BIOLÓGIA ANGOL NYELVEN - oktatas.hu...Biológia angol nyelven — emelt szint Azonosító jel: I. Ferns and Cycads 9 points The picture on the left shows a maidenhair spleenwort (aranyos

Biológia angol nyelven emelt szint — írásbeli vizsga 1211

Azonosító jel:

BIOLÓGIA ANGOL NYELVEN

EMELT SZINTŰ

ÍRÁSBELI VIZSGA

2012. május 15. 8:00

Az írásbeli vizsga időtartama: 240 perc

Pótlapok száma Tisztázati Piszkozati

NEMZETI ERŐFORRÁS MINISZTÉRIUM

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írásbeli vizsga 1211 2 / 20 2012. május 15.

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. 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.

A D A D C B D correct acceptable wrong



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. 3. 4. 5. 6. 7. 8. 9. total



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.

1. 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?

………………………………………………………………………………………………

j f m á m j j a s o n d

növényi plankton tömege

állatiplankton tömege

hőmérséklet növényi tápanyagtartalom

Figure 1 mass of phytoplankton

mass of zooplankton

temperature concentration of plant nutrients

relative value



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)

………………………………………………………………………………………………

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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: ………………………………………………………………………...

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……………………………………………………………………………………………….

1. 2. 3. 4. 5. 6. total

Figure 2



U C A G

U UUU phe UUC phe UUA leu UUG leu

UCU ser UCC ser UCA ser UCG ser

UAU tyr UAC tyr

UAA STOP UAG STOP

UGU cys UGC cys

UGA STOP UGG trp

U C A G

C

CUU leu CUC leu CUA leu CUG leu

CCU pro CCC pro CCA pro AAG pro

CAU his CAC his CAA gln CAG gln

CGU arg CGC arg CGA arg CGG arg

U C A G

A

AUU ile AUC ile AUA ile

AUG met

ACU thr ACC thr ACA thr ACG thr

AAU asn AAC asn AAA lys AAG lys

AGU ser AGC ser AGA arg AGG arg

U C A G

G

GUU val GUC val GUA val GUG val

GCU ala GCC ala GCA ala GCG ala

GAU asp GAC asp GAA glu GAG glu

GGU gly GGC gly GGA gly GGG gly

U C A G

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.

………………………………………………………………………………………………

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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.



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. 3. 4. 5. 6. 7. total



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 B) epithelium C) epidermis D) conducting tissue E) connective tissue

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. 3. 4. 5. 6. 7. total



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

1. giant amoeba 2. Chlamydomonas (single-celled green algae) 3. baker’s yeast 4. 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. 3. 4. 5. 6. 7. total

Eukaryotes Heterotrophs

Chemotrophs

Animals with pseudotissue



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 0.24 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 6.6 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.



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. 3. 4. 5. 6. 7. total



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 and IB i B) aa and IB i C) Aa and IB i D) Aa and IB IB E) AA and IB 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? (2 points)

from his father: ……………………….. 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. 3. 4. 5. total



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 phosphatides myosin water

carotene urea fibrinogen chlorophyll* insuline 1. a) Peptide. ………………………………………………… 2. b) Non-peptide…………………………………………….. 5. 2. a) It occurs in the blood plasma.. ………………………… 3. b) It is active within the cell. ………………………………. 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:

5. a) Its biological role is attributed to its peculiar polarity features..........................................................................

6.

b) Its biological role can be explained by its system of conjugated double bonds …………………………….

7.

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. ……………..

E:

b) When functioning, the hydrophobic parts surround a apolar liquid droplet, while its hydrophilic parts point towards the water. ………………………………………

F:

7. a) Its molecule also contains Mg2+ ion …………………… G:

b) It contains no Mg2+. ………………………………. H:

*chlorophyll = the molecule which is responsible for absorbing light in photosynthesis.

A B C D E F G H total



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!

1. What is the name of the part indicated by letter ‘E’?

…………………………………………………

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).

D

gall bladder

kidney

liver

A

B

E

C

Y = true for all

X = true for none



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. 3. 4. 5. 6. 7. total



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’, Rh-negative blood.

3. A mother with such blood group can have Rh incompatibility developing in the course of her first pregnancy.

4. A mother with such blood group can have Rh incompatibility developing in the course of her second pregnancy.

5. 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.

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. 3. 4. 5. 6. 7. Esszé total



X. B Nature Conservation and Genetics 20 points Extinction threshold 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)

……………………………………………………………………………………………….

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6. Explain why the decrease in genetic diversity causes the adaptability of the population to decline. (2 points)

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7. Name one factor which increases the genetic diversity of the population while the average biological fitness of the individual is not decreasing.

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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:

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Maximum

score Achieved

score I. Ferns and Cycads 9 II. Life in the Mediterranean Sea 9 III. Breaking the Genetic Code 11 IV. Microscopic Investigation 9 V. Levels of Organisation and Nutritional Types 7 VI. Enforced Diet 8 VII. Pedigree Analysis 9 VIII. Molecule Identification 8 IX. Vessels and Organs 10

Total score of the test: 80 X. Optional essay or problem solving task 20

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: ………………………………….. Dátum/Date …………………………………..

Documents

BIOLÓGIA ANGOL NYELVEN - oktatas.hu...Biológia angol nyelven — emelt szint Azonosító jel: I. Ferns and Cycads 9 points The picture on the left shows a maidenhair spleenwort (aranyos