters.) () The weight loss of the population had a profound effect on the incidence rate of several diseases and their

mortality rate. In 2002, compared to 1997, the death rate for diabetes fell by 51%, for heart attack by one third and for stroke by one fifth. All this is due to the falling rates of obesity According to an international survey published in The Lancet medical journal the average blood sugar level of mankind in 2008 was 5.46 millimole per liter (the normal range in a litre is between 4 and 6.1) and this figure had risen by 024 millimole since 1980 The number of diabetics has increased from 153 million in 1980 to 347 million in 2008 (during this time the population of the world has increased from 4.5 billion to 66 billion) Based on newspaper article by Csilla Molnár: Üzemanyag-korlátozás (Magyar Nemzet, 2011. jul 09) 1. Calculate by how much the average body mass of Cubans decreased between 1990 and 1995. Let us suppose that the average height remained constant all along: 170 cm 2. While the number of diabetics decreased in Cuba, the world average was on the increase Calculate the

percentage by which the rate of diabetics increased in the world between 1980 and 2008. 3. Sugar is carried in the blood mainly in the form of glucose How many grams of glucose on average did the blood of a healthy adult man contain in 2008? A C = 12, A O = 16, AH = 1. írásbeli vizsga 1211 10 / 20 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: 4. In the course of sustained starvation, the body mainly converts its protein reserves into glucose. Name a hormone which promotes this process (Let us ignore the unexpected stress effect during this time.) . 5. Obesity mainly increases the incidence rate of type II diabetes In this type of disease the permanently much higher level of blood sugar results in a level of insulin slightly higher than the normal. What might be the reason for this? Write the letters of the correct answers into the boxes. (2 point) A) Because of the higher level of blood sugar the insulin secretion of the pancreas is slightly

increased. B) Slightly increased level of insulin results in a manifold increase in the blood sugar level. C) High blood sugar level causes the insulin sensitivity of the target cell receptors to increase. D) In spite of high insulin level, the rate of blood sugar uptake is low due to the low number of insulin receptors in the membrane of the target cells. E) In spite of high blood sugar level, the rate of blood sugar uptake is low due to the low number of blood sugar level detecting receptors in the membrane of the target cells. 6. In the case of type II diabetes there is often also an increased level of glucagon in the blood of diabetics. Glucagon has an antagonistic effect to insulin High levels of insulin, however, decrease the rate of glucagon synthesis. What might cause the level of glucagon to increase in diabetic patients? A) Increased inhibitory effect of high glucose levels. B) Increased inhibitory effect of high insulin levels. C) The receptors inhibiting glucagon synthesis

do not work. D) The receptors promoting glucagon synthesis do not work. E) Glucagon has the effect of lowering blood sugar level in diabetics. 7. The 'enforced diet' might not only have beneficial effects Deficiency diseases can also arise due to low amounts of essential amino acids. Describe what is meant by the term essential amino acid. 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 11 / 20 7. total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: VII. Pedigree Analysis 9 points The family tree shows the inheritance of a dominant disease (let us indicate the allele which causes it with letter‘A’) and the inheritance of the ABO blood group system (IA, IB and i). The two genes coding for these characteristics are located on the same chromosome. The manifestation of the disease is marked by shading the symbols solid black, while the blood groups are shown by the appropriate letters next to the symbols 1. Find the genotype of the woman

marked by I/2 in the family tree Write the letter of the correct answer into the empty box. A) AA B) aa C) Aa D) Aa E) AA and and and and and IB i IB i IB i IB IB I B IB 2. Give the genotype of the man marked by I/1 by using the same type of symbols as in question 1. 3. If the woman II/1 marries a healthy man with blood group AB, what phenotypes can their children have? (2 points) . 4. Which alleles did the man II/2 inherit from his mother and father? from his father: . (2 points) from his mother: 5. The II/2-II/3 married couple had a child who suffered from the disease and had blood group ‘O’. a) Which alleles did she inherit from her mother? . b) Which alleles did she have to get from her father? . c) Name the process which can explain that the father passed on these two alleles at the same time (let us rule out the possibility of a point mutation). . 1. 2. írásbeli vizsga 1211 3. 4. 5. total 12 / 20 2012. május 15 Azonosító jel: Biológia angol

