Views: 152

CHEMICALS OF LIFE

1. Define the term ‘reducing sugars’?

Reducing sugar contains aldehydic or ketonic group in the hemiacetal and hemiketal forms and can reduce
Tollen’s reagent or Fehlmg’s solution.

2. Name the basic unit molecules that comprise

• Nucleic acid

– nucleotides

• Polysaccharides

-Monosacharides

• Proteins

Amino acids

• Sucrose

Monosacharides

3. Outline the reducing properties of  the reducing sugars

These sugars  readily undergo oxidation reactions (i.e. lose an electron or gain an oxygen atom) in presence of copper ions  contained in benedict or Fehling’s solution by causing it to reduce from Cu²⁺ to Cu solid  hence producing  a characteristic orange  or brick- red  colour change.

4. Explain the procedure for preparation of  fehlings solution

Fehling’s solution is a chemical reagent used to differentiate between water-soluble carbohydrate and ketone functional groups, and as a test for reducing sugars and non-reducing sugars, supplementary to the Tollens’ reagent test.

Fehling’s solution is prepared by combining two separate solutions: Fehling’s A, which is a deep blue aqueous solution of copper(II) sulfate, and Fehling’s B, which is a colorless solution of aqueous potassium sodium tartrate (also known as Rochelle salt) made strongly alkali with sodium hydroxide. These two solutions, stable separately, are combined when needed for the test because the copper(II) complex formed by their combination is not stable: it slowly decomposes into copper hydroxide in the alkaline conditions. 

Solution A; Dissolve 34.6gms. Of pure copper sulphate in distilled water and dilute to 500ml. (blur)

Solution B: Dissolve 173gms. of sodium potassium tartrate and 30gms.of pure sodium hydroxide in water and dilute to 500ml. alternately, dissolve 121gms. of pure sodium hydroxide and 93.1gms.of pure tartaric acid in water, then dilute the solution to 500ml. ( colorless ).Mix equal volumes of solutions A and B immediately before use, and then use as the reagent.

5. (a) State four  properties of disaccharides
   

Disaccharides are carbohydrates composed of two monosaccharide units joined together by a glycosidic bond. Here are four properties of disaccharides:

1. Sweetness: Disaccharides, like monosaccharides, generally have a sweet taste. The degree of sweetness can vary depending on the specific disaccharide. For example, sucrose, commonly known as table sugar, is relatively sweet.

2. Solubility: Disaccharides are soluble in water. They can easily dissolve in water to form solutions, although the solubility may vary among different disaccharides. For instance, lactose, found in milk, is less soluble than sucrose.

3. Stability: Disaccharides are relatively stable molecules. They are not easily broken down under normal physiological conditions. However, some disaccharides, such as lactose, can be broken down by specific enzymes in the body, leading to intolerance in individuals lacking these enzymes.

4. Energy Source: Disaccharides serve as an important source of energy in organisms. When digested, disaccharides are broken down into their constituent monosaccharides, which can then be further metabolized to release energy through cellular respiration.

It’s important to note that different disaccharides have unique properties and functions. Examples of common disaccharides include sucrose (glucose + fructose), lactose (glucose + galactose), and maltose (glucose + glucose).

• Identify two examples of disaccharides
  
• Sucrose
• Galactose
• Maltose

6. (a)Name the polysacharide that result  from hydrolysis of sucrose

Glucose and fructose

• State any two functions of pentose
  

• Pentoses are a type of monosaccharide that contain five carbon atoms. Two functions of pentoses are:

  • Energy Production: Pentoses, such as ribose and deoxyribose, play essential roles in energy production within cells. They are key components of important molecules involved in cellular respiration and energy metabolism. Ribose, for example, is a component of adenosine triphosphate (ATP), the primary energy currency of cells.
  • Nucleic Acid Structure: Pentoses are integral components of nucleic acids, which include DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Ribose is a component of RNA, where it forms the backbone of the RNA molecule. Deoxyribose, a modified form of ribose, is a component of DNA, where it provides the structural framework for the genetic code.

  These functions highlight the importance of pentoses in cellular processes related to energy production and genetic information storage and expression.

7. Explain the chemical reactions of  fehlings solution  on reducing sugars

The active reagent is a tartrate complex of Cu²⁺, which serves as an oxidizing agent. The tartrate serves as a ligand. Fehling’s solution can be used to distinguish aldehyde vs ketone functional groups. The compound to be tested is added to the Fehling’s solution and the mixture is heated. Aldehydes are oxidized, giving a positive result, but ketones do not react, unless they are α-hydroxy ketones. The bistartratocuprate(II) complex oxidizes the aldehyde to a carboxylate anion, and in the process the copper(II) ions of the complex are reduced to copper(I) ions. Red copper(I) oxide then precipitates out of the reaction mixture, which indicates a positive result i.e. that redox has taken place (this is the same positive result as with Benedict’s solution).

Fehling’s test can be used as a generic test for monosaccharides and other reducing sugars (e.g., maltose). It will give a positive result for aldose monosaccharides (due to the oxidisable aldehyde group) but also for ketose monosaccharides, as they are converted to aldoses by the base in the reagent, and then give a positive result.^[8]

Fehling’s can be used to screen for glucose in urine, thus detecting diabetes. Another use is in the breakdown of starch to convert it to glucose syrup and maltodextrins in order to measure the amount of reducing sugar, thus revealing the dextrose equivalent (DE) of the starch sugar.

Formic acid (HCO₂H) also gives a positive Fehling’s test result, as it does with Tollens’ test and Benedict’s test also. The positive tests are consistent with it being readily oxidizable to carbon dioxide.

The net reaction between an aldehyde and the copper(II) ions in Fehling’s solution may be written as:

RCHO + 2 Cu²⁺ + 5 OH⁻ → RCOO⁻ + Cu₂O + 3 H₂O

or with the tartrate included:

RCHO + 2 Cu(C₄H₄O₆)₂²⁻ + 5 OH⁻ → RCOO⁻ + Cu₂O + 4 C₄H₄O₆²⁻ + 3 H₂O

9. Outline the emulsion procedure for food test  and the expected results

This test is done to show the presence of lipids in a substance. The substance is first dissolved in ethanol. This solution is then dissolved in water. If lipids are present in the mixture, it will precipitates and forms an emulsion.

