How does the rate of alcoholic fermentation (in cm3 min-1), measured in terms of volume of CO2 produced depends on the type of the sugar unit-sucrose, glucose, fructose and starch used as substrate, determined using gas collection method?
Being an inquirer, I have always wondered and questioned the facts and phenomenon which I came across. The journey of this Extended Essay started from one of my industrial visit to the alcohol manufacturing industry. During my secondary school Science classes, I was aware of the fact that alcohol is produced from the process of alcoholic fermentation where a sugar unit, chemically termed as poly hydroxy aldehydes or ketones undergoes degradation to produce ethanol and carbon dioxide. This is a bio-chemical reaction and is thus carried out in presence of the enzyme zymase which is secreted by the micro-organism yeast. Ethanol is the main ingredient for all the different varieties of alcohol which are consumed as beverages across the world. However, there is still so much of variation in the price, taste and flavour of these beverages depending on the category they belong to like-whiskey, rum, beer and so on. One of the main reason behind this difference is the stages of fractional distillation that are carried out in varying the percentage composition of the ethanol within the liquor. Another major reason is the use of additives to enhance the taste, texture and flavour. But the main fact that strike me is the changes in the type of raw materials that are used for the production. Some of them are made from rice grains, some from fruits and some from other fruits or plant parts which are rich in carbohydrates. The question that intrigued me was that how the change of raw material would actually affect the industrial production of alcoholic beverages.
From my discussion with the engineers and researchers in the factory, I got to know that the percentage of ethanol in the final product depends on the type of raw material initially chosen as the substrate. This triggered more interest in me. If all the raw materials are sugar units, then what might be the key factor which controls the rate or the amount of the alcohol produced? Further research and reading of some articles brought me to the fact that carbohydrates or sugar units can exist in various forms and types. The type of sugar unit in rice grain and that in grapes are not the same. This difference in the nature of the substrate used causes variations in the kinetics of the process of alcoholic fermentation and thus the speed at which the alcohol is produced. Eventually, varying the nature of substrate, the amount of ethanol generated from the same mass of substrate can also be varied. This brought me to the idea of investigating how the rate at which alcoholic fermentation occurs might depend on the type of substrate chosen.
Alcoholic fermentation is a biochemical reaction where glucose is converted into ethanol and carbon dioxide in presence of the micro-organism yeast.
Glucose + Yeast ----🡪 Ethanol + Carbon dioxide
Glucose is a mono saccharide, a poly hydroxy aldehyde, a carbohydrate. There are many other compounds that belongs to the same class. The major carbohydrates includes – sucrose, galactose, lactose, fructose, maltose, starch and many more. Starch is the compound in which stores the plant food. Plants produces glucose through the process of photosynthesis, glucose units polymerizes into a large molecule starch and thus the plant stores the food within the cells in the form of starch. All these types of carbohydrates can also undergo alcoholic fermentation like glucose.
Carbohydrates or specifically complex carbohydrates or polysaccharides are basically polymers of glucose. Starch is also a polymer of glucose, many glucose units linked with each other through glycosidic bonds. Sucrose is a dimer of glucose as it is glucose and fructose joined together through a glycosidic linkage. Sucrose is also known as a disaccharides as it contains two sugar units. The term “saccharides” refers to sugar units. Depending on the number of sugar units present, the carbohydrates are classified into various categories – mono saccharides (only one sugar unit), disaccharides (two sugar units) and polysaccharides (more than two sugar units). All of these saccharides unit can be hydrolysed into simple sugar units and can thus participate in the reaction of alcoholic fermentation.
Alcoholic fermentation is the biological method of producing alcohol. As the name suggests, the sugar units are fermented (allowed to decompose in the presence of micro-organisms) and produce the alcohol ethanol. Along with ethanol, carbon dioxide is also produced as a by-product.
This is an enzyme controlled reaction. The microorganism yeast added produces the enzyme zymase which takes part in this reaction. Like other enzymatic reactions, this reaction is also very slow without the enzyme, specific in nature (cannot be performed with any enzyme other than zymase), has an optimum temperature and pH where the production is maximum and the rate is also high. As ethanol is produced as one of the product, the population of the yeast is found to decay after a certain point. The presence of ethanol makes the medium toxic and does not allow the yeast to reproduce or multiplicate. Finally, the yeast dies and cannot produce the enzyme anymore. This eventually causes the reaction to stop and the production ceases down. This is a disadvantage of the process of alcoholic fermentation that bothers the industrialists. The yield of ethanol produced becomes low as the substrates are not used up any more once the yeast population dies.
The rate of alcoholic fermentation can be monitored by monitoring the level of carbon dioxide produced. As carbon dioxide is one of the product, if the amount of carbon dioxide generated is plotted against time, the graph obtained can be used to calculate the rate of the reaction.
The aim of the investigation is to understand the effect of the nature of the substrate on the rate of alcoholic fermentation. The different substrates that are used in this investigation are- glucose, sucrose, fructose and starch. These substrates vary both in their structural, constitutional as well as molecular composition. Glucose, galactose and fructose are mono saccharide, sucrose is a disaccharides with two sugar units and starch is a poly saccharides with more than two sugar units. Thus, the number of sugar units can also be considered as an alternate independent variable of this investigation- mono saccharides (glucose, fructose and galactose), disaccharides (sucrose) and polysaccharides (starch). Again the molecular formula of the substrates are also different. The molecular formula of the different substrates are given below.
