Paper Section A Section B BIOLOGY - Trial Paper Collection · PDF fileAnswer all the questions in this section ... Discuss the extent to which the angiospermophytes and filicinophytes

For exam. use 1 Scale to

Name:

Section A : [40 marks]

SMJK Perempuan China Pulau Pinang

STPM Trial Examination 2009

BIOLOGY Paper 2

Time : 2 Yo hours

Answer all the questions in this section.

100% Paper 1 x2

Paper 2: Section A Section B

Ting. U6S_

1. Table 1.1 below shows the relative amounts of bases adenine, thymine, guanine and cytosine in DNA from different organisms.

Source Adenine Thymine Guanine Cytosine Bacterium 23.8 23.1 26.8

126

.3

Maize 26.8 27.2 22.8 1

23.2

Drosophila 30.7 29.5 19.6 20.2

Chicken 28.0 28.4 22.0 21.6

Human 29.3 30.0 20.7 20.0

Table 1. 1

a) Analyze the table given.

i) Based on the infor'T'ation from the table, what can you deduce about the ratio of the bases. (1)

ii) State the importance of your deduction to the structure of DNA. (2)

I

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Diagram 1 shows a protein synthesis process occurring in an eukaryotic cell.

x

:)l fl r J .~.

':1.:1 tt'l' in(' ac d ,"-Iel \ ·l ctl")i;)ni"'C'

y Diagram 1

b) On the diagram, label the following :

i) 5' and 3' end of DNA and mRNA. (1)

ii) Transcribing (sense) and non-transcribing (antisense) strands. (1)

c) Define processes X and Y (2)

d) Explain what happens in steps I, II and III. (3)

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2. Diagram 2 shows an outline of the main stages in the Calvin cycle.

RuBP

GP (PGA)

regeneration of

RuBP

Diagram 2

TP (PGAL)

a) State precisely where the Calvin cycle occurs. (1) ______________ _

b) Complete Diagram Z to show where the following occur: (5)

i) CO, fixation ii ATP is converted to ADP iii Reduced NADP is oxidized iv Glucose is synthesised

c) Enzyme X IS needed in the b) i) process above .

i) Identify enzyme X. (1) _________ __________________ _

ii) Explain the relationship between the low efficiency of photosynthesis in C, plants during low concentration of C02 with reference to enzyme X. (3)

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3. In an investigation into the flowering of a variety of tobacco, different light and dark treatments were carried out.

Diagram 3 shows six different light and dark treatments given to the said lant over several 24-hour eriods.

Peri ",i11O II r

8 Ie 10 20 24

A Non- Ilowerin g

B rlowering

<I L-___ _ !'i 011- t1 0 I>. eri n g.

R : red light FR : far red light

Diagram 3

Ii 4

D

E

f

Pcriod/hollr

12 10 ,0 24 .. '

Critical night lellb>th

G; ) Predict, for treatment D, E and F shown in Diagram 3, whether flowering would occur in the plant In each case, give a reason for your answer. (6)

Treatment Prediction Reason

D

E

F

~ . The graph~below shC'\'>' the effect of the concentration of two phytohormones (plant hormones), A and B on the growth of a species of plant.

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E E

60

50 -

.c 40

!30~ o I 1:: 20 .., ~ ,

I O~ o -t""'--;,--.I----,!-r·--r---i-..

10 4 10 :! 10 ' Hi ' 10' l C~

Phytohormone A concentration! ~, ~ Inrn '

b) Identify phytohormones A and B (1)

A: ------------------------ B:

60

50 E E 1! 40

Lo Lo ' ~J

.. ' - -+-- ..--o r-r I ,--..

o 0.2 0.4 0 .6 O.B 1.0 1.2 PhytOhormone B concentra Tloni Jig mm '

------------------------c) Explain how farmers can use phytohormones A and B to produce seedless fru its. (3)

4. Diagram 4 represents a female human cell with only the sex chromosome pair is shown in the cell.

, , \,' \, .'

