To investigate some of the properties of immobilised enzymes.

Introduction

The four-session practical exercise is designed to give some insight into the preparation and properties of immobilised enzymes. 

It consists of three parts:

• Preparation and properties of covalently-immobilised a-amylase. 
• Preparation and properties of non-covalently-immobilised a-amylase.
• The use of a packed-bed and stirred tank reactor.

Before starting work, read through the Methods and Results sections. This practical is very demanding and must be approached with thought and care. It will be necessary to retain samples of the soluble enzyme before and after coupling (as well as the immobilised enzyme, of course) for protein and activity assay (see Appendix A) in order to determine the amount of enzyme coupled. You are expected to work in teams of three with a named team leader. You must plan and organise your experiments carefully, for which marks will be awarded. a-amylase is an enzyme produced and purified from Bacillus. It hydrolyses the a-1-4 links in starch randomly along its structure (i.e. it is an endo-glycosidase). It cannot hydrolyse a-1-6 links. Complete hydrolysis of starch by a-amylase produces a mixture of short glucose oligomers (e.g. maltose, maltotriose), some limit dextrin containing a-1-6 links but relatively little glucose. The quality of hydrolysed starches is given in terms of its dextrose equivalent (DE), which equals the percentage of the starch that is hydrolysed. (’Dextrose’ is another word for glucose). 

a-Amylase is assayed by the creation of new reducing (terminal; equivalent in reducing power to glucose) sugar by the catalysed hydrolysis of soluble starch.

In this practical, a-amylase is immobilised by means of covalent and non-covalent binding to solid supports. The amount of enzyme attached to the supports is determined and the activities of the immobilised enzymes are compared to that of the free (non-immobilised) enzyme. Packed bed reactors containing the immobilised enzymes are prepared and their ability to hydrolyse starch compared. 

Plan
During week 1 you should

• Prepare covalently-immobilised a-amylase
• Prepare non-covalently-immobilised a-amylase
• Construct standard curves for protein and reducing sugar (see Appendix A for details). Draw these before week 2 when they will be required.

During week 2 you should 

• Wash the covalently-immobilised a-amylase
• Wash the non-covalently-immobilised a-amylase
• Assay samples from the preparations; see ‘Assays’ later

During weeks 3 and 4 you should run the stirred tank and packed bed reactors and analyse their products.

Covalent Immobilisation of a-amylase (Methods Enzymol. 44, pp. 98-99)

Week 1. Weigh out 1.0 g Enzacryl AA gel. (This is a gel based on a polyacrylamide matrix with aromatic amino groups present as the reactive moieties). Add to 50 ml ICE-COLD 2 M HCl, stir at 0°C, and gradually add 40 ml ICE-COLD 2% sodium nitrite solution (Together these produce nitrous acid, HNO₂).

NaNO₂ + HCl HNO₂ + NaCl

Keep the beaker surrounded by ice during addition. This should take about 10 - 15 minutes. (This creates the reactive diazonium groups from the aromatic amino groups. If these are allowed to warm up, they decompose to give nitrogen gas with the loss of their specific reactivity)

Stir for another 15 minutes, then filter the gel on a filter paper disc on a Buchner funnel, by suction. Wash with 200 ml ICE-COLD 20 mM phosphate buffer, pH 7.0, 0.1 mM CaCl₂ (a-amylase buffer). Add the buffer in 25 ml batches, and KEEP IT COLD. At this stage, the amino groups of the gel matrix should be diazotized. Quickly scrape gel off the filter into a test tube. Add 5 ml of an ICE-COLD solution of a-amylase (2 mg/ml in phosphate buffer). Cap, label (’Cov‘) swirl in an ice bath for about an hour and leave in the fridge until next week. (This allows the diazo groups to covalently couple to the tyrosine phenolic groups on the enzyme)

Non-Covalent Immobilisation of a-amylase

Week 1. Weigh out 0.6 g dry phenolic resin (invented and patented by M. F. Chaplin, J Chem Soc., Perkin 1, 1979, pp 2144-2153), suspend in 20 ml 20 mM K phosphate pH 7.0, 0.1 mM CaCl2 (a-amylase buffer) for 10 minutes. Filter and re-suspend in 5 ml of 2 mg/ml a-amylase in the 20 mM phosphate buffer. Cap, label (’Non‘) swirl for 30 min and leave in the fridge until next week

N.B.: Keep a solution of the free enzyme (0.5 ml 2 mg/ml) similarly capped in the fridge as a comparison for both ‘Cov‘ and ‘Non‘ above (label ‘Enz‘).

Covalent Immobilisation of a-amylase

Filter the gel (’Cov‘), using a fluted filter paper, into a test tube; use 5 ml of 20 mM K phosphate buffer pH 7.0 to aid this process. Keep about 2 ml of the filtrate for assay of unbound protein and activity, (Label it ‘Cov-supernatant‘)

Wash gels to remove any free enzyme, using 3 x 20 ml batches of 20 mM potassium phosphate/500 mM NaCl pH 7.0 (’high salt buffer’). Let the gel damp-dry briefly between each 20 ml portion of buffer. Wash once more in the same buffer without NaCl, and re-suspend the gel in 5 ml in 20 mM K phosphate (pH 7.0). Label it ‘Cov-immobilised‘ and refrigerate until next week.

Non-Covalent Immobilisation of a-amylase

Filter the gel (’Non‘), using a fluted filter paper, into a test tube; use a further 5 ml of 20 mM K phosphate buffer pH 7.0 to aid this process. Keep about 2 ml of the filtrate for assay of unbound protein and activity, (Label it ‘Non-supernatant‘), .

Wash gels to remove any free enzyme, using 3 x 20 ml batches of 20 mM K phosphate buffer pH 7.0. Re-suspend in 5 ml of this buffer. Label it ‘Non-immobilised‘ and refrigerate until next week.

At this stage During Week 2 you should have the following samples:

• Cov-supernatant: unbound a-amylase; left over from covalent immobilisation (2 ml of not more than 1.0 mg/ml)
• Cov-immobilised: covalently bound a-amylase; 1 g dry gel containing not more than 10 mg enzyme.
• Non-supernatant: unbound a-amylase; left over from non-covalent immobilisation (2 ml, not more than 1.0 mg/ml).
• Non-immobilised: non-covalently bound a-amylase; 0.6 g dry gel containing not more than 10 mg enzyme.
• Enz: stored free a-amylase (0.5 ml of 2.0 mg/ml). Some of this should be refrigerated for next week.

Assays (See Appendix for details of the procedures)

Dilute samples ‘cov-supernatant‘ and ‘non-supernatant‘ 1:20 v/v and sample ‘enz‘ 1:40 v/v. Assay these diluted samples of ‘cov-supernatant‘, ‘non-supernatant‘ and ‘enz‘ for protein content (Assay 1) and a-amylase activity by production of reducing equivalents (Assay 3). Ensure that you record how you dilute these samples in your notebook.