Glucose – C6H12O6
Fructose – C6H12O6
Sucrose – C12H22O11
Starch – (C6H10O5)n
Thus, based on the number of C content- there are three categories:
Glucose and fructose– 6 C each
Sucrose – 12 C
Starch – n number of C
Hence, technically under the banner of types of substrate as the main independent variable, the investigation aims to compare the effect of nature of substrate on the rate of alcoholic fermentation in three perspectives:
Mono saccharides (Glucose, Fructose)
Disaccharides (Sucrose)
Poly saccharides (Starch)
Number of C atoms in the substrate chosen:
The dependent variable is the rate of alcoholic fermentation. In alcoholic fermentation, CO2 is produced as one of the major by-product along with ethanol. Thus, monitoring the amount of CO2 produced in cm3 as a function of time, the rate of the reaction can be measured. The method of upward displacement of water has been used to collect the gas and the volume of water displaced has been recorded using an inverted graduated measuring cylinder. The volume of gas liberated has been recorded at regular intervals of time using a stop-watch and then a scatter plot of volume of gas liberated versus time has been obtained. A linear trend line has been used in the scatter plot thus obtained and the rate has been calculated from the gradient of the trend line thus obtained.
Rate = \(\frac{Volume\ of\ gas\ liberated\ in\ cm^3}{Time\ in\ minutes}\)
= gradient of the scatter plot of volume of gas versus time in cm3 min-1
Temperature: Alcoholic fermentation is an enzymatic reaction. Thus, at high temperature, the enzyme will lose the shape and gets denatured while at low temperature, the rate of the reaction will be extremely less. Thus, it is essential to carry out the reaction at an optimum temperature. The enzyme zymase secreted from the micro-organism yeast works best at an optimum temperature of 38.00 ℃. A water bath was used to control this temperature.
Mass of substrate used: The rate of any bio-chemical reaction depends on the initial concentration of the substrate used. More the concentration of the substrate used, faster the rate and thus more the amount of product obtained. Thus, it is important to use the same mass of substrate in all trials and keep the concentration of the substrate at a constant level to obtain fair and accurate result. To do this, 5.00 ± 0.01 g of the substrate was used in all trials. A digital mass balance was used for this.
Mass and type of yeast used: The yeast is the micro-organism that produces the enzyme zymase which acts as a catalyst in this reaction. More the amount of yeast added, more the enzyme produced, more the catalytic power and thus faster the rate of the reaction. Thus, it is essential to use the same mass of yeast in all trials. To control this, 2.50 ± 0.01 g of yeast was used in all trials. The production of the enzyme and the biological variety of the enzyme made will also depend on the type of the yeast added. Thus, in all cases, the same type of yeast was used- Baker’s yeast.
Surface area of the reactants: The rate of any biochemical reaction especially enzyme catalysed ones depend on the surface area of the reactants used. Enzyme catalysis follows adsorption mechanism. The enzymes are adsorbed on the surface of the substrate, the enzyme-substrate complexes are formed and then the products are generated which are desorbed from the surface area of the substrates. Thus, in all trails, the surface area was kept constant by using the same glass beaker – 100 cm3 of capacity to contain the reactants.
Apparatus | Quantity | Least count | Absolute error |
---|---|---|---|
Digital mass balance | 1 | 0.01 g | ± 0.01 g |
Thermometer | 1 | 1.0 ℃ | ± 0.5 ℃ |
Water bath | 1 | --- | --- |
Spatula | 1 | --- | --- |
Graduated measuring cylinder – 100 cm3 | 1 | 1.00 cm3 | ± 0.50 cm3 |
Trough | 1 | --- | --- |
Bent tube | 1 | --- | --- |
Beaker – 100 cm3 | 1 | --- | --- |
Cork and lid | 1 | --- | --- |
Soft tissues | 1 | --- | --- |
Equation of trend line:
y = 8.0029 x + 36.752
Rate of alcoholic fermentation = \(\frac{Change\ of\ volume\ (∆V)}{Change\ of\ time\ (∆t)}=\frac{∆y}{(∆x)}\) = gradient = co - efficient of x = 8.0029 cm3 min-1
Equation of trend line:
y = 8.6857 x + 37.905
Rate of alcoholic fermentation = \(\frac{Change\ of\ volume\ (∆V)}{Change\ of\ time\ (∆t)}=\frac{∆y}{∆x}=\) gradient = co - efficient of x = 8.6857 cm3 min-1
Equation of trend line:
y=6.0857 x + 37.905
Rate of alcoholic fermentation \(=\frac {change of volume∆V} {change of time∆t} \) = gradient = co - efficient of x = 6.0857 cm3 min-1
Equation of trend line:
y = 0.2657 x + 37.905
Rate of alcoholic fermentation \(=\frac {change of volume∆V} {change of time∆t} \) = gradient = co - efficient of x = 0.2657 cm3 min-1