BRITISH BIOLOGY OLYMPIAD 2013

ROUND TWO QUESTION PAPER

Time allowed 1 1/2 hours

Instructions:

Answer all questions on the answer sheet.

Write your names, school and centre number on the top of the answer sheet.

Answers written in margins or on the question paper will not be marked.

Do all rough work on the question paper.

Use black ink or black ball-point pen.

You may use a calculator.

When you have completed the test and checked your answers, return all test materials to the invigilator.

Information:

The paper consists of 53 questions scoring 89 marks.

The marks for questions are shown in brackets.

Answers consist of one or more letters or numerals, ticks and crosses or plus and minus signs or the results of numerical calculations.

Good luck.

BBO Round 2 - 2013

Q1 (3 marks)Urine production is the result of continuous filtration of plasma through the kidneys. Indicate in the table on the answer sheet, true statements about the mammalian kidney with a tick and false statements with a cross.

a. The kidneys have a direct effect on blood pressure. b. The kidneys help regulate total blood volume in circulation.c. The loops of Henle remove water, ions and nutrients from the blood.d. Those able to excrete the most hyperosmotic urine, such as the kangaroo rats living in

the desert, have relatively short loops of Henlee. The kidneys partner the lungs in controlling the pH in plasma.f. The kidneys help maintain blood pH by excreting hydrogen ions and reabsorbing

bicarbonate ions as needed.g. The kidneys dispose of volatile acids produced in metabolism.h. Ammonia (NH3) is produced in proximal tubule cells during acidosis.i. The glomerular filtration rate is affected by blood pressure.j. The kidneys produce ADH (antidiuretic hormone).

Q2 (2 marks)The European (freshwater) eel usually obtains oxygen by gills but can spend long periods of time out of water using dermal gaseous exchange. The graph below shows the level of blood saturation by oxygen and oxygen supply through different organs when the eel was removed from the water. (arbitrary units)

:A, B, C, D

Match the following statements (I to IV) to the corresponding lines (A D) in the table on the answer sheet.

I. Total blood saturation by oxygenII. Oxygen supply through gillsIII. Oxygen supply through skinIV. Oxygen supply from air bladder

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BBO Round 2 - 2013

Q3 (4 marks)Fish are specially adapted for aquatic life in different depths (e.g. surface, middle, bottom) of the water column and various special habitats (e.g. sea grass beds, rock crevices). Their swimming speeds are also partly dependent on their body morphology. Match, in the table on the answer sheet, the fish (A H, not drawn to scale) with their respective habitats and indicate the two fastest swimmers and the two slowest swimmers.

Q4 (2 marks)Tom ran after a thief and caught him after an 80m chase. Which of the following biochemical/ physiological changes did his body undergo during the chase? Indicate the correct answers with a tick and incorrect answers with a cross.

a. The cardio-vascular system leapt into action, with the heart pump rate going from 4/5 litres up to around 20 litres a minute, arteries constricted to maximize pressure around the system whilst the veins opened out to ease return of blood to the heart.

b. Fat from fatty cells and glucose from the liver were metabolized to create instant energy.c. Blood vessels to the kidney and digestive system constricted, to effectively shut down

systems that were not essential for the moment. d. Blood vessels to the skin constricted reducing any potential blood loss. Sweat glands

opened, providing an external cooling liquid to the over-worked system. e. Endorphins, body's natural pain killers, were released.

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BBO Round 2 - 2013

Q5 (2 marks)Mark the following statements related to the grey matter and white matter in the brain as correct (tick) or incorrect (cross)

a) Grey Matter is grey in colour because of the nuclei present in nerve cell bodies, whereas myelin is responsible for the white appearance of the white matter.

b) Grey matter occupies 60% of the brain while white matter fills the rest.c) White matter controls functions that the body is unaware of such as temperature, blood

pressure and heart rated) The senses of the body (speech, hearing, feelings, seeing and memory) and control of the

muscles, are part of the grey matters functione) Grey matter consumes 94% of the total oxygen used by the brain in humans

Q6 (1 mark)Carbon particles were injected into the blood of mice. In which one of the following places would you most likely find the particles?