Although you know the concentration of protein in sample ‘enz‘ (2 mg/ml), you will probably get a different value as determined in the Dye-binding assays due to the different standard protein (bovine serum albumin not a-amylase) used. Use this value to ‘correct’ the protein concentrations (i.e. if the apparent Dye-binding concentration of sample ‘enz‘ is 1.5 mg/ml then all final protein concentrations as determined by the Dye-binding method should be multiplied by the factor 2.0/ 1.5).

Tabulate the protein concentration (mg/ml, uncorrected and corrected) and activity (mmol reducing sugar released/min/ml and mmol reducing sugar released/min/mg) of the diluted and original undiluted samples cov-supernatant, non-supernatant and enz. By allowing for the volumes of solutions used in the binding (5 ml) and filtering (another 5 ml), tabulate also the total protein content of the supernatants. 

This Table will allow you to calculate:

1. the (corrected) weight of a-amylase protein not bound to each immobilisation matrix from the protein concentrations in the unbound residual enzyme samples cov-supernatant and non-supernatant. Make sure that you allow for the dilutions and the final volume of the wash solution (10 ml) and the correction for the use of the bovine serum albumin standard.
2. the weight of protein bound to each immobilisation matrix, by subtracting the (corrected weight of) protein not bound (from above) from the amount added (5 ml x 2 mg/ml = 10 mg).
3.  the percentage of the enzyme protein that was added that is bound to each immobilisation matrix.
4. the specific activity of the original a-amylase solution used (sample ‘enz‘); Note that the specific activity equals the activity of one mg a-amylase protein. the units are in mmol reducing sugar released per min per mg of protein).
5. the specific activity of the unbound a-amylase solutions left after each of the immobilisation processes. Note that these would be expected to be identical to the specific activity of the original a-amylase solution used (sample ‘enz‘) unless some denaturation occurred in the immobilisation process.

Note: you have to dilute the samples until they contain about 50 mg/ml protein so that your determinations are in the right range for the assays. Do not forget to allow for these dilutions when you determine the protein content and a-amylase activity of the original solutions.

Prepare two small packed bed reactors containing all of the covalently and non-covalently immobilised a-amylase (’cov-immobilised‘ and ‘non-immobilised‘). Do not allow them to run dry. (Note that if they are allowed to develop an air lock, they will not flow and must be repacked) Run 5 ml of 1% starch in 20 mM K phosphate pH 7.0, 0.1 mM CaCl₂. through each column.

 Remove the gels (’cov-immobilised‘ and ‘non-immobilised‘) from the columns. Using all of the samples of immobilised enzymes determine (separately) their activity in a stirred reactors (beaker) containing 50 ml of 1.0% w/v starch in 20 mM K phosphate pH 7.0, 0.1 mM CaCl2. Withdraw samples at intervals (e.g. 1 min, 5 min, 10 min, etc.) to determine their reducing sugar content. 

From the assay of the stirred tank and packed bed reactors you should calculate :

1. the concentration of reducing equivalent in the partially-hydrolysed starch samples (mmoles of reducing equivalent produced per ml per reactor).
2. the productivity of the reactors (mmoles of reducing equivalent produced per minute per reactor), 
3. the fractional conversions, X (X = moles of reducing equivalents produced/moles of potential glucose units in the starch solution); Note that to calculate the number of moles of potential glucose in the 1% starch, solution, the apparent M.Wt of potential glucose is 180 - 18 = 162, as water is necessary to release the glucose; the complete hydrolysis of 162 g of starch produces one mole (180 g) of glucose, Also remember that the number of moles in a sample is the weight in grams divided by the weight of one mole (i.e. weight/ M.Wt). Also, note that because a-amylase cannot hydrolyse limit dextrins, maltose, maltotriose or maltotetraose, the highest value expected for the fractional conversion, X, is about 0.2.
4. the dextrose equivalent DE of the products (in this case the fractional conversion, X, expressed as a percentage), 
5. the activity of the immobilised enzymes (mmoles of reducing equivalent produced per minute per g resin).
6.  the specific activity of the immobilised enzymes (mmoles of reducing equivalent produced per minute per mg enzyme) by using the known amount of protein immobilised (determined previously).
7. the effectiveness factors for the immobilised enzymes. Note that the effectiveness factor is the specific activity of the immobilised enzyme divided by the specific activity of an equal quantity of the free enzyme (calculated previously).

Assay samples ‘cov-immobilised‘, ‘non-immobilised‘ and ‘enz‘ for a-amylase activity by loss of iodine reactive material (Assay 4). This reaction may be very rapid with excess free enzyme. For the free and immobilised enzymes estimate the % digestion of the starch when the iodine reactive material has been used up, by comparing these results with your specific activity results from the production of reducing equivalents.

1. determine the relative specific activities of the immobilised enzymes compared with the free a-amylase. Use the reciprocals of the times needed to decolourise the blue starch-iodide divided by the amounts of a-amylase protein present.
2. compare these specific activities with those calculated earlier. Explain your results on the basis that starch molecules, once next to an immobilised a-amylase, have difficulty diffusing away due to their bulk. Thus, immobilised a-amylase is expected to produce some starch that is completely hydrolysed before other starch molecules are hydrolysed at all, whereas free a-amylase hydrolyses all starch molecules roughly equally.

 Practical: Appendices

Assays
Note that all assays should be done in duplicate, where possible. Ensure all the cuvettes are clean by checking their absorption against each other at the assay wavelength before use. 

1 Dye-binding Protein Assay 
1.5 ml of protein sample solution (0 - 50 mg/ml) is mixed with 1.5 ml Coomassie blue reagent (0.6% dye in dilute perchloric acid). Use 1.5 ml distilled water plus 1.5 ml Coomassie blue reagent as blank to zero the spectrophotometer. Read the absorbency at 620 nm. 

A standard curve is prepared by using the stock solution of bovine serum albumin (BSA, 50 mg/ml) using at least four data points in duplicate. e.g. 0.4 ml stock + 1.1 ml water (= 0.4/1.5 x 50 mg/ml = 13.3 mg/ml), 0.8 ml stock + 0.7 ml water, etc. N.B. only the concentration within the 1.5 ml ’sample’ solution is relevant; the (constant) amount of reagent added is not relevant for sample concentration calculations

2 Reducing Sugar Assay 
2.0 ml of DNS reagent (ready prepared; 3,5-dinitrosalicylic acid and sodium potassium tartrate dissolved in dilute sodium hydroxide) is added to sample (200 ml, 0.2 ml), containing 0 - 2 mg reducing sugar (i.e. 0 - 10 mg/ml). The tube is placed in a boiling water bath and the solution heated at 100°C for 5 minutes. Rapidly cool in ice to room temperature. Use 0.2 ml distilled water plus 2.0 ml DNS reagent, heated as above, as blank to zero the spectrophotometer. Read absorbency at 570 nm. A standard curve is prepared by using the stock solution of maltose (10 mg/ml) using at least four data points in duplicate. e.g. 0.05 ml stock + 0.15 ml water (= 0.05/0.2 x 10 mg/ml = 2.5 mg/ml), 0.1 ml stock + 0.1 ml water, etc. N.B. only the concentration within the 0.2 ml ’sample’ solution is relevant; the (constant) amount of reagent added is not relevant for sample concentration calculations. You are reminded that the M.Wt. of maltose is 342 and maltose contains a single reducing group (i.e. 342 g maltose contains one mole of reducing group/equivalent). For your graphs, you must calculate the molar concentration of reducing groups in the standard maltose solutions.

