Danho
ZIMSEC A Level · 9190/2 · N2006

Biology Paper 2 November 2006

Questions
59
Total marks
119
Syllabus code
9190/2

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Questions
59
Pass mark
36
Sit this paper

Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Section A

Section A, Question 1

[2 marks]amino acids and protein structure
Fig. 1.1 shows two amino acids, one with a hydrogen atom as its side chain and one with a benzene ring carrying a hydroxyl group. When these two molecules are joined together,
  1. Athe two side chains react with one another and a molecule of carbon dioxide is released from the pair
  2. Bthe two carboxyl groups react with one another and a molecule of hydrogen is released from the pair
  3. Cthe amino group of one reacts with the carboxyl group of the other and a molecule of water is released
  4. Dthe amino group of one reacts with the amino group of the other and a molecule of water is taken in
[1 marks]amino acids and protein structure
Name the bond formed when the carboxyl group of one amino acid reacts with the amino group of another and a molecule of water is released.

Answer this when you sit the paper.

[2 marks]amino acids and protein structure
The tertiary structure of a globular protein is held in shape by
  1. Athe covalent bonds that hold each amino acid residue to the one that follows it
  2. Bhydrogen bonds, ionic bonds and disulfide bridges between the side chains of the residues
  3. Cpeptide bonds and glycosidic bonds between the residues at either end of the chain
  4. Dester bonds between the side chains and the phosphate groups attached to the chain
[2 marks]amino acids and protein structure
The main features of a fibrous protein are that
  1. Aits chains are short and identical, it forms crystals in the cell and it carries oxygen
  2. Bits chains are coiled into a compact ball, it dissolves readily in water and it acts as an enzyme
  3. Cits chains are twisted into long strands, it is insoluble in water and it has a structural role
  4. Dits chains are branched like a polysaccharide, it dissolves in lipids and it stores energy

Section A, Question 2

[2 marks]enzyme activity
In Fig. 2.1, curve C shows sucrase activity rising steadily as the factor X is increased and then levelling off at a constant value. For curve C, the factor X is
  1. ApH
  2. Bsubstrate concentration
  3. Ctemperature
  4. Dinhibitor concentration
[2 marks]enzyme activity
According to the induced fit hypothesis, what happens as a sucrose molecule approaches the active site of sucrase before the complex forms?
  1. AThe enzyme splits into two subunits, each of which binds to one half of the sucrose molecule.
  2. BThe active site is already an exact fit for sucrose and stays completely rigid as the substrate enters it.
  3. CThe sucrose molecule changes shape to match the rigid active site of the enzyme it is approaching.
  4. DThe active site changes shape slightly so that it moulds itself closely round the sucrose molecule.
[2 marks]enzyme activity
Raising the temperature well above its optimum reduces the activity of sucrase because
  1. Athe sucrose molecules move too fast to be caught by the active sites of the enzyme molecules
  2. Bthe enzyme reacts with the water in the solution and is used up as the reaction proceeds
  3. Cthe hydrogen bonds holding the tertiary structure break, so the active site loses its shape
  4. Dthe peptide bonds of the chain are hydrolysed, so the enzyme is digested into its amino acids

Section A, Question 3

[2 marks]recombinant DNA and genetic engineering
A plasmid used as a vector in genetic engineering can be described as
  1. Aa small circle of DNA in a bacterium that replicates separately from the main chromosome
  2. Ba loop of messenger RNA that carries the gene from the bacterium out into the culture medium
  3. Ca protein coat that surrounds the bacterial chromosome and protects it from enzyme attack
  4. Da length of linear DNA held inside the nucleus of the bacterium and copied at cell division
[3 marks]recombinant DNA and genetic engineering
The primary structure of a useful protein is known. A gene coding for that protein can be built by
  1. Areading the amino acid sequence, working out the messenger RNA codons for it and assembling the complementary DNA
  2. Breading the amino acid sequence and joining the amino acids end to end to make a strand of DNA of the same length
  3. Ccutting the protein into short lengths with a restriction enzyme and joining those lengths into a circle with ligase
  4. Dtreating the protein with DNA polymerase, which copies the order of the amino acids straight onto a sugar backbone
[2 marks]recombinant DNA and genetic engineering
In Fig. 3.1, step C is the return of the recombinant plasmid into a bacterial cell. A difficulty at this step is that
  1. Athe bacterial cell wall has to be removed completely first, after which the cell cannot divide any further
  2. Bonly a small proportion of the treated bacteria take up a plasmid, so the successful cells must be identified
  3. Cthe plasmid is digested by the restriction enzyme as soon as it enters, so no foreign gene survives the step
  4. Dthe bacterium already carries the same gene, so the new plasmid is rejected before it reaches the cytoplasm
[2 marks]recombinant DNA and genetic engineering
Bacteria carrying a recombinant plasmid are grown in a fermenter and multiply. A benefit of this to humans is that
  1. Athe bacteria can be released into a patient's blood, where they take over the work of the faulty organ
  2. Bthe bacteria break the plasmid down and release the sugars stored inside it for use in the food industry
  3. Cthe bacteria become resistant to antibiotics, which can then be used to treat a much wider range of illness
  4. Dhuman insulin can be made in quantity, so a diabetic is not dependent on hormone extracted from animals

