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Biológia angol nyelven emelt szint — írásbeli vizsga 1813

EMBERI ERŐFORRÁSOK MINISZTÉRIUMA

Azonosító jel:

BIOLÓGIA ANGOL NYELVEN

EMELT SZINTŰ ÍRÁSBELI VIZSGA

2019. május 14. 8:00

Időtartam: 300 perc

Pótlapok száma Tisztázati Piszkozati

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Biológia angol nyelven emelt szint

1813 írásbeli vizsga 2 / 24 2019. május 14.

Azonosító jel:

Important information

Read the following instructions carefully before you start your work. Your written paper consists of two parts. For answering correctly the tasks (I-VIII.) compulsory for everyone you are awarded 80 points. The last task (IX.) 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 do 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!

When answering open-ended questions, you need to write technical terms, a few words, a whole sentence or several sentences or an essay. 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. Points cannot be given for contradictory answers given to the same question. Each correct answer is awarded 1 point, unless otherwise indicated. Use black or blue pen. Do not write anything into the grey boxes.

Good luck for your work!

A D A D C B D correct acceptable wrong



Azonosító jel:

I. Blood circuits 8 points In the picture the circulatory system of four vertebrate groups are shown in the form of schematic diagrams. The vessels shaded black carry deoxygenated, while the unshaded vessels oxygenated blood. The dotted stretches of vessels indicate mixed blood in them. The animals marked ‘A’ and ‘C’ have external fertilisation. The body of animals ‘B’ and ‘D’ are covered by keratinised epithelium and animal ‘D’ nourishes its young with her milk.

Write the appropriate letters into the empty boxes. If a statement is not true for any of the groups put ‘E’ in the box.

1. Most of the oxygen carried by its blood is dissolved in the blood plasma.

2. The blood is not mixed in its four-chambered heart.

3. The muscles receive oxygenated blood from its respiratory system rather than directly from its heart.

4. The gas exchange taking place in the pulmonary circuit only partially meets the oxygen demand of the animal.

5. Its offspring rely on the oxygen contained in its mother’s blood.

6. Oxygen is taken up into its blood in the lungs during both inhalation and exhalation.

7. Its body temperature is internally regulated due to its efficient metabolism.

8. It has one blood circuit.

1. 2. 3. 4. 5. 6. 7. 8. Total

A

D C

B



Azonosító jel:

II. Nitrogen sources 12 points According to the classic model of the nitrogen cycle plants take up inorganic nitrogen produced as a result of organic matters being broken down by decomposing bacteria (the process also called mineralisation).

1. Give the name of an inorganic nitrogen-containing ion readily available for plants.

...............................................................................................................................................

2. The rate of microbial breakdown processes is slow in the tundra region. Write the letter of the statement which best explains it.

A) The tundra is mainly inhabited by animals with constant body temperature. B) The tundra is characterised by low species number and large population size. C) In winter the precipitation falls in the form of snow. D) The soil is frozen for most of the year only its surface thaws during the short summer. E) There are no microbes in the tundra soil.

3. Which inorganic nitrogen compound is produced as a result of the activity of decomposing

bacteria? Write the letter of the correct answer into the box.

A) NO2–

B) NH3 C) NO3

– D) amino acids E) urea

4. What role is played by the nitrifying bacteria in the nitrogen cycle?

A) They fix atmospheric nitrogen. B) They reduce the organic nitrogen content of the soil. C) They reduce the nitrate ion to atmospheric nitrogen. D) They oxidise ammonia to nitrate ion. E) They are in symbiosis with leguminous plants.

5. The nitrifying bacteria are capable of manufacturing their own organic molecules relying

on an inorganic substance. Which nutritional group do they belong to?

A) photoautotrophic B) photoheterotrophic C) chemoautotrophic D) chemoheterotrophic

6. In order to ensure their nitrogen supply, leguminous plants establish a symbiotic

relationship with a unique group of bacteria. Give the common name of these bacteria. ...............................................................................................................................................

7. In what way do the bacteria living in the root nodules of leguminous plants benefit the plant? Write your answer on the dotted line.

...............................................................................................................................................