Process 

• Add the food sample to 2 cm3 of ethanol, shake well.
• Allow to settle in a test tube rack for 2 minutes for food to dissolve in ethanol.
• Empty any clear liquid into a test tube containing 2 cm3 of distilled H2O.
• A MILKY-WHITE EMULSIONis a positive result: lipid is present.
• If the mixture remains clear, there are no fats present in the sample

Explanation

• Lipidsare insoluble in water and soluble in ethanol (an alcohol).
• After lipids have been dissolved in ethanol and then added to H2O,they will form tiny dispersed droplets in the water. This is called an emulsion.
• These droplets scatter light as it passes through the water so it appears white and cloudy

10. Explain the principle behind benedict test

Benedict’s Test  is used to test for simple carbohydrates. The Benedict’s test identifies reducing sugars (monosaccharide’s and some disaccharides), which have free ketone or aldehyde functional groups. Benedict’s solution can be used to test for the presence of glucose in urine. 

Some sugars such as glucose are called reducing sugars because they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction. When reducing sugars are mixed with Benedicts reagent and heated, a reduction reaction causes the Benedicts reagent to change color.  

As a resuly,the solution changes to orange red/ brick red. This reaction is caused by the reducing property of simple carbohydrates. The copper (II) ions in the Benedict’s solution are reduced to Copper (I) ions, which causes the color change.

The color varies from green to dark red (brick) or rusty-brown, depending on the amount of and type of sugar.

 The red copper(I) oxide formed is insoluble in water and is precipitated out of solution. This accounts for the precipitate formed. As the concentration of reducing sugar increases, the nearer the final colour is to brick-red and the greater the precipitate formed. Sometimes a brick red solid, copper oxide, precipitates out of the solution and collects at the bottom of the test tube.

Complex carbohydrates such as starches DO NOT react positive with the Benedict’s test unless they are broken down through heating or digestion (try chewing crackers and then doing the test). Table sugar (disaccharide) is a non-reducing sugar and does also not react with the iodine or with the Benedict Reagent. Sugar needs to be decomposed into its components glucose and fructose then the glucose test would be positive but the starch test would still be negative.

11. Benedict’s quantitative reagent contains potassium thiocyanate and is used to determine how much reducing sugar is present. This solution forms a copper thiocyanate precipitate which is white and can be used in a titration. The titration should be repeated with 1% glucose solution instead of the sample for calibration
12. Composition and Preparation of Benedict’s Solution

Benedict’s solution is a deep-blue alkaline solution used to test for the presence of the aldehyde functional group, – CHO.

Anhydrous sodium carbonate = 100 gm

Sodium citrate – 173 gm

Copper(II) sulfate pentahydrate = 17.3 gm

One litre of Benedict’s solution can be prepared from 100 g of anhydrous sodium carbonate, 173 g of sodium citrate and 17.3 g of copper(II) sulfate pentahydrate.

Procedure of Benedict’s Test

Approximately 1 ml of sample is placed into a clean test tube.

2 ml (10 drops) of Benedict’s reagent (CuSO4) is placed in the test tube.

The solution is then heated in a boiling water bath for 3-5 minutes.

Observe for color change in the solution of test tubes or precipitate formation.

Result Interpretation of Benedict’s Test

If the color upon boiling is changed into green, then there would be 0.1 to 0.5 percent sugar in solution.If it changes color to yellow, then 0.5 to 1 percent sugar is present.If it changes to orange, then it means that 1 to 1.5 percent sugar is present. If color changes to red,then 1.5 to 2.0 percent sugar is present. And if color changes to brick red,it means that more than 2 percent sugar is present in solution.

Positive Benedict’s Test: Formation of a reddish precipitate within three minutes. Reducing sugars present. Example: Glucose     Negative Benedict’s Test: No color change (Remains Blue). Reducing sugars absent. Example: Sucrose.

13. Outline buret test

The Buret Test for proteins using involves testing for the presence of the peptide bond. Biuret reagent is a copper-based reagent that turns purple when bound to protein in an alkaline solution.The more peptide bonds present, the greater the intensity of the purple colour, indicating a higher protein concentration.

The presence of protein can also be detected using Millon’s reagent. Millon’s reagent reacts with tyrosine amino acids, common to most proteins, and results in the formation of a reddish-brown precipitate when heated.

14. The table below summarises the major tests and their expected results in the presence and absence of protein.

Test reagent	Positive result	Negative result
Biuret reagent	Violet/purple colour	Blue colour
Millon’s reagent	Red-brown colour	White colour

15. Explain the preparation of  aminoacid mixture  in chromatographic analysis of protein 

In chromatographic analysis of proteins, an amino acid mixture can be prepared as a reference standard to identify and quantify individual amino acids in the protein sample. Here’s an overview of the preparation process:

• Selection of Amino Acids: Determine which amino acids are of interest for analysis. The selection may depend on the specific protein being studied or the research objectives. Common amino acids used in the mixture include alanine, glycine, leucine, lysine, phenylalanine, proline, serine, threonine, and valine, among others.

• Weighing and Measuring: Weigh the desired amount of each amino acid using an analytical balance. The weights should be accurate and precise. The quantities of amino acids should be determined based on the desired concentration and the intended application of the mixture.

• Dissolving Amino Acids: Dissolve each amino acid separately in an appropriate solvent. Typically, a suitable solvent such as distilled water or a buffered solution (pH adjusted to the desired range) is used. It is important to ensure that each amino acid is completely dissolved to obtain a clear and homogeneous solution.

• Mixing and Adjusting Concentration: Combine the individual amino acid solutions in the desired ratios to create the amino acid mixture. The concentrations of the amino acids can be adjusted based on the intended application and the desired concentration range for analysis. Dilutions or additional solvent may be required to achieve the desired concentrations.

• Filtration and Storage: Filter the amino acid mixture through a fine filter or syringe filter to remove any particulate matter or undissolved material. This ensures a clear and debris-free solution. Store the prepared amino acid mixture in a suitable container, such as an amber glass vial, to protect it from light and degradation. It is recommended to store the mixture at a low temperature, such as in a refrigerator or freezer, to maintain stability and prolong shelf life.

The prepared amino acid mixture can then be used as a reference standard in chromatographic analysis, such as high-performance liquid chromatography (HPLC) or gas chromatography (GC), to compare and identify the amino acid composition of the protein sample under investigation.

