BiologyMad A-Level Biology

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Chapter 1 - Transmission of Genetic Information In Chapter Questions [back to top]
1.	The chromatids shown in Fig. 1.2 have an identical pattern of dark and light bands because each chromatid has the same sequence of genes (although each gene might a number of different ‘versions’ or ALLELES).
2. a	An event that occurs in BOTH prophase of mitosis and prophase 1 of meiosis is that the chromosomes shorten and condense and can now be seen with suitable staining.
b	In prophase 1 of meiosis the homologous chromosomes come together with their partner (to form pairs) whereas in mitosis no such pairing occurs.
3.	The homologous chromosomes are separated during the first meiotic division.
4.	A human male with 23 pairs of chromosomes can produce 2²³ different possible sperms - this equals 8,388, 608 (or about 8.4 million!) different sperms! (P.S. this is called INDEPENDENT ASSORTMENT)
5.	Cross-over of genetic material between homologous chromosomes would greatly add to the number of genetic differences occurring in daughter cells. If considered together with independent assortment this will produce an almost infinite number of genetically different daughter cells.
6.	Terms explained: HAPLOID: This is a term which refers to cells in which the nuclei contain a single representative of the homologous chromosomes. Thus in human with 46 chromosomes (23 homologous pairs) a haploid cell has 23 chromosomes. The letter n is often used to represent the haploid number of chromosomes. (where 2n would be diploid). HOMOGAMETIC: This is the sex which produces gametes with the same sex chromosome i.e. the female in humans as all the gametes are X (the heterogametic sex is male as half the gametes are X and half are Y).
7.	The DOMINANT allele is the one that always shows its effect in the phenotype.
8.	Snapdragons with the genotype RW are pink as the R (red) and W (white) alleles are CODOMINANT and each produces an intermediate effect in the phenotype giving a pink flower.
9. a	I^o is recessive - it is only expressed in the phenotype when it is in a homozygous state (i.e. I^oI^o ); it is ‘masked’ if I^A or I^B are present.
b	The alleles I^A and I^B are codominant because when they occur together (as in I^AI^B ) they produce a new phenotype that has expressed characteristic produced by a combination of each of these alleles.
10.	The question as set is unclear – it is not certain what the genotypes to be considered actually are! The F2 ratios would be (if both parents are heterozygous): (a) 3 : 1 (b) 9 : 3 : 3 : 1
11.	The reason that haemophilia is never passed on from a father to his son is that he passes his Y chromosome to his son and this is genetically ‘empty’. The haemophilia allele is linked to the X chromosome. (THIS WOULD GO TO HIS DAUGHTERS).
12. a	individual 4 must be X^HX^h as they are a combination of a fusion between X^H and X^h as shown on the Punnett square.
b	This cannot be answered – there is no individual no. 10.

Chapter 2 - Variation and Selection In Chapter Questions [back to top]
1.	A quadrat frame thrown over the shoulder is NOT a suitable sampling method because (quite apart from the danger!) it is not truly free of bias. There might be a temptation to throw it in a desired direction.
2.	The data is reasonably normally distributed with approximately equal values lying above and below median of 3.8. There are few values at the extremes of 2.9 and 4.5.
3.	A normal distribution curve that is thin has a smaller standard deviation than one which is broad.
4.	Yes cells that produce white hairs in Himalayan rabbits will possess the black coat colour gene; the gene is switched on (or expressed) if the skin with the cells in falls below 34^oC.
5.	The genotypes that would be expected are: AABB AABb AAbb AaBB AaBb Aabb aaBB aaBb aabb
6.	A gene pool is a term to describe the total genes present in all the organisms of one species in an interbreeding (or potentially interbreeding) population.
7.	If the frequency of l allele will be 0.4 (40%) if the L allele is 0.6 (60%) as p +q = 1 (100%).
8.	The Hardy-Weinburg principle tells us that that the frequency of a particular allele will stay constant from one generation to the next (if certain conditions remain constant e.g. that there is no selection for or against the allele or that the population size is large enough to avoid genetic drift). It also indicates that (subject to certain assumptions( that the sum of two alleles of a particular gene will have a value of 1 (where 1 = 100% of the population).
9.	Hardy-Weinburg calculations should always start with the homozygous recessive as there is only one possible genotype, unlike the dominant phenotype where there will be two genotypes.
10.	In fig. 2.12 the LOWER curve has the smallest standard deviation.
11.	The evolution of warfarin resistance in rats is an example of disruptive selection.
12.	Snails with banded yellow shells are at an advantage compared to shells with unbanded yellow shells, in a mixture of grass and shrubs. This is because they are better camouflaged (cryptic appearance) in a mixture of yellow herbage where the ‘lines and bands’ present closely match the banding on the snail.
13.	The sickle cell allele would be reduced in malaria-free parts of the world because the sickle cell allele Hb^S would not confer an advantage in the heterozgote Hb^SHb^A,, which is the case in the presence of malaria.

