Samacheer Kalvi Class 12 Chemistry Solutions Chapter 14 Biomolecules

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules

12th Chemistry Guide Biomolecules Text Book Back Questions and Answers

Part - I Text Book Evaluation

I. Choose the Correct Answer

 

Question 1. Which one of the following rotates the plane polarized light towards left?
(a) D(+) Glucose
(b) L(+) Glucose
(c) D(-) Fructose
(d) D(+) Galactose
Answer: (c) D(-) Fructose
In simple words: D(-) Fructose is a sugar that rotates plane-polarized light in a counter-clockwise direction, which is to the left. This property is known as levorotation.

🎯 Exam Tip: Remember that D and L prefixes relate to the configuration (structure) of the molecule, while (+) and (-) refer to the direction of optical rotation. These are not always directly related.

 

Question 2. The correct corresponding order of names of four aldoses with configuration given below Respectively is,
(a) L-Erythrose, L-Threose, L-Erythrose, D-Threose
(b) D-Threose, D-Erythrose, L-Threose, L-Erythrose
(c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer: (d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
In simple words: This question tests your knowledge of how different aldose sugars are named based on their specific arrangements of atoms. Each name (like D-Erythrose or L-Threose) refers to a unique structural form.

🎯 Exam Tip: Familiarize yourself with the Fischer projection formulas for common aldoses and ketoses to correctly identify their D/L configuration and names.

 

Question 3. Which one given below is a non-reducing sugar?
(a) Glucose
(b) Sucrose
(c) Maltose
(d) Lactose
Answer: (b) Sucrose
In simple words: Sucrose is a sugar that cannot reduce other substances because its important functional groups are used in the bond between its parts. Other sugars listed can reduce other substances.

🎯 Exam Tip: Remember that reducing sugars have a free anomeric carbon (aldehyde or ketone group) that can open to form an aldehyde. Non-reducing sugars have this carbon involved in a glycosidic bond.

 

Question 4. Glucose (HCN) \( \rightarrow \) Product (hydrolysis) \( \rightarrow \) Product (HI + Heat) A, the compound A is
(a) Heptanoic acid
(b) 2-Iodohexane
(c) Heptane
(d) Heptanol
Answer: (a) Heptanoic acid
\[ \text{CHO}\left(\text{CHOH}\right)_4\text{CH}_2\text{OH} \underset{\text{HCN}}{\longrightarrow} \text{CH}(\text{CN})\left(\text{CHOH}\right)_4\text{CH}_2\text{OH} \underset{\text{H}_3\text{O}^+}{\longrightarrow} \text{COOH}\left(\text{CHOH}\right)_5\text{CH}_2\text{OH} \underset{\text{HI + Heat}}{\longrightarrow} \text{CH}_3\left(\text{CH}_2\right)_5\text{COOH} \]In simple words: When glucose reacts with HCN, then undergoes hydrolysis, and finally reacts with HI and heat, the end product formed is Heptanoic acid. This reaction increases the carbon chain length and then reduces the functional groups.

🎯 Exam Tip: Remember that reaction with HCN followed by hydrolysis extends the carbon chain by one carbon, and strong reduction with HI converts hydroxyl groups to methylene groups, ultimately leading to a carboxylic acid or alkane, depending on conditions.

 

Question 5. Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the presence of little hydrochloric acid, it becomes levorotatory. Reason: Sucrose hydrolysis gives unequal amounts of glucose and fructose. As a result of this change in sign of rotation is observed.
(a) If both assertion and reason are true and reason is the correct explanation of assertion
(b) If both assertion and reason are true but reason is not the correct explanation of assertion
(c) If assertion is true but reason is false
(d) If both assertion and reason is false
Answer: (a) If both assertion and reason are true and reason is the correct explanation of assertion
In simple words: Sucrose rotates light to the right, but after breaking down with acid, it rotates light to the left. This happens because the glucose and fructose produced rotate light differently and in unequal amounts, causing the overall direction to change.

🎯 Exam Tip: For assertion-reason questions, first determine if each statement is true individually, then check if the reason correctly explains the assertion. Hydrolysis of sucrose produces invert sugar, where the combined rotation becomes negative.

 

Question 6. The genetic information flows from
(a) Amino acids Protein DNA
(b) DNA Carbohydrates Proteins
(c) DNA RNA Proteins
(d) DNA RNA Carbohydrates
Answer: (c) DNA RNA Proteins
In simple words: Genetic information moves from DNA to RNA, and then from RNA to make proteins. This is called the central dogma of molecular biology.

🎯 Exam Tip: This sequence represents the central dogma: DNA replication, DNA transcription to RNA, and RNA translation to protein. Understanding this flow is key in molecular biology.

 

Question 7. In a protein, various amino acids linked together by
(a) Peptide bond
(b) Dative bond
(c) \( \alpha \)-Glycosidic bond
(d) \( \beta \)-Glycosidic bond
Answer: (a) Peptide bond
In simple words: Amino acids, which are the building blocks of proteins, are connected to each other using special links called peptide bonds. This bond forms by removing a water molecule.

🎯 Exam Tip: Remember that peptide bonds are amide linkages formed between the carboxyl group of one amino acid and the amino group of another, through a dehydration reaction.

 

Question 8. Among the following, the achiral amino acid is
(a) 2-ethylalanine
(b) 2-methyl glycine
(c) 2-hydroxymethylserine
(d) Tryptophan
Solution:
\[ \text{H}_2\text{N}-\underset{\text{CH}_2\text{OH}}{\overset{\text{OH}}{\text{C}}}-\text{COOH} \]Answer: (c) 2-hydroxymethylserine
In simple words: An achiral amino acid means it does not have a chiral carbon, which is a carbon atom bonded to four different groups. While most amino acids have a chiral center, certain specific derivatives can be achiral.

🎯 Exam Tip: Glycine is the simplest achiral amino acid because its central carbon is bonded to two hydrogen atoms. For other named amino acids, check if their structure allows for a plane of symmetry or if the central carbon has two identical groups.

 

Question 9. The correct statement regarding RNA and DNA respectively is
(a) the sugar component in RNA is arabinose and the sugar component in DNA is ribose
(b) the sugar component in RNA is 2'-deoxyribose and the sugar component in DNA is arabinose
(c) the sugar component in RNA is arabinose and the sugar component in DNA is 2'-deoxyribose
(d) the sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose
Answer: (d) the sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose
In simple words: RNA has a sugar called ribose, while DNA has a slightly different sugar called 2'-deoxyribose, which is ribose missing an oxygen atom. This difference makes DNA more stable.

🎯 Exam Tip: This is a fundamental difference between RNA and DNA. Remember that ribose has a hydroxyl group at the 2' carbon, while deoxyribose has a hydrogen atom there.

 

Question 10. In aqueous solution amino acids mostly exist in,
(a) \( \text{NH}_2\text{-CH(R)-COOH} \)
(b) \( \text{NH}_2\text{-CH(R)-COO}^- \)
(c) \( \text{H}_3\text{N}^+\text{-CH(R)-COOH} \)
(d) \( \text{H}_3\text{N}^+\text{-CH(R)-COO}^- \)
Answer: (d) H3N+-CH(R)-COO-
In simple words: In water, amino acids exist as zwitterions, meaning they have both a positive charge on the amino group and a negative charge on the carboxyl group, making the molecule electrically neutral overall.

🎯 Exam Tip: Understand the concept of zwitterions: at physiological pH, the amino group is protonated (positive) and the carboxyl group is deprotonated (negative), forming an internal salt.

 

Question 11. Which of the following is not produced by the body?
(a) DNA
(b) Enzymes
(c) Hormones
(d) Vitamins
Answer: (d) Vitamins
In simple words: Our bodies can make DNA, enzymes, and hormones, but we cannot make most vitamins ourselves. We must get vitamins from the food we eat.

🎯 Exam Tip: Vitamins are essential nutrients that our body needs but cannot synthesize on its own, so they must be obtained through diet. This is a key characteristic that distinguishes them from other biomolecules.

 

Question 12. The number of \( \text{sp}^2 \) and \( \text{sp}^3 \) hybridised carbon in fructose are respectively
(a) 1 and 4
(b) 4 and 2
(c) 5 and 1
(d) 1 and 5
Answer: (d) 1 and 5
In simple words: In the common ring form of fructose, one carbon atom has a double bond-like structure (sp2 hybridized), and five other carbon atoms have only single bonds (sp3 hybridized). Fructose forms a furanose ring.

🎯 Exam Tip: To determine hybridization, count the number of regions of electron density (bonds and lone pairs) around each carbon. One double bond counts as one region for \( \text{sp}^2 \), while four single bonds count as four regions for \( \text{sp}^3 \).

 

Question 13. Vitamin B2 is also known as
(a) Riboflavin
(b) Thiamine
(c) Nicotinamide
(d) Pyridoxine
Answer: (a) Riboflavin
In simple words: Vitamin B2 is also known by its chemical name, Riboflavin. It plays an important role in energy production and cell growth.

🎯 Exam Tip: It's helpful to know the common names and chemical names for important vitamins, especially the B-complex vitamins, as questions often use both.

 

Question 14. The pyrimidine bases present in DNA are
(a) Cytosine and Adenine
(b) Cytosine and Guanine
(c) Cytosine and Thiamine
(d) Cytosine and Uracil
Answer: (c) Cytosine and Thiamine
In simple words: In DNA, the smaller ring-shaped bases are cytosine and thymine. These pair up with the larger purine bases.

🎯 Exam Tip: Remember the two types of nitrogenous bases: Purines (Adenine, Guanine) are larger, and Pyrimidines (Cytosine, Thymine in DNA; Cytosine, Uracil in RNA) are smaller.

 

Question 15. Among the following L-serine is
(a) \( \text{pka 12.10 NH}_2 \quad \text{H}_2\text{N}-\overset{\text{H}}{\overset{\text{N}}{\text{C}}}-\text{COOH} \quad \text{NH}_2 \quad \text{pka 9.00 OH} \quad \text{pka 2.03} \)
(b) \( \text{pka 6.04} \quad \overset{\text{H}}{\overset{\text{N}}{\text{H}}}-\overset{\text{N}}{\text{C}}-\text{NH}_2 \quad \text{pka 9.09 OH} \quad \text{pka 1.70} \)
(c) \( \text{HO} \quad \text{pka 9.05 NH}_2 \quad \text{OH} \quad \text{pka 2.13} \)
(d) \( \text{OH} \quad \text{pka 8.96 NH}_2 \quad \text{OH} \quad \text{pka 2.20} \)
Answer: (c) HO \( \text{pka 9.05 NH}_2 \quad \text{OH} \quad \text{pka 2.13} \)
In simple words: L-serine is an amino acid that has a hydroxyl group in its side chain. The given option (c) correctly shows the zwitterionic structure of L-serine with its specific pKa values.

🎯 Exam Tip: Identifying L-amino acids involves understanding Fischer projections and looking for the specific configuration at the alpha-carbon, with the amino group on the left. Serine is characterized by its \( \text{CH}_2\text{OH} \) side chain.

 

Question 16. The secondary structure of a protein refers to
(b) hydrophobic interaction
(c) sequence of a-amino acids
(d) \( \alpha \)-the helical backbone
Answer: (d) \( \alpha \)-the helical backbone
In simple words: The secondary structure of a protein describes how parts of the protein chain fold into regular shapes like alpha-helices or beta-sheets. These shapes are formed by hydrogen bonds.

🎯 Exam Tip: Secondary structure involves local folding patterns like alpha-helices and beta-sheets, stabilized by hydrogen bonds between the amide hydrogen and carbonyl oxygen of the polypeptide backbone.

 

Question 17. Which of the following vitamins is water-soluble?
(a) Vitamin E
(b) Vitamin K
(c) Vitamin A
(d) Vitamin B
Answer: (d) Vitamin B
In simple words: Vitamin B is a group of vitamins that dissolve in water, which means your body doesn't store them in large amounts. This is different from fat-soluble vitamins like A, D, E, and K.

🎯 Exam Tip: Remember that water-soluble vitamins (B and C) are not stored extensively and need regular replenishment, while fat-soluble vitamins (A, D, E, K) are stored in the body's fatty tissues.

 

Question 18. Complete hydrolysis of cellulose gives
(a) L-Glucose
(b) D-Fructose
(c) D-Ribose
(d) D-Glucose
Answer: (d) D-Glucose
In simple words: When cellulose, a complex carbohydrate found in plant cell walls, is completely broken down, it yields only D-Glucose units. This makes D-Glucose its fundamental building block.

🎯 Exam Tip: Cellulose is a polysaccharide made up exclusively of D-glucose units linked by \( \beta \)-1,4 glycosidic bonds. Its complete hydrolysis always yields D-glucose.

 

Question 19. Which of the following statement is incorrect?
(a) Ovalbumin is a simple food reserve in the egg-white
(b) Fibrinogen are involved in blood clotting
(c) Denaturation makes the protein more active
(d) Insulin maintains the sugar level in the human body.
Answer: (c) Denaturation makes the protein more active
In simple words: Denaturation changes a protein's natural shape, usually causing it to lose its activity, not become more active. The other statements are correct facts about proteins.

🎯 Exam Tip: Denaturation is the process where a protein loses its specific three-dimensional shape and, consequently, its biological activity. This can be caused by heat, pH changes, or chemicals.

 

Question 20. Glucose is an aldose. Which one of the following reactions is not expected with glucose?
(a) It does not form an oxime
(b) It does not react with the Grignard reagent
(c) It does not form osazones
(d) It does not reduce tollens reagent
Answer: (b) It does not react with the Grignard reagent
In simple words: Glucose reacts with many things because it has an aldehyde group. But it typically does not react with Grignard reagents because its aldehyde group is in equilibrium with its cyclic hemiacetal forms, which are less reactive to Grignard reagents.

