Help on AS Chapter Questions

Indge, Rowland and Baker (2000) A New Introduction to Biology 
Hodder & Stoughton, London

Provided by: Ian White

cover

Ch.1 -  Answers to in-chapter questions
          
Answers to exam questions
          
Answers to assignment questions

Ch. 2 - Answers to in-chapter questions
          
Answers to exam questions
          
Answers to assignment questions

Ch. 3 - Answers to in-chapter questions
          
Answers to exam questions
          
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Ch. 4 - Answers to in-chapter questions

Ch. 5 - Answers to in-chapter questions
          
Answers to exam questions
          
Answers to assignment questions

Ch. 6 - Answers to in-chapter questions  

Ch. 7 - Answers to in-chapter questions  
          
Answers to exam questions
          
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Ch. 8 - Answers to in-chapter questions
          
Answers to exam questions
          
Answers to assignment questions

Ch. 9 - Answers to in-chapter questions
          
Answers to exam questions
          
Answers to assignment questions

Ch. 10 - Answers to in-chapter questions
                
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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 - Cells and Cell Structure

In Chapter Questions  [back to top]
 

1.  a It has a cell wall. (cellulose in plants, murein in bacteria, chitin in fungi)

     b

Many options!  It has a nucleus; chromosomes; linear (not ccc) DNA; membrane-bound organelles; mitochondria; endoplasmic reticulum (e.r.); 80S ribosomes (not 70S); much larger (20-100m, not about 1.0 m).
 
2. 10 x 40 = 400. 
 
3. Magnification is how much bigger an object appears; it does not give more detail!  Resolution is the minimum distance between two objects that allows them to be separated.  This relates directly to detail – higher resolution is always better!  Resolution is determined by the quality of the lenses; by how close the object is to the lenses (nearer is better); and by the wavelength of the radiation used – the maximum resolution being about ½ the wavelength.  Thus light microscopes are limited to about 0.2m (= 200nm) and transmission electron microscopes to < 0.1 nm, so they can resolve individual atoms!  N.B.  For AS there is no such thing as an electron microscope (e.m.)!  There are,  however, transmission e.m. and scanning e.m.!
 
4. Because the electron beam would be scattered by the molecules in air.  The result of this is that living cells (which need water) can never be observed in an e.m. (of either sort!)
 
5.  a Because they have not been stained and so won’t show up;
 
     b Because the resolution of the light microscope used is not good enough.
 
6.  a  Because water flows down the water potential gradient, it would otherwise enter membrane-bound organelles and cause them to swell up and burst (or shrink)
 
     b Because enzymes would be denatured at different pH values and so the activity of some organelles would be affected (though they might look the same)
 
     c Enzymes are denatured by heat, so the same point as b) above; also the process of homogenisation generates quite a lot of heat (friction) and so you want to prevent the proteins being ‘cooked’
 
7. Only whole cells and nuclei would have been removed, so all other organelles would remain.  Remove the supernatant and re-centrifuge at 20,000G for 20 minutes, then remove the supernatant and re-centrifuge again at 100,000G for 60 minutes.  Discard the supernatant this time, resuspend the pellet and repeat the last step (this ‘washes’ the pellet and ensures it is 99% pure!)
 
Chapter 1 - Cells and Cell Structure

Answers to Exam Questions [back to top]

1.   a
Feature Epithelial cell from small intestine Prokaryotic cell
Golgi apparatus a r
Mitochondria a r
Nuclear Envelope a r
Plasmid r a
Ribosomes a a
  
     b

i.   
With a transmission e.m., the resolution is about 1000 times better than that of the
light microscope, due to the shorter wavelength of the electron beam.  The better the resolution, the more fine detail can be observed.
ii   
The treatment of the cell that is necessary before it can be studied in any form of e.m. leads to the formation of many artefacts, or distortions of reality.
2.   a Nuclear pore.
      b Y = 80s ribosomes: synthesis of proteins: this is where mRNA is translated into protein
     c Z = Golgi apparatus: proteins are modified after synthesis (often by adding carbohydrate groups).  Proteins are also surrounded with membrane, prior to secretion from the cell (e.g. hormones, digestive enzymes)

Length = 20mm ÷ 30,000 = 0.67μm

Chapter 1 - Cells and Cell Structure

Answers to Assignment Questions [back to top]

1. RBC’s contain uniformly distributed haemoglobin.  Centre of cells is paler; thus centre must be thinner.
2. 8mm ÷  1000 = 8.0 μm
3. Evaporation of water from within the cells reduces their volume; cell membrane remains the same length, so ‘wrinkles’ appear

OR

Evaporation of water from the plasma around the cells lowers its water potential.  This causes water to leave the cells by osmosis, thus causing the cells to shrink in volume and the ‘wrinkles’ to appear as above.

4. It contains lipid.  This is dissolved by the detergent, causing the cells to burst.  This is why detergents are very good disinfectants – they dissolve bacterial cell membranes and disable those viruses that rely on a lipid coat to penetrate the host cell
5.   a [left] ‘dog’s bone’
[right] ‘oval’
[horizontal] ‘ring doughnut’
      b In this scanning e.m. electron micrograph (book is wrong!) the cells are all at different angles and so all have different shapes.
6. Great resolution and the obviously false colours.?
7.
Conclusion Evidence
The plasma membrane allows water molecules to pass through it The RBC’s in a smear that is left to dry slowly appear smaller and crinkled around the edges
Many of the molecules in the cytoplasm are of the same type The RBC’s in the picture are a uniform dark colour
RBC’s do not use oxygen and cannot respire aerobically There are no mitochondria present
RBC’s are biconcave in shape The appearance of the cells when seen with a scanning e.m.
RBC’s do not contain DNA, so they cannot make proteins There is no nucleus present
 

Chapter 2 - Getting In and Out of Cells

In Chapter Questions  [back to top]

1. Raw materials - oxygen; glucose; amino-acids; lipids; mineral ions; water etc.

Waste products – Carbon dioxide; urea

2. Membranes are made of organic molecules (proteins and phospholipids).  These are made of elements with low atomic numbers and so do not absorb electrons and do not show up well in transmission e.m. photomicrographs.  In addition, the membrane is fluid and so the image will only show one moment (which might not be typical).
3. Hydrophilic = ‘water-loving’ so found where the water is (outside)

Hydrophobic = ‘water-hating’ so found where there is no water – inside the membrane.  Note that it is possible for the centre of any large molecule (e.g. protein) to be ‘dry’ even though the molecule itself is in water – just like the inside of a submarine is dry.