15. Write the structure of the product obtained when glucose is oxidised with nitric acid.

12. Write a reaction which shows that all the carbon atoms in glucose are linked in a straight chain.

On prolonged heating with HI, it forms n-hexane, shows that all the six carbon atoms are linked in a straight chain :

12. Where does the water present in the egg go after boiling the egg?
    Denaturation of proteins is a process that changes the physical and biological properties of proteins without affecting the chemical composition of protein. In an egg, denaturation of protein is the coagulation of albumin present in the white of an egg. When an egg is boiled in water, the globular proteins present in it change to a rubber like insoluble mass which absorbs all the water present in the egg by making hydrogen bond with it.
13. What are three types of RNA molecules which perform different functions?

                 m-RNA, t-RNA, r-RNA

20. What is a glycosidic linkage?

The two monosaccharide units are joined together through an etheral or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage..

16. What are the products of hydrolysis of sucrose?
    

Invert sugar: An equimolar mixture of glucose and fructose is obtained by hydrolysis of sucrose in presence of an acid such as dil. HC1 or the enzyme invertase or sucrase and is called invert sugar.

17. Write the name of linkage joining two amino acids.

           Peptide linkage joins two amino acids.

18. Name the deficiency diseases resulting from lack of Vitamins A and E in the diet.

 Deficiency of Vitamin A causes Xerophthalmia and deficiency of Vitamin E causes Sterility.

19. Name one water soluble vitamin which is a powerful antioxidant.
    Water soluble vitamin : Vitamin C
20. Using a biochemical equation , give products of hydrolysis of lactose.

Lactose on hydrolysis with dilute acids gives an equimolar mixture of D-glucose and D-galactose.

Lactose on hydroloysis with dilute acids gives an equimolar mixture of D-glucose and D-galactose.

21. Define a ‘Peptide linkage’.

Peptide linkage : It is an amide linkage formed between – COOH group of one α-amino acid and NH₂ group of the other α-amino acid by loss of a molecule of water. – CO – NH – bond is called Peptide linkage.

22. (ii) Glycosidic linkage : The two monosaccharide units are joined together through an etheral or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
23. Name the four bases present in DNA. Which one of these is not present in RNA?

The four bases present in DNA are :

• Adenine (A)
• Guanine (G)

• Cytosine (C)

• Thymine (T)

In RNA, Thymine (T) is absent. It has Uracil (U) in place of Thymine.

24. Name the bases present in RNA. Which one of these is not present in DNA?

The four bases present in RNA are :

• Purines – Adenine (A) and Guanine (G)
• Pyrimidines – Uracil (U) and Cytosine (C)
• Uracil is not present in DNA.

25. Explain the following terms :
    1. Invert sugar: An equimolar mixture of glucose and fructose obtained by hydrolysis of sucrose in presence of an acid such as dil. HCl or the enzyme invertase or sucrase is called invert sugar.
    2. Polypeptides: They are formed when several molecules of a-amino acids are joined together by peptide bonds.

26. Name the products of hydrolysis of sucrose. Why is sucrose not a reducing sugar?
    

The hydrolysis of sucrose, which is a disaccharide composed of glucose and fructose, results in the formation of two monosaccharides: glucose and fructose. Sucrose is broken down into its constituent monosaccharides through the process of hydrolysis, which involves the addition of water.

The chemical equation for the hydrolysis of sucrose is as follows:

Sucrose + H2O → Glucose + Fructose

Sucrose is not considered a reducing sugar because it lacks a free or potentially free aldehyde or ketone group. In sucrose, the glucose and fructose units are linked by a glycosidic bond between the anomeric carbon of glucose and the hydroxyl group of fructose. This glycosidic bond is formed between the anomeric carbon of glucose (C1) and the hydroxyl group of fructose (C2 or C6). As a result, the anomeric carbon of both glucose and fructose is involved in this glycosidic linkage, rendering them non-reducing.

In order for a sugar to be classified as a reducing sugar, it must have a free or potentially free aldehyde or ketone group. This allows the sugar to undergo oxidation reactions, reducing other substances in the process. However, since sucrose lacks a free aldehyde or ketone group, it does not exhibit reducing properties.

27. What are essential and non-essential amino acids in human food? Give one example of each type.

Essential amino acids : Amino acids which the body cannot synthesize are called essential amino acids.

Example : Valine, leucine etc. Therefore they must be supplied in diet.

Non-essential amino acids : Amino acids which the body can

synthesize are called non-essential amino acids. Therefore, they may or may not be present in diet.

Example : Glycine, alanine etc.

27. Explain the following

a.Nucleoside : A nucleoside contains only two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 9-position of the purine moitey is linked to C₁ of the sugar (ribose or deoxyribose) by a β-linkage.

1. Nucleotides : A nucleotide contains all the three basic components of nucleic acids, i.e. a phosphoric acid group, a pentose sugar and a nitrogenous base. These are formed by esterification of C₅‘ – OH of the sugar of the nucleoside with phosphoric acid.

29. What is essentially the difference between α-form of glucose and β-form of glucose? Explain.

The α-form and β-form of glucose are two stereoisomers of glucose, which differ in the spatial arrangement of their hydroxyl groups around the anomeric carbon (C1) of the glucose molecule. The difference between the α-form and β-form of glucose is determined by the position of the hydroxyl group attached to the anomeric carbon.

 

 In the α-form of glucose:

The hydroxyl group (-OH) attached to the anomeric carbon (C1) is oriented in the opposite direction (trans configuration) compared to the -CH2OH group at the fifth carbon (C5).

The -OH group at C1 is positioned below the plane of the glucose ring, while the -CH2OH group at C5 is positioned above the plane of the ring.

 In the β-form of glucose:

 The hydroxyl group (-OH) attached to the anomeric carbon (C1) is oriented in the same direction (cis configuration) as the -CH2OH group at the fifth carbon (C5).

Both the -OH group at C1 and the -CH2OH group at C5 are positioned above the plane of the glucose ring.

The difference in spatial arrangement of the hydroxyl groups gives rise to distinct three-dimensional structures for the α-form and β-form of glucose. This difference in structure leads to variations in the physical and chemical properties of the two forms.