🎯 Exam Tip: Aldehydes generally react with Grignard reagents. However, the cyclic hemiacetal form of glucose, which is dominant in solution, is less reactive towards Grignard reagents compared to simple aldehydes. This is a common exception to remember.

 

Question 21. If one strand of the DNA has the sequence ‘ATGCTTGA', then the sequence of complementary strand would be
(a) TACGAACT
(b) TCCGAACT
(c) TACGTACT
(d) TACGRAGT
Answer: (a) TACGAACT
In simple words: In DNA, adenine (A) always pairs with thymine (T), and guanine (G) always pairs with cytosine (C). So, to find the complementary strand, you just swap each base for its partner.

🎯 Exam Tip: Remember the base pairing rules: A-T and G-C. Always write the complementary strand in the 5' to 3' direction relative to the given 3' to 5' strand, or vice-versa.

 

Question 22. Insulin, a hormone chemically is
(a) Fat
(b) Carbohydrates
(c) Protein
(d) Carbohydrates
Answer: (c) Protein
In simple words: Insulin is a hormone made of protein. It helps control blood sugar levels in the body.

🎯 Exam Tip: Hormones can be proteins, steroids, or amines. Insulin is a classic example of a protein hormone, while others like estrogen are steroids.

 

Question 23. \( \alpha \)-D (+) Glucose and \( \beta \)-D (+) glucose are
(a) Epimers
(b) Anomers
(c) Enantiomers
(d) Conformational isomers
Answer: (b) Anomers
In simple words: Anomers are special types of isomers of sugars that only differ in their structure at the anomeric carbon (the carbon that used to be the carbonyl carbon in the open-chain form). Alpha and beta forms of glucose are anomers.

🎯 Exam Tip: Anomers are cyclic monosaccharide isomers that differ only in the configuration at the anomeric carbon (C1 for aldoses, C2 for ketoses). Epimers differ at any chiral center other than the anomeric carbon.

 

Question 24. Which of the following are epimers?
(a) D(+)-Glucose and D(+)-Galactose
(b) D(+)-Glucose and D(+)-Mannose
(c) Neither (a) nor (b)
(d) Both (a) and (b)
Answer: (d) Both (a) and (b)
In simple words: Epimers are sugars that are different in their structure at only one carbon atom, other than the anomeric carbon. Both pairs, D(+)-Glucose and D(+)-Galactose (at C4), and D(+)-Glucose and D(+)-Mannose (at C2), show this single-carbon difference.

🎯 Exam Tip: Galactose is a C4 epimer of glucose, and mannose is a C2 epimer of glucose. Understanding these specific relationships helps in distinguishing various monosaccharides.

 

Question 25. Which of the following amino acids is achiral?
(a) Alanine
(b) Leucine
(c) Proline
(d) Glycine
Answer: (d) Glycine
In simple words: Glycine is the only amino acid that does not have a chiral carbon, meaning its alpha-carbon is attached to two hydrogen atoms, making it symmetrical and achiral.

🎯 Exam Tip: The alpha-carbon in an amino acid is chiral if it is bonded to four different groups. Glycine has two hydrogen atoms on its alpha-carbon, making it achiral.

 

II. Short Answer

 

Question 1. What type of linkages holds together monomers of DNA?
Answer:
1. The monomers of DNA are held together by phosphodiester linkages.
2. Phosphoric acid creates phosphodiester bonds between adjacent nucleotides. These bonds are crucial for the molecule's structural integrity.
3. The sugar-phosphate linkage forms the strong backbone of each DNA strand.
In simple words: DNA's small units are joined by special links called phosphodiester bonds. These bonds are made by phosphoric acid and form the main support structure of the DNA molecule.

🎯 Exam Tip: Remember that phosphodiester bonds connect the 3'-hydroxyl group of one deoxyribose to the 5'-phosphate group of the next, forming the sugar-phosphate backbone of DNA.

 

Question 2. Give the differences between primary and secondary structure of proteins
Answer:

In simple words: Primary structure is just the straight order of amino acids, like letters in a word. Secondary structure is how these amino acids start to fold into simple, repeated patterns like spirals (alpha-helix) or pleated sheets (beta-sheet).

🎯 Exam Tip: Clearly distinguish that primary structure is defined by covalent peptide bonds, while secondary structure is defined by hydrogen bonds within the backbone, forming local repetitive structures.

 

Question 3. Name the Vitamins whose deficiency cause
1. rickets
2. scurvy
Answer:
1. Rickets is caused by a deficiency of Vitamin D.
2. Scurvy is caused by a deficiency of Vitamin C.
In simple words: Not enough Vitamin D causes weak bones, known as rickets. Not enough Vitamin C causes scurvy, which affects gums and skin.

🎯 Exam Tip: Remember these classic deficiency diseases and their associated vitamins. Vitamin D is essential for calcium absorption, and Vitamin C for collagen synthesis.

 

Question 4. Write the Zwitter ion structure of alanine
Answer:
\[ \text{H}_3\text{N}^+-\underset{\text{CH}_3}{\text{CH}}-\text{COO}^- \]In simple words: The zwitterion form of alanine shows both a positive charge on the amino group and a negative charge on the carboxyl group. This is how it typically exists in water.

🎯 Exam Tip: When drawing a zwitterion, ensure the amino group is protonated \( \left(\text{NH}_3^+\right) \) and the carboxyl group is deprotonated \( \left(\text{COO}^-\right) \) at neutral pH.

 

Question 5. Give any three difference between DNA and RNA
Answer:

In simple words: DNA stores genetic plans and has deoxyribose sugar and thymine. RNA carries out these plans and has ribose sugar and uracil instead of thymine. They are found in different parts of the cell too.

🎯 Exam Tip: Key differences include the sugar component (deoxyribose in DNA, ribose in RNA), one of the pyrimidine bases (thymine in DNA, uracil in RNA), and their typical cellular locations and primary functions.

 

Question 6. Write a short note on peptide bond
Answer:
A peptide bond is a chemical link that joins amino acids together to form proteins.
1. Amino acids are connected covalently by peptide bonds.
2. The carboxyl group of the first amino acid reacts with the amino group of the second amino acid. This reaction forms an amide linkage between the two amino acids, releasing a water molecule.
3. This amide linkage is specifically called a peptide bond.
4. When amino acids link this way, the resulting compound is known as a dipeptide (if two amino acids join) or polypeptide (if many join).
5. A short chain of amino acids is a polypeptide, while a long chain is a protein.
\[ \text{H}_2\text{N}-\text{CH}_2-\text{COOH} \quad (\text{Glycine}) \quad + \quad \text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{COOH} \quad (\text{Alanine}) \quad \xrightarrow{-\text{H}_2\text{O}} \quad \text{H}_2\text{N}-\text{CH}_2-\text{CO}-\text{NH}-\text{CH}(\text{CH}_3)-\text{COOH} \quad (\text{Glycylalanine Dipeptide}) \]In simple words: A peptide bond is like a special glue that connects amino acids to build proteins. It forms when the acid part of one amino acid joins with the amine part of another, releasing water. This bond is very strong.

🎯 Exam Tip: Remember that a peptide bond is an amide bond, formed by a condensation reaction between the carboxyl group of one amino acid and the amino group of another, releasing water.

 

Question 7. Differentiate between Hormones and Vitamins
Answer:

In simple words: Hormones are chemicals made by our body to send messages and control functions like growth. Vitamins are chemicals we need from food because our body can't make them, and they help with many important processes.

🎯 Exam Tip: Hormones are endogenous signaling molecules, while vitamins are exogenous essential nutrients. This distinction is key to understanding their roles in the body.

 

Question 8. Write a note on the denaturation of proteins.
Answer:
Denaturation is the process where a protein loses its natural three-dimensional structure without breaking its basic amino acid chain.
1. Proteins usually have a specific three-dimensional shape formed by interactions like disulfide bonds, hydrogen bonds, hydrophobic interactions, and electrostatic interactions.
2. These interactions can be disrupted when a protein is exposed to high temperatures, certain chemicals (like urea), changes in pH, or altered ionic strength. This disruption leads to the loss of the protein's higher-order structure.
3. The process of a protein losing its higher-order (secondary, tertiary, or quaternary) structure, while keeping its primary structure intact, is called denaturation. Once denatured, a protein usually loses its biological function.
In simple words: Denaturation is when a protein changes its natural folded shape, often due to heat or chemicals. When its shape changes, it can no longer do its job properly.

🎯 Exam Tip: Focus on understanding that denaturation primarily affects secondary, tertiary, and quaternary structures, leaving the primary structure (amino acid sequence) intact. It generally leads to loss of biological activity.

 

Question 9. What is reducing and non-reducing sugars?
Answer:

In simple words: Reducing sugars have a free aldehyde group that can react, while non-reducing sugars have this group locked in a bond and cannot react in the same way.

🎯 Exam Tip: Focus on the "free aldehyde/ketone group" for reducing sugars and "glycosidic bond involvement" for non-reducing sugars.

 

Question 10. Why carbohydrates are generally optically active?
Answer: Carbohydrates are usually optically active. This is because they have one or more carbon atoms that are attached to four different groups, called chiral carbons. The number of possible mirror-image forms (optical isomers) of a carbohydrate can be calculated using the formula \( 2^n \), where \( n \) is the number of chiral carbons. This allows them to rotate plane-polarized light.
In simple words: Carbohydrates can twist light because they have special carbon atoms called chiral carbons. The more these special carbons, the more ways they can twist light.

🎯 Exam Tip: Remember that the presence of chiral carbons is key to optical activity, and the formula \( 2^n \) determines the number of optical isomers.

 

Question 11. Classify the following into monosaccharides, oligosaccharides and polysaccharides.
(i) starch
(ii) fructose
(iii) sucrose
(iv) lactose
(v) maltose
Answer:

Monosaccharides are simple sugars that cannot be broken down further. Fructose is an example. Oligosaccharides are made of a few simple sugar units linked together. Sucrose, lactose, and maltose are examples, each made of two sugar units (disaccharides). Polysaccharides are large, complex carbohydrates made of many sugar units. Starch is a common polysaccharide.
In simple words: Monosaccharides are single sugars. Oligosaccharides are a few sugars joined. Polysaccharides are many sugars joined.

🎯 Exam Tip: Learn common examples for each carbohydrate type to classify them correctly.

 

Question 12. How are vitamins classified?
Answer: Vitamins are classified based on how well they dissolve in water or fat. There are two main groups:

1. Water-soluble vitamins: These vitamins dissolve easily in water. Examples include vitamins from the B group and Vitamin C. They are not stored in the body for long.
2. Fat-soluble vitamins: These vitamins dissolve in oils or fats. Examples include Vitamin A, D, E, and K. They can be stored in the body's fatty tissues.

In simple words: Vitamins are sorted by if they dissolve in water or fat. Water-soluble ones (like B, C) don't stay in the body long. Fat-soluble ones (like A, D, E, K) can be stored in the body.

🎯 Exam Tip: Knowing which vitamins are water-soluble and which are fat-soluble helps understand how the body stores and uses them.

 

Question 13. What are hormones? Give Examples.
Answer: Hormones are special organic substances made by one part of the body. They are released into the blood and travel to other parts, causing specific changes or responses. They act as messengers between cells. Almost every body process is controlled by one or more hormones. Examples include insulin, epinephrine (adrenaline), estrogen, and androgen. These chemicals ensure the body functions smoothly.
In simple words: Hormones are body messengers made in one place and sent through the blood to tell other parts what to do. Insulin is one example.

🎯 Exam Tip: Emphasize "organic substances," "secreted into bloodstream," and "physiological response" as key definitions, along with examples.

 

Question 14. Write the structure of all possible dipeptides which can be obtained from glycine and alanine.
Answer: When glycine and alanine combine, two different dipeptides can form, depending on which amino acid provides the carboxyl group and which provides the amino group for the peptide bond.
(i) Glycylalanine: This forms when the carboxyl group of glycine reacts with the amino group of alanine. The structure is \( \text{H}_2\text{N} - \text{CH}_2 - \text{CO} - \text{NH} - \text{CH}(\text{CH}_3) - \text{COOH} \).
(ii) Alanylglycine: This forms when the carboxyl group of alanine reacts with the amino group of glycine. The structure is \( \text{H}_2\text{N} - \text{CH}(\text{CH}_3) - \text{CO} - \text{NH} - \text{CH}_2 - \text{COOH} \). A peptide bond is formed by the elimination of a water molecule between the carboxyl group of one amino acid and the amino group of another.
In simple words: Glycine and alanine can join in two ways to make two different dipeptides: glycylalanine and alanylglycine. The order of joining matters.

🎯 Exam Tip: Draw the peptide bond carefully, ensuring the correct amino acid contributes its carboxyl and amino groups for each possible dipeptide.

 

Question 15. Define enzymes.
Answer: Enzymes are special proteins found in living cells that help speed up (catalyze) many biochemical reactions. For example, they help digest food, extract energy, and build new molecules. Enzymes are known for being very specific, meaning each enzyme usually works on only one or a few types of reactions. They are essential for life processes.
In simple words: Enzymes are like tiny helpers, usually proteins, that make chemical reactions in our bodies happen much faster. They are very picky about which reaction they help.

🎯 Exam Tip: The key terms for defining enzymes are "proteins," "biocatalysts," "speed up reactions," and "specific in their action."

