Class 11 Bio Botany · Chapter 8

Samacheer Class 11 Bio Botany - Biomolecules

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Chapter-wise textbook exercise answers for Biomolecules with validation-aware solutions.

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Sections in this chapter
II. a. Extracellular enzymes 1II. Choose the wrong answer. 2III. But A : T, need not be equal to G : C 17III. Co- Enzymes: 4III. Match The Following And Find The Correct Answer. 5III. Non-reversible/ Irreversible Inhibitors 2IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer. 4IV. Nucleoside 3Part I 5Part II 33VI. Assertion & Reason – Find Out The Correct Answer. 12
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1II. a. Extracellular enzymes1 questions
Q.5Explain the three types of Co-Factors.v
Solution

II. Prosthetic groups:
Eg. Vit B 2 (Riboflavin) – & Flavin adenine dinucleotide (FAD) Kreb’s cycle
III. Co- Enzymes:
These are co-factors but don’t remain attached to enzymes Eg. NAD, NADP, Co-enzyme A, ATP etc.

Answer:

II. Prosthetic groups:
Eg. Vit B 2 (Riboflavin) – & Flavin adenine dinucleotide (FAD) Kreb’s cycle
III. Co- Enzymes:
These are co-factors but don’t remain attached to enzymes Eg. NAD, NADP, Co-enzyme A, ATP etc.

2II. Choose the wrong answer.2 questions
Q.2This has nothing to do with the structure of a. Cytosine b. Pyrimidine c. Adenine d. Thyaminev
Solution

c. Adenine

Answer:

c. Adenine

Q.3Choose the right answer a) Amylose – linear unbranched polymer of with 20% starch b) Amylopectin – a polymer with some 1,6 linkages that give it a linear structure c) Inulin – Polymer of galactose d) Amino acid – Here a basic structure of carbon linked to a basic amino groupv
Solution

d) Amino acid – Here a basic structure of carbon linked to a basic amino group
III. Match The Following And Find The Correct Answer.

Answer:

d) Amino acid – Here a basic structure of carbon linked to a basic amino group
III. Match The Following And Find The Correct Answer.

3III. But A : T, need not be equal to G : C17 questions
Q.1Classify Polysaccharides.v
Solution

Polysaccharides have more than 10 monosaccharides
They can be divided in to
* Homopolysaccharides
* Hetero polysaccharides
1. Homopolysaccharides
2. Heteropolysaccharides
a. Starch
a. Peptidoglycan
b. Glycogen
b. Hyaluronic acid
c. Cellulose
c. chondroitin
d. Chitin
d. keratan sulphate
e. Inulin
e. Agar Agar

Answer:

Polysaccharides have more than 10 monosaccharides
They can be divided in to
* Homopolysaccharides
* Hetero polysaccharides
1. Homopolysaccharides
2. Heteropolysaccharides
a. Starch
a. Peptidoglycan
b. Glycogen
b. Hyaluronic acid
c. Cellulose
c. chondroitin
d. Chitin
d. keratan sulphate
e. Inulin
e. Agar Agar

Q.1How will you identify the presence of glucose in a given food sample?v
Solution

Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu 2+ ions to Cu + ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).
This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Answer:

Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu 2+ ions to Cu + ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).
This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Q.2Why do we call Glucose and Fructose Isomers -Discuss.v
Solution

Glucose and Fructose have a same molecular formula, C 6 H 12 0 6 but different structural formulas- so they are known as Isomers.

Answer:

Glucose and Fructose have a same molecular formula, C 6 H 12 0 6 but different structural formulas- so they are known as Isomers.

Q.2What is protein denaturation.v
Solution

Exposure to heat causes atoms to vibrate violently distrust hydrogen and ionic bonds. There is the loss of 3D structure protein become elongated, disorganised strands. Soaps, detergents, acid, alcohol and some disinfectants disrupt the interchain bond cause the molecule to be non-functional.

Answer:

Exposure to heat causes atoms to vibrate violently distrust hydrogen and ionic bonds. There is the loss of 3D structure protein become elongated, disorganised strands. Soaps, detergents, acid, alcohol and some disinfectants disrupt the interchain bond cause the molecule to be non-functional.