Importantly, the α-form and β-form of glucose can interconvert between each other through a process known as mutarotation. In an aqueous solution, glucose exists in an equilibrium mixture of the α-form and β-form, with the proportions of each form depending on the specific conditions (temperature, concentration, etc.). This mutarotation allows glucose to exhibit different properties and reactivity depending on the specific form present at a given time.

 In a-α-glucose, the OH group at C₁ is towards right while in p-D-glucose, the OH group at C₁ is towards left.

29. Distinguish between  primary structure and secondary structure of proteins.

The primary structure and secondary structure of proteins are two levels of organization that describe different aspects of the protein’s structure. Here’s a distinction between the two:

Primary Structure:

• The primary structure of a protein refers to the linear sequence of amino acids in the polypeptide chain.
• It is determined by the specific sequence of amino acids, which are linked together by peptide bonds.
• The primary structure provides the foundation for all higher levels of protein structure and plays a crucial role in determining the protein’s overall shape, function, and interactions.
• The primary structure is unique to each protein and can be determined through techniques like protein sequencing or genetic analysis.

Secondary Structure:

• The secondary structure of a protein refers to the local folding or coiling of the polypeptide chain.
• It is primarily stabilized by hydrogen bonding interactions between amino acids along the backbone of the protein.
• The two most common types of secondary structure are alpha helices and beta sheets.
• In an alpha helix, the polypeptide chain is coiled in a spiral shape, with hydrogen bonds formed between nearby amino acids.
• In a beta sheet, the polypeptide chain forms a flat, sheet-like structure with hydrogen bonds between different segments of the chain.
• The secondary structure plays a critical role in determining the overall 3D shape of the protein and is important for its stability and function.

In summary, the primary structure of a protein refers to the linear sequence of amino acids, while the secondary structure refers to the local folding or coiling patterns in the protein’s polypeptide chain. The primary structure is determined by the specific sequence of amino acids, while the secondary structure is primarily stabilized by hydrogen bonding interactions.

28. Explain what is meant by the following :

pyranose structure of glucose

The six membered ring containing 5 carbon atoms and one oxygen atom because of its resemblance with pyron is called the pyranose form.

29.Write the main structural difference between DNA and RNA. Of the four bases, name those which are common to both DNA and RNA.

DNA	RNA
1. The sugar present in DNA is 2-deoxy-(-) ribose.	1. The sugar present in RNA is D-(-) ribose.
2. DNA contains cytosine and thymine as pyrimidine bases.	2. RNA contains cytosine and uracil as pyrimidine bases.
3. DNA has double standard α-helix structure.	3. RNA has single stranded α-helix structure.

 

29. The base which are common to both DNA and RNA are :

• Adenine (A)
• Guanine (G)
• Cytosine (C)

30. Name the only vitamin which can be synthesized in our body. Name one disease that is caused due to the deficiency of this vitamin.
    1. Vitamin that can be synthesized ‘.Vitamin B₁₂
    2. Disease due to the deficiency of Vitamin B₁₂: Pernicious anaemia.
31. State two functions of carbohydrates.

Carbohydrates such as glucose, starch, glycogen etc. provide energy for functioning of living organisms.

Carbohydrates, especially cellulose in the form of wood is used for making furniture, houses etc. by us.

32. Answer the following questions:

Why are vitamin A and vitamin C essential for us? 

Because deficiency of vitamin A and vitamin C causes night blindness and scurvy respectively.

33. Amino acids may be acidic, alkaline or neutral. How does this happen?

Amino acids can be broadly classified into three classes i.e. acidic, alkaline and neutral amino acids depending on the number of —NH₂ group and — COOH group.

Acidic amino acids : Those a-amino acids such as aspartic acid, asparagine and glutamic acid which contain two -COOH groups and one -NH₂ group are called acidic amino acids.

Alkaline or Basic amino acids : Those a-amino acids such as lysine, arginine and histidine which contain two -NH₂ groups and one -COOH group, are called basic amino acids.

Neutral amino acids : Those a-amino acids such as glycine, alanine, valine etc. which contain one -NH₂ and one – COOH group, are called neutral amino acids.

34. Explain how concentrated ascorbic acid  in unknown sample is determined

The concentration of ascorbic acid in an unknown sample can be determined through various analytical methods. One commonly used method is titration. Here’s a general outline of how concentrated ascorbic acid in an unknown sample can be determined using titration:

• Preparation of Standard Solution: Prepare a standard solution of ascorbic acid with a known concentration. This can be done by accurately weighing a known amount of pure ascorbic acid and dissolving it in a suitable solvent, such as distilled water. The concentration of the standard solution should be within the range expected for the unknown sample.

• Titrant Selection: Select an appropriate titrant that reacts specifically with ascorbic acid and can be easily measured. One commonly used titrant is iodine solution (I2) or iodate solution (IO3-) in the presence of an acid.
• Titrating the Unknown Sample: Take a known volume of the unknown sample and transfer it to a titration flask. Add a few drops of indicator solution to the flask. The indicator should change color at the endpoint of the titration, indicating the completion of the reaction between ascorbic acid and the titrant.
• Titration Procedure: Begin titrating the unknown sample by slowly adding the titrant solution from a burette into the flask containing the sample. The titrant reacts with the ascorbic acid in the sample, leading to a color change in the indicator. The titration should be carried out until the color change indicates the endpoint of the reaction. The endpoint is the point at which all the ascorbic acid in the sample has reacted with the titrant.
• Calculation of Concentration: The concentration of ascorbic acid in the unknown sample can be determined by calculating the amount of titrant used during the titration. Using the known concentration of the standard solution and the volume of titrant required to reach the endpoint, you can calculate the concentration of ascorbic acid in the unknown sample using stoichiometry and the balanced equation of the reaction.

It is important to note that the specific procedure and details may vary depending on the exact titration method used and any additional reagents or steps required. It is recommended to follow an established and validated procedure or consult a relevant scientific reference for accurate and precise determination of ascorbic acid concentration.