3 Assay of a-amylase by production of reducing equivalents 
Add 0.8 ml 20 mM K phosphate (a-amylase buffer) to 0.2 ml soluble enzyme in phosphate buffer (containing about 10 mg amylase). Note that the enzyme solutions must be diluted before they are assayed). Pre-incubate for about 4 minutes at 37°C. Add 1.0 ml, 1% starch in phosphate buffer (pre-warmed to 37°C). Incubate for exactly 5 minutes at 37°C. Stop the reaction by removing 0.2 ml of the incubated mixture and adding this to 2 ml of DNS reagent. 

The tube should be placed in a boiling water bath and the solution heated at 100°C for 5 minutes to develop the reducing sugar assay colour. Rapidly cool in ice to room temperature and read absorbency at 570 nm. Use 0.1 ml buffer plus 0.1 ml starch plus 2.0 ml DNS reagent, heated as above, as a blank to zero the spectrophotometer. Note that the reducing sugars in only 0.2 ml of the 2.0 ml in the 37°C incubation mixture is used in the reducing sugar assay and allowance should be made for this when calculating the amount of reducing sugar produced by the enzyme in the 0.2 ml original sample.

4 Assay of a-amylase by loss of iodine reactive material 
Make a mixture of 0.1 ml buffer plus 0.1 ml starch for use as blank. Add one drop to one drop of K phosphate containing 0.05% iodine. A blue coloration will be observed. 

Free enzyme assay: Add 0.8 ml K phosphate (20 mM, pH 7, ‘a-amylase buffer’) to 0.2 ml enzyme (containing about 10 µg free amylase). Incubate for 4 minutes at 37°C. Add 1 ml 1% starch (pre-warmed to 37°C) and incubate at 37°C. At known times (e.g. 0, 30 s, 1, 2, 5, 10 min etc), remove 1 drop and drop into 1 ml K phosphate containing 0.05% iodine. 

Immobilised enzyme assay: Add 1.0 ml K phosphate (20 mM, pH 7, ‘a-amylase buffer’) to half the immobilised enzyme. Incubate for 4 minutes at 37°C as above. Add 1 ml 1% starch. Keep the immobilised enzymes agitated. At known times (e.g. 0, 30 s, 1, 2, 5, 10 min etc), remove one drop and drop into one drop of K phosphate containing 0.05% iodine. In both assays, blue coloration will be observed while macromolecular starch is still present. The enzyme activity is inversely proportional to the time taken. If no blue colour is observed in the first samples, repeat the assay as either (1) the reaction has already occurred at to rapid a pace, or (2) you forgot to add the enzyme/ iodine/starch/etc.

OUTLINE THE MAJOR FEATURES OF MENELIAN GENETICS. DISCUSS THE ROLE OF CHROMOSOMES IN EXPLAINING THESE PRINCIPLES.

