Help on A2 Chapter Questions

Indge, Rowland and Baker (2000) Further Studies in Biology 
Hodder & Stoughton, London

Provided by: Ian White

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A2 Biology, Specification A A2 Human Biology, Specification A

Ch.1 -  Answers to in-chapter questions

Ch. 2 - Answers to in-chapter questions

Ch. 3 - Answers to in-chapter questions

Ch. 4 - Answers to in-chapter questions

Ch. 5 - Answers to in-chapter questions
           Assignment
           Exam Questions

Ch. 6 - Answers to in-chapter questions

Ch. 7 - Answers to in-chapter questions  

Ch. 8 - Answers to in-chapter questions

Ch. 9 - Answers to in-chapter questions  
   
        Assignment
          
Exam Questions

Ch. 10 - Answers to in-chapter questions

Ch. 11 - 

Ch. 12 - Answers to in-chapter questions

Ch. 13 - Answers to in-chapter questions
            
Assignment
            
Exam Questions  
 

Ch. 7 - Answers to in-chapter questions

Ch. 8 - Answers to in-chapter questions

Ch. 9 - Answers to in-chapter questions

Ch. 10 - Answers to in-chapter questions

Ch. 11 - Answers to in-chapter questions

Ch. 12 - Answers to in-chapter questions

Ch. 13 - Answers to in-chapter questions

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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

Io is recessive  - it is only expressed in the phenotype when it is in a homozygous state (i.e. IoIo ); it is ‘masked’ if IA or IB are present.

     b

The alleles IA and IB are codominant because when they occur together (as in IAIB ) 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 XHXh as they are a combination of a fusion between XH and Xh 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 34oC.

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 HbS would not confer an advantage in the heterozgote HbSHbA,, which is the case in the presence of malaria.


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 4 - Numbers and Diversity

In Chapter Questions  [back to top]

1.

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.

2.

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

3.

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

N1  x  N2

    Nm

                 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

4.

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.

5.

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.

6.

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.

7.

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!).

8.

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).

9.

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]

1.

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. 

2.

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)

       b

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)

       c

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

4.

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.

5.

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

6.

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.

7.

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

8.

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.

9.

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.

       b

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.

       b

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.

3.

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.

       b

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

5.

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

       b

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

      50        

= 0.004%.

1.25 x 106

Thus, the energy left at the 6th 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).]

        c

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

       d

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).

       e

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.]

        f

 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.

       b

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.

       b

Respiratory Quotient = (R.Q.)

CO2 out

=

 52 

= 0.65

O2 in 

 80

        c

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 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 5oC 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.


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 30oC to 40oC, after which it remains more or less constant, up to an air temperature of 45oC. 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 40oC, 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-3 all 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

    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.

ii       Another incorrect statement. The table shows that slightly more water is produced than the mass of fat respired (1.13g water from 1.06g fat). It also shows that the volume of water evaporated from the lungs is greater still (1.80g). But this water would have been evaporated anyway. The figure missing from the table is the amount of water produced from an ‘energy-equivalent’ mass of carbohydrate (glycogen) so that the net gain of water from fat metabolism can be seen. The animal needs energy and needs to conserve water. Fat is a better energy source than carbohydrate, since the hydrogen in the latter is already fully oxidised so it is ‘water neutral’. What a useless question!

Titicaca

Lake Titicaca is the world's highest navigable lake, lying at 3,810 m above sea level in the Andes, astride the border between Peru to the west and Bolivia to the east. The name Titicaca has been translated as ‘Rock of the Puma’ or ‘Crag of Lead’. It lies in a vast basin (about 58,000 square km in area) that comprises most of the Altiplano (High Plateau) of the northern Andes. On the Bolivian shore some of the highest peaks in the Andes rise to more than 6,400 m.

Titicaca is the second largest lake in South America, covering some 8,300 square km and extending in a northwest-to-southeast direction for a distance of 190 km. It is 80 km across at its widest point. A narrow strait, Tiquina, separates the lake into two bodies of water, averaging between 140 and 180 m deep, but the bottom tilts sharply toward the Bolivian shore, reaching a depth of 280 m in the NE corner.