 

Question 16. Write the structure of \( \alpha -D-(+) \) glucopyranose
Answer: The \( \alpha \)-D-(+)-glucopyranose is a cyclic six-membered structure of glucose, resembling pyran. In this Haworth projection, the C1 carbon (anomeric carbon) has its hydroxyl group pointing downwards, which defines the alpha (\( \alpha \)) configuration. The \( \text{CH}_2\text{OH} \) group is oriented upwards. The ring includes five carbons and one oxygen atom.
\[ \begin{array}{c} \text{CH}_2\text{OH} \\ | \\ \text{H} - \text{C} - \text{O} \\ | \\ \text{HO} - \text{C} - \text{H} \\ | \\ \text{H} - \text{C} - \text{OH} \\ | \\ \text{H} - \text{C} - \text{OH} \\ | \\ \text{CH}_2\text{OH} \end{array} \] (Note: This is a linear projection. For the actual cyclic structure, the C1 hydroxyl group is below the plane for alpha anomer. The cyclic structure of \( \alpha \)-D-(+)-glucopyranose is a six-membered ring (pyranose ring) formed by five carbon atoms and one oxygen atom. The hydroxyl group on the anomeric carbon (C-1) is in the \( \alpha \) (alpha) position, meaning it is pointing downwards relative to the plane of the ring (or in the trans position to the \( \text{CH}_2\text{OH} \) group at C-5). All other hydroxyl groups are positioned to give the specific glucose configuration.)
In simple words: \( \alpha \)-D-(+)-glucopyranose is a sugar that forms a ring with six parts, including five carbon atoms and one oxygen. The "alpha" part means a specific hydroxyl group on the first carbon is pointing down. This cyclic form is how glucose usually exists.

🎯 Exam Tip: When drawing cyclic structures, pay attention to the positions of the hydroxyl groups (above or below the ring) as they determine the alpha or beta anomer.

 

Question 17. What are different types of RNA which are found in cell?
Answer: There are three main types of RNA found in cells, each with a specific role in protein synthesis:

• Ribosomal RNA (rRNA): This type forms a major part of ribosomes, which are the cell's protein-making factories.
• Messenger RNA (mRNA): This carries genetic information from DNA in the nucleus to the ribosomes in the cytoplasm, acting as a blueprint for protein assembly.
• Transfer RNA (tRNA): This helps bring specific amino acids to the ribosome, matching them to the mRNA sequence during protein production.

In simple words: Cells have three main types of RNA: rRNA for making ribosomes, mRNA for carrying genetic messages, and tRNA for bringing building blocks to make proteins.

🎯 Exam Tip: Remember the full names and primary functions of each type of RNA (rRNA, mRNA, tRNA) related to protein synthesis.

 

Question 18. Write a note on formation of \( \alpha - \) helix
Answer: The alpha-helix (\( \alpha \)-helix) is a common, stable secondary structure in proteins. In an \( \alpha \)-helix, the chain of amino acids twists into a right-handed spiral. This spiral shape is held together by hydrogen bonds that form between the oxygen of a carbonyl group in one amino acid and the hydrogen of an amino group four amino acids further along the chain. These hydrogen bonds are crucial for its stability. The side parts of the amino acids stick out from the helix. Each full turn of the helix contains about 3.6 amino acid units and is roughly 5.4 angstroms long. Proline is known as an "helix breaker" because its rigid structure prevents it from easily fitting into the helix, causing a bend or disruption.
In simple words: An \( \alpha \)-helix is a spiral shape in proteins, like a coiled spring. Hydrogen bonds hold it together, forming between amino acids that are four positions apart. This is a very common and strong shape for proteins.

🎯 Exam Tip: Highlight "right-handed helical structure," "hydrogen bonding (n to n+4 residues)," and "3.6 residues per turn" as essential points.

 

Question 19. What are the functions of lipids in living organisms?
Answer: Lipids play many vital roles in living organisms:

• They are a key part of cell membranes, helping cells keep their shape and control what goes in and out.
• Triglycerides, a type of lipid, serve as a major energy storage for animals, providing more energy per gram than carbohydrates or proteins.
• They form protective layers, especially in aquatic animals, to keep them safe and insulated.
• Lipids in connective tissues cushion and protect internal organs from damage.
• They are crucial for absorbing and moving fat-soluble vitamins (A, D, E, K) around the body.
• Lipids are needed to activate certain enzymes, like lipases, which break down fats.
• They also work as emulsifiers, helping to mix fats with water during digestion and metabolism.

In simple words: Lipids are super important! They make up cell walls, store a lot of energy, protect organs, help absorb vitamins, and are key for many body processes.

🎯 Exam Tip: Remember at least five major functions, categorizing them by structural, energy storage, protective, and regulatory roles.

 

Question 20. Is the following sugar, D-sugar or L-sugar?
\[ \begin{array}{c} \text{CHO} \\ | \\ \text{OH} - \text{C} - \text{H} \\ | \\ \text{OH} - \text{C} - \text{H} \\ | \\ \text{OH} - \text{C} - \text{H} \\ | \\ \text{CH}_2\text{OH} \end{array} \quad \begin{array}{c} \text{CHO} \\ | \\ \text{HO} - \text{C} - \text{H} \\ | \\ \text{HO} - \text{C} - \text{H} \\ | \\ \text{HO} - \text{C} - \text{H} \\ | \\ \text{CH}_2\text{OH} \end{array} \quad \begin{array}{c} \text{CHO} \\ | \\ \text{HO} - \text{C} - \text{H} \\ | \\ \text{CH}_2\text{OH} \end{array} \]
Answer: The sugar shown is an L-sugar. This classification is based on the configuration of the hydroxyl group (\( \text{OH} \)) on the chiral carbon furthest from the carbonyl group. In D-sugars, this hydroxyl group is on the right, while in L-sugars, it is on the left. The specific structure provided matches the configuration of L-glyceraldehyde at its reference carbon, confirming it as an L-sugar.
In simple words: This is an L-sugar. We know this because the \( \text{OH} \) group on its lowest chiral carbon points to the left, just like in L-glyceraldehyde.

🎯 Exam Tip: To determine if a sugar is D or L, look at the chiral carbon farthest from the aldehyde or ketone group. If its -OH group is on the right, it's D; if it's on the left, it's L.

 

Part - II - Additional Questions

 

Question 1. Polyhydroxy aldehydes or ketones are called
(a) Vitamins
(b) Enzymes
(c) Carbohydrates
(d) Lipids
Answer: (c) Carbohydrates
In simple words: Sugars and starches are called carbohydrates, which are just molecules with many alcohol parts and a key aldehyde or ketone part.

🎯 Exam Tip: Remember the basic definition of carbohydrates: they are polyhydroxy aldehydes or polyhydroxy ketones.

 

Question 2. The general molecular formula of Carbohydrates is
(a) \( \text{C}_n(\text{H}_2)_2n \)
(b) \( \text{C}_n(\text{H}_2\text{O})_n \)
(c) \( \text{C}_n(\text{H}_2\text{O})_{2n} + 2 \)
(d) \( \text{C}_n(\text{H}_2\text{O})_{2n} - 2 \)
Answer: (b) \( \text{C}_n(\text{H}_2\text{O})_n \)
In simple words: The basic recipe for carbohydrates is that they have carbon and then hydrogen and oxygen in the same ratio as water molecules.

🎯 Exam Tip: The formula \( \text{C}_n(\text{H}_2\text{O})_n \) highlights why carbohydrates are sometimes called "hydrates of carbon."

 

Question 3. Green leaves of plants, during photosynthesis synthesise
(a) Vitamins
(b) Enzymes
(c) Carbohydrates
(d) Lipids
Answer: (c) Carbohydrates
In simple words: Plants make carbohydrates (sugars) in their green leaves using sunlight, water, and air. This is how they get food and energy.

🎯 Exam Tip: Photosynthesis is the process plants use to make their food, and that food is essentially carbohydrates.

 

Question 4. During photosynthesis, carbondioxide and water are converted into
(a) Sucrose
(b) Fructose
(c) Maltose
(d) Glucose
Answer: (d) Glucose
In simple words: Photosynthesis takes carbon dioxide and water and turns them into glucose, which is a simple sugar.

🎯 Exam Tip: Glucose is the direct product of photosynthesis, though it can be further processed into other sugars or complex carbohydrates.

 

Question 5. The number of optical isomers of a carbohydrate is given by the formula \( 2^n \), where n is the number of
(a) Carbon atoms
(b) Chiral Carbon atoms
(c) achiral carbon atom
(d) hydroxy group
Answer: (b) Chiral Carbon atoms
In simple words: The formula \( 2^n \) tells us how many different forms a carbohydrate can have that can twist light. The 'n' in the formula stands for the number of carbon atoms that are "chiral," meaning they are attached to four different things.

🎯 Exam Tip: Understanding chiral centers is fundamental to stereochemistry and optical isomerism in carbohydrates.

 

Question 6. Which among the following is not a monosaccharide?
(a) Fructose
(b) Erythrose
(c) Maltose
(d) Ribose
Answer: (c) Maltose
In simple words: Maltose is not a single sugar unit; it's made of two sugar units joined together. Fructose, erythrose, and ribose are all single sugar units.

🎯 Exam Tip: Memorize common examples of monosaccharides (glucose, fructose, ribose), disaccharides (maltose, sucrose, lactose), and polysaccharides (starch, cellulose) to easily distinguish them.

 

Question 7. The medicinal value of a drug is measured in terms of its
(a) Deoxyribose
(b) Goldnumber
(c) Therapeutic index
(d) Equilibrium constant
Answer: (c) Therapeutic index
In simple words: How good and safe a medicine is can be measured by its "therapeutic index." This number tells us the difference between how much medicine works and how much is harmful.

🎯 Exam Tip: The therapeutic index is a critical concept in pharmacology, indicating a drug's safety profile; a higher value is generally preferred.

 

Question 8. Fructose is a
(a) aldopentose
(b) ketopentose
(c) aldohexose
(d) ketohexose
Answer: (d) ketohexose
In simple words: Fructose is a type of sugar that has six carbon atoms and a ketone group, which is a specific kind of chemical group.

🎯 Exam Tip: Understand the naming conventions: "aldo-" means aldehyde, "keto-" means ketone, and "-pentose" or "-hexose" refers to the number of carbons (five or six, respectively).

 

Question 9. Which is known as blood sugar?
(a) Glucose
(b) fructose
(c) sucrose
(d) maltose
Answer: (a) Glucose
In simple words: Glucose is often called "blood sugar" because it's the main sugar our body uses for energy, flowing in our blood.

🎯 Exam Tip: Glucose is the central energy molecule in human metabolism, making its common name "blood sugar" very fitting.

 

Question 10. Sucrose undergoes hydrolysis when boiled with dil.\( \text{H}_2\text{SO}_4 \) to give
(a) Glucose
(b) fructose
(c) both (a) & (b)
(d) maltose
Answer: (c) both (a) & (b)
In simple words: If you heat sucrose with a little acid, it breaks down into two simpler sugars: glucose and fructose.

🎯 Exam Tip: Sucrose is a disaccharide made of one glucose unit and one fructose unit. Hydrolysis separates these two.

 

Question 11. The number of asymmetric carbon atoms present in glucose are
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4
In simple words: Glucose has four special carbon atoms that are "asymmetric," meaning they are attached to four different things. This helps give glucose its unique shape.

🎯 Exam Tip: To find asymmetric carbons, identify carbons bonded to four different groups. In the linear form of glucose, carbons 2-5 are chiral.

 

Question 12. Glucose is
(a) dextro rotatory
(b) laevorotatory
(c) optically inactive
(d) non-reducing sugar
Answer: (a) dextro rotatory
In simple words: Glucose can spin light to the right. Scientists call this "dextrorotatory."

🎯 Exam Tip: The "D" in D-glucose refers to the configuration at its lowest chiral center, while the "(+)" or "dextro-" refers to its optical rotation. These are not always directly related but happen to align for glucose.

 

Question 13. Glucose [O] (\( \text{Br}_2/\text{H}_2\text{O} \)) \( \rightarrow \) A (Con \( \text{HNO}_3 \)) \( \rightarrow \) B
(a) Saccharic acid, Gluconic acid
(b) Gluconic acid, Saccharic acid
(c) Glycollic acid, Tartaric acid
(d) Tartaric acid, Glycollic acid
Answer: (b) Gluconic acid, Saccharic acid
In simple words: First, mild oxidation of glucose gives gluconic acid. Then, strong oxidation of gluconic acid makes saccharic acid. This shows different parts of glucose can be oxidized.

🎯 Exam Tip: Remember that bromine water is a mild oxidizing agent that only affects the aldehyde group of glucose, while concentrated nitric acid is a strong oxidizing agent that oxidizes both the aldehyde and the primary alcohol groups.

 

Question 14. The isomers called anomers differ in the configuration of
(a) C1 carbon
(b) C2 carbon
(c) C4 carbon
(d) C5 carbon
Answer: (a) C1 carbon
In simple words: Anomers are sugar molecules that are almost identical but differ only in how the atoms are arranged around their first carbon atom, called the anomeric carbon.

🎯 Exam Tip: Anomers are a type of stereoisomer unique to cyclic sugars, defined by the orientation at the anomeric carbon (C1 for aldoses, C2 for ketoses).

 

Question 15. The cyclic structure of glucose is similar to
(a) furan
(b) pyran
(c) pyridine
(d) pyrrole
Answer: (b) pyran
In simple words: Glucose, when it forms a ring shape, looks a lot like a chemical called pyran. Both have a ring with five carbons and one oxygen.

🎯 Exam Tip: The terms "pyranose" (six-membered ring) and "furanose" (five-membered ring) are used to describe the cyclic forms of sugars based on their resemblance to pyran and furan, respectively.

 

Question 16. Sugars differing in configuration at an asymmetric centre are known as
(a) anomers
(b) epimers
(c) enantiomers
(d) diasteromers
Answer: (b) epimers
In simple words: Epimers are like sugar twins that are different in just one spot where atoms are arranged.

🎯 Exam Tip: Epimers are a specific type of stereoisomer. Make sure to distinguish them from anomers (differ at anomeric carbon) and enantiomers (non-superimposable mirror images).

