Free until 14 Nov · 80% off for the first 100 · Pricing

Biology 0610/43 May/June 2019: question paper and mark scheme

Cambridge IGCSE Biology (0610), Paper 4 Theory (Extended), May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

0610_s19_qp_430610_s19_ms_430610/43/M/J/19

Paper
Paper 4 Theory (Extended)
Total marks
80
Time
1 hour 15 minutes
Questions
6
Code on the paper
0610/43/M/J/19
0610_s19_ms_43.pdf

0610/43 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 113 marks

Question

Bacteria are classified in the Prokaryote kingdom. (a) State two features of animal cells that are not found in bacteria. 1 …… 2 …… [2] (b) The bacterium Bacillus megaterium was grown in the laboratory fermenter shown in Fig. 1.1. air lock water sterile air magnetic stirrer bacteria, source of nitrogen and glucose Fig. 1.1 (i) Explain why a source of nitrogen and glucose were added to the fermenter. nitrogen …… glucose …… [2] (ii) Suggest why it is important to stir the contents of the fermenter continuously. …… [3] (c) Samples were taken from the fermenter at frequent intervals and the number of live bacteria was determined. The results are shown in Fig. 1.2. 800 C 700 600 500 number of live bacteria D 400 / million per cm3 B 300 200 100 A 0 0 10 20 30 40 50 60 time / hours Fig. 1.2 Describe and explain what happens to the number of live bacteria shown in the stages labelled A, B, C and D in Fig. 1.2. …… [6] [Total: 13]

Mark scheme

1(a) nucleus ; 2 mitochondria ; (rough) endoplasmic reticulum ; AVP ; 1(b)(i) nitrogen 2 for making, amino acids / proteins ; glucose for respiration / source of energy ; 1(b)(ii) for even / uniform, distribution / concentration, of contents / AW ; 3 ensure more access of bacteria with, medium / AW ; prevent settling / clumping of, contents / AW ; so oxygen / glucose / nutrients, dissolves ; so that growth is not limited ; so bacteria can absorb substances (more easily) ; bacteria need oxygen for respiration ; 1(c) description: 6 at least two correctly named phases ; any two phases described correctly ; i.e. A: level, B: increasing, C: level, D: decreasing (numbers of live bacteria) correct relevant data quote for any phase ;; A is the lag, phase / stage: bacteria, growing / dividing, slowly ; bacteria adjusting to conditions / absorbing resources ; B is the exponential / log, phase / stage: resources are, not limiting / AW ; bacteria, growing / dividing, rapidly ; growth / birth, rate greater than death rate ; C is the stationary, phase / stage: growth / birth, rate equal to death rate ; D is the death, phase / stage: death rate greater than growth, birth / rate ; C or D: many bacteria are dying ; concentration of resources decreasing / resources become limiting ; space becomes limiting / AW ; (named) toxins building up ;

Question 214 marks

Question

(a) State the word equation for photosynthesis. …… [2] (b) Scientists investigated the effect of light intensity on the rate of photosynthesis in the leaves of eucalyptus trees at two different concentrations of carbon dioxide, A and B. The results are shown in Fig. 2.1. rate of photosynthesis / μmol per m2 per s 50 45 B 40 35 30 25 A 20 15 10 5 0 0 250 500 750 1000 1250 1500 1750 2000 light intensity / arbitrary units Key: A carbon dioxide concentration 140 ppm B carbon dioxide concentration 1000 ppm Fig. 2.1 (i) Suggest and explain why the scientists kept the temperature of the leaves at 20 °C while they recorded results. …… [2] (ii) Calculate the percentage increase in the rate of photosynthesis at a light intensity of 1250 arbitrary units when the carbon dioxide concentration was increased from 140 ppm to 1000 ppm. Show your working and give your answer to the nearest whole number. …… % [3] (iii) Describe the effect of increasing light intensity on the rate of photosynthesis when the concentration of carbon dioxide was 140 ppm. …… [3] (iv) Explain the effect of increasing light intensity on the rate of photosynthesis when the concentration of carbon dioxide was 1000 ppm. Use the term limiting factor in your answer. …… [4] [Total: 14]

Mark scheme

2(a) carbon dioxide + water ; 2 glucose + oxygen ; 2(b)(i) temperature is a factor that affects the rate of photosynthesis ; 2 reference to kinetic energy ; idea of effect of temperature, on enzymes / diffusion rate (of carbon dioxide) ; idea that temperature is a variable that should be standardised ; AVP ; 2(b)(ii) 74 ;;; 3 2(b)(iii) rate (of photosynthesis) increases and, reaches a plateau / AW ; 3 rate (of photosynthesis) increases until 1750 (a.u) / 25 µmol per m2 per s ; any comparative use of figures for rate ; 2(b)(iv) light intensity is the limiting factor, at all light intensities used / AW ; 4 because rate of photosynthesis does not level off (even at high light intensities) ; carbon dioxide / temperature / chlorophyll / another factor, was not a limiting factor ; correct reference to (light) energy ; light is absorbed by chlorophyll ; AVP ;