4. Cholesterol will reduce the movement of the membrane proteins which are essential for molecules to cross the membrane – particularly by active transport.  Note that cholesterol is essential in our diet (we cannot make it) and that without it our membranes would not function. It is only when excess cholesterol is deposited within our blood vessels that a problem occurs.
5. If the concentration difference across tissue B was much greater than the concentration difference across tissue A; OR if tissue A had a higher concentration  inside it; OR if the temperature of tissue B was higher, allowing the membrane to be more fluid AND, since the molecules will have more kinetic energy, they will collide more often with the membrane and so tend to cross it more readily
6. B is the slowest; A has a shorter distance across it (thus faster diffusion); C has a much larger surface area (microvilli), thus faster diffusion.
7.  a Lipid-soluble molecules are soluble in the hydrophobic core of the membrane and so can cross it (how they ever reached the cell membrane in the first place, given that there is a very wet solution (blood or tissue fluid) outside the cell is a moot point!)
     b Small molecules (mainly gases) can diffuse between the various molecules that make up the membrane.
8.    Easy questions!  Water always flows DOWN the water potential gradient, so it will flow from solution P to solution Q and from solution S (pure water) to solution R.
Chapter 2 - Getting In and Out of Cells

Answers to Exam Questions [back to top]

1    a i.    Messed up question!  Pressure potential is positive, so the net water potential is: -10 +3 = -7Mpa

Water always flows down the water potential gradient, so water will flow into the cell top right (-12Mpa) from both the other cells.

ii.    Pure water is 0, so any solution must have less water in it, i.e. a negative water potential.
      b i.    Because glycerol is equally soluble in the lipid membrane and in the biological membrane
i.    Because the biological membrane pumps sodium ions as part of its active transport system.  It also has protein channels that allow specific ions through, which the synthetic membrane lacks
2.   a i.    level of detail visible in the mitochondria;
     
level of detail visible in the nuclear envelope;
     
the nuclear pores are visible
     
the microvilli are clearly visible
ii.   A bacterial toxin has damaged the membrane (it actually blocks the sodium pump channels), destroying the microvilli, thus reducing the surface area available for absorption and so reducing water uptake from the gut (i.e. you get the squits!)
      b Slow down absorption due to the reduced surface area, as described above.  Glucose absorption is an active process, so the damage to the microvilli will have a profound effect; you pump water out (with sodium ions) and then allow then to re-enter ‘dragging’ glucose with them.  If the reabsorption cannot take place, you both dehydrate and also lose salt, leading to an erratic heartbeat, severe cramps and death.
Chapter 2 - Getting In and Out of Cells

Answers to Assignment Questions [back to top]

1.

A = Ultra-filtration of the blood caused by the high blood pressure in the glomerulus, in which 10% of the water and an equivalent quantity of soluble molecules with a RMM below 48,000 (haemoblobin is 44,000, hence a simple solution of haemoglobin would not be effective as blood)

B = re-absorption of 90% of the water and 100% of the glucose (by active transport) and amino-acids, means that by the time the fluid enters the Loop of Henlé it contains few organic molecules of use to the body.

2.   a The only process that could ensure that all the glucose is re-absorbed, is active uptake
      b Microvilli give a large surface area; mitochondria provide the ATP (from aerobic respiration) to allow active uptake (and / or active secretion of glucose to maintain the concentration gradient into the cell).
3.   a Since the concentration of urea in the solution in the artificial kidney is virtually zero, urea will leave the blood by diffusion and be carried away.
      b The concentration of ions in the kidney machine is set at the optimum level for the body; ions will thus leave the blood until they reach equilibrium, i.e. the optimum level
4. Because the Visking tubing (or its equivalent) has pore that are too small to allow proteins and RBC’s to pass through it., whilst still allowing small molecules such as urea, water and ions such as Na+ to pass across.
5. This will give a larger surface area and so, by Fick’s Law, the rate of diffusion will be faster and so the time for complete dialysis reduced.
6. ‘continuous’ = goes on all the time
‘ambulatory’ = the patient can walk around
‘peritoneal’ = uses the peritoneum as the dialysis membrane
‘dialysis’ = the process by which excess chemicals are removed from the blood
7. If the concentration in the dialysis fluid is greater than that in the blood, then water will flow into the fluid, from the blood, by osmosis

Chapter 3 - Biological Molecules

In Chapter Questions  [back to top]

1.  a Glucose; amino acids (all 20 of them);  
     b Starch and protein  
     c Starch, proteins and fats (the latter depending on the length of the fatty acid chains – waxes yes, oils no
2. Maltose = glucose (C6) + glucose (C6) – water, so 12 carbons in all.  
3. C12H22O11  This comes up a lot, so learn it (same formula for sucrose and lactose too!)  
4. The proportions of amylose and amylopectin will vary; the position of the 1: 6 branches within the amylopectin will vary; the lengths of the individual molecular chains will vary.  All are only made of one monomer - a glucose - and so we usually say there is only one sort of starch.  
5. Women are lighter so eat less in total; they worry more about their weight, so eat less when dieting; they like chocolate more!  Men eat more sandwiches and chips!  
6. Very strange cell – particularly as bacterial cell walls contain several polysaccharides!  However, there are 20 different amino acids and the sequence in which they are arranged makes each protein different (like the 26 letters in our alphabet can make many words and an infinite number of different sentences).  This is known as the protein’s primary structure and is determined by the order of the bases in the gene (DNA and mRNA) which coded for it.  Since neither DNA nor RNA are branched molecules, nor are proteins (our sentences are linear too!).  Since polysaccharides are made of only one monomer (usually a glucose), there is usually considered to be only one polysaccharide.  
7. One between each pair, so 8 in all
8. N- terminus means the amino-acid at one end of the chain with a free amine group
C- terminus means the amino acid at the other end of the chain with a free carboxyl group.  This is important in protein sequencing in the laboratory and in digestion in the small intestine, both of which use enzymes (exopeptidases) which can only attack one end or the other of the peptide chain
9. d -only – the others all involve covalent bonds, formed by condensation reactions.  
10. Influenza (‘flu) is an RNA virus and so mutates rapidly.  Similar viruses are also common in other animals, particularly pigs.  Thus new mutants and new hybrids with the animal viruses are constantly being produced.  Each has different antigens, so each will need different antibodies to react with it.  ‘Flu vaccines therefore need to be given each year and are designed to target the scientists’ best guess as to what will be the current strains a year or so ahead.  
11.

An unsaturated fatty acid is one containing C=C double bonds (this has the effect of putting a ‘kink’ in the tail of the fatty acid and so makes the membrane more fluid.  Hence they are found in fish and plant oils, where the organisms’ operating temperature needs fluid membranes at 10- 20oC.  ‘Poly’ = many; in this case more than one C=C bond.  Given the (largely misleading) advertising messages we see daily, cholesterol is also a polyunsaturated fat!  

12. A phospholipid has two fatty acid chains, a triglyceride has three; a phospholipid has a phosphate group, a triglyceride does not.  This means that phospholipids are polar molecules, whilst triglycerides are not.  
13. One C=C double bond, so it is monounsaturated.  
14. The fastest a substance can move is to be level with the solvent front, so the top number ( numerator) is always smaller than the bottom number (denominator)!  

Chapter 3 - Biological Molecules

Answers to Exam Questions  [back to top]

1.   a Glycerol = 3; glucose = 6
      b i.    There are 20 different amino acids, each with a different ‘R’ group.  The ‘core’ of the amino acid has just 2 carbons; the balance (0 to 9) are found in the side chain.
i.    Starch molecules are made of varying numbers of amylose and amylopectin molecules and each of these has a variable number of glucose sub-units.  Starch varies – ask any cook!
      c Each time a glucose molecule is joined to another (by a 1:4 glycosidic bond), a molecule of water is released.  Hence a condensation reaction.
2.   a Place the spots on a line about 2 cm from one end of the paper. 
Include reference substances as well as the unknown(s).
Place in a tank of solvent and leave to run for some hours.
Examine and mark the solvent front when removed from the solvent tank.
Calculate the Rf values of the spots (and compare with the reference spots) to identify the unknown molecules.
Check the Rf values against data tables for the same solvent to confirm the identification.
      b i.    Because this represents the total radioactivity, rather than its peak emission
ii.   Since the radioactivity was introduced in the fatty acids through 14C, it follows that the more carbon atoms in the fatty acid, the more radioactive it will be.