 

Question 17. D mannose and D-glucose are
(a) C-2 epimers
(b) C-3 epimers
(c) C-4 epimers
(d) C-5 epimers
Answer: (a) C-2 epimers
In simple words: D-mannose and D-glucose are very similar sugars, but they are different at their second carbon atom.

🎯 Exam Tip: Visualize or draw the open-chain Fisher projections of D-glucose and D-mannose to confirm the difference at the C2 position.

 

Question 18. D-glucose and D-galactose are
(a) C-2 epimers
(b) C-3 epimers
(c) C-4 epimers
(d) C-5 epimers
Answer: (c) C-4 epimers
In simple words: D-glucose and D-galactose are types of sugar that are almost the same, but they have a different arrangement of atoms only at their fourth carbon.

🎯 Exam Tip: Know the specific chiral carbons where common epimers like glucose/mannose (C2) and glucose/galactose (C4) differ.

 

Question 19. In our body galactose is coverted into glucose by
(a) hydrolysis
(b) mutrarotation
(c) fermentation
(d) epimerisation
Answer: (d) epimerisation
In simple words: Our body changes galactose into glucose using a special process called epimerisation, which is like flipping the atoms at one specific carbon.

🎯 Exam Tip: Recognize epimerisation as a key metabolic pathway for interconverting sugars that are epimers, like glucose and galactose.

 

Question 20. Fructose is
(a) dexto rotatory
(b) laevo rotatory
(c) optically inactive
(d) disaccharide
Answer: (b) laevo rotatory
In simple words: Fructose is a sugar that makes polarized light turn to the left, which is called "laevorotatory."

🎯 Exam Tip: Be careful not to confuse D/L configuration with dextro/laevo rotation; they are determined differently. Fructose is D-fructose but is levorotatory.

 

Question 21. Glucose and fructose are
(c) Functional isomers
(d) tautomers
Answer: (c) Functional isomers
In simple words: Glucose and fructose are "functional isomers" because they have the same atoms but different main chemical groups; glucose has an aldehyde group, and fructose has a ketone group.

🎯 Exam Tip: Functional isomers have the same molecular formula but different functional groups, leading to different chemical properties.

 

Question 22. Fruit sugar is
(a) glucose
(b) fructose
(c) ribose
(d) erythrose
Answer: (b) fructose
In simple words: Fructose is also called "fruit sugar" because it's found in many fruits and is very sweet.

🎯 Exam Tip: Fructose is recognized as the sweetest natural sugar and is abundant in fruits.

 

Question 23. Equal amount of glucose and fructose is termed as
(a) Fruit sugar
(b) blood sugar
(c) invert sugar
(d) cane sugar
Answer: (c) invert sugar
In simple words: When you have the same amount of glucose and fructose mixed together, it's called invert sugar. This happens when sucrose breaks down.

🎯 Exam Tip: Invert sugar is commonly used in food industries because it is sweeter than sucrose and less prone to crystallization.

 

Question 24. Sucrose is converted into glucose and fructose by the enzyme
(b) invertase
(c) lactase
(d) fructase
Answer: (b) invertase
In simple words: The enzyme called invertase helps break down sucrose sugar into two simpler sugars, glucose and fructose.

🎯 Exam Tip: Enzymes are highly specific; knowing which enzyme acts on which substrate (e.g., invertase on sucrose) is important.

 

Question 25. Partial reduction of fructose with sodium amalgam and water produces mixture of
(a) D-Glucose and D mannose
(b) Sorbitol and Mannitol
(c) Glycollic acid and Tartaric acid
(d) D-Glucose and D-Galactose
Answer: (b) Sorbitol and Mannitol
In simple words: When fructose is partially reduced using sodium amalgam and water, it turns into a mix of two sugar alcohols: sorbitol and mannitol. This happens because a new chiral center is created.

🎯 Exam Tip: Understand that the reduction of ketoses can create new chiral centers, leading to mixtures of diastereomeric alditols.

 

Question 26. Sorbitol and Mannitol are
(a) enantiomers
(b) tautomers
(c) epimers
(d) functional isomers
Answer: (c) epimers
In simple words: Sorbitol and mannitol are called epimers because they are exactly alike except for how one specific hydroxyl group is placed on their second carbon atom.

🎯 Exam Tip: Remember that epimers differ in configuration at only one chiral center.

 

Question 27. Which of the following statement incorrect?
(a) Nucleoside + Phosphate \( \rightarrow \) Nucleotide
(b) Nucleoside + Base \( \rightarrow \) Nucleotide
(c) Sugar + Base \( \rightarrow \) Nucleoside
(d) n Nucleoside \( \rightarrow \) Polynucleotide
Answer: (b) Nucleoside + Base \( \rightarrow \) Nucleotide
In simple words: A nucleoside plus a base does not make a nucleotide. A nucleotide is formed when a nucleoside is combined with a phosphate group, or a sugar combines with a base to form a nucleoside.

🎯 Exam Tip: Know the fundamental building blocks and their combinations in nucleic acids: Sugar + Base = Nucleoside; Nucleoside + Phosphate = Nucleotide.

 

Question 28. The number of asymmetric carbon atom present in fructose is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
In simple words: Fructose has three carbon atoms that are attached to four different groups, making them asymmetric or chiral. These chiral carbons are important for its optical activity.

🎯 Exam Tip: Remember to identify asymmetric carbon atoms by checking if they are bonded to four different groups.

 

Question 29. Fructose forms a five membered ring similar to
(a) Furan
(b) Pyran
(c) Pyridime
(d) Pyrrole
Answer: (a) Furan
In simple words: Fructose can create a ring shape that looks like furan, a five-membered ring with one oxygen atom. This furanose structure is common for fructose in solutions.

🎯 Exam Tip: Distinguish between furanose (five-membered ring with oxygen) and pyranose (six-membered ring with oxygen) forms for different sugars.

 

Question 30. The general formula of disaccharides is
(a) \( C_n(H_2O)_{2n} \)
(b) \( C_n(H_2O)_n \)
(c) \( C_n(H_2O)_{n-1} \)
Answer: (c) \( C_n(H_2O)_{n-1} \)
In simple words: Disaccharides are made by joining two simple sugar units, and when they link, one molecule of water is removed. So, their general formula shows one less water molecule compared to the sum of two monosaccharide formulas.

🎯 Exam Tip: Recall that the formation of a disaccharide from two monosaccharides involves the loss of one water molecule through a glycosidic linkage.

 

Question 31. In disaccharides, two monosaccharides are linked by
(a) hydrogen bond
(b) ionic bond
(c) peptide linkage
(d) glycosidic linkage
Answer: (d) glycosidic linkage
In simple words: Two simple sugar molecules connect to form a disaccharide through a special bond called a glycosidic linkage. This bond is formed when a water molecule is taken out between them.

🎯 Exam Tip: Understand that glycosidic linkage is a key feature of carbohydrate polymers, connecting sugar units.

 

Question 32. Honey is a mixture of
(i) glucose
(ii) fructose
(iii) sucrose
(iv) maltose
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (i (ii) & (iii)
(d) (i), (ii), (iv)
Answer: (c) (i), (ii), (iii)
In simple words: Honey naturally contains glucose, fructose, and a small amount of sucrose. These are the main sugars that give honey its sweet taste and properties.

🎯 Exam Tip: Remember that honey is primarily composed of monosaccharides (glucose and fructose) with a minor proportion of disaccharides like sucrose.

 

Question 33. Which carbon atoms of \( \alpha \)-D-glucose and \( \beta \)-D-fructose are joined together
(a) C1 and C2
(b) C1 and C4
(c) C1 and C5
(d) C2 and C4
Answer: (a) C1 and C2
In simple words: When \( \alpha \)-D-glucose and \( \beta \)-D-fructose link up to form sucrose, they connect at the first carbon (C1) of glucose and the second carbon (C2) of fructose. This specific connection makes sucrose a non-reducing sugar.

🎯 Exam Tip: Knowing the specific carbon atoms involved in glycosidic linkages helps understand the structure and properties (like reducing or non-reducing) of disaccharides.

 

Question 34. The disaccharide present in milk of mammals is
(a) glucose
(b) lactose
(c) maltose
(d) sucrose
Answer: (b) lactose
In simple words: Lactose is a special type of sugar found in milk from all mammals. It is often called "milk sugar" because of this.

🎯 Exam Tip: Associate lactose directly with milk and remember it is a disaccharide made of glucose and galactose.

 

Question 35. Which among the following is a non-reducing sugar
(a) Lactose
(b) Maltose
(c) Sucrose
(d) Glucose
Answer: (c) sucrose
In simple words: Sucrose is called a non-reducing sugar because its anomeric carbons (the carbon atoms involved in forming the ring and potentially acting as a reducing agent) are both tied up in the glycosidic bond. This means they cannot open up to form an aldehyde group that would allow them to act as a reducing agent.

🎯 Exam Tip: Remember that reducing sugars have a free anomeric carbon that can open to form an aldehyde group, while non-reducing sugars do not.

 

Question 36. On hydrolysis lactose gives
(i) galactose
(ii) glucose
(iii) fructose
(iv) maltose
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (a) (i) & (ii)
In simple words: When lactose breaks down with water, it splits into two smaller sugar units: galactose and glucose. This process happens in our bodies when we digest milk.

🎯 Exam Tip: Know the hydrolysis products of common disaccharides: lactose yields glucose and galactose, sucrose yields glucose and fructose, and maltose yields two glucose units.

 

Question 37. Which is produced during digestion of starch by the enzyme \( \alpha \) – amylase?
(a) Glucose
(b) Maltose
(c) Lactose
(d) Sucrose
Answer: (b) Maltose
In simple words: When our bodies digest starch, the enzyme \( \alpha \)-amylase breaks it down into smaller sugar units, mainly maltose. Maltose is a disaccharide made of two glucose units.

🎯 Exam Tip: Remember the role of \( \alpha \)-amylase in starch digestion and its primary product, maltose.

 

Question 38. Which carbon atoms of \( \alpha \)-D glucose are linked to form maltose?
(a) C1 and C2
(b) C1 and C3
(c) C1 and C4
(d) C3 and C4
Answer: (c) C1 and C4
In simple words: Maltose is formed when two glucose molecules connect. The first carbon (C1) of one glucose molecule joins with the fourth carbon (C4) of the other glucose molecule. This specific linkage is important for its structure.

🎯 Exam Tip: Understand the \( \alpha \)-(1,4) glycosidic linkage that forms maltose from two glucose units.

 

Question 39. \( \alpha \)-D-glucose and \( \beta \)-D-fructose are joined together to form which sugar?
(a) glucose
(b) cellulose
(c) sucrose
(d) fructose
Answer: (c) sucrose
In simple words: When one molecule of \( \alpha \)-D-glucose and one molecule of \( \beta \)-D-fructose combine, they form the common table sugar known as sucrose. This joining happens by losing a water molecule.

🎯 Exam Tip: Identify sucrose as the disaccharide formed from glucose and fructose, noting its common name as table sugar.

 

Question 40. Starch contains
(a) 80% amylopectin and 20% amylose
(b) 80% amylose and 20% amylopectin
(c) 50% amylopectin and 50% amylose
(d) 80% glucose and 20% fructose
Answer: (a) 80% amylopectin and 20% amylose
In simple words: Starch is made up of two types of molecules: amylose and amylopectin. Most of the starch (about 80%) is amylopectin, which is branched, while a smaller part (about 20%) is amylose, which is a straight chain.

🎯 Exam Tip: Remember the two main components of starch, amylose (linear) and amylopectin (branched), and their approximate proportions.

 

Question 41. Starch is a polymer of glucose molecules linked by
(a) \( \alpha \) (1, 2) glycosidic bonds
(b) \( \alpha \) (1, 3) glycosidic bonds
(c) \( \alpha \) (1, 4) glycosidic bonds
(d) \( \alpha \) (1, 5) glycosidic bonds
Answer: (c) \( \alpha \) (1, 4) glycosidic bonds
In simple words: Starch is a long chain of many glucose units. These glucose units are mainly connected by \( \alpha \)-(1,4) glycosidic bonds, which form a straight or slightly branched chain.

🎯 Exam Tip: Focus on the \( \alpha \)-(1,4) glycosidic linkage as the primary bond in starch, responsible for its polymeric structure.

 

Question 42. Number of \( \alpha \)-D glucose molecules joined by \( \alpha \) (1,4) glycosidic bonds in amylose and amylopectin are respectively
(a) 200 and 1000
(b) up to 4000 and 10,000
(c) up to 1000 and 400
(d) up to 400 and 1000
Answer: (b) up to 4000 and 10,000
In simple words: Amylose, a part of starch, can have up to 4000 glucose units linked by \( \alpha \)-(1,4) bonds. Amylopectin, the branched part, is much larger and can have up to 10,000 glucose units connected in the same way, plus additional branch points. Both structures are crucial for energy storage.

🎯 Exam Tip: Recognize that amylose and amylopectin are both polymers of glucose but differ greatly in the number of glucose units and branching. Amylopectin's size is much larger.

 

Question 43. With iodine solution amylose and amylopectin give respectively
(a) blue and red colour
(b) violet and red colour
(c) blue and purple colour
(d) violet and yellow colour
Answer: (c) blue and purple colour
In simple words: When you add iodine solution to starch, amylose turns blue. Amylopectin, due to its branched structure, gives a purple or reddish-purple color. These color changes are used to tell them apart.

🎯 Exam Tip: Remember the distinctive iodine test colors for amylose (blue) and amylopectin (purple) as a common way to differentiate them.

 

Question 44. Major constituent of plant cell walls is
(a) starch
(b) cellulose
(c) glycogen
(d) amylose
Answer: (b) cellulose
In simple words: Cellulose is the main material that makes up the strong walls of plant cells. It provides structure and support to plants, allowing them to stand upright.