Section A, Question 4

[2 marks]monohybrid and sex-linked inheritance
Free earlobes are dominant to attached earlobes. A mother has attached earlobes, and her son and her daughter both have free earlobes. The genotypes of the mother and of the two children are
  1. Athe mother is heterozygous, and one child is homozygous dominant while the other is heterozygous
  2. Bthe mother is homozygous recessive, and both children are heterozygous for the earlobe gene
  3. Cthe mother is heterozygous, and both of the children are homozygous dominant for that gene
  4. Dthe mother is homozygous dominant, and both children are homozygous recessive for that gene
[2 marks]monohybrid and sex-linked inheritance
Free earlobes are dominant to attached earlobes. A father with free earlobes and a mother with attached earlobes have two children, both with free earlobes. The father's genotype
  1. Amust be homozygous recessive, since attached earlobes appear somewhere in the family line
  2. Bmust be homozygous dominant, since every one of his children has the dominant phenotype
  3. Cmust be heterozygous, since his wife has attached earlobes and passes a recessive allele on
  4. Dcannot be settled from this evidence, since a homozygous or a heterozygous father fits it
[2 marks]monohybrid and sex-linked inheritance
Red green colour blindness is caused by a recessive allele carried on the X chromosome. A colour blind man marries a woman who is neither colour blind nor a carrier. Their children will be
  1. Asons who are carriers of the allele and daughters who show colour blindness in every case
  2. Bsons with normal colour vision and daughters who are carriers of the recessive allele
  3. Csons who are colour blind and daughters who have normal colour vision and are not carriers
  4. Dsons and daughters who are all colour blind, since the father passes his allele to every child
[2 marks]monohybrid and sex-linked inheritance
Red green colour blindness is far more common in men than in women because
  1. Aa man has two X chromosomes, so he has twice the chance of inheriting the recessive allele
  2. Bthe allele is carried on the Y chromosome, which is passed from a father to each of his sons
  3. Ca man's Y chromosome carries a second copy of the allele, which reinforces the first copy
  4. Da man has one X chromosome, so a single recessive allele on it is expressed in his phenotype

Section A, Question 5

[1 marks]meiosis and genetic variation
State the term for a pair of homologous chromosomes lying together, each already made of two chromatids, during prophase I of meiosis.

Answer this when you sit the paper.

[1 marks]meiosis and genetic variation
During prophase I of meiosis, the chiasma between chromatids of a homologous pair functions to
  1. Ajoin the two chromatids of one chromosome at the point where the DNA was copied
  2. Bseparate the two chromosomes of the pair so that each can be drawn into its own cell
  3. Chold the pair together while lengths of DNA are exchanged between the chromatids
  4. Dattach the chromosome to the spindle so that it can be pulled towards a pole of the cell
[3 marks]meiosis and genetic variation
In a sexually reproducing population, genetic variation between individuals arises from
  1. Arepeated mitosis of the body cells, the growth of the embryo and the ageing of both parents
  2. Bsemi conservative replication of DNA, the action of enzymes and the diet of the two parents
  3. Ccrossing over in prophase I, independent assortment at metaphase I and random fertilisation
  4. Dcrossing over in prophase I, independent assortment at metaphase I and DNA replication before it
[1 marks]meiosis and genetic variation
Chromosomes occur as homologous pairs in a diploid cell because
  1. Aone member of each pair came from the mother and one from the father at fertilisation
  2. Beach chromosome is copied during interphase, which gives two chromatids joined together
  3. Ceach pair carries a duplicate set of genes in case the first set is damaged by a mutation
  4. Dthe chromosomes have to lie in pairs before the spindle fibres are able to attach to them

