Biology Higher and Standard level

Biology

Higher and Standard level

Specimen papers 1A, 1B and 2

For first examinations in 2025

CONTENTS

Biology higher level paper 1A specimen question paper

Biology higher level paper 1A specimen markscheme

Biology higher level paper 1B specimen question paper

Biology higher level paper 1B specimen markscheme

Biology higher level paper 2 specimen question paper

Biology higher level paper 2 specimen markscheme

Biology standard level paper 1A specimen question paper

Biology standard level paper 1A specimen markscheme

Biology standard level paper 1B specimen question paper

Biology standard level paper 1B specimen markscheme

Biology standard level paper 2 specimen question paper

Biology standard level paper 2 specimen markscheme

SPEC/4/BIOLO/HPM/ENG/TZ0/XX

0000 – 6001

Specimen Paper

2 hours [Paper 1A and Paper 1B]

Biology

Higher level

Paper 1A

Instructions to candidates

y Do not open this examination paper until instructed to do so.

y Answer all questions.

y For each question, choose the answer you consider to be the best and indicate your choice on

the answer sheet provided.

y A calculator is required for this paper.

y The maximum mark for paper 1A is [40 marks].

y The maximum mark for paper 1A and paper 1B is [75 marks].

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1. What allows the movement of water under tension in the xylem?

A. Adhesion of water molecules to dissolved mineral salts

B. Cohesion of water molecules due to hydrogen bonding

C. Adhesion between water molecules due to uneven sharing of charges

D. Cohesion between water molecules and other polar substances

2. The diagram shows the elements present in two organic molecules, W and X.

Which molecules could W and X be?

Nitrogen

Hydrogen

Carbon

Oxygen

Phosphorus

Molecule W Molecule X

A. monosaccharide amino acid

B. nucleic acid triglyceride

C. phospholipid protein

D. triglyceride fatty acid

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Turn over

3. Which property of DNA explains how genetic information can be replicated accurately?

A. Complementary base pairing

B. The double helical shape

C. 5’ – 3’ bonding in the sugar-phosphate backbone

D. The ability of DNA to bind to histones

4. Which molecules are produced during the hydrolysis of a triglyceride molecule?

A. Water and glycerol

B. Fatty acids and glycerol

C. Water and fatty acids

D. Water and lipids

5. The structure of monomers affects the structure and function of the polymers they form. Which row

describes the structural features of polysaccharides made from alpha-glucose and beta-glucose?

Monomer Polymer Shape of polymer

A.

alpha-glucose

beta-glucose

starch

cellulose

unbranched, straight

branched, helical

B.

alpha-glucose

beta-glucose

starch

cellulose

branched, helical

unbranched, straight

C.

alpha-glucose

beta-glucose

cellulose

starch

branched, helical

unbranched, straight

D.

alpha-glucose

beta-glucose

cellulose

starch

unbranched, straight

branched, helical

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6. The diagram shows the structure of insulin.

Gly Ile Val Glu Gln

Cys Cys

Ala Ser Val

Cys

Ser

Leu

Tyr Gln

Leu

Glu

Asn

Tyr

Cys

Asn

Phe

Val

Asn

Gln His

Leu

Cys

Gly Ser

His

Leu

Val

Glu

Ala

Leu

Tyr

Leu

Val

Cys

Gly Glu

Arg Gly

Phe

Tyr Thr Pro

Lys

Ala

S

S S

S

From the diagram, what can be concluded about the structure of insulin?

A. It is composed of two polypeptide chains stabilized by disulfide bonds.

B. It is a simple protein composed of one continuous polypeptide chain.

C. It is a fibrous protein.

D. Its molecules do not display quaternary structure.

7. The diagram represents the metabolic pathway that converts threonine to isoleucine.

O

OH

CH

C

O

–

+

H N

3

CH

3

O

C

O

–

+

H N

3

CH

3

CH

3

CH

2

CH

Threonine

deaminase

Threonine

Isoleucine

E

2

E

3

E

4

E

5

What is a feature of this pathway?

A. The pathway is regulated by positive feedback.

B. Isoleucine is a competitive inhibitor.

C. Isoleucine is a substrate for threonine deaminase.

D. Isoleucine is an allosteric inhibitor.

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Turn over

8. What is a common feature of enzymes?

A. They all react with substrates.

B. They all decrease the rate of reaction.

C. They are all secreted from cells in vesicles.

D. They all bind to the active site of their substrate.

9. What is a reason that Taq polymerase is a suitable enzyme for use in the polymerase chain

reaction (PCR)?

A. It can work at a wide range of pH.

B. It works at higher temperatures than most enzymes.

C. It can separate two strands of DNA.

D. It allows DNA to be replicated without the use of primers.

10. Which are examples of non-coding DNA?

A. Dominant and recessive alleles

B. Promoters and telomeres

C. Oncogenes and tumour suppressor genes

D. Introns and exons

11. The table shows the mRNA codons for three amino acids.

Valine Threonine Proline

GUU ACU CCU

GCC ACC CCC

GCA ACA CGA

GCG ACG CCG

Which substitution mutation of a base triplet on a DNA strand will lead to the same polypeptide

being formed at translation?

A. TGA to TCA

B. CGT to CTA

C. CAA to CGA

D. GCT to GGA

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12. Which is a feature of phloem sieve tube cells?

A. Numerous chloroplasts

B. No nucleus

C. Lignified walls

D. No cytoplasm

13. How does the Miller-Urey experiment contribute to an explanation of the origin of life?

A. It shows how phospholipids form protocells in specific laboratory conditions.

B. It explains how organic molecules arise from inorganic ones under certain

environmental conditions.

C. It explains the synthesis of RNA, recreating deep sea vent conditions in the laboratory.

D. It shows how the last universal common ancestor (LUCA) evolved from vesicles.

14. The image shows a group of enveloped viral particles.

X

What is the most likely composition of the structure labelled X?

A. Membrane derived from the host cell

B. Viral DNA

C. Viral cell walls

D. Viral enzymes

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Turn over

15. The diagram shows the movement of ions that can occur across the membrane of a neuron.

Outside

Na

3Na

2K

K

2K

3Na

Inside

ATP

ADP + Pi

+

From the diagram, what can be deduced about the movement of sodium ions?

A. They are actively pumped out and some re-enter by facilitated diffusion.

B. They are actively pumped out and some re-enter by simple diffusion.

C. They diffuse out of the cell along with potassium ions.

D. There is a net movement of sodium ions into the cell.

16. In multicellular animals, embryonic stem cells have the ability to differentiate into a range of cells

with different functions. What is the term used to describe cells with this property?

A. Pluripotent

B. Multipotent

C. Totipotent

D. Specialized

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17. The image shows a glycoprotein embedded in the phospholipid bilayer of a cell membrane.

Which feature of this glycoprotein can be deduced from the image?

A. It is a peripheral protein that allows attachment to neighbouring cells.

B. It is a channel protein that provides hydrophilic channels for carbohydrate transport.

C. It is an integral protein that may be involved in cell recognition.

D. It is a transport protein that increases the permeability of the membrane to glucose.

18. What activates a tyrosine kinase pathway in a cell following binding to a receptor?

A. Odorant molecules

B. Epinephrine

C. Taste molecules

D. Insulin

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Turn over

19. The graph shows the effect of methylation on the expression of MT1E, a gene involved in the

control of prostate cancer development. Patients with a reduced expression of this gene are more

likely to develop prostate cancer.

0

2

-2

-4

-6

4

6

8

MT1E

Relative expression

Methylated

Unmethylated

Key:

What are effects of MT1E methylation?

A. It reduces transcription of MT1E, increasing the risk of prostate cancer.

B. It increases translation of MT1E, reducing the risk of prostate cancer.

C. It reduces replication of MT1E, reducing the risk of prostate cancer.

D. It increases the chances of mutation in proto-oncogenes, increasing the risk of

prostate cancer.

20. What is a function of histones?

A. Supercoiling of DNA during binary fission in prokaryotes

B. Synthesis of proteins

C. Formation of microtubules during mitosis

D. Condensation of DNA

21. What is the cause of positive phototropism?

A. Increased concentration of auxin on the side of the stem closest to the light

B. Degradation of auxin on the side of the stem closest to the light

C. Increased concentration of auxin on the side of the stem furthest from the light

D. Degradation of auxin on the side of the stem furthest from the light

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22. Which role does positive feedback play in fruit ripening?

A. The production of ethylene leads to fruit ripening, which stops the production of ethylene.

B. The production of ethylene leads to fruit ripening, which causes more ethylene production.

C. The production of RuBP leads to fruit ripening, which stops the production of RuBP.

D. The production of RuBP leads to fruit ripening, which causes more RuBP production.

23. The karyogram shows a chromosome abnormality in a human female.

1 2 3 4 5

6 7 8 9 10 11 12

13 14 15 17 1816

23

19 20 21 22

What could cause the abnormality to arise?

A. Failure of homologous chromosomes to separate during gamete formation

B. Chromosome 21 replicating after fertilization

C. A gene mutation of the mother’s chromosomes during cell division

D. Polyploidy occurring during anaphase I of meiosis in both parents

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Turn over

24. The image shows an organism from the phylum Arthropoda.

A dichotomous key was used to classify this arthropod. To which group does it belong?

Dichotomous key:

1. antennae pointing forwards . . . . . . . . . . . . . . . . . . . . . . . . . . go to 2

antennae pointing sideways . . . . . . . . . . . . . . . . . . . . . . . . . . go to 3

2. branched antennae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.

unbranched antennae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.

3. one pair of legs in each body segment . . . . . . . . . . . . . . . . . . . . . . C.

 twopairsoflegsineachbodysegment .....................D.

25. Scientists studied differences in the base sequences of a gene found in five animal species

in order to determine their evolutionary relationships. A cladogram was constructed based on

these differences.

Species V

Species W

Species X

Species Y

Species Z

What can be deduced from the cladogram?

A. There are only three clades shown.

B. Morphological differences between V and Z increase with time.

C. There is only one difference between the amino acid sequences of X and Y.

D. Y and Z have a more recent common ancestor than W and X.

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26. What is always a consequence of the evaporation of water from mesophyll cells in leaves of a

healthy plant?

A. Plasmolysis occurs in mesophyll cells.

B. Photosynthesis stops.

C. Stomata close to reduce transpiration.

D. Water moves up the stem in the xylem.

27. The diagram shows where the exchange of substances between blood and tissue fluid occurs in a

capillary bed.

Direction of blood flow

Region X

Capillaries

Cells

Tissue fluid

What explains the movement of solutes between blood and tissue fluid at region X?

A. Blood plasma has a higher concentration of solutes than tissue fluid.

B. Tissue fluid has a more negative water potential than blood plasma.

C. Hydrostatic pressure is higher in blood than in tissue fluid.

D. The permeability of capillary walls is highest at region X.

28. How is involuntary peristalsis in the intestine directly controlled in humans?

A. By the endocrine system

B. By the central nervous system (CNS)

C. By the sympathetic nervous system

D. By the enteric nervous system (ENS)

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Turn over

29. Which cell is a component of the innate immune system?

A. T lymphocyte

B. Phagocyte

C. B lymphocyte

D. B memory cell

30. How would the body respond to a rise above normal body temperature?

Type of feedback Response

A. positive increased secretion from sweat glands

B. positive spread limbs to increase surface area

C. negative vasodilation of skin blood vessels

D. negative shivering

31. Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X

chromosome. An X chromosome will be present in only half of the male gametes. A recessive

mutation is induced by radiation in the X chromosome of a male. In which generation will the effect

of this radiation appear?

A. F1 females

B. F1 males

C. F2 females

D. F2 males

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32. The number of hummingbird species in South America has increased continuously as shown in

the graph.

Z

0

50

100

150

20 15 5 010

Time before present / 10

6

years

Number of species

What could be a reason for the increase in the number of hummingbird species during the period

labelled Z?

A. Food and shelter are plentiful and there is little intraspecific competition.

B. Closely related species can coexist indefinitely in the same niche.

C. A wide range of unoccupied potential niches exists, leading to adaptive radiation.

D. There are few limiting factors on the population size of hummingbirds.

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Turn over

33. Polyploidy has been a cause of rapid speciation in some plant genera, such as Helianthus. Which

observation is evidence that speciation has occurred?

A. A polyploid plant reproduces asexually.

B. A polyploid plant produces male and female gametes.

C. Fertile offspring are produced when a polyploid plant crosses with a diploid plant.

D. Fertilization can occur between polyploid individuals.

34. Yersinia pestis is a bacterium that caused an outbreak of bubonic plague in the 14th century. It

normally produces ATP in the presence of oxygen but can still produce ATP if oxygen is absent.

Which term describes this characteristic?

A. Facultative respiration

B. Facultative anaerobe

C. Obligate anaerobe

D. Obligate aerobe

35. What are common features of holozoic nutrition and saprotrophic nutrition?

Mostly fungi and bacteria

Secretion of enzymes to

digest food

Ingestion of food

particles

A.

✓ ✓ ✗

B.

✓ ✗ ✓

C.

✗ ✓ ✗

D.

✗ ✗ ✗

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36. The preferred temperature ranges for three species of trout that are found together in freshwater

lakes in North America are shown.

Lake trout

(Salvelinus namaycush)

6–13 °C 12–18 °C 11–23 °C

Brown trout

(Salmo trutta)

Rainbow trout

(Oncorhynchus mykiss)

What can be deduced from this information?

I. The niches for all three species overlap.

II. Brown trout have the most specialized niche.

III. Lake trout can avoid competition by living in colder water.

A. I only

B. II only

C. I and III only

D. I, II and III

37. Which is a density-independent limiting factor for a kangaroo?

A. A forest fire

B. Predation

C. Climate change

D. Eutrophication

38. Black walnut (Juglans nigra) secretes the chemical juglone into the soil surrounding its roots.

Juglone inhibits cell respiration in other species of plants. What does this example illustrate?

A. Mutualism

B. Intraspecific competition

C. Allelopathy

D. Parasitism

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39. What assumption is made when using the Hardy-Weinberg equation for calculating changes in

allele frequencies in a population?

A. The population size is large.

B. Natural selection is taking place.

C. Mutations are occurring in the population.

D. There is variation among the phenotypes of the population.

40. Domestic dogs (Canis familiaris) have evolved from grey wolves (Canis lupus). Evidence suggests

that the domestication of dogs first occurred around 30 000 years ago. Which best describes the

evolution giving rise to the domestic dog?

A. The wolf produced offspring in large numbers which underwent natural selection.

B. Variations in the wolf population that resembled modern dogs favoured wolf survival.

C. Wolves showing favourable traits were selected for breeding.

D. Dogs were better suited to changes in the natural environment than wolves.

Disclaimer:

Content used in IB assessments is taken from authentic, third-party sources. The views expressed within them belong to their

individualauthorsand/orpublishersanddonotnecessarilyreecttheviewsoftheIB.

References:

14. Murphy,F.andWhiteld,S.,n.d.Herpes virus image under the microscope. [image online] Available at: <https://

www.bbvaopenmind.com/en/science/research/virus-inspired-science/> [Accessed 3 July 2020].

17. University of Delaware, n.d. [Model 2]. [image online] Available at: <http://www1.udel.edu/chem/white/C527/

TransmembraneProteins.pdf> [Accessed 28 July 2020].

19. Demidenko, R., Daniunaite, K., Bakavicius, D., Sabaliauskaite, R., Skeberdyte, A., Petroska, D., Laurinavicius, A.,

Jankevicius, F., Lazutka, J.R. and Jarmalaite, S., 2017. Decreased expression of MT1E is a potential biomarker of

prostate cancer progression. Oncotarget, 8, pp.61709–61718.

23. Hill, M.A., 2020. Embryology Trisomy 21. [image online] Available at: <https://embryology.med.unsw.edu.au/

embryology/index.php/Trisomy_21> [Accessed 28 July 2020].

24. Gschmeissner, S., n.d. Centipede underside, SEM. [image online] Available at: <https://www.sciencephoto.com/

media/374228/view/centipede-underside-sem> [Accessed 15 July 2020].

32. McGuire, J.A., Witt, C.C., Remsen Jr., J.V., Corl, A., Rabosky, D.L., Altshuler, D.L. and Dudley, R., 2014.

MolecularPhylogeneticsandtheDiversicationofHummingbirds.Current Biology, 24(8), pp.910–916.

36. Mepps.com Staff, n.d. Favored temperature range. [image online] Available at: <https://www.mepps.com/mepps-

tactics/article/how-water-temperature-and-oxygen-affect-shing/418>[Accessed8December2020].

SPEC/4/BIOLO/HPM/ENG/TZ0/XX/M

2 pages

Markscheme

Specimen paper

Biology

Higher level

Paper 1A

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1. B 16. A 31. D 46. –

2. C 17. C 32. C 47. –

3. A 18. D 33. D 48. –

4. B 19. A 34. B 49. –

5. B 20. D 35. C 50. –

6. A 21. C 36. C 51. –

7. D 22. B 37. A 52. –

8. A 23. A 38. C 53. –

9. B 24. C 39. A 54. –

10. B 25. D 40. C 55. –

11. D 26. D 41. – 56. –

12. B 27. C 42. – 57. –

13. B 28. D 43. – 58. –

14. A 29. B 44. – 59. –

15. A 30. C 45. – 60. –

Candidate session number

SPEC/4/BIOLO/HP1/ENG/TZ0/XX

0000 – 6002

Specimen Paper

2 hours [Paper 1A and Paper 1B]

Biology

Higher level

Paper 1B

Instructions to candidates

y Write your session number in the boxes above.

y Do not open this examination paper until instructed to do so.

y Answer all questions.

y Answers must be written within the answer boxes provided.

y A calculator is required for this paper.

y The maximum mark for paper 1B is [35 marks].

y The maximum mark for paper 1A and paper 1B is [75 marks].

12EP01

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Answer all questions. Answers must be written within the answer boxes provided.

1. Beetroot (Beta vulgaris) tissue was cut into pieces of similar size and shape. The masses

of the individual pieces were measured. Three pieces of beetroot were placed in each of six

sucrose solutions of different concentrations.

After three hours, the pieces of beetroot were removed from the solutions, surface dried

and their masses recorded. The mean percentage change in mass of the beetroot in each

solution was calculated and displayed in the graph.

30.0

20.0

10.0

0

-10.0

0.2 0.4 0.6 0.8 1.0

Concentration of sucrose

solution / mol dm

-3

Change in mass / %

-20.0

-30.0

Concentration of sucrose solution / mol dm

–3

Change in mass / %

0.0 28.0

0.2 5.0

0.4 –9.0

0.6 –19.0

0.8 –28.0

1.0 –30.0

(This question continues on the following page)

12EP02

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Turn over

(Question 1 continued)

(a) (i) State the independent variable in this experiment. [1]

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

(ii) Outline the reason that beetroot pieces of similar size and shape are needed in

order to obtain reliable results. [2]

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

(b) Suggest a reason for expressing the changes in mass as percentages. [1]

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

(c) Explain the effect of the 0.0 mol dm

–3

sucrose solution on the water potential of the

beetroot pieces. [3]

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

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2. In an experiment investigating the secretion of proteins by gland cells, researchers supplied

radioactively labelled amino acids to a sample of pancreas cells that secrete digestive enzymes.

The graph shows the relative abundance of radioactively labelled proteins in three different

organelles during the period after the cells were exposed to the radioactively labelled amino acids.

0

10

20

30

40

50

60

70

80

90

0 7 37 57 11717

Incubation time after exposure to

radioactively labelled amino acids / minutes

Administration of radioactively labelled amino acids

Location of

radioactively

labelled

proteins / %

X

Golgi apparatus

Key:

Secretory

vesicles

(a) Suggest a reason for using amino acids that were radioactively labelled. [1]

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

(b) (i) X is a membrane-bound organelle. Identify X. [1]

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

(This question continues on the following page)

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Turn over

(Question 2 continued)

(ii) Explain the role of ribosomes located on X. [2]

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

(iii) Outline the evidence provided by the data that molecules containing radioactively

labelled amino acids move from X to the Golgi apparatus. [1]

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

(c) Gland cells that secrete large volumes of fluid typically have many aquaporins in their

plasma membranes. Suggest a reason for this. [1]

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

(d) After synthesis, globular proteins assume their final tertiary structure. Explain the

relationship between the sequence of amino acids and the tertiary structure of

globular proteins. [4]

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

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3. Rhizobium is a genus of bacteria that live in the roots of flowering plants in the family

Fabaceae. The plants form nodules surrounding a population of Rhizobium. The bacteria

fix nitrogen from the air into nitrogen compounds, especially ammonium, which are made

available to the host plant.

The graph shows changes in the diameter of root nodules for plants grown in a nitrate-

enriched environment and a control group grown in a nitrate-free environment.

10 15 20

0

2

4

6

Time / days

nitrate-enriched environmentcontrol

Mean root nodule

diameter / mm

Key:

(a) Outline the concept of mutualism with respect to this example. [2]

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

(This question continues on the following page)

12EP06

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Turn over

(Question 3 continued)

(b) Distinguish between the cells in Rhizobium and Fabaceae. [2]

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

(c) Suggest a reason that mutualistic relationships often involve species from distantly

related groups of organisms. [1]

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

(d) Analyse the results of the experiment. [3]

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

12EP07

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SPEC/4/BIOLO/HP1/ENG/TZ0/XX

4. Before the structure of DNA was known, researchers analysed the ratios of the DNA bases

found in cells of different species. The tetranucleotide hypothesis, formulated in 1910,

proposed that DNA was made up of equal amounts of adenine, guanine, cytosine and thymine.

This hypothesis was revised following the publication of Chargaff’s paper in 1951, from which

data is shown in the table.

Ratios

Source A:G T:C A:T G:C purines:pyrimidines

Ox 1.29 1.43 1.04 1.00 1.10

Human 1.56 1.75 1.00 1.00 1.00

Chicken 1.45 1.29 1.06 0.91 0.99

Salmon 1.43 1.43 1.02 1.02 1.02

Wheat 1.22 1.18 1.00 0.97 0.99

Yeast 1.67 1.92 1.03 1.20 1.00

(a) State one example of a purine. [1]

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

(b) Explain how these results falsify the tetranucleotide hypothesis. [2]

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

(c) Using these data, evaluate the hypothesis that the ratio of purines to pyrimidines in

DNA in all organisms is 1.00. [1]

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

(This question continues on the following page)

12EP08

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SPEC/4/BIOLO/HP1/ENG/TZ0/XX

(Question 4 continued)

(d) Discuss how these results are consistent with the existence of a last universal common

ancestor (LUCA). [2]

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

(e) Discuss the evidence of the relationship between genome size and the complexity

of organisms. [4]

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

12EP09

Disclaimer:

Content used in IB assessments is taken from authentic, third-party sources. The views expressed within them belong to their

individual authors and/or publishers and do not necessarily reect the views of the IB.

References:

2. Jamieson, J.D. and Palade, G.E., 1967. Intracellular transport of secretory proteins in the pancreatic exocrine cell.

Journal of Cell Biology, 34, pp.597–615.

3. image

Ninjatacoshell, 2009. Medicago italica root nodules 2. [image online] Available at: <https://commons.wikimedia.

org/wiki/File:Medicago_italica_root_nodules_2.JPG> [Accessed 16 July 2020].

graph

Fujikake, H., Yamazaki, A., Ohtake, N., Sueyoshi, K., Matsuhashi, S., Ito, T., Mizuniwa, C., Kume, T., Hashimoto,

S., Ishioka, N.-S., Watanabe, S., Osa, A., Sekine, T., Uchida, H., Tsuji, A. and Ohyama, T., 2003. [Soybean root

nodule growth]. Journal of Experimental Botany, 54(386), pp.1379–1388. reference redacted.

4. Chargaff, E., 1951. Structure and function of nucleic acids as cell constituents. Federation proceedings, 10(3),

pp.654–659.

12EP10

Please do not write on this page.

Answers written on this page

will not be marked.

12EP11

Please do not write on this page.

Answers written on this page

will not be marked.

12EP12

SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

10 pages

Markscheme

Specimen paper

Biology

Higher level

Paper 1B

– 2 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

This markscheme is confidential and for the exclusive use of

examiners in this examination session.

It is the property of the International Baccalaureate and must not

be reproduced or distributed to any other person without the

authorization of the IB Global Centre, Cardiff.

– 3 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

The following are the annotations available to use when marking responses.

Annotation

Explanation

Shortcut

Correct point (automatically awards 1 mark when stamped)

These are annotations which can be used to show which marking point was

used to award a mark. It is easier to use these than to pick up the tick stamp

and then the text box (they each automatically award 1 mark when stamped)

Quality marks awarded for clarity and structure (these each automatically

award 1 mark when stamped)

Pointer (use when you want to delete an annotation or change colour)

Benefit of the doubt

Error carried forward

Irrelevant, a significant amount of material that does not answer the question

Contradiction

Omission/incomplete

Too vague

No working shown

Unclear

– 4 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

Annotation

Explanation

Shortcut

This is a dynamic annotation; it can be used to surround work

This is a dynamic, vertical wavy line that can be expanded (for instance, to highlight a section of

irrelevant work)

This is a dynamic, horizontal wavy line that can be expanded (for instance, to highlight a section of

irrelevant work)

Valid part (to be used when more than one element is required to gain the mark

eg: drawings)

Same as

Or words to that effect

Advantage / pro (to identify elements in an unclear discussion when pairs are required)

Disadvantage / con (to identify elements in an unclear discussion when pairs are required)

Difference (to identify elements in an unclear comparison)

Similarity (to identify elements in an unclear comparison)

Highlight, stamp and drag out to highlight an area of the script

Text box used for additional marking comments. It can be linked to a specific tick if that is appropriate

Seen; to be stamped on parts of a question or option which have been left blank

0

Zero; to be used when a question part is not worthy of credit. Awards zero for the question part

You must make sure you have looked at all pages. Please put the annotation on any blank page, to indicate that you have seen it.

– 5 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

General Marking Instructions

1. Follow the markscheme provided, award only whole marks and mark only in RED.

2. Make sure that the question you are about to mark is highlighted in the mark panel on the right-hand side of the screen.

3. Where a mark is awarded, a tick/check (

ü

) must be placed in the text at the precise point where it becomes clear that the candidate deserves

the mark. One tick to be shown for each mark awarded.

4. Sometimes, careful consideration is required to decide whether or not to award a mark. In these cases use RM™ Assessor annotations to support

your decision. You are encouraged to write comments where it helps clarity, especially for re-marking purposes. Use a text box for these additional

comments. It should be remembered that the script may be returned to the candidate.

5. Personal codes/notations are unacceptable.

6. Where an answer to a part question is worth no marks but the candidate has attempted the part question, use the “ZERO” annotation to

award zero marks. Where a candidate has not attempted the part question, use the “SEEN” annotation to show you have looked at the question.

RM™ Assessor will apply “NR” once you click complete.

7. If a candidate has attempted more than the required number of questions within a paper or section of a paper, mark all the answers. RM™ Assessor

will only award the highest mark or marks in line with the rubric.

8. Ensure that you have viewed every page including any additional sheets. Please ensure that you stamp “SEEN” on any additional pages that are

blank or where the candidate has crossed out his/her work.

9. Mark positively. Give candidates credit for what they have achieved and for what they have got correct, rather than penalizing them for what they

have got wrong. However, a mark should not be awarded where there is contradiction within an answer. Make a comment to this effect using a

text box or the “CON” stamp.

Assistant Examiners (AEs) will be contacted by their team leader (TL) through RM™ Assessor, by e-mail or telephone – if through RM™ Assessor

or by e-mail, please reply to confirm that you have downloaded the markscheme from IBIS. The purpose of this initial contact is to allow AEs to raise

any queries they have regarding the markscheme and its interpretation. AEs should contact their team leader through RM™ Assessor or by e-mail

at any time if they have any problems/queries regarding marking. For any queries regarding the use of RM™ Assessor, please contact

[email protected].

– 6 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

Subject Details: Biology HL Paper 1B Markscheme

Candidates are required to answer all questions in Paper 1B. Maximum total = 35 marks.

1. Each row in the “Question” column relates to the smallest subpart of the question.

2. The maximum mark for each question subpart is indicated in the “Total” column.

3. Each marking point in the “Answers” column is shown by means of a semicolon (;) at the end of the marking point.

4. A question subpart may have more marking points than the total allows. This will be indicated by “max” written after the mark in the “Total” column.

The related rubric, if necessary, will be outlined in the “Notes” column.

5. An alternative word is indicated in the “Answers” column by a slash (/). Either word can be accepted.

6. An alternative answer is indicated in the “Answers” column by “OR”. Either answer can be accepted.

7. An alternative markscheme is indicated in the “Answers” column under heading ALTERNATIVE 1 etc. Either alternative can be accepted.

8. Words inside brackets ( ) in the “Answers” column are not necessary to gain the mark.

9. Words that are underlined are essential for the mark.

10. The order of marking points does not have to be as in the “Answers” column, unless stated otherwise in the “Notes” column.

11. If the candidate’s answer has the same “meaning” or can be clearly interpreted as being of equivalent significance, detail and validity as

that in the “Answers” column then award the mark. Where this point is considered to be particularly relevant in a question it is emphasized

by OWTTE (or words to that effect) in the “Notes” column.

12. Remember that many candidates are writing in a second language. Effective communication is more important than grammatical accuracy.

13. Occasionally, a part of a question may require an answer that is required for subsequent marking points. If an error is made in the first marking point

then it should be penalized. However, if the incorrect answer is used correctly in subsequent marking points then follow through marks should be

awarded. When marking, indicate this by adding ECF (error carried forward) on the script.

14. Do not penalize candidates for errors in units or significant figures, unless it is specifically referred to in the “Notes” column.

– 7 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

Question

Answers

Notes

Total

1.

a

i

sucrose concentration

1

1.

a

ii

a. surface area;

b. varies with size (and shape)

OR

is a variable that affects osmosis

OR

would have affected the rate of osmosis/diffusion/movement of water into/out of the

tissue

OR

is a variable that needs to be controlled;

2

1.

b

initial masses/size of beetroot pieces were (slightly) different

OR

(relative changes in mass allow for a) fair comparison between pieces at different

concentrations with (slightly) different initial mass

1

1.

c

a. water enters the pieces of beetroot tissue;

b. (water enters) by osmosis;

c. pressure potential rises;

d. increasing the water potential (of the tissue);

e. until an equilibrium is reached / no further net movement of water;

f. the cells/tissue is/are turgid;

Accept ψ

s

for solute potential, ψ

p

for

pressure potential and ψ

for water

potential

.

3

– 8 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

Question

Answers

Notes

Total

2.

a

to track their movement / make them detectable

1

2.

b

i

rER

1

2.

b

ii

a. production of polypeptides/proteins/translation;

b. (production of proteins that are) insoluble/bound for export/for insertion in the

membrane;

2

2.

b

iii

high levels of radioactive amino acids appear first in X and later in Golgi

1

2.

c

a. water diffuses (through the plasma membrane) via aquaporins

OR

aquaporins allow water movement/secretion of fluid;

b. the rate of water transport by aquaporins is not very high (so many are required);

1

2.

d

a. the sequence of amino acids determines the tertiary structure of a protein;

b. tertiary structure is the folding of a polypeptide chain;

c. R groups of amino acids can be polar or non-polar;

d. the protein folds due to interaction with water/the aqueous environment;

e. (the protein folds due to) R group interactions;

f. example of R group interaction H/covalent/ionic bond/disulphide bridges;

4

– 9 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

Question

Answers

Notes

Total

3.

a

a. symbiotic relationship between two (unrelated) species

OR

to the benefit of both species;

b. bacteria get organic molecules AND plant gets nitrates;

2

3.

b

Rhizobium

Fabaceae

prokaryote

eukaryote;

circular DNA

linear DNA/chromosomes;

membrane bound organelles

absent

(accept a specific example)

membrane bound organelles

present;

(accept a specific example)

70S ribosomes

80S ribosomes;

small

large;

Some form of distinction MUST be

made.

e.g. ‘Rhizobium is small’ would be

insufficient. However, ‘smaller’ implies

a distinction

2

3.

c

distantly related species are more likely to have different characteristics/metabolisms

(so are more likely to provide for each others’ requirements)

1

3.

d

a. with additional nitrate, nodule size does not change/remains small;

b. without additional nitrate, nodule size increases (over time);

c. (suggests that) nitrate inhibits nodule growth;

d. when nitrate is scarce, nodule growth ensures nitrate supply from a larger bacterial

population;

e. (suggests that) mutualism is affected by environmental conditions;

3

– 10 – SPEC/4/BIOLO/HP1/ENG/TZ0/XX/M

Question

Answers

Notes

Total

4.

a

adenine/guanine

1

4.

b

a. the only ratios that are 1:1 are between A:T and C:G;

b. ratios of A:G and T:C are not 1:1 in any of the experimental organisms;

2

4.

c

overall ratio of purines:pyrimidines in all experimental organisms is 1:1 but yeast G:C ratios

represent an anomaly

1

4.

d

a. same four bases found across a variety of organisms;

b. pattern of ratios in the data is common / comparable across the organisms shown

OR

same complementary base pairing;

c. this suggests a common ancestor;

2

4.

e

Evidence for a relationship

a. eukaryote genomes (tend to be) larger than prokaryote genomes;

b. most DNA sequences in higher eukaryotic genomes are non-coding;

c. introns occur within eukaryotic genes;

Evidence against a relationship

d. no relationship between genome size and complexity within the eukaryotes;

e. example given; e.g. Amoeba proteus genome 100x larger than Homo sapiens genome;

4

Candidate session number

SPEC/4/BIOLO/HP2/ENG/TZ0/XX

0000 – 6003

27 pages

Specimen paper

2 hours 30 minutes

Biology

Higher level

Paper 2

Instructions to candidates

y Write your session number in the boxes above.

y Do not open this examination paper until instructed to do so.

y Section A: answer all questions.

y Section B: answer two questions.

y Answers must be written within the answer boxes provided.

y A calculator is required for this paper.

y The maximum mark for this examination paper is [80 marks].

28EP01

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0000 – 6003

28EP02

Section A

Answer all questions. Answers must be written within the answer boxes provided.

1. Squirrels (small mammals) have evolved several defences against predators. One method

adopted by California ground squirrels (Otospermophilus beecheyi) against rattlesnakes

(Crotalus oreganus) is to chew moulted rattlesnake skins, then lick their own fur so they take

on the scent of a rattlesnake.

An experiment tested the amount of time rattlesnakes spent studying and flicking their tongue

at filter papers soaked in different liquids. The snakes were released from chamber C into

foraging area F and observed for 30 minutes. There were two filter papers in area F. One

was a control paper soaked in distilled water. The other was soaked in one of three types of scent:

• squirrel scent

• a mixture of squirrel and rattlesnake scent

• rattlesnake scent.

C

F

(Position changed

at each trial)

Filter paper

(This question continues on the following page)

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SPEC/4/BIOLO/HP2/ENG/TZ0/XX

0000 – 6003

Turn over

28EP03

(Question 1 continued)

The graphs show the results of the experiment.

Scented paper

100

80

60

40

20

0

300

200

20

40

60

80

120

140

160

180

220

240

260

280

100

0

Water paper

Rattlesnake

only

Squirrel

only

Squirrel and

rattlesnake

Rattlesnake

only

Squirrel

only

Squirrel and

rattlesnake

Number of tongue flicks

Time studying filter paper / s

Key:

Type of scent

[Source: adapted from Clucas, B., Owings, D.H. and Rowe, M.P., 2008. Proc Biol Sci, 275(1636), pp.847–852.

https://doi.org/10.1098/rspb.2007.1421 redacted.]

(a) Calculate the difference in mean time spent studying squirrel-scented filter paper with

and without rattlesnake scent. [1]

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

(b) Outline the conclusions that can be drawn from the control results. [2]

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

(This question continues on the following page)

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28EP04

(Question 1 continued)

(c) The positions of the filter papers were changed for every trial. Outline one reason that

this was necessary. [1]

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

(d) Suggest a reason for rattlesnake scent lowering the chances of snake attacks

on squirrels. [1]

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

The researchers observed that when squirrels saw a rattlesnake, they waved their tail from

side-to-side so that they were more visible to the rattlesnakes (tail-flagging). The graph

shows the probability of a snake strike at different distances on tail-flagging and non-tail-

flagging squirrels.

Probability of strike

0

0.2

0.4

0.6

0.8

1.0

Distance / cm

5 10 15 20 25 30 35

Key:

Tail-flagging Non-tail-flagging

[Source: adapted from Barbour, M.A. and Clark, R.W., 2012. Proceedings of the Royal Society B, [e-journal]

https://doi.org/10.1098/rspb.2012.1112 redacted.]

(This question continues on the following page)

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Turn over

28EP05

(Question 1 continued)

(e) Compare and contrast the effect of tail-flagging with non-tail-flagging on strike probability. [2]

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

(f) Estimate the distance between squirrel and snake where there is a 50 % chance of a

strike at a tail-flagging squirrel. [1]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm

(g) Suggest a reason for the behaviour of snakes towards tail-flagging squirrels. [1]

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

(This question continues on the following page)

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SPEC/4/BIOLO/HP2/ENG/TZ0/XX

0000 – 6003

28EP06

(Question 1 continued)

When played a recording of the call of a red-tailed hawk (Buteo jamaicensis), foraging grey

squirrels (Sciurus carolinensis) in the wild became more vigilant and looked up from their

feeding more often. In the habitat of the squirrels, there are also sounds made by songbirds

which chatter to each other when there is no imminent threat. The charts show the response

of the squirrels in the 30 seconds immediately after the scare of hearing hawk sounds and

after 3 minutes, both with and without bird chatter playing.

No chatter Chatter No chatter Chatter

2.0 40

30

20

10

0

1.6

1.2

0.8

0.4

0

Response after 3 minutes

Response immediately after hearing hawk sounds

Number of lookups

in 30 seconds

Key:

Percentage of time spent

vigilant in 30 seconds / %

[Source: adapted from Lilly, M.V., Lucore, E.C. and Tarvin, K.A., 2019. PLoS ONE, 14(9): e0221279. https://doi.org/10.1371/

journal.pone.0221279. redacted.]

(h) State the conditions that led to the squirrels making the least mean number of lookups. [1]

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

(This question continues on the following page)

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SPEC/4/BIOLO/HP2/ENG/TZ0/XX

0000 – 6003

Turn over

28EP07

(Question 1 continued)

(i) Discuss briefly whether the sound of bird chatter represents a significant difference in

the vigilance of squirrels immediately after they have heard the sound of a hawk. [2]

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

(j) Using the data in the bar charts, analyse the effect of bird chatter on the behaviour

of squirrels. [3]

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

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SPEC/4/BIOLO/HP2/ENG/TZ0/XX

0000 – 6003

28EP08

2. The cytoplasm in eukaryotic cells is compartmentalized into membrane-bound organelles, as

seen in this electron micrograph of a plant cell.

(a) State two membrane-bound organelles that are common to both plant and animal cells. [2]

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

(b) (i) State one organelle that is found in a plant cell but not in an animal cell. [1]

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

(ii) State one organelle that is common to prokaryotic and eukaryotic cells. [1]

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

– 9 –

SPEC/4/BIOLO/HP2/ENG/TZ0/XX

0000 – 6003

Turn over

28EP09

3. The diagram shows the movement of auxin across the plasma membrane of a plant cell as

well as the position of auxin efflux carriers.

plasma membrane

auxin efﬂux carriers

auxin diffusion

(a) (i) Annotate the diagram, using an arrow, to show the direction of the net flow of

auxin through the cell. [1]

(ii) Identify an environmental factor that causes auxin efflux carriers to be

concentrated on one side of a cell. [1]

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

(b) (i) Predict the difference that could be seen using a microscope between plant

meristematic cells grown in media with and without auxin. [1]

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

(ii) State the hormone that interacts with auxin to integrate shoot growth. [1]

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4. Multicellular organisms need efficient exchange surfaces where molecules can move

between tissues and their surroundings. In mammals, gas exchange takes place in the

alveoli of the lungs.

capillary

alveolus

O

2

O

2

O

2

O

2

CO

2

CO

2

CO

2

CO

2

(a) Describe how the exchange of gases is brought about between the alveoli and the blood. [3]

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(Question 4 continued)

The images show the lungs of a human and the gills of a fish taken with a light microscope.

Alveoli in the lungs of a human Gills from a ﬁsh

(b) Describe two features common to the exchange surfaces of alveoli and gills. [2]

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(This question continues on page 13)

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(Question 4 continued)

The image shows cells from the wall of a proximal convoluted tubule of a rat kidney.

(c) Outline the visible features of these cells that adapt them for reabsorption of materials

from the glomerular filtrate. [2]

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5. Predation may result in a cyclic pattern of population changes, as for example in the case of

the lynx, Lynx canadensis (predator), and the hare, Lepus americanus (prey).

Year

1900 192518751850

0

40

80

120

160

12

9

6

3

0

Hare population / thousands

Lynx population / thousands

(a) The predator benefits from a predator-prey relationship by obtaining food. Suggest how

the hare population could benefit from this relationship. [1]

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(b) Outline the factors that contribute to the carrying capacity of prey organisms in a

predator-prey relationship. [2]

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(Question 5 continued)

(c) (i) Outline how the hare population size can be estimated using the

mark-release-recapture method. [2]

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(ii) State one assumption that is made when using this method to estimate

population size. [1]

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6. In humans, sex is determined genetically and human populations have approximately equal

numbers of males and females in each generation.

(a) Outline how sex is determined in humans. [2]

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

(b) Down syndrome is a genetic condition in which the cells of an individual contain three

copies of chromosome 21. Explain how non-disjunction leads to trisomy 21. [3]

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(Question 6 continued)

(c) The pedigree chart shows the inheritance of hemophilia in a human family over four

generations. Symbols represent the phenotype of the individual.

Key:

unaﬀected female

aﬀected female aﬀected male

unaﬀected male

I

II

III

IV

Explain the evidence from this chart that the condition may be sex-linked. [2]

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Please do not write on this page.

Answers written on this page

will not be marked.

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7. Two alleles of a gene influence body colour in Drosophila. The dominant allele B results in a

brown body colour while the recessive allele b codes for black body. In addition, two alleles of

another gene determine wing shape; normal wings (V) are dominant over vestigial wings (v).

In a dihybrid cross between a double heterozygote for body colour and wing shape (BbVv)

and a homozygous recessive for both traits (bbvv), the following offspring were obtained:

• brown body, normal wings: 29

• brown body, vestigial wings: 8

• black body, normal wings: 10

• black body, vestigial wings: 33

(a) Identify the recombinant phenotypes amongst the offspring. [1]

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

(b) Explain whether these results fit the predicted Mendelian ratios for this cross. [2]

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

(c) Outline the statistical test that could be used to provide more evidence of the type

of inheritance. [2]

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

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Section B

Answer two questions. One additional mark is available for the construction of your answers for each

question. Answers must be written within the answer boxes provided.

8. There are a number of advantages to specificity.

(a) Explain how vaccination can result in specific immunity to a viral disease. [7]

(b) Outline the role of membrane proteins in the movement of specific ions at specific

times in the transmission of nerve impulses. [4]

(c) Describe the barriers that exist to hybridization between species. [4]

9. Various mechanisms can lead to inhibition in biological systems. A toxin is a substance

capable of disrupting metabolic processes in organisms.

(a) Describe how toxins such as DDT might concentrate in the bodies of birds. [4]

(b) Toxins often act as inhibitors. Compare and contrast competitive and non-competitive

enzyme inhibition. [7]

(c) Rotenone is a naturally occurring toxin that blocks the electron transport chain in

insects and fish. Outline the consequences of exposure to a toxin like rotenone for

cell respiration. [4]

10. Changes in the form of biological molecules lead to changes in their function.

(a) Outline the role of UV radiation as a mutagen. [4]

(b) Explain how the function of DNA is linked to its molecular structure. [7]

(c) Describe the use of gene knockout technology in research. [4]

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

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

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

References:

2. Electron micrograph of cells from the leaf of a corn plant, n.d. [image online] Available at: <http://innolearn.weebly.

com/bio-cell.html> [Accessed 9 December 2020].

3a. Smith, R.S., 2008. The Role of Auxin Transport in Plant Patterning Mechanisms. PLoS Biol, 6(12): e323. https://doi.

org/10.1371/journal.pbio.0060323.

4b. Left image:

[Alveoli in the lungs of a human], n.d. [image online] Available at: <http://medcell.med.yale.edu/histology/respiratory_

system_lab.php#slides> [Accessed 9 December 2020].

Right image:

[salt water fish gills histology], n.d. [image online] Available at: <https://br.pinterest.com/pin/255297872605284426/>

[Accessed 9 December 2020].

4c. Transmission electron micrograph of the S1 segment of a rat proximal tubule, n.d. [image online] Available at:

<https://abdominalkey.com/anatomy-of-the-kidney/> [Accessed 9 December 2020].

5. Wani, n.d. [Hare and lynx populations]. [image online] Available at: <https://www.cleanpng.com/png-cat-snowshoe-

hare-eurasian-lynx-lotkavolterra-e-3913514/> [Accessed 29 July 2020].

28EP28

SPEC/4/BIOLO/HP2/ENG/TZ0/XX/M

20 pages

Markscheme

Specimen paper

Biology

Higher level

Paper 2

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This markscheme is confidential and for the exclusive use of

examiners in this examination session.

It is the property of the International Baccalaureate and must not

be reproduced or distributed to any other person without the

authorization of the IB Global Centre, Cardiff.

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The following are the annotations available to use when marking responses.

Annotation

Explanation

Shortcut

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These are annotations which can be used to show which marking point was

used to award a mark. It is easier to use these than to pick up the tick stamp

and then the text box (they each automatically award 1 mark when stamped)

Quality marks awarded for clarity and structure (these each automatically

award 1 mark when stamped)

Pointer (use when you want to delete an annotation or change colour)

Benefit of the doubt

Error carried forward

Irrelevant, a significant amount of material that does not answer the question

Contradiction

Omission/incomplete

Too vague

No working shown

Unclear

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Annotation

Explanation

Shortcut

This is a dynamic annotation; it can be used to surround work

This is a dynamic, vertical wavy line that can be expanded (for instance, to highlight a section of

irrelevant work)

This is a dynamic, horizontal wavy line that can be expanded (for instance, to highlight a section of

irrelevant work)

Valid part (to be used when more than one element is required to gain the mark

eg: drawings)

Same as

Or words to that effect

Advantage / pro (to identify elements in an unclear discussion when pairs are required)

Disadvantage / con (to identify elements in an unclear discussion when pairs are required)

Difference (to identify elements in an unclear comparison)

Similarity (to identify elements in an unclear comparison)

Highlight, stamp and drag out to highlight an area of the script

Text box used for additional marking comments. It can be linked to a specific tick if that is appropriate

Seen; to be stamped on parts of a question or option which have been left blank

0

Zero; to be used when a question part is not worthy of credit. Awards zero for the question part

You must make sure you have looked at all pages. Please put the annotation on any blank page, to indicate that you have seen it.

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General Marking Instructions

1. Follow the markscheme provided, award only whole marks and mark only in RED.

2. Make sure that the question you are about to mark is highlighted in the mark panel on the right-hand side of the screen.

3. Where a mark is awarded, a tick/check (

ü

) must be placed in the text at the precise point where it becomes clear that the candidate deserves

the mark. One tick to be shown for each mark awarded.

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your decision. You are encouraged to write comments where it helps clarity, especially for re-marking purposes. Use a text box for these additional

comments. It should be remembered that the script may be returned to the candidate.

5. Personal codes/notations are unacceptable.

6. Where an answer to a part question is worth no marks but the candidate has attempted the part question, use the “ZERO” annotation to

award zero marks. Where a candidate has not attempted the part question, use the “SEEN” annotation to show you have looked at the question.

RM™ Assessor will apply “NR” once you click complete.

7. If a candidate has attempted more than the required number of questions within a paper or section of a paper, mark all the answers. RM™ Assessor

will only award the highest mark or marks in line with the rubric.

8. Ensure that you have viewed every page including any additional sheets. Please ensure that you stamp “SEEN” on any additional pages that are

blank or where the candidate has crossed out his/her work.

9. Mark positively. Give candidates credit for what they have achieved and for what they have got correct, rather than penalizing them for what they

have got wrong. However, a mark should not be awarded where there is contradiction within an answer. Make a comment to this effect using a

text box or the “CON” stamp.

Assistant Examiners (AEs) will be contacted by their team leader (TL) through RM™ Assessor, by e-mail or telephone – if through RM™ Assessor

or by e-mail, please reply to confirm that you have downloaded the markscheme from IBIS. The purpose of this initial contact is to allow AEs to raise

any queries they have regarding the markscheme and its interpretation. AEs should contact their team leader through RM™ Assessor or by e-mail

at any time if they have any problems/queries regarding marking. For any queries regarding the use of RM™ Assessor, please contact

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Subject Details: Biology HL Paper 2 Markscheme

Candidates are required to answer all questions in Section A and two out of three questions in Section B. Maximum total = 80 marks.