Gregor Mendel was an Augustinian nineteenth century monk who, due to a series of momentous experiments, is now widely regarded as the forefather of genetics.  Mendel studied the inheritance of seven contrasting characteristics of the species Pisum sativum, more commonly known as the garden pea.  Each of the variables that Mendel experimented with were discontinuous;
there were no intermediate forms.  For example, one of the variables was length of stem, which was always either tall or short.  From his experiments Mendel was able to draw solid conclusions about the inheritance of characteristics in organisms.  With the advancements in genetics since his time we are now able to explain Mendel’s principles in terms of chromosomes and genes. Understanding of these exact terms did not exist in Mendel’s lifetime.  However, Mendel’s principles still form the basis of modern day genetics.
In his first series of experiments Mendel allowed Pisum to self-fertilise for several generations, so that he knew that these pea plants were purebred.  He then cross-fertilised plants which were purebred for contrasting characteristics.  For example, he crossbred pure-bred dwarf Pissum with pure-bred tall Pissum.  He carried out reciprocal crosses.  Even though these plants obviously showed many characteristics he only looked at one characteristic at a time.  In collecting the results of his experiments, Mendel recorded the numbers of individuals in each class in the progeny, this established the ratios of the contrasting characters of many subsequent generations.  In the F1 generation all the plants were tall.  Mendel then left the F1 generation plants to self-fertilise.  In the F2 generation there were both tall and dwarf plants in an approximate ratio of 3:1.  The same ratio was found in the F3, F4, and F5 etc. generations.  Mendel realised that because the ‘dwarf’ characteristic had disappeared in the F1 and had then reappeared in the F2, the controlling factor for ‘dwarf’ had remained intact and undiluted from one generation to another.  It is never expressed, however, in the presence of a factor for ‘tall’.  He understood that there must be two independent factors for ‘dwarf’ and ‘tall’.  Mendel comprehended that the 3:1 ratio was the product of the binomial expression derived from randomly combining two pairs of unlike elements.
We now know Mendel’s ‘factors’ to be genes found on homologous pairs of chromosomes in the nucleus of the cell.  There are two or more forms of each gene known as alleles.  In Mendel’s experiments the allele in pea plants for ‘tallness’ was dominant and the allele for ‘dwarfness’ was recessive.  A pure breeding ‘tall’ plant is homozygous for the ‘tall’ allele and a pure breeding ‘dwarf’ plant is homozygous for the ‘dwarf’ allele.  This means that when cross-pollinated the ‘tall’ parent can only pass on gametes containing ‘tall’ alleles and the ‘dwarf’ parent can only pass on gametes containing the ‘dwarf’ allele.  As one allele from each gamete combines to form the gene at fertilisation, when a ‘tall’ parent and a ‘dwarf’ parent are crossed, all the offspring must have one ‘tall’ allele and one ‘dwarf’ allele on the gene that codes for length of stem.  However, as the ‘tall’ allele is dominant, only this allele was expressed and therefore was shown to be present in all of the F1 generations.  Still, when the F1 generation was left to self-fertilise there was a ‘dwarf’ allele present on the genes of all of the plants.  There is a fifty percent chance of each allele being in a gamete, so half of the gametes of the F1 generation contained the ‘dwarf’ allele.  Therefore if a ‘dwarf’ allele containing gamete and another ‘drawf’ allele containing gamete were to combine the offspring would be homozygous recessive.  If a recessive and dominant allele were to combine the offspring in the F2 generations would be identical to their parents still carrying the ‘dwarf’ gene but only expressing the ‘tall’ gene as it is dominant.  If A were to stand for the ‘tall’ allele and a for the ‘dwarf allele, if the F1 generation (all being Aa) are allowed to self-fertilise then the offspring would be AA, Aa, aA and aa in a genotypic ratio of 1:2:1 giving the phenotypic ratio of 3:1, which Mendel observed.  This can be summed up in Mendel’s first law, which states that ‘The characters of an organism are controlled by pairs of alleles which separate in equal numbers into different gametes as a result of meiosis.’
Mendel also studied the simultaneous inheritance of two characteristics.  In one experiment he traced the inheritance of seed colout and texture in Pisum.  First he crossed a pure-breeding variety having roundd and yellow seeds with another pure-breeding variety having wrinkled and green seeds.  All the F! generation had round, yellow seeds, thus showing these to be the dominant traits.  When self-pollinated, the plants which grew from the F1 seeds were round and yellow, round and green, wrinkled and yellow and wrinkled and green in a ratio of 9:3:3:1 respectively.  This is a dihybrid ratio.  Mendel thus established that dissimilar pairs of factors that combined in a hybrid could separate from one another and come together in all possible combinations in subsequent generations.  Mendel did not express his discovery as a law.  However, with the information that we now have , his discovery can be stated as his second Law which states that: ‘Two or more pairs of alleles segregate independently of each other as a result of meiosis, provided the genes concerned are not linked by being on the same chromosome.’
The behaviour of chromosomes in meiosis explains how independent segregation occurs.  The alleles which determine the two pairs of contrasting characteristics are located on different pairs of homologous autosomes.  Because the chromosomes of one pair separate independently of the other pair, the alleles segregate independently.  At anaphase I in meiosis the pairs of homologous chromosomes pass to opposite poles of the cell.  At Anaphase II the centromeres of the chromosomes break in two and the cromatids are pulled, centromere first, towards opposite poles of the cell, thus becoming four separate sister chromatids.  So each of the chromatids separate randomly and independently, thus explaining Mendel’s second law.
Drosophilla have been widely used in genetics research.  Drosophila was introduced into genetics research by an American biologist, Thomas Hunt Morgan who established the chromosome theory of heredity-the theory that Mendel’s factors are actually the linear series of genes on a chromosome.  Morgan’s work established the truth of Mendel’s interpretation of his experiments.  Using Drosophila Morgan went on to determine sex linkage, involving the characters that are controlled by genes on sex-determining chromosomes, crossing over, which is the exchange of genes between chromosomes as a result of chiasmata formed during meiosis.  He also discovered chromosome maps which show the relative positions of many genes on the four chromosomes of Drosophila.
Gregor Mendel, who remained an undiscovered genius for his lifetime prepared the foundations on which modern day genetic are built.  His ratios can now be explained in terms of chromosomes and the biochemical processes which take place within cells.  Today there are many dire warnings about genetics; dead men are becoming fathers through their frozen sperm, little girls are infused with modified viruses whose infectious qualities have been replaced with healthy genes that the girls lacked at birth and the whole encyclopaedia of the human genome is read, one piece of DNA after another, in perfect sequence, telling us with dreadful accuracy what it means to be normal.  On the other hand research is being carried out into genetic diseases, which could save the lives of millions of humans that are alive today and those which have yet to be born.  With the discoveries of genetics, which are being made in this rapidly advancing field, comes knowledge along with its associates power and danger.  None of these amazing things might be happening today if a monk in a Moravian monastery had not had a particular joy for growing peas in a greenhouse.

In this section I will look at all the factors involved in the rate of reproduction of yeast cells, bring them together at the end of the section and deduct a scientific conclusion as to what I would expect to happen.

Yeast Cell Respiration

Yeast cells, like any other organism, need a substrate to respire.

Respiration in general

In respiration, they use the substrate to produce ATP for cell metabolism.  The overall equation for respiration using glucose is:

C₆H₁₂O₆ + 6O₂ Þ 6CO₂ + 6H₂O

Therefore, the more glucose available, the greater the rate of respiration and the easier it is for cell metabolism to take place.  The more detailed version of the respiratory pathway is shown in Figure 1.  However, one cannot merely assume that a greater rate of respiration can occur with a higher level of glucose concentration because it is a living organism involved.  This means that high levels of glucose concentration could cause the yeast cells to become crenated and therefore the reproductive mechanisms would not function because of the physiological strain caused by the low pressure potential.  This would mean that the initial rate of reproduction would be much less than that of cells in solutions of glucose concentrations on the limit of the cell’s pressure potential.  However, this need not be a worry because I will not use glucose of such a high concentration. Also, because it is an organism, other factors like waste products and method of uptake need to be taken into account. These are discussed below.

Glucose catabolism in yeast

In the case of yeast, something else also needs to be considered.  This is that it is an anaerobe.  This means that yeast cells first respire the glucose anaerobically and then break down the ethanol produced to manufacture ATP aerobically using the available oxygen in the solution.  In yeast, anaerobic respiration is called fermentation. These organisms are called facultative aerobes, because when oxygen is present, they respire aerobically, but if oxygen is absent, their respiration is purely anaerobic.

In anaerobic respiration, as in aerobic respiration, the first step consists of breaking down glucose into pyruvic acid (see Box 1). However, the two processes differ considerably in the details of this step.  In anaerobic respiration, there is no Krebs cycle.  The end products of anaerobic respiration

are usually ethyl alcohol and carbon dioxide, as in yeast cells, or lactic acid, as in muscle cells.  Another difference between the two methods of respiration is the amount of energy released.  In aerobic respiration, 32 ATP molecules of energy are released per glucose molecule (see Figure 1).  In anaerobic respiration, only two ATP molecules of energy are released for each glucose molecule.As the oxygen supply starts to fall short of the demand by the increasing population, it is less easy for the yeast cells to oxidise the ethanol and this collects as a toxic by-product of anaerobic respiration.  If left for long enough, this would in the end caused a rapid decline in the population because as the oxygen in the solution is used up, the level of ethanol will increase with the population, and, with time, will reach a level of toxicity where it actually kills off the population.

Transport of Glucose into Cells

In order to understand how glucose is transported into the yeast cell, it is necessary to study glucose and the barrier that it has to cross, the outer cell membrane, as well as the implications of this.

Glucose

The structure of glucose is shown on the right. The many hydroxyl (OH) groups present result it being very polar. It is therefore only soluble in polar substances such as water. It will not dissolve in non-polar hydrophobic environments.