34. Differentiate between fibrous proteins and globular proteins.

Globular Proteins	Fibrous Proteins
Globular proteins have almost spheroidal shape due to folding of the polypeptide chain.	Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres.
2. Globular proteins are soluble in water.	2. Fibrous proteins are insoluble in water.
3. Globular proteins are sensitive to small changes of temperature and pH. Therefore they undergo denaturation on heating or on treatment with acids/bases	3. Fibrous proteins are stable to moderate changes of temperature and pH.
4. They possess biological activity that’s why they act as enzymes.	4. They do not have any biological activity but serve as chief structural material of animal tissues.
Example: Maltase, invertase etc., hormones (insulin) antibodies, transport agents (haemoglobin), etc.	Example: Keratin in skin, hair, nails and wool, collagen in tendons, fibroin in silk etc.

 

36. What is meant by the denaturation of a protein?
    Denaturation of protein  is the process that renders a protein to loose its biological function  due  to coagulation of globular protein under the influence of change in temperature, change in pH etc.. this process causes the native shape of the protein is destroyed and biological activity is lost and the formed protein is called denaturated proteins and the phenomenon is denaturation.
    What are the different types of RNA found in cells of organisms ? State the functions of each type.

Different types of RNA found in the cell are :

Messenger RNA (mRNA): carries the message of DNA for specific protein synthesis. 

,Ribosotnal RNA (rRNA) : provides the site for protein synthesis., Transfer RNA (t-RNA): transfers amino acids to the site of protein synthesis.

37. Name two components of starch.
    1. Amylose and amylopectin.
38. How are hormones and vitamins different in respect of their source and functions?

Hormones and vitamins are both essential substances for the proper functioning of the body, but they differ in terms of their source and functions. Here’s a comparison between hormones and vitamins:

Source:

• Hormones: Hormones are chemical substances produced by specialized cells or glands in the body, such as the endocrine glands. These glands release hormones into the bloodstream, which then travel to target tissues or organs to regulate various physiological processes.
• Vitamins: Vitamins are organic compounds that are essential for the normal functioning of the body, but they cannot be synthesized in sufficient quantities by the body itself. They are obtained from external sources, primarily through the diet. Vitamins are found in various foods, including fruits, vegetables, grains, dairy products, and meat.

Functions:

• Hormones: Hormones act as chemical messengers in the body and play a crucial role in regulating and coordinating various physiological processes. They help maintain homeostasis, control growth and development, regulate metabolism, influence mood and emotions, and control reproductive functions. Hormones are involved in numerous processes, including metabolism, stress response, blood sugar regulation, sexual development, and the functioning of various organs and tissues.
• Vitamins: Vitamins are organic compounds that are necessary for various biochemical reactions and metabolic processes in the body. They act as coenzymes or cofactors, assisting enzymes in carrying out their specific functions. Vitamins play a vital role in energy production, immune function, cell growth and repair, bone health, blood clotting, antioxidant defense, and many other processes. Each vitamin has specific functions and is required in different amounts by the body.

In summary, hormones are produced by specialized cells or glands in the body and act as chemical messengers to regulate physiological processes. They are synthesized within the body. On the other hand, vitamins are obtained from external sources, primarily through the diet, and they are essential for various metabolic processes and biochemical reactions in the body. Both hormones and vitamins are crucial for maintaining proper health and functioning, but they differ in their sources and specific roles in the body.

39. Define the following terms :

Anomers: A pair of stereoisomers which differ in configuration only around C₁ are called anomers. Two isomers are said to be anomers if the isomerisation in the molecule is at first carbon.

40. (ii) What one difference between a-helix and P-pleated sheet 

One key difference between α-helix and β-pleated sheet is the arrangement of the polypeptide chain in three-dimensional space.

In an α-helix:

• The polypeptide chain is coiled in a right-handed helical structure.
• The backbone of the polypeptide forms the inner part of the helix, while the side chains extend outward.
• The hydrogen bonds form between the carbonyl oxygen of one amino acid and the amide hydrogen of an amino acid four residues down the chain.
• The α-helix structure is stabilized by intramolecular hydrogen bonding.
• The helix is compact and relatively rigid.

In a β-pleated sheet:

• The polypeptide chain folds back and forth, creating a sheet-like structure.
• The backbone of the polypeptide runs in an alternating pattern, with adjacent segments going in opposite directions.
• The hydrogen bonds form between the carbonyl oxygen of one chain and the amide hydrogen of an adjacent chain.
• The β-pleated sheet structure is stabilized by intermolecular hydrogen bonding.
• The sheet can be parallel, with the adjacent strands running in the same N-to-C terminal direction, or anti-parallel, with the adjacent strands running in opposite directions.
• The sheet structure can be either planar or twisted.

Overall, the key distinction lies in the different arrangements of the polypeptide chain in the α-helix and β-pleated sheet. The α-helix is a coiled structure, while the β-pleated sheet forms a sheet-like arrangement.

40. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Vitamins are classified into two types:

Water insoluble vitamins : These are fat soluble substances E.g. Vitamin A, D, E and K.

Water soluble vitamins : These include Vitamin B-Complex and Vitamin C (except B₁₂).

Vitamin K or phylloquinone is responsible for the coagulation of blood.

42. Write the structural difference between starch and cellulose.
    1. Starch contains the β-D-glucose as its monomer units while cellulose contains β-D- glucose as its monomer units
       (ii) What type of linkage is present in Nucleic acids?
    2. Phosphodiester linkages are present in Nucleic Adds
43. Give one example each for fibrous protein and globular protein. 
    (i) Globular protein : All enzymes and hormones like insulin.
    Fibrous protein : Keratin in skin.
44. What are enzymes? Describe their functions. Name two diseases which are caused due to deficiency of enzymes.
    Enzymes are protein molecules which acts catalyst in Biochemical Reactions (biocatalyst). They increase the rate of biochemical reactions. For example : Zymase, Invertase etc.
    Two diseases due to deficiency of enzymes are : Anemia, Gauchea’s disease.
45. Explain the significance of lock and key model of enzyme action

The lock and key model only allows one type of specific substrate to form a substrate-activesite complex with each specific type of enzyme. This is due to their complementary shapes, as only one shape and hence one type of substrate can fit into an enzyme’s active site.