nyelven emelt szint VIII. Molecule Identification 8 points The molecules listed below play a range of important roles in living organisms. By applying a method used in plant identification, work out step by step which compounds are indicated by letter A, B.etc Write the name of the compounds (groups of compounds) into the table. There are two odd ones out among the ones listed below, which do not fit anywhere. The compounds: bile salts haemoglobin carotene urea phosphatides fibrinogen myosin chlorophyll* insuline Peptide. Non-peptide. It occurs in the blood plasma. It is active within the cell. 2. 5. 3. 4. 3. a) The change in its structure is an important step in blood clotting. A: b) It plays no role in blood clotting . B: 4. a) It is an enzyme which breaks down ATP C: b) It is a transport protein D: 1. a) b) 2. a) b) 5. a) Its biological role is attributed to its peculiar polarity features. b) Its biological role can be explained by its system of conjugated

double bonds . 6. a) In biological systems its molecules are oriented so that its hydrophobic parts point towards each other, while its hydrophilic parts point towards the water. b) When functioning, the hydrophobic parts surround a apolar liquid droplet, while its hydrophilic parts point towards the water. 2+ water 6. 7. E: F: 7. a) Its molecule also contains Mg ion G: b) It contains no Mg2+. H: *chlorophyll = the molecule which is responsible for absorbing light in photosynthesis. A B írásbeli vizsga 1211 C D E F 13 / 20 G H total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: IX. Vessels and Organs 10 points The figure shows the schematic picture of the blood supply of the kidney and the liver. After studying the picture answer the following questions! C liver 1. What is the name of the part indicated by letter ‘E’? gall bladder A D kidney B Y = true for all E X = true for none For which duct shown in the figure

are the following statements true? Answer with the letters of the figure above. If the statement is true for none, write X into the box If it is true for all, write Y. 2. The blood flowing through this part left the right ventricle last on its way out of the heart. 3. The blood flowing through this part is carried to the left atrium first as it gets to the heart. 4. Urine is flowing through this part 5. This part contains a digestive enzyme 6. It contains steroids (compound with steroid frame) írásbeli vizsga 1211 14 / 20 2012. május 15 Azonosító jel: Biológia angol nyelven emelt szint 7. Decide which of the following statements are true Write the letters of the correct statements into the boxes. (4 points) A) Letter A and B indicate one artery each. B) Letter C points to a vein. C) The pH value of the liquid flowing in part ‘A’ is higher than in part ‘B’ because the higher concentration of carbon dioxide in part ‘B’ shifts the pH towards the acidic extreme.

D) The sugar concentration of the liquid flowing in part ‘A’ in a running man can be higher than in part ‘B’ because some of the sugar is taken up by the tissue cells from the blood flowing through the tissue. E) The liquid flowing in the duct indicated by letter ‘D’ has an acidic pH because it contains bile acids/salts. F) Following a tough jogging, the lactic acid content of the liquid flowing through part ‘B’ is lower than in part ‘C’ because in the skeletal muscles the lactic acid is converted into glucose. G) Blood is carried into the liver through both an artery and a vein. 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 15 / 20 7. total 2012. május 15 Azonosító jel: Biológia angol nyelven emelt szint Optional tasks Write the letter of the chosen task into the box. (A or B) X. A Body Fluids 20 points The human blood 7 points A) ‘A’, Rh-positive blood group B) ‘B’, Rh-negative blood group C) True for both D) True for neither Write the