Chapter 3 - Biological Molecules

Answers to Assignment Questions  [back to top]

1. To allow them to be compared
2.   a All the carbon will have been oxidised to CO2 gas and so will not be in the ash
      b The ash will contain minerals and vitamin residues.
3. Bar chart
4.
34.4 – 10.8 x 100 = 218.5%
     10.8
(i.e. their weight has more than doubled in a little over a week)
5.   a They contain only C-C bonds and no C=C bonds.
      b Unsaturated are in bold:
    
Vegan =  39 + 68 + 166 + 52 + 12 + 313 +317 + 15 = 982
     Control = 33 + 80 + 276 + 108 + 36 + 353 + 69 + 8 = 963

Unsaturated concentration: 
       Vegan concentration = 325 ÷ 982 x 100 = 33.1%
       Control concentration = 497 ÷ 963 x 100 = 51.6%
Saturated concentration:
      
Vegan = 657 or 66.9%
     
Control = 466 or 48.4%7

      c Control group produce roughly equal quantities of both saturated and unsaturated fats in their milk
Vegan group produce roughly twice as much saturated fat as unsaturated fat. 

Given that plant fats tend to be unsaturated, this shows that the fats produced in the breast milk are synthesised in the body and not directly related to those in the diet.  There is also the recent discovery that omega-3 fatty acids are linked to a number of aspects of brain activity and development.  On the basis of the information in this table, it would seem that the two groups of babies will receive very different quantities of this essential nutrient.  What the effect of this on their children will be, we can only guess!


Chapter 4 - Enzymes

In Chapter Questions [back to top]

1. Protein synthesis (DNA ® mRNA = transcription, mRNA ® protein = translation)  
2. Human body temperature is, of course, not constant.  It varies throughout the day, rises when exercising or when ill, and a woman’s average temperature varies around the time of ovulation.  By being warm-blooded, mammals are able to remain active throughout the year – though at the cost of needing a much larger energy intake than a comparably sized reptile.  Crocodiles can feed weekly or less; a tiger needs to eat daily!  In cold winter climates, when energy supplies are limited, some mammals hibernate to conserve energy; others migrate.  A constant temperature means that enzymes (and other processes too) can be optimised for that temperature.  37oC is a compromise.  Higher temperatures mean faster reactions (good) but at the cost of faster enzyme denaturation (bad).
3. The hydrogen bonds within each albumen molecule (the main protein in egg white) break when heated, allowing the molecules to uncoil and become entangled.  They then reform between the molecules on cooling, ensuring that the egg white remains coagulated.  Since the proteins now disperse light, the cooked white is white, whilst raw albumen allows the light to pass through easily.  Albumen coagulation is the principle of making sponge cakes and custard (amongst other sauces) – and custard is the basis of ice-cream (yum, yum!).  A similar process happens when meat is exposed to acid – you can therefore ‘cook’ meat or fish by marinating it in acid (e.g. lemon juice).  Ever eaten gravilax?
4. The buccal cavity (or mouth, as we scientists call it) has a pH of about 8.0 to ensure that the tooth enamel is strengthened and more calcium can be deposited between meals.  On entering the stomach, gastric juice (pH 2.0) denatures it immediately (hence amylase has to work very quickly).  Being a protein, amylase will then be attacked by gastric protease (or pepsin) and broken down.
5. Competitive – normally similar in shape to the substrate (therefore chemically related); bind to active site; complementary to the shape of the active site; normally only bind temporarily; causes inhibition proportional to the relative concentration of the inhibitor (i.e. more substrate speeds things up again).

Non-competitive – binds elsewhere on the enzyme than the active site, distorting the shape of the active site and stopping it binding with the substrate and forming an enzyme-substrate complex.  Normally permanent in its effect, it has no chemical connection to the substrate.  Heavy metals and many disinfectants and poisons work in this way.  Inhibition is proportional to the quantity of inhibitor and so adding extra substrate has no effect.

Some allosteric enzymes have evolved to require a molecule similar to a non-competitive inhibitor before they can act i.e. the inhibitor has now become a promotor.


Chapter 5 - Gas Exchange

In Chapter Questions  [back to top]

1. If external they would be very vulnerable to infection, physical damage or attack and becoming clogged with dust etc.  In addition, the water losses from such a large area would be prohibitive.
2. Gas exchange is a process of diffusion, which can only take place in solution in living things.
3. It would steadily decrease until it reached zero.  The oxygen affinity of haemoglobin varies from species to species, so the actual concentration achieved would vary too.
4. Because when you fainted the autonomic processes of the body would ‘resume normal service’ and you would start to breathe again.
5. Hyperventilation means breathing too quickly or too deeply – this is the opposite of holding one’s breath – so the question makes no sense!  You feel light headed if you do this (it can happen in a panic attack) due to the very low concentration of carbon dioxide in the blood raising the pH above normal.  If you breathe pure oxygen another effect happens – tunnel vision – leading to death (particularly if flying a WW II aircraft and you forgot to switch the oxygen off as you descended to land).
6. 15 x 0.5 = 7.5 litres/min.  You extract about 25% of the oxygen, so consume about 0.35 litres/min of oxygen.

Chapter 5 - Gas Exchange          

Answers to Exam Questions  [back to top]

1.


3 features that increase the rate of diffusion:
      i. thin alveolar walls;
      ii. large surface area;
      iii. transport system to maintain diffusion gradient.

2.  a i.  Manv small alveoli (= more surface area than a few, larger ones & walls are thin so more alveoli can be accommodated).
ii.  2 processes ensuring a difference in concentration maintained:
                   -
a ventilation mechanism occurs;
                   - a transport system removes gas (removes O2 to body; brings CO2 to lungs).
     b Exhaled air from mouth to mouth resuscitation is still valuable because:
          i.
it contains 15% oxygen;
         ii. high levels of CO2 are present which may stimulate the inspiratory centre of the medulla.
3.  a Nerve A (intercostal and phrenic nerves) stimulates the external intercostal muscles (& diaphragm).
     b

Nerve B (vagus nerve) causes relaxation of the diaphragm & external intercostal muscles. (\air out of lungs).            
                                   
  

     c The medulla inspiratory centre receives messages from the carotid and aortic bodies. If CO2 levels rise - it increases the breathing rate. 
     d i.  The type of breathing for person in Column X is fast and shallow.
ii.   Although Column X shows a faster breathing rate, it is delivering less oxygen to the body (& conversely removing less CO2) as it is the tidal volume that is of most significance - as there is about 150 cm3 of 'dead' air space in the trachea and bronchi that needs to be deducted from the tidal volume before the alveolar ventilation rate can be calculated. 

It is this rate that is of most significance in explaining O2 delivery into the body. Thus it is possible to breathe less frequently (as in Column Y) but more deeply (\bigger tidal volume) and get a bigger increase in the alveolar ventilation rate than with faster, shallower breathing.