46. Describe any five types of enzyme inhibition

Enzyme inhibition refers to the process by which the activity of an enzyme is reduced or inhibited. There are various types of enzyme inhibition, including:

• Competitive Inhibition: In competitive inhibition, an inhibitor molecule competes with the substrate for binding to the active site of the enzyme. The inhibitor molecule is structurally similar to the substrate and can bind to the active site but does not undergo the catalytic reaction. By occupying the active site, the inhibitor prevents the substrate from binding and inhibits the enzyme’s activity.
• Non-competitive Inhibition: In non-competitive inhibition, the inhibitor molecule binds to a different site on the enzyme known as the allosteric site. This binding induces a conformational change in the enzyme, which affects its active site and reduces its catalytic activity. Importantly, the inhibitor can bind to the enzyme regardless of whether the substrate is bound or not.
• Uncompetitive Inhibition: Uncompetitive inhibition is a type of inhibition where the inhibitor molecule binds to the enzyme-substrate complex rather than the free enzyme or the free substrate. This binding alters the enzyme’s active site, preventing the release of the product and inhibiting the enzyme’s activity.
• Mixed Inhibition: Mixed inhibition occurs when the inhibitor molecule can bind to both the free enzyme and the enzyme-substrate complex, but with different affinities. The inhibitor can bind to either the free enzyme or the enzyme-substrate complex, resulting in different outcomes. Mixed inhibition can lead to a decrease or increase in the enzyme’s activity, depending on the relative affinities of the inhibitor for the enzyme and the enzyme-substrate complex.
• Irreversible Inhibition: Irreversible inhibition refers to the inhibition process in which the inhibitor forms a covalent bond or irreversible interaction with the enzyme, permanently inactivating it. This type of inhibition is typically irreversible and requires the synthesis of new enzymes to restore the enzyme’s activity.

It’s worth noting that enzyme inhibition can be an important mechanism for regulating enzyme activity and controlling metabolic pathways in living organisms. The type of inhibition observed depends on the specific characteristics of the inhibitor and its interaction with the enzyme

47. Explain how shape of enzyme is related to its function

The shape of an enzyme is very important because it has a direct effect on how it catalyzes a reaction. An enzyme’s shape is determined by the sequence of amino acids in its structure, and the bonds which form between the atoms of those molecules.

48. Explain end-point inhibition of enzymes

End product inhibition is also known as feedback inhibition. Here the end product of a metabolic pathway inhibits the enzyme involved in early reactions.

49. State the characteristics of enzymes

They Speed up chemical reactions.

They are required in minute amounts.

They are highly specific in their action.

They are affected by temperature.

They are affected by pH.

Some catalyze reversible reactions.

Some require coenzymes.

They are inhibited by inhibitors.

50. State the role of each of the following chemicals in life
    1. Starch  and glycogen

Starch and glycogen are both storage polysaccharides (polymers made up of glucose monomers) and thus act as a store for energy in living organisms. Starch is a storage polysaccharide in plants and glycogen is the storage polysaccharide for animals. Cellulose is found in plant cell walls and helps gives plants strength. All polysaccharides are made up of glucose monomers, but the difference in the properties of these substances can be attributed to the way in which the glucose molecules join together to form different structures.

1. Cellulose

Cellulose is the most abundant natural biopolymer. The cell wall of plants is mostly made of cellulose; this provides structural support to the cell. Wood and paper are mostly cellulosic in nature.

Cellulose is made up of glucose monomers that are linked by β 1-4 glycosidic bonds .This gives cellulose its rigidity and high tensile strength—which is so important to plant cells. While the β 1-4 linkage cannot be broken down by human digestive enzymes, herbivores such as cows, koalas, buffalos, and horses are able, with the help of the specialized flora in their stomach, to digest plant material that is rich in cellulose and use it as a food source. In these animals, certain species of bacteria and protists reside in the rumen (part of the digestive system of herbivores) and secrete the enzyme cellulase. The appendix of grazing animals also contains bacteria that digest cellulose, giving it an important role in the digestive systems of ruminants. Cellulases can break down cellulose into glucose monomers that can be used as an energy source by the animal. Termites are also able to break down cellulose because of the presence of other organisms in their bodies that secrete cellulases. Cellulose  serve various functions in different animals. Arthropods (insects, crustaceans, and others) have an outer skeleton, called the exoskeleton, which protects their internal body parts. This exoskeleton is made of the biological macromolecule chitin, which is a polysaccharide-containing nitrogen. It is made of repeating units of N-acetyl-β-d-glucosamine, a modified sugar. Chitin is also a major component of fungal cell walls; fungi are neither animals nor plants and form a kingdom of their own in the domain Eukarya.

1. Cholesterol

Cholesterol is a wax-like substance, found only in animal source foods.  Triglycerides, LDL, HDL, VLDL are different types of cholesterol found in the blood cells. Cholesterol is an important lipid found in the cell membrane. It is a sterol, which means that cholesterol is a combination of steroid and alcohol. In the human body, cholesterol is synthesized in the liver. These compounds are biosynthesized by all living cells and are essential for the structural component of the cell membrane. In the cell membrane, the steroid ring structure of cholesterol provides a rigid hydrophobic structure that helps boost the rigidity of the cell membrane. Without cholesterol, the cell membrane would be too fluid. It is an important component of cell membranes and is also the basis for the synthesis of other steroids, including the sex hormones estradiol and testosterone, as well as other steroids such as cortisone and vitamin D.

51. State the advantages of using electrophoresis  in the separation and characterization of proteins

Electrophoresis is a laboratory technique used to separate macromolecules in a fluid or gel based on their electric  charge, binding affinity, and size under an electric field.. An electric current is used to move molecules to be separated through a gel. The  pores in the gel work like a sieve, allowing smaller molecules to move faster than larger molecules. The conditions used during electrophoresis can be adjusted to separate molecules in a desired size range.

 Anaphoresis is the electrophoresis of negative charge particles or anions whereas cataphoresis is electrophoresis of positive charge ions or cations. Electrophoresis has a wide application in separating and analysing biomolecules such as proteins, plasmids, RNA, DNA, nucleic acids.

Charged macromolecules are placed in the electric field move towards the negative or positive pole based on their charge. Nucleic acid has a negative charge and therefore it migrates towards the anode.

52. Define the terms
    1. activation energy

All chemical reactions require a certain minimum amount of energy to take place. This energy is known as the activation energy . Enzymes lower the energy of activation thus speeding up chemical reactions enzymes work by lowering the activation energy for a reaction, thus dramatically increasing the rate of the reaction.