letter of the correct answer into the box next to each statement. There is only one correct answer to each question. 1. Blood group antigens are found in the blood plasma. 2. It coagulates when exposed to a blood serum deriving from an ‘O’, Rhnegative blood. A mother with such blood group can have Rh incompatibility developing in the course of her first pregnancy. A mother with such blood group can have Rh incompatibility developing in the course of her second pregnancy. On the surface of their red blood cells there are only two types of antigens with regard to the ABO and Rh blood group system. 3. 4. 5. 6. In case of emergency its red blood cells can be transferred into a person with blood group‘AB’, Rh-negative. 7. Its blood group can be determined by a double homozygous genotype, too. Tissue circulation 13 points Discuss the human tissue fluid in the form of a brief essay. In your answer include the following points: • From what, where and how (as a result of

which physical and chemical circumstances/conditions) is it formed? • What is the characteristic feature of its composition straight after it is formed and how is it changing later on? • What is the function of the tissue fluid? • Where does it flow from the tissues and which physical and chemical factors are the main driving forces of its streaming? 1. 2. írásbeli vizsga 1211 3. 4. 5. 6. 16 / 20 7. Esszé total 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: X. B Nature Conservation and Genetics Extinction threshold 20 points 9 points The following text is about the dangers threatening small populations. Fill in the gaps with suitable expressions. In populations decreasing in size mating between closely related individuals, in other word (1). and the random elimination of different gene variants, the process referred to as (2). will sooner or later inevitably occur The mating of relatives decreases the biological fitness of each

particular individual, whereas a decrease in genetic diversity causes the adaptability of the population to decline. In such populations the mortality rate is (3). and the number of births per capita (birth rate) is (4). As a result, the population is getting smaller and smaller and the accelerating process can lead to extinction. 5. Give your reason why the mating between close relatives can lead to a decrease in biological fitness of the offspring. (2 points) . . . . 6. Explain why the decrease in genetic diversity causes the adaptability of the population to decline. (2 points) . . . . 7. Name one factor which increases the genetic diversity of the population while the average biological fitness of the individual is not decreasing. . írásbeli vizsga 1211 17 / 20 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: Means of nature conservation – essay 11points Discuss the basic principles, the means and aims of protecting our natural assets. In your

essay include the following points: • The basic idea of biological diversity (on a genetic and ecosystem level), its economic and medical significance. • At least three anthropological effects, which result in populations decreasing in size or the extinction of species. • Concrete objectives of two international agreements in nature conservation. • Connection between consumer habits in Hungary and species richness in ecosystems in distant lands. • An example for active and sustainable nature conservation (positive effects of regular human intervention). You can write your essay on page 18-19. 1. 2. 3. 4. 5. 6. 7. Essay total Essay: írásbeli vizsga 1211 18 / 20 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: írásbeli vizsga 1211 19 / 20 2012. május 15 Biológia angol nyelven emelt szint Azonosító jel: Maximum score Achieved score 9 9 11 I. Ferns and Cycads II. Life in

the Mediterranean Sea III. Breaking the Genetic Code IV. Microscopic Investigation V. Levels of Organisation and Nutritional Types VI. Enforced Diet VII. Pedigree Analysis VIII. Molecule Identification IX. Vessels and Organs 9 7 8 9 8 10 80 Total score of the test: 20 X. Optional essay or problem solving task Total score of the written examination: 100 marking teacher Date: . elért pontszám egész számra kerekítve / achieved score rounded to an integer programba beírt egész pontszám / integer score written into the programe Feladatsor / Test Választható esszé vagy problémafeladat / Optional essay or problem solving task javító tanár / marking teacher jegyző / administrator Dátum/Date: . írásbeli vizsga 1211 Dátum/Date . 20 / 20 2012. május 15 ÉRETTSÉGI VIZSGA 2012. május 15 Biológia angol nyelven emelt szint Javítási-értékelési útmutató