A) In the glomerular filtrateB) In the adipose tissue of the liver C) In monocytes of the spleenD) In plasma cells of the bone marrowE) In the loops of Henle

Q7 (1 mark)Professor Jones is running an experiment with his graduate students in his physiology lab. One group of students drinks a litre of water, another group drinks a litre of coffee, and a third group drinks a litre of concentrated salt solution. The volume of urine produced by all individuals in the three groups is measured over a period of several hours. At the end of the monitoring period, which group will have produced the greatest volume of urine and which group the least?

A) Those who drank coffee will produce the most urine, while those who drank water will produce the least

B) Those who drank coffee will produce the most urine, while those who drank the salt solution will produce the least

C) Those who drank the salt solution will produce the most urine, while those who drank water will produce the least

D) Those who drank the salt solution will produce the most urine, while those who drank coffee will produce the least

E) There will be no difference between the three groups

Q8 (1 mark)Multiple sclerosis is an autoimmune disorder in which the bodys immune system attacks and destroys the myelin of its own nervous system. Which one implication does this damage have for the nervous system?

A) The cell bodies of nerve cells can no longer reach action potential because the receptors that take up sodium have been damaged.

B) Degraded myelin molecules block receptor proteins in the postsynaptic membrane.C) Lack of myelin decreases production of acetylcholine disrupting muscular coordination.D) Axons conduct nervous impulses less effectively because their insulating sheaths have been

damaged.E) Because the corpus callosum is made of myelin, the right and left sides of the brain are no

longer able to communicate.

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BBO Round 2 - 2013

Q9 (1 mark)The human thyroid gland is controlled by a negative feedback mechanism. The hypothalamus secretes thyrotropin-releasing hormone (TRH). TRH stimulates the anterior pituitary to secrete thyroid stimulating hormone (TSH). TSH induces the thyroid gland to secrete thyroxine. What is the next step in the control system?

A) Thyroxine will inhibit the secretion of TRH.B) Thyroxine will cause the bodys basal metabolic rate and body temperature to drop.C) The hypothalamus will secrete a thyroid-inhibiting hormone that slows down production ofD) thyroxine.E) Thyroxine will stimulate the increased production of TSH.F) The thyroid gland will respond to the rising level of thyroxine and slow down its production.

Q10 (2 marks) In the figure, the letter in each box represents an organ or tissue.

Match each organ or tissue (a e) on the answer sheet to the correct box in the figure.

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BBO Round 2 - 2013

Q11 (3 marks)Match plant structures (1 10) with the corresponding function (A J) in the table on the answer sheet.

Plant cell / Tissue structure Function(s) / Feature(s)

1 Thylakoid membranes A An intercellular communication network

2 Vascular cambium BStorage of water, digestive enzymes and other

inorganic and organic substances

3 Central vacuole C Production of new plant tissues/organs

4 Plasmodesmata D Modified parenchyma cell without nucleus

5 Apical meristem ESmall opening in the surface of an ovule,

through which the pollen tube penetrates.

6 Periderm F Mechanical support

7 Sieve tube G Presence of electron transport proteins

8 Trichome H Production of secondary vascular tissues

9 Secondary cell wall I Secondary protective tissue

10 Micropyle J Protection and absorption

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Q12 (2 marks)Study the graph below and determine which of the statements (a to g) are correct. Indicate correct answers with a tick and incorrect ones with a cross.

a) It is a photosynthetic O2 response curve.

b) Point A is light saturation point.

c) Point B is light compensation point.

d) C is the maximal photosynthetic rate.

e) Plants stop growth when they grow under the irradiance greater than the value shown at point B.

f) Respiration rate is greater than photosynthetic rate when plants are grown under the light

below the value shown at point A.

g) Plants grow (accumulate biomass) when their growth light environments are higher than the

photon flux shown at point A.

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BBO Round 2 - 2013

Q13 (1 mark)Give the letter(s) of the factor(s) shown below which regulate(s) the output of the light-independent Calvin cycle in plants.

A) Concentration of CO2 in the leafB) Concentration of O2 in the leafC) Amount of photorespirationD) Temperature

Q14 (2 marks)Lettuce seeds were treated with various concentrations of coumarin and then placed in a range of environmental conditions to germinate. The graph below shows the results of these investigations.

From the results shown, which of the following statements are correct (tick) and which are not (cross)?