Section A, Question 6

[2 marks]transport in plants
Fig. 6.1 shows a leaf section with the xylem vessel of the vascular bundle lying below the palisade cells. Water reaches the palisade cells because
  1. Aevaporation from the mesophyll lowers the water potential there, so water moves from the xylem down a gradient
  2. Bthe palisade cells actively pump water inwards from the xylem, using the ATP made in their own mitochondria
  3. Cthe guard cells force water upwards through the leaf whenever the stomata are closed during the night
  4. Dwater is pushed out of the xylem by the pressure of the sucrose being loaded into the phloem beside it
[3 marks]transport in plants
Sucrose made in the palisade cells of a leaf reaches the phloem when
  1. Ait dissolves in the xylem sap and is carried down the stem to the roots, where it enters the phloem for the return journey
  2. Bit is converted to starch in the mesophyll, and the starch grains are then carried whole along the sieve tubes to the roots
  3. Cit passes into the air spaces of the leaf as a vapour and condenses on the outside of the phloem sieve tubes below
  4. Dit diffuses through the walls and the cytoplasm of the mesophyll and is then actively loaded into the companion cells and sieve tubes
[1 marks]transport in plants
Name the cell that lies alongside a sieve tube element in phloem tissue and supplies the ATP used to load sucrose into it.

Answer this when you sit the paper.

[2 marks]transport in plants
The spongy mesophyll tissue of a dicotyledonous leaf functions mainly to
  1. Aprovide a rigid layer of thickened cells that holds the leaf blade flat and out towards the sunlight
  2. Bprovide a waterproof layer that stops water being lost from the underside of the leaf during the day
  3. Cprovide the tissue in which most of the leaf's starch is stored between one growing season and the next
  4. Dprovide large air spaces through which carbon dioxide diffuses to the cells and water vapour leaves the leaf

Section A, Question 7

[2 marks]ecology, niche and energy flow
In ecology, the niche of an organism is
  1. Athe group of organisms of one species that live in the same area and breed with one another
  2. Bthe physical place in which the organism lives, described by the depth of water and the type of soil
  3. Cthe role the organism plays in its community, including what it feeds on and what feeds on it
  4. Dthe total number of organisms of that species that a particular habitat is able to support
[1 marks]ecology, niche and energy flow
State the ecological term for all the populations of different species living and interacting together in one habitat.

Answer this when you sit the paper.

[2 marks]ecology, niche and energy flow
Energy enters the community of a river when
  1. Athe flatworms among the pebbles break down dead material and release the energy stored in it
  2. Bthe water plants trap sunlight and convert it into chemical energy in the compounds they make
  3. Cthe fish take in dissolved oxygen through their gills and use it to release energy from food
  4. Dthe mud at the bottom of the river absorbs heat during the day and releases it again at night
[2 marks]ecology, niche and energy flow
Roughly a tenth of the energy in one trophic level of a food chain reaches the level above it because
  1. Amost of it is lost as heat in respiration, and more is left in the parts not eaten or not digested
  2. Bmost of it is used up in building the bodies of the consumers, which store it as fat and protein
  3. Cmost of it is recycled by the decomposers and returned to the producers at the start of the chain
  4. Dmost of it is reflected from the surface of the water before the plants are able to absorb any of it
[1 marks]ecology, niche and energy flow
Fig. 7.1 shows a river holding oxygenating plants such as water milfoil and hornwort, dragonfly larvae that climb the tall emerging vegetation, and fish. A possible food chain in this river is
  1. Awater milfoil to dragonfly larva to fish
  2. Bwater milfoil to fish to dragonfly larva
  3. Cdragonfly larva to water milfoil to fish
  4. Dfish to dragonfly larva to water milfoil