Chapter 5 - Gas Exchange 

Answers to Assignment Questions [back to top]

1.  a Will be errors in finding an animal's surface area by:
     
i. removing skin as it may stretch and not lie entirely flat where it bulges over the body;
    
ii. cylinder and cones give only a rough, approximate guide.
2   a
Length of one side Total surface area Volume   Surface area / Volume ratio
1  62 1 3  6 : 1
2 242  83  3 : 1
3 542  273  2 : 1
4  962    643   1.5 : 1
5  1502   1253  1.2 : 1
6 2162 2163  1: 1
     b  A  true      B  not true     C  true      D  true
3   a Log scales keep the graph size manageable ­  (shortens 'x' axis length considerably).
     b  i   10 cm3                      ii   105.5 g
4. Elephants' ears have a large surface area for blood cooling.
5. Large reptiles cannot live in cooler regions because they have a relatively small external body surface area in relation to their volume - they need the surface area to absorb heat as they are ectotherms ( = "cold blooded animals").
6. Real size =  Size of photograph 
                    
Magnification  

A microvillus length is 30 mm on the  photograph
(= 30,000mm), but in reality it is 3 mm, so, applying the formula­:

3 =    30,000       and      3 x mag = 30,000         \ mag =   30,000     = 10,000 x
         mag.                                                                           3

7.  If a 25 mm length of small intestine is taken, it contains 6 villi, thus:


\ each microvillus has length (each side + top) of 30 + 30 + 4 = 64.
With 6 microvilli = 6 x 64 = 384 mm \ the increase in surface area = 384 ¸ 25 = 15 times more surface area
 


Chapter 6The Heart and Circulation  

In Chapter Questions  [back to top]

1. Oxygen: source = lungs, sink = muscles;

Glucose: source = small intestine, sink = muscles

2. Pulse points require an artery to pass over a bone near to the skin.  The pressure within, and diameter of, an artery changes with each beat of the heart; that of the blood in a vein is much lower and constant.
3. 

Artery: blood spurts up to the ceiling (!), blood bright red in colour

Vein: blood pours out more slowly, dull red in colour.

Before you faint or call an ambulance, place a clean cloth over the wound and press hard; get the victim to raise their arm as high as possible.  If the cloth gets saturated, just add another one, leaving the first in place.  Victim should also be given hot, sweet drink (tea!) and kept warm to reduce shock.

4.  a High blood pressure will result in more fluid being squeezed out of the capillaries at the artery end and will tend to reduce absorption at the venous end of the capillary bed.  Hence a build-up of tissue fluid (called odema) most obvious on the ankles of old ladies (old men, very wisely, keep their legs covered by trousers!)
     b This will reduce the water potential of the blood and so will reduce the reabsorption of fluid at the venous end of the capillary bed, resulting, again, in odema as above.  This might occur in some types of diabetes, in particular.
5. Epithelium (found in all organs); smooth muscle (both longitudinal and circular); fibrous tissue (collagen)
6.  a hepatic portal vein (high after a meal, low in between)
     b hepatic vein (blood has been through two capillary beds in two different organs)
7.  a they have no mitochondria, which is the sole site of aerobic respiration
     b they have no nucleus, rough endoplasmic reticulum, 80s ribosomes and source of large quantities of ATP (i.e. mitochondria) all of which are needed for protein synthesis.
8. hepatic vein; (posterior) vena cava; right atrium, right ventricle; pulmonary artery; pulmonary vein; left atrium, left ventricle; aorta; renal artery.
9.

Exactly the same.  Liquids are not compressible and blood transmits the pulse uniformly and simultaneously.

10. a A (pressure in ventricles > in atria)
    b C (pressure in ventricle < in aorta)
11. a this enables the atria to fully empty before the ventricles begin to contract.  It also puts an upper limit on the heart rate, thus ensuring that the heart never becomes anaerobic.
     b this ensures that the ventricles both contract from the bottom up, thus ensuring that they fully empty with each beat.
12. 5500 ¸ 70 = 78.57 cm3  
13. The heart responds to two types of stimulus – nerves and hormones.  Adrenaline still works!

Chapter 7 - Enzyme Technology

In Chapter Questions  [back to top]

1.

Enzymes are specific, so the product(s) made by them are pure.  They work best under benign conditions, so saving the cost of fuel and they are not dangerous, so extensive safety measures (such as might be needed with strong acids or high pressures) are avoided.  Since enzymes are proteins, any contamination of the product is unlikely to be toxic.  They are readily denatured by heat, so pasteurisation (or similar) of the end-product will prevent them continuing to work, potentially degrading the product.  Because they are natural, their product is the natural form of the product, i.e. any glucose made will be 100% a-, or 100% b- glucose; synthetic glucose would be a 50:50 mixture of the two. 

This is the basis of much drug-testing of athletes.  If the breakdown products in their blood or urine are 100% of one form, their origin is natural; if a 50:50 split, then the chemical was produced synthetically.  They could be cheating either way, of course!

2.

Because as they are viruses and can replicate very quickly, they would rapidly kill the bacteria in the fermenter. – visit http://www.cellsalive.com/phage.htm for good animations of a bacteriophage infecting an E.coli cell

3. A biosensor could be used to monitor the level of glucose in the diabetic’s blood.  If connected to a transducer, the electrical signal could be connected to a device which automatically injected the correct quantity of insulin into the patient.  With the widespread use of ‘Humulin’ (fast-acting human insulin) it is more important to keep the levels of insulin and glucose within a narrow range.  All attempts to do this with an automatic syringe have so far failed, due to infection or blockage of the canula used (in place of a needle) to inject the insulin.  The modern approach is now to inject the correct amount of insulin to control the actual carbohydrate intake and activity levels of the diabetic, rather than controlling the diet to allow for fixed volumes of insulin.  This makes for a more natural life-style and less variation in blood glucose levels.
4. A transducer or use a pH probe, attached to a read-out of some kind.
5. Enzymes, normally used in an industrial or laboratory process, which are trapped in some form of matrix so that they do not leach into and contaminate the product(s).
6. A stain of biological origin, i.e. one based on protein (needs proteases), lipid (removed by detergents anyway, but might need lipase) or carbohydrates (polysaccharides, since sugars are water-soluble ; this will need a range of carbohydrases).  Thus a ‘cocktail;’ of enzymes is necessary to remove them all.
Chapter 7 - Enzyme Technology

Answers to Exam Questions  [back to top]

1.   a An enzyme that is normally secreted by the organism and so works outside the cell.  Such enzymes (including those of digestion) are normally more stable and robust than the much more common intracellular enzymes.  They are therefore favoured for industrial use.
      b i.    This is to enable the bacteria to grown and reproduce rapidly – for which protein is essential
ii.    This is to induce the bacteria to produce large quantities of the desired extracellular protease as they attempt to make up for the lack of readily available protein
      c

-   Other bacteria would use up the growth medium without producing any useful product. 
-    They might contaminate the product and make it valueless – or even toxic. 
-    They might out-compete the desired microbe and, by the ‘competitive exclusion principle’, completely replace the desired microbe in the fermenter.
-    They might be poisonous or hazardous to the employees at the factory (either directly as a pathogen, or by inducing an allergic reaction).