A) Coumarin is an inhibitor of seed germination B) Light tends to reverse the effects of coumarin C) Germination is independent of light and temperature D) At higher temperatures light is less effective in overcoming the effects of coumarin

Q15 (2 marks)The AGAMOUS (AG) gene is involved in flower development. Plant mutants without a functional AG would produce flowers with only receptacle, sepals and petals. A scientist generated a transgenic plant harbouring a green fluorescence protein (GFP) gene driven by the AG promoter in a wild type background that produces normal flowers. Give the letter(s) of the following flower parts in which you are likely to observe strong GFP fluorescent signals?

A) Receptacle B) SepalC) PetalD) StamenE) Carpel

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Q16 (1 mark)Phytochromes exist in two isoforms, Pr and Pfr. In darkness, they are synthesized as Pr form, then turned into Pfr form after absorbing red light (most effective at 666 nm). When irradiated with far red light, Pfr transforms back to Pr. According to the description above, give the letter(s) of the following which is/are likely to be the absorption spectrum/a of phytochrome?

Q17 (1 mark)Give the letter(s) of the following which is/are of benefit to the plant of C 4 photosynthesis as compared to C3 photosynthesis.

A) C4 photosynthesis needs fewer light quanta to fix one mole of CO2.B) C4 photosynthesis can proceed at a much lower CO2 concentration than C3 photosynthesis.C) Plants with C4 photosynthesis are more economic in water use.D) Plants with C4 photosynthesis require fewer types of minerals.

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(B)

(A)

(C) (D)

(E)

BBO Round 2 - 2013

Q18 (4 marks)Bateson and Punnett (1908) studied the flower colour and pollen grain shape in the sweet pea, Lathyrus odoratus, which is related to the garden pea, Pisum sativum, which Mendel studied.They crossed a true-breeding purple-flowered plant that had long pollen grains with a true-breeding red-flowered plant that had round pollen grains, and tabulated the following results for the F2 progeny:

Phenotype ObservedPurple flowers, long pollen grains 296Purple flowers, round pollen grains 19Red flowers, long pollen grains 27Red flowers, round pollen grains 85Total number of progeny 427

a ) (1 mark)If the genetic traits are assorted independently, what phenotype ratio would you expect to see?

b) (2 marks)On the answer sheet fill in the expected values for the respective phenotype and test for independent assortment by calculating the chi squared value.

df Chi squared values1 3.8412 5.9913 7.8154 9.4885 11.070

Table: chi squared values for p = 0.05

c) (1 mark)Indicate likely explanation(s) with a tick and inappropriate explanation(s) with a cross for the above observation.

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Q19 (3 marks)Some allele combinations can result in a particular mental disorder in humans. The Table shows the enzyme activities of different genotypes (reported as percentage of the normal activity).

Allele 2Allele 1

R231X P292L R407W IVS-12 E290K R158Q R271Q Y424C

R231X

BBO Round 2 - 2013

Q21 (3 marks)A number of nutritional mutant strains were isolated from wild-type red bread mould Neurospora crassa that responded to the addition of certain supplements in the culture medium by growth (+) or no growth (0). Given in the Table below are the responses for single-gene mutants.

Strain

Supplements added to minimal culture medium

Citrulline (1) Glutamic semialdehyde (2) Arginine (3) Ornithine (4) Glutamic acid (5)

A + 0 + 0 0B + + + + 0C + 0 + + 0D 0 0 + 0 0

a) (2 marks)Indicate the order of the five metabolites within the metabolic pathway by giving the numbers.

b) (1 mark)Indicate the strain that is blocked at each of the four steps in the metabolic pathway by giving the letters.

Q22 (2 marks)For a particular mammal a gene for hair colour is expressed as follows. A1A1 = lethal, A1A2 = gray, A2A2 = black; a second gene for hair length is expressed as: B1B1 = long hair, B1B2 = short hair, B2B2 = very short hair (fuzzy). If parents that are both A1A2B1B2 are crossed:

a) What is the fraction of adult offspring that is expected to be gray and fuzzy?

b) In the case when fuzzy is also a lethal trait, what is the fraction of adult progeny expected to be black and short?

Q23 (1 mark)What is the most frequent origin of Down's syndrome?