Section A, Question 8

[2 marks]the menstrual cycle and hormonal control
In Fig. 8.1, PROCESS 1 is bracketed under the early ovary stages that respond to FSH, PROCESS 2 is marked where the egg is shed from the ovary, and PROCESS 3 where the structure left behind afterwards breaks down. Processes 1, 2 and 3 are
  1. Aovulation, growth of the follicle and degeneration of the corpus luteum
  2. Bgrowth of the follicle, implantation of the embryo and the shedding of the uterus lining
  3. Cfertilisation of the egg, ovulation and the repair of the lining of the uterus wall
  4. Dgrowth of the follicle, ovulation and degeneration of the corpus luteum
[2 marks]the menstrual cycle and hormonal control
In Fig. 8.1, HORMONE 2 is secreted by the growing follicle before the egg is shed, and HORMONE 1 is secreted by the corpus luteum that forms afterwards. Hormones 1 and 2 are
  1. Afollicle stimulating hormone and luteinising hormone respectively
  2. Bluteinising hormone and follicle stimulating hormone respectively
  3. Coestrogen and progesterone respectively
  4. Dprogesterone and oestrogen respectively
[1 marks]the menstrual cycle and hormonal control
Name the structure that develops from the remains of a follicle after the egg has been released, and that secretes progesterone.

Answer this when you sit the paper.

[2 marks]the menstrual cycle and hormonal control
A rising level of oestrogen in the blood inhibits the secretion of FSH by the anterior pituitary. This negative feedback is important because
  1. Ait stops the corpus luteum breaking down, so progesterone is secreted all through the month
  2. Bit stops the egg being fertilised, so an embryo cannot implant while the follicle is growing
  3. Cit stops further follicles being stimulated, so usually one egg alone is released in a cycle
  4. Dit stops the uterus lining thickening, so the endometrium stays thin throughout the cycle
[3 marks]the menstrual cycle and hormonal control
An egg is fertilised and the embryo implants in the wall of the uterus. The effect on the menstrual cycle is that
  1. Athe oestrogen level falls to zero at once, so the uterus lining is broken down and rebuilt around the implanted embryo
  2. Bthe embryo secretes a hormone that keeps the corpus luteum alive, so progesterone stays high and the lining is not shed
  3. Cthe corpus luteum breaks down on schedule, so progesterone falls and a lighter than usual menstruation still takes place
  4. Dthe anterior pituitary secretes extra FSH, so a second follicle ripens and the ovarian cycle continues alongside the pregnancy

Section A, Question 9

[2 marks]classification
The taxonomic groups species, genus, family, order, class and phylum, arranged from the largest and most inclusive group down to the smallest, are
  1. Aspecies, genus, family, order, class, phylum
  2. Bclass, phylum, order, family, genus, species
  3. Cphylum, class, order, family, genus, species
  4. Dphylum, order, class, family, species, genus
[2 marks]classification
Organisms are placed in the same species when they
  1. Ashare common features and can interbreed to produce offspring that are themselves fertile
  2. Bshare common features and live in the same habitat, feeding on the same kind of food
  3. Cshare common features and can interbreed, though their offspring are sterile like the mule
  4. Dshare common features and have the same number of chromosomes in each of their body cells
[3 marks]classification
Organisms are placed in the Kingdom Fungi on the grounds that they
  1. Ahave cells without a nucleus or membrane bound organelles, and take in food across the whole of the cell surface
  2. Bhave a body of many cells with no cell walls at all, store fat as an energy reserve and take food into a gut
  3. Chave a body of thread like hyphae with chitin in the walls, store glycogen and feed by absorbing digested food
  4. Dhave a body of thread like hyphae with cellulose in the walls, store starch and make their own food by photosynthesis