More than 25 rivers empty into Titicaca; the single outlet empties only 5 percent of the lake's water; the rest is lost by evaporation under the fierce sun and strong winds of the dry Altiplano.

Titicaca's level fluctuates seasonally and over a cycle of years. It was formerly believed that Titicaca was drying up, but modern studies refute this, indicating a more or less regular cycle of rise and fall. The waters are limpid and only slightly brackish, with surface temperatures averaging 14º C.  Fish life consists principally of two small species of killifish and a catfish. In 1939, trout were introduced, while a large frog (Telmatobius) inhabits the shallower regions of the lake.

Forty-one islands, some of them densely populated, rise from Titicaca's waters. The largest, Titicaca Island, lies just off the tip of the Copacabana Peninsula in Bolivia.

Ruins on the shore and on the islands attest to the previous existence of one of the oldest civilizations known in the Americas, antedating the Christian era. The chief site is at Tiahuanaco, Bolivia, at the southern end of the lake. On Titicaca Island ruins of a temple mark where, according to the tradition of the Incas, the legendary founders of the dynasty, Manco Capac and Mama Ocllo, were sent down to Earth by their father, the Sun, emerging from the depths of Lake Titicaca to found their empire.

With lofty snow-capped peaks along its far shores, the vast blue lake at 3,800m is one of the Andes' most enchanting scenes.


Chapter 10 - Gas Exchange

In Chapter Questions  [back to top]

1.

There is no question 1!

2.    a

when the buccal cavity (= red) line is below the horizontal ‘0’ line.

       b

when the opercular cavity (= blue) line is below the horizontal ‘0’ line.

       c

when the two lines cross

       d

whenever the blue line is below the red line

       e

whenever the blue line is below the red line.

3.   

During the short period when the blue line is above the red line.

4.

They prevent the trachea collapsing when the pressure inside is lower than outside (i.e. most of the time – the insect uses internal pressure to ‘stiffen’ its exoskeleton in much the same way that a plant cell inflates itself with water.

5.     a

Note the key word ‘from’!  At noon photosynthesis will be at its maximum, so oxygen will be diffusing out of the stomata.  The leaf will also be transpiring at its maximum (unless it is a cactus, which closes its stomata during the day to conserve water), water vapour will be leaving the leaf.

       b

At night the stomata will be shut tight, so little or nothing will be leaving the leaf.   (In practice, a little water vapour and carbon dioxide from respiration will diffuse out)


Chapter 12 -  Digestion and Diet

In Chapter Questions  [back to top]

1.

Faeces is not waste products of biochemical reactions and so are not excreted from the body. Faeces is made of undigested food, bacteria and dead cells from the lining of the gut – a process known as egestion. 

2.

A particular amylase enzyme works at the particular place in a starch molecule which has the complimentary shape to that of the active site of the amylase enzyme. Hence the specific 3D shape due to tertiary level structure of the protein determines the activity of the enzyme.

3.

Endopeptidases are first released and then exopeptidases as this is more efficient since the action of endopeptidases is to hydrolyse peptide bonds within the protein molecule thereby creating more terminal bonds for the exopeptidases to hydrolyse.  If exopeptidases were the only enzymes secreted the protein digesting process would be much slower as the amino acids would be removed two at a time, one from each end of the protein molecule.

4.    a

When a molecule of triglyceride is broken down by lipase 3 molecules of fatty acids and one molecule of glycerol are formed. Therefore a total of 4 molecules.

       b

When a molecule of triglyceride is broken down by lipase to form monoglyceride and fatty acids one monoglyceride molecule is formed and two fatty acid molecules giving a total of 3 molecules.

5.

A mutualistic relationship is a relationship between two organisms where both benefit. The protozoans living in the gut of the termite engulf small particles of food material from the termite by phagocytosis and hydrolyse the cellulose by secreting enzymes. The protozoans produce glucose. The termites use the glucose as a respiratory substrate but because the microbes respire anaerobically they also produce fatty acids which are absorbed and used by the termites, both as a respiratory substrate and as a structural component of their exoskeleton.