A) An excess of DNA ligase, causing abnormal DNA replication.B) A tandem duplication of DNA during meiotic cell division.C) A point mutation in the gene encoding a histone.D) A frameshift mutation during foetal development.E) Nondisjunction of parental chromosomes during meiosis.

Q24 (1 mark)A population of laboratory rats in Hardy-Weinberg equilibrium displays a polymorphism for coat colour controlled by two alleles at a single locus. The B allele codes for black coat colour and is completely dominant to the allele b for white coat colour. There are 200 rats in the population: 32 are white and 168 are black. How many of the black rats are heterozygous (Bb) at the coat-colour locus?

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Q25 (2 marks)A cell in the G1 phase has two homologous pairs of chromosomes. It undergoes a mitotic division, followed by meiosis. At the end of meiosis 2, what is the sum of all the products of these divisions in all cells? Write the appropriate numbers in the table on the answer sheet.

Q26 (1 mark)Which statement about evolution is correct?

A) Analogous structures may be used to infer evolutionary relationships between organisms.B) The existence of vestigial structures, which are structures found in organisms but not used,

is consistent with evolutionary theory.C) As it is a theory, evolution has not yet been subject to a vast amount of experimental

verification.D) The development of complex structures (e.g. the eye) by accident is consistent with

evolution by natural selection.E) Natural selection can provide a population with new source of alleles.

Q27 (1 mark)You have a summer job in a lab breeding fruit flies. You are given your first mating pair of flies, both of which have grey bodies and normal wings. You are asked to start a population of flies that share these characteristics with their parents. But upon mating the two flies, you end up with a large amount of variation, as described in the table below. Give the letter(s) of the statement(s) which could explain the outcome of your cross.

Traits Number of offspringGrey body, normal wings 45

Black body, shrivelled wings 5Grey body, shrivelled wings 15

Black body, normal wings 15

A) The genes for body colour and wing shape assort independently.B) Both parents are heterozygous for both genesC) One parent is heterozygous for both genes and the other is heterozygous for the wing shape

geneD) You have achieved a rare resultE) Each parent is heterozygous for one gene

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BBO Round 2 - 2013

Q28 (1 mark)Which is the one correct statement concerning the geographic distribution of animals or plants in a population?

A) most often uniform, providing each individual with the maximum availability of space and resources.

B) most often clumped, with individuals gathered around resources such as food or mates.C) determined by the amount of daylight as compared to hours of darkness.D) random in most species.E) spatially uniform only when competition for food resources is low.

Q29 (1 mark)The cryptic shape, cryptic colouration, and cryptic behaviour of a leaf-eating insect are most likely to be evidence of which one kind of biological relationship?

A) MutualismB) Interspecific competitionC) CommensalismD) PredationE) Intraspecific competition

Q30 (2 marks)A large forest is cleared. The land is rapidly colonized by species with which of the following characteristics? Write the number(s) of the characteristic(s) that apply.

1. long lifespan2. rapid reproduction3. fast growth4. strong dispersal ability5. strong defence against natural enemies or predators.

Q31 (2 marks)A large proportion of angiosperms are pollinated by animals. Assign the following flower descriptions (I to V) to the most likely pollinator (a to e).

I. Flower white, open during night, intense fragrance, nectar hidden in long, tight tubes.II. Flower often with ultraviolet colouring pattern, open during daytime, pleasant fragrance.III. Flower large and coarse, bright red, open during daytime, no fragrance but large amounts of

nectarIV. Flower large and coarse, far opened, open during night, intense fragrance, large amounts of

nectarV. Flower reddish brown, no nectar, smell of rotten flesh

a. batsb. birdsc. beesd. fliese. moths

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BBO Round 2 - 2013

Q32 (1 mark)Hay is boiled in water and cooled. Some pond water, containing only heterotrophic protozoa, is added to it and kept in the dark for a long time. Write on the answer sheet the letter or letters of the following which is/are correct.

A) Heterotrophic succession of protozoa will occur with increase in total biomass.B) The energy of the system is maximum at the beginning.C) Succession will occur, eventually reaching a steady state in which energy flow is

maintained.D) The ecosystem may undergo succession but finally all organisms will die or go into resting

stages.