2.   a An enzyme that is attached to an inert material (or contained within a porous inert ‘bead’) and used in an industrial process.  The advantage is that the enzyme can be re-used and is more stable; the process is therefore cheaper.  In this particular case, the advantage is that the enzyme will not contaminate the milk and therefore make it unsaleable.
      b In any question like this the answer is always the same. “At low temperatures the molecules have less kinetic energy and so collide with each other less frequently and with insufficient force to overcome the activation energy.  As a result, fewer enzyme-substrate complexes are formed and the reaction proceeds more slowly”.  [On the other hand, the enzyme will denature more slowly and so last longer before it needs to be replaced – important if it is very expensive to replace.]
Chapter 7 - Enzyme Technology

Answers to Assignment Questions  [back to top]

 

There is no prepared answer to this assignment!

 


Chapter 8 - DNA and Protein Synthesis

In Chapter Questions [back to top]

1. A nucleic acid (DNA or one of the three forms of RNA) is a long, linear molecule (a polymer).  Nucleotides are the monomers from which they are built.
2.
DNA RNA
DNA is normally double-stranded RNA - normally single-stranded (though it usually coils back on itself to create double-stranded regions, showing that it has complementary sequences within the same molecule
DNA has thymine RNA has uracil  
confined to the nucleus (extra-chromosomal DNA is found in mitochondria and chloroplasts) RNA is found throughout the cell (and mitochondria and chloroplasts too!)
DNA may be circular (as in prokaryotes) RNA is linear
DNA in eukaryotes is associated with chromosomes and their proteins (especially histones) RNA is mobile and no so protected
DNA is a source of genetic code only RNA can (very rarely) act as an enzyme too
DNA has the pentose deoxyribose RNA has the sugar ribose  
DNA is billions of base-pairs long RNA is much shorter – mRNA may be only 100 bases
DNA viruses rarely mutate and can readily be immunised against (e.g. measles) RNA viruses mutate rapidly and are hard to immunise against (e.g. HIV)
DNA is copied at mitosis and meiosis only RNA whenever proteins are being synthesised
DNA can be traced back in a continuous lineage to the first cell on Earth RNA only exists for a fraction of the life-time of the cell

 

3. A common question! a) TGG CTG ; b) UGG CUG
4. AUG = start; GGG = alanine; CAC = histidine; UGC = cysteine; GUA = valine
5. UGA will pair with ACU, which is the codon for the amino-acid threonine
6. The question is completely misguided; it is essential to the DNA, not to the cell!  A cell has only a limited life and, except as a zygote, if it dies it can readily be replaced.  It is true that a mutated cell may malfunction to such an extent that the individual’s survival is put at risk (e.g. cancer) but the occurrence of such mutations is very rare.  However, the DNA in our cells extends back in an unbroken line to the origin of life on Earth; thus if an individual molecule mutates, it loses part of its ancestry and can never recover.  True, if the host organise dies as a result of the mutation then the DNA loses out too (unless reproduction is already completed) It is true, too, that without such mutations, evolution could not occur and the wonderful diversity of life on Earth would not exist.  Thus it might be more accurate to say that it is in everything’s best interests that DNA has a small, but inevitable, mutation (or error) rate!  It is arrogant in the extreme to claim that, in our mere 70 or so years on the planet, we are somehow more important to the overall scheme of things than the molecule to which we owe our very existence and which gave us (and everything else) life itself.  That has been around for well over 4500,000,000 years! (or 1 million times longer than Stonehenge or the Pyramids!)
7. A mutation on chromosome 12 to the PAH (phenylalanine hydroxylase) gene. See http://www.yourgenesyourhealth.org/pku/cause.htm for an animated explanation!

Chapter 8 - DNA and Protein Synthesis

Answers to Exam Questions [back to top]

1.   a i    The phosphate group of each nucleotide is attached to the deoxyribose (pentose) sugar of the next (by condensation reactions – which link the 3’ and 5’ carbons)
ii    Hydrogen bonds – three between G & C; two between A & T.
      b

DNA code:      T          C         G         A         C         A         T          G         A
mRNA code:   a          g         c         u         g         u         a          c         u  

      c     Assuming no introns are present, 450 ÷ 3 = 150 amino-acids – a very average-sized protein
ii    3 bases  ≡ 1 codon ≡ 1 amino-acid
2.   a i    In DNA, A = 26% therefore T = 26%; A + T = 52%.  The percentages of G & C are equal, so, in this case, G + C = 100 – 52 = 48%; therefore G = 24% and C = 24%
ii    See above – note that the percentages given for the mRNA are of no interest whatsoever – we do not know what the sequence was, thus the percentages will be all over the place!
      b i     In RNA (all types), thymine is absent and is replaced with Uracil.
ii    [Assuming it is not yet another misprint] DNA genes contain introns and these are removed (probably in the nucleolus) before the mRNA leaves the nucleus.  mRNA is made by complementary base pairing, so the percentage  of adenine in mRNA corresponds to the percentage of thymine in the DNA.
3.   a

or

      b The sequence of the bases (= codons) on the mRNA determines the sequence of the amino-acids in the protein – its primary structure.  On the ribosome, a tRNA brings a particular amino-acid into position next to another; the anticodon at the base of its central loop pairs with the codon on the mRNA.  The main function of the ribosome seems to be to hold these two very large molecules in the correct position long enough for a peptide bond to form between the two amino-acids.

Chapter 8 - DNA and Protein Synthesis

Answers to Assignment Questions [back to top]

1. Because it is a chain of simpler molecules (nucleotides) joined together (by condensation reactions) with covalent bonds.
2.   a 42 = 16
      b A codon consists of three nucleotides.  Therefore, there are 43 = 64 different possible combinations.  Since there are only 20 (common) amino-acids, there are more than enough codons to code for them all.  Indeed, the code is said to be degenerate because of this excess.  It is the sequence of the amino acids in the protein (its primary structure) that determines how it will fold up and therefore its final shape.  Since this, in turn, determines whether the protein will work effectively, it is important to get it right! 

DNA (and RNA) can be almost any length, so there is no difficulty about arranging the codons in any order, just as the letters in the alphabet can be arranged in any order to make different words and sentences.

3.   a As Watson and Crick well knew, Adenine pairs with Thymine (A = T) and Guanine pairs with Cystosine (G ≡ C).  This is the only possible combination, since a purine (A and G) always pairs with a pyrimidine (C and T and Uracil).  You need to remember these, so remember – in DNA, those with a ‘Y’ (cYtosine and thYmine), are the pYrimidines! They also happen to be the smaller nucleotides (i.e. tinY!)
      b They are antiparallel, with one complete turn for every 10 nucleotides.
4.    Because A = T is exactly the same size (and shape) as G ≡ C.
5.

The model you make is only 2-dimensional, whereas in DNA the nucleotides are 3- dimensional, though the base pairs are ‘flat’ molecules.
It is not possible to reproduce the hydrogen bonds linking the two strands together.
It is not possible to reproduce the covalent bonds forming the ‘backbone’ of DNA.