6.    

The rumen is used to produce enzymes from micro organisms just as a fermenter does in:
a.   
the temperature is closely monitored and controlled (stays within 38 - 42 degrees Celsius)
b.    
the pH is prevented from changing by the presence of phosphate and hydrogen carbonate buffers in saliva as the fermenters have inorganic buffers.
c.   
the contents of the rumen are mixed by peristaltic action as the contents of fermenters are                   mixed with paddles.  
d.   
Gas is released by belching so that pressure does not build up: valves are used to release gas from fermenters.

7.

Urea from the liver will enter the plasma of the blood by diffusion and leave the liver via the hepatic vein and return to the heart. From the heart the blood will be pumped around the body and so some of the urea will reach the salivary glands where it diffuses into the saliva and can be secreted.

8.

The small intestine is long and circular in cross section so its internal surface area for absorption is large. The surface of the epithelial cells adjacent to the lumen of the gut is greatly expanded by the presence of folds called villi. Each villus has its surface area still more increased by the presence of microvilli.  The total surface area in a human gut is about 50 m3.

9.

By the time food is halfway along the small intestine it will have received digestive juices from:

  • Salivary glands

  • Gastric glands in the stomach

  • Pancreas

  • Liver

  • Intestinal glands of the duodenum and ileum

10.

If the stomach of a newborn mammal does not secrete hydrochloric acid it will not provide the correct pH for the activation of pepsinogen and so pepsin will not be formed and proteins will not be digested.  If trypsin is not present in the small intestine then again proteins will not be digested into amino acids and so the new born will not be able to absorb amino acids.  However, proteins can be absorbed directly through the epithelial cells by pinocytosis.  The large surface area of the small intestine means that there can be many pinocytotic channels present.  This allows for the uptake of protective proteins (e.g. antibodies) from the mother’s milk in the days immediately after birth.

11.

If digestive juices are secreted when no food is present in the gut then water, hydrochloric acid, proteases, lipases, carbohydrases, bile, sodium hydrogen carbonate would all be wasted.

12.   a

The diameter of the pupil changes as a result of the nervous system because the change needs to be quick as high light levels could damage the retina and light levels can change suddenly.

       b

Hormones are more likely to control the concentration of a chemical within the blood since they travel in the blood to the target organs and only a gradual change is needed over the whole body – it is part of homeostasis.

13.

Women in the age range 15 - 50 years menstruate and lose blood - and hence red blood cells - on a regular basis and need to have an input of iron so that they can form haemoglobin – the major component of red blood cells.

14.

Graph should show a small drop in mass with each moult, but an increase in length in a ‘step-like’ pattern with times of rapid growth and times of no growth at all as the exoskeleton prevents growth.

i.e.                    not             


Chapter 13 -  Nervous system and Receptors

In Chapter Questions  [back to top]

1.

Motor neurones are also known as effector neurones as they carry the impulse to the effector (muscle or gland) where the response to the stimulus occurs.

2.

Motor, relay and sensory neurones all possess a cell body containing a nucleus, synaptic knobs. Dendrons and dendrites are present in all three but the sensory neurone does not have these coming off the cell body.

3.

The membrane of a neurone is said t0 be polarised because there are different charges on its inside and outside.

4.    a

The resting potential is -70mV

       b

The action potential is +40 m V

       c

The action potential (when the membrane is not at +40mV) lasts for 4 milliseconds.

5.

The neurotransmitter is released from the pre-synaptic knob by exocytosis.

6.

When pressure on a Paccinian corpuscle is released, it springs into an elongated shape at right angles to the original deformation of shape so that other stretch mediated channels are open and another action potential is produced this time informing the person that the pressure has been released, thus giving more time to recover if first the elongated shape and then the round shape have to form and so allowing more time for repolarisation.

7.

Light is refracted at both surfaces of the cornea (covered with conjunctiva), and the lens (both outer and inner surfaces).