Q33 (1 mark) An ecologist is comparing the growth of a herbaceous plant species growing in two different sites A and B. To compare the populations from the two sites, she has harvested 30 individuals from each site, then measured the root length, root biomass, and shoot biomass of each individual. A summary of those measurements are as follows:

Location Mean root length (cm) Mean root biomass (g) Mean shoot biomass (g)

Site A 27.2 + 0.2 348.7 + 0.5 680.7 + 0.1Site B 13.4 + 0.3 322.4 + 0.6 708.9 + 0.2

Based on the data presented, which one of the following statements is likely to be true?

A) Soil water availability is lower in Site B than in Site A.B) Plant productivity is higher in Site A than in Site B.C) Soil water availability is lower in Site A than in Site B.D) Soil nutrient availability is lower in Site B than in Site A.

Q34 (3 marks) Nitrogen, as a mineral nutrient, has the greatest effect on plant growth. The atmosphere contains nearly 80% nitrogen gas (N2), yet plants have to be provided ammonium salts or nitrates as fertilizers for optimum growth and yield. Certain nitrogen-fixing bacteria (rhizobia, cyanobacteria, etc.) can convert atmospheric N2 into ammonia using nitrogenase by the following reaction:

N2 + 8 e- + 8 H+ + 16 ATP 2 NH3 + H2 + 16 ADP + 16 Pi

Such bacteria can be used as biofertilizers in agriculture. In soil, ammonia is protonated to ammonium (NH4+). This, in turn, is converted to nitrate (NO3--) and then to N2 gas by the action of nitrifying and denitrifying bacteria, respectively. Plants require nitrogen mainly in the form of nitrate, which is exported from roots to shoots, reconverted to ammonium and assimilated as amino acids.

Cont.

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BBO Round 2 - 2013

Q34 cont.For each of the following three questions write the one letter that applies.

(a) Plants do not themselves fix N2, because:

A) It is easily available from the soil.B) They lack the nitrogenase enzyme complex.C) The process has a very high requirement of ATP per mole of N2 fixed.D) Hydrogen evolved in the process is deleterious to plants.

(b) Processes related to nitrogen conversion to different chemical forms in the soil, carried out by the nitrogen-fixing bacteria, nitrifying bacteria and denitrifying bacteria can be, respectively, described as:

A) reduction, oxidation and oxidation.B) reduction, oxidation and reduction.C) reduction, reduction and oxidation.D) oxidation, oxidation and reduction.

(c) Based on the given information, which type of soil bacteria will NOT be beneficial for plants?

A) Nitrogen-fixing bacteriaB) Nitrifying bacteriaC) Denitrifying bacteriaD) Nitrifying and denitrifying bacteria

Q35 (1mark)Vervet monkeys (Cercopithecus aethiops) warn fellow monkeys by producing unique warning signals according to the type of predators such as eagles, leopards and snakes. Depending on the type of signals, monkeys in the group choose the appropriate method to escape. A newborn Vervet monkey is capable of producing all of these signals but it does not know which signal should be used in each case. If a baby monkey produces the signal for eagles when a sparrow is flying over, adult monkeys look up at the sky and then ignore the signal. However, if an eagle is indeed hovering, the entire group joins in the warning. Sometimes, baby monkeys are punished by their mothers for producing wrong signals.

Which of the following learning types is/are associated with the warning signal development in baby monkeys?

A) Imprinting

B) Associative learning

C) Problem solving

D) Social learning

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Q36 (2 marks)Fruit flies usually find food by following the odour of ethanol produced from fruits. The fruit also serves as the place for male and female flies to mate and reproduce. The graph below shows the relationship between the number of allured flies and the concentration of ethanol.

Based on the graph, indicate the correct statement(s) with a tick and incorrect statement(s) with a cross.

A) The male/female ratio in the number of flies occupying the food source varies depending on the ethanol concentration.

B) The number of mating animals would be the lowest when ethanol concentration is 9.C) The competition between males would be most severe when ethanol concentration is 7.D) The number of laid eggs would be the highest when ethanol concentration is around 6

and 7.E) The number of attracted flies would be the highest when ethanol concentration is 6.

Q37 (1 mark)When an ant in a colony dies, the live ants will throw the dead ant out of the anthill. If a live ant from the colony, Ant X, is sprayed with a chemical characteristic of dead ants, the live ants will repeatedly throw Ant X out of the anthill until the chemical on Ant X wears off.