Chapter 9  - The Cell Cycle

In Chapter Questions [back to top]

1. Replication – using the enzyme DNA polymerase (or replicase)
2. hard to see.  The nuclear envelope and nucleolus appear to have disappeared and the chromosomes are in the process of dehydrating, shortening, coiling and becoming clearly visible.
3. Two cones, back-to-back (<>)
4.  a Larger surface area, allowing easy access for the RNA polymerase that is needed to synthesise mRNA at the start of protein synthesis.  TO allow the DNA to ‘uncoil’ as part of this process.
     b More easy to move around and more stable, thus less likely to damage the DNA in the process.  Note that the DNA is ‘inert’ during this process and so no new mRNA can be synthesised; in effect, protein synthesis is suspended during mitosis.
5. A = late anaphase or early telophase
B = early telophase
C = metaphase
D = late telophase
The correct order is thus C; A; B; D
Neither Prophase nor Interphase are shown.
6. G1 ; S; G2 Note that mitosis really begins at the S stage of interphase; it is just that nothing is visible until the start of prophase!
7. There is no question 7
8. There is no question 8
9. 23 + 23 = 46.
Chapter 9 - The Cell Cycle

Answers to Exam Questions [back to top]

1.  a i.   'S' (=synthesis) stage of interphase 
ii.
   anaphase - by far the shortest
     b i.    part D - mitosis in root tip
ii.
  
either Toluidine Blue (good) or Aceto-orceine (smelly)
iii  
chromosomes (nuclei) stain greeny-blue or maroon (red wine colour).  The stain is taken up by the chromosomes - it makes them more visible.
2.  a i.   prohase 
ii.
  S stage of interphase 
     b i.   see 1bii
ii.  – see table:

Stage in cell cycle

Number of cells seen on slide

% of time in cell cycle spent in metaphase

Interphase 147 60 [(147 ÷244) x 100 = 60%]
Prophase 68 28 [68 ÷244) x 100 = 28%]
Metaphase 10 4   [(10 ÷244) x 100 = 4%]
Anaphase 4 [(4 ÷244) x 100 = 2%]
Telophase 15 [(15 ÷244) x 100 = 15%]
Total: 244
3.  a

Sperms and eggs contain half the genetic material of normal body (= somatic) cells.  From this you cannot say whether DNA or proteins carry the genetic code.

     b i.   live capsulated - Only in the presence of the capsulated form do the mice die.
ii. 
The live Pneumococcus bacteria were ‘transformed’ by the dead capsulated ones – probably, he thought, by absorbing some factor from them (i.e. DNA)
     c i.   The DNA core (from the T2-phage), which was incorporated into the host  cell nuclear body (cccDNA)  
ii.  
It contains phosphorous. The DNA gets incorporated into the host cell, but the outer protein coat of the virus does not.  Without knowing the rest of their experiment, you cannot deduce how this shows that it is the DNA that is the site of the genetic code and not the protein! [You need to realise that the virus reproduced inside the host cell- and killed them - and that viruses have only two components – the protein (in the coat) and the DNA (in the core).  Thus, for the virus to replicate, it need only inject its DNA, and that the protein coat is not involved in replication (i.e. is not the genetic material].

Chapter 9 - The Cell Cycle

Answers to Assignment Questions [back to top]

1.   a Animal cell as: no cell wall or large central vacuole. 
      b The chromatids are being pulled apart hence anaphase
2.   a 9mm
      b

Using formula... actual length = length of photo/image
                                                    
magnification

                       9mm = 9000μm               

                            X =     9000
                                      1000

                                = 9μm

3. 14 chromosomes, but only 5 units of DNA in each nucleus
4.   a none - RBC's have no nucleus
      b yes they are actively dividing 
      c no - the epidermal  cells are dead; it is the dermal cells, deep in the skin that divide
5.
Stage No. of cells at stage %
Interphase 52 79
Prophase 5 8
Metaphase 3 5
Anaphase 2 3
Telophase 4 6
TOTAL: 66 100
6.   a interphase: as greatest number of cells at this stage where synthesis is taking place 
       b
Stage No. of cells   at stage % of cells at stage Time in 24hours spent at that stage
Prophase 5 5/14 x 100 = 35.7 % 24 x 0.357 = 8.6hr
Metaphase 3 3/14 x 100 = 21.4% 24 x 0.214 = 5.1hr
Anaphase 2 2/14 x 100 = 14.3% 24 x 0.143 = 3.4hr
Telophase 4 5/14 x 100 = 28.6% 24 x 0.286 = 6.9hr

Total:

14   24 hours
The results would probably be best displayed as a pie chart

Chapter 10 - Gene Technology

In Chapter Questions [back to top]

1. The complementary code would be: TAG CTG GGA TCT.
2.  a It is usually easier to start with mRNA rather than the DNA, since it then becomes a needle in a bale rather than a haystack!  Reverse transcriptase is use to transcribe (copy) the RNA into DNA, so that it ca then be inserted into the target cell
     b This is one of a class of enzymes that are used to cut the DNA at a specific sequence.  The ones of use to scientists cut the DNA with a short overlap, leaving so-called ‘sticky ends’ which, will hybridise together when the two strand of DNA are mixed together.  Providing the same enzyme is used on both DNA molecules, this means that the host and target DNA will join together when mixed, merely needing DNA ligase to permanently join the two pieces of DNA together.
3. See answer to 2 b)
4. After the insertion process is complete, only about 1 in a million cells will have been transformed i.e. will have absorbed the human insulin gene in an active form.  To separate these cells would be impossible, but if they also are resistant to the antibiotic ampicillin, then growing the new cells in a medium laced with this antibiotic will kill all untransformed cells, leaving the desired cells to be cultured.
5. Cows which have been injected with DNA from another organism, to make them secrete a desirable protein in their milk
6. Patients with CF have a defective (recessive) gene.  A ‘puffer’, similar to that used by asthma patients could be used to administer the normal gene to their lung cells and so enable them to produce the missing protein.  This has been tried, using a virus as a vector, but the patients subsequently developed cancer, suggesting that the virus ‘replicate’ gene had triggered the host cell to divide too – not good!  Note that it is illegal to attempt to genetically modify any cells that could be passed on to your offspring – genetically modifying you may be OK (I hear no-one objecting!), but to affect, not only your children but your entire future gene-line is not acceptable!

See  http://www.ygyh.org/cf/whatisit.htm for an animated full description of CF

Chapter 10 - Gene Technology    

Answers to Exam Questions  [back to top]

1. a

For ‘may’ read ‘must’.  By using the same enzyme, the ‘sticky ends’ on each length of DNA will be the same and so they will join up when mixed together.

    b DNA Ligase.
    c

Use a plasmid that has antibiotic resistance genes on it. 
Those (few) bacteria which take up the plasmid will thus become resistant to the antibiotic. 
Grow the bacteria in a culture medium with the antibiotic in it. 
Those bacteria which have been ‘transformed’ will survive;
the rest will die, thus selecting the ones you want.

2.  a

Remove the cell wall from the sunflower cells. 
Extract their DNA and cut it with a restriction endonuclease. 
Cut the other DNA with the same enzyme. 
Mix the two DNA samples together. 
Treat with DNA ligase to join the fragments together. 
Incubate with DNA polymerase to get enough ‘hybrid’ DNA to do anything useful with!

     b

Without Gene 2, there would be no extra amino-acids in the grass leaf cells, which is what the sheep eat
Without Gene 3, all the extra protein would be digested before the sheep could absorb it – by the bacteria which live there!

Chapter 10 - Gene Technology

Answers to Assignment Questions  [back to top]

1   a

Unlike an enzyme, it contains a short sequence (10 amino acids) that repeats over and over again.  It does not have a definite tertiary structure, but continually changes shape.

     b

The definite tertiary structure of an enzyme and its unique sequence means that the tertiary form of each enzyme is unique and each enzyme has a perfectly-formed active site.  The mussel protein has a variable shape, and so can squeeze into small cracks, whilst the repeating sequence contains tyrosine, which can cross-link to make the protein glue ‘set’.