🎯 Exam Tip: Cellulose is a polysaccharide (complex carbohydrate) and a primary structural component of plant cell walls, unlike starch which is for energy storage.

 

Question 45. In cellulose, glucose molecules are linked by
(a) \( \alpha \) (1, 2) glucosidic bond
(b) \( \alpha \) (1, 4) glycosidic bond
(c) \( \beta \) (1, 2) glycosidic bond
(d) \( \beta \) (1,4) glycosidic bond
Answer: (d) \( \beta \) (1,4) glycosidic bond
In simple words: Cellulose is built from many glucose units connected by \( \beta \)-(1,4) glycosidic bonds. This specific type of bond makes cellulose strong and tough, which is why it's great for plant cell walls.

🎯 Exam Tip: Note the crucial difference between \( \alpha \) and \( \beta \) glycosidic bonds; \( \beta \)-(1,4) linkages are characteristic of cellulose and make it indigestible by many organisms.

 

Question 46. Cotton is almost pure
(a) glucose
(b) starch
(c) cellulose
(d) amylose
Answer: (c) cellulose
In simple words: Cotton fibers are made almost entirely of cellulose. This is why cotton is strong and useful for making clothes and other textiles.

🎯 Exam Tip: Remember that cotton is a natural source of nearly pure cellulose, highlighting the prevalence of this polysaccharide.

 

Question 47. Gun cotton is
(a) cellulose acetate
(b) cellulose nitrate
(c) ethyl cellulose
(d) cellulose bromide
Answer: (b) cellulose nitrate
In simple words: Gun cotton is a powerful explosive made by treating cellulose with nitric acid. It's essentially a modified form of cellulose used for propellants.

🎯 Exam Tip: Understand that gun cotton is a derivative of cellulose, specifically cellulose nitrate, known for its explosive properties.

 

Question 48. Animal starch is
(a) glucose
(b) glycogen
(c) amylose
(d) amylopectin
Answer: (b) glycogen
In simple words: Glycogen is the main way animals store glucose for energy, just like plants use starch. It's often called "animal starch" because of this similar role.

🎯 Exam Tip: Glycogen is a highly branched polysaccharide and the primary storage form of glucose in animals, found mainly in the liver and muscles.

 

Question 49. Protein are polymers of
(a) \( \alpha \)-amino acids
(b) \( \beta \)-amino acids
(c) \( \alpha \)-hydroxy acids
(d) \( \beta \)-hydroxy acids
Answer: (a) \( \alpha \)-amino acids
In simple words: Proteins are large molecules made from many small units called amino acids. Specifically, they are built from \( \alpha \)-amino acids linked together in long chains.

🎯 Exam Tip: Remember that all naturally occurring proteins are polymers of \( \alpha \)-amino acids, where the amino group and carboxyl group are attached to the same carbon atom (the \( \alpha \)-carbon).

 

Question 50. The amino acids that can be synthesised by humans are called
(a) essential amino acids
(b) non-essential amino acids
(c) polar amino acids
(d) non-polar amino acids
Answer: (b) non-essential amino acids
In simple words: Our bodies can make non-essential amino acids on their own. This means we don't need to get them from the food we eat.

🎯 Exam Tip: Distinguish between essential amino acids (must be obtained from diet) and non-essential amino acids (synthesized by the body).

 

Question 51. At isoelectric point an amino acid exists as
(a) positive ion
(b) negative ion
(c) zwitter ion
(d) protein
Answer: (c) zwitter ion
In simple words: At its isoelectric point, an amino acid has both a positive and a negative charge within the same molecule, but its total net charge is zero. This special form is called a zwitterion.

🎯 Exam Tip: The isoelectric point (pI) is the pH at which an amino acid exists predominantly as a zwitterion, with equal positive and negative charges, resulting in a net zero charge.

 

Question 52. The amino acid which contains achiral carbon is
(a) alanine
(b) gylcine
(c) phenyl alanine
(d) valine
Answer: (b) glycine
In simple words: Glycine is the only amino acid that does not have a chiral carbon. This is because its \( \alpha \)-carbon is attached to two hydrogen atoms, making it symmetrical.

🎯 Exam Tip: Remember that a chiral carbon is bonded to four different groups. Glycine's central carbon is bonded to two hydrogen atoms, so it is achiral.

 

Question 53. Which is optically inactive?
(a) alanine
(b) glycine
(c) phenyl alanine
(d) valine
Answer: (b) glycine
In simple words: Glycine is optically inactive because it does not have a chiral carbon atom. Without a chiral carbon, it cannot rotate plane-polarized light.

🎯 Exam Tip: Optical activity is directly linked to the presence of chiral centers in a molecule; achiral molecules like glycine are optically inactive.

 

Question 54. A peptide bond is
(a) -C-NH\( _2 \)
(b) -C-O-NH-
(c) -C-NH-
(d) -CO-NH\( _2 \)
Answer: (c) -C-NH-
In simple words: A peptide bond is the link that connects amino acids together to form proteins. It is made by joining the carboxyl group of one amino acid to the amino group of another, releasing a water molecule.

🎯 Exam Tip: Recognize the specific structure of a peptide bond as an amide linkage (-CO-NH-) formed by a dehydration reaction.

 

Question 55. Number of peptide bonds present in a dipeptide is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
In simple words: A dipeptide is made of two amino acids joined together. Because there are only two units, there is only one link, or one peptide bond, connecting them.

🎯 Exam Tip: A dipeptide means two amino acid units, which always results in exactly one peptide bond between them.

 

Question 56. In proteins \( \alpha \)-helix and \( \beta \)-strands are two most common substructures present in
(a) Primary structure
(b) Secondary structure
(c) Tertiary structure
(d) Quaternary structure
Answer: (b) Secondary structure
In simple words: The \( \alpha \)-helix and \( \beta \)-strands are the most common ways a protein chain folds into smaller, repeating patterns. These patterns form the protein's secondary structure, stabilized by hydrogen bonds.

🎯 Exam Tip: Remember that secondary structures (alpha-helix and beta-sheet) describe the local folding patterns of the polypeptide chain, stabilized by hydrogen bonds between backbone atoms.

 

Question 57. Enzymes are
(a) antibodies
(b) transporters
(c) bio catalysts
(d) receptors
Answer: (c) bio catalysts
In simple words: Enzymes are special proteins that speed up chemical reactions in living things without being used up themselves. They act like helpful tools for our body's processes.

🎯 Exam Tip: Define enzymes as biological catalysts, emphasizing their role in accelerating biochemical reactions without being consumed.

 

Question 58. Which among the following is water soluble vitamin?
(a) Vitamin A
(b) Vitamin B
(c) Vitamin D
(d) Vitamin
Answer: (b) Vitamin B
In simple words: Vitamin B is a group of vitamins that dissolve easily in water. Our bodies don't store them much, so we need to get them regularly from our diet.

🎯 Exam Tip: Classify vitamins into water-soluble (B and C) and fat-soluble (A, D, E, K), understanding the implications for storage and daily intake.

 

Question 59. Which is essential for blood clotting?
(a) Vitamin A
(b) Vitamin B
(c) Vitamin D
(d) Vitamin K
Answer: (d) Vitamin K
In simple words: Vitamin K plays a very important role in helping our blood clot. Without enough Vitamin K, our blood might not clot properly, leading to excessive bleeding.

🎯 Exam Tip: Link Vitamin K directly to its primary function in coagulation, ensuring sufficient intake for healthy blood clotting.

 

Question 60. The chemical name of vitamin B6 is
(a) Niacin
(b) Biotin
(c) Pyridoxine
(d) Folic acid
Answer: (c) Pyridoxine
In simple words: Pyridoxine is the chemical name for Vitamin B6. This vitamin is important for many body functions, including brain development and immune health.

🎯 Exam Tip: Familiarize yourself with the common names and corresponding chemical names for important vitamins.

 

Question 62. The pyrimidine bases present in DNA are
(i) Cytosine
(ii) Thymine
(iii) Uracil
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (a) (i) & (ii)
In simple words: DNA contains two main pyrimidine bases: cytosine and thymine. These bases pair up with purine bases to form the genetic code.

🎯 Exam Tip: Remember the four bases in DNA: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T). Uracil (U) replaces Thymine in RNA.

 

Question 63. A nucleotide is derived from a nucleoside by the addition of a molecule of
(a) ribose
(b) deoxy ribose
(c) phosphoric acid
(d) purine
Answer: (c) phosphoric acid
In simple words: A nucleoside is a sugar joined to a base. To turn it into a nucleotide, a phosphate group (from phosphoric acid) is added to the sugar part.

🎯 Exam Tip: Understand the building blocks of nucleic acids: Base + Sugar = Nucleoside; Nucleoside + Phosphate = Nucleotide.

 

Question 64. Which of the following is an intercellular signalling molecule?
(a) Enzymes
(b) Hormone
(c) Protein
(d) Vitamin
Answer: (b) Hormone
In simple words: Hormones are like messengers that cells use to talk to each other across the body. They travel through the bloodstream and tell other cells what to do.

🎯 Exam Tip: Hormones are vital for intercellular communication, regulating various physiological processes by transmitting signals between cells.

 

Question 65. Which among the following is a steroidal hormone?
(a) insulin
(b) epinephrine
(c) estrogen
(d) adenine
Answer: (c) estrogen
In simple words: Estrogen is a type of hormone that is made from cholesterol, which gives it a specific structure called a steroid. It plays a big role in female reproductive health.

🎯 Exam Tip: Steroidal hormones are lipid-soluble and derived from cholesterol, characterized by a distinct four-ring structure.

 

Question 66. The sweetest of all known sugars is
(a) glucose
(b) sucrose
(c) fructose
(d) ribose
Answer: (c) fructose
In simple words: Fructose is naturally found in fruits and honey, and it is known to be the sweetest among all natural sugars. This is why fruits taste so sweet.

🎯 Exam Tip: Remember fructose as the sugar with the highest sweetness intensity, commonly found in fruits and honey.

 

Question 1. (i) D-glucose is so named because the H and OH on C5 carbon are in the same configuration as the H and OH on C2 carbon in D-Glycerldehyde.
(ii) Dextro rotatory compounds rotate the plane of polarised light in anti clock wise direction.
(iii) Laevo rotatory compounds rotate the plane of polarised light in clockwise direction.
(iv) The D or L isomers can either be dextro or laevo rotatory compounds.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Correct statements:
Answer: (d) (i) & (iv)
In simple words: D-glucose gets its "D" name because its structure at carbon C5 matches D-Glyceraldehyde, meaning the OH is on the right. Also, D or L forms don't always determine how light rotates; some D-sugars can still be levorotatory. The direction of rotation is not directly related to D/L configuration.

🎯 Exam Tip: D/L configuration refers to the arrangement around the furthest chiral carbon from the aldehyde/ketone group, while dextrorotatory/levorotatory describes experimental optical rotation.

 

Question 2. (i) During hydrolysis of sucrose the optical rotation of the reaction mixture changes from levo to dextro.
(ii) Honey bees have the enzyme invertase that catalyzes the hydrolysis of sucrose to glucose and fructose.
(iii) In sucrose C1 of \( \alpha \)-D-glucose is joined to C2 of \( \beta \)-D-fructose
(iv) Sucrose is a reducing sugar.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Correct statements:
Answer: (b) (ii) & (iii)
In simple words: Honey bees use an enzyme called invertase to break down sucrose into glucose and fructose. Also, in sucrose, the C1 carbon of \( \alpha \)-D-glucose is indeed linked to the C2 carbon of \( \beta \)-D-fructose. The optical rotation actually changes from dextro to levo during sucrose hydrolysis, making the first statement incorrect. Sucrose is a non-reducing sugar, making the fourth statement incorrect as well.

🎯 Exam Tip: Understand that the "inversion of sugar" refers to the change in optical rotation from positive to negative after sucrose hydrolysis, and remember that sucrose is a non-reducing sugar.

Question 3.
(i) Fibrous proteins are linear molecules similar to fibres.
(ii) Fibrous proteins are generally soluble in water and are held together by disulphide bridges and weak intermolecular hydrogen bonds.
(iii) Globular proteins have an over all spherical shape.
(iv) Globular proteins are usually insoluble in water and have many functions including catalysis.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer: (b) (i) & (iii)
In simple words: Fibrous proteins look like long threads, and globular proteins are round in shape. Both statements describe how these proteins appear.

🎯 Exam Tip: Remember that fibrous proteins are often insoluble and have structural roles, while globular proteins are usually soluble and perform functional roles like enzymes.

 

Question 4.
i) Enzymes activate a reaction by increasing the activation energy by destabilizing the transition state.
ii) Many biochemical reactions are catalyzed by special proteins called enzymes.
iii) Sucrase enzyme catalyses the hydrolysis of lactose into glucose and galactose
iv) Enzymes are specific.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer: (c) (ii) & (iv)
In simple words: Enzymes are special proteins that speed up many body reactions. Also, each enzyme usually only works on one specific type of molecule or reaction, like a special key for a lock.

🎯 Exam Tip: Always remember that enzymes are biological catalysts, meaning they help reactions happen faster in living things without being used up themselves. Their specificity is a key feature.

III. Pick out the InCorrect Statement

 

Question 1.
i) Mono-saccharides are carbohydrates that cannot be hydrolysed further and are called simple sugars.
ii) Glucose is a ketohexose.
iii) Glucose contain one primary alcoholic group and four secondary alcoholic group.
iv) Glucose is levulose
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer: (c) (ii) & (iv)
In simple words: Glucose is actually an aldohexose, not a ketohexose, and its other name is dextrose, not levulose (which is fructose). So, statements (ii) and (iv) are incorrect.

🎯 Exam Tip: Distinguish between aldohexoses (like glucose) and ketohexoses (like fructose) based on their aldehyde or ketone functional groups. Remember common names like dextrose for glucose and levulose for fructose.

