Effect of nature of substrate on rate of alcoholic fermentation

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:

• Nature of substrate – Glucose, fructose, sucrose and starch
• Types of saccharides –

Mono saccharides (Glucose, Fructose)

Disaccharides (Sucrose)

Poly saccharides (Starch)

• Number of C atoms in the substrate chosen:

6 C atoms – (Glucose and Fructose)

12 C atoms – Sucrose

More than 12 C atoms – Starch

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.

• Laboratory protective clothing was used.
• Safety gloves were used to avoid contact with the reagents and chemicals.
• Safety mask was used as the recent COVID protocols and also not to inhale any of the chemicals.
• Any eatables were not allowed inside the laboratory.
• All unused chemicals were returned back to the laboaratory technician for reuse.
• To dispose all the unused materials safely, a waste bin was kept and all of them were diluted to a large extent before being disposed.

• Take a 100 cm³ clean and dry glass beaker.
• Place the beaker on the water bath.
• Set the temperature of the water bath at 38.0 ℃.
• Place a clean and dry watch glass on the top pan digital mass balance.
• Set the reading of the mass balance to 0.00 ± 0.01 g using the ‘Tare’ button.
• Transfer the baker’s yeast powder into the watch glass using a spatula until the balance reads 2.50 ± 0.01 g.
• Transfer the weighed yeast powder completely from the watch glass to the beaker.
• Wash the watch glass completely using distilled water.
• Place the watch glass on the top pan digital mass balance.
• Set the reading of the balance to 0.00 ± 0.01 g using the ‘Tare’ button.
• Transfer the glucose powder from the reagent bottle to the watch glass until it reads 5.00 ± 0.01 g.
• Transfer the weighed glucose powder from the watch glass to the glass beaker.
• Add distilled water in the beaker till the mark of 100 cm³ using a graduated measuring cylinder.
• Cover the beaker with a lid and cork through which a bent tube is passed. The outlet of the tube ends in a trough filled with water with an inverted measuring cylinder of capacity of 100 cm³ in it.
• The stop-watch was started and the reading of the graduated measuring cylinder was noted down.
• Repeat step - 15 as the stop-watch reads 2.00 ± 0.01 mins, 4.00 ± 0.01 mins, 6.00 ± 0.01 mins, 8.00 ± 0.01 mins and 10.00 ± 0.01 mins.
• Repeat steps 1-16 for four more times.
• Repeat steps 1-17 for other types of sugar units- fructose, galactose, sucrose and starch.

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

As there are 4 categories in this investigation – four different types of substrates and the groups are mutually exclusive of each other, an ANOVA test is done.

Null hypotheses: The mean of all the four groups is identical.

Alternate hypotheses: The mean of all the four groups is not identical.

Total number of values (N) = 6 + 6 + 6 + 6 = 24

Number of values in each category (n) = 6

Number of categories (a) = 4

Calculating the degrees of freedom (df)

df_between = a - 1 = 4 - 1 = 3

df_within = N - a = 24 - 4 = 20

For Group-Glucose:

50.0 + 53.0 + 57.4 + 70.8 + 95.6 + 133.8 = 460.60

For Group-Fructose

50.0 + 53.0 + 63.0 + 82.0 + 101.0 + 139.0 = 488.00

For Group-Sucrose

50.0 + 51.6 + 53.6 + 71.4 + 92.0 + 107.4 = 426.00

For Group-Starch

50.0 + 50.0 + 50.8 + 51.4 + 52.0 + 52.4 = 306.60

For all values,

Total (T)

= 50.0 + 53.0 + 57.4 + 70.8 + 95.6 + 133.8 + 50.0 + 53.0 + 63.0 + 82.0 + 101.0 + 139.0 + 50.0 + 51.6 + 53.6 + 71.4 + 92.0 + 107.4 += 50.0 + 50.0 + 50.8 + 51.4 + 52.0 + 52.4 = 1681.20

\(=\sum\) (a_i)² = (460.60)² + (426.00)² + (426.00)² + (306.60)² = 725775.92

\(=\sum \) Y² = (50.0)²+...... +(52.4)² = 134986.8

Sum of square between (SS _between)

\(\frac{\sum(a_i)^2}{n} - \frac{T^2}{N}=\frac{725775.92}{6}-\frac{(1681.20)^2}{24} \) = 134986.80- 120962.65 = 14,024.15 

Statistic value (F) = \(\frac{\frac{SS_{between}}{df_{between}}}{\frac{SS_{within}}{df_{within}}}=\frac{\frac{105671.74}{3}}{\frac{14024.15}{20}}=\frac{35223.91}{701.20}\) = 50.23

At df_between = 3  and df_within = 20,  the critical value of F is 2.9747

Since the calculated statistic value of F (50.23) is greater than the critical value of F (2.9747), the null hypotheses is rejected and the alternate hypotheses is accepted.

Thus, it can be claimed that the rate of alcoholic fermentation differs from one type of substrate to another.

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?