Q.3Tabulate other sugar compoundsv
Solution

Other Polysaccharides
Structure
Functions
Inulin
Polymer of fructose
It is not metabolised in the human body and is readily filtered through the kidney
Hyaluronic acid
Hcteropolymcr of d glucuronic acid and D-N acetyl glucosamine
11 accounts for the toughness and flexibility of cartilage and tendon
Agar
Mucopolysaccharide from red algae
Used as solidifying agent in culture medium in laboratory
Heparin
Glycosamino glycan contains variably sulphated disaccharide unit Drcscnt in liver
Used as an anticoagulant
Chondroitin sulphate
Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid
Dietary supplement for treatment of osteoarthritis
Kcratan sulphate
Sulphated glycosaminoglycan and is a structural carbohydrate
Acts as cushion to absorb mechanical shock

Answer:

Other Polysaccharides
Structure
Functions
Inulin
Polymer of fructose
It is not metabolised in the human body and is readily filtered through the kidney
Hyaluronic acid
Hcteropolymcr of d glucuronic acid and D-N acetyl glucosamine
11 accounts for the toughness and flexibility of cartilage and tendon
Agar
Mucopolysaccharide from red algae
Used as solidifying agent in culture medium in laboratory
Heparin
Glycosamino glycan contains variably sulphated disaccharide unit Drcscnt in liver
Used as an anticoagulant
Chondroitin sulphate
Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid
Dietary supplement for treatment of osteoarthritis
Kcratan sulphate
Sulphated glycosaminoglycan and is a structural carbohydrate
Acts as cushion to absorb mechanical shock

Q.4Classify enzyme reactions:v
Solution

II. a. Extracellular enzymes
Enzymes secreted outside & work externally Eg digestive enzymes. b. Intracellular Enzymes
Remain within cells & work there Eg Insulin.

Answer:

II. a. Extracellular enzymes
Enzymes secreted outside & work externally Eg digestive enzymes. b. Intracellular Enzymes
Remain within cells & work there Eg Insulin.

Q.5Distinguish between Waxes & Steroids.v
Solution

Waxes
Steroids
Esters formed between long-chain alcohol and saturated fatty acids
Fur feathers, fruits, leaves, skin, and insect exoskeleton are waterproofed with a coating of wax.
Complex compounds found in cell membrane & animal hormones. Eg – Cholesterol
It reinforces the structure of the life cell membrane in animal cells also in Mycoplasma.

Answer:

Waxes
Steroids
Esters formed between long-chain alcohol and saturated fatty acids
Fur feathers, fruits, leaves, skin, and insect exoskeleton are waterproofed with a coating of wax.
Complex compounds found in cell membrane & animal hormones. Eg – Cholesterol
It reinforces the structure of the life cell membrane in animal cells also in Mycoplasma.

Q.6Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.v
Solution

The non-polar aliphatic R group has 6 amino acids Glycine, Alanine, Valine, Leucine, Methionine & Isoleucine.
Structure of Glvcine Alanine Valine

Answer:

The non-polar aliphatic R group has 6 amino acids Glycine, Alanine, Valine, Leucine, Methionine & Isoleucine.
Structure of Glvcine Alanine Valine

Q.7Distinguish between Macronutrients & Micronutrients.v
Solution

Macronutrients:
* Nutrients required in larger quantities for plant growth are called Macronutrients.
* e.g. Potassium and Calcium
Micronutrients:
* Nutrients required in trace amount for plant growth are called Micronutrients
* e.g. Zinc and Bora

Answer:

Macronutrients:
* Nutrients required in larger quantities for plant growth are called Macronutrients.
* e.g. Potassium and Calcium
Micronutrients:
* Nutrients required in trace amount for plant growth are called Micronutrients
* e.g. Zinc and Bora