Section B

Section B, Question 10

[3 marks]membrane transport
Glucose enters a red blood cell far faster than its size and solubility would allow it to cross the phospholipid bilayer. This is because
  1. Ait is taken in by endocytosis, the membrane folding inwards round each glucose molecule as it arrives
  2. Bit passes through a channel or carrier protein, moving down its concentration gradient without ATP being used
  3. Cit passes through a carrier protein that is driven by ATP, so it can be moved against its concentration gradient
  4. Dit dissolves in the hydrophobic tails of the bilayer and diffuses straight across the membrane between them
[3 marks]membrane transport
Sodium ions are pumped out of a nerve cell into a tissue fluid that is already richer in sodium than the cell is. This transport
  1. Ais facilitated diffusion, because the ions pass through a protein that spans the whole of the membrane
  2. Bis simple diffusion, because ions are small enough to pass between the phospholipid molecules freely
  3. Cis osmosis, because the ions follow the water that moves out of the cell into the surrounding fluid
  4. Dis active transport, because the ions move against the concentration gradient using ATP from respiration
[2 marks]membrane transport
A root hair cell takes up nitrate ions from soil water in which nitrate is very dilute. Treating the root with a respiratory inhibitor stops the uptake, which shows that
  1. Athe nitrate is taken in through the cell wall, which is why the inhibitor has an effect on the cell
  2. Bthe nitrate is taken in by active transport, since the process depends on ATP from respiration
  3. Cthe nitrate is taken in by osmosis, since water and the ions dissolved in it move in together
  4. Dthe nitrate is taken in by simple diffusion through the gaps between the phospholipid molecules
[2 marks]membrane transport
The cell surface membrane plays a part in cell signalling because
  1. Ait carries glycoprotein receptors with shapes complementary to particular hormone molecules
  2. Bit carries pores large enough for whole hormone molecules to pass freely into the cytoplasm
  3. Cit carries a layer of cellulose on which the hormones of the blood are able to settle and act
  4. Dit carries the chromosomes on its inner surface, where a hormone can act on them directly
[2 marks]membrane transport
Membranes inside a cell, rather than at its surface, are important because
  1. Athey store the cell's genetic information, which is copied each time the whole cell divides in two
  2. Bthey supply the cell with glucose, which they manufacture from the carbon dioxide it produces
  3. Cthey form the spindle on which the chromosomes are separated at each division of the nucleus
  4. Dthey divide the cell into compartments and hold enzymes in place, as on the cristae of a mitochondrion

Section B, Question 11

[2 marks]the genetic code and transcription
The genetic code is described as a triplet code because
  1. Athree separate strands of DNA have to be read together before a protein can be made
  2. Ba sequence of three bases in the DNA specifies one amino acid in the polypeptide chain
  3. Cthree different bases are found in DNA, and each of them specifies one amino acid
  4. Deach amino acid is joined to three others when the polypeptide chain is being built
[2 marks]the genetic code and transcription
The genetic code is described as degenerate because
  1. Amost amino acids are specified by more than one triplet of bases in the coding sequence
  2. Bsome triplets of bases specify more than one amino acid, depending on where the reading starts
  3. Cthe code has decayed over evolutionary time, so parts of it no longer specify anything at all
  4. Deach triplet of bases overlaps the next one, so a single base is read as part of two triplets
[1 marks]the genetic code and transcription
How many bases of a DNA molecule code for one amino acid of a polypeptide chain?

Answer this when you sit the paper.

[3 marks]the genetic code and transcription
During transcription in the nucleus of a cell,
  1. ADNA polymerase unwinds the whole molecule and builds a second complete copy of both strands
  2. Bthe ribosome moves along the DNA and joins amino acids together in the order the bases give
  3. Ctransfer RNA copies the base sequence of the DNA and carries that copy out of the nuclear pore
  4. DRNA polymerase unwinds a length of the double helix and builds messenger RNA against one strand
[2 marks]the genetic code and transcription
A DNA template strand reads C A T. The messenger RNA codon transcribed from it is
  1. AC A U
  2. BG T A
  3. CG U A
  4. DU A C
[2 marks]the genetic code and transcription
Before a messenger RNA molecule made in the nucleus of a eukaryotic cell can be translated,
  1. Athe introns are cut out and the exons joined together, and the molecule leaves through a nuclear pore
  2. Bthe exons are cut out and the introns joined together, and the molecule leaves through a nuclear pore
  3. Cthe molecule is converted back into DNA by reverse transcriptase before it can leave the nucleus
  4. Dthe molecule is wound round histone proteins so that it is compact enough to pass out of the nucleus

Section B, Question 12

[3 marks]excretion and osmoregulation
The counter current mechanism in the loop of Henle works because
  1. Athe ascending limb takes water in from the medulla while the descending limb, running the other way, pumps sodium and chloride ions inwards
  2. Bthe two limbs carry filtrate in the same direction, so the ions pumped out of one limb are taken straight back into the other
  3. Cthe blood in the vasa recta flows in the same direction as the filtrate, which keeps the ions moving along the loop
  4. Dthe ascending limb pumps sodium and chloride ions into the medulla while the descending limb, running the other way, loses water to it
[2 marks]excretion and osmoregulation
The high solute concentration built up in the medulla of a kidney matters because
  1. Athe collecting duct passes through it, so water leaves the filtrate by osmosis and the urine is concentrated
  2. Bthe collecting duct passes through it, so salts are forced back into the filtrate and the urine is made dilute
  3. Cthe glomerulus lies within it, so the pressure there is high enough to force the filtrate out of the blood
  4. Dthe loop of Henle ends there, so the urea in the filtrate is broken down before the urine reaches the bladder
[1 marks]excretion and osmoregulation
Name the hormone that makes the walls of the collecting duct more permeable to water when the blood becomes too concentrated.