, ' I

~X ; , Xiii

i ! ! 1 \ J ' . ...-

Diagram 4

a) State what is meant by non-disjunction, (1)

5

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b) The cell mentioned above undergoes meiosis. If non-disjunction of sex chromosomes occurs only during meiosis I, draw and label the end product of meiosis I. (1)

c) If non-disjunction of the sex chromosomes occurs only during meiosis II , draw two types of gametes that may form. (1)

d) With reference to b) , if the gamete with the lesser chromosomes is fertili zed by a normal X sperm, i) draw the result of the fertilization . (Tip : autosomes need not be drawn) (1) i) state the abnormali ties (syndrome) that will be inherited by the progeny produced. (1) ii) State ON E major characteristic that will be observed in the progeny. (1)

i) Result of fertilization:

! )

ii)) Abn ormality :

iii) Major Ch2racteristic :

I e) In another unrelated case, referring to c) If the gamete w ith the most chromosomes is fertilized by a normal Y sperm,

I) draw the resu lt of the fertilization. (Tip: autosomes need not be drawn) (1) ii) state the abnormalities (syndrome) that will be inherited by the progeny produced. (1)

6

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iii) State ONE major characteristic that wi ll be observed in the progeny. (1)

i) Result of fertiliza tion :

( )

I ii ) Abnormality :

iii) Major Characteristic:

I

e) Non-disjunction is sometimes induced in plants. Give one reason why th is is done. (1)

Section B: (60 marks)

Answer any four questions from this section.

5. a) 'Prokaryotic and eukaryotic cells are different with respect to s ize & structural organization'. Explain what is meant by this statement. (5)

b) Discuss the extent to which the angiospermophytes and filicinophytes are adapted to life on land . (1 0)

6. a) In mice, coat colour is determined by the action of genes at several different loci. A mutant allele at one coat-colour locus produces an unpigmented coat. The usual allele at this locus produces a pigmented coat. Pure-breeding pigmented mice were crossed with pure-breeding unpigmented mice. Their offspring were mated.

i) Draw a genetic diagram to explain these crosses up to F2, using A for normal pigmentation alle le and a for the unpigmented allele. (4)

In the F2 generation, the actua l numbers were 141 pigmented and 59 unpigmented mice .

ii) Run and detail a:\'.2 (chi-square) test to determine if these results supports your pred iction from a i). Use the following 'Ie probability table to help you . (6)

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nip .80 .70 .50 .30 .20 .10 .05 .02 .01

1 .064 .148 .455 1.074 1.642 2.706 3.841 5.142 6.635 , 2 .446 .713 1.386 2.408 3.219 4.605 5.991 7.824 9.210

3 \ 1005 1.424 2.366 3.665 4.642 6.251 7.815 9.837 11 .341

4 1.649 2·195 3.357 4.878 5.989 7.779 9.488 11 .668 13.277

b) Name three pharmaceutical products obta ined by using recombinant DNA technology. (3)

c) Why is it that patients using recombinant DNA pharmaceutical products would unlikely to suffer from allergic reactions? (2)

7. a) Using adrenaline as an example, explain how non-steroid hormone acts. (11)

b) Suggest how chemical compounds based on the following act to prevent pregnancy. (4)

i) oestrogen ii ) progesterone

8. a) Exp lain how resting potential and action potential is generated on an axon. (8)

b) State the mechanism present in the nerves that allow only unidirectional impulses and prevents fatigue. (7)

9. a) Explain briefly the following ecological terms:

i) Biotic potential (3) ii) Environmental resistance (3) iii) Carrying capacity (3)

b) Describe how the population size of a named animal in a named habitat can be estimated. (6)

10. Three types of natural selection can occur in any natural population. Using suitable examples, discuss all three. (15)

Prepared by-: ... :::l-- . LEE wAI YE(r

KElVA PANITIA BIOlOGI SMJK PEREMPUAN CHINA

PUlAU PINANG

Approved by f\ 0 TaYYOkeCheV -\

Guru Kanan Sains dan M" tt: :~::­!?MJK PEREMPUAN CHINA , p;r, .'-.. G

8

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SMJK Perempuan China Pulau Pi nang STPM Trial Exam 2009 : Biology Prepared by Madam Lee Wai Vee

Paper 2:

SUGGESTED ANSWER

I. a) i) amount of A consistent with amount ofT, while C consistent with G for all the organisms. (.1 m)

ii) The importance of the ratios A to T and G to C : -Adeni)le (a double-ringed structure) always linked by two hydrogen bonds to its complementary base, thymine in a ratio of 1: 1 -Gu 4nine is always linked by three hydrogen bonds to its complementary base cytosine in a ratio 1:1

(1 m)

this will give a stable structure for the double helix (as the width between the two (1m) polynucleotides is consistent)

b) Non.trj"scribing strand

5'~3'

5·~,~3·

mANA

aa :;: amino aCid Mel = Methionine

b) i) 1 m, ii) t m

B

~aa,~ eJlIJl6J

G T A lANA

c) X: Transcription is the synthesis of messenger RNA (mRNA) containing a hase sequence complementary to a certain section of DNA representing a cistron or gene. (1 m)

Y: Translation is a process in which the sequence of bases on the mRNA is translated on the ribosome to become a sequence of amino acids in the polypeptide. (1m)

d) I: activation of amino acids occurs/ tRNA with specific anticodon is attached to specific amino acid /specific aminoacyl-tRNA complex is formed (1m)

, II: specific amino acid-tRNA complex with specific anticodon attaches complementarily to codon 011 the A site of ribosome (1 m)

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, III: Elongation o\!polypeptide occurs through the condensation of amino acids at the tips of amino acid-tRNA complex on the A site and P site of the ribosome. (1 m)

2. a) stroma of chloroplast (1 m)

b) (Sm)

RuBP

'\ regenerntion ~OfRUBP

TP (PGAL)

---. Glucose

r ATP

(Dr

c) i) X: RuBP carboxylase (1 m) ii)

• oxygen act as a competitive inhibitor for C02 to attach to the active site of RuBP carboxylase

• if C02 concentration is low, the affinity of RuBp carboxylase towards C02 is reduced. • Photorespiration occurs, resulting in a lower yield in the plants. (3m)

3. a) Trealment Prediction Reason

The far· red light fl ash immediately after the red·light cancels the red·light effect. D Flowering (1m) Plant still experience critical night length needed. (1 m)

Red-light is fl ashed after the far-red light flash so it is not cancelled off. Dark E Non- flowering period is interrupted and shortened. (1 m)

(tm) ". When the plant is alternately exposed to red and far red light, the plant wi ll

F Flowering (1 m) respond to the last fl ash of light, whichever it is.

Plant still experience criti cal night length needed. (1m)

b) A: auxin B: gibberellin BOTH correct 1m (Note: DO NOT GIVE Y, m)

c) Parthellocarpy (1): When IAA (auxin)/ gibberellin is applied to the stigma ofaflower (1) it can cause fruit development Wit/lOlit f ertilisation producing seedless fruits. (1)

4. a) The failure of a pair or pairs of homologous chromosomes or sister chromatids to

separate (and to move to opposite poles of the cell) during cell division. (1 m)

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b)

c)

no x H lj x

I \ / \ I I I I U )

n = 22 + XX

"

d) (3m) i) Result of fertilization: n

xV U

44.XO

ii)Abnormality: Turner syndrome

iii) Major Sterile female (ovary Characteristic development is not

complete) .,

I)

n = 22 + 0

Both correct - 1 m

i)

V X I\ ! \

U n = 22 + 0 n = 22 + X

Any two: Both correct -1m

e) (3m) i) Result of

1fTI(h fertilization :

X ~\X 1'1 Y

~~ 44 +XXY

ii) Abnormality: Klinefelter syndrome

iii) Major Sterile male, development of Characteristic: feminine body characteristics

such as enlarged breasts.

To produce polyploidy plant species / to increase the number of chromosomes in the cells. Polyploidy plants are usually larger, tend to grow faster and are more robust than diploid plants of similar type. (either one: 1m)

Section B: 5. a) Size Prokayotic cell Eukaryotic cell BOTH correct

Range of size Small (0.5-10.0 J,lm) in Usually larger (10-100 J,lm) 1m diameter in diameter

Structural

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Nucleus No distinct nucleus; only A dinstinct membrane- 1m diffused area of bound nucleus nucleoplasm wi th no nuclear envelope

Chromosomes No chromosomes; has Present in nucleus :pircular strands of DNA in the nucleoid region

Ribosomes 70S 80S Membrane None Present Any 3 (3m) bound organ'elles Cell wall Made up of peptidoglycan Made up of polys ace aha ride

\ called cellulose (5)

b)Angiospermophytes and fi licinophytes -adapted to life on land th ro ugh thei r morphological structures & reproductioll methods.