The Fluid Mosaic Model of the phospholipid bilayer

Biological membranes are bilipid layers. Lipid bilayers are fluid, and individual phospholipids diffuse rapidly throughout the two-dimensional surface of the membrane.  This is known as the fluid mosaic model of biological membranes (fluid because the molecules move around, mosaic because it includes proteins, cholesterol, and other types of molecules besides phospholipids).  In a real cell, the membrane phospholipids create a spherical three-dimensional lipid bilayer shell around the cell.  However, they are often represented two-dimensionally as:

Each  represents a phospholipid.  The circle, or head, is the glycerol and negatively charged phosphate group and the two tails are the two highly hydrophobic hydrocarbon chains of the phospholipid.  The tails of the phospholipids orient towards each other creating a hydrophobic environment within the membrane.  This leaves the charged phosphate groups facing out into the hydrophilic environment.  The membrane is approximately 5 nm thick.  This bilipid layer is semi permeable, meaning that some molecules are allowed to pass freely (diffuse) through the membrane.  The lipid bilayer is virtually impermeable to large molecules, relatively impermeable to small polar molecules and charged ions, and quite permeable to lipid soluble low molecular weight molecules.  Its substantial permeability to water molecules is not well understood.  The lipid bilayer is impermeable to medium and large-sized polar molecules.  These molecules depend on hydrophilic intrinsic protein channels to assist their transport across the barrier by a process called facilitated diffusion.

Facilitated Diffusion

 Molecules that can diffuse through the membrane do so at differing rates depending upon the number of channels available for them to pass through the hydrophobic interior of the membrane bilayer.Facilitated diffusion utilises membrane protein channels in the membrane to allow charged molecules (which otherwise could not diffuse across the cell membrane because it is non-polar) to freely diffuse in and out of the cell.  These channels are the way in which medium-sized polar molecules like glucose may pass into the cell.  The number of protein channels available limits the rate of facilitated transport, whereas the speed of diffusion is dependent only on the concentration gradient. This means that there is a limit to the rate of reproduction of yeast cells, even in ideal circumstances, determined by the rate of facilitated diffusion (see Figure 2). The mechanism is not yet fully understood.

The Reproduction of Yeast Cell

The way in which yeast cells reproduce also has an effect on their rate of reproduction.Figure 3 shows the two most commonly studied yeast species, namely budding yeast Saccharomyces cerevisiae (see A and B) and fission yeast Schizosaccharomyces pombe (see C, D, and E).

Budding yeasts (the ones that I will be using) have an oval shape.  They multiply by forming buds, which grow in size, and, finally, pinch off from the mother cell.  The digestive organelles, called vacuoles, can be identified inside the cells (see F).

Fission yeasts are the rod-shaped ones.  They form a septum in the middle of the cell (see C and D) and divide evenly.

Figure 4 a) shows the basic stages of cell replication. The budding cell is the same size as the original for simplicity. The actual shape is shown in Figure 4 b). The chromosomes replicate themselves and then position themselves with their centromeres attached to the spindle.  Their copies then separate and move along the spindle to the polar body on opposite sides of the cell.  The bud vacuoles and bud organelles start to form and cytoplasmic

division can then take place.This is what I expect to happen in terms of population.  At first when the yeast cells are put into the glucose solution, they take a while to acclimatise. They may have to produce the RNA to produce the right enzymes, or take a while to reach to the reproductive stage in their cell cycle.

This is called the lag phase.  Once they have done this, the rate of increase in population should get greater as the population of cells increases. This is because if each cell reproduces at a certain rate, the more cells there are, the greater the total rate of reproduction.

However, there are plenty of reasons why the population may not reproduce to its full capacity.

A constant doubling time is a sign of a culture that is growing without constraints like low temperature, lack of food, space or water and without the accumulation of waste products. 

Population Dynamics

Here is my mathematical explanation of population growth

Suppose that the culture is not significantly depleting its nutrient reserves or, more accurately, that progress through the cell cycle rather than nutrient uptake is rate limiting.  Then the mean reproduction time in this first stage of population growth should be constant for any cohort of cells.  Adding glucose would ensure that cell cycle progress continues to be rate limiting until space becomes a limiting factor.  The equation for the population is just:

dN/dt=rN((K-N)/K)

Where r is a constant (the intrinsic rate of increase, i.e. the number of reproductive events per unit time per yeast cell), t the time they have been multiplying for, K the maximum number of individuals that the particular environment can support and N is the yeast population number.  The solution to this equation is

((K-N)/K)N = N₀2^(t/T)

Where T is the doubling time (normally quoted as 30-60 minutes in ideal conditions) and N₀ is the population at time zero. The introduction of K into the equation means that the effect of the environment in slowing down growth to a steady state (or decline) is taken into account.  K is also described as the carrying capacity for the environment.

If N>K, growth rate is negative. If K>N, graph of rate is positive, therefore population size tends to settle out at around K=N, that is, population size adjusts to the carrying capacity of the environment. When K=N, the population growth rate is zero.  Population size becomes stable. This produces an exponential curve, which levels off, as shown in Figure 5.

However, when a biotic or abiotic factor limits the size of the population, either by two putting limits on the sustainable limits or by slowing down its growth through increased death rate, the exponential increase curve of the population size curves off to a pattern in a fairly short time and can continue in this stable phase for a fairly long time.  The factor that causes the eventual decline of the population can be a variety of things: a rapid change in the anaerobic conditions, a build up of toxins produced by the organisms themselves or the introduction of a new predator or disease etc.

This would produce a population curve throughout the duration of existence of a species (for a simple case like the one if we are looking at) that looked something like Figure 5.This graph only applies if the number living organisms are counted (and not only the number of the cells in a solution) because, for example at stage c), cells are still being produced, its just that the mortality rate is equal to the natality rate but the total number of cells (live + dead) in the solution will still be increasing.

Summary

This is a summary of all evidence so far collected to support a prediction.

1. Yeast cell respiration

Yeast cells require glucose for metabolism. They respire it anaerobically, then oxidise the ethanol produced. Lack of oxygen causes the build-up of toxic ethanol, which can lead to their death. The larger the population, the greater an effect this will have, leading to the population levelling out as mortality rate increases to mortality rate. This causes the transition from the exponential phase to the stationary phase.

1. The phospholipid bilayer

Yeast cells gain glucose from the medium in which they are suspended, and because glucose is a polar molecule, the rate of its transport across the membrane depends on the availability of hydrophilic protein channels. An increase in glucose concentration has less of an effect at higher glucose concentrations. Therefore, when the rate of reproduction relies on rate of diffusion, it follows the same trend with concentration as facilitated diffusion does.

1. The reproduction of yeast cells

The rate of reproduction can be limited by progress through the cell cycle (in the lag and exponential phases) or, if the level of glucose is low enough, can be limited by the rate at which glucose can be obtained. A higher glucose level could therefore increase the rate of reproduction.