• The bar graph above (Figure -12) displays the rate of alcoholic fermentation for different types of sugar units as calculated above in the data processing section from the gradient of the scatter plot of volume of gas versus time. The rate is measured in cm³ min^-1. The magnitude of the rate are 8.0029, 8.6857, 6.0857 and 0.2657 for glucose, fructose, sucrose and starch respectively.
• Thus, it can be claimed that the rate of alcoholic fermentation depends on the type of the sugar unit chosen as substrate. The rate is maximum for fructose and minimum for starch.
• Sucrose and fructose have a higher rate than sucrose which has a much higher rate of fermentation than starch. Glucose and fructose are monosaccharides while sucrose is a disaccharide and starch is a polysaccharide. Thus, it can be claimed that mono saccharides are better as a substrate for alcoholic fermentation than disaccharides followed by poly saccharides.
• Both glucose and fructose have the same number of C and the same molecular formula yet the rate for fructose is higher than the rate for glucose. This indicates that apart from molecular formula or molecular mass, the chemical composition or the type of functional group in the substrate plays an important role in the mechanism of the reaction. Glucose is a polyhydroxy aldehyde and fructose is a polyhydroxy ketone. Thus, it is possible that the difference in rate is due to the change in the functional group present in them.
• Again, glucose and fructose have 6 C, sucrose has 12 C and starch has n number of C atoms. Thus, it can also be claimed that as the number of C units in the substrate increases, the rate of alcoholic fermentation decreases. This can be because of the fact that higher the C content, more the hydrophobicity, lower the solubility of the substrate in the reaction medium. In simple words, glucose and fructose is most soluble in water while starch is least soluble in water. If the substrate is not able to dissolve itself within the aqueous medium, the contact of the substrate with the enzyme is inhibited and thus the formation of enzyme substrate complex is also hindered and obstructed which can eventually reduce the rate of alcoholic fermentation. Larger the size of the substrate, lower the solubility of the substrate in the aqueous medium, lesser the amount of substrate that is acted upon by the enzyme and thus lower the production of alcohol.
• The rate for starch is significantly lower than the other three sugar units. This can be explained based on the fact that the starch contains multiple sugar units linked with each other through the glycosidic linkage. The first step of alcoholic fermentation is to break the linkages to release single sugar units on which the enzyme can work. Usually, in the real life the same task is performed by salivary amylase which breaks down the complex carbohydrate structures into small and simple sugar units on which enzyme can act upon during digestion. As starch has multiple sugar units and thus there are many linkages within it. It is difficult for the enzyme to act on it and thus carry out fermentation.

• The investigation considers the range of independent variable. Four different types of sugar units have been used.
• The data processing is coherent and leads to a conclusion.
• The interpretation of the graph has been done in multiple perspectives to answer the research question completely.
• The rate of the reaction has been computed graphically which leads to a more reliable result.

• The digital mass balance has been used to record the mass of the yeast added as well as the sugar units used. The mass balance has an instrumental error associated with it. This will introduce systematic error in the investigation and thus reduce the accuracy of the data collected. To improve this, the digital mass balance is calibrated using standard values of mass.
• The gas has been collected by the method of downward displacement of water. This process has a methodological limitation. The gas that has been collected may be mixed with water vapor. Thus, the volume of the gas as recorded will be actually lower than what has been actually measured. To improve this, the level of carbon dioxide can be measured using a probe in ppm. This will reduce the inaccuracy in the volume / amount of the gas measured.
• The water bath has been used to maintain the temperature. However, it is impossible to ensure that the temperature remained the same throughout the reaction. Thus, variations of temperature during the reaction may cause inaccuracy in the result obtained. To improve this, the investigation should have been done in an isothermal condition. This can be done only by using a thermostat.
• As the gas has been passed through a cork and then through a bent tube, there is a possibility for the gas to escape through. This will again make the values of volume recorded lower than actual. To improve this, the use of Vernier Logger pro and a carbon dioxide sensor can be done.
• In Figure - 8 to Figure - 11, the trend line does not pass through the origin. Ideally, at time = 0.00 mins, the volume of gas liberated must be zero. Thus, the line must pass through the origin. However, the value of positive intercept indicates that there was carbon dioxide already produced even before the volume of gas was recorded for the first time; the reading of volume is 50.00 cm³ at 0.00 mins in Figure - 3. This can be due to the time lag between the instant the yeast and the sugar unit comes in contact with each other and the time the stop-watch was started or the reading was recorded. This is a human error and a methodological limitation. To improve this, the value at 0.00 mins can be subtracted from the values at other values of time and then the graph can be plotted.
• There are multiple sources of random error in the investigation. To improve them, repeated trials has been taken and an average value has been used.

Apart from the nature of substrate, the rate of alcoholic fermentation will also depend on the pH of the medium. As the reaction is an enzymatic reaction and enzymes are pH sensitive; changes structure on change of pH, thus pH plays a key role in the rate of any enzymatic biochemical reaction. I would like to perform the alcoholic fermentation of glucose at various pH values in the same methodology, obtain a scatter plot of
volume versus time and then calculate the rate from there. Buffer tablets can be used to vary the pH.

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