Azonosító jel:

8. How do leguminous plants benefit the bacteria living in their root nodules? Write your answer on the dotted line. ...............................................................................................................................................

In bogs the rate of mineralisation is not fast enough to meet the nitrogen demands of plants. Some plant species capture insects and break them down to ensure their nitrogen supply. The common sundew (Drosera rotundifolia, harmatfű) shown in the picture got its name from the dew-like droplets covering its leaves which are sticky and contain digestive juices to capture and digest small insects.

9. Which molecules of the insect’s body contain nitrogen? Write the letters of the correct

answers in the boxes. (2 points)

A) proteins B) glycogen C) fats D) nucleic acids E) carbohydrates

10. As a result of what type of chemical reaction does the sundew gain soluble nitrogen-

containing compounds from the macromolecules of the insects?

A) hydrolysis B) condensation C) oxidation D) reduction E) photosynthesis

Scientific studies conducted in the last few years have shown that some plants living in nitrogen poor soils are able to leave out the last step of bacterial decomposition and can take up organic nitrogen-containing molecules like amino acids in addition to their inorganic nitrogen sources. Their roots contain amino acid transporting proteins (transporters), whose synthesis is triggered by the low level of available nitrogen. 11. Briefly describe the plan of an experiment which is designed to show that the gene of the

amino acid transporting protein is also present in plants making use of inorganic nitrogen sources.

...............................................................................................................................................

...............................................................................................................................................

...............................................................................................................................................

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Total



Azonosító jel:

III. About genetic diversity 8 points ‘Genetic diversity is maintained by various factors. By means of recombination the genetic material of the cell is rearranged during the formation of gametes (in animals and in humans) giving rise to new combinations and thereby renewing the genetic makeup (…) In addition to the mechanisms mentioned above, the genetic diversity also arises from the random errors in DNA copying (…) Over and above the errors which are either repaired or cause diseases and deaths there are a large number of copying errors which go unnoticed in the genetic material of subsequent generations.’

(Based on the article of Soós Noémi és Kusza Szilvia) 1. Name the type of cell division during which the ‘rearrangement’ or recombination of the

genetic material mentioned in the excerpt takes place. ...............................................................................................................................................

2. During the formation of human gametes the rearrangement of the genetic material takes place in two steps. Summarise the main points of the recombination events taking place in each step.

a) During the prophase of the 1st stage of cell division:

...............................................................................................................................................

b) During the metaphase of the 1st stage of cell division:

...............................................................................................................................................

...............................................................................................................................................

3. What does the author of the excerpt mean by the random copying error of the DNA? Write the letter of the correct answer into the box.

A) selection B) mutation C) condensation D) modification E) genetic drift

4. Based on the features of the genetic code give one reason why the copying errors of the

DNA could remain unnoticed for an extended period of time and are not expressed in the structure of the proteins produced. ...............................................................................................................................................

...............................................................................................................................................

5. Give another reason why changes in the DNA could result in changes in the order of amino acids and still the function of the protein encoded remains unaffected. ...............................................................................................................................................

...............................................................................................................................................



Azonosító jel:

The diversity of the population provides the basis for its adaptation. ‘The DNA molecule of the mitochondrion is one of the nucleic acid molecules of the cell which is widely used in population genetic studies. One of the reasons for this is that it is rather conservative relative to the nuclear DNA, which means that the mitochondrial DNA has a lot lower rate of copying errors than the nuclear DNA’. The genes of mitochondrial DNA code for proteins involved in biochemical processes taking place in mitochondria. 6. Which processes can be damaged directly if copying errors occur in the mitochondrial

DNA? Write the letters of the correct answers in the boxes.