8.

The neurotransmitter released by rods has an inhibitory effect, i.e. there is a reduction in the amount of transmitter and when the level reaches a threshold low level then an impulse is generated.  

9.

If there is a deficiency of vitamin A and retinal is a derivative of vitamin A then there will be a deficiency of retinal and therefore formation of rhodopsin will be hindered.  Hence, night vision (vision in dim light) will be very poor.

10.

A hot plate is not dropped if the temperature sensors in the skin send an impulse along the sensory neurone in the arm to a relay neurone in the grey matter of the spinal cord, then up the spinal cord, along another relay neurone to the brain (cerebrum) where the brain makes a decision to override the pain in order to safely place the hot plate on a surface.  Therefore this is not an innate reflex.


Chapter 13 - Nervous System and Receptors

Answers to Assignment Questions [back to top]

1.    a

An action potential is created when the sodium channels in the membrane of the nerve open to allow a sudden influx of sodium ions.  (Normally, of course, the cell actively pumps out sodium ions, creating a ‘resting potential’ of around – 92 mV (c. 1/10th volt) within the cell.)  ‘Voltage-sensitive’ refers to the fact that the gate only opens when a stimulus above a certain critical value reaches it.

      b.

Each ion has a different diameter (particularly when associated with water molecules as in the human body).  The ion channel protein will only allow ions of the appropriate charge and diameter to enter.  It is not 100% accurate: strontium is accepted as a substitute for calcium, which can be a problem if the strontium is radioactive (Sr90) – this then becomes concentrated in bone tissue leading to cancer.

2.     a

Tetrodotoxin. [An aminoperhydroquinazoline poison]{Tetrodotoxin}
will block sodium channels so that they will not open to allow the action potential to be generated.  This will have the effect of incapacitating the nerve, leading to paralysis (motor neurones) or numbness (sensory neurones).

 

Tetrodotoxin (or)
Octahydro-12-(hydroxymethyl)-2-imino-5,9:7,
10a-dimethano- 10aH-(1,3) dioxocino(6,5-a)
pyrimidine-4,7,10,11,12-pentol 

 

3.

The large number of mitochondria immediately next to the pre-synaptic membrane.  Presence of what look like microvilli (?), implying secretion.

4.

Bungaratoxin is a polypeptide, so it will be digested in the stomach. It is too large to be absorbed intact (unless you have gastric ulcers).

5.     a

Bungaratoxin would be injected in low dose into a large animal (e.g. horse).  A few days later another injection would be made.  These stimulate the horse into producing large quantity of the specific antitoxin, which can then be extracted from a citrated and centrifuged sample of horse blood.  When these antibodies are mixed with a fluorescent dye (attached other than to the toxin binding site), they can then be injected into a hapless iddy-bitty mouse that had earlier been injected with bungaratoxin (you know he really did have a bad day!)  This animal is then killed (swing round by the tail and bang head on edge of bench – no, not you, the mouse!) and the neuromuscular junctions examined.  The fluorescent antibodies will only bind to the bungaratoxin, which had itself attached to the neuromuscular junctions.

       b

No information in part (a) is of any relevance that I can see.  Breathing involves two main sets of muscles, the intercostals and the diaphragm.  Both sets are under the control of the autonomic nervous system i.e. they need nervous stimulation before they will contract.  If krait venom (bungaratoxin) has the effect of blocking the secretion of neurotransmitter from the neuromuscular junctions, then these muscles will be paralysed.  The heart would not be affected, so the individual would suffocate, or be eaten alive, but paralysed....hmmm!

6.

Black widow spider venom.  The supply of acetylcholine in the pre-synaptic vesicles is limited.  If it is all released under the influence of the toxin, then there will be no more available and neurotransmission will cease.

Organophosphate insecticides.  These stimulate the parasympathetic nervous system, by binding to acetylcholinesterase and so leading to acetylcholine remaining in the synapse for far longer than normal.