1. Environmental constraints on the population

These, strictly speaking, should include availability of oxygen and glucose and the build up of toxins. Their effect is the same: when the number of yeast cells is sufficient, they have the effect of either putting constraints on the size of the population, or increasing the mortality rate to the natality rate. The effect of either is the levelling-off of the [living population]-[time] graph. A limitation to the experiment arises if one of these other than glucose concentration affects the population growth rate.

Prediction

During the times for which I plan to run my experiment, I only expect to see the period from the lag phase to the end of the exponential phase. Not many cells will be dying during these initial phases (although possibly a few during the lag if they do not successfully adapt), so the total count will be approximately equal to the living population. The rate of growth will therefore follow the pattern:

Plan

This is the experimental procedure that I plan to follow when I carry out the practical experiment.  It is in chronological order. Decisions are backed up with reasoning.

Apparatus:

10 x 100 cm³ conical flasks with cotton wool plugs

Haemocytometer slide with cover slip   

1ml teat pipette

Stop clock

Water bath

25 cm³ and 50 cm³ glass pipettes with rubber bulbs

Diagram:

Chronological order of experiment

1)       Set up the apparatus as shown above

§         The water bath should be at 40 degrees C (313K)

§         The stop clock should be able to measure hours

2)       Put 70 ml of glucose solution in the 100cm³ conical flask

§         This solution is made by simple dilution

§         Use a Burette for this

§         The concentrations that I will use are:

3)       The controls that I will use for this experiment are water (de-ionised), glucose and yeast cells that have been subjected to boiling hot water.

4)       Leave the solution in the water bath to warm up for 10 minutes or until it is at the same temperature as the water bath

5)       Put 1ml of yeast solution into each conical flask

§         Swirl the container that the yeast solution is in before taking any out in case it has settled

§         This should be freshly made 1% solution

§         Use a 1ml pipette

§         Is a sample will need to be accurately diluted for this

§         Check and note down the actual temperature of the glucose solution before adding the yeast solution in the table above

§         Start the stop clock at beginning

§         Put cotton wool plug in

6)       Extract a 1ml sample using a pipette

§         Use a 1ml pipette

§         Do this after 24hrs and 48 hrs,

7)       Count the yeast cells and calculate their concentration using a the haemocytometer:

I.                     Make sure both the slide and the cover slip are dry and clean

II.                   Place cover slip on the middle of the slide, making sure it is lengthways on so that the are longer side of the cover slip is parallel to the long this side of the slide

III.                  Press they cover slip on to the slide of the haemocytometer until rainbow coloured “Newton’s rings” form

IV.                Place the slide under a microscope and focus the microscope so that the detailed part of the grid (square type A) can be seen clearly in the middle- see Figure 6.

V.                  Place drop of the sample solution in the one of the grooves sticking out from under the cover slip, on the side on which the cells are going to be counted

• Count the number of cells in 6 of the 25 grids (B-type Squares (the grids with three lines around them- see Figure 6)) in the main area (A-type Square) and divide this by six

• Don’t count B-type Squares with larger than seven in
• Count cells from top left C-type Square (see Figure 6).
• Count the C-type Squares in the B-type Squares line by line, as you would read a book.
• Count the six B-type squares in the A-type Square in the same way.
• Count budding cells that can be seen as two lobes as two cells.
• Record the tally of cells per line of C-type cells

VI.                This is the number of cells in 0.004mm³ of the solution

VII.               Wipe haemocytometer slide with cotton wool soaked in ethanol (evaporates quickly) in-between use and check for contamination before next and every use.

8)       Multiply this by 1/.004 (or by 2.5 and move the decimal place 2 places in this directionà)

• This is the amount in 1 mm³ and can be calculated later

9)       Compare this to the original value found in step 2)

10)   Calculate K for that concentration and fill this in on the table

• The equation is: K =(2/(c24 /c48)*60*60*4*12)/100000
• Where c24/c48 are concentration of cells after 24 hours/concentration of cells after 48 hours (measured in cells/ml. These are calculated in step
• This can also be calculated later
• This calculates the doubling time in 10⁵ seconds.

11)   Run one additional experiment with 0.57143% (30/40) glucose, taking a cell count every thirty minutes.  Record the results for this in the time/population table

Reasons for apparatus

These are the explanations for each choice made above. 

1)       100cm³ conical flask with cotton wool plug

I chose this because it holds larger volume than a test tube/ boiling tube and sits easily in the water bath.  The cotton wool plugs prevents gas exchange.

2)     Haemocytometer

This is the most accurate way of measuring cell content of the solution: actually counts the number one in a precise volume. Other methods of population measurement would be: density of solution, change in mass and changes in the acidity of the solution due to increase in carbon dioxide concentration. These are all indirect ways of measuring population growth.  The haemocytometer method actually measures the amount of yeast cells per unit volume.  It also involves taking a small sample out, rather than having to measure the mass/density of the whole solution, which would mean removing the entire experiment from of the water bath, which would disrupt the rate of or reproduction.

3)     Slide with cover slip

The cover slip makes sure that the correct volume is on the slide and that the cells are visible (in focus).

4)     1ml pipette

This is most accurate for measuring out the 1ml of yeast solution needed.  This is because it is graduated to a higher degree of accuracy than a wider measuring vessel could be - if the lines are closer together it is harder to get the level in the pipette exactly right and if the same difference in height corresponds to the smaller volume, the effect of human error is reduced. This reduces the percentage error for yeast solution added

5)     Burette

Burettes are very accurate for the same reason as explained above.  It is also very easy to adjust to the height/volume of the liquid using the rubber bulb and valves at the top.  This means that the measurements will be to a higher degree of accuracy, which in turn reduces the percentage error.

6)     Stop clock

This is to make sure that all the times are correct therefore no guessing or estimation

Also it can be stopped and the time that the sample is removed from the culture and it can be checked later after one has concentrated on the cell count.

7)      Water bath

This is to keep the solution at a constant temperature and to optimise the temperature for of the yeast cells to reproduce so that the limiting factor is the glucose concentration.

8)      Concentrations using ratios

This enables me to easily gain highly accurate dilutions of my glucose solution because it to eliminates human error due to taking measurements beyond the accuracy of the measuring vessels or the human eye or using several measuring vessels.  The percentage concentrations are irrelevant as these can easily be plotted on a graph afterwards and change anyway when the yeast culture is added.  This will reduce the percentage error.

Fair Testing

This is the summary of the steps that I will take to ensure that this is a valid scientific experiment.

Ø      Keep the temperature constant by using a water bath at 30ºC

Ø      Check the actual temperature with a thermometer and record on table

Ø      Use ratios for my dilutions

Ø      Use a haemocytometer because it is the most accurate method of measurement known to me

Ø      Count the number in five different squares of the main grid on the haemocytometer slide and average them

Ø      Do a base count to check the concentration of the solution with which I start.