1211 BIOLÓGIA ANGOL NYELVEN EMELT SZINTŰ ÍRÁSBELI ÉRETTSÉGI VIZSGA JAVÍTÁSI-ÉRTÉKELÉSI ÚTMUTATÓ NEMZETI ERŐFORRÁS MINISZTÉRIUM Biológia angol nyelven emelt szint Javítási-értékelési útmutató Instructions for marking the advanced written paper 1. Always use red ink 2. Put a tick next to each correct answer Each correct tick is worth one point Half a point is not to be given. If a question which is worth two points is answered correctly, put two ticks. 3. If there is a correct answer to a question which is not included in the marking scheme you can still accept it. The same principle applies to synonyms (e g: platelets trombocytes) 4. The equally acceptable, alternative answers are separated with a backslash (/) in the marking scheme. 5. At the end of each task add up the points in the grey boxes 6. Mark each correct answer with a tick on the margin in the optional essay In the marking scheme only the most important terms and expressions are included

in the correct logical order. You can also accept the answers if they are logically built up even if this logic is different from the one given in the marking scheme, unless it is otherwise stated in the given task. In the end, please add up the points of the correct answers and write them into the appropriate boxes of the summary chart. 7. In the essay, points can only be given for answers responding to the guiding questions 8. If the candidate started answering both optional tasks (A and B) act according to the instructions given under the heading of ‘Important Information’. 9. If the candidate is expected to formulate an answer in full sentences (eg reasoning and explanations) only grammatically correct sentences can be accepted. Do not subtract points for spelling mistakes but do not accept ambiguous sentences either. Good luck for your work. írásbeli vizsga 1211 2/8 2012. május 15 Biológia angol nyelven emelt szint Javítási-értékelési útmutató Marking

scheme I. Ferns and Cycads 9 points This task is based on the following chapter of the Detailed Requirements: 3.41 Source of Figures: original Each correct answer is 1 point. 1. 2. 3. 4. 5. 6. 7. 8. 9. C A D A D C B C D II. Life in the Mediterranean Sea 9 points This task is based on the following chapter of the Detailed Requirements: 2.11, 212, 223, 5.11 és 541 Source of Graph No1: Mojetta, Angelo: A Földközi-tenger Alexandra, 2005 nyomán Source of Graph No2: 1. The surface waters cool down, their density increases, as a result, the deeper and warmer waters rise to the surface. Alternative wording is also acceptable 1 point 2. Phosphorous: H2PO4- / HPO4 2- (the less soluble PO4 3- is also acceptable) 1 point Nitrogen: NO31 point 3. For the terminal oxidation / They are involved in the electron transport 1 point 4. The main source of each element is the slowly sinking organic matter of plant origin, it takes time for them to break down. Or: In the illuminated upper water

layers the plants extract nitrogen and phosphorous and accumulate them in their bodies. Other explanation(s) with similar contents can also be accepted. 1 point 5. In winter: the nutrients are available everywhere (the water is mixing) 1 point In summer: the nutrients are more abundant in deep water layers and the far too strong light on the surface inhibits photosynthesis / in layers deeper than that the illumination is very weak. (The answer should include the interaction between both the light intensity and the source of nutrients. The point can only be given if both conditions are mentioned) 1 point 6. Reasons for growth: increasing temperature / illumination / availability of (still) abundant nutrient supply. (Two conditions are to be named) 1 point Reasons for decline: depletion of nutrient supply (because the mixing process ceases to operate / the nutrients accumulate in the living organisms) / inhibitory effect of the increasing intensity of sunlight/illumination (which is