 

Question 2.
i) Amino acids obtained through diet are called non-essential amino acids.
ii) Amino acids have amphoteric behaviour.
iii) Iso electric point is the pH value at which the Zwitter ion of amino acid won't move towards an electrode.
iv) pH above the isoelectric point the amino acid is positively charged.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer: (d) (i) & (iv)
In simple words: Amino acids we get from food are called essential because our body can't make them. Also, if the pH is higher than the isoelectric point, the amino acid will have a negative charge, not a positive one.

🎯 Exam Tip: Understand the difference between essential and non-essential amino acids. Also, remember that above its isoelectric point, an amino acid has a net negative charge, while below it, it has a net positive charge.

 

Question 3.
i) In the α-helix sub structure, the amino acids are arranged in a left handed helical structure.
ii) Each turn of an α- helix contains about 6.3 residues and is about 4.5 A long.
iii) In the α-helix sub structure, the amino acids are arranged in a right handed helical structure.
iv) Each turn of an α- helix contains about 3.6 residues and is about 5.4 A long.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer: (a) (i) & (ii)
In simple words: The alpha-helix structure in proteins is actually right-handed, not left-handed. Also, each turn of the helix has about 3.6 amino acid parts and is about 5.4 Angstroms long, not 6.3 residues and 4.5 Angstroms.

🎯 Exam Tip: Key features of the α-helix include its right-handed twist, about 3.6 amino acid residues per turn, and a pitch (length of one turn) of approximately 5.4 Å (Angstroms).

 

Question 4.
i) Each vitamin has a specific function in the system, mostly as co-enzymes.
ii) Water soluble vitamins can be stored in fatty tissues and livers.
iii) Excess of fat soluble vitamins will be excreted through urine and not stored in our body.
iv) Vitamins are small organic compounds that can not be synthesised by our body.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: Water-soluble vitamins are not stored much in the body and are usually flushed out. On the other hand, fat-soluble vitamins can be stored in fatty parts of the body, so having too much can be harmful.

🎯 Exam Tip: Understand the key difference: water-soluble vitamins (like B and C) are generally not stored and need regular intake, while fat-soluble vitamins (A, D, E, K) are stored in the body's fat and liver, meaning excessive intake can be toxic.

IV. Assertion & Reason

 

Question. 1. Assertion : Thymine pairs with Adenine whereas Cytosine pairs with Guanine in DNA molecule. Reason: The hydrogen bonding between bases of two strands is highly specific.
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of assertion.
c) Assertion is true and reason is false.
d) Both assertion and reason are false
Answer: (a) If both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: The statement that Adenine always pairs with Thymine, and Cytosine with Guanine in DNA, is true. This happens because the hydrogen bonds between these specific pairs are very strong and fit together perfectly.

🎯 Exam Tip: Remember Chargaff's rules: the amount of A equals T, and the amount of C equals G. This specificity is crucial for DNA's stable double helix structure and accurate replication.

 

Question. 2. Assertion (A): Glycyl alanine is a dipeptide Reason (R): In glycyl alanine the amino group of glycine react with the carboxyl group of alanine to form a peptide linkage.
a) Both A and R are correct, R explains A.
b) Both A and R are correct but R does not explain A.
c) A is correct but R is wrong.
d) A is wrong but R is correct
Answer: (c) A is correct but R is wrong.
In simple words: Glycyl alanine is indeed a dipeptide made from glycine and alanine. However, when forming glycylalanine, the carboxyl group of glycine reacts with the amino group of alanine to create the bond. The reason given describes the opposite reaction.

🎯 Exam Tip: For dipeptides, the name indicates the order of amino acids from the N-terminus (left) to the C-terminus (right). In 'glycylalanine', glycine is the N-terminal residue, and alanine is the C-terminal residue. The peptide bond forms between the carboxyl group of the N-terminal amino acid and the amino group of the C-terminal amino acid.

 

Question. 3. Assertion (A): A nucleotide is derived from a nucleoside by the addition of a molecule of phosphoric acid. Reason (R) : A nucleoside is a molecule without the phosphate group.
a) Both A and R are correct, R explains A.
b) Both A and R are correct but R does not explain A.
c) A is correct but R is wrong.
d) A is wrong but R is correct
Answer: (a) Both A and R are correct, R explains A.
In simple words: A nucleoside is made of a sugar and a base. When you add a phosphate group to it, you get a nucleotide. This is true because a nucleoside does not have a phosphate group on its own.

🎯 Exam Tip: Remember the basic building blocks: Base + Sugar = Nucleoside. Base + Sugar + Phosphate = Nucleotide. The addition of a phosphate group is the defining difference.

 

Question. 4. Assertion (A) : The specific association of bases A and T, C and G of the two chains of the double helix is known as complementary base pairing. Reason (R): There are 10.5 base pairs per turn of the helix of DNA.
a) Both A and R are correct, R explains A.
b) Both A and R are correct but R does not explain A.
c) A is correct but R is wrong.
d) A is wrong but R is correct
Answer: (b) Both A and R are correct but R does not explain A.
In simple words: It is true that DNA bases pair up specifically (A with T, C with G). It is also true that there are about 10.5 base pairs in one full twist of the DNA ladder. However, the number of base pairs per turn doesn't explain why they pair up specifically.

🎯 Exam Tip: While both statements are facts about DNA structure, ensure the reason directly explains the assertion. The number of base pairs per turn describes the physical dimension of the helix, not the chemical principle of base pairing.

V. Match the following

 

Question 1. Type of Sugar Example

Answer:
i. – Glyceraldehyde
ii. – Dihydroxy acetone
iii – Erythrose
iv – Erythrulose
v – Ribose
vi - Ribulose
vii - Glucose
viii - Fructose
In simple words: This list shows the correct pairings between different types of sugars based on their structure and functional groups, and their specific examples. For instance, aldotetrose is erythrose, and ketohexose is fructose.

🎯 Exam Tip: To master sugar classification, understand the terms 'aldo' (aldehyde group) and 'keto' (ketone group), and the prefixes 'triose', 'tetrose', 'pentose', 'hexose' (number of carbon atoms).

 

Question 2.

Answer:
i. - Beri - beri
ii. - Cheilosis
iii - Pernicious anaemia
iv. -Depression
v. – Scurvy
In simple words: This list correctly links each vitamin with the specific health problem caused by not having enough of it. For example, not enough Thiamine (Vitamin B1) causes Beri-beri, and a lack of Ascorbic acid (Vitamin C) causes Scurvy.

🎯 Exam Tip: Learning the specific deficiency diseases for each vitamin is important for understanding their roles in the body. Focus on the main, classic deficiency symptoms.

 

Question 3.

Answer:
i. – Thiamine
ii. - Riboflovin
iii. - Niacin
iv. – Pyridoxine
V. - Folic acid
In simple words: This table correctly matches each B vitamin with its scientific name. For instance, Vitamin B1 is Thiamine, and Vitamin B9 is Folic acid.

🎯 Exam Tip: Memorizing the chemical names of vitamins, especially the B-complex vitamins, is crucial. It helps in understanding their chemical structure and biochemical roles.

 

Question 4. Biomolecule Example

Answer:
i. – heparin
ii. – histidine
iii. – phylloquinone
iv. – androgen
v. – Folic acid
In simple words: This section lists different types of important biological molecules and provides examples for each, such as heparin for polysaccharides and histidine for amino acids. It helps us remember what belongs to which group.

🎯 Exam Tip: Categorizing biomolecules by their type (e.g., polysaccharide, amino acid, vitamin, hormone) and knowing common examples for each category is fundamental in biochemistry. Understand what defines each group.

VI. Two Mark Questions

 

Question 1. How do plants synthesise glucose?
Answer: Plants make glucose through a complex process called photosynthesis. During photosynthesis, the green leaves of plants use sunlight to change carbon-dioxide and water into glucose and oxygen. This glucose can then be used by the plant or converted into other carbohydrates for storage. Plants are amazing at converting light energy into chemical energy this way.
\[ 6CO_2 + 6H_2O \xrightarrow{Sunlight} C_6H_{12}O_6 + 6O_2 \]
In simple words: Plants make their food, glucose, by using sunlight, water, and carbon dioxide in a process called photosynthesis. They also release oxygen while doing this.

🎯 Exam Tip: For photosynthesis, remember the key reactants (carbon dioxide, water, sunlight) and products (glucose, oxygen). The balanced chemical equation is important to show complete understanding.

 

Question 2. What is muta rotation?
Answer: The specific rotation of α-D-glucose is \( 112^\circ \) and for β-D-glucose, it is \( 18.7^\circ \). When a pure form of either of these sugars is dissolved in water, they slowly change into each other through an open-chain structure. This interconversion continues until they reach a balance, with a constant specific rotation of \( +52.5^\circ \). This phenomenon, where the optical rotation of a sugar solution changes over time until it becomes constant, is called mutarotation.
In simple words: Mutarotation is when the light-bending ability of a sugar solution slowly changes until it settles at a fixed value. This happens because the sugar is changing between its different forms in the water.

🎯 Exam Tip: Key terms for mutarotation are 'specific rotation', 'α-D-glucose', 'β-D-glucose', 'open-chain structure', and 'equilibrium'. Mentioning the change in optical rotation over time is essential.

 

Question 3. What are epimers? What is epimerisation?
Answer: Sugars that are different in their arrangement around just one specific carbon atom (that is not the anomeric carbon, the C1 carbon in aldoses) are called epimers. For example, D-glucose and D-mannose are epimers because they only differ at the C2 carbon. The process where one epimer changes into another is known as epimerisation. This change usually needs special enzymes to happen. For example, in our bodies, galactose is converted into glucose this way.
In simple words: Epimers are sugars that are almost the same but have a different layout around only one carbon atom. Epimerisation is the process where one of these sugars changes into its epimer.

🎯 Exam Tip: Focus on the "one chiral carbon" distinction to define epimers accurately. Providing a clear example like glucose and mannose (C2 epimers) helps illustrate the concept.

 

Question 4. What is inversion of sucrose?
Answer: Sucrose, which is dextrorotatory, can be broken down by heating it with dilute sulfuric acid (\( H_2SO_4 \)) or by using the enzyme invertase. This process, called hydrolysis, yields equal amounts of D-glucose and D-fructose. Glucose is dextrorotatory, but fructose is strongly laevorotatory, meaning it rotates light to the left much more than glucose rotates it to the right. Because of this, the total optical rotation of the solution changes from right-handed to left-handed. The new mixture of glucose and fructose is called invert sugar, and the change in optical rotation is known as the inversion of sucrose.
\[ C_{12}H_{22}O_{11} + H_2O \xrightarrow{H_2SO_4 \text{ or Invertase}} C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)} \]
In simple words: Inversion of sucrose is when sucrose sugar is broken down into glucose and fructose. The solution starts by bending light to the right, but after breaking down, it bends light to the left. The new sugar mix is called invert sugar.

🎯 Exam Tip: Key points are the hydrolysis of sucrose, the products (glucose and fructose), and the change in optical rotation from positive to negative. Mentioning the enzyme invertase or acid catalysis is also important.

 

Question 5. Fructose is prepared commercially by the hydrolysis of Inuiin in acidic medium
Answer: Fructose is often prepared for commercial use by breaking down a substance called inulin. This breakdown happens when inulin is treated with acid. Inulin is a natural polymer of fructose units. This process releases many fructose molecules.
\[ (C_6H_{12}O_5)_n + n H_2O \xrightarrow{H^+} n C_6H_{12}O_6 \text{ (Fructose)} \]
In simple words: Fructose is made for sale by breaking down inulin using acid. Inulin is like a long chain of fructose molecules, and acid helps to cut them apart.

🎯 Exam Tip: Remember that inulin is a polymer of fructose and its hydrolysis yields fructose. The acidic medium acts as a catalyst for this breakdown.

 

Question 6. Why sucrose is called as invert sugar?
Answer: Sucrose is called invert sugar because of what happens when it is broken down. When sucrose is hydrolysed (broken down with water), it produces equal amounts of glucose and fructose. Sucrose itself rotates plane-polarized light to the right (dextrorotatory). Glucose also rotates light to the right, but fructose rotates light very strongly to the left (laevorotatory). Because fructose's leftward rotation is stronger than glucose's rightward rotation, the overall solution changes from rotating light to the right to rotating it to the left. This change in the direction of optical rotation is called inversion. Hence, the resulting mixture of glucose and fructose is referred to as invert sugar.
\[ \text{Sucrose} \xrightarrow{\text{Invertase}} \text{Glucose} + \text{Fructose} \]
In simple words: Sucrose is called invert sugar because when it breaks down, the way it bends light changes from right to left. This "inversion" happens because the fructose it creates is very strong at bending light to the left.

🎯 Exam Tip: The key concept for invert sugar is the change in optical rotation from positive to negative after sucrose hydrolysis, due to the higher laevorotatory power of fructose compared to the dextrorotatory power of glucose.

 

Question 7. Write about the structure of sucrose?
Answer: Sucrose is a disaccharide made from one molecule of α-D-glucose and one molecule of β-D-fructose. These two monosaccharide units are joined together by a glycosidic bond between the C-1 carbon of α-D-glucose and the C-2 carbon of β-D-fructose. This particular bond is known as an α-1, 2-glycosidic bond. Because the anomeric carbons of both glucose (C1) and fructose (C2) are involved in forming this bond, sucrose does not have a free aldehyde or ketone group. This means that sucrose is a non-reducing sugar. The specific linkage gives sucrose its unique properties.
In simple words: Sucrose is made of two smaller sugars, glucose and fructose, linked together. The special bond joining them uses up the parts that usually make other sugars reactive, so sucrose itself isn't a "reducing sugar".

🎯 Exam Tip: Remember the specific linkage (α-1, 2-glycosidic bond) in sucrose, and how it leads to sucrose being a non-reducing sugar because both anomeric carbons are involved in the bond.