A) Calvin cycle (carbon dioxide reduction) B) the synthesis of cell membrane receptor proteins C) terminal oxidation D) photolysis (in the light phase) E) citric acid cycle

1. 2. 3. 4. 5. 6. Total

IV. Insulin resistance 14 points Insulin resistance is one of the commonest ailments of our modern society, which, if left untreated, could easily lead to type 2 diabetes. In the physiological background of both lies the fact that the insulin which is properly produced is not recognised by the cells and as a result, the insulin cannot exert its effect. The result of the resistance is that glucose accumulates in the blood. Insulin resistance can be diagnosed by glucose tolerance test. The person tested should consume a certain amount of sugar (adults are to drink a solution containing 75g glucose) and the rate at which glucose appears in the blood and then the rate at which glucose is cleared from the blood are tested. During the investigation blood samples are taken several times including one before the consumption of glucose followed by several subsequent ones (30, 60, 90 and 120 minutes after the consumption of glucose). The glucose and insulin concentrations of each sample are measured and then a time-blood sugar and a time-insulin graph is plotted. 1. Name the organ and the type of cells which produce insulin.

...............................................................................................................................................

2. Explain how the glucose transport across the membrane of skeletal muscle cells is affected by insulin in a healthy organism.

...............................................................................................................................................

3. What is meant by the expression that ‘the insulin is not recognised by the cells’ in case of insulin resistance? Write the letter of the correct answer in the box.

A) Insulin is not produced. B) Insulin cannot be released into the blood. C) Insulin cannot break down the blood sugar level. D) Insulin cannot be taken up and broken down by the cells. E) Insulin cannot bind to the insulin receptor.



Azonosító jel:

The changes in the blood sugar level and insulin level measured in three people during their glucose tolerance tests are shown in the two graphs. (µU/ml is a unit expressing concentrations.)

Changes in the level of insulin during glucose tolerance test:

Changes in the blood sugar level during glucose tolerance test:

4. Write the letter of the correct explanations in the boxes based on the graphs and your

knowledge. (2 points)

In case of insulin resistance …

A) the blood sugar level remains high because the cells cannot take up glucose. B) the blood sugar level remains low because insulin cannot exert its effect on the cells. C) the start of insulin production is delayed. D) the level of insulin falls because the cells cannot take up glucose. E) the insulin level remains high because the high level of blood sugar continuously

induces the production of glucose.

bloo

d su

gar l

evel

(mm

ol/l)

idő (perc)

insu

lin le

vel (

µU/m

l)

30 60 120

cukorbeteg

idő (perc) 120 60 30 90

Type 2 diabetic

Insulin resistant person

Healthy person

Type 2 diabetic

time (minutes)

time (minutes)

Insulin resistant person

Healthy person



Azonosító jel:

5. Write the letter of the correct explanations in the boxes based on the graphs and your

knowledge. (2 points)

In a healthy organism …

A) the high blood sugar level triggers the production of insulin. B) due to the insulin produced the rate of glucose absorption into the blood is increased. C) due to the insulin produced the blood sugar level drops. D) the blood sugar level remains high because the insulin level falls. E) the insulin is continuously produced to ensure the glucose supply of the cells.

6. Name one of the typical risk factors of type 2 diabetes.

...............................................................................................................................................

One of the characteristic symptoms of diabetes is frequent urination and the production of large quantities of urine. Fill in the gaps of the following explanation. (One word can be used more than once.) Due to the high blood sugar levels the sugar content of the (7) ……………………. produced

in the renal corpuscles is (8) ……………………., therefore the sugar cannot be fully

(9) ………………………. from the renal tubules into the blood. Because of the sugar

left in the (10) …………………….. the osmotic suction force of the solution is

(11) ……..…………….., which retains (12) ……………………. and thereby increases the

volume of urine excreted.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Total



Azonosító jel:

V. The wild apple 11 points The wild apple (Malus sylvestris, vadalma) is rare in Hungary, the mature trees often live so far apart that due to lack of cross pollination there is a possible danger of inbreeding. It thrives on forest edges where it receives the amount of sunlight required to bloom and bear fruit. Due to intensive forest management and its weak competitive ability it has been forced to live outside the closed forest. In the spring during daytime its strong smelling flowers are pollinated by bees, while in the mornings and evenings as well as on cold days the pollination is carried out by wasps and at night by moths. Its seeds are consumed by squirrels and are also stored for the winter. Around sixty years of age its heartwood is attacked by heart rot* (decay of the heartwood) caused by fungi and, as a result, the aging wild apple trees provide hiding and breeding sites for hole nesting birds and small mammals. Its seedlings and young shoots (sarj) are preferentially consumed and damaged by wild boars. The cultivated apple, with increasingly higher number of varieties and larger area of cultivation, has continuously hybridised with the wild apple therefore the genome of their progenies have become mixed to varying degrees. In the near future it will no doubt have an important role to play in resistance-breeding since it is resistant to frost and a range of diseases including powdery mildew, apple mosaic virus and apple scab (Venturia inaequalis, varasodás). Because of its climate tolerance its significance as a stock in grafting and in propagation of cultivated apple varieties is likely to increase in the future. Ancient members of the apple genus first arose in Southeast Asia and some of its representatives were known to have existed since the end of the Cretaceous period in the tropical and subtropical mountains at the time. Its way of pollination might have been the same as that of the modern apple species. During the last glaciers, the Pleistocene, they underwent substantial species diversification and spread widely. According to recent molecular biological investigations it was at that time that the wild apple evolved from Malus orientalis (kaukázusi alma) and making its way through Eastern Europe and the Danube-Dnieper basin it has conquered nearly the whole of Europe.

(Based on the article by Dr. Bartha Dénes) heartwood: non-functioning part of the xylem of woody stems (geszt) resistance: the ability to fight off a disease After carefully reading the excerpt, answer the questions. 1. Choose the correct statement with regard to the reproduction of the wild apple.

A) The dioecious trees live far away from each other and therefore inbreeding can occur. B) Its bisexual flowers are not self-pollinating. C) At the time its ancestors evolved in the Cretaceous period there were no insects on

Earth. D) The activity of pollinating insects can be explained by temporal niche

partitioning. E) The sperm cells are transferred to the stigma by insects.

2. Write down the two forms of the vegetative reproduction and propagation of the wild apple mentioned in the text. (2 points)

…………………………………… , ……………………………………



Azonosító jel:

3. Name the type of population interactions the wild apple establishes with the following

living organisms. (3 points)

powdery mildew: ..................................................................................................................

mature moths: .......................................................................................................................

wild boar: .............................................................................................................................. 4. Why is the survival of the species threatened? Write the letters of the two correct answers

in the boxes. A) Because of its hybridisation with the cultivated apple varieties. B) Because of the inbreeding due to the low number of its population. C) Because of the climatic changes taking place. D) Because of the apple mosaic virus disease. E) Because of the heart rot caused by fungi attacking aging trees.

5. The heart rot of aging trees is caused by xylophagous (farontó) fungi. Which statements are true for these fungi? (2 points)

A) They are competitors of the wild apple. B) They gain carbon compounds by breaking down the cellulose content of the tree trunk. C) Their digestive enzymes act outside the fungal hyphae. D) The cellulose molecules of the heartwood are built into their cell wall. E) Their enzymes are capable of reducing the cellulose found in the phloem of the tree

trunk. 6. What was established through the molecular genetic studies?

A) The geographic distribution of the wild apple following the Pleistocene has genetic

causes. B) The wild apple is able to form hybrids with the cultivated apple varieties. C) The wild apple has an important role to play in the maintenance of forest biodiversity. D) The wild apple is related to the Malus orientalis.

1. 2. 3. 4. 5. 6. Total



Azonosító jel:

VI. The ‘scissors’ of genetic engineering 11 points Even bacteria are known to have natural enemies called bacteriophages (phages for short) which take advantage of the bacteria to spread. 1. Put the events of the phage infection in the correct order.

A) Relying on the enzymes, cellular organelles and basic materials of the bacterial cell, the phage makes multiple copies of its DNA and synthesises its coat proteins.

B) The bacteriophage gets attached to the cell wall of the bacterial cell. C) The bacterial cell is burst open or lysed with the help of enzymes encoded by the

phage particles, which allows the phage particles to spread out. D) The replicated phage DNA molecules and coat proteins are assembled into new phage

particles ready to infect other bacteria. E) The phage injects its DNA molecule into the bacterial cell.

2. Which one came first during evolution: the phage or the bacterium? Give your reason for your answer based on your knowledge about the origin of viruses.

...............................................................................................................................................

...............................................................................................................................................