Question 312 marks

Question

Cotton, Gossypium hirsutum, is grown for the fibres that form within the fruits after fertilisation, as shown in Fig. 3.1. Fibres from the fruits of cotton plants are used in the textile industry. fibres Fig. 3.1 Cotton plants have been genetically engineered to produce a protein that is toxic to the caterpillars of several insect pests. This gives the cotton plants resistance to the pests. The cry gene for pest resistance was isolated from the bacterium Bacillus thuringiensis and inserted into the cells of cotton plants as shown in Fig. 3.2. cry gene A A bacterial DNA plasmid cotton plant cell not to scale Fig. 3.2 (a) An enzyme cuts the cry gene from the DNA of B. thuringiensis. (i) State the name of the enzyme that cuts DNA. …… [1] (ii) State the name of the regions labelled A on Fig. 3.2. …… [1] (iii) Explain how the DNA is inserted into the plasmid. …… [3] (b) The plasmids containing the cry gene are inserted into the cells of cotton plants. Outline how the cells of cotton plants use the cry gene to make the toxic protein. …… [4] (c) Outline the advantages to farmers of growing genetically engineered cotton plants that contain the toxic protein. …… [3] [Total: 12]

Mark scheme

3(a)(i) restriction (enzyme) ; 1 3(a)(ii) sticky ends ; 1 3(a)(iii) plasmid is cut with the same (restriction) enzyme (as DNA) ; 3 to form, sticky ends / region A, that are complementary to sticky ends of, gene / cry ; reference to base sequences (being complementary / AW) ; (DNA) ligase used to join plasmid and, cry / gene ; AVP ; e.g. formation of recombinant DNA / plasmid 3(b) gene / cry, codes for the sequence of amino acids in, (toxic) 4 protein ; DNA / gene / cry, is copied / transcribed, to form mRNA ; cry / gene / DNA, remains in nucleus ; mRNA moves from nucleus to, cytoplasm / ribosome ; mRNA passes through ribosomes / AW ; ribosomes assemble amino acids (into protein molecules) ; ribosomes make proteins ; AVP ; e.g. order of amino acids determined by base sequence of, mRNA / DNA / gene 3(c) to kill / harm / deter, caterpillars / insects / pests ; 3 higher yield / reduce losses due to, (certain) caterpillars / insects / pests ; higher quality of cotton ; use less insecticide ; (so) less pollution ; (so) less money spent on, insecticides / spraying ; (so) less risk to workers’ health (from using insecticides) ; (so) less chance of insecticide resistance ; AVP ;

Question 417 marks

Question

Mammals have a double circulation. (a) State what is meant by the term double circulation. …… [1] (b) Table 4.1 shows some information about the functions of the components of blood. Complete Table 4.1. Table 4.1 function type of cell production of antibodies phagocyte promotes blood clotting transports oxygen [4] (c) Blood is transported in arteries and veins. Complete the drawings of the cross-sections of an artery and a vein to show the differences between these two types of blood vessel. Label the lumen in each drawing. artery vein [2] (d) A diagram of a mammalian heart and associated blood vessels is shown in Fig. 4.1. A L B C K J D E H F G Fig. 4.1 (i) Sketch arrows on Fig. 4.1 to show the pathway taken by deoxygenated blood from the heart towards the lungs. [2] (ii) Table 4.2 contains statements about the structures visible in Fig. 4.1. Complete Table 4.2 by: • stating the name of each structure • identifying the structure with the corresponding letter from Fig. 4.1. Table 4.2 statement name of structure letter from Fig. 4.1 chamber that creates the highest blood pressure blood vessel containing blood with the highest concentration of oxygen structure that prevents blood going from ventricle to atrium structure that prevents backflow of blood from artery to ventricle chamber that receives blood from vena cava [5] (e) Mammals also have a lymphatic system. Outline the functions of the lymphatic system. …… [3] [Total: 17]

Mark scheme

4(a) one loop to lungs / pulmonary circulation, and one loop to rest of the 1 body / systemic circulation ; blood flows through heart twice, for one (complete) circuit / to get back to the same point ; 4(b) lymphocyte / AW ; 4 engulf/ digest / kill / destroy, bacteria / pathogens ; platelet ; red blood cell ; 4(c) wall of artery thicker than wall of vein ; 2 lumen labelled in both drawings ; 4(d)(i) arrow(s) start in right-hand side of heart in correct direction ; 2 arrow(s) point upwards inside pulmonary artery ; 4(d)(ii) 5 one mark per row statement name of structure letter from Fig. 4.1 chamber that left ventricle F; creates the highest blood pressure blood vessel pulmonary vein C containing blood / aorta /A; with the highest concentration of oxygen structure that atrioventricular valve E; prevents blood going from ventricle to atrium structure that semilunar valve K; prevents backflow of blood from artery to ventricle chamber that right atrium J; receives blood from vena cava 4(e) tissue fluid drains (into lymphatic vessels) ; 3 transports tissue fluid ; back into the blood / circulatory system ; contains, lymphocytes / antibodies (in lymph nodes) ; defence against infection / provide (active) immunity / AW ; (lacteals) absorbs /transports, fats / fatty acids ; from small intestine / duodenum / ileum ; AVP ;