2   a

because the body would attack it and render it useless; some patients would develop an allergic reaction; ‘moules frites’ are VERY yummy – shame if they could no longer eat them!

     b

To repair wounds after surgery.
To attach minute probes / sensors permanently in the body

3.  a 10 amino acids means 10 codons means 30 nucleotides
     b 5 different amino acids means 5 different t-RNA’s
     c i.    Because, once the amino-acid sequence is known, by reading back from the genetic code, we can deduce one possible sequence for the RNA
ii.    Most amino acids have more than one possible codon; thus, with 10 amino acids, the actual sequence is unlikely to be the same as the manufactured one.
4   a

Because it takes place after the polypeptide has left the ribosome (on the rough endoplasmic reticulum), on the smooth e.r.  This can also take place in the Golgi body, and may even occur after the protein has left the cell – as in digestive proteases, such as trypsin and chymotrypsin.

     b i.    Bacteria have ribosomes and so, even though they are 70s and not 80s, they can translate the mRNA sequence faithfully into the protein.
ii.    Bacteria have no endoplasmic reticulum, and so are unable to carry out post- translational modification.
     c

Because tobacco plants do have smooth e.r. and so should be able to modify the protein.

5.  a So there is no risk of people eating the GM- plants and getting ill and/or becoming allergic to mussels and / or it will reduce public protests.
     b

To reduce the risk of the GM plants cross-pollinating with wild plants the mussel gene becoming widely dispersed in the environment.  AND public protests!


Chapter 11 - Immunology and Forensic Biology

In Chapter Questions [back to top]

1. Planet Earth only has one genetic code!  However, different gene sequences will alter the sequence of the amino-acids in the polypeptide (its primary structure).  This, in turn, alters the position of the atoms in the molecule and so alters the bonds that can be formed (hydrogen and disulphide bridges).  Since proteins always fold up in the most stable way, these bonds determine the shape of the molecule and so its function and efficiency.  
2. Because the antigens in the donor blood (A & B; on the RBC’s membrane – they are short carbohydrate chains) are compatible with the antibodies in the recipient.  It does not matter if the antibodies (and other proteins) in the donor’s plasma are incompatible with the antigens (on the RBC’s) in the recipient’s blood, since they will be diluted too much to cause agglutination.  
3. Anticoagulants are proteins which block the host’s clotting mechanism.  Leeches feed by sucking blood (along with those charming little airborne rodents, Vampire bats and those mobile malaria menaces – female mosquitoes).  Therefore, to prevent their mouthparts becoming clogged up with clotted blood, they all have an anticoagulant in their saliva.  One of the best known of these is heparin, which is widely used to treat stroke and post-coronary thrombosis patients, to prevent further unwanted thromboses.  The dose, however, is critical and is affected by diet and once started, must be continued for the rest of your life.  Too much heparin causes bleeding under the skin - the patient gets bruises all over their body – and so most patients prefer aspirin, which is not quite as effective but has fewer side-effects.  Warfarin – another anticoagulant - used to be used as a rat poison, but modern rats are largely resistant so nowadays a chemical, which affects vitamin uptake, is used.  This takes several days to take effect, which means that the rat cannot learn to avoid the bait – once it begins to feel unwell, it’s too late!  
4. Mutations in genes may cause a change to the phenotype of the individual and so affect its ability to survive and reproduce.  This means that there will be selection pressure, usually to eliminate the mutation, since most mutations are harmful, to a greater or lesser extent (e.g. ginger hair, where such individuals cannot make melanin in their skin and so burn easily in the sun).  Mutations in ‘nonsense’ DNA have no such selection pressure. 

Since mutations occur and a constant, known, rate, this means that these regions of DNA represent a mutation ‘clock’ which can be used to estimate how long ago two species diverged.  In humans mitochondrial DNA has been used to estimate when Eve lived and where she came from (about 50,000 years ago, Eastern Africa); similar mutations to the Y chromosome have suggested that Adam may have had a different origin…..!  

5. A single cell’s DNA is now enough….! However, the reasoning behind the question is that most blood cells are RBC’s and so have no nucleus and hence no DNA, whilst sperm cells all contain DNA.   
6. Those that are common to mother and child.  Reading from the bottom of the child’s DNA profile, they are: father, father, father, mother, father, mother.  
7. In both profiles, Suspect 1 is the only one to have any DNA bands matching with those found at the crime scene.  
8. There only a limited number of blood groups but an almost infinite number of possible DNA fingerprints – which, apart from members of the same family are unique.  
9. a.   210 = 1024               b) 225 = 33554432  

Chapter 12 - Crop Plants

In Chapter Questions [back to top]

1. Mash up grain in water.  Add Biuret reagent.  Pale blue ® lilac if peptide bonds (i.e. protein) present.
2. When the paddy field is flooded, they drown; when it is drained any hydrophytes (= water-plants) dry out and die.  The paddy fields are a rich source of eels too, a fish very rich in oil, which may explain why Orientals on their traditional diet have a very low incidence of heart attacks.
3. There are several.  Aerobic respiration produces a net 38 ATP’s per glucose molecule, anaerobic respiration only 4 (of which 2 are used up); anaerobic respiration takes place in the cytoplasm and can only use carbohydrate as a carbon source, whilst aerobic respiration takes place in the mitochondria and can use carbohydrate, fat or protein as the source.  Anaerobic respiration is quicker (no waiting for oxygen to diffuse) and builds up lactic acid in the muscles, making us tired (and so preventing us doing tissue damage through over-exercise.  Pass the TV remote, someone….!).
4. High temperatures in July are associated with high sunlight levels.  Since the higher the light level the more oxygen produced in cyclic photophosphorylation (the light reaction of photosynthesis – see Module 5!), the more slowly the carbon fixation will be since the key enzyme RUBISCO (see Module 5 again!) is inhibited by the oxygen released.  Thus higher temperatures select in favour of C4 plants.  Interestingly, rice, though tropical and needing a mean temperature of 20oC to grow, is not a C4 plant.
5. Without knowing all the parameters, this is tricky!  Enzymes work best at their optimum temperature and for photosynthesis this is about 35 - 40ºC.  However, photosynthesis is a multi-stage reaction, with two distinct processes; the splitting of water using light energy, which takes place using chlorophyll and is known as the light reaction, and the trapping and combining of carbon dioxide with other organic compounds, in a cyclical series of reactions known as the light independent reaction.  During the night, it is the absence of light which stops photosynthesis, but the reactions of respiration continue both day and night.  So the growth of the plant is the result of “photosynthesis – respiration” or net photosynthesis.  So, for maximum growth, a plant needs long days and short nights – exactly what it gets in a temperate climate summer.  Since both respiration and photosynthesis are affected by temperature, warm days and cool nights are also indicated – again, just what happens in a UK summer (we hope!)

Photosynthesis during the day is affected by many factors, but the limiting factor is nearly always lack of CO2.  For this reason, the optimum temperature for net photosynthesis is that which slows down respiration to the point where photosynthesis can use all the available CO2, but respiration is not immediately using up all the products!  So, in an atmosphere enriched with CO2 to around 0.1 – 0.25% (1000 – 2500ppm), or 3-8 times normal, photosynthesis is optimum at 35 – 40oC; but if the air has just the normal concentration of CO2 then 25oC is nearer the optimum – and there would be little change in growth rate (= net photosynthesis) over the range 18 – 40oC, assuming a constant night temperature.  But hot days usually mean warm nights too, when respiration is much quicker!  For this reason, the highest yields of arable crops in the UK are produced in Scotland, not, as you might at first think, on the South coast.