Ø      Use pipettes just large enough to hold the quantities that I am at measuring to reduce the percentage errors

Ø      Use a stop clock to ensure accurate timing

Ø      Use 70 millilitres of glucose solution so that the dilutions are easy and the samples large enough to sufficiently reduce the percentage error.

Ø      Plug the conical flask so that the rate of respiration is not relative to the rate of diffusion of oxygen into the solution, but rather that the gases diffuse at a constant rate

Analysis of results

This is what I will do with my results and how I will record them.

I will record my results in Table 1, shown overleaf.

These are the calculations required to fill the table in:

• Averages: sum of entries/number of entries
• Glucose concentration: [glucose]/total volume in beaker x100
• Concentration of cells per cm³: [average cells per B type square]
• Number of reproductions = [yeast cells] / [yeast cells previously]

          2

• Percentage increase: [yeast cells] - [yeast cells previously] x100

   [yeast cells previously]

• Population increase rate: number of reproductions / [yeast cells previously]

time taken

I will plot these graphs of my results:

• Population increase every half hour over a period of three hours.
• Initial rate of population increase (R1) against glucose concentration
• Secondary rate of population increase (R2) against glucose concentration
• R1 and R2 against glucose concentration (on one graph for comparison)
• The initial and secondary percentage population increase against glucose concentration
• Comparison graphs for the initial and final percentage increase, and for the initial and final rate graphs

I will also statistically analyse my results using standard deviation

I will apply this test to each “group” of results I get- i.e. all three repeats of the six readings at each concentration will be analysed together. Spreadsheet’s standard deviation and average functions will calculate the standard deviation as a percentage of the average of each group of data by dividing the standard deviation by the mean and multiplying by 100. The formula for standard deviation used is:

Where             S = the sum of

n = the number of values

x = a value of the data (the number of cells in a B-type square in this case).

Limitations

These are the factors that I expect to be the greatest limitations on my experiment:

1.         Cells dying.  This can’t be helped and is a limitation because it stops me from counting the actual living population.  For the purposes of this experiment to it doesn’t matter too much because I will still get a similar increase in population with increase in glucose concentration.

2.       The vessel containing the solution.  Because these reactions are taking place in heterogeneous state, namely substances in solution reacting, the liquid must be in a container.  This in itself is limiting because a population needs space to grow.  It is therefore possible that the population will be limited by the space factor therefore resulting in the experiment only being valid for shorter periods of time and lower concentrations.  I have tried to avoid this limitation by using a low initial concentration of yeast cells (approximately 0.01-0.02%)

3.       Denaturation of the enzymes at high temperatures.  This makes it impossible to have a higher temperature than around 40 degrees Celsius.  This means that when the other conditions are favourable, temperature could become the limiting factor.  I have therefore not used very high concentrations of glucose.

4.       Accumulation of toxins.  As mentioned earlier, ethanol is produced by the reduction of acetaldehyde.  If there is not sufficient oxygen present this will not get broken down and can cause some, and eventually all of the yeast cells, to die.  I have avoided this in my plan (to the greatest extent possible) by only running the experiment over a short period.

Precautions

I must take the following precautions during the execution of this experiment:

1.         Watch out for parallax error- always take measurements at eye level.

2.       Wipe up any spillage immediately after they occur.

3.       Check the temperature of the water baths every time I take a sample out and adjust it if it is too high/low.

4.       Make sure that I don’t force any more solution out of the bottom of the Burette than naturally drops out when the release valve is open- the Burette is graduated to take the leftover amount into account.

5.       Shake the test tubes before extracting a sample to ensure even distribution as the cells may settle out.

Implementing

This section is about my actual practical experimentation.

Modifications

These are the changes that I made to my plan in its implementing:

6.       I had to use test tubes instead of conical flasks for incubation of my cultures in the water bath.  This was due to the fact that the conical flasks floated.

7.       I had to leave extracted samples in the fridge between taking them out of the test tube and doing the cell count because it took so long to count all the cells (approximately three or four hours).  This meant that the samples all came out at the same time and stopped reproducing at the same time so the population of each sample remained the same until they were counted.  I put all the samples in the fridge at the same time (or at negligible intervals on a scale of 24 hours- 60 seconds) and so they cooled down at the same rate and stopped propagating at the same time.

8.       When I checked the samples after 24 hours, they had not increased significantly enough to be sure of a pattern when I checked the percentage increase. I therefore decided to check them again at 72 rather than 48 hours after the cells were introduced. This also reduces the percentage inaccuracies in timing.

9.       I found that the population increased too slowly to get results of population increase against time results in three hours. I therefore did not include this extension in the investigation.

Results

My table of results is on the following page. The values used for the graphs plotted are printed in red.

Errors and Limitations

These are limitations or experimental errors that I think have affected my results the most and are responsible for the anomalies pointed out in the next section:

10.    Only being able to use 50cm³ in my boiling tubes.  The smaller the volume used, the larger the percentage errors are if the measurements are made to the same degree of accuracy.

11.       Dead cells. I did not expect the culture to reach the stationery phase and could only tell that it had because of the decrease in rate of reproduction. If I could have measured only the number of living cells, I could have been sure when the culture had reached this stage.

12.     Accumulation of toxins.  This had an effect because I ran the experiment for 72 hours, by which time the concentration of ethanol had built up to a toxic level.

13.     Uneven distribution of cells. Although I shook the test tubes, the distribution was not perfect as the cells clumped together and shaking did not separate the clumps. As these clumps were spread through the solution, the probability of only one or two being in a 0.04 cm³ sample is fairly high, but whether or not there is a clump in the particular B-type square that I count is random, so sometimes there was and sometimes not, leading to huge fluctuations in cell counts.

Anomalies

The results discussed in the following section seemed obviously faulty. They can be identified on Graph 5 (the highlighted results). The anomalous results were highlighted on the table and the figures in blue on the table were not included in the calculated averages. The averages not including the anomalies were then used to plot the graphs (apart from in the case of Graph 5).

Graph 5

14.     The results highlighted in Graph 5 at 0.78% glucose (purple and yellow highlighting on the table)

15.     The R2 result above the others at 0.39% glucose and the R1 result below the others at 0.78% (blue numbers on table)

Interpretation of Results

Here are the observations that can be made from the graphs, and a brief explanation of its significance.

16.     The graph of R1  (Graph 1) showed that there was already an increase after 24 hours. It showed an increase in reproductive rate with glucose concentration that curved off at higher concentrations. This was only produced when some anomalous results were not included (as explained above). The pattern is explained in the conclusion.

17.     The graph of R2 (Graph 2) showed a definite curve of the same type as the R1 graph.

18.    