Chapter 4 - Numbers and Diversity

In Chapter Questions [back to top]

Pesticides applied to farmland in North America could lead to an increase in pesticide residues in Antarctic penguins if the pesticides were persistent (in other words they were chemicals like DDT that do not readily breakdown in the environment into harmless residues). In such situations the pesticides enters FOOD CHAINS and WEBS. Pesticides are taken up by producers or primary consumers and the pesticide accumulates in the tissues (often in fatty tissues). As these trophic levels are consumed the pesticide passes to the next trophic level; as the next level consumes many organisms from the level below the pesticide builds to higher levels. Pesticide residues reach their highest levels in top predators (tertiary & quaternary consumers). In this example if the North American pesticide entered a river then it could be taken up by migratory fish (like salmon that enter the sea) or migratory birds that could fly to the southern hemisphere. Alternativly the pesticide could enter the sea and so get into marine food webs. If this happened it could accumulate in the fish prey of the penguins.

Quadrat placement methods:
      1.    not truly at random;
      2.    is at random;
      3.    is at random.

In the pond the number of fish in the whole population is estimated using the formula:

N1 x N2

                 Where:       N1 = no. of individuals marked & released
                                  N2 = no. of individuals caught in the second sample
                                  Nm = no. of marked individuals in the second sample
                 Thus population size :
                                                =    56 x 48
                                                          17
                                                 =    2320
                                                         17
                                estimated population size = 136

The mark-release-recapture technique would be unsuitable in this case because:
1. the after marking the released birds would not mix thoroughly into the population as they occupy territories which exclude other blue tits;
2. the population is changing because as it is the breeding season new, young birds will be entering the population.

To estimate the index of diversity for seaweeds two areas of data would be required:
1. the number of different species;
2. the number of individuals of each species.

The vertically growing underground stem of marram is a valuable adaptation in mobile dunes because the more sand that is blown onto the top of the plant, the more vigorously the stem grows; this will maintain stop the plant being swamped with overlying sand.

The index of diversity in mobile dunes will be 0 and so it indicates the lowest possible diversity for a habitat (short of no species at all!).

The plants that grow after the departure of the seals represent secondary succession; presumably the death of tussock grass and its temporary replacement by another species is a cyclic annual event, therefore the process has happened in the same places before (probably many times).

The diversity of shrubs in an old hedge is almost certainly going to be greater than in a younger hedge. Interestingly the number of woody plant species present in a hedge is used as a crude measure of the hedge age; the more species present the older the hedge.

Chapter 5 - Energy Transfer

In Chapter Questions [back to top]

If animals fed on their own young then there would be no young to grow and reproduce and carry on the species and so the species would die out. If animals fed on young then there would not be enough energy transferred as the young are small and few in number. The animals would have to expend energy in mating and producing young and would not even get an equal amount of energy in return as energy would be lost by the developing young in respiration as heat.

In short organisms at a trophic level need to obtain energy from the trophic level below, which has a greater overall energy content than the level above it. This allows for the respiration energy losses of the higher trophic level.

ATP sounds as if it is a small molecule (it has a short name!!) but it contains a ribose sugar (pentose therefore one less C than glucose), a purine base and three phosphate groups so it is actually larger. Therefore the answer is false.

3. a

The electrons from the photolysis of water go to chlorophyll a in photosystem II in order to restabilise it (replacing the electrons lost to P.S.I)

Electrons from photosystem I the electrons combine with hydrogen ions from water to form hydrogen atoms and then reduce NADP.
(extension detail beyond syllabus: or the electrons come back to photosystem I in cyclic photophosphorylation and in so doing form more ATP)

Electrons from photosystem II go to photosystem I in non-cyclic photophosphorylation and produce ATP

Dark reactions imply that an absence of light is a requirement for this process however, light is irrelevant for this process as long as the products of the light reaction are present.