now too strong) in the surface waters. Reference to the idea that as the temperature of the water increases its gas dissolving capacity decreases is also acceptable. (Two conditions are to be indentified to receive the point) 1 point írásbeli vizsga 1211 3/8 2012. május 15 Biológia angol nyelven emelt szint III. Javítási-értékelési útmutató Breaking the genetic code 11 points This task is based on the following chapter of the Detailed Requirements: 6.11-2 1. If there were base duplets it would only be possible to code for 42=16 amino acids while there are 20 (21) amino acids occurring in living things. (43=64 possible arrangements are enough to code for that many) 1 point 2. ribosomes 1 point 3. phosphoric acid / phosphate group 1 point sugar / ribose / pentose 1 point 4. (poli)condensation / ester formation 1 point 5. Eight different combinations (AAA, AAU, AUA, UAA, AUU, UAU, UUA, UUU) 1 point They can code for six different amino acids, 1 point because

there is one STOP codon among them, 1 point and one amino acid can be coded for by two triplets. 1 point 6. the ratio of leucine to izoleucine: 1:2 1 point 4 7. 6 (= 1296) giving the mathematical expression is sufficient, the calculation is not reqired) 1 point IV. Microscopic investigation 9 points This task is based on the following chapter of the Detailed Requirements: 1.1; 343 1. A and D 2 points 2. the magnification of the objective lens is to be multiplied by the eye piece lens (the answer: multiplying the magnifications of the lenses is not to be accepted)1 point 3. Epidermis 1 point 4. A and C 2 points 5. Meristematic tissue/cambium 1 point 6. No, because stomata were only to be found in the lower epidermis Or: No, because there were palisade mesophyll cells towards the upper surface and spongy mesophyll cells towards the lower epidermis. 1 point 7. No, because the leaflets of the moss plant do not contain tissues 1 point írásbeli vizsga 1211 4/8 2012. május 15

Biológia angol nyelven emelt szint Javítási-értékelési útmutató V. Levels of Organisation and Nutritional Types 7 points This task is based on the following chapter of the Detailed Requirements: 2.21, 32 és 33 Eukaryotes Heterotrophs Animals with pseudotissue 5. 7. 1. 6. 3. 4. 2. Chemotrophs Each correctly placed answer is 1 point. VI. Enforced Diet 8 points The source of the text: Molnár Csilla: Üzemanyag-korlátozás Magyar Nemzet, 2011.július 09 1. x : (17)2 = 15 ; which means that x = 1.5 (17)2 = 435 kg-is the amount by which the average body mass decreased. 1 point The point can only be given if the calculation is clearly shown. 2. In 1980 Total number of people: 4500 million 100% Number of diseased: 153 million 3.4 % In 2008 Total number of people: 6600 million 100% Number of diseased: 347 million 5.26 % As a result the proportion of diabetics has grown by 1.86% The point can only be given if the calculation is clearly shown. 1 point 3. 1 mole

glucose is 180 g In 1 liter of blood there is 5.46 millimole = 546 018 = 09828 g In 5 liters of blood: 4.91 g glucose 1 point Point can only be given if the calculation is clearly demonstrated. The calculations in different orders are also acceptable. 4. the hormone of the adrenal cortex regulating sugar metabolism (eg cortisol) / glucocorticoids Glucagon is also acceptable. 1 point írásbeli vizsga 1211 5/8 2012. május 15 Biológia angol nyelven emelt szint 5. A and D Javítási-értékelési útmutató 1+1 point, a total of 2 points 6. C 1 point 7. The human body is unable to manufactue these from other substances / The body has to take them up ready-made in the food. 1 point VII. Pedigree analysis 9 points This task is based on the following chapter of the Detailed Requirements: 6.21 1. C 2. aa and ii 3. healthy with either blood group A or B 1 point 1 point 1 point 1 point 4. from the father: a and i; from the mother: A and IB 5. a) a and i; b) A and i; c) there