Answer this when you sit the paper.

[2 marks]excretion and osmoregulation
Amino acids taken in beyond what the body needs cannot be stored, and are dealt with when
  1. Athe liver removes the amino group as ammonia and converts it to the less toxic urea
  2. Bthe kidney removes the amino group as urea and converts it into ammonia for excretion
  3. Cthe liver converts the whole amino acid into urea, which is then stored in the gall bladder
  4. Dthe muscles convert the surplus amino acids into lactate, which is breathed out as a vapour
[2 marks]excretion and osmoregulation
Metabolic waste products have to be removed from the body because
  1. Asubstances such as carbon dioxide are used up by the tissues faster than the body can make them
  2. Bsubstances such as ammonia are toxic, and a build up would alter blood pH and damage enzymes
  3. Csubstances such as ammonia are insoluble, and a build up would block the capillaries of the tissues
  4. Dsubstances such as urea are needed by the gut bacteria, which cannot make them for themselves
[2 marks]excretion and osmoregulation
Keeping the water potential of the blood within narrow limits is important because
  1. Awater moves into or out of cells by osmosis, so cells would swell and burst or shrink otherwise
  2. Bwater is the substrate of respiration, so a cell short of water cannot release energy from glucose
  3. Cwater carries the oxygen dissolved in it, so a change in its potential stops oxygen reaching the cells
  4. Dwater holds the chromosomes apart in the nucleus, so a change in its potential prevents cell division

Section B, Question 13

[2 marks]respiration and lung disease
The term oxygen debt refers to
  1. Athe oxygen used by the liver in making glucose from the fats stored in the body during a fast
  2. Bthe extra oxygen taken in after exercise to deal with the lactate that built up in the muscles
  3. Cthe extra oxygen taken in during exercise to allow the muscles to respire aerobically throughout
  4. Dthe oxygen carried by haemoglobin that is not released to the tissues while a person is at rest
[2 marks]respiration and lung disease
The lactate that has built up in the muscles after a sprint is dealt with when
  1. Athe muscle converts it into protein, which is used to repair the fibres damaged during the sprint
  2. Bthe kidney filters it from the blood, so that it leaves the body dissolved in the urine as it forms
  3. Cthe blood carries it to the liver, where some is oxidised and the rest is converted back to glucose
  4. Dthe blood carries it to the lungs, where it is broken down and breathed out as carbon dioxide gas
[2 marks]respiration and lung disease
Emphysema arises in a long term smoker when
  1. Athe cilia lining the bronchi grow through the alveolar walls and break them into a few large spaces
  2. Bphagocytes drawn into the lungs release elastase, which destroys the elastic tissue of the alveolar walls
  3. Cthe alveoli fill with tissue fluid, which sets hard and prevents the walls from stretching during a breath
  4. Dthe smoke reacts with the mucus in the airways to form a solid layer over the surface of the alveoli
[2 marks]respiration and lung disease
Chronic bronchitis develops in a smoker because
  1. Athe smoke thins the walls of the bronchi, so they collapse whenever the person breathes out hard
  2. Bthe smoke raises the temperature of the airways, so the bacteria living in them multiply much faster
  3. Cthe smoke destroys the cilia and irritates the lining, so extra mucus is made and is not swept away
  4. Dthe smoke destroys the goblet cells, so no mucus is made and the airways dry out and become inflamed
[3 marks]respiration and lung disease
Lung cancer arises in a smoker when
  1. Athe nicotine in the smoke narrows the arteries of the lung, so the cells behind the blockage grow out of control
  2. Bcarcinogens in the tar cause mutations in the genes that control cell division, so a tumour grows in the airway lining
  3. Cthe tar coats the alveoli so thickly that oxygen can no longer reach the blood, and the starved cells begin to divide
  4. Dcarbon monoxide in the smoke binds to the haemoglobin, and the cells it passes are turned into cancer cells
[1 marks]respiration and lung disease
State the general term for a chemical, such as those in tobacco tar, that causes mutations leading to cancer.

Answer this when you sit the paper.

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The answers, and why they are the answers

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