Biology Specimen Booklet

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Cambridge Pre-U Specimen Papersand Mark Schemes

Cambridge International Level 3Pre-U Certificate in

BIOLOGY

For use from 2011 onwards

w w w . X t r e m e P a p e r s . c o m





www.cie.org.uk/cambridgepreu 1

Cambridge Pre-U Specimen Papers and Mark Schemes

Specimen Materials

Biology (9790)Cambridge International Level 3Pre-U Certificate in Biology (Principal)

For use from 2011 onwards

QAN 500/380/72

www.cie.org.uk/cambridgepreu 1



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Cambridge Pre-U Specimen Papers and Mark Schemes

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Syllabus Updates

This booklet of specimen materials is for use from 2011. It is intended for use with the version of thesyllabus that will be examined in 2013, 2014 and 2015. The purpose of these materials is to provideCentres with a reasonable idea of the general shape and character of the planned question papers inadvance of the first operational examination.

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Copyright © University of Cambridge Local Examinations Syndicate 2011



This document consists of 31 printed pages and 1 blank page.

© UCIE 2011 [Turn over

For Examiner's Use

Section A

21

22

23

24

25

26

Total

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONSCambridge International Level 3 Pre-U CertificatePrincipal Subject

BIOLOGY 9790/01

Paper 1 Structured Questions For Examination from 2013

SPECIMEN PAPER

2 hours 30 minutesCandidates answer on the Question Paper.

No Additional Materials are required.

READ THESE INSTRUCTIONS FIRST

Write your name, Centre number and candidate number on all the work you hand in.Write in dark blue or black pen.You may use a pencil for any diagrams, graphs or rough working.Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all questions.

Section ATwenty questions for which you must choose what you consider to be the rightanswer and record your choice in the appropriate space provided. Marks will notbe deducted for any wrong answers. Write your answers in the spaces providedon the Question Paper.

Section BWrite your answers in the spaces provided on the question paper.

The number of marks is given in brackets [ ] at the end of each question.



2

© UCIE 2011 9790/01/SP/13

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Use

Section A

1 The resolving power of a microscope depends on the wavelength used by the system.

Table 1.1 shows the wavelengths and resolving powers of three types of microscope.

Table 1.1

type of microscope wavelength / µ m resolving power / µ m

light microscope 0.8 0.4

ultra-violet microscope 0.2 0.1

electron microscope 0.005 0.0025

Table 1.2 gives details of four biological structures which are investigated using

microscopes. The ticks ( ) and crosses ( ) indicate whether or not each structure can beclearly seen with each microscope.

Table 1.2

chloroplast,length 5 µ m

Escherichiacoli bacterium,

length 2 µ m

ribosome,diameter

25 nm

thickness of plasma

membrane10 nm

light microscope

ultra-violetmicroscope

electron microscope

Which row correctly completes Table 1.2 to show which structures can be clearly seenwith an ultra-violet microscope?

chloroplast,length 5 µ m

Escherichia coli bacterium,

length 2 µ m

ribosome,diameter

25 nm

thickness of plasma

membrane10 nm

A

B

C

D

answer [1]





4

© UCIE 2011 9790/01/SP/13

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4 An action potential arrives at the synaptic knob increasing the permeability of themembranes to …… 1……, which diffuse in and cause vesicles to move to the pre-synapticmembrane and fuse with it.

……2…… occurs and the …… 3…… moves across the synaptic cleft by …… 4…… andattaches to receptors on the post-synaptic membrane, causing …… 5…… channels to

open and a post-synaptic potential to be generated.

Which words correctly complete the numbered gaps?

1 2 3 4 5

A acetylcholine endocytosis transmitter substance

activetransport calcium ion

B calcium ions exocytosis acetylcholine diffusion sodium ion

C calcium ions exocytosistransmitter substance

activetransport sodium ion

D sodium ions endocytosis acetylcholine diffusion calcium ion

answer [1]

5 The diagram shows the major components of the lac operon.

3

2

1

DNA

Which statement is correct?

A 1 is a ribosome, 2 is a t-RNA molecule and 3 is a phosphorylated amino acid, theactivator, so lactose-digesting enzyme can be made.

B 1 is mRNA polymerase, 2 is β-galactosidase, the inducer and 3 is the repressor, solactose-digesting enzyme cannot be made.

C 1 is mRNA polymerase, 2 is the repressor and 3 is lactose, the inducer, so lactose-digesting enzyme can be made.

D 1 is the repressor, 2 is a β-galactosidase molecule and 3 is lactose, the promoter, solactose-digesting enzyme can be made.

answer [1]



5

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6 Small samples from crime scenes can be genetically profiled (DNA finger printed).

Which are necessary parts of a successful genetic profiling process?

crime scene sample PCR ethidium bromideand X-rays

A red blood cells

B saliva

C semen

D skin cells

Key

= used= not used

answer [1]

7 A symbiont may be defined as a species in which individuals live in a long-term, intimateand beneficial relationship with hosts of a different species. As the name suggests,endosymbionts live within their hosts.

Which statement provides evidence that mitochondria and chloroplasts areendosymbionts?

A Proteins encoded by the nucleus are exported to these organelles.

B Their inner membrane has different structure from other intracellular membranes.

C They are surrounded by double membrane.

D They contain their own ribosomes.

answer [1]



6

© UCIE 2011 9790/01/SP/13

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8 Approximately half of the total protein in a pea seed consists of the storage protein vicilin.

• Each molecule of vicilin is made up of three identical polypeptides.

• Each polypeptide is made up of two β-pleatedsheet regions with linking α-helix regions, folded

into the shape shown to the right.

• This allows the three polypeptides to packtogether into a compact, flat storage molecule,as shown below.

Which row correctly describes the structure of vicilin?

primary structure secondarystructure tertiary structure quaternary

structure

Aamino acid

sequence of onepolypeptide

α-helix andβ-pleated sheetregions of each

polypeptide

association of threepolypeptides

folding of eachpolypeptide

Bamino acid

sequence of onepolypeptide


polypeptide


association of threepolypeptides

C association of threepolypeptides

amino acidsequence of one

polypeptide


polypeptide


D association of threepolypeptides

amino acidsequence of one

polypeptide



polypeptide

answer [1]

β-pleated sheet regions

α -helix regions



7

© UCIE 2011 9790/01/SP/13 [Turn over

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9 The graphs represent the frequency of alleles in species X, Y and Z during and after selection.

allelefrequency

allelefrequency

X Y Z

range of characteristic(s) regulated by alleles

range of characteristic(s) regulated by alleles

characteristics selected for

characteristics selectedagainst

duringselection

after selection

key

In which species does evolution take place?

A X only

B Y only

C Y and Z

D none of X, Y nor Z

answer [1]



8

© UCIE 2011 9790/01/SP/13

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Use

10 Curve X shows the oxygen dissociation curve for human haemoglobin. Under certainconditions this curve becomes displaced to the right. This is termed the Bohr effect and isshown by curve Y.

100

80

60

40

20

00 2 4 6 8 10 12 14

partial pressure of oxygen / kPa

percentagesaturation of haemoglobin

Y

X

Which change is responsible for the Bohr effect?

A a decrease in the partial pressure of oxygen

B a decrease in the temperature of the blood

C an increase in pH of the blood

D an increase in the partial pressure of carbon dioxide

answer [1]



9

© UCIE 2011 9790/01/SP/13 [Turn over

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11 Which statements correctly describe the structure and function of prokaryote ribosomes?

1 Prokaryote ribosomes are smaller than eukaryote ribosomes and sediment at70 S.

2 A prokaryote ribosome consists of two subunits, one of 50 S and one of 30 S.

3 In prokaryotes, ribosomes translate mRNA in the same cellular compartment in whichit is transcribed.

4 In prokaryotes, ribosomes can begin translating mRNA before its synthesis has beencompleted.

5 A prokaryote ribosome can accommodate only one amino acyl-tRNA at a time.

A 1, 2, 3, 4 and 5

B 1, 2, 3 and 4 only

C 1, 3 and 5 only

D 2, 4 and 5 only

answer [1]



10

© UCIE 2011 9790/01/SP/13

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12 A snake venom causes death by leading to paralysis of muscles. It exerts its effect atsynapses.

The statements below were put forward by scientists as possible explanations for theeffects of this venom.

1 It interferes with the binding of neurotransmitter vesicles to the membranes.

2 It binds with neurotransmitter receptor sites.

3 It blocks calcium and sodium channels.

4 It destroys the myelin sheath of the neurone.

5 It binds with neurotransmitter.

Which statements should be investigated further?

A 1, 2, 3 and 5 only

B 2, 4 and 5 only

C 4 only

D 1, 2, 3, 4 and 5

answer [1]

13 What happens during the light-dependent reactions of photosynthesis?

1 ADP is hydrolysed.

2 ADP is phosphorylated.

3 ATP is hydrolysed.

4 ATP is phosphorylated.

5 NADP is oxidised.6 NADP is reduced.

A 1 and 5 only

B 2 and 6 only

C 1, 4 and 5 only

D 2, 3 and 6 only

answer [1]



11

© UCIE 2011 9790/01/SP/13 [Turn over

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14 One of the many recessive mutations of the CFTR gene changes one amino acid in theregion of the CFTR protein that binds ATP. The graph shows the effect of differentconcentrations of ATP on normal and mutant CFTR proteins.

100

80

60

40

20

0

percentageof functioningion channels

normal CFTR

mutant CFTR

concentration of ATP

Which correctly describes individuals who are homozygous for this mutation?

1 Their CFTR protein cannot bind ATP and cannot act as an ion channel.

2 Their CFTR protein binds ATP less readily than normal CFTR protein.

3 They produce CFTR protein that must bind ATP to function as an ion channel.

4 They produce a mixture of normal and mutant CFTR protein, both of which can act asan ion channel.

A 1 only B 2 only C 2 and 3 only D 2 and 4 only

answer [1]



12

© UCIE 2011 9790/01/SP/13

For Examiner's

Use

15 Many plants are not fertilised by pollen from their own flowers. This is known asself-incompatibility. In any individual species a single gene, the S gene, is responsible andit may have many different alleles.

If a pollen grain has an S allele which matches an allele in the genotype of the stigma thenthe pollen grain fails to germinate or the pollen tube fails to grow through the style.

The genotype of the stigma of a flower is S 3S 4.

Which pollen grains would germinate?

S 1

1

S 1S 3

2

S 3

3

S 2S 4

4

pollen grain

A 1 only B 2 and 4 only C 3 only D 3 and 4 only

answer [1]

16 Molecules can be transported in several ways.

1 cohesion/tension

2 diffusion

3 mass flow

4 osmosis

Which row shows the correct method of transport?

answer [1]

into capillaries in phloem out of stomata in xylem

A 1 2 3 4

B 2 1 4 3

C 3 4 1 2

D 4 3 2 1



13

© UCIE 2011 9790/01/SP/13 [Turn over

BLANK PAGE



14

© UCIE 2011 9790/01/SP/13

Questions 17 to 20

The graphs on the opposite page show features of the survival and reproductive success of adultmales and females of four different species:

A red deer, Cervus elaphus

B Bewick’s swan, Cygnus columbianus

C dwarf mongoose, Helogale parvula

D southern elephant seal, Mirounga leonina

Column 1 shows the age-specific survival which is the probability that adult animals of different ageswill survive for a further year.

Column 2 shows survivorship curves. A survivorship curve shows the proportion of a population thatsurvives to different ages.

Column 3 shows the mean annual reproductive success which is the number of offspring producedby adult males and females of different ages.

Study the graphs and then for each question identify the most appropriate graph that matches thestatements in questions 17 to 20. Record your answer to each question by using a letter for theappropriate species and a number for the appropriate graph, e.g. A1, B2, C3, etc.

17 A species with a higher proportion of females than males in the oldest age groups.

answer [1]

18 A species with a mortality rate above 60% among juveniles.

answer [1]

19 A species in which both males and females have the same effective breeding period.

answer [1]

20 A polygynous species.

answer [1]



15

© UCIE 2011 9790/01/SP/13 [Turn over

1.00

0.75

0.50

0.25

00 5 10 15

age / years a g e s p e c i f i c

s u r v i v a l

1.00

0.75

0.50

0.25

00 5 10 15

age / years

a g e s p e c i f i c s u r v i v a l

1.00

0.75

0.50

0.25

00 5 10 15

age / years a g e s p e c i f i c s u r v i v a l

1.00

0.750.50

0.25

00 5 10 15

age / years a g e s p e c i f i c s u r v i v a l

0.80

0.40

00 5 10 15

age / years p r o p o r t i o n o

f o r i g i n a l

p o p u l a

t i o n s u r v i v i n g

p r o p o r t i o n o

f o r i g i n a l

p o p u l a t i o n s u r v i v i n g

p r o p o r t i o n o f o r i g i n a l

p o p u l a t i o n s u r v i v i n g

p r o p o r t i o n o f o r i g i n a l

p o p u l a

t i o n s u r v i v i n g

0.50

0.25

00 6 12 18

age / years

0.40

0.20

00 5 10 15

age / years

0.40

0.20

00 5 10 15

age / years

2.4

1.8

1.2

0.6

0

1.2

0.9

0.6

0.3

0

18

12

6

0

543210

0 6 12 18age / years

m e a n n u m

b e r o

f

o f f s p r i n g p e r

i n d i v i d u a l

m e a n n u m

b e r o f


i n d i v i d u a

l

m e a n n u m

b e r o f


i n d i v i d u a

l

m e a n n u m b e r o

f

o f f s p r i n g p e r i n

d i v i d u a l

0 5 10 15

age / years

0 2 4 6 8 10age / years

0 5 10 15age / years

A red deer

species1

age-specific survival2

survivorship3

mean annualreproductive success

B Bewick’sswan

C dwarf mongoose

D southern

elephant seal*

key=females=males



16

© UCIE 2011 9790/01/SP/13

For Examiner's

Use

Section B

Answer all the questions.

21 ATP is a nucleotide that performs many essential roles in prokaryotic and eukaryotic cells. It is considered to be the major ‘energy currency’ of cells.

Fig. 21.1 shows the structure of ATP.

O–

O–

P O

O

O–

P O

O

O–

P OO

OH OH

N

N

N

N

H2N

O

Fig. 21.1

(a) Describe the features of a molecule of ATP that make it suitable for its role.

[3]



17

© UCIE 2011 9790/01/SP/13 [Turn over

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Use

(b) (i) Explain why ATP is said to be an ‘energy currency’.

[1]

(ii) Describe an example of ATP acting as an ‘energy currency’.

[2]

(c) The enzyme ribonucleotide reductase (RNR) is needed for DNA synthesis. Theenzyme catalyses the reaction in which adenosine diphosphate is converted todeoxyadenosine diphosphate (dADP).

ADP

reduced NADP NADP

dADP + H 2ORNR

(i) State how adenosine diphosphate differs from deoxyadenosine diphosphate.

[1]

(ii) Suggest how dADP is used in the synthesis of DNA.

[2]



18

© UCIE 2011 9790/01/SP/13

For Examiner's

Use

(d) Excess adenosine is deaminated to deoxyinosinol in a reaction catalysed by theenzyme, adenosine deaminase (ADA), which consists of one polypeptide.

Fig. 21.2 shows a ribbon model of ADA.

Fig. 21.2

Describe the structure of the enzyme, ADA, as shown in Fig. 1.2. You may addlabels to the diagram to help your answer if you wish.

[3]



19

© UCIE 2011 9790/01/SP/13 [Turn over

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A deficiency of ADA is a cause of severe combined immunodeficiency syndrome(SCID).

Children with non-functional adenosine deaminase are at risk of infections as a toxicproduct builds up inside T lymphocytes (T cells) and kills these cells.

(e) Outline the roles of T lymphocytes in the immune system.

[3]

(f) Gene therapy has been used to treat SCID.

Explain the problems encountered in using gene therapy as a treatment for geneticdiseases, such as SCID.

[4]

[Total: 19]



20

© UCIE 2011 9790/01/SP/13

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Use

22 Skin cancer cells may be grown in culture and examined using the technique of immunofluorescence in which antibodies are used to attach fluorescent dyes to specificmolecules within the cells.

Fig. 22.1 is an immunofluorescent light micrograph of skin cancer cells. The DNA in thelarge cell nuclei is stained blue. These nuclei are typical of cells undergoing division incancer. Proteins in the cytoplasm are stained green.

There are two cells in the process of dividing. Each of these cells has two areas stainedbright yellow, labelled A on Fig. 22.1.

Fig. 22.1

(a) (i) Suggest why proteins in the cytoplasm of the non-dividing cells in Fig. 22.1 arenot evenly distributed.

[1]

(ii) Suggest the identity of the two areas stained yellow in the dividing cells andoutline their function.

[3]

A



21

© UCIE 2011 9790/01/SP/13 [Turn over

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(b) Before the skin cancer cells could be stained with antibodies, the cells had to be fixedand treated with a mild detergent to increase the permeability of the cell surfacemembranes.

(i) State why it is necessary to increase the permeability of the cell surfacemembranes before staining cells using the technique of immunofluorescence.

[1]

(ii) State and explain two advantages of using immunofluorescence in studying thechanges that occur in cells during cell division.

advantage 1

advantage 2

[4]

[Total: 9]



22

© UCIE 2011 9790/01/SP/13

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Use

23 Fig. 23.1 is a photomicrograph of the lower epidermis of the leaf of an oleander, Neriumoleander . Fig. 23.2 is a photomicrograph of the lower epidermis of the leaf of a privet,Ligustrum vulgare . Both photomicrographs are to the same scale.

Fig. 23.1 Fig. 23.2

(a) State two ways, visible in Fig. 23.1 and Fig. 23.2, in which the epidermis of oleander differs from the epidermis of privet. In each case explain how oleander is adapted tosurvive severe drought conditions.

1

2

[4]



23

© UCIE 2011 9790/01/SP/13 [Turn over

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The concentration of three ions, potassium, chloride and phosphate, were determined inguard cells of closed and open stomata. Fig. 23.3 shows these concentrations measuredin arbitrary units which are the same for all three ions.

7

6

5

4

3

2

1

0

concentrationof K + ions /arbitrary units

concentrationof C l – ions andPO 4

3– ions /arbitrary units

key

concentration of:K + ionsC l – ionsPO 4

3– ions

2

1

0guardcells

nuclei

regionwith thick

cuticle

openstomatal

pore

Fig. 23.3

(b) Suggest a possible mechanism, that can be supported by the data in Fig. 23.3, toaccount for the changes in stomatal aperture.

[4]



24

© UCIE 2011 9790/01/SP/13

For Examiner's

Use

(c) Many plant cells have cytoplasmic connections (plasmodesmata) betweenneighbouring cells, but these are absent from guard cells.

Explain how this helps guard cells function efficiently.

[2]

[Total: 10]

24 Fig. 24.1 summarises the reactions which occur in the Calvin cycle.

compound A

enzyme B

step C

two molecules of glycerate 3-phosphate

CO 2

glucose

Fig. 24.1

(a) Where, precisely, in a plant cell do the reactions shown in Fig. 24.1 take place?

[1]



25

© UCIE 2011 9790/01/SP/13 [Turn over

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(b) Name

(i) compound A

[1]

(ii) enzyme B

[1]

(c) Calculate the proportion of carbon atoms from glycerate 3-phosphate moleculeswhich are incorporated into glucose. Show your working.

[2]

(d) Some biologists describe enzyme B as ‘the most important enzyme in our biosphere’. Explain why they might hold this opinion.

[2]



26

© UCIE 2011 9790/01/SP/13

For Examiner's

Use

(e) The Calvin cycle is part of the light-independent reactions of photosynthesis. Thesereactions continue when a plant is moved from light conditions to dark conditions, butonly for a very short time.

With reference to Fig. 24.1, explain why this is the case.

[4]

(f) Evidence suggests that the earliest eukaryotic cells did not carry out photosynthesis.

Explain how some eukaryotic cells are thought to have become photosynthetic.

[4]

[Total: 15]



27

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25 Fig. 25.1 shows a European starling, Sturnus vulgaris .

Fig. 25.1

(a) Outline the aspects of the biology of S. vulgaris that must be considered whendescribing the niche of this species.

[4]



28

© UCIE 2011 9790/01/SP/13

For Examiner's

Use

The relative contribution of individual species to the biodiversity of communities has beeninvestigated for particular species. Fig. 25.2 shows the effect of different densities of thegastropod mollusc, Littorina littorea , on the number of species of algae in tidal rock poolson the Eastern seaboard of the USA. L. littorea grazes on algae.

14

12

10

8

6

4

2

0

number of species of algae

0 50 100 150

density of L. littorea / number m –2

Fig. 25.2

200 250

(b) Describe and explain the results shown in Fig. 25.2.

[5]



29

© UCIE 2011 9790/01/SP/13 [Turn over

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The effects of removing all the individuals of a species of predatory starfish on a rockyshore community were investigated. Fig. 25.3 is part of a food web for this community andshows the changes that occurred. The minus symbol (–) indicates that the numbers of individuals in the species concerned decreased; the plus symbol (+) indicates that thenumbers of individuals in the species concerned increased.

predatory starfish

dog whelk1 species +

chitons

2 species –

limpets

2 species –

mussel

1 species +

acorn barnacles

3 species –

goose barnacle

1 species +

Fig. 25.3

(c) Explain how the results of the study support the idea that the predatory starfish isa keystone species.

[3]

[Total: 12]



30

© UCIE 2011 9790/01/SP/13

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26 The gene for colour vision in humans is sex-linked. The recessive allele, ch , causesred-green colour blindness. The gene for the ABO blood group system is onchromosome 9. There are three alleles, I A , I B and I o and four possible phenotypes.

Fig. 26.1 shows the inheritance of these two genes in a family.

1 2

A B

1 2 3 4

1 2 3 4

O A B A

5 6 7

O A A AB O O B

I

II

III

female

male

affected unaffectedKey for colour blindness

Fig. 26.1

(a) State the genotypes of the following people in the family shown in Fig. 26.1.

I -2

II -1

II -7 [3]



31

© UCIE 2011 9790/01/SP/13 [Turn over

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(b) With reference to Fig. 26.1, explain why

(i) the grandfather ( I -1) is colour-blind, but neither of his sons is colour-blind

[2]

(ii) one grandson ( III -3) has inherited colour-blindness but the other ( III -1) has not

[2]

(iii) there are four phenotypes in the ABO blood group system.

[3]



32

Copyright Acknowledgements:

Question 22 Figure 22.1 Immunofluorescent LM of skin cancer cells, © Nancy Kedersha/Science Photo LibraryQuestion 25 Figure 25.1 Starling, Stumus vulagris , © Jeroen Stel (rspb.images.com)

Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Everyreasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, thepublisher will be pleased to make amends at the earliest possible opportunity.

University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

© UCIE 2011 9790/03/SP/2013

Nail-patella syndrome is a rare autosomal dominant trait that affects fingernails,toenails, elbows and kneecaps. The locus of the gene for nail-patella syndrome, Np / np ,is 10 map units from the ABO locus on chromosome 9.

A man with nail-patella syndrome and blood gr ou p AB has a family of five children withhis wife who does not have the syndrome and is blood group O.

Three children do not have the nail-patella syndrome and are blood group A.

Two children have nail-patella syndrome and are blood group B.

(c) State the genotypes of the father and the mother.

father

mother

[3]

(d) Explain why there is a small probability of these parents having a child with bothblood group A and nail-patella syndrome.

[2]

[Total: 15]



This document consists of 8 printed pages.



BIOLOGY 9790/01

Paper 1 Structured Questions For Examination from 2013

SPECIMEN MARK SCHEME

2 hours 30 minutes

MAXIMUM MARK: 100



2

© UCIE 2011 9790/01/SM/13

Section A

QuestionNumber Key Question

Number Key

1 C 11 B2 A 12 A3 C 13 B4 B 14 C5 C 15 A

6 D 16 D7 A 17 A28 B 18 C29 B 19 B3

10 D 20 A3/D3





4

© UCIE 2011 9790/01/SM/13

(f) ref to somatic gene therapy ; inserting genes into cells means that treatment is short-lived ; idea of inappropriate immune response to viral vectors ;gene inserted into the wrong place inducing a tumour ;another problem associated with gene being inserted in, wrong place / into another gene ; child receiving treatment for SCID developed leukaemia ;

further detail regarding treatment for SCID ; credit a case study ; [max 4]

[Total: 19]

22 (a) (i) protein forms, fibres / (micro)filaments / cytoskeleton ;ref to distribution of endoplasmic reticulum in cytoplasm ;

AVP ; [max 1]

(ii) spindle apparatus / spindle fibres ; Accept spindle / microtubules / tubulin / centrioles /microtubule organising centres / MTOCs [1]

function to max 2 attach to chromosomes / kinetochores ;detail of, elongation / structure / shortening, of microtubules ;for movement of chromosomes ;during mitosis ;

Accept if centrioles given as identity forms poles of the cell ;organises the spindle ; [max 2]

(b) (i) antibody molecules too large to pass through membrane ; [1]

(ii) locate position of specific, proteins / structures ; antibody molecules have complementary shape to target, proteins / structures ;

can see distribution of, proteins / structures, in light microscope ; do not need to prepare sections for the electron microscope ; easier to look at a large number of cells than in EM ;

higher degree of specificity than using other staining techniques ; idea of variable regions of antibodies giving greater specificity ; [2 + 2]

[Total: 9]



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23 (a) Oleander

lower stomatal density / AW ; less water vapour lost through stomatal transpiration / described ;

stomata in pits / stomata below leaf surface / sunken stomata ;

longer diffusion pathway for water vapour / ref to boundary layer / ref vapour pressure deficit(VPD) at stomatal opening ;

hairs / trichomes, in pits / around stomata ; ref slower air movement / stagnant air / ref VPD at stomatal opening ;

thicker cuticle ; less evaporation from leaf surface / epidermis ; A less water loss from leaf surface /epidermis [2 + 2]

(b) increase in [K + ] when stoma is open ;

comparative data quote ; values similar for both guard cells ;

active transport of K + inwards ;further details of K + pump ;chloride ions diffuse in ; lowers, solute potential / water potential ; water enters by osmosis ;phosphate values very similar ;used in ATP synthesis ; [max 4]

(c) ref to symplast ; K+ would diffuse out of guard cells (to adjacent cells) ; other substance, lost / shared / AW ; e.g. malatefurther explanation ; e.g. higher rate of active transport would be required

AVP ; e.g. further detail [max 2]

[Total: 10]

24 (a) stroma of the chloroplast ; [1]

(b) (i) ribulose bisphosphate / RuBP ; [1]

(ii) rubisco / ribulose bisphosphate carboxylase (oxygenase) ; [1]

(c) award two marks for the correct answer (1/6 or eq) with or without working

RuBP = 5, glycerate 3-phosphate = 3, glucose = 6 ; 1/6 / eq ; [2]



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(d) ref to carbon fixation ; key role in carbon cycle ; only / main, route into food chains for carbon ; the major route out of the atmosphere for carbon dioxide ; [max 2]

(e) ATP and, NADPH 2 / reduced NADP ;produced in the light-dependent stage ;production stops after dark ;are required for step C ;will rapidly be used up after dark ; [max 4]

(f) ref to endosymbiosis (in correct context) ; pre-existing prokaryotes could already photosynthesise ;ref to cyanobacteria as putative ancestral chloroplast ;uptake of prokaryotes into other prokaryotes ; to give a symbiotic, union / community ;

transfer of some genes to host cell nucleus ; retention of other genes in chloroplast ; idea of so it became an obligate symbiosis ; [max 4]

[Total: 15]

25 (a) biotic factors affecting / abiotic factors affecting / description of, habitat ;trophic level / what it feeds on ; adaptations for feeding / foraging method ;time / places, where it feeds ;where it roosts ;whether / when, living, individually / in small groups / in flocks ;where / when, it reproduces ;predator / what feeds on it ; parasites ; competitors ;

AVP ; AVP ; [max 4]

(b) number of species is low when few Littorina ; suggest successful competition by few species of algae ; environment unsuitable for both / AW ;

maximum number of algal species when Littorina is at 150 m –2 ; Littorina grazes most competitive species reducing their effect ;

smallest number of algal species at, highest Littorina density / 250 m –2 ;

ref overgrazing ; AVP ; [max 5]



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(c) definition of keystone species a species whose presence and role within an ecosystem has a disproportionate effect onother organisms within the system / AW ;

removal of the species has profound effects on the, community / ecosystem ;

removal of predatory starfish reduces numbers of seven species ; increases numbers of three species ;

reduces, species richness / biodiversity ; presence keeps other predators in check ;

AVP ; [max 3]

[Total: 12]

26 (a) Accept any sensible symbols Accept without X and Y chromosomes but male must indicate absent allele by using a dash

or by putting in a Y chromosome

I BI

o Ch Ch / IB

Io Ch ch / I

BI

o XCh XCh / IB

Io XCh XCh ;

I o I o Ch ch / I o I o XCh XCh ; I B I o Ch – / I B I o XCh Y ; [3]

(b) (i) fathers pass on X chromosome to their daughters / fathers never pass on X chromosome to their sons ; mother has (at least one) dominant allele and this has been passed on tothe sons ; [2]

(ii) III-3 has inherited colour blindness from mother who is a carrier ; she has inherited Xch from, her father / I-1 ; other grandson / III -1, cannot inherit Xch through the male line ; [max 2]

(iii) multiple alleles / 3 alleles at this locus but each person diploid so can only have 2 ; gives 6 different genotypes / genotypes listed ; codominance between I

A and IB , so gives AB ;

dominance between I A / I B and I o , so means I A I o is same phenotype asI A I A / I B I o is same phenotype as I B I B ; [max 3]

(c) one mark for each genotype, one mark for giving notation for linkage

father I

A np / (IA np) ( I

B Np) I B Np

mother I o np / (I o np) ( I o np) I o np [3]



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(d) loci are linked so I A and np are likely to be inherited together ; so if blood type A, likely to be free of the disease ; 5% / small, chance, of I A and Np ; as a result of crossing over between loci in father ; [max 2]

[Total: 15]



This document consists of 25 printed pages and 3 blank pages.


For Examiner's Use

Section A

Section B

8

9

10

Total


BIOLOGY 9790/02

Paper 2 Long Answer For Examination from 2013

SPECIMEN PAPER

2 hours 45 minutesCandidates answer on the Question Paper.

No Additional Materials are required.


Write your Centre number, candidate number and name on the work you hand in.Write in dark blue or black pen.You may use a soft pencil for any diagrams, graphs or rough working.Do not use staples, paper clips, highlighters, glue or correction fluid.

Section AAnswer all questions.Write your answers in the spaces provided on the Question Paper.

Section BAnswer all questions.Write your answers in the spaces provided on the Question Paper.

Section CAnswer one question.

Write your answers on the Question Paper. Separate answer paper will beavailable if required.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or partquestion.



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Section A

Data analysis

1 Fig. 1.1 shows an American eel, Anguilla rostrata , which lives for part of its life in the riversand mountain streams of the Eastern USA. Adult fish migrate to the Atlantic Ocean when

they are ready to breed. After breeding the adults die.

Young eels migrate from the sea back to the rivers and streams and may live for five yearsor more before reaching the stage when they are ready to breed.

This species of fish has become rare in mountain streams over recent years.

Fig. 1.1

As part of a long-running study to find out more about the biology and behaviour of A. rostrata , mark-release-recapture was used to estimate the population size in onemountain stream in Virginia. Very young eels were not marked.

Table 1.1 shows the results of the mark-release-recapture.



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Table 1.1

year total number

of fishcaught

number of fish marked

total number of fish

captured oneyear later

number of marked fishrecaptured

one year later

populationestimate

2000 334 279 352 98 1002

2001 352 226 290 57 1150

2002 290 149 180 25 1073

2003 180 76 232 11 1603

2004 232 116 184 21 …….

2005 184 72

Totals 1572 918

(a) (i) Calculate the population estimate for the year 2004 to complete Table 1.1.

Show your working.

Write your answer in Table 1.1. [2]





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The annual growth of the eels was also measured. Fig. 1.2 shows a box-whisker plot of theresults for growth in length and growth in mass of eels in one stream that were marked withtags and then recaptured from 2000 to 2005.

60

50

40

30

20

10

0

increase inlength / mm

80

60

40

20

0

increase inmass / g

2000-2001 2001-2002 2002-2003

year

2003-2004 2004-2005

Fig. 1.2

The line in each box represents the median; the top and bottom of each box show 25 th and75 th percentiles; the ‘whiskers’ show the 10 th and 90 th percentiles.



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(b) (i) Describe the results shown in Fig. 1.2.

[3]

(ii) Explain the advantage of plotting box-whiskers to show these data rather than bar charts or histograms.

[3]



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(c) A. rostrata is not officially recognised as an endangered species, but its numbers are indecline.

Discuss the limitations of the results of this study in terms of providing sufficientinformation to inform the conservation of A. rostrata in Virginia.

[4]

(d) Numbers of eels are not as high in other mountain streams in the area. It is thought thisis due to dams downstream that are barriers to migration from the sea.

Suggest how the study might be extended to see if this is the case.

[2]

[Total: 20]



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2 In an investigation into pollen release from Timothy grass, Phleum pratense , the number of pollen grains released into the atmosphere was sampled at hourly intervals, on threeconsecutive days. Traps sited just above the level of the leaves were used to do this.

The wind speed and the relative humidity were recorded at the times of sampling.

The results of the investigation are shown in Fig. 2.1

700

600

500

400

300

200

100

02

1

0

100

80

6040

20

02400 1200 2400 1200 2400 1200 2400

numberof pollengrainscollectedper hour

% relativehumidity

windspeed/ ms –1

J uly 10 J uly 11 J uly 12

Fig. 2.1



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(a) Use the data in Fig. 2.1 to make conclusions about the factors that determine therelease of pollen from Timothy grass.

[6]

(b) Discuss how confident you can be that your conclusions about release of pollen fromTimothy grass are valid.

[4]

[Total: 10]



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3 The technique of polyacrilamide gel electrophoresis (PAGE) is used to separate andidentify proteins. One method of PAGE involves treating proteins with an ionic detergent todissociate proteins into their constituent polypeptide subunits. Sodium dodecyl sulfate(SDS) is often used for this. Proteins treated with SDS have a uniform net charge on eachpolypeptide so that during electrophoresis they are separated only on the basis of their relative molecular mass.

After treatment with SDS, proteins are placed in wells ( A to F) cut into the polyacrilamidegel. A dye is added to each sample to show the progress of the samples across the gel. Acurrent is applied to the gel and when the dye reaches a point towards the end of the gel,the current is switched off.

The relative mobility of each polypeptide is calculated as follows:

frontdyebytravelleddistancebandepolypeptidbytravelleddistance

Six proteins were analysed with SDS-PAGE and the results are shown in Fig. 3.1.

The relative molecular mass of one protein, lactate dehydrogenase, was unknown.

well

dye front

Fig. 3.1

A DCB FE



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(a) (i) Calculate the relative mobility of the proteins in sample wells A, B, C, E and F andadd your calculated values to the spaces in Table 3.1.

Space for working

Write your answers in Table 3.1 [2]

Table 3.1

well protein relative molecular mass relative mobility

A carbonic anhydrase 29 000 …………………

B human albumin 68 000 …………………

C lactate dehydrogenase …………………… …………………

D myoglobin 17 200 1.00

E egg albumin 43 000 …………………

F transferrin 77 000 …………………



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(ii) Use the graph paper provided to draw a graph of the relative molecular massplotted against the relative mobility of proteins A, B, D, E and F. [4]

(b) Complete Table 3.1 by using your graph to find the relative molecular mass of lactatedehydrogenase.

Explain how you arrived at your answer.

Write the relative molecular mass of lactate dehydrogenase in the space in Table 3.1 [2][Total: 8]



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Section B

Read the passage carefully and answer all the questions in the spaces provided.

You are advised to spend no more than 50 minutes on this section.

Type 2 diabetes – the growing threat

Diabetes mellitus currently affects at least 2.5 million people in the UK and is a condition inwhich the body is unable to maintain a normal blood glucose concentration. Many people whohave no experience of diabetes think that the more common form is type 1, requiring insulininjections. Yet this is not the case. By far the more common is type 2, which representsapproximately 85–90% of cases, and is on the increase. Originally thought of as affecting older people it is becoming increasing common among the young. It is thought that obesity is animportant risk factor. There is no entirely successful way of treating type 2 diabetes although itcan be managed by control of diet, appropriate exercise and the use of medication.

Those with the condition, at least initially, produce insulin normally but certain body cells

develop insulin resistance. This means that they do not respond to the hormone by taking upglucose from the blood rapidly enough to maintain a normal blood glucose concentration. Thepermeability of cell membranes is dependent on the presence of transporter protein molecules.Table 5.1 provides information about two types of such transporters, GLUT and SGLT.Table 5.1 distinguishes four types (isoforms) of GLUT.

Table 5.1

transporter group

type of mechanism isoform mainly present in further information

GLUT1 all cells low-level basal glucoseuptake required tosustain respiration

GLUT2

cells in small intestinelining, in the liver and

in cells of kidneytubules

in the kidney tubulethese transport glucose

from cells lining thenephron into capillaries

GLUT3 neurones probably main glucosetransporter in neurones

GLUT(glucose

transporters)

facilitateddiffusion

GLUT4 adipose cells andstriated muscle cells(skeletal and cardiac)

insulin-controlledglucose transporter

SGLT(sodium-glucoselinked

transporters)

secondaryactive

transportalong sodium

gradient

cells lining the proximaltubule of nephrons

transport glucosedirectly from

glomerular filtrate intocells lining nephron



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Insulin is produced by the β cells of the islets of Langerhans within the pancreas. When theinsulin concentration of the blood is low, GLUT4 molecules are removed from the cellmembranes of adipose cells and skeletal muscle cells into vesicles in the cytoplasm. Except inthe case of type 2 diabetes, an increase in blood insulin concentration means that insulincombines with specific sites on the cell surface membrane. This causes the GLUT4 moleculesto be restored to the membrane, making it permeable to glucose. When blood insulin

concentration falls the GLUT4 molecules are removed from the membrane into cytoplasmicvesicles again.

On the onset of type 2 diabetes the patient’s cells become insulin-resistant. Initially thepancreas responds by producing extra insulin. This only partially alleviates the problem of insulin resistance and, in time, overworking of the pancreatic β cells leads to their death andsubsequently a reduction in insulin production. At this stage the patient may need to receiveinsulin injections, although this offers only a partial solution.

5 (a) After a meal, blood glucose concentration rises above the target concentration(4.5–5.5 mmol dm –3 ) at which it is normally maintained by homeostasis.

With the help of Table 5.1, outline how the glucose concentration is reduced to normalin a person who does not have diabetes.

[3]

(b) In the space below draw a simple labelled diagram showing how protein transporter molecules may form part of a cell surface membrane.

[4]



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(c) Explain how the uptake of glucose by cells in the proximal convoluted tubule differsfrom its uptake by liver cells.

[3]

(d) (i) Suggest a mechanism by which the glucose transporter GLUT4 is restored to themembrane when insulin binds to the cell surface membrane.

[1]

(ii) To what extent might the removal of GLUT4 from a muscle cell surface membranerender it impermeable to glucose?

[1]

[Total: 12]



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6 Table 6.1 presents the results of an experiment comparing rates of glucose production by agroup of people with type 2 diabetes and a control group without the condition, during 23hours of fasting.

Table 6.1

glucose productioncontrols / µ mol kg

body mass –1 min –1

type 2 diabetic patient /µ mol kg body mass –1

min –1 significance level

total glucoseproduction 8.9 ± 0.5 11.1 ± 0.6 p < 0.05

glucose fromhydrolysis of

glycogen in the liver 2.8 ± 0.7 1.3 ± 0.2 p < 0.05

glucose fromgluconeogenesis 6.1 ± 0.5 9.8 ± 0.7 p < 0.01

(a) Discuss the conclusions which can be drawn from the data in Table 6.1.

[6]



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(b) In non-diabetic individuals, the blood glucose concentration in the renal vein is onlyslightly lower than in the renal artery.

Explain why one might expect the glucose concentration of the blood in the renal veinto be much lower than in the renal artery and suggest why, in fact, the concentrationsare almost identical.

[4]

[Total: 10]



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7 It is possible to restore insulin secretion in a diabetic patient by transplanting a pancreas or isolated islets of Langerhans, but limited donor organs and risks involved restrict thesetherapies to a small proportion of diabetics. Recent experiments suggest that it may, infuture, be possible to treat diabetes with adult stem cells from the patient’s own bonemarrow.

Explain why such an approach, once perfected, is more likely to offer, at least initially, atreatment for type 1 rather than type 2 diabetes and discuss why this approach may bepreferable to the use of transplants or embryonic stem cells.

[8]

[Total: 8]

[Total for Section B: 30]



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Section C

Answer one question on the lined paper that follows.

You are advised to spend no more 50 minutes on this section of the examination. Credit will begiven for answers that draw from a wide range of syllabus material and also for evidence of

reading around the subject.

8 ‘There is no evolutionary advantage in being multicellular’.

Discuss this view.

9 All living organisms need to synthesise ATP. Explain the similarities and differencesbetween organisms in the ways in which this is achieved.

10 Why do people get heart disease and what should be done about it?



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[30]



28


Question 1 Figure 1.1 American Eel © Andrew J. Martinez/Science Photo Library.



© UCIE 2011 9790/02/SP/13

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BIOLOGY 9790/02

Paper 2 Long Answer For Examination from 2013


2 hours 45 minutes

MAXIMUM MARK: 120



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Section A

1 (a) (i) two marks for the correct answer written into Table 1.1

if incorrect, one mark for working

Lincoln index: N = population estimate n1 = number marked and releasedn2 = total number recaptured m2 = number marked among those recaptured

N =m2

n2n1 ×

N =21

184116 ×

1016 ; ; [2]

(ii) numbers of marked fish recaptured in, 2003 / 2004, are small ;estimates based on small numbers are unlikely to be accurate ;over a hundred / many, very young eels caught and not marked ;proportion of those not marked increases from 2000 to 2005 ;total number of fish caught one year on is not adjusted for very young fish ;so population is underestimated ;cannot be repeated within each year to see if estimate is reliable ;

AVP ; [max 3]

(iii) marking may injure animal ;alters behaviour / makes it prone to predation / less able to feed / AW ;

marks may be lost ;chances of catching fish may vary if marked ;chances of catching fish vary with their age ; A ‘trap happy’ / ‘trap shy’some fish easier to catch than others ;fish unlikely to mix thoroughly in streams ;activity of fish may depend on, environmental conditions / AW ;effect of deaths ;effect of migration, into / out, of streams ;

effect of any named limitation giving, underestimate / overestimate, of population size ;

AVP ;

R effect of ‘births’ (spawning occurs at sea) [max 3]



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(b) (i) most increase in, length / mass, in fourth group ;wide range of results ; especially for increase in mass ;50% of fish in sample are between 25 th and 75 th percentile ;median increase in mass remained roughly constant (except 4 th group) ;

comparative data quote ; AVP ; [max 3]

(ii) bar chart may plot the mean (and SD / SE) ;b+w gives more information about the range of results in a sample ;position of median, shows skewness of data / AW ;can show, outliers / anomalous results ;useful if data are not normally distributed ;easier to compare data from different categories than using histograms ;

AVP ; [max 3]

(c) no data on migration to and from the sea ;no information on reproduction of eels ;data is only about eels, not about food supply / habitat / niche / AW ;no information on age structure ;no standard against which to compare data on growth ;no information on likely causes of death ;no information on, behaviour / movement, of eels during each year ;

AVP ; AVP ; [max 4]

(d) catch eels as they migrate upstream ;mark and release ;recapture at stations below and above the dams ;use, radiotagging / transmitters / AW, to follow movement of eels ;

AVP ; [max 2]

[Total: 20]



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2 (a) majority / most, pollen released between midnight and midday ; very little / none, released between midday and midnight ; ref to, regular / diurnal, pattern ; most pollen released at 0700 each day ; ref to figures for maximum release ; e.g. 500 to 700 pollen grains per hour most pollen released when wind speed low ;

maximum pollen released when relative humidity high / ora ;steep decrease in pollen release as relative humidity falls ; ref to figures in support ; very little pollen released when wind speed higher ; ref to data for, wind speed / humidity ; [max 6]

(b) idea that cannot be sure from the results whether timing or abiotic factors are more importantin determining release of pollen ;

repeated pattern of release

only data from three days ;no information on how many, sites / plants ;data not in form of mean grains collected per hour so not repeated ;not carried out by other people / no data on reproducibility ;

link between wind speed and humidity

no data with, constant wind speed / wind speed higher at night than during the day ;no data with, constant relative humidity / humidity higher early part of day / AW ;

idea that correlation is not cause and effect ;

not an experiment as no factors have been controlled ; no hypothesis to test ; data not analysed statistically therefore impossible to assign level of significance / confidencein the conclusions ; [max 4]

[Total: 10]



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3 (a) (i)

well protein relative molecular relative mobility

A carbonic anhydrase 29 000 0.86

B human albumin 68 000 0.38

C lactate dehydrogenase 0.72

D myoglobin 17 200 1.00

E egg albumin 43 000 0.62

F transferrin 77 000 0.36 ;;

5 correct = 2 3 or 4 correct / correct working but incorrect answers = 1

0, 1 or 2 correct with no correct working = 0 [2](ii) x-axis for relative mobility, y axis for relative molecular mass, sensible scales ;

axes labelled appropriately ; points plotted correctly ; straight line – not extending beyond first and last point ; [4]

(b) co-ordinates on graph explained or shown on graph ; answer = approx 34 000 ; [2]

[Total: 8]



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4 Planning Task

P = defining the problemM = methods

Analysis, conclusions and evaluation

D = Interpretation of data or observations and identifying sources of error C = Drawing conclusionsE = Suggesting Improvements and evaluation

Sections Expected answer Mark

Pdefiningtheproblem

Hypothesis or prediction ;e.g. rate of uptake of glucose is faster than rate of uptake of maltose / K m for uptake of glucose is lower / transport protein has a higher affinity

Theory to support hypothesis or prediction ;e.g. glucose is a smaller molecule / does not require to be hydrolysed byenzyme / ref to production of maltase inside yeast cell

Outline of strategy and justification / evaluation ;e.g. method of following the uptake of glucose and maltose separatelytaking samples at intervals and calculating uptake

this could be awarded at the end of the plan

method of determining (the concentration) of glucose at intervals ;

method of determining (the concentration) of maltose at intervals ;(semi) quantitative Benedict’s solution

At least two control variables ;e.g. temperature, concentration of yeast, pH, volumes used, pre-treatment of glucose

Risk assessment ;ref to hazard and precaution

some points may be taken from a diagram or a flow or sequence diagram [max 6]



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Pmethods

use range of concentrations of glucose and maltose ;and / or use range of concentrations of yeast suspension ;to find suitable concentrations to make comparisondilution table(s) included ;

yeast mixed with glucose and maltose solutions ;equilibration in water bath ;staggered start ;

samples taken at stated intervals ;filtered to remove yeast ;method of finding concentration of sugars described ;calculate quantity of sugars absorbed knowing initial concentration ;details of calculation ; uncertainty / precision, of results ;

plot results and take gradient to give initial rate ;calculate mass of sugar absorbed per unit time ;

repeats / replicates ;calculate, standard deviation / standard error / 95%CI ;

plot overall graph as a line graph ;state that answer is where rate of uptake becomes constant ;find K m ;find ½ K m ;discussion of affinity of transport proteins ;

use, suitable named statistical test ; e.g. t-test / z-test / ANOVA

AVP ; AVP ; [max 16]

[Total: 22]



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Section B

5 (a) conversion to glycogen in liver and muscle ; conversion to, fat / lipid / fatty acids / triglycerides in adipose tissue ; GLUT4 in adipose and muscle ; GLUT2 in liver ;

uptake by cells for respiration by all cells due to, GLUT1 / neurones GLUT3 / nephrons /kidney GLUT2 and SGLT ; [max 3]

(b) lipid bilayer with heads and tails correctly orientated ;labelled, lipids / fatty acids and glycerol / phosphoglycerides ; transporter molecule passing through the bilayer ;labelled appropriately ; must be drawn as a channel protein [4]

(c) liver cells facilitated diffusion and PCT cells active transport ;

active transport against concentration gradient and facilitated diffusion with the gradient ; active transport requires, respiratory / metabolic energy, ATP, whereas facilitated diffusiondoes not ; (in the case of SGLT / secondary active transport) the ATP / metabolic energy pumpscreates the sodium gradient which causes the glucose molecules to move passively ; (in the case of SGLT) re-absorption of glucose from the proximal tubule is against a steepdiffusion gradient / needs to take place quickly as filtrate is passing rapidly along thenephron ; [max 3]

(d) (i) by exocytosis / vesicles fuse with (cell surface) membrane ; [1]

(ii) still some permeability due to GLUT1 ; [1]

[Total: 12]



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6 (a) more glucose is produced by a starving person with type 2 diabetes thanby a person without the condition / AW ; perhaps because the diabetic is less able to regulate the blood sugar concentration / at amore advanced stage of starvation / reduced fat reserves ; in type 2 diabetes, the main source / a greater proportion (of glucose) is fromgluconeogenesis ;

gluconeogenesis is the production of glucose from amino acids / proteins ; relatively small amount from glycogen as almost used up (by this stage) ; less from glycogen with type 2 diabetes ; as less to start with ; correct reference to statistical significance in discussion of data ; correct explanation of p < 0.05 ;reference to ± as (possibly), variation / standard deviation / standard error / indication of good agreement within replicates ; [max 6]

(b) one would expect glucose to be used up in respiration in the kidney ; many mitochondria / high rates of respiration in kidney ;

providing ATP for sodium-potassium pumps ; ref to, selective reabsorption / active transport ; lack of difference could be due to, gluconeogenesis / amino acids converted to glucose ; could be a great deal of gluconeogenesis in kidney / gluconeogenesis must be, equal tokidneys’ glucose consumption through respiration ; [max 4]

[Total: 10]

7 insulin secreting cells are only found in, pancreas / islets of Langerhans ; reference to β cells as the source of insulin ; use stem cells to replace, dysfunctional β cells / insulin-producing cells ; so stem cells would only need to be introduced into a specific part of the body ; whereas (in the case of type 2) cells all over the body are dysfunctional ; and so cannot be replaced / much more difficult to replace them all ; use of stem cells may be preferable to transplants as less invasive / fewer side-effects / no needto wait for suitable donor ; use of patients own adult stem cells (instead of transplants or embryonic stem cells) avoidsimmunological rejection / need for immunosuppressant drugs ; use of adult rather than embryonic stem cells avoids ethical issues about sourcing ; in (advanced cases) of type 2 diabetic patients where pancreas is deteriorating β cell transplantmay be of benefit but not a complete solution ; need to understand more about what causes type 2 ; [max 8]

[Total: 8]



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Section C

Marking Strategy

Sequence of maker activities for each essay

1. Familiarise yourself with the expected content.

2. Read through the essay.

3. Write marginal notes on script, highlight evidence of breadth, exemplification and argumentationas well as major and minor errors of fact and irrelevant material.

4. Apply the general descriptions for

• Breadth• Argumentation• Communication

• Spelling, punctuation and grammar.5. Match the content of the essay with a descriptor for Scientific Content (20, 16, 12, 8, 4, 0 as

appropriate) and then decide whether

• all sub-descriptors at that level have been met so that the full mark for that level can beawarded

• three out of the four sub-descriptors have been met so that intermediate marks can beawarded (18, 14, 10, 6, 2)

• one or two of the sub-descriptors at that level have been met so that the full mark for thelevel below can be awarded.

Marks should be written at the end of the essay as followsB = ………

A = ………C = ………S = ………SC = ………Total = ………

Breadth Maximum 3 marks

Mark Descriptors

Candidate has:

3 given a balanced account including most of the relevant topic areas and selected a widerange of facts, principles, concepts and / or examples pertinent to the title

2 given a fairly balanced account including some of the relevant topic areas and selectedsome of the appropriate facts, principles, concepts and / or examples pertinent to the title

1 given an account including a few of the relevant topic areas and selected a few of theappropriate facts, principles, concepts and / or examples pertinent to the title

0 given an account that relies on one topic area alone and selected a few of the appropriatefacts, principles, concepts and / or examples pertinent to the title.





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Scientific Content Maximum 20 marks

Mark Descriptors

The candidate :

a recalls and consistently uses all facts and principles (relevant to the essay)

b shows sound understanding of all principles and concepts

c writes accurately with no major errors, very few minor errors20

d gives detail fully in keeping with that expected of candidates at the end of a programme of study designed to prepare candidates for university

a recalls and consistently uses most facts and principles (relevant to the essay)

b shows sound understanding of most principles and concepts

c writes accurately with no major errors, few minor errors16


a recalls and consistently uses some facts and principles (relevant to the essay)

b shows sound understanding of some principles and concepts

c writes some material accurately with not more than one major error, some minor errors

12


a recalls some facts and principles (relevant to the essay)

b shows some understanding of some principles and concepts

c writes some material accurately with more than one major error or many minor errors8

d gives some detail appropriate for that expected of candidates at the end of a programmeof study designed to prepare candidates for university

a recalls a few facts and principles (relevant to the essay)

b shows limited understanding of a few principles and concepts

c writes material including many errors some of which may be major errors4

d gives a little detail appropriate for that expected of candidates at the end of a programmeof study designed to prepare candidates for university



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a recalls no relevant facts and principles

b shows no understanding of relevant principles and concepts

c writes irrelevant material or includes many major errors0

d gives no detail appropriate for that expected of candidates at the end of a programme of study designed to prepare candidates for university

Expected content

For each of the questions, guidance is given as to the kind of content from the syllabus that may beappropriate to answering the question. Some candidates will include all of these areas and othersmay write in more detail about these or may include other relevant topics, in each case reflecting thecandidate’s reading-around the subject and personal research and other interests. Some topics bothin the candidate’s answers and in the following expected content may not be directly on the syllabus,but it is important to credit such responses where they are given and thus they are included here.





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therefore most recently evolved life forms should be superior

this is a flawed argument because natural selection operates on all species all the time

therefore current life forms have equal status in terms of success / can only judge on basis of future possibilities

could consider further the particular example of humans

humans have more control over environment than any other organism

they are a product of an evolutionary trend towards greater complexity

perhaps control over environment may be greater evolutionary advantage than adaptation tochange

9 All living organisms need to synthesise ATP. Explain the similarities and differences

between organisms in the ways in which this is achieved.

Candidates should avoid lengthy descriptions of respiration and photosynthesis but should attempt to highlight underlying similarities and difference in whatever it is they arecomparing.

Comparisons can be made between production of ATP in glycolysis and Krebs cycle by substrate-linked phosphorylation and production in chloroplast and mitochondria by chemiosmosis.

Further points can be made by considering chemiosmosis in prokaryotes. Wider considerationcan be included by discussing energy sources in different forms of nutrition.

The following syllabus sections are most directly relevant: 1.1, 1.5, 2.2, 4.2.

Chemotrophs and phototrophs

source of energy to make ATP may be chemical (chemotrophs) or light (phototrophs)

all animals, fungi and most bacteria are chemotrophic

all plants, algae and some bacteria are phototrophic

most phototrophs are photosynthetic, using carbon dioxide as a source of carbon

aerobic and anaerobic respiration

all organisms / cells make ATP as a result of respiration

respiration is oxidation of a chemical to release energy which is used to make ATP

in aerobic respiration oxygen is final electron acceptor, not in anaerobic

a few bacteria are obligate anaerobes, but most organisms can do both

anaerobic is less efficient than aerobic in terms of ATP per molecule of glucose, but can be morerapid and is a useful supplement when ATP becomes limiting, e.g. when oxygen shortage





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inorganic molecules sometimes used as electron acceptors by prokaryotes e.g.chemoautotrophic bacteria (chemosynthesis)

e.g. nitrifying bacteria – energy from oxidation of inorganic substances during respiration(ammonia and nitrite)

ammonium Nitrosomonas nitrite + ATP Nitrobacter nitrate + ATP

ref to the nitrogen cycle

also Rhizobium , Sulfolobus , some sulfur bacteria, deep sea hydrothermal vents, methanogens(Archaebacteria)

photosynthesis

ATP manufacture as a result of aerobic respiration and light-dependent reactions of photosynthesis. Harnessing energy from electron flow / redox reactions on membranes in

specialised organelles and requires a hydrogen (electron) donor (e.g. glucose, water, hydrogensulfide).

ATP also made during light-dependent reactions of photosynthesis

underlying similarity with respiration is that ATP made on membranes in specialised organelle byprocess of chemiosmosis as a result of electron flow from electron donor to electron acceptor

organelle = chloroplast

thylakoids equivalent to cristae

energy source is light (not oxidation of chemicals) – photophosphorylation

water = electron donor to PSII (electron donor usually organic molecule in respiration)

cyclic and non-cyclic photophosphorylation

in cyclic, PSI (chlorophyll) is electron donor and acceptor

final electron acceptor for non-cyclic is NADP (compare NAD in respiration)

prokaryotes, algae, C3 and C4 plants

photosynthetic bacteria do not have chloroplasts but do have membranes, e.g. blue greenbacteria

algae have simpler membrane systems than plants (no true grana)

C4 plants have larger grana than C3 plants for more efficient use of light energy

more ATP needed to drive photosynthesis than C3 plants

photoheterotrophs

photoheterotrophs a special case – organic source of carbon but use light as source of energy

e.g. ocean planktonic bacteria



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10 Why do people get heart disease and what should be done about it?

Answers should be divided between discussion of risk factors for heart disease and ways inwhich the community (governments, health authorities, etc.) and individuals can reduce incidence

/ prevalence of heart disease and the methods of the treatment for those with the disease

The following syllabus section is most directly relevant: 3.1.

risk factors for heart disease

e.g. age, ethnicity, sex (males), heredity, smoking, lack of exercise, diet, obesity, diabetes, highblood pressure

heart disease and its aetiology

coronary heart disease

description of supply of oxygenated blood to heart by coronary arteries

damage to coronary arteries, e.g. by high blood pressure

role of LDLs in transporting fatty acids and cholesterol to organs

LDLs accumulate in wall of coronary artery lead to plaque

atheroma / atherosclerosis

occlusion of lumen / blood flow becomes uneven / plaque bursts

increased chance of blood clotting in artery

ref to angina / heart attack

HDLs transport cholesterol to liver

reduce chances of plaque developing

explanation of effects of risk factors

oestrogen provides protection

heredity, e.g. familial hypercholesterolaemia ; platelet glycoprotein receptor gene involved inblood clotting ; apolipoprotein E (APOE) involved in lipoprotein particles

treatments

inserting stents

angioplasty

coronary by-pass surgery

heart transplants



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discussion of pros and cons of different treatments

preventive medicine

screen for those at risk

discussion of thresholds for determining people who need intervention to reduce risk

intervention could be determined by individual (e.g. diet, weight loss, etc.) or a medicalintervention, such as drug treatment

statins to lower blood cholesterol concentrations

ref to mode of action: inhibit the enzyme HMG-CoA reductase involved in production of cholesterol in liver

warfarin used as an anticoagulant

ref to mode of action: vitamin K antagonist (inhibits enzyme that recycles oxidised vitamin K to itsreduced form after it has participated in the carboxylation of e.g. prothrombin and factor VII)

β blockers to reduce blood pressure

ref to mode of action: block receptor sites for adrenaline and noradrenaline in heart

ref to control of the heart by sympathetic nervous system and endocrine system

actions that can be taken by individuals to reduce risk

improve diet, reduce weight, have regular check-ups, take exercise, stop smoking

actions that can be taken by community to reduce incidence / prevalence of heart disease

idea that prevention is cheaper than treatment, but untargeted programmes are often not costeffective

provide information to, those at risk / whole population

provide facilities for people to take exercise

advertise about dangers of choices that put people at risk, e.g. smoking

provide funding for drug treatments, e.g. statins

fund research to evaluate effectiveness of different preventative measures



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© UCLES 2011 [Turn over


BIOLOGY 9790/03

Paper 3 Practical Examination For Examination from 2013

SPECIMEN CONFIDENTIAL INSTRUCTIONS

2 hours 30 minutes

Great care should be taken to ensure that any confidential information given does not reach thecandidates either directly or indirectly.



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Instructions for preparing apparatus

These instructions give details of the apparatus and materials required by each candidate for thispaper. Sufficient information is given to permit the Centre to set up and test the apparatus andmaterials so that the candidates can be fairly assessed. No access to the question paper ispermitted in advance of the examination .

If a candidate breaks any of the apparatus, or loses any of the material supplied, the matter should berectified and a note made on the supervisor’s report.

Candidates must be provided with a microscope with:

• Low-power objective lens, e.g. ×10 (equal to 16 mm or 2/3”)

• High-power objective lens, e.g. ×40 (equal to 4 mm or 1/6”)

• Eyepiece graticule fitted within the eyepiece and visible in focus at the same time as thespecimen.

Each candidate should have sole, uninterrupted, use of the microscope for at least 35 minutes.Supervisors are advised to remind all candidates that all substances in the examination should betreated with caution. Pipette fillers and safety goggles should be used when necessary.

HEALTH AND SAFETY

Supervisors are advised to remind candidates that all substances in the examination should betreated with caution. Only those tests described in the question paper should be attempted.

In accordance with the COSHH (Control of Substances Hazardous to Health) Regulations, operativein the UK, a hazard appraisal of the examination has been carried out.

Attention is drawn in particular to certain materials used in the examination. The following codes areused where relevant.

C = corrosive substance F = highly flammable substanceH = harmful or irritating substance O = oxidising substanceT = toxic substance N = dangerous to the environment

The attention of Supervisors is drawn to any local regulations relating to safety, first-aid and disposalof chemicals.

‘Hazard Data Sheets’, relating to materials used in this examination, should be available from your chemical supplier.



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Instructions to Supervisors

Each candidate must be provided with the following apparatus and materials for Section A only.

To be supplied by the Centre.

Question 1

Each candidate will require, for a period of 90 minutes:

(i) 100 cm 3 of full fat milk in a beaker labelled milk .

(ii) 50 cm 3 of sodium carbonate solution in a beaker labelled sodium carbonate solution .This must be prepared as follows:

Add 95 cm 3 distilled or de-ionised water to 0.3 g of anhydrous sodium carbonate, stir todissolve and then add distilled or de-ionised water to make up to 100 cm 3.

[F] (iii) 10 cm3

of phenolphthalein solution in a beaker or dropping bottle labelledphenolphthalein . This solution is flammable.

(iv) 20 cm 3 of 1% lipase solution in a beaker labelled lipase solution . This must be preparedas follows:

Dissolve 1 g of lipase in 50 cm 3 of cold distilled or de-ionised water and stir thoroughly.Make up to 100 cm 3 with more water.

The solution must be prepared just before the examination.

(v) 20 cm 3 of 5% bile salts (sodium tauroglycocholate) solution in a beaker labelled 5% bilesalts solution . This must be prepared as follows:

Add 95 cm 3 distilled or de-ionised water to 5.0 g sodium tauroglycocholate, stir to dissolveand then add distilled or de-ionised water to make up to 100 cm 3.

(vi) 50 cm 3 of distilled or deionised water provided in a beaker labelled water .

(vii) Twelve test-tubes (e.g. 12 × 1.4 cm); test-tube rack or racks; one bung to fit all test-tubes;test-tube holders; one dropping pipette; glass rod; permanent marker pen or other suitableway to label glassware.

(viii) Five small beakers, e.g. 50 cm 3 or 100 cm 3 for candidates to prepare their bile saltssolutions.

(ix) Stopwatch, stop clock or bench timer.

(x) 2 × 10 cm 3 syringes; 2 × 2 cm 3 syringes and 2 × 1 cm 3 syringes.

(xi) One 600 cm 3 beaker to act as a water bath; thermometer.Candidates must not use thermostatically-controlled water baths, but they should haveaccess to one to collect water at 55 – 60 °C.

(xii) A beaker of water labelled washing water .

(xiii) Beaker for waste water labelled waste .



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(xiv) Paper towels.

Centres are advised to have stocks of the milk, lipase solution, sodium carbonate solution and bilesalts solution available for candidates.

Extra supplies of test-tubes and other glassware should also be available should candidates require

them.

Sodium tauroglycocholate may obtained from science education suppliers.



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Question 2

Each candidate must have sole use of a microscope for 35 minutes.

(i) Hand lens, e.g. ×10.

(ii) Slides R1 and R2 .

R1 is a transverse section of the spinal cord of a small mammal. R2 is a transversesection of a small mammal cerebellum. Suitable prepared slides may be obtained fromscience education suppliers.

(iii) Candidates must be provided with a microscope with:

• Low-power objective lens, e.g. ×10 (equal to 16 mm or 2/3”)

• High-power objective lens, e.g. ×40 (equal to 4 mm or 1/6”)

• Eyepiece graticule fitted within the eyepiece and visible in focus at the same time asthe specimen.

(iv) Plastic ruler (mm / cm).



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This document consists of 13 printed pages and 3 blank pages.


For Examiner's Use

Section A

Section B

Total


BIOLOGY 9790/03

Paper 3 Practical Examination For Examination from 2013

SPECIMEN PAPER

2 hours 30 minutes

Candidates answer on the Question Paper.

Additional Materials: As listed on the Confidential Instructions.


Write your Centre number, candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use a pencil for any diagrams, graphs or rough working.Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all questions.

Section AWrite your answers in the spaces provided on the Question Paper.

Section BWrite your answers in the spaces provided on the Question Paper.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.



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For Examiner's

Use

Section A

Answer all the questions in the spaces provided.

1 You are recommended to spend no longer than 90 minutes on question 1.

You should read through the whole of this question carefully and then plan your use of thetime to make sure that you finish all the work that you would like to do.

The enzyme lipase catalyses the hydrolysis of ester bonds in triglycerides. Its activity isaffected by the presence of bile salts.

Full fat milk will be used as the source of triglycerides.

You are to investigate the effect of different concentrations of bile salts on the rate of hydrolysis of triglycerides in milk.

You are provided with a 1% solution of lipase and a 5% solution of bile salts.

Proceed as instructed in steps 1 to 9 .

1 Use the syringes and the small beakers to prepare a number of different concentrationsof bile salts using the 5% bile salts solution and water provided. You will need amaximum of 10 cm 3 of each bile salts solution.

(a) Complete the table below to show how you have prepared the different solutions.

final concentration of bilesalts /%

volume of 5% bile saltssolution / cm 3 volume of water / cm 3

[3]

(b) Suggest suitable controls for this experiment.

[2]



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For Examiner's

Use

2 Label test-tubes with the concentrations of bile salts you have prepared and for your control or controls.

3 Prepare the labelled test-tubes with milk, sodium carbonate solution, bile salts solutionsand phenolphthalein as necessary using the following quantities where appropriate:

•

5.0 cm3

of milk• 5.0 cm 3 of sodium carbonate solution• 1.0 cm 3 of the appropriate bile salts solution• three drops of phenolphthalein

4 Put a bung into each test-tube in turn and invert twice so that the contents are auniform pink colour.

5 Put some warm water in a beaker to act as a water bath. The beaker should be abouthalf-full. Adjust the temperature of the water to 50 oC (+/ – 2 oC).

6 Place the test-tubes prepared in step 3 into the water bath.

7 Stir the lipase solution with the glass rod provided. Put 2.0 cm 3 of the lipase solutioninto the same number of labelled clean test-tubes as you used in step 2, and placethem in the water bath.

8 Prepare the space on page 5 to record your results.

(c) (i) The pink colour of the phenolphthalein will fade over time.

State how this is used to gain information about the rate of hydrolysis of the lipid inthe milk by lipase.

[3]

(ii) Measure and record the temperature in the water bath at appropriate points duringthe experiment in the space below.

[1]



5


For Examiner's

Use

9 After the test-tubes have been in the water bath for at least five minutes , add thelipase solution to the test-tubes containing the milk and bile salts solutions.

Immediately after adding the lipase solution insert a bung into each test-tube andinvert twice to mix the contents.

(d) You should record your results to show the effect of bile salts on the rate of triglyceridehydrolysis by lipase in a logical way in the space below.

Record and justify any further decisions that you make about your investigation in thespace below the table.

[8]

[3]



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For Examiner's

Use

(e) Plot a graph of your results on the graph paper provided.

[5]



7


For Examiner's

Use

(f) Describe and explain the pattern of results shown by your graph.

[10]



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For Examiner's

Use

(g) Identify the limitations and sources of error in this investigation.

Explain how you would improve the method you used in this investigation to give moreaccurate and reliable results.

[10]

[Total: 45]



9


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For Examiner's

Use

Section B

Answer all the questions in the spaces provided.

2 You should not take longer than 60 minutes to complete question 2.

You should read through the whole of this question carefully and then plan your use of thetime to make sure that you finish all the work that you would like to do.

R1 is a transverse section of the spinal cord of a small mammal.

(a) (i) Make a low power plan drawing of R1 .

Label your plan drawing.

[6]

(ii) Use a ruler to measure the actual size of the specimen on slide R1 and the size of your drawing between the same points. Put a line on your drawing to show the sizethat you have measured. Calculate the magnification of your drawing.

Show your working.

magnification …………….………………… [2]



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For Examiner's

Use

(b) Use the high power lens of your microscope to locate a cell body of a motor neurone inR1 .

Make a labelled drawing to show the cell body. Annotate your drawing to indicate the functions of the structures you have drawn.

Use the eyepiece graticule and slide micrometer to measure the diameter of the cellbody. Indicate the actual diameter on your drawing and show how you have derivedyour answer.

[8]

(c) Slide R2 is a transverse section of part of the brain of a small mammal.

Compare, using a hand lens and your microscope, the structure and appearance of R1 and R2 .

Present your comparison as a table in the space below.

[5]



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For Examiner's

Use

(d) Fig. 2.1 is an electron micrograph that shows a cross section of a neurone.

Fig 2.1

(i) Describe the appearance of the section of the neurone. You may use drawings or diagrams to illustrate your answer.

[5]



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For Examiner's

Use

(ii) Explain how the structural features you describe in (i) are related to the function of the neurone.

[4]



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For Examiner's

Use

(e) Fig. 2.2 is an electron micrograph that shows a junction between two neurones in thebrain.

Fig 2.2

Identify structures A and B and relate the appearance of these structures to their function. You may use the space opposite for any diagrams you may wish to draw to

illustrate your answer.

A

B



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For Examiner's

Use

[5]

[Total: 35]



16


Question 2 Figure 2.1 Myelinated neuron, Road not taken © Wikimedia Commons.Question 2 Figure 2.2 Synapse nerve junction, © Thomas Deerinck, NCMIR/Science Photo Library



© UCLES 2011 9790/03/SP/2013

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BIOLOGY 9790/03

Paper 3 Practical Examination For examination from 2013


2 hours 30 minutes

MAXIMUM MARK: 80



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Section A

Question Sections Indicative material Mark

1 (a) MMO

Decisionmaking

at least five different concentrations of bile salts ;could include 0%control (water) included ;dilutions agree with concentrations chosen ; [3]

(b) MMODecisionmaking

0% / water ;use boiled lipase ; [2]

(c) (i) MMO

Decision

making

idea of found end point when pink colour just no longer visible ;indicates when pH decreases to certain level ;

as fatty acids neutralise sodium carbonate / AW ; [3]

(ii) MMOCollection

temperature within range 50 ± 2 °C at every one of at leastthree readings ; [1]

MMOCollection

Presentationof dataRDO

at least five results obtained and recorded in seconds ; times vary across tubes so that lower concentrationsgenerally have longer times ; monotonic sequence of times vs. concentration ; replicates and means included ;

data recorded as a single table ; table includes columns for raw data (bile salts concentration,time taken) and calculated values (rate) ;

appropriate column headings with units in columnheadings ; e.g. bile salts concentration (%), time taken (s), rate (s –1 )independent variable (bile salts concentration) in left handcolumn ; results recorded to same degree of precision within eachcolumn ; [7 max]

(d)

Display of calculationand reasoning

rates calculated and given to appropriate significant figures ; [1]



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MMODecisionmaking

accept three separate decisions even if not justified

use of tube without phenolphthalein as colour comparator ;to identify end point ;

ref to including bile salts in colour comparator ;

as bile salts give colour to milk ;

use replicates ;to check on reliability / repeatability ;

R accuracy / precision

AVP ;; e.g. when to start timer [max 3]

(e) Presentationof data

Graph

line graph, bile salts concentration on horizontal axis ;ecf if time plotted, not rate

axes scaled correctly using at least half the graph paper ;axes titles and units – rate ( ecf from the table) andconcentration ;points plotted accurately ;appropriate line that is not extrapolated beyond highestconcentration ;if rate plotted line starts at the origin

R if broken axis A not at origin if time plotted [5]

(f) Analysis of data andconclusions

Description of patterns andtrends

increase in, rate / activity, with increase in concentration of bile salts ; A ref to decrease in time as ecf comparative data quote ; % bile salts and rate/time at twodifferent concentrations

ref to shape, e.g. straight line / exponential / plateau ;ref to anomalous result(s) ; A ‘no anomalous results’ [max 3]

Analysis of data andconclusions

Makingconclusionsdrawing ontheoreticalknowledgeandunderstanding

bile salts emulsify fats ;bile salts promote formation of micelles ;ref to hydrophilic and hydrophobic ends of each molecule ;

increase surface area of, globules / AW ;

effectively increase substrate concentration ;lipase can only act on the surface of globules ;not water soluble ;hydrolysis / breakage, of ester bonds ;release of fatty acids (and glycerol) ;higher concentration of bile salts results in, moreemulsification / higher substrate concentration ;

AVP ; [max 7]



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(g) Evaluation of procedures and data

Identifying limitations and sources of error

Suggesting improvements

reliability only one sample per concentration /no repeats / not enough repeats /

should have been repeated ;

ref to at least three samples, mean /standard deviation / standard error ;

end point / timing

end point difficult to judge ;so that end point may not have beenthe same in each case ;

stated problem with timing ; note that stopwatch should be started beforemixing

e.g. times all overestimates as startedstop watch before adding lipaserates therefore underestimates ;

use colour standard ; R colorimeter

ref to improved timing method ; R havesomeone else to start the stopwatch

way to slow down the reaction e.g. lower temperature / more milk ;

set up separately / staggered start ;

indicator ref to drops of phenolphthalein beinginaccurate / AW ;use set volume of phenolphthalein ;colour changes over a range of pH ;

use, pH meter / pH probe and data logger /more sensitive indicator ;

record time to reach constant pH ;

precision in preparation

stated problem with syringe(s) ;A air bubbles / precision explainedR liquid in nozzle

ref to, uncertainty / percentage error ;

use, graduated pipette(s) / burette /micropipette ;

temperature problem with maintaining constanttemperature ;data quote from (c) (ii) ;rate of reaction / activity, depends ontemperature ;

use thermostatically-controlled water bath ;

results ref to anomalous results ;difficult to identify line of best fit / AW ;ref to, range / error, bars ;not enough intermediateconcentrations to determine trend ;not wide enough range of concentrations ;

ref to discard / repeat ;

use SD / SE / 95%CI as error bars ;

stated intermediate concentrations ;

use concentrations of bile salts > 5%

[10]

[Total: 45]







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(e) Analysis of data andconclusions

Interpretationof data and

observations

A – presynaptic (neurone) ;B – postsynaptic (neurone) ;accept sensory and motor / interneurone

synaptic vesicles in A ;contain neurotransmitter ;

impulses only travel in one direction across synapses / AW ;synaptic, gap / cleft ;

mitochondria to provide energy ; AVP ; [max 5]

[Total: 35]



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