What is the best behavioral explanation of the ant colony?

A) The ants are exhibiting a negative taxis triggered by the chemical.B) The other ants can learn only through trial and error. C) The ants exhibit a learned behaviour. D) The live ants continue the behaviour until they have been habituated.E) The chemical acts as a sign stimulus for a fixed action pattern.

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Q38 (2 marks)Male fiddler crabs use their enlarged claws chelipeds (major chelipeds) for signalling (e.g. fighting for burrows, waving at females, etc.). A student studied male-female interactions by using mirrors to reflect two different-sized images of the same waving male crab to females. Mirror combinations used in the experiment were: 10x : 3x (Treatment I), 3x : 1x (Treatment II) and 10x : 1x (Treatment III). Ten waving males were presented to 20 females in three trials for each treatment. She recorded the percentage of females (Graph A) and time taken by each female to approach each reflection (Graph B) for each treatment as well as whether the male was right or left-handed (Graph C).

Larger image; Smaller image; Right-handed male; Left-handed male

Indicate correct conclusion(s) that can be drawn about the interactions between male and female crabs with a tick, incorrect conclusion(s) with a cross and the statement(s) that cannot be concluded with a dash ().

a. Female fiddler crabs generally prefer larger males.b. In mate-choice selection, male handedness is an important criterion.c. Males that wave faster generally attracted more females.d. The mean time taken for females to make a choice differed between Treatments II and III.e. An obvious difference in cheliped size of males may be necessary before females become

more decisive.

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Q39 (1 mark)Which of the following is NOT a characteristic of all mammalian species?

A) Dorsal nerve cordB) Endothermic metabolismC) LungsD) Mammary glandsE) Placenta

Q40 (2 marks) The schematic diagram below represents group-in-group relationships. The T3 taxon, represented by the largest circle, includes three T2 taxa. Each of these three T2 taxa has one T1 taxon, represented by circles filled with dots; the dots represent individuals.

According to above scheme, assign the correct taxa from the options given below to each of the circles. Fill in your answers by writing the appropriate number in the table. Marks will only be awarded if the entire table is correctly filled.Options:

i. Annelidaii. Lepidoptera

iii. Polychaeta iv. Mollusca v. Orthoptera

vi. Insectavii. Arthropoda

viii. Crustacea

ix. Gastropodax. Arachnida

xi. Lumbricus (earthworm) xii. Hirudo (leech)

xiii. Gryllus (cricket) xiv. Unio (freshwater mussel) xv. Euscorpias (scorpion)

xvi. Daphnia (water flea)

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Q41 (2 marks)The Table below shows the genetic codes of amino acids.

Some viruses (e.g. tobacco mosaic virus (TMV)) have RNA sequences that contain a "leaky" stop codon. In TMV 95% of the time the host ribosome will terminate the synthesis of the polypeptide at this codon but the rest of the time it continues past it.

The following sequences show part of a mRNA from TMV. Indicate the sequence(s) that may result in two polypeptides in the indicated frame with a tick and those that will not with a cross.

A. 5-AUG-UCU-UGU-CUU-UUC-ACC-CGG-GGG-UAG-UAU-UAC-CAU-GAU-GGU-UAA-3B. 5-AUG-ACC-CGG-GGG-UUU-CUU-UUC-UAG-UAU-GAU-CAU-GAA-GGU-UGU-UAA-3C. 5-AUG-CUU-UUC-UCU-UAU-UAG-CAU-GAU-GGU-UGU-ACC-CGG-GGG-CCC-UAA-3D. 5-AUG-CAU-GUU-CUU-UUC-UCU-UAU-UGU-GGU-UGU-ACC-CGG-GGG-UUC-UAA-3E. 5-AUG-CAU-GAU-GGU-UGU-ACC-CGG-GGG-UAG-CUU-UUC-UCU-UAU-UGC-UAA-3F. 5-AUG-UCU-UAU-UGG-CAU-GAU-GGU-UGU-CUU-UUC-ACC-CGG-GGG-AAA-UAA-3

Q42 (1 mark)Give the number(s) of the following which is/are true about endosymbiosis.