The bacteria, similarly to other living things, can defend themselves against the intruders. They produce what is called restriction enzymes which are able to cut the phage DNA at particular sites called cutting sites characterised by specific base sequences. Cutting sites tend to be 4-6 base pair long and are repeated several times along the phage DNA. The bacterial cell attaches methyl groups to short stretches of the bacterial DNA with the same base sequence as the cutting sites to prevent the restriction enzyme from binding to them and cutting the bacterial DNA. With the help of this mechanism the bacterial cell can protect its own DNA. 3. The restriction enzymes prevent the multiplication of phage particles in the bacterial cell.

Which of the following answers support this finding? (2 points)

A) The phage is prevented from attaching to the surface of the bacterial cell. B) The bacterial cell kills itself to prevent the phage particle multiplying. C) The phage DNA encodes for the phage protein coat and once cut up, the DNA is no

longer able to serve as a template for synthesising mRNA. D) Once the phage DNA is cut up it is no longer able to replicate itself. E) The restriction enzymes inhibit the enzymes of the bacterial cell.



Azonosító jel:

The restriction enzymes ‘borrowed’ from the bacteria have become indispensable tools of gene technology since they allow the DNA to be cut at specific sites. The enzymes generating so called sticky ends at the cutting site are of particular importance. 4. Explain the use of sticky ends in gene

technology based on the diagram.

……………………………………………..

……………………………………………...

……………………………………………..

………………………………………………………………………………………………

The restriction enzymes also work according to the general principles of enzyme catalysis summarised briefly below. Fill in the gaps of the following explanation with the words listed below. One word can be used more than once. (6 points)

protein, nucleic acid, order of amino acids, base sequence,

active site, substrate, activating energy, reaction heat

The enzymes are (5) …………………. molecules, which are characterised by a specific

(6) ………………………, which in turn specifies the spatial structure of the enzyme. An

important part of this spatial structure is the (7) ………………………, which is a characteristic

groove on the surface of the enzyme. The (8) ………………….. of the enzyme catalysed

reaction fits accurately into the (9) ………………………... according to the key and lock

analogy. This specific binding enables the enzyme to catalyse a chemical reaction by lowering

the (10) ………………………….. of the reaction.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Total

isolated gene

digestion by the restriction enzyme

cutting site

plasmid vector



Azonosító jel:

VII. ‘Below in the valley the gardens still flower …’ 5 points

(Petőfi: End of September) A particular plant has three known phenotypes: white, pale yellow and yellow. The colour of the flower is controlled by one gene (A), which has two alleles: A1 and A2. The colour of the flower has intermediate inheritance. The plant has a wild type variety as well. 1. Write down the genotype of the existing phenotypes. (1 point)

white: .........................................................

pale yellow: ...............................................

yellow: .......................................................

2. The gardener would like to produce plants with only pale yellow flowers. What phenotypes he should cross to produce exclusively pale yellow progeny?

...............................................................................................................................................

There is a wild population of this plant species in a closed valley existing in genetic equilibrium. 68% of the plants have yellow flower. 3. What percentage of the population is white and pale yellow? Write down your calculations

to support your answer. (3 points)

1. 2. 3. Total



Azonosító jel:

VIII. The optic tract 11 points The diagram at the end of this task demonstrates the two eyes and the optic nerves / tracts originating from them. The dashed lines show the light rays entering the eyes. The diagram is greatly simplified because in addition to the lenses the light rays are also refracted on the two surfaces of another refracting medium before entering the vitreous humour. 1. Name this refracting medium!

...............................................................................................................................................

2. If the distance between the arrow and the eye changes the eye can adjust to the changes within certain limits (distance accommodation). Give the name of the muscle involved in focusing the light rays to produce a sharp image.

...............................................................................................................................................

3. If the arrow moves laterally to one side, the eye balls follow the movement. Name the layer of the eye the muscles moving the eyes are attached to.

...............................................................................................................................................

4. In a given moment the eye can only see sharply parts of the arrow from which the light rays fall on a particular area of the retina. What is the name of this area of the retina? Which cell type of the retina allow for sharp vision? (2 points)

...............................................................................................................................................

...............................................................................................................................................

...............................................................................................................................................