ATP is also needed in the light independent reaction in order to provide a phosphate group to convert ribulose phosphate to ribulose bisphosphate.

Less than 2% of the energy of light is converted to chemical potential energy. 2% of 18800kJ per m² per day is 94kJ per m² per day.

pyruvate + coenzyme a + NAD è acetylcoenzymeA + carbon dioxide + reduced NAD

One molecule of glucose produces:
                           2 reduced NAD from glycolysis
                           2 reduced NAD from the link reaction
                           6 reduced NAD from Krebs cycle
Therefore one molecule of glucose produces 10 molecules of reduced NAD.

The RQ of protein is 0.9, if 5.4 cm² of oxygen is consumed in one minute then 5.4 x 0.9 cm² of carbon dioxide will be produced. 4.86 cm² of carbon dioxide will have been produced in one minute.

10.

The RQ of the locust would change from 1 to 0.7 in the first fifteen minutes of flight as the glucose (carbohydrate) is used up and triglyceride (lipid) is used instead as a respiratory substrate.

11.

Sundew would occupy both trophic level 1 (producer) and trophic level 3 (secondary consumer) as it digests aphids (animals that digest sap from phloem).

12.

10% of 10% of energy would be transferred from primary consumer to the tertiary consumer. that is 1% of the energy in the primary consumer would be transferred to the tertiary consumer.

13.

A young animal would transfer chemical potential energy in food into chemical potential energy in tissues most efficiently, as they are still growing. At maturity there is relatively less growth and proportionately more food is used for respiration of reproduction

Chapter 5 - Energy Transfer

Answers to Assignment Questions [back to top]

1. a

Absorbed by the algal cells in the Calvin cycle of photosynthesis and incorporated into an organic carbon compound.

A product of photosynthesis – this includes carbohydrates, lipids, proteins and vitamins.

2. a

If the objective is to allocate molecules between the two organisms in the lichen, then an inability to distinguish accurately between them renders the whole experiment worthless.

The fungal cells use carbohydrate in respiration, thus turning an organic molecule into carbon dioxide and water, both of which are inorganic and will ‘leak’ out of the lichen. Thus no trace of these molecules will be left behind.

Water potential is proportional to the concentration of solutes in the solution, i.e. to the number of soluble molecules present in the cell. Polyols, being soluble, will therefore help to lower the water potential and thus make water absorption more effective. [They are also hygroscopic, and thus aid water retention and the initial uptake of water. They are widely used as artificial sweeteners, the most common being sorbitol and mannitol]

4. a

Cyclic photophosphorylation takes place on the thylakoid membranes. Thus damage to these membranes will reduces ATP synthesis within the chloroplasts. Secondly, the main products of photosynthesis are respiratory substrates (e.g. glucose), so anything which reduces photosynthesis will consequently affect ATP production from respiration too.

ATP and NADPH are both required in the Calvin cycle to reduce GP (or Phospo-Glyceric Acid, PGA) to GP (or PhosphoGlyceraldehyde, PGAld).

Thin Layer Chromatography (T.L.C.) on two samples of chlorophyll – one from a sample of lichen in a clean area, and one from an area known to be polluted. The solvents used (petroleum ether) are VERY inflammable and so suitable precautions must be taken! Calculate the Rf values for each sample. This is the distance moved by the pigment divided by the distance moved by the solvent front. Thus the maximum value is 1.0 and it has no units. If the two samples have different Rf values for the 4 or 5 pigments that are normally visible, then the molecules must have a different structure. Small molecules tend to have larger Rf values.

Chapter 5- Energy Transfer

Answers to Exam Questions [back to top]

1. a

i 6250	x 100 = 0.5%
12.5 x 10⁵

NOTE: 12.5 x 10⁵ is NOT in ‘standard form’. The correct number is 1.25 x 10⁶

ii Wrong wavelength (frequency) for photosynthesis.
    Reflected at surface
    Absorbed by water / suspended solids in the water
    Cannot all be used due to other limiting factors – e.g. [CO₂]; temperature

After 3 trophic levels the energy left for the organisms in the 4^th level will be:

50	= 0.004%.
1.25 x 10⁶

Thus, the energy left at the 6^th level would be <1% of that, i.e. around 0.00004%. This remaining energy is just not enough to support the needs of what would inevitably be a large animal.
[N.B. Better ‘niches’ exist for them and so there is no need to be at trophic level 6 in the food chain. A shark, obviously, does not think like this and simply eats what comes along and so must inevitably eat some of its food as level 6 (and above).]