1. allosteric enzymes

Allosteric enzymes  are  enzymes that attach or fit or bind  to specific active sites on  molecules in the cellular environment. The sites form weak, noncovalent bonds with these molecules, causing a change in the conformation of the enzyme. This change in conformation translates to the active site, which then affects the reaction rate of the enzyme. Allosteric interactions can both inhibit and activate enzymes and are a common way that enzymes are controlled in the body.

53. Compare and contrast ionic, covalent, and hydrogen bonds. How do these bonds work? Do they form between atoms, molecules, or ions? What is the role of electrical charge? How strong are the bonds, and how do they relate to the dissolving properties of water?The three kinds of bonds (ionic, covalent, and hydrogen) can be ranked in terms of strength: covalent bonds are the strongest, followed by ionic bonds, and finally, hydrogen bonds as the weakest. Ionic bonds occur between two differently charged ions, or atoms that have gained or lost electrons. What holds them together is the fact that opposites attract: electrons are stolen rather than shared in the ionic bond. In contrast, in covalent bonding, the electrons are shared between the two atoms. In hydrogen bonds, the weakest kind of bond, the partial charge of the hydrogen atom is what attracts it to the slightly negative charge of another atom. The hydrogen bond is important in biologically significant macromolecules such as DNA, RNA, and proteins. In this case, no electrons are exchanged or shared. The best explanation of hydrogen bonding is in the case of the water molecule. Because hydrogen is such a poor electrophile, when it is covalently bound to oxygen, instead of evenly sharing this single electron, the electron spends more time around the oxygen atom. Therefore, the hydrogen becomes slightly positively charged while the oxygen becomes slightly negatively charged. These opposite charges on separate but adjacent molecules can then work to create hydrogen bonds between the molecules. Because of this special property of water, ionically bound molecules can easily dissolve in aqueous solution, as their negative and positively charged atoms can associate with the slightly negative and slightly positive charges on the water molecule.
54. What kind of food would you want to eat for breakfast before running a mile in gym class? Why? If you were a rodent preparing for hibernation, or a bird preparing for an autumn migration, what type of biomolecule would you want to use for energy storage? Why?

Before running a mile in gym class, you would want to eat a simple carbohydrate. This would be a good idea because a mile is not a very long distance, and what you would need is quick energy. If you were preparing for hibernation, however, or a long distance migration, you would want to use fat for storage because fat stores, per gram, twice as much energy as carbs or protein, and can therefore last the distance.

55. During the winter, many species of amphibians and reptiles go into a hibernation-like state called overwintering. They eat little or no food, slow their metabolism and breathing rate dramatically, and most of their bodily functions shut down. Interestingly, many of these species choose to spend the winter underwater. Why might it be advantageous to overwinter at the bottom of a pond rather than on land?it might be advantageous to overwinter in water because of water’s heat of fusion. This fact means that you would be less likely to freeze to death during your overwintering.
56. Why are fats and waxes solid at room temperature, whereas oils are liquids? With your answer to that question in mind, how do you think the phospholipids of coldwater fish compare to warmwater fish? Would you expect to see differences in the amounts of saturated and unsaturated fatty acids?
    Fats and waxes are solid at room temperature because they are made of saturated fatty acids. In contrast, oils are mostly unsaturated fatty acids. Saturated fatty acids are called as such because they do not have many double or triple bonds in their structure; each carbon is bound to four other atoms, and this fact makes it quite easy to stack the fatty acid molecules together closely. Therefore, saturated fats are solid at room temperature. In contrast, unsaturated fatty acids have several kinks in their structure and therefore, are liquid at room temperature. We would expect the phospholipids of cold water fish to be more heavily unsaturated compared to the phospholipids of warm water fish, due to their adaptation to temperature.
57. humans, blood pH is maintained fairly precisely around 7.4, and this is largely accomplished with the help of bicarbonate ion (HCO₃^–) and carbonic acid (H₂CO₃), both of which are floating around in the blood. If blood suddenly becomes acidic for some reason, bicarbonate turns into H₂CO₃. If blood suddenly becomes too basic, some H₂CO₃turns into HCO₃^–. What’s going on here? Using your knowledge of acids, bases, and buffers, explain how pH is maintained at a constant level in this system.

pH refers to the concentration of hydrogen in a solution. A buffer is a molecule that can help to control the pH of a solution by taking up a hydrogen or releasing one into solution, depending on the surrounding pH. If blood becomes too acidic, that fact means that there is an excess of hydrogen ions in the blood.  The bicarbonate ion can then take up the excess hydrogen ion and become H₂CO₃.  If, in contrast, the blood becomes too basic, that fact means that there is a lack of hydrogen ions, and bicarbonate can then release its extra hydrogen into solution, becoming HCO₃^–.

58. Compare and contrast proteins and nucleic acids. What do these biomolecules have in common? In what ways are they different? Be sure to consider both their structures and their functions.Proteins and nucleic acids are both important biological macromolecules. They both have carbon, nitrogen, and hydrogen, but are made of different molecular components. Proteins are made of amino acids, and nucleic acids are made of nucleotides. They are similar in that, during the synthesis of the macromolecules, these molecular components are covalently bound to each other, making a string of nucleotides or a string of amino acids. Proteins and nucleic acids are also similar in that both have tertiary structure to them. In proteins, the kinds of folds are alpha helices and pleated sheets, and in nucleotides, helices are formed by the DNA molecule, and a similar kind of structure is formed by the single-stranded RNA molecule. They are different in their functions. Proteins are synthesized from DNA, with an RNA intermediate. 
59. State the use of the following  substances in  ascending paper chromatography during amino acid isolation

Solvent

Chromatography is used to separate mixtures of substances into their components.

They all have a stationary phase (a solid, or a liquid supported on a solid) and a mobile phase (a liquid or a gas). The mobile phase  is usually a solvent or gas  that flows through the stationary phase and carries the components of the mixture with it. Different components travel at different rates.  In paper chromatography, the stationary phase is a very uniform absorbent paper. The mobile phase is a suitable liquid solvent or mixture of solvents.

1. Ninhydrin

Ninhydrin is also used in amino acid analysis of proteins. Most of the amino acids, except proline, are hydrolyzed and react with ninhydrin.

1. Define the term ‘ relative front’ as used in chromatography

RF value (in chromatography)  refers to the distance travelled by a given component divided by the distance travelled by the solvent front. For a given system at a known temperature, it is a characteristic of the component and can be used to identify components. For example, the photosynthetic pigments of an organism and the metabolites of a drug excreted in the urine can be identified by their RF values in paper or thin-layer chromatography.