W-conotoxin from cone shells.  By preventing Ca2+ ions from crossing the pre-synaptic membrane, the vesicles containing the neurotransmitter are no longer stimulated to fuse with the membrane and so remain intact.  No neurotransmitter = no nerve impulse!

For further information on neurotoxins visit: http://faculty.washington.edu/chudler/toxin1.html 

Chapter 13 - Nervous System and Receptors

Answers to Exam Questions [back to top]

1.     a

i      After 1 millisecond, the Sodium ions can flow freely into the neurone, whilst potassium ions can only leave the cell more slowly; these ions have a positive charge and so the membrane potential becomes rapidly, but briefly, positive.

ii     After 1.5 milliseconds the membrane becomes impermeable to Sodium ions again; potassium ions are still able to cross the membrane until 4 milliseconds;  the resting potential is restored by the ‘sodium pump’ actively pumping sodium ions out of the cell, whilst the inflow of the balancing potassium ions is prevented.

       b

During such periods, the cell will be actively pumping Sodium ions at a rate considerably above its normal resting rate; this requires energy (in the form of ATP) which the cell obtains from respiration – hence intense nervous activity = higher metabolic rate. 

2.    a

i    Innate.

ii    If the worm is in its burrow, dim light will come from the entrance and so moving towards it would enable the worm to feed and reproduce.  Bright light would indicate daytime and the worm avoids this [u-v light paralyses worms, hence a worm caught on the surface on a sunny summer morning is unable to return to the safety of its burrow.  Thus the early bird…

       b

Earthworms do not have a waterproof skin as they breathe through it; thus in dry conditions rolling into a ball deep underground reduces their surface area and it is less likely that the soil will dry out completely.

       c

These giant axons run the whole length of the worm and allow a rapid contraction of the longitudinal muscles in every segment.  The effect is to shorten and widen the worm, jamming it into its burrow and moving it as far as possible from any danger (e.g. a not so early bird….)

3.     a

Light falling on the light-sensitive pigment rhodopsin (made up of opsin and cis- retinal) breaks it down into trans- retinal which can no longer bind to opsin, which is therefore released. This causes the cell membrane to become more permeable to both Na+ and K+ ions, making the cytoplasm more negative as the K+ ions leave the cell.  This is called hyperpolarisation and causes a generator potential to be produced in the cell.  [Re-synthesising rhodopsin requires ATP and takes up to 20 minutes to complete, hence we cannot see well in the dark for some time after the lights go out – just as well, really!]

       b

Normally the voltage-sensitive Na+ gates in the membrane are closed and so the whole length of the neurone has a resting potential (-92mV). When stimulated, these gates open and Na+ ions flood in, causing a rapid depolarisation – an action potential.  This lasts only momentarily; then the Na+ gates close and the K+ gates open.  This restores the resting potential, but disturbs the adjacent membrane, in turn causing the Na+ gates there to open… thus the action potential spreads along the surface of the neurone.  This process can be speeded up many hundreds of times by insulating the neurone (by Schwann cells).  This causes the impulse to leap from the gap between one Schwann cell and the next (a Node of Ranvier).  This is rather like ‘skipping’ stones over the surface of a pond (Barnes Wallis’ so-called ‘bouncing bomb’ (= Highball) used in the famous 617 Squadron ‘Dambuster Raid’ of May 1942 used the same principle).

        c

i      Summation occurs when a number of stimuli, each below the critical threshold, are received in such a short space of time that they trigger an action potential.  In the diagram, there are 3 rod cells connected to one bipolar neurone, so stimulating more than one rod cell weakly (or one cell repeatedly) will cause the axon to be triggered and a flash of light to be seen.

1. (a)    (i)      ii     A synapse has two different membranes. The pre-synaptic membrane releases the transmitter substance (e.g. acetyl choline) whilst the post-synaptic membrane is sensitive to this chemical and responds by triggering an action potential.  Since the two membranes on each side of the synapse can each only carry out a single function, it follows that a ‘reverse impulse’ is impossible (rather like a diode in an electrical circuit.).

 

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Last updated 20/06/2004