1. Both plastid and lysosome are products of endosymbiosis2. Eukaryotic cells could engulf other eukaryotic cells to establish a symbiotic relationship3. Cyanobacteria are ancestors of plastids and mitochondria4. Cyanobacteria lost their chlorophyll b gene in endosymbiosis.5. Flagella of some eukaryotic cells are derived from cyanobacteria

Q43 (1 mark)Suppose the lac repressor of E. coli is mutated so that it never binds to the operator. Which one of the following is true?

A) Glucose digesting enzymes are never producedB) Lactose digesting enzymes are never producedC) Lactose digesting enzymes are always producedD) The result depends on the concentration of glucoseE) The result depends on the concentration of lactose

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Q44 (2 marks)There are various mechanisms by which a cell can commit suicide a phenomenon known as apoptosis. One of the mechanisms is triggered by reactive oxygen species. The outer membrane of the mitochondrion normally expresses a protein Bcl-2 on its surface. Another protein Apaf-1 binds Bcl-2. Reactive oxygen species cause Bcl-2 to release Apaf-1 and a third protein Bax to penetrate the mitochondrial membrane, releasing cytochrome c. The released cytochrome c forms a complex with Apaf-1 and caspase 9. This complex sequentially activates many proteases that digest cellular proteins.

Four possible conditions for cells are listed below. I. The cell has expressed a mutant form of Apaf-1 that always binds Bcl-2.

II. The cell does not express Bcl-2 at all.III. The cell over expresses a form of Bcl-2 that is targeted to the cell membrane only.IV. A chemical which extends the half life of Bcl-2 is added to the cell.

Below are listed three possible fates of these cells after they are exposed to reactive oxygen species. a) The cell resists apoptosis.b) The cell is forced towards apoptosis.c) The fate of the cell cannot be predicted.

On the answer sheet, match the fates a,b &c with the conditions I to IV

Q45 (1 mark)Which one of the following statements describes the function of a molecular chaperone?

A) Transports rRNA from the nucleus to the cytoplasm.B) Acts as an energy source during the polymerization of amino acids into a polypeptide.C) Acts as a carrier molecule and bring "activated" monomers to a polymer for incorporation.D) Binds to specific structures on the polypeptide in order to assist the folding of a protein into

its correct three-dimensional shape.E) Unfolds proteins with the incorrect three-dimensional shape and refolds them into the proper

shape.

Q46 (1 mark)An anticodon nucleotide sequence of five successive tRNAs involved in protein synthesis was analysed, yielding the following content:

A G C T U40% 27% 13% 0% 20%

What was the corresponding anti-sense strand of original DNA?

A G C T UA 20% 13% 27% 40% 0%B 40% 27% 13% 20% 0%C 60% 27% 13% 0% 0%D 20% 13% 27% 0% 40%E 40% 27% 13% 0% 20%

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Q47 (1 mark)Which one of the following graphs shows the relative change in the amount of mitochondrial DNA of a cell undergoing mitosis?

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Q48 (1 mark)Mammalian plasma membranes are characterized by the presence of different types of phospholipids (SM, PC, PE, PS and PI). The graph below shows the percentage distribution of each phospholipid across the plasma membrane of human erythrocytes.

Indicate the correct statement(s) with a tick and incorrect statement(s) with a cross. The numbers indicated are approximate figures.

A) It can be concluded that membranes, in general, are asymmetric.B) 24% of the total membrane phospholipids contain SM and 4% contain PI.C) 80% of the inner total membrane phospholipids contain PE and 16% contain PC.D) Most PC is confined to the outer surface of the erythrocytes while most of the PE and PS are

confined to the inner surface of the erythrocytes.

Q49 (1 mark)You extract RNA from liver cells and then carry out an agarose gel electrophoresis of the liver RNA. The RNA fragments are then transferred to an RNA-binding membrane (nitrocellulose or nylon) using capillary action. Next, you hybridize a probe for gene X to the RNA on the membrane.

Which one of the following statements regarding your experiment is true?

A. You are trying to determine how many copies of Gene X are in liver cellsB. You are trying to determine if Gene X is expressed in liver cellsC. You are trying to determine the chromosomal location of Gene XD. You are trying to determine whether Gene X has a mutant sequence

Q50 (1 mark)Give the letter(s) of the statement(s) about both lysosomes and peroxisomes which is/are correct.