must have been recombination / crossing over . 1 point 1 point 1 point 1 point 1 point VIII. Molecule identification 8 points This task is based on the following chapter of the Detailed Requirements: 1.3; 211; 213; 2.15; 461; as well as on the competent use of plant identification keys: 341 Each correct answer is 1 point. A) fibrinogen B) insuline C) myosin D) haemoglobin E) phosphatides F) bile salts G) chlorophyll H) carotene IX. Vessels and Organs 10 points This task is based on the following chapter of the Detailed Requirements: 2.13, 432, 442, 4.71-2 Figure: original 1. 2. 3. 4. 5. 6. 7. ureter X X E X Y B, C, D, G írásbeli vizsga 1211 1+1+1+1 = 6/8 1 point 1 point 1 point 1 point 1 point 1 point 4 points 2012. május 15 Biológia angol nyelven emelt szint Javítási-értékelési útmutató X. A Body fluids 20 points 1. the human blood 1. D 2. C 3. D 4. B 5. A 6. B 7. C Each correct answer is 1 point. 7 points 2. Tissue circulation 13 points As the

tissue fluid is formed • The blood plasma is 1 point • filtered through 1 point • the arterial end of the capillaries 1 point • because it is here that the blood pressure is higher than the osmotic pressure of the blood proteins. 1 point Composition: • When it is formed it is blood plasma without proteins 1 point • into which white blood cells and 1 point • immune protein (produced by the white blood cells) can get. 1 point (Metabolic products of the cells and/or substances deriving from degraded cells are also acceptable) Function: • Transport/exchange of substances between the blood and the cells • Immunity 1 point 1 point Streaming: • Some of it is taken back into the capillaries (towards the venal end) 1 point • because here the blood pressure is lower than the osmotic pressure of the blood plasma (proteins) 1 point • Some of it gets into the lymph capillaries 1 point • where the lymphatic streaming is ensured by the (suction force of) the heart / lower

pressure of the venal system / difference in pressure 1 point X. B Nature conservation and genetics 20 points This task is based on the following chapter of the Detailed Requirements: 5.3, 543, 55, 6.31, 632 pontjai alapján készült Extinction threshold (9 points) 1. 2. 3. 4. 1 point 1 point 1 point 1 point inbreeding genetic drift increasing decreasing írásbeli vizsga 1211 7/8 2012. május 15 Biológia angol nyelven emelt szint Javítási-értékelési útmutató 5. • • 6. • • The alleles causing diseases / abnormalities / mortality are mostly recessive (therefore their effect is not expressed in heterozygote condition), 1 point the genetic make up of close relatives is similar, therefore among their offspring the probability of homozygous recessive conditions / the occurrence of diseased or abnormal individuals is higher. 1 point The adaptability of the population is based on the increasing proportions of individuals with more favourable phenotype /

traits (alternative wording is also acceptable), 1 point If the diversity is small at the beginning of the process, selection does not have much to select from (alternative wording is also acceptable). 1 point 7. immigration / mutation Means of nature conservation – essay • • 1 point (11 points) o The idea of genetic diversity at the level of genes (e.g number, frequency of alleles) o at the level of ecosystem (e.g species richness of communities / diversity of communities in bioms Example for economic and medical significance: o crossing with wild growing species / sources of raw materials for medicines o natural enemies of pests / tourism Or other two good example Destructive anthropological effect: o hunting and fishing / habitat distruction (monoculture, cities, industry) / habitat fragmentation o trade and commerce / pollution of the environment (DDT, pesticides) o introduction of new species Any three examples or their concrete cases to be named. • International

agreements on(their names are not required but their objectives are): o trade in endangered species (Washington Convention, CITES) Species richness / biological diversity (Rio Convention o protection of Wetlands (Ramsar Convention) protection of cultural and natural heritage (World heritage) The aims of any two international conventions in nature conservation are to be discussed. • Identifying one connection between consumer habits and the ecosystem of distant lands E.g: Coffee consumption – increasing areas of coffee plantation in the tropics Air conditioning – Due to increased energy consumption the ice coverage is melting, shrinking of polar bear habitats. • Any example for active nature conservation practice: E.g assisted reproduction and repatriation of endangered species / Habitat reconstruction, /Grazing and grass cutting on mountain pastures. írásbeli vizsga 1211 8/8 2012. május 15