60. (a) Name water soluble vitamins

There are nine water-soluble vitamins: the B vitamins — folate, thiamine, riboflavin, niacin, pantothenic acid, biotin, vitamin B6, and vitamin B12 — and vitamin C. Deficiency of any of these water-soluble vitamins results in a clinical syndrome that may result in severe morbidity and mortality.

• Name lipid soluble vitamins

Vitamins A, D, E, and K are called the fat-soluble vitamins, because they are soluble in organic solvents and are absorbed and transported in a manner similar to that of fats.

• State two roles of vitamin A

Vitamin A is important for normal vision, the immune system, reproduction, and growth and development. Vitamin A also helps your heart, lungs, and other organs work properly. Carotenoids are pigments that give yellow, orange, and red fruits and vegetables their color.

61. Define  enzyme

Enzymes are biological molecules that catalyze (i.e., increase the rates of) chemical reactions. All known enzymes are proteins. They are high molecular weight compounds made up principally of chains of amino acids linked together by peptide bonds. Enzymes can be denatured and precipitated with salts, solvents and other reagents. They have molecular weights ranging from 10,000 to 2,000,000.

62. Describe
    1. enzyme regulation

Enzymes can be regulated by other molecules that either increase or reduce their activity. Molecules that increase the activity of an enzyme are called activators, while molecules that decrease the activity of an enzyme are called inhibitors.

There are many kinds of molecules that block or promote enzyme function, and that affect enzyme function by different routes.

1. Competitive vs. noncompetitive

An inhibitor may bind to an enzyme and block binding of the substrate, for example, by attaching to the active site. This is called competitive inhibition, because the inhibitor “competes” with the substrate for the enzyme. That is, only the inhibitor or the substrate can be bound at a given moment.

In noncompetitive inhibition, the inhibitor doesn’t block the substrate from binding to the active site. Instead, it attaches at another site and blocks the enzyme from doing its job. This inhibition is said to be “noncompetitive” because the inhibitor and substrate can both be bound at the same time.

1. Cofactors and coenzymes

Many enzymes don’t work optimally, or even at all, unless bound to other non-protein helper molecules called cofactors. These may be attached temporarily to the enzyme through ionic or hydrogen bonds, or permanently through stronger covalent bonds.Common cofactors include inorganic ions such as iron , magnesium , zinc etc Coenzymes are a subset of cofactors that are organic (carbon-based) molecules. The most common sources of coenzymes are dietary vitamins. Some vitamins are precursors to coenzymes and others act directly as coenzymes. For example, vitamin C, which acts as a coenzyme for several enzymes.

63. Classify enzymes on the basis of the  reactions they catalyze , giving an example in each case

According to the naming conventions  by the International Union of Biochemistry and Molecular Biology, enzymes are generally classified into six main family classes and many sub-family classes. These enzymes are identified using the EC numbers; each enzyme is described by a sequence of four numbers preceded by “EC”. The first number broadly classifies the enzyme based on its mechanism. The top-level classification is

1. EC 1 Oxidoreductases: catalyze oxidation/reduction reactions
2. EC 2 Transferases: transfer a functional group (e.g. a methyl or phosphate group)
3. EC 3 Hydrolases: catalyze the hydrolysis of various bonds
4. EC 4 Lyases: cleave various bonds by means other than hydrolysis and oxidation
5. EC 5 Isomerases: catalyze isomerization changes within a single molecule
6. EC 6 Ligases: join two molecules with covalent bonds.

64. Explain four effects of temperature  on enzyme controlled reactions

Like most chemical reactions, the rate of an enzyme-catalyzed reaction increases as the temperature is raised. A ten degree Centigrade rise in temperature will increase the activity of most enzymes by 50 to 100%. Variations in reaction temperature as small as 1 or 2 degrees may introduce changes of 10 to 20% in the results. In the case of enzymatic reactions, this is complicated by the fact that many enzymes are adversely affected by high temperatures.

Because most animal enzymes rapidly become denatured at temperatures above 40·C, most enzyme determinations are carried out somewhat below that temperature.

Over a period of time, enzymes will be deactivated at even moderate temperatures. Storage of enzymes at 5·C or below is generally the most suitable. Some enzymes lose their activity when frozen.

65. Explain the factors affecting  enzyme activity

Temperature

In humans, enzymes function best at 37 °C .This is the optimum temperature. At very high temperatures proteins denature; this means that the hydrogen, hydrophobic and electrostatic interactions that result in the protein’s three-dimensional shape break down, unravelling the protein into its primary structure, a long chain of amino acids. When a protein is denatured, the shape of its active site, as well as the rest of the protein shape is altered. The substrate can no longer fit in the active site of the enzyme and chemical reactions cannot take place. Low temperatures can slow down or even inactivate enzymes, as low temperature means less available kinetic energy, so that even the lower energy of activation that the enzyme allows is not available. The first graph shows the effect of temperature on enzyme activity.

pH

Enzyme activity is sensitive to pH. Enzymes have an optimum pH as shown on the graph, but they can function effectively within a pH range. The effectiveness of the enzyme falls sharply when the pH is outside its optimum range. An enzyme can become denatured when exposed to a pH outside its pH range, as pH affects the charge on some amino acids, and therefore affects the electrostatic interactions holding the tertiary structure together. The second graph shows the effect of pH on enzyme activity. The optimal pH and temperature for an enzyme will be determined by the kind of living thing it is found in. The enzymes in the human body have an optimum temperature of 37 °C. Bacteria that live in compost heaps have enzymes with an optimal range in the 40’s, and bacteria called hyperthermophiles (lovers of very high temperatures) that live in hot springs have enzymes with optimum temperatures above 80 °C.

Figure 1.20: Graphs showing the effect of temperature and pH on enzyme activity respectively.

In the investigation that follows, the effect of temperature on catalase enzyme activity will be investigated. Hydrogen peroxide is potentially toxic and so living tissues contain an enzyme named catalase to break it down into non-toxic compounds, namely water and oxygen. You will study the effect of the enzyme catalase on the breakdown of hydrogen peroxide. You will further examine the effect of pH and temperature on enzyme activity.