Decomposed by bacteria and fungi etc. and recycled as simple nutrients (N, P, K) OR covered in mud and, eventually, formed into fossil fuels.

i Electrons are raised to higher energy levels (orbitals). A pair of electrons are eventually lost to Photosystem I or Photosystem II and are replaced from hydrogen in water (split by photolysis)

ii No ATP and no NADPH will be made. These are the basic raw materials (together with CO₂) for the Calvin Cycle, so without them the absorption of CO₂ will cease. Note these chemicals are required to reduce GP to TP (PGA to PGAld).

Glucose molecules contain a lot of energy, but they are relatively inert and so this energy is not readily available to the reactions within the cell. In addition, if it were all to be released at once, it would ‘fry’ the cell. ATP molecules contain less energy, but it is readily released in a useable form.

[Note that all forms of respiration are around 40% efficient, i.e. 40% of the energy in glucose is stored in ATP and the other 60% is lost as heat – petrol engines are around 20% efficient, electric motors, 90%. Glycolysis is also 40% efficient, but only glucose can be respired anaerobically.]

i Respiration is the primary source of ATP (= energy) for all the cell’s chemical reactions (endergonic or anabolic). When [ADP] is low, [ATP] must be high. This is an example of ‘feedback inhibition’ or ‘end-point inhibition’, such that the [ATP] in the cell is maintained at a constant level (= homeostasis). In extreme conditions of energy shortage, ADP can be further broken down to AMP + Pi, releasing a small amount of useable energy. The AMP so formed is the most powerful stimulant to cellular respiration. In muscles, ATP can also be replenished from a store of phosphocreatin without needing respiration at all. This allows sudden bursts of movement without the muscle running out of energy and is the reason why body-builders and weightlifters take creatin supplements.

ii In the First World War, munitions workers (almost all female) sometimes died from an uncontrollable fever, associated with dramatic weight loss. The explosives used to fill shells (picric acid) splashed on to their skin and was absorbed. Inside their body, it caused ‘uncoupling’ of the twin processes of ‘oxidation’ (i.e. movement of electrons down the cytochrome chains on the cristae of the mitochondria) and ‘phosphorylation’ (i.e. the synthesis of ATP from ADP and Pi by the movement of H⁺ions through the ATPase.) DNP ‘punches holes’ in the inner membrane of the mitochondria, allowing the H⁺ ions trapped between the inner and outer membranes to ‘leak’ back other than via the ATPase (= ‘stalked particles’). It is because this cannot normally happen that the two processes of oxidation and phosphorylation are normally said to be ‘tightly coupled’, so that each NADH molecule leads to exactly 3 molecules of ATP and every molecule of FAD leads to two molecules of ATP.

In the presence of DNP this means that most of the energy is released as heat and little, if any, is stored as ATP. The body therefore respires even quicker, leading to rapid weight loss and a high fever. The poor women in the munitions factories had to lie in ice-baths to lose heat and prevent their deaths – otherwise DNP would be a perfect ‘slimming pill’! Amphetamines (= ’Ecstasy’ or MDPA) were used in the 1960’s for they have a similar effect, altering the brain’s ‘thermostat’.

2. a

Homogenise in ice-cold Ringer’s solution (= isotonic salt/glucose solution). Centrifuge at 500+ g for 20 mins and collect the supernatant ( = liquid). Repeat the centrifugation having re-suspended the sample, to ‘wash’ it of whole cells, at 1000 g and collect the pellet. Check sample under the transmission electron microscope to ensure purity.

i Incorporated in ATP in cyclic photophosphorylation.

ii Light independent reactions are where CO₂ uptake occurs. Line 2 of the table shows that intact chloroplasts absorb 52,000 units of CO₂, whilst the membrane fraction alone absorbs just 40. Thus the stroma is the key area of the chloroplast.

3. a

Folding increases Surface Area. This is the site of the ATPase (‘stalked particles’), thus greater surface area = more ATPase.