A h t n210spermopilYI es : Morphological structu res : - specia lised cells for efficient water transport, i.e. xylem vessels '. Reproduction methods: - pollination using wind or insect to fertilise the ovum in the ovary, reduce the need forI - seed dispersal: animal, wind or water dispersal, splitting dependence on water mechanism

- seed survival: seed can remain dormant when the conditions on land are unfavourable for germination (eg. - end os perms can provide nutrients for the growing embryo

(Sm) F ilicinophytes

" -

Morphol02ical structures : - vessels for water transport - have rhizomes with adventit ious roots which help to absorb water quickly from the soil. - leaves are usually large fronds consisting of pinnules and rachis. Reduce water loss - pinnules reduce water loss through transpiration - young fern leavl;\s are often coiled to reduce exposed surface area Reproduction methods: - through spore production produced by sporophyte which is the reduce the need forI dominant phase. dependence on water - spores are easily dispersed through the wind.

- spores survival: spores can survive for long periods without requiring a lot of water.

(Sm)

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6. a)

Pure-breeding homozygous recessive unpigmented mouse ""-

aa

Puro-b'''d,n', (\

F1 PHEN OTYPES

all pigmented mice

Aa Aa : 4 ••••••••• • ",

homozygous dominant ® ~igmented mouse a ®

AA/@.Aa Aa .. ~~, ~ "'- @ I-"A""""a~~ A'~ " " A'a" ':

Cross within F1 mice and produce r gametes :

@ AA

® Aa

3 pigmented: 1 unpigmente~

F2 PHENOTYPES

AA Aa

Aa aa

Genetic fork diagram with correct format also accepted. (4m)

b) To test if the F2 result of 141pigmented and 59 unpigmented mice supports the 3: 1 ratio predicted from a), chi-square value is calculated using the following formula:

(0 - E)2 • Chi-square ('1'2 ) = L --------

E

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• I = the sum of • 0 = observed results

• E = ex~ ected results Phcnolypic Predicled Observed (0) Expecled (E) (0- E) (0 - E)- (0 - E)z cUlegory ratio E

Pigmented 3 141 150 -9 81 0.54 mice

Unpigmented I 59 50 9 81 1.62 mice

" Chi-square value, )' 2.16 Tab le: 2m

Correct chi-square value (2.16) : 1m - The number of degree of freedom, n is one less than the number of phenotypic ratio, therefore n =1 (I m)

-from the chi-square table given, chi-square value of2.16 when n=l, corresponds to a probability between 0.10 to 0.20 (I m)

- by convention, a probability of > 0.05 is not rejected, the deviation is not significant therefore the results supports the 3: I prediction in ai). (I m)

b) Insulin, human growth hormone (HGH), Tissue Plasminogen Activator (TPA), erythropoietin, interferon , (1 - I-antitrypsin (A AT) ANY 3 (3m)

c) - product of tn; nscription & translation from the real human gene (I m) - compound identical to naturally-produced secretions (1m)

7. a)

• Adrenaline: peptide / non-steroid hormone Not soluble in lipid & cannot diffuse through plasma membrane. Acts as 1st messenger& binds to a specific receptor protein in the plasma membrane of liver cells. The hormone-receptor complex formed increases the affinity of receptor to bind with and activate G protein. Once G protein is activated, it moves to stimulate the enzyme adenylyl cyclase (located on the plasma membrane) This enzyme would then convert ATP into cAMP (cyclic adenosine monophosphate) cAMP acts as a secondary messenger (which then activates other enzyme) the cascade effect occurs to produce a large amount of cAMP from a small a mount of adrena lin e. cAM P activates inactive protein kinase A into active protein kinase A via the cascade effect the active protein kinase A then activates phosphorylase kinase via the cascade effect .

Phosphorylase kinase then activates glycogen phosphorylase also via the cascade effect.

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Glycogen phosphorylase catalyses the phosphorylation of glycogen into glucose-I-phosphate.

The glucose-I-phosphate that is formed goes through a series of enzyme catalysed ,'eactions to form glucose. Glucose produced in liver cells are released into the blood.

(11m) b) Chemical compounds based on oestrogen or progesterone can be used in oral contraceptive pills to prevent pregnancy.

i) Oestrogen: High level of this hormone inhibit the production of gonadotrophic hormone namely FSH, from the pituitary gland. (l m) Development of follicles in the ovary will be inhibited. (1 m)

i ii) Pregesterone : High level of this hormone inhibit the production of gonadotrophic hormone narr.dy LH from the pituitary gland. (1 m) The absence of LH will prevent ovulation from occurring (1m)

8. a) Res ting potential refers to the difference in charge (electrical potential difference) between the inside & the outside of the cell membrane. The resting potential difference across the neurone membrane is generated by the following events: maintain ed by :

• T he sodium & potassium voltage gated channels are both closed. (l m) • While tire sodium-potassium pump (Na+/K+ ATPase) is always working. For every 2 K+

pump into the cell, 3 Na+ will be actively transported out of the cell. (1 m) • T he axon membrane is more permeable to K+ than Na+ due to the presence of more K+

non- gated, voltage dependent channels than Na+ non-gated channels. More K+ can diffu se ('leak') out of the cell faster than Na+ can diffuse back in. (I m)

• The nett result is : the outside of the membrane is positive compared to th e inside of the axon. The .cell membrane of the neurone is said to be polarised. (I m)

Formation of action potential • When a stimulus ( electric current) reaches a resting neurone: 1. some sodinm voltage gated channels in the stimulated region of the axon membrane open.

( I m) -> Na+ will move into the axon (via facilitated diffusion) down an electrochemical gradient. • When the potential difference across the membrane reaches a threshold value (-40 to -50

mY), many more sodium voltage gated channels open. (example of +ve feedback mech.) (l m)

• The sodium-potassium pump is inactive (l m) • The rapid diffnsion of Na+ leads to a sudden increase in the cell's potential difference

which becomes positive (+40 mY). (l m) • This reversal in the potential difference is known as depolarisation and an action

potential is generated. b) After an actio,; potential for about 5 to 10 ms that portion of axon undergoes a refractory period.

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• Purpose o(re(ractorv period: Allows impulses to move only in one direction & limits the frequency at which successive impulses can pass along an axon.

• During the absolute refractory period (which lasts about 1 ms) (I m) • the axon membrane is unable to respond to another stimulus, however strong it is. (1 m) • An action potential cannot be produced due to a conformational change in voltage-gated

sodium channels which are still in a closed, inactive state. This also prevents the action potential from moving backwards. (1 m)

• Following the absolute refractory period, there is a relative refractory period (which lasts around 5 ms). (1 m)

• During this period, the resting potential is gradually restored by the Na+f K+ pump and the relative permeability of membrane to facilitated diffnsion of ions is also restored fA new action potential can then be produced if the stimnlus is greater than the usual one. O~ .

• The refractory period is also a limiting factor in the speed of conduction of the nerve impulse. (lm)

• Under the All or nothing law, an action potential can only be generated after the threshold value is exceeded. (1m)

• After the threshold is reached, the size of the action potential produced remain s constant and is independent of the intensity of the stimulus. This is the all or nothing response. All action potentials are of the same amplitude. (1 m)

(any 7)

9.a) i) Biotic Potential:

• Maximum rate of increase per individual under ideal (optimal) conditions, that is when population density is low & resources are plentiful. 0 m)

• The actual capacity of a natural population will only reach biotic potential if all the individuals in tbe population survive and reproduce at the maximum rate. (l m)

• The biotic potential is opposed by environmental resistance o OR

In nature, bioJic potential is reached if the environmental resistance is absent. (1 m)

ii) Environmental resistance • refers to the biotic and abiotic factors that oppose tbe achievement of biotic potential by

a population. • Can be in the form of: Availahility of food & space, competition, predation and

parasitism • Limits the size of a popnlation

iii) Carrying capacity • the maximum stable population size a particular environment can support. (I m) • The area occupied by a population has limited resources and this limits the population

growth by maintaining an equilibrium between the natality rate and the mortality rate. • Population size increases until it reaches saturation within the carrying capacity of its

ecosystem.

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• Carrying capacity is a dynamic value (not constant) which changes in response to environm~ntal changes (any 2)

b) Named animal must be appropriate. (reject sessile animals, ego garden snails, caterpillar etc or those not restricted geographically) (I m) for Caplllre-mark-release-recaptlire method: (I m)

• Individuals are trapped in an area and captured, marked with a tag, recorded, and then released. (I m)

• After a period of time has elapsed, traps are set again, and individuals are captured and identified. (I m)

The.population size is estimated using the following expression: \

Population size = Number marked x Number caught during second trapping Number of marked individuals recaptured

" (1m) •• This estimate of population relies on the assumptions that:

10.

• it is only applicable to a population whose movement is restricted geographically. • organisms mix randomly within the population. • marking does not hinder the movement of the organisms or make them conspicuous to

predators. • sufficient time must elapse between capture and recapture to allow random mixing. • changes in population size as a result of immigration, emigration, births and death are

negligible.

(any I)

• In natural,5election: 1. Stab ilising selection favours intermediate phenotypic forms. 2. Directional selection favours one extreme phenotype of the population. 3. Disruptive selection favours the selection of the two extreme forms & the gradual

elimination of the intermediate forms. (name all three correctly: 1 m )

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Stabilisin g selection

• Intermediate forms are favored and ~ I I , extremes are eliminated M ·,6

• These indh-idu aIs have greater survival 'Eiii W'M'W -- .., & reproductive success. ,,"-

IWWW I " 8. .11 . • Extreme forms are less adaptive and 5;;; WWWWW are eliminated.

z£

• Eg. Newborn Birthweight: R:lIn g l o f nlul C tlr th. hl h t1m. I

'btl 1. < 2.27 kg (5 pounds)& > 4.54 kg (10 MI pounds) ,

(4m) WW'W W'M'W

\ R4r1g. or ", . Iu l c tlr th, h i tat t me 2

M M MI

WMW R:lng e o ' .... ILI.' tlr tl1 ' hl h t t ml ' (1m) ,

Directional selection

• Allele frequencies shift in one direction.

I~ v. • As environmental conditions change, ~ 6 " .-individua ls of a population with one of '5 fti n_ 'M''M'W the extreme phenotypes is at an :;:8.

0 • IM'M'W advantage & is selected. . ;;;

• Directional selection pressure increases j .s

WMMWW the chances of the advantageous alleles Ri nge o f 'I.IUI I for til' trait at lime I

to be passed on to the next generation. i • Selection pressure decreases the I'MW chances of disadvantageous alleles being transmitted. M'M'VV

• Eg. Peppered moths in industrial M'M'VV melan ism Range Of "V. lul . tor ti l' nit at 11m Ii 2

-, (4m) VV WW

'M'VW 'M'WV

Range Of 'Ialue . fOr tr lt n it ,i tim, 3 (I m)

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,-~~---.---~-.------------------,------------------------------- -Disruptive selection • Forms at both ends of the range of

variation are favored • Intermediate forms are selected against • After many generations, disruptive

selection may cause the formation of two separate gene pools and the fo r mation of new species.

• Eg: I. The soils in mines are often

' contaminated with high [ I of toxic metals such as lead, copper or zinc.

2. Certain modified polygenes in the grasses confer variation in phenotype such as tolerance to toxic metals.

3. These grasses can grow from surrounding uncontaminated soil but their ability to grow on uncontaminated soil decreases.

4. Toxic metals in the soil act as a natural selection agent and confer selective advantage on 'lnetal-tolerant grasses growing on contaminated soil

5. After several generations, more toxic metal-tolerant grasses are found growing on contaminated soil, and more non­tolerant grasses growing on uncontaminated soil (3m)

~no & ot walut 1. tlr the hi tat 1m. I

Rong 0 of value 1. br tilt hli ht 1m. a ( I m)

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