5. The diagram does not show the location of the blind spot. Explain where the name of this part of the eye comes from.

...............................................................................................................................................

6. The distance of the object viewed is judged by detecting the angle included between the visual axis of the two eyes. Finish the following sentence explaining the phenomenon.

The closer an object of a given size appears, the .................................................................

...............................................................................................................................................

7. Judging the distance of a given object is helped by looking at the surroundings of the object as well. Describe another way of judging the distance.

...............................................................................................................................................

...............................................................................................................................................

...............................................................................................................................................

...............................................................................................................................................



Azonosító jel:

8. It was a Japanese military physician who first observed that damages to the optic tract

causes varying parts of the visual field to be lost even if the eyes remain intact. The diagram shows the location of three possible damages marked x, y and z. Which damage causes the loss of the right visual field of both eyes.

A) Damage at location ‘x’. B) Damage at location ‘y’. C) Damage at location ‘z’. D) Damage at location ‘x’ or ‘z’. E) Damage at location ‘y’ or ‘z’.

9. The neurons of the optic tract synapse at part of the brain marked ‘A’. Give the name of

this part of the brain.

...............................................................................................................................................

10. The optic tract radiates to an area of the cerebral cortex not shown in the diagram. Within which lobe is this area located?

...............................................................................................................................................

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Total

A



Azonosító jel:

IX. A Optional task — Respiration 20 points Lung function test 10 points ‘— I don’t understand. What am I for you?(…) — A tiny piece of life. Do you know who we are? Two little funnels having the same air flowing through them’

(Jókai Anna) One of the most common methods to diagnose respiratory illnesses is the so called lung function test (spirometry) which is designed to test the lung capacities, how obstructed the airways are and the healthy state of the breathing mechanism. The patient is asked to breath into the plastic mouthpiece of the apparatus and the flow rate and other parameters of his breathing are measured by computer aided methods. The diagram below show the lung capacity graph recorded for a person before and during the lung function test. The asterisk (*) marks the beginning of the investigation, while the ‘1s’ indicates a period of 1 second. 1. Based on the diagram describe what the patient was supposed to do during the lung

function test.

...............................................................................................................................................

...............................................................................................................................................

During the investigation several physiological parameters typical of the breathing mechanism are recorded such as vital capacity (VC) or FEV1 values marked in the diagram.

2. Mark the vital capacity with a curly bracket ( }) and by using the VC abbreviation.

3. Explain the meaning of the FEV1 value based on the diagram.

...............................................................................................................................................

*

time

volu

me



Azonosító jel:

lung volume (dm3)

flow

rate

(dm

3 /s)

normal breathing

lung volume (dm3)

I.

II.

flow

rate

(dm

3 /s)

4. During spirometry other graphs describing the flow parameters of breathing are also

recorded. One such graph recorded during the lung function test described above is shown below. Is it possible to read the following data off the graph? If it is, then write their values in the rectangles and if it is not, then cross out the rectangle. (3 points)

a) Maximum flow rate during expiration:

b) Vital capacity:

c) Tidal volume:

With the help of the lung function test various diseases and abnormal conditions of the airways and lungs can be recognised. These ailments can be divided into two main groups:

• lung diseases (e.g. asthma) caused by the narrowing of airways (obstructive diseases); • lung diseases caused by the reduced ability of the lung to expand (restrictive diseases).

The two main groups of lung diseases produce characteristic and distinctively different changes in the pattern of their lung function graphs. The diagram shows the abnormal patterns in relation to the normal lung function graph. (There is a continuous transition between the two main types of diseased conditions and cases with mixed features can also occur. The tasks and the graphs refer to the characteristic cases.) Identify graph I. and II. as being typical for either obstructive or restrictive diseases and compare the two types of lung diseases in the following tasks.



Azonosító jel:

Compare the two groups of lung diseases. Write the letter of the correct answer in the boxes next to the statements.

A) Typical of diseases caused by the narrowing of airways. B) Typical of diseases caused by the reduced expandability of the lungs. C) Typical of both of them. D) Typical of neither of them.