Comparison of R1 and R2 in Graph 6 shows that the population, if at all, only had a short lag phase, in the order of about 1-5 hours. The population then increased drastically, as Graph 6 shows. The fact that the initial reproductive rate R1 is greater than R2 shows that by the time I measured the population again at 72 hours after inoculation shows that the population had already reached the stationary phase by this time. Therefore, on average over the second and third 24-hour periods, the population’s net increase rate is lower than during the exponential phase. This is explained more in the conclusion.19.     Graph 3 and Graph 4 show the percentage increase over the first 24 hours and the next 48 hours, consecutively. A hand-drawn version of Graph 3 was the basis for the decision to leave the culture for another 48 hours rather than just another 24. It did not show a clear trend, which I put down to the fact that the population had not increased sufficiently for the limiting factor to have an effect.

Conclusion

In this section I will summarise what my results show and explain why this occurred. I will then decide whether it matches my prediction.

Summary

• The population had a short lag phase, followed by a rapid growth phase, then the population slowed down again as it reached a stationary phase (Graph 6).

For example, compare the growth rates R1 and R2 at 0.98% glucose concentration: R1 is 1.11E-05; R2 is 0.78E-05. The culture slowed down greatly before the time R2 was taken (72 hours after inoculation).

When the yeast culture was introduced, the conditions did not change much from those in the working culture. They may have had to produce the RNA required to synthesise respiratory enzymes to catabolise glucose. As soon as this was done, the cells were in an optimum position to reproduce- the only limiting factor was the rate of progress through the cell cycle. As all the cells can be assumed to be reproducing, the more cells there are at this point, the greater the macroscopic population increase rate. The growth is therefore exponential. As the glucose gets used up, the cells do not receive glucose by diffusion as quickly as before, as the concentration gradient from the medium to the cytoplasm is not as great, and glucose concentration becomes rate limiting. This is explained below. When some cells get no nutrients, they die. The more there are, the more die, so the population plateaus off and a stationary phase begins. See Figure 5.

• The growth was most rapid at the higher concentrations, and negligible at 0% concentration.

During the exponential phase already, the cultures in the most dilute media will be limited by the rate of diffusion into the cells. The more dilute the solution, the lower the rate of diffusion of glucose into the cell so the slower the rate of reproduction. The populations in the most dilute solutions will therefore increase less rapidly. At 0% glucose the yeast cells had no respiratory substrate so couldn’t reproduce.

• The effect of an increase in glucose concentration decreased at higher concentrations.

For example, compare the results for R1: at 0.2% the rate is 7.07E-06, at to 0.39% it is 9.26E-06 whereas at 0.78% it is 1.09E-05, at 0.98% 1.11E-05. The percentage difference between the rates at lower concentrations is much larger than that at higher ones.

Glucose is a polar molecule and is taken into cells by facilitated diffusion. The rate of facilitated diffusion depends on the number of hydrophilic protein channels on the outer membrane. If there is an excess of channels then the rate of diffusion will increase proportionally to the concentration gradient. However, the greater the utilisation of the channels, the more often a glucose molecule will occupy the channel thus preventing a second molecule from passing through. The rate increase with concentration increase therefore decreases as the number of channels utilised increases. The rate of diffusion therefore curves off at higher concentrations. See Figure 2.

Comparison to prediction

My prediction was: “During the times for which I plan to run my experiment, I only expect to see the period from the lag phase to the end of the exponential phase. Not many cells will be dying during these initial phases (although possibly a few during the lag if they do not successfully adapt), so the total count will be approximately equal to the living population. The rate of growth will therefore follow the pattern:

“

I ran my experiment for longer than I had planned. The culture therefore reached the beginning of the stationary phase. The first sentence was therefore probably a correct estimation. The prediction of the rate of growth in relation to the concentration of glucose was supported by my results. I can therefore conclude that my results did support my qualitative prediction and the rate of population increase does follow the pattern of facilitated diffusion rate.

Evaluation

In this section I will consider the values from the standard deviation test and the number of anomalies, and decide how confident I can be that the results were not purely by chance.

The values obtained are shown on the last column of the table. The greatest standard deviation from the average is 39% at 0.784% glucose. This, however takes the results that I called anomalous into account- hence the high value. Apart from this, the highest value is 30%. This is not very high considering that this is the deviation of the raw data. There were few results that seemed to vary significantly to be recognised as anomalies. These were all mentioned earlier.

More generally, my results are fairly consistent, as well as being consistent with the prediction, which is an indication that they are reliable. Some of the finer points of the progress of the population with time remain to be conclusively proven, but the primary quantitative prediction as to the effect of glucose concentration on the rate of reproduction appears to be conclusively and reliably proven.

Succession

This is the process in which communities of plant and animal species are replaced over time by a series of different and usually more complex communities. It is driven by interspecific competition and an important aspect of it is that at each of the seres until the climax community is reached, populations alter the environment in ways that encourage their direct competitors. The climax community is normally some kind of woodland.

Diversity

This is defined as the number of different species in a particular area.

Heather Moorland

Heather moors are the driest and most widespread of the three types of moorland. They are mostly areas of herbaceous heath* frequently created by the burning down or destruction of woodland (particularly birch Betula and Scots pine Pinus sylvestris) which would otherwise populate the area. They are populated typically by only a few species of heather (e.g. Dorset Heath- Erica ciliaris L., Bell Heather- Erica cinerea L., Common (Ling) Heather- Calluna vulgaris (L.)), mosses (e.g. Feather Moss- Hypnum jutlandicum, Star Moss/Hair Moss- Polytrichum commune) and grasses (e.g. Mat Grass- Narda stricta).

*Strictly this is a misnomer as the dominant heather Calluna vulgaris is an evergreen undershrub, but the term may be conveniently used for areas of low vegetation

This was done primarily for animals to graze on, but nowadays they are valued for their game (grouse). In order to maintain the conditions best suited to heather, and prevent the eventual succession to woodland (e.g. silver birch Betula pendula), it must be periodically (every ten or so years) burned. This has the effect of:

• Burning any developing tree (or shrubs like gorse) shoots (which may be forming due to decline of heather density in the degenerate phase -see below)
• Improving the germination of heather seeds which can benefit from short “heat treatment” and thrive in the moisture- and nutrient-rich blanket
• Increasing the acidity of the soil making it ideal for heather and too harsh for many other plants

Also, after about 15 years, heather’s ability to regenerate vegitatively declines and it is therefore necessary to grow new stems, unless heather is to decline.

 After the heather is burned, secondary succession occurs. Because heather has the advantages described above, although there may be many species to start with, it takes over eventually as the other plants competing for the same niche are not as successful. The growth phases of Calluna, the primary species, are shown on the left.After this, the natural series continues with different shrubs like gorses Ulex minor and U. galli colonising the area, and eventually into woodland, probably silver birch Betula pendula.

Prediction

I therefore predict that, as succession occurs on heather moorland (see limitation point 3), there will be a fairly short period of high diversity before the heather establishes itself. Then the diversity will fall, as heather’s competitors