6. So that light is not the limiting factor.  Plants need roughly equal amounts of both red and blue light, so it must not only be bright but also balanced.
7. This would almost certainly be true, though the Victorians had been using piles of horse manure to achieve the same effect for many, many years.  The CO2 produced by the respiration of the pigs, and, even more, by the decomposition of their waste, would mean that the plants could photosynthesise more rapidly.  Assuming that this took place in the winter (pigs get sunstroke, so it would not be wise to do this in the summer months!), their heat would also both reduce his heating bills and help the plants grow better.

The only caveat would be that pig waste is particularly rich in ammonia, which is very toxic to plants and so the fumes from the pig manure would not only be very smelly but also be toxic to his plants.  The Victorians overcame this problem by moving the manure into their glasshouses only after it had ceased smelling – about 7-14 days after production!  This technique is still extensively used today – but for growing mushrooms, not plants.

8. Nitrogen (in the form of Nitrate, NO3).  In its absence they cannot make their main protein, chlorophyll, so their leaves go yellow and growth is stunted.
9. Some other factor was limiting OR the concentration of nitrate became toxic OR disease became promoted due to the extra, sappy, growth of the plants.
10. Water.  Followed by cellulose, in the decaying plant material.  Organic manures are usually made on site and cost little apart from labour costs; indeed, as a way of disposing of toxic waste they can even save the farmer money!  Since the cellulose etc. assists the soil to retain water, (it is known as humus), the large volumes of organic fertilisers that must be used are not really a problem – they actually benefit the soil.
11. a % loss for rice = 42.2%; total crop is 445.7 million tons so loss is 188 million tons.
     b Fruit crops are generally long-lived and tall (think walnuts and olives!) or grown under cover (tomatoes and strawberries).  In the former case the crop will be unaffected by low-growing annual weeds; in the latter there are no weeds!
12. This shows how statistics can be used to mislead!  Africa is not a major food producer – indeed, with the Sahara and other deserts and erratic rainfall, much of the continent is unable to feed itself using the subsistence agriculture commonly used.  In contrast, the industrial agriculture of Europe, North America, and much of Russia and China (and the intensive agriculture of Indo-China) is extremely efficient and produces a surplus of the major crops which can be used of offset starvation in much of Africa.  Here, losses due to weeds (and all other pests too) are low.  Thus, when reading statistics, it is essential to know the size of the sample.  5% of a lot is still sufficient; 100% of nothing will not fill your belly!
13. When each plant competes with its neighbour (intra-specific competition) for water, minerals and light, the yield from each plant will go down.  Interestingly, the overall yield from the land may go up, since there are more individuals present.  Whether the farmer makes a bigger profit will depend on the extra cost of the seed, fertiliser, irrigation and spraying that the higher crop density will require.
14. Aphids do some damage since they suck sap from the phloem and so reduce the plants’ available sucrose (= energy).  However, their real menace to the farmer comes from the fact that they are vectors of plant virus diseases.  Since plants neither sneeze nor move about much, virus diseases rely on animal vectors to spread them from host to host.  Since the peach-potato aphid is the more mobile, it will be the better vector of disease.
15. This is not really true!  Systemic insecticides are absorbed by the plant and transmitted all over it in the phloem sap.  The insecticide is not toxic by contact but is absorbed from the insects’ gut (most are organophosphates, or nerve poisons).  Thus insects that land on the surface of the plant are unaffected (e.g. ladybirds), whilst those that suck nectar are unaffected too, since nectar does not contain the poison (e.g. butterflies).  Small sucking insects (e.g. aphids) soon succumb; larger, biting, insects (e.g. caterpillars) rarely eat enough sap to poison themselves – remember they weigh perhaps as much as 10,000 aphids!

Marketed as better for the environment, it matters little to the ladybird if it is poisoned by DDT or starves to death because there are no aphids for it to eat!  In addition, whilst in days of old we could wash apples to remove the poisons, all the washing (yes, and peeling too!) will not remove the residuals of systemic poisons, the long-term effects of which we do not yet know....

16. If pesticides are stable, they remain in the environment for a long time.  If, in addition, they are not readily excreted by animals, then they tend to build up in the host animal as you move up the food chain (known as bio-accumulation).  For this reason, modern pesticides are designed to break down after, at most, a few weeks.  This also reduces the speed at which resistance to the pesticide becomes a problem and reduces the minimum time between spraying and harvest, to ensure that the residuals left in the crop are below permitted levels.  Since each sparrow-hawk eats many more than 19 small birds in its lifetime (that’s about a couple of days-worth!), it would appear that this insecticide is being broken down.  Remember, WE are at the top of the food-chain!  In the late 1960’s human breast milk had well above the permitted levels of DDT, which has now been found in the fat even of seals, polar bears and penguins at both poles!    

Chapter 13 - Manipulating Reproduction

In Chapter Questions [back to top]

1. Ovulation.  The female ovum is fertile for just 12 hours after ovulation.
2. By remaining thick and vascular, the endometrium ensures that the zygote (by now grown into a ball  of cells, or blastocyst) has sufficient living cells to ensure that the embryo’s presence is detected and that it will receive sufficient nutrients and oxygen at the commencement of the next 38 weeks of its development, during which time it will be totally dependent on the placenta for its needs.
3. When our body temperature is too high, muscular activity is reduced (you feel lethargic), blood supply to the skin increases (you go red, or flushed); the skin hairs lie flat and sweating increases, the evaporation of which will remove much heat from the body.  These are physiological changes.  In addition, we remove warm clothes, seek shade (or the pool!) and tend to wear clothes that are lighter in colour and thinner. We also drink more, cool, drinks. These are behavioural changes and are equally important in thermoregulation.  Think about snakes (ugh), which cannot sweat (only horses can do that – gentleman perspire, ladies glow...), yet snakes can maintain a core body temperature more even than any mammal can.
4. Only cells with the right receptors can respond to LH, since they work with the hormone in much the same way as enzymes do with their substrate (think ‘locks and keys’).  Each receptor is specific.
5. Freeze embryo of pedigree animal after growing it in a laboratory for several days, so allowing it to be split into several identical ‘twins’.  Wait until its clones’ identity (male or female) is known and then wait long enough to see if the animal is a real winner.  If so, keep the embryos for yourself and clean up at Smithfields Show for several years to come!  If not, pass on the embryo to a charity who will fly it to a third-world country and use it to improve the quality of the local herd at minimal cost.  Defrost the embryo on arrival at the farm, insert into the womb of a cow known to be in oestrus. Probably give a hormone injection to improve the implantation rate and wait 11 months for the calf to be born!
6.

During pregnancy, as long as the corpus luteum is intact and functional, it will be producing progesterone to ensure that pregnancy progresses.  Progesterone will feedback to the pituitary, where it will inhibit FSH production (hence no follicles developing) and LH production (hence no ovulation).  Without those two pituitary hormones, there are no eggs released and no additional pregnancy is possible.  The oestrogen is made in the ovary and organises the development of the endometrium and of the ovum (eggs!).  

   

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Last updated 20/06/2004