Respiratory Quotient = (R.Q.)	CO₂ out	=	52	= 0.65
	O₂ in		80

i R.Q. = 1.0 means aerobic respiration of carbohydrate. Maize uses starch as its main energy store and germination requires oxygen as it is a process requiring lots of energy.

ii    The main respiratory substrate varies over a 24 hour period:
                   Aerobic respiration of carbohydrate          = 1.0
                   Aerobic respiration of fat                         = 0.7
                   Anaerobic respiration of carbohydrate      = > 1.0

Chapter 8 - Liver and Kidney In Chapter Questions [back to top]
1.	Terrestrial organisms have limited supplies of water and need to conserve water for their own metabolic purposes: ammonia requires water for its elimination. Ammonia is very soluble in water and also very toxic.
2.	If a person has 4 litres of blood and the kidneys receive 1 litre of blood per minute, then in one hour the kidneys receive 60 litres of blood. This means the total volume of blood passes through the kidneys 15 times in an hour.
3. a	Large proteins are not filtered out of the blood into the filtrate as they are too big to pass through the basement membrane in the Bowman's capsule. Proteins with a RMM < 68,000 are filtered. Haemoglobin has a RMM of 65,000, so a solution of Haemoglobin would not work as a blood substitute and blood in the urine is a sign of kidney damage.
b	The concentrations of glucose, urea and sodium ions in the blood plasma and in the filtrate in the renal capsule are the same. This is because these are small molecules and so easily pass through the basement membranes.
4.	There would be reduced blood pressure so less fluid would be forced through the basement membrane by ultrafiltration. With little or no fluid passing through into the Bowman’s capsule, the kidneys would effectively shut down. They would also get greatly reduced supplies of glucose and oxygen in the arterial blood, so their respiration would be lower.
5. a	When there is over 50mg per 100 cm³ in the blood then it is excreted in urine.
b	As glucose is removed from the filtrate by active transport into the cells of the first convoluted tubule and transported out of the cells by facilitated diffusion, the rate of its reabsorption is limited by the number of proteins in the membrane acting as carriers and the time available. The concentration of glucose can be so high that there are not enough proteins in the membrane to carry out their uptake.
6.	If a person suffers from diarrhoea then more water will be lost via faeces (up to 10 litres/day) and little or none in sweat and urine. The amount of water lost in breath would be the same. This causes problems with the build-up of toxins in the blood and with thermoregulation (patients usually have a very high temperature as well)
7.	It is not a good idea to drink saltwater as it contains ions which raise the concentration of the blood (lower its water potential) and increases the pressure of the blood. It stimulates the thirst centre of the grain and makes the person more thirsty. Instead, drink a little seawater adapt successfully.
8.	Metabolic water is formed by metabolic processes, e.g. respiration produces water as an end product and by condensation reactions (= anabolism) e.g. amino-acids being joined together in protein synthesis taking place in the body.