 

Question 8. Write notes on lactose.
Answer: Lactose is a disaccharide, which means it's a sugar made from two simpler sugar units. It is found naturally in the milk of mammals, which is why it is often called "milk sugar." Lactose is formed by linking two monosaccharides: β-D-galactose and β-D-glucose. These two units are connected by a β-1, 4-glycosidic bond. One of the aldehyde carbons is still free and not involved in the glycosidic bond. This is why lactose retains its ability to act as a reducing sugar. Many people have trouble digesting lactose because they lack the enzyme lactase.

In simple words: Lactose is a sugar found in milk, made from galactose and glucose joined together. It is called a reducing sugar because it still has a part that can react, like an open end.

🎯 Exam Tip: Remember lactose's common name (milk sugar), its constituent monosaccharides (β-D-galactose and β-D-glucose), the type of glycosidic bond (β-1, 4), and its reducing nature.

 

Question 9. Write a note on maltose?
Answer: Maltose is known as malt sugar because it comes from malt. Sprouting barley is a main source of malt. The enzyme alpha-amylase helps make maltose when starch is digested. Maltose is made of two alpha-D-glucose units. These units are joined by an alpha-1,4-glycosidic bond. This bond connects the anomeric carbon of one glucose unit to the C-4 carbon of the other unit. Because one of the glucose units still has its aldehyde group, maltose acts as a reducing sugar.
In simple words: Maltose is a sugar that comes from malt. It is made of two glucose units connected in a special way. It can act as a reducing sugar.

🎯 Exam Tip: Remember that maltose is a disaccharide formed from two glucose units, and its reducing property is key in many biochemical tests.

 

Question 10. What are Polysaccharides? How are they classified?
Answer: Polysaccharides are large molecules made from many small sugar units (monosaccharides) joined by glycosidic bonds. They are the most common type of carbohydrates. Unlike simple sugars, they do not taste sweet and are often called non-sugars. These molecules can be arranged in straight lines or in branched chains. There are two main kinds:
(i) **Homopolysaccharides:** These are made from only one type of simple sugar unit. Examples include starch, cellulose, and glycogen.
(ii) **Heteropolysaccharides:** These are made from more than one different type of simple sugar unit. Examples include hyaluronic acid and heparin.
In simple words: Polysaccharides are big sugars made of many small sugar units. They don't taste sweet and can be straight or branched. They are split into two types based on how many kinds of sugar units they contain.

🎯 Exam Tip: To differentiate between homopolysaccharides and heteropolysaccharides, always identify if they are built from one type of monomer or multiple types.

 

Question 11. Write about starch.
Answer: Plants use starch to store energy. It is a vital food reserve for them. Common foods like potatoes, corn, wheat, and rice are excellent sources of starch. Starch is a large molecule (polymer) made of many glucose units connected by alpha-1,4-glycosidic bonds. It can be divided into two parts: amylose, which dissolves in water, and amylopectin, which does not. Starch typically has about 20% amylose and 80% amylopectin. Amylose is an unbranched chain of up to 4000 alpha-D-glucose molecules, linked by alpha-1,4-glycosidic bonds. Amylopectin is a branched molecule with up to 10,000 alpha-D-glucose units. These units are linked by alpha-1,4-glycosidic bonds in the main chain and by alpha-1,6-glycosidic bonds at the branch points. New branches containing 24 to 30 glucose molecules form at these alpha-1,6 linkages. When mixed with an iodine solution, amylose turns blue, while amylopectin turns purple.
In simple words: Starch is how plants store energy, found in foods like potatoes. It's made of many glucose units. It has two parts, amylose and amylopectin, which react differently with iodine.

🎯 Exam Tip: Remember the iodine test for starch (blue with amylose, purple with amylopectin) as a common identification method.

 

Question 12. Write briefly about cellulose.
Answer: Cellulose is the main material found in the cell walls of plants. It gives plants their structure. For example, cotton is almost entirely made of pure cellulose. When cellulose is broken down (hydrolysed), it produces D-glucose molecules. It is a long, unbranched chain polysaccharide where glucose units are connected by beta-1,4-glycosidic bonds. Cellulose is used to make many products like paper, various fibers, rayon, and even explosive gun cotton.
In simple words: Cellulose is a plant material that makes up cell walls and cotton. It is a long chain of glucose units. We use it to make paper and other products.

🎯 Exam Tip: Distinguish cellulose from starch by its beta-1,4-glycosidic linkages, which make it largely indigestible for humans.

 

Question 13. Name the Vitamins whose deficiency cause
1. rickets
2. scurvy
Answer: Here are the vitamins whose deficiencies cause specific conditions:
1. **Rickets**, which leads to soft and weak bones, is caused by a lack of **Vitamin D**.
2. **Scurvy**, a disease that causes bleeding gums and weakness, results from a deficiency of **Vitamin C**.
In simple words: Not enough Vitamin D causes rickets, and not enough Vitamin C causes scurvy.

🎯 Exam Tip: Focus on linking each vitamin with its primary deficiency disease, as this is a common direct recall question type.

 

Question 14. What are essential and non – essential amino acids?
Answer: Amino acids are the building blocks of proteins. They are categorized based on whether our bodies can produce them or if they must be obtained from our diet. This classification is vital for understanding nutritional requirements.

In simple words: Essential amino acids must come from food, while non-essential ones can be made by our bodies.

🎯 Exam Tip: Remember a few examples for each category to illustrate your understanding, as these often appear in questions.

 

Question 15. What is isoelectric point?
Answer: The **isoelectric point** is the specific pH level at which an amino acid carries no overall electrical charge. At this pH, the positive and negative charges on the molecule balance each other out, making it electrically neutral. This unique pH value affects how the amino acid behaves in solutions.
In simple words: The isoelectric point is a pH where an amino acid has no total electrical charge, meaning positive and negative charges are equal.

🎯 Exam Tip: Clarify that at the isoelectric point, the amino acid exists as a zwitterion and does not move in an electric field.

 

Question 16. What are Zwitter ions?
Answer: In water, the acidic carboxyl group of an amino acid can give a proton to its basic amino group. This creates a molecule with both a positive and a negative charge on different parts. Even though it has both positive and negative charges, the overall molecule remains electrically neutral. This type of molecule is also called amphoteric, meaning it can act as both an acid and a base. Such molecules are known as **Zwitter ions** and are represented as \( \text{H}_3\text{N}^{+}-\text{CH}(\text{R})-\text{COO}^{-} \). They are important for understanding how amino acids act in biological systems.
In simple words: Zwitter ions are amino acid molecules that have both a positive and a negative charge at the same time, making them overall neutral.

🎯 Exam Tip: The key characteristic of a zwitterion is its overall neutrality despite having distinct positive and negative charges.

 

Question 17. What are proteins? How are they classified?
Answer: Proteins are large molecules made from long chains of smaller units called amino acids. They are vital for almost every process in living organisms. Proteins are grouped into two main types based on their shape and structure:

In simple words: Proteins are long chains of amino acids. They are sorted into two types: fibrous proteins, which are long and tough, and globular proteins, which are round and do many jobs.

🎯 Exam Tip: Focus on the shape and solubility differences to easily distinguish between fibrous and globular proteins.

 

Question 18. What is complementary base pairing in DNA?
Answer: In the DNA double helix, specific bases from one strand connect with specific bases on the opposite strand using hydrogen bonds. This forms flat pairs of bases. This highly specific pairing ensures DNA's stability. Adenine (A) always forms two hydrogen bonds with Thymine (T). Guanine (G) always forms three hydrogen bonds with Cytosine (C). This precise matching of bases is called **complementary base pairing**. Any other type of base pairing would weaken the DNA's double helix structure.
In simple words: Complementary base pairing means that in DNA, A always joins with T, and G always joins with C. They connect using hydrogen bonds to keep the DNA stable.

🎯 Exam Tip: Remember the specific number of hydrogen bonds for each base pair: A-T has two, G-C has three.

 

Question 19. Name the biological functions of nucleotides.
Answer: Nucleotides play several important roles in living things. They are the building blocks of DNA and RNA. They act as **energy carriers**, with ATP (adenosine triphosphate) being the main example, providing energy for cellular processes. They also serve as **co-enzymes**, like coenzyme A, \( \text{NAD}^{+} \), and \( \text{FAD} \), which are crucial helpers in many metabolic reactions. Furthermore, nucleotides function as **chemical messengers**, such as cyclic AMP (cAMP), which helps cells communicate and respond to signals.
In simple words: Nucleotides are important in living things. They carry energy (like ATP), help enzymes work (as co-enzymes), and send messages inside cells (like cAMP).

🎯 Exam Tip: When listing functions, provide a key example for each to demonstrate practical understanding.

 

Question 20. Name the vitamins whose deficiency cause
(i) Pellagra
(ii) Beri – Beri
(iii) Night blindness
Answer: Here are the vitamins whose deficiencies cause specific conditions:
(i) **Pellagra**, which involves sun-sensitive skin problems (dermatitis), is caused by a lack of **Vitamin B3 (Niacin)**.
(ii) **Beri-Beri**, a disease affecting nerves and heart, results from a deficiency of **Vitamin B1 (Thiamine)**.
(iii) **Night blindness**, making it hard to see in dim light, is caused by not having enough **Vitamin A (Retinol)**.
In simple words: Lack of Vitamin B3 causes Pellagra, lack of Vitamin B1 causes Beri-Beri, and lack of Vitamin A causes night blindness.

🎯 Exam Tip: Memorize the common name of each vitamin (e.g., Niacin for B3, Thiamine for B1, Retinol for A) along with its deficiency disease.

 

Question 21. What is glycosidic linkage?
Answer: A **glycosidic linkage** is a type of chemical bond that joins two monosaccharide units together to form a disaccharide or a polysaccharide. This bond involves an oxygen atom. It forms when the hydroxyl group \( ( -OH ) \) of the anomeric carbon of one sugar molecule reacts with a hydroxyl group of another sugar molecule, releasing a molecule of water in the process. This bond is fundamental to the structure of all complex carbohydrates.
In simple words: A glycosidic linkage is the bond that connects two sugar units together, like a bridge between them. It is made when two sugar molecules lose a water molecule.

🎯 Exam Tip: Recognize that glycosidic linkages are formed through dehydration (removal of water) and broken by hydrolysis (addition of water).

VII. Three Mark Questions

 

Question 1. Write about the configuration of carbohydrates.
Answer: Most carbohydrates are **optically active**, meaning they can rotate plane-polarized light. This is because they contain one or more chiral carbon atoms. The number of possible optical isomers for a carbohydrate can be calculated using the formula \(2^n\), where \(n\) is the number of chiral carbons. Fischer developed a projection formula to show the three-dimensional structure of carbohydrates. This formula relates their structure to the two enantiomeric forms of glyceraldehyde. Carbohydrates are labeled as **D- or L-configuration** based on the orientation of the hydroxyl group on their penultimate (second-to-last) carbon atom. For instance, D-glucose is named this way because the hydrogen and hydroxyl groups on its C5 carbon are arranged similarly to the C2 carbon of D-glyceraldehyde. Separately, compounds can be **dextrorotatory** (rotate light clockwise, denoted as \(+\)) or **laevorotatory** (rotate light anti-clockwise, denoted as \(-\)). It is important to note that D/L configuration does not directly tell you if a compound is dextro or laevo rotatory.
In simple words: Carbohydrates can twist light because they have special carbon atoms. We name them D or L based on their structure. They can also be called (+) for clockwise light twist or (-) for counter-clockwise twist.

🎯 Exam Tip: Clearly differentiate between D/L configuration (based on a reference molecule like glyceraldehyde) and \(+/- \) optical rotation (an experimentally observed property).

 

Question 2. Explain the cyclic structure of glucose.
Answer: Glucose can exist in two different crystal forms, each with its own melting point (419 K and 423 K). To explain this, scientists suggested that a hydroxyl group within the glucose molecule reacts with its own aldehyde group. This reaction forms a ring-shaped structure called a **cyclic hemiacetal**. This cyclization process changes the carbon atom of the aldehyde group, which was previously non-chiral (achiral), into a chiral carbon atom. This new chiral carbon can have two different orientations, leading to two distinct isomers. These two isomers, which differ only in the configuration at the C1 carbon, are known as **anomers**. They are specifically called the alpha-form and beta-form of glucose. The cyclic structure of glucose resembles pyran, a six-membered ring containing five carbon atoms and one oxygen atom. Therefore, the cyclic forms of glucose are referred to as **glucopyranose**.
In simple words: Glucose can form a ring shape when a part of its chain reacts with another part. This makes a new special carbon atom with two possible forms, called anomers. The ring looks like a molecule called pyran.

🎯 Exam Tip: Focus on explaining the formation of the hemiacetal ring and the creation of the anomeric carbon (C1) as key steps.

 

Question 3. Write about the cyclic structure of fructose.
Answer: Fructose, similar to glucose, also exists in a cyclic (ring) structure in solution. However, unlike glucose which forms a six-membered ring, fructose typically forms a **five-membered ring**. This ring structure is similar to that of furan, a common organic compound. Because of this similarity, the cyclic forms of fructose are called **furanose** forms. When fructose is a part of larger sugar molecules, such as sucrose (a disaccharide), it is usually found in this furanose ring form.
In simple words: Fructose also makes a ring shape, but it's a five-sided ring like furan. So, we call these fructose rings "furanose" forms.

🎯 Exam Tip: Remember that fructose forms a five-membered furanose ring, which distinguishes it from the six-membered pyranose ring of glucose.