5. Pneumothorax belongs to this group.

6. They involve (substantially) reduced rate of air flow during expiration.

7. They can be caused by the damaged or impeded functioning of breathing muscle.

8. The condition characterised by graph II. The asthma – essay 10 points Describe the relationship between the process of inflammation and the clinical picture of asthma. In your essay include the following points. 1. Describe the general symptoms of inflammation and explain how these symptoms are

induced. (6 points) 2. Discuss which parts of the airways are affected and how they change in asthmatic patients

and thereby what symptoms are caused. (2 points) 3. The short term handling of symptoms arising in an asthma attack includes the inhalation of

drugs (e.g. asthma sprays) affecting the autonomic nervous system. Explain how the working of the autonomic nervous system is to be changed in order to restore the lung functions quickly. Discuss the effect of these substances on the airways. (2 points)

You can write your essay on pages 22-23.

1. 2. 3. 4. 5. 6. 7. 8. Essay Total

healthy airway asthmatic state asthma attack



Azonosító jel:

IX. B Optional task — Food chain 20 points Two unicellular organisms 10 points Scientists investigated the coexistence of Escherichia coli, a heterotrophic bacterium and the eukaryotic unicellular Colpoda feeding on bacteria. In the apparatus (so called chemostat) a continuous glucose supply was ensured. In the first series of experiments (marked by triangles in the diagram) the bacteria were supplied with a liquid medium of low glucose concentration (0.025 mg/ml), while in the second series of experiments (marked by circles in the diagram) the bacteria were given a more concentrated sugar solution (0.5 mg/ml). At the beginning of the experiment (0-1 day) there were only E. coli bacteria present in both chemostats. Calpoda unicellular eukaryots were only added to both media on the 1st day. The graphs show the density (number of cells per millilitre) of Calpoda unicellular eukaryots and E. coli bacteria as well as the glucose concentration of the medium (mg/ml). In the first series of experiments the glucose concentration was kept too low to be plotted on the graph.

unicellular eukaryote Colpoda

E. coli bacteria

glucose mg/ml

Cell density 1/ml

time (days)



Azonosító jel:

After studying the graph carefully write the appropriate letter in the box next to the statements.

A) It is true for day 0-1 of the experiment. B) It is true for days 4-9 of the experiment. C) It is true for both periods of the experiment. D) It is not true for either period of the experiment.

1. There was competition taking place between the two unicellular species.

2. Growing in the presence of low glucose concentration, the bacteria used up almost all the glucose.

3. The density of the bacteria depended upon the amount of glucose supplied per unit time.

4. The density of the bacteria depended upon the unicellular eukaryote Colpoda.

5. The density of E. coli bacteria was higher than that of the unicellular eukaryote Colpoda

6. The number of unicellular eukaryote Colpoda depended on the amount of glucose supplied per unit time.

7. The more glucose they obtained per unit time, the higher the E. coli’s rate of reproduction (biological production) became.

8. Regardless of the bacterial density, the unicellular eukaryote Colpoda population consumed the same amount of bacteria whether supplied with low or high concentrations of glucose.

9. Each day more glucose was supplied than the unicellular organisms could consume in the same time period.

10. Almost all the glucose supplied per unit time was consumed by the unicellular eukaryote Colpoda.

Ecological pyramid – essay 10 points Include the following points in your essay: 1. Explain the difference between pyramids of number, pyramids of biomass and pyramids

of productivity. (3 points) 2. Identify the nutritional types of organisms found at the different levels of a terrestrial

ecological pyramid according to their energy sources. (2 points) 3. Explain the meaning of the gradually smaller sized rectangles in an ecological pyramid of

productivity as well as in a pyramid of number with regard to consumers. Give your reason why the same pattern is not seen in a pyramid including parasites. (5 points)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Essay Total



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Azonosító jel:

score maximum achieved I. Blood circuits 8 II. Nitrogen sources 12 III. About genetic divesity 8 IV. Insulin resistance 14 V. The wild apple 11 VI. The scissors of genetic engineering 11 VII. ’Below in the valley the gardens still flower …’ 5 VIII. The optic tract 11 Total score of the test: 80 IX. Optional essay and problem solving task 20

Total score of the written examination: 100

date marking teacher __________________________________________________________________________

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