A) they contain digestive enzymes.B) they are vesicle-like in structure.C) they are formed by the Golgi apparatus.D) they are bound by a single phospholipid bilayer membrane.E) they are able to (internally) break down macromolecules internally.

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BBO Round 2 - 2013

Q51 (1 mark) Which one of the following statements describes the effect of ageing (senescence) on telomerase activity?

A) Ageing cells gradually lose their ability to edit introns from transcribed mRNA sequences, leading to failures in protein synthesis.

B) Chromosomes gradually decrease in length because normal DNA synthesis cannot complete replication at the end of the lagging strand.

C) The cells ability to repair mistakes made during DNA replication decreases, allowing the number of base substitutions in the genome to increase.

D) DNA replication in the 3' to 5' direction is unaffected, but replication in the 5' to 3' direction slows down, decreasing the rate at which dying cells can be replaced.

E) DNA can no longer completely uncoil, making replication and gene expression less efficient.

Q52 (2 marks)A gene regulatory protein X controls cell proliferation. Protein X is found in the cytosol and has no typical nuclear localisation signal (NLS). When cells are treated with a specific growth hormone, protein X re-localizes from the cytoplasm into the nucleus where it activates the transcription factors involved in cell proliferation. Recently, a protein (Y) that interacts with protein X has been identified in unstimulated cells. To investigate the function of protein Y, a mutant lacking the gene encoding protein Y was generated. Fractionation of cells from the wild type and mutant produced membrane (M), cytoplasmic (C), and nuclear (N) fractions for each cell type. Proteins extracted from each fraction were separated by SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis) and analysed by Western blotting for the presence of proteins X and Y.

On the basis of the results shown above, which of the following statement(s) concerning protein Y is/are plausible?

A. In the absence of growth hormone, protein Y associates with protein X, and the X/Y complex is subjected to a degradation pathway.

B. In the presence of growth hormone, protein Y interacts with protein X, and the complex remains in the cytoplasm.

C. Protein X interacts with protein Y in the absence of growth hormone. Upon growth hormone treatment, protein X is released from protein Y and re-localises to the nucleus.

D. Protein Y is a membrane-associated protein and re-localises with protein X to the nucleus upon the growth hormone treatment.

E. Protein Y is one of the nuclear import proteins and the growth hormone does not induce protein Y to translocate protein X to the nucleus.

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BBO Round 2 - 2013

Q53 (3 marks)Neil is dissecting a signal transduction pathway (illustrated in the following figure) that leads to oncogenesis in cancer cells, in the hope that he can find inhibitors to block the signaling pathway and use them as chemotherapy drugs for cancer treatment.

In the following three questions indicate correct answers with a tick and incorrect ones with a cross.

a) Components of signal transduction, including A, B and C, usually are activated through phosphorylation or dephosphorylation reactions. What are the mechanisms by which proteins A , B and C are phosphorylated or dephosphorylated?

A) Receptors may contain enzyme domains which can catalyse phosphorylation/dephosphorylation reactions.

B) Enzymes that participate in phosphorylation/dephosphorylation reactions may exist in the cytoplasm.

C) Proteins A, B and C may contain enzyme domains which can catalyse phosphorylation/dephosphorylation reactions.

D) Phosphorylation or dephosphorylation may not be mediated through enzymatic reactions. E) A phosphate group is transferred from the receptor to protein A. F) The phosphate group can only be provided by H3PO4.

b) Which of the following experiments can prove that the signal transduction pathway is BC, but not CB?

A) Adding an A antagonist will activate B.B) Adding an A agonist will activate C.C) Adding a B agonist will activate C.D) Adding a B antagonist will activate C.E) Increasing the expression level of B will generate more of the active C molecules.F) Cell response can be seen when B antagonist and active C molecules are added into the cell.

c) If this is a highly activated signal transduction pathway in cancer cells, in which of the following processes may the signaling pathway be involved?

A) Inhibiting cell divisionB) Inhibiting cell differentiationC) Hypomethylation of some tumor suppressor genesD) Activating the transcription of an oncogeneE) Arresting the cell cycle at S phaseF) Inhibiting the expression of some DNA repair gene

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BRITISH BIOLOGY OLYMPIAD 2013ROUND TWO QUESTION PAPERQ32 (1 mark)