Chapter 9 - Homeostasis In Chapter Questions [back to top]
1.	Neither Neither substance contains much water, so the question refers to the composition of the two foods. Chocolate contains a mixture of fat and sugar (sucrose). The enzymes that digest these are produced in the pancreas and thus found in the small intestine, so there will be little absorbed from the stomach. The fat in chocolate will ‘line’ the stomach and slow down the absorption of substances by creating a barrier to water-soluble sugars. This can be used to slow down the uptake of alcohol too. A large fatty meal slows down uptake (thus getting you drunk slower); little or nothing to eat and a fizzy drink make the alcohol reach the bloodstream fastest. This gives rise to the observation that champagne does not give you a hangover – in fact you get as drunk on less booze, thus less hangover! If any student wishing me to demonstrate this, I will be happy to oblige!! Toast contains mainly starch and protein and the enzymes that digest these chemicals are found in the mouth and stomach as well as in the pancreatic secretions (remember – the pancreas produces every digestive enzyme!). Since the original food molecules are too big to be absorbed directly, they can only be absorbed (assimilated = absorbed and used) once they have been digested. So there can be some absorption of maltose from the stomach. Hence toast faster. N.B. This is a dumb question. Chocolate invariably contains either invert syrup or glucose (since they are cheap) and so these hit the bloodstream very quickly indeed. In any case, sucrose can be absorbed directly from the stomach, so the basic premise of the question is wrong.
2. a	b cells of pancreas (too high); a cells of pancreas (too low)
b	b cells secrete insulin; a cells secrete glucagon. Both are transported in the bloodstream to the responding cells – mainly in the liver and muscles.
c	Mainly liver and muscle cells, but most cells in the body respond to some extent.
Chapter 9 - Homeostasis Answers to Assignment Questions [back to top]
1. a	Some innate action which improve the organisms chances of survival. e.g. ‘Spadefoot toads burrow deep into the soil.’ [Line 27] and ‘Juvenile green toads basking in the sun’ [Line 33]
b	An action internal to the organism - in effect, some form of homeostatic mechanism – that improves survival or efficiency. e.g. ‘Can absorb water through their skin’ [Line 11] and ‘Oxygen and carbon dioxide can also pass readily through it’ [Line 13] and ‘Increased permeability of the bladder...makes up for water lost by evaporation’ and ‘they absorb water through the skin more readily...’[Line 23]
2. a	Diffusion in living things only occurs in solution. Thus the gases have to dissolve in water before they can be exchanged with the external environment. Having a moist skin will permit this to occur; if the skin were dry, no gases could dissolve and so no exchange would be possible.
b	Loose folds will increase the surface area of the skin and so allow faster and more efficient diffusion (Fick’s Law). At high altitude, the partial pressure of oxygen is lower than at sea-level and so this is an evolutionary adaptation to low oxygen levels. The atmosphere will also be drier and colder, so the Titicaca toad must also have other adaptations too. [see sheet for further useless facts about this wretched body of water]
3.	When the bladder is permeable, water will move freely by osmosis from the bladder back into the animal’s body when the water potential is low, i.e. when the animal is drying out.
4. a	Water potential will fall. Note that eventually the water that remains is attached to the soil particles by ionic and chemical forces and no solution of soil water remains.
b	Urea in the spadefoot’s body fluids will lower their water potential. Since the animal’s skin is permeable to water, osmosis will occur. The more concentrated the body fluids, the more likely water can be absorbed from the surrounding soil; in extreme drought, water loss will certainly be reduced.
5.	Basking in the sun will lead to a rise in temperature. This will tend to lead to increased evaporation of water from the skin, which will have a cooling effect. The amphibians have long realised this, of course, and pump blood through their skin to remove heat quickly and so ensure that the surface of the animal is kept cool and so evaporation is minimised. Furthermore, the contents of the cells on the outside of the skin become more concentrated, so reducing the tendency of water to be lost by evaporation; finally, a layer of mucilage on the outside of all amphibians reduces water loss, particularly when it dries out. Finally, by basking on a rock which has been exposed to the sun already, they are able to absorb heat through their lower surface which, being pressed tightly against the rock, prevents evaporation. All in all, an extraordinarily stupid question (if there were no evolutionary advantages to basking, amphibians would not do it).
6. a	The internal temperature of the frog remains just below that of the surrounding air from 30^oC to 40^oC, after which it remains more or less constant, up to an air temperature of 45^oC. This is caused by the conduction of heat from the air to the frog – an ectotherm.
b	This is possible because water loss through the skin is minimal until 40^oC, when it rises sharply cooling the frog by the loss of latent heat – so frogs sweat too! Line 39 suggests a waxy layer, so probably this melts at this temperature.
Chapter 9 - Homeostasis Answers to Exam Questions [back to top]
1. a	Straight line – over ‘glucose reabsorbed’ line and parallel to ‘glucose excreted. i.e. add the two values together.
b	Only in the proximal convoluted tubule. (only place apart from gut to have microvilli)
c	The body reabsorbs glucose to avoid energy loss. The normal concentration of glucose in the blood is 150 – 180 mg 100cm^-3. From 0 – 200 mg 100cm^-3all the glucose is reabsorbed i.e. no glucose in normal urine. From 200 - 380 mg 100cm^-3, the rate of glucose re-absorption levels off, as the active uptake of glucose in this region of the nephron is becoming saturated. From 400 mg 100cm^-3 upwards, the nephron is fully saturated and all glucose over 400 mg 100cm^-3 is excreted. The proximal convoluted tubule is relatively short and is the only part of the nephron in which active uptake of soluble organic molecules occurs.
2. a	i Water released as a result of metabolic processes e.g. respiration (glucose + oxygen ® carbon dioxide + water).
	ii Food is wet (lettuce >90% water, even seeds are 15 – 20% water)
b	i Water has the formula H₂O. Lipids have large quantities of hydrogen in their molecule and relatively little oxygen. Thus, when oxidised, the mass of water released is considerably greater than the mass of hydrogen in the original lipid.