Q.8Define Activation energy?v
Solution
  • The minimum quantity of energy the reactants must possess in order to undergo a specified reaction is known as Activation energy.
  • Energy being the biocatalysts reduce the activation energy, thereby help the reaction occurs.
  • The rate of reaction increases if activation energy decreases.
Answer:
  • The minimum quantity of energy the reactants must possess in order to undergo a specified reaction is known as Activation energy.
  • Energy being the biocatalysts reduce the activation energy, thereby help the reaction occurs.
  • The rate of reaction increases if activation energy decreases.
Q.9Differentiate Between DNA & RNA.v
Solution

DNA
RNA
Double-stranded
Single-stranded
The genetic material in almost all living
organism except for RNA virus
Not genetic material except RNA virus
2 types
Prokaryotic DNA is circular
Eukaryotic DNA is linear
3 types
m RNA
t RNA
r RNA
Controls all aspects of a cell
plays important role in protein synthesis

Answer:

DNA
RNA
Double-stranded
Single-stranded
The genetic material in almost all living
organism except for RNA virus
Not genetic material except RNA virus
2 types
Prokaryotic DNA is circular
Eukaryotic DNA is linear
3 types
m RNA
t RNA
r RNA
Controls all aspects of a cell
plays important role in protein synthesis

Q.9Distinguish between Primary metabolite & Secondary metabolite.v
Solution

Between Primary metabolite & Secondary metabolite:
* Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
* Secondary metabolites do not show any direct function in growth and development of organisms. Example: Ricin, gums.

Answer:

Between Primary metabolite & Secondary metabolite:
* Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
* Secondary metabolites do not show any direct function in growth and development of organisms. Example: Ricin, gums.

Q.10Give examples for Secondary metabolites.v
Solution

Secondary metabolites:
* Pigments – Carotenoids/Anthocyanins
* Alkaloids – Morphine, codeine
* Essential oil – Lemongrass oil, Rose oil
* Toxins – Abrin & ricin
* Lectins – Concanavalin. A
* Drugs – Vinbiastine, curcumin
* Polymeric substances – Rubber, gums, cellulose

Answer:

Secondary metabolites:
* Pigments – Carotenoids/Anthocyanins
* Alkaloids – Morphine, codeine
* Essential oil – Lemongrass oil, Rose oil
* Toxins – Abrin & ricin
* Lectins – Concanavalin. A
* Drugs – Vinbiastine, curcumin
* Polymeric substances – Rubber, gums, cellulose

Q.11How will you test reducing sugar?v
Solution

Substrate
Reagent
Result
1. Glucose is taken in a test tube (Aldehyde)
An alkaline solution of copper di sulfate (Benedicts’ reagent) added & heated
So brick-red precipitate of copper oxide is formed (i.e.) (Cu + is reduced to Cu + ) Aldehyde is oxidised to COOH group.

Answer:

Substrate
Reagent
Result
1. Glucose is taken in a test tube (Aldehyde)
An alkaline solution of copper di sulfate (Benedicts’ reagent) added & heated
So brick-red precipitate of copper oxide is formed (i.e.) (Cu + is reduced to Cu + ) Aldehyde is oxidised to COOH group.

Q.12Draw the structure of a basic amino acid.v
Solution

3 Marks

Answer:

3 Marks

Q.13Why do some people have curly hair?v
Solution

Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Answer:

Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Q.14Deoxyribose (C 5 H 10 O 4 ) is not a carbohydrate – Discuss.v
Solution

Carbohydrates are hydrates of carbon Deoxy ribose is a carbohydrate, but its formula C 5 H 10 O 4 -does not apply the general formula of Carbohydrate (C 2 H 20 )X formula has Carbohydrates formula.
5 Marks Questions

Answer:

Carbohydrates are hydrates of carbon Deoxy ribose is a carbohydrate, but its formula C 5 H 10 O 4 -does not apply the general formula of Carbohydrate (C 2 H 20 )X formula has Carbohydrates formula.
5 Marks Questions

4III. Co- Enzymes:4 questions
Q.6Tabulate the uses of enzymesv
Solution

Enzyme
Source
Application
Bacterial protease
Bacillus
Biological detergents
Bacterial glucose isomerase
Bacillus
Fructose- Syrup manufacture
Fