Biology 2024 Oct/Nov | Past Paper

Q1

Short Answer 1 mark

The olive plant, Olea europaea, is grown in many parts of the world. The fruits of the plant (olives) and the oil that can be obtained from the fruits (olive oil), provide food for humans. Triglycerides are the main type of lipid in olive oil. They are synthesised in the olive plant from glycerol and fatty acids. Scientists can analyse samples of dif ferent olive oils to identify:

Q1(a)

MCQ 1 mark

Table 1.1 shows some details of the five most com mon fatty acids found in samples of olive oil produced by olive plants grown in different regions in Portugal. Table 1.1 Key C : D = number of carbon atoms : number of double bonds in the hydrocarbon chain X = missing detail fatty acid percentage of total fatty acid content C : D chemical structure oleic acid 55.0-83.0 18 : 1 CH (CH CH=CH(CH palmitic acid 7.5-20.0 16 : CH (CH linoleic acid 3.5-21.0 18 : CH (CH (CH=CH)CH (CH=CH)(CH stearic acid 0.5-5.0 18 : CH (CH palmitoleic acid 0.3-3.5 16 : CH (CH CH=CH(CH

C

D = number of carbon atoms : number of double bonds in the hydrocarbon chain

C

D chemical structure

Q1(i)

Short Answer 1 mark

Table 1.1 shows the C : D values for oleic acid. In Table 1.1, write the values for D for each of the four other fatty acids listed.

Q1(ii)

Short Answer 1 mark

In the first column of Table 1.1, draw a circle around each of the fatty acids that can be described as saturated.

Q1(iii)

Short Answer 1 mark

State the detail of chemical structure, represented by X, which is missing from Table 1.1.

Q1(iv)

Short Answer 1 mark

The analysis of the triglycerides present in the different samples of olive oil showed that: present in all olive oils. With reference to Table 1.1 and to the structure of triglycerides, suggest ex planations for these observations.

Q1(b)

Short Answer 1 mark

Glycerol is soluble in water. Triglycerides are insoluble in water . Explain why water is a good solvent for some substances such as glycerol, but is a poor solvent for substances such as triglycerides.

Q1(c)

Short Answer 1 mark

Phloem is the plant tissue responsible for the transport of organic substances, such as fatty acids, from one area of a plant to another . The tissue is composed of more than one type of cell. Name the type of cell that forms the transport vessels of phloem tissue. [Total: 8] * 0000800000003 *

Q2

Short Answer 1 mark

People who become infected with human immunodeficiency virus (HIV) are at risk of developing HIV/AIDs, particularly if antiretroviral therapy (ART) is not available.

Q2(a)

Short Answer 1 mark

In people infected with HIV , the use of ART also helps to reduce transmission of the virus to uninfected people. Outline two control methods, other than ART, that can be used to reduce the transmission of HIV. In people with HIV/AIDs, a serious lung disease known as pneumocystis pneumonia can result from infection by an opportunistic pathogen known as Pneumocystis jirovecii. Fig. 2.1 shows P . jirovecii cells in one stage of their life cycle, as seen using a light microscope at a magnification of × 600. 0.4 μm

Q2(b)

Short Answer 1 mark

Define magnification.

Q2(c)

Short Answer 1 mark

Fig. 2.1 shows that P . jirovecii is a unicellular organism. Although the cells of many species of bacteria are the same size as those of P . jirovecii, research concluded that the organism is a eukaryote and is not a bacterium. In 1988, analysis of ribosomal RNA (rRNA) resulted in P . jirovecii being classified as a fungus.

Q2(i)

Short Answer 1 mark

Studies of the structure of P . jirovecii have identified that the cell wall is made of Explain why this feature helped scientists to confirm that P . jirovecii is not a bacterium.

Q2(ii)

Short Answer 1 mark

Scientists have identified other features of the cell structure of P . jirovecii. Some of these are listed in Table 2.1. Complete each row of Table 2.1 so that the table shows: Table 2.1 structural feature of P. jirovecii function bacterial cells ribosomes protein synthesis smooth endoplasmic reticulum Golgi body modification of proteins and lipids aerobic respiration [3] * 0000800000005 *

Q2(d)

Short Answer 1 mark

P. jirovecii can adhere (attach) to squamous epithelial cells of the alveoli and to the network of fibrous proteins that support the alveolar wall, known as the extracellular matrix (ECM). Examples of proteins in the ECM are elastin and collagen. Adhesion (attachment) of P . jirovecii to alveolar epithelial cells and the ECM stimulates the growth of its population.

Q2(i)

Short Answer 1 mark

Cell surface glycoproteins known as gpA glycoproteins are essential in allowing P. jirovecii cells to adhere to alveolar epithelial cells and ECM proteins. Suggest how a gpA glycoprotein is able to adhere to alveolar epithelial cells and ECM proteins.

Q2(ii)

Short Answer 1 mark

One consequence of the pneumonia that results from P . jirovecii infection is a decrease in the quantity of oxygen that is delivered to body tissues. Explain why a severe P . jirovecii infection results in a decrease in the quantity of oxygen that is delivered to body tissues. With reference to the mechanism of action of caspofungin, explain how the drug may be useful to treat cases of pneumonia caused by P . jirovecii. [Total: 17] * 0000800000007 *

Q3

Short Answer 1 mark

During transcription, base pairing occurs between nucleotides. Fig. 3.1 is a diagram to show complementary base pairing between a DNA nucleotide and an RNA nucleotide. Only the base pair is shown in molecular detail.H H ribose

Q3(a)

Short Answer 1 mark

Explain why Fig. 3.1 does not include any phosphodiester bonds. Y ou may add labels and annotations to Fig. 3.1 if you wish. [Total: 5] * 0000800000008 *

Q4

Short Answer 1 mark

Adult stem cells are undif ferentiated cells that are found in most animal tissues. Adult stem cells can divide by mitosis throughout their lifespan to form identical stem cells

Q4(a)

Short Answer 1 mark

Mitosis is important for the repair of tissues. Explain what is meant by repair of tissues.

Q4(b)

Short Answer 1 mark

Uncontrolled cell division is a characteristic feature of tumour formation from a dif ferentiated cell. Describe other features of tumour formation from a fully dif ferentiated cell.

Q4(c)

Short Answer 1 mark

Telomeres prevent loss of genes. Adult stem cells have chromosomes with long telomeres. Explain why long telomeres are an advantage to cells that carry out many cell cycles. Haematopoietic stem cells (HSCs) are adult stem cells that are located in the bone m arrow of bones. HSCs have a role in the formation of blood cells. Fig. 4.1 is an outline summary showing the formation of some of the dif ferent types of blood cell that can be formed from HSCs. The first stage is the division of HSCs to produce progenitor cells. These cells are also able to divide by mitosis, but are not stem cells.Key = progenitor cells HSCs self- renewal HSCs immature red blood cells mature red blood cells platelets mature neutrophils monocytes immune response mature B-lymphocytes mature T-lymphocytes (T-helper and T-killer cells) immature B-lymphocytes immature T-lymphocytes immature neutrophils immature monocytes megakaryocytes CMP cells GMP cellsMEP cells Pre-T cellsPre-B cells CLP cells

Q4(d)

Short Answer 1 mark

With reference to Fig 4.1, explain why GMP cells, which are progenitor cells, cannot be described as haematopoietic stem cells (HSCs).

Q4(e)

Short Answer 1 mark

Fig. 4.1 shows that monocytes dif ferentiate into cell type X, which has a similar function to neutrophils. Name cell type X.

Q4(f)

Short Answer 1 mark

Cell type Y shown in Fig. 4.1 releases molecules with antigen binding sites. Name the molecules released by cell type Y. [Total: 13] * 0000800000011 *

Q5

Short Answer 1 mark

Malaria is an infectious disease caused by the protoctist, Plasmodium. As part of its lifecycle, Plasmodium infects human red blood cells. Researchers can compare haemoglobin from the red blood cells of a healthy person with haemoglobin from a person with malaria.

Q5(a)

Short Answer 1 mark

Throughout the world, most deaths from malaria are caused by P . vivax and P . falciparum. Name one other species of Plasmodium that causes malaria. Plasmodium

Q5(b)

Short Answer 1 mark

In the laboratory , oxygen at dif ferent partial pressures can be bubbled through a solution of haemoglobin to determine the percentage saturation of haemoglobin at each partial pressure. A graph constructed from the results is known as an oxygen dissociation curve. Fig. 5.1 is an oxygen dissociation curve for normal adult haemoglobin in humans. partial pressure of oxygen / kPa percentage saturation of haemoglobin with oxygen 0 2 4 6 8 10 12 14 Fig. 5.1 * 0000800000012 *

Q5(i)

Short Answer 1 mark

In the experiment used to obtain the results shown in Fig. 5.1, the temperature and pH were standardised. Explain what the researchers would consider when deciding which temperature and pH to use in the experiment.

Q5(ii)

Short Answer 1 mark

Using a dif ferent, more rapid technique, researchers compared the haemoglobin contained in red blood cells of a healthy person with the haemoglobin of a person with malaria who had been infected with P . vivax. By analysing the results, the researchers concluded that the oxygen dissociation curve of a person with malaria would be shifted to the right. With reference to Fig. 5.1, explain how a shift to the right of the ox ygen dissociation curve would af fect oxygen loading in the lungs, and unloading in respiring tissues, in a person with malaria.

Q5(c)

Short Answer 1 mark

A red blood cell that is infected with Plasmodium cannot carry out its function as ef fectively as a normal red blood cell. Describe how the size and structure of a red blood cell is related to its function, other than the fact that it contains a very large number of haemoglobin molecules. [Total: 10] * 0000800000014 *

Q6

Short Answer 1 mark

The transport of water from the soil solution to the xylem of roots occurs by the apoplast and symplast pathways. Mineral ions can be transported dissolved in water.

Q6(a)

Short Answer 1 mark

Describe the transport of water from the soil solution to the endodermis of roots by the apoplast pathway and explain why this pathway cannot continue at the endodermis.

Q6(b)

Short Answer 1 mark

Researchers investigated the mechanism of transport used for the uptake of potassium ions ) into root epidermal cells at different concentrations of K in the soil solution. Complete Table 6.1 to provide information about the two dif ferent transport mechanisms that were identified by the researchers. Table 6.1 net movement of membrane protein needed (yes or no) ATP used (yes or no) name of transport mechanism against the concentration gradient down the concentration gradient [3] [Total: 7] * 0000800000015 * The boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply official endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country , territory, or area or any of its authorities, or of the delimitation of its frontiers or boundaries. reasonable ef fort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE * 0000800000016 *

Board	CAIE
Level	A Level
Subject	Biology
Code	`9700`
Year	2024
Session	Oct/Nov
Q&A	35 questions

Biology 2024 · Oct/Nov

Questions & Model Answers 35 Qs

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