Chapter 3 - Evolution and Classification of Species In Chapter Questions [back to top]
1.	Biologists are undecided about the biological status of the bandro because it lives in isolation from other lemur species in papyrus reed beds. It could be a sub-species or a separate species. The fact that it has several local and two scientific names adds to the confusion; is it two separate species, a single species or merely a sub-species of another lemur? Breeding data, DNA fingerprinting and / or protein electrophoresis studies would help to clarify the situation, but the data is not provided in the chapter.
2.	Homologous chromosomes pair at prophase 1 of meiosis.
3.	The barriers that led to the evolution of some many different species of fruit fly in the Hawaiian islands were areas of sea between the different islands but also cooled larva flows that separated different blocks of forest on the islands (the blocks of forest are isolated by sufficient distance to prevent gene flow between isolated fruit fly breeding groups or DEMES).
4.	Allopatric speciation is where new species arise from groups of the species that have been physically separated by some physical barrier; sympatric speciation occurs by isolation mechanisms that do not involve physical separation (e.g. different courtship behaviours).
5.	The binomial name for tomato is Lycopersicon esculentum (the generic name followed by the specific name – BINOMIAL as it has TWO names).
6.	Euglena has a eukaryotic cell structure as internal membrane bound organelles like mitochondria and chloroplast (the nucleus can be seen but is not labelled).
7.	There are many nuclei concentrated in the tip of the fungal hypha because this is where growth occurs. It is therefore an area where protein synthesis will occur and so if the nuclei are present m-RNA will have a shorter distance to travel to reach the rough ER.
8.	An organism with cells containing nuclei and a cell wall is eukaryotic. The kingdoms that have such cells include PROTOCTISTA (not all have a cell wall, but some, such as algae, do), FUNGI and PLANTAE.

Chapter 6 - Decomposition and Recycling In Chapter Questions [back to top]
1.	The iron cycle would look almost exactly the same as the diagram given on p 115.
2.	A maize plant would take up carbon: a As CO2 from the air through the stomata (in leaves) and lenticels (in stem); b NOT as hydrogen carbonate from the soil; c NOT as carbonate (which is mostly insoluble) from the soil; d NOT as organic molecules, these would give the plant CO2 when these decayed, so releasing the gas to the atmosphere.
3.	The trend in atmospheric CO2 would rise between 1990 and 2000 as a result of increasing world industrial growth, energy generation increases and greater transport (more cars, planes etc).
4.	An increase in CO2 levels may lead a farmer to gain a smaller crop profit because: Although the yield might rise, the quality of the crop might fall; Increased weed growth might cause crop problems and extra costs to control; The quantity of biomass which is not ‘marketable yield’ might produce higher disposal costs.
5.	Differences between nitrification and denitrification: The process of denitrification requires energy input whereas nitrification releases energy; denitrification breaks molecules down in complexity whereas nitrification increases molecular complexity.
6.	ATP is required in nitrogen fixation in order to provide the energy for the reduction of nitrogen (N) to ammonia (NH3).

Chapter 7 - Uptake and Loss of Water in Plants In Chapter Questions [back to top]
1. a	Provides more space for water to flow freely.
b	End walls would block the flow of water (and interrupt the continuous column of water which is essential for transpiration to work).
c	This strengthens the walls and allows them to withstand the suction pressure i.e. it stops the cells collapsing inwards. A second effect is that it allows the plant to support a greater weight and so grow taller (e.g. trees). Being waterproof, a layer of lignin also kills the cells.
d	These allow water to flow sideways and so bypass any blockages in the water stream perhaps caused by insect attack. Because of their ‘self-sealing’ structure, they also prevent water leaking out of the stem when it is damaged.
2.	Because water is ‘pulled’ up the stem by the loss of water molecules from above. Without cohesion, no upward movement of water molecules would occur.
3.	At night, the stomata close and the water loss by evaporation (into cooler and more humid air) is greatly reduced. The volume of water entering the plant through the roots remains much the same, so root pressure forces water out of the stomata and the leaves drip. This is known as guttation.
4.	At midday the sun is at its strongest. The insolation will tend to raise the temperature of the leaf well above that of the surrounding air –is quite common. Normally the plant loses enormous quantities of heat from the evaporation of water – 1000 litres per day is normal for a large tree! This cooling effect is most marked in a wood or forest where the air temperature can be 5^oC or more below that outside. For this reason, trees are widely planted in the streets and car-parks of sunny cities to provide natural ‘air-conditioning’. They also absorb dust from the air. With water loss reduced or stopped, the temperature of the leaf will rise and might cause the enzymes to denature even faster!
5.	GCSE question – high wind, high temperature, daylight (= open stomata) and low humidity.