 

Question 4. What is a peptide bond? How is it formed?
Answer: A **peptide bond** is a special type of amide linkage that connects two amino acids. These bonds are fundamental to building proteins. It forms through a **condensation reaction** between two amino acids. Specifically, the carboxyl group \( ( -COOH ) \) of one amino acid reacts with the amino group \( ( -NH_2 ) \) of another amino acid. During this reaction, a molecule of water is removed. The resulting compound formed by two amino acids linked by a peptide bond is called a **dipeptide**. When many amino acids join this way, they form a polypeptide chain.
In simple words: A peptide bond is like a special glue that joins two amino acids. It forms when one amino acid's acid part meets another's amino part, and water is removed.

🎯 Exam Tip: Highlight that a peptide bond is an amide linkage and its formation involves the loss of a water molecule (dehydration synthesis).

 

Question 5. Differentiate tertiary and quaternary structure of proteins
Answer: Proteins have different levels of structural organization. The **tertiary structure** describes the full three-dimensional shape of a single polypeptide chain, while the **quaternary structure** describes how multiple polypeptide chains fit together to form a larger protein complex. These structures are crucial for the protein's function.

In simple words: Tertiary structure is the 3D shape of one protein chain. Quaternary structure is how many protein chains fit together to make a bigger protein.

🎯 Exam Tip: Remember that tertiary structure is about *one* polypeptide chain's 3D fold, while quaternary structure involves *multiple* polypeptide chains interacting.

 

Question 6. What are lipids? How are they classified?
Answer: The term "**lipids**" comes from the Greek word 'lipos,' which means fat. Lipids are a diverse group of organic molecules that are crucial for life. They are characterized by their **insolubility in water** and their **solubility in organic solvents** like chloroform and methanol. Lipids are major building blocks of **cell membranes**. They also serve as an important **energy source** for living systems, providing two to three times more energy per gram than carbohydrates or proteins. Lipids are classified into three main types: **Simple lipids, Compound lipids, and Derived lipids**.
In simple words: Lipids are fats that don't mix with water. They are important for cell membranes and give a lot of energy. They come in simple, compound, and derived types.

🎯 Exam Tip: The defining characteristic of lipids is their insolubility in water, which is due to their largely non-polar nature.

 

Question 7. How are hormones classified?
Answer: Hormones are chemical messengers that are categorized based on the distance they travel to reach their target cells. This classification helps understand their diverse roles in regulating bodily functions. The three main types are:

In simple words: Hormones are grouped by how far they travel. Endocrine hormones travel far, paracrine act on nearby cells, and autocrine act on themselves.

🎯 Exam Tip: Distinguish between the three types by the distance and target cells they influence, and provide specific examples for each.

 

Question 1. What are carbohydrates? How are they classified?
Answer: **Carbohydrates** are organic compounds that are typically defined as polyhydroxy aldehydes or polyhydroxy ketones. They are the body's primary source of energy. Their general molecular formula is often represented as \( \text{C}_\text{n}(\text{H}_2\text{O})_\text{n} \). Carbohydrates are broadly classified based on their structure and whether they can be broken down into simpler sugars:
* **Monosaccharides:** These are the simplest sugars and cannot be broken down further by hydrolysis. They include:
* **Aldoses:** Monosaccharides that contain an aldehyde \( ( -CHO ) \) functional group (e.g., Glucose).
* **Ketoses:** Monosaccharides that contain a ketone \( ( -C=O- ) \) functional group (e.g., Fructose).
* They are also classified by the number of carbon atoms:

* **Disaccharides:** These sugars yield two monosaccharide units upon hydrolysis (e.g., Sucrose).
* **Polysaccharides:** These complex carbohydrates yield a large number of monosaccharide units upon hydrolysis (e.g., Starch). They are further divided into:
* **Homopolysaccharides:** Composed of only one type of monosaccharide unit (e.g., Starch).
* **Heteropolysaccharides:** Composed of more than one type of monosaccharide unit (e.g., Hyaluronic acid, Heparin).
In simple words: Carbohydrates are sugars, the main energy source, usually with formula \( \text{C}_\text{n}(\text{H}_2\text{O})_\text{n} \). They are grouped into simple sugars (monosaccharides), two-sugar units (disaccharides), and many-sugar units (polysaccharides).

🎯 Exam Tip: When classifying carbohydrates, clearly define each group (monosaccharide, disaccharide, polysaccharide) by its ability to hydrolyze and provide common examples for each type.

 

Question 2. Elucidate the structure of glucose.
Answer: Glucose has the molecular formula \( \text{C}_6\text{H}_{12}\text{O}_6 \). Many experiments helped to figure out its exact structure:

• When glucose is heated with concentrated hydroiodic acid (HI) and red phosphorus at 373 K, it forms a mixture of n-hexane and 2-iodohexane. This shows that the six carbon atoms in glucose are connected in a straight line.
• Glucose reacts with hydroxylamine (\( \text{NH}_2\text{OH} \)) to form an oxime, and also reacts with hydrogen cyanide (HCN) to form a cyanohydrin. These reactions prove that a carbonyl group (\( \text{C=O} \)) is present in glucose.
• Glucose gets oxidized when treated with bromine water, forming gluconic acid. This reaction specifically shows that an aldehyde group (\( \text{-CHO} \)) is present in glucose, as bromine water is a mild oxidizing agent that only affects aldehydes.

• When glucose is oxidized with a strong oxidizing agent like concentrated nitric acid (\( \text{HNO}_3 \)), it forms saccharic acid. This indicates the presence of a primary alcoholic group (\( \text{-CH}_2\text{OH} \)) at one end of the molecule.

• Glucose also reduces Tollens' reagent and Fehling's solution, which further confirms the presence of an aldehyde group.
• Glucose reacts with acetic anhydride to form pentaacetate, which shows that it has five hydroxyl groups (\( \text{-OH} \)) that are attached to different carbon atoms.
• The glucose molecule is referred to as D(+) glucose, indicating its specific stereochemistry and optical activity.

Based on all these reactions, the straight-chain structure of D(+) Glucose can be written as:

   CHO
   |
H-C-OH
   |
HO-C-H
   |
H-C-OH
   |
H-C-OH
   |
  CH2OH

In simple words: Glucose is a simple sugar with a straight chain of six carbon atoms. It has one aldehyde group and five alcohol groups, which gives it its unique chemical properties. This open-chain structure explains most of its reactions.

🎯 Exam Tip: When elucidating structures, always link each experimental observation directly to a specific functional group or structural feature to build a clear and logical argument.

 

Question 3. Elucidate the structure of fructose.
Answer: Fructose, like glucose, has the molecular formula \( \text{C}_6\text{H}_{12}\text{O}_6 \). Its structure can be determined by several chemical tests:

• When fructose is heated with concentrated hydroiodic acid (HI) and red phosphorus, it produces n-hexane and 2-iodohexane. This result shows that all six carbon atoms in fructose are linked together in a straight chain.
• Fructose reacts with hydroxylamine (\( \text{NH}_2\text{OH} \)) to form an oxime, and with hydrogen cyanide (HCN) to form a cyanohydrin. These reactions confirm the presence of a carbonyl group (\( \text{C=O} \)) in the molecule.
• Fructose reacts with acetic anhydride in the presence of pyridine to form a pentaacetate. This indicates that fructose contains five hydroxyl groups (\( \text{-OH} \)).
• Unlike glucose, fructose does not get oxidized by bromine water. This proves that fructose does not have an aldehyde group.
• When fructose undergoes partial reduction with sodium amalgam and water, it produces a mixture of two epimeric alcohols: sorbitol and mannitol. This specific reaction confirms that fructose has a keto group, likely at the C-2 position, and also explains the formation of a new asymmetric carbon at C-2 during reduction.

When oxidized with nitric acid, fructose forms glycollic acid and tartaric acid, which have fewer carbon atoms than fructose. This observation, along with the reduction products, confirms the presence of a keto group at C-2 and primary alcoholic groups at C-1 and C-6.

The straight-chain structure of fructose is:

  CH2OH
   |
   C=O
   |
HO-C-H
   |
H-C-OH
   |
H-C-OH
   |
  CH2OH

In simple words: Fructose is a sugar that has its carbon atoms in a straight line, similar to glucose. However, fructose has a ketone group, not an aldehyde group, and it also has five alcohol groups. This structure makes it behave differently in some chemical tests.

🎯 Exam Tip: Remember that fructose is a ketohexose, meaning it contains a ketone functional group and six carbon atoms, which distinguishes it from aldohexoses like glucose.

 

Question 4. Explain the composition and structure of nucleic acids.
Answer: Nucleic acids are very important molecules found in the nucleus of cells. They are responsible for passing on traits from one generation to the next in every living thing. Chromosomes, which are found in cells, are made up of proteins and these nucleic acids.

Nucleic acids are large polymers made of smaller units called nucleotides. There are two main types:
(i) Deoxyribonucleic acid (DNA)
(ii) Ribonucleic acid (RNA)

These molecules store genetic information in every organism.

Composition and Structure:
When DNA and RNA are broken down (hydrolyzed), they each yield three main components:

1. Nitrogenous Base: These are organic compounds that come from two main parent compounds: pyrimidine and purine.
2. Pentose Sugar: A five-carbon sugar.
3. Phosphate Group: A group containing phosphorus and oxygen.

Pentose Sugar:
DNA contains 2'-deoxy-D-ribose, while RNA contains D-ribose. In nucleotides, both types of pentose sugars exist in a beta-furanose (closed five-membered ring) form.

Phosphate group:
The phosphate group forms a phosphodiester bond, which links nucleotides together to create the long chains of DNA and RNA. The number of phosphate groups helps classify nucleotides as mononucleotides, dinucleotides, or trinucleotides. The molecule without the phosphate group is called a nucleoside.

In simple words: Nucleic acids like DNA and RNA are made of building blocks called nucleotides. Each nucleotide has a nitrogen base (like adenine or cytosine), a five-carbon sugar, and a phosphate group. These blocks link up to form long chains that carry all our genetic information.

🎯 Exam Tip: Remember the key differences in bases (Thymine in DNA vs. Uracil in RNA) and sugars (Deoxyribose in DNA vs. Ribose in RNA) as they are common distinguishing features.

 

Question 5. Explain the double strand helix structure of DNA.
Answer: The double helix structure of DNA, discovered by Watson and Crick in 1953, is a very important model in science. Here's how it works:

• DNA has a 3-dimensional structure made of two anti-parallel helical chains. These chains wind around a central axis to form a right-handed spiral.
• The outer part of the helix, called the backbone, is made of alternating deoxyribose sugar and phosphate groups. This part is hydrophilic, meaning it likes water, and faces outwards towards the surrounding water.
• The purine and pyrimidine bases are stacked inside the double helix. This arrangement helps reduce repulsion between the charged phosphate groups.
• The two strands are not perfectly aligned, creating two gaps along the helix: a wider major groove and a narrower minor groove on the surface.
• Each full turn of the helix contains about 10.5 base pairs and measures approximately 3.4 nanometers (34 Ångströms). The distance between stacked bases is about 0.34 nanometers (3.4 Ångströms).
• The bases on opposite strands are held together by hydrogen bonds, forming specific pairs. Adenine (A) always pairs with Thymine (T) using two hydrogen bonds. Guanine (G) always pairs with Cytosine (C) using three hydrogen bonds. This exact pairing is called complementary base pairing.
• Any other type of base pairing would make the double helical structure unstable.
• The DNA double helix is kept strong by two main forces: hydrogen bonding between complementary base pairs and stacking interactions between the flat bases.

In simple words: DNA looks like a twisted ladder, called a double helix. Its sides are made of sugar and phosphate, and the rungs are made of pairs of bases (A with T, and G with C) held together by weak bonds. This twisted shape keeps our genetic information safe and organized.

🎯 Exam Tip: Focus on remembering the key features: right-handed helix, anti-parallel strands, sugar-phosphate backbone, complementary base pairing (A-T, G-C), and the role of hydrogen bonds in stability.

 

Question 6. What are the differences between DNA and RNA?
Answer: DNA and RNA are both types of nucleic acids, but they have several key differences that determine their roles in living organisms. Here is a comparison of their main features:

In simple words: DNA is usually a double-stranded, stable molecule that stores genetic instructions for a long time and can make copies of itself. RNA is typically a single-stranded, less stable molecule that helps carry out those instructions and is made from DNA.

🎯 Exam Tip: For comparison questions, create a table to clearly list the differences point-by-point, which helps in scoring full marks.

 

Question 7. Explain DNA fingerprinting.
Answer: DNA fingerprinting is a method that helps identify individuals based on their unique DNA patterns. It was first developed by Sir Alec Jeffreys in 1984. Here’s how it works and its significance:

• Every person's DNA fingerprint is unique, just like actual fingerprints.
• DNA can be collected from tiny samples like blood, saliva, or hair.
• This method can find small variations in human DNA that are specific to each person.
• The collected DNA is cut into many smaller pieces using special enzymes called restriction enzymes.
• These DNA fragments have different lengths and are then separated by size using a technique called gel electrophoresis.
• After separation, the DNA fragments are moved from the gel to a nylon sheet, a process known as blotting.
• Next, the fragments go through auto-radiography. This involves exposing them to DNA probes, which are small pieces of synthetic DNA that are radioactive and designed to stick to specific DNA fragments.
• When an X-ray film is placed over the nylon sheet, dark marks appear wherever a radioactive probe has attached.
• The resulting pattern of marks is unique for each person and can be compared with other samples.
• DNA fingerprinting works because of small differences in DNA sequences (even single base-pair changes) between individuals.
• This method is now widely used and accepted in legal cases around the world.

In simple words: DNA fingerprinting uses tiny bits of our DNA to create a unique pattern, like a barcode, that belongs only to us. This helps scientists and police identify people from small samples, such as a drop of blood or a strand of hair.

🎯 Exam Tip: Focus on the key steps: DNA extraction, fragmentation by restriction enzymes, gel electrophoresis, blotting, probing with radioactive markers, and visualization, emphasizing its application in identification.