Question 514 marks

Question

Sickle-cell anaemia is an inherited disease. Fig. 5.1 is a photomicrograph of some blood cells from a person who has sickle-cell anaemia. a sickle-shaped red blood cell Fig. 5.1 (a) Explain how red blood cells become sickle-shaped. …… [3] (b) Some people who have sickle-cell anaemia have parents who do not have sickle-cell anaemia. Explain how people with sickle-cell anaemia inherit the disease. …… [4] (c) Sickle-cell anaemia is most common in areas of the world where the infectious disease malaria is found. Some species of the genus Plasmodium cause malaria in humans. (i) Define the term species. …… [2] (ii) The distribution of sickle-cell anaemia is the result of natural selection. Explain the distribution of the sickle-cell allele in human populations. …… [5] [Total: 14]

Mark scheme

5(a) caused by a mutation ; 3 change in, DNA / base sequence ; of gene for haemoglobin ; (causes) a different sequence of amino acids ; (so) abnormal haemoglobin produced ; AVP ; 5(b) both parents carry the, recessive allele / allele for (sickle cell) 4 anaemia ; both parents are heterozygous ; half the gametes of both parents have the recessive allele ; people / children, who are homozygous recessive have (sickle cell) anaemia ; there is a, 0.5 × 0.5 / 0.25, chance of, being homozygous recessive / having (sickle cell) anaemia ; 5(c)(i) (group of) organisms that can reproduce ; 2 to produce fertile offspring ; 5(c)(ii) people with sickle cell anaemia / heterozygotes / carriers / are 5 resistant to, malaria / AW ; people with (homozygous) sickle cell anaemia are, less likely to survive / die of sickle cell disease ; people who are heterozygous / have one copy of the sickle cell allele, are more likely to survive / have selective advantage ; idea that they are more likely to breed ; pass on allele for sickle cell ; so increase in frequency of sickle cell allele (in population) ; selective advantage for sickle cell only exists where, (mosquitoes carrying) malaria are present ; AVP ; ref. to evolution / adaptation to (local) conditions

Question 610 marks

Question

Fig. 6.1 is a photomicrograph of part of a cell from the pancreas that produces enzymes that are released into the small intestine. S P R Q ×20 000 Fig. 6.1 (a) (i) Structure Q is part of the nucleus of the cell. State one function of a nucleus. …… [1] (ii) State the names of the structures labelled P and R in Fig. 6.1. P …… R …… [2] (b) The structure labelled S transports enzymes to the cell membrane for release into the pancreatic duct. These structures contain molecules of amylase, trypsin and lipase. Complete the sentences with the most appropriate words. Enzymes are made of protein and act as …… because they increase the rate of chemical reactions, but are not changed in those reactions. Amylase speeds up the digestion of …… to …… Trypsin continues the chemical digestion of protein begun by the enzyme …… in the stomach. The optimum pH for pancreatic enzymes is greater than pH 7. Bile is produced by the …… and enters the small intestine, where it …… stomach acid to provide the appropriate pH. Bile also breaks down fat by …… to increase the surface area for the action of lipase. [7] [Total: 10]

Mark scheme

6(a)(i) stores / contains, DNA / chromosomes / genes ; 1 controls the cell ; AVP ; 6(a)(ii) P – endoplasmic reticulum / (rough) ER / ribosome ; 2 R – mitochondrion / mitochondria ; 6(b) catalysts ; 7 starch ; maltose / glucose ; pepsin ; liver ; neutralises ; emulsification ;

Checked your answers against these? Let AI mark the whole paper, on your own working.

Mark my 0610/43

How this paper is marked

Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.

Check your answers against it

Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.

0610/43 in other series

Find another Biology paper

Every IGCSE Biology mark scheme

Question papers and mark schemes © Cambridge University Press & Assessment. CIE Buddy is independent and not connected to Cambridge.

0610/43 questions

Where can I find the 0610/43 May/June 2019 mark scheme?

On this page, free: the mark scheme for IGCSE Biology 0610/43, May/June 2019 (0610_s19_ms_43), and the question paper (0610_s19_qp_43), to read or download, with each question's marking points written out underneath.

How many marks is 0610/43 out of?

Biology 0610/43 for May/June 2019 is out of 80 marks, with 1 hour 15 minutes to answer it, across 6 questions.

Can I get my 0610/43 answers marked?

Yes. Upload your answers to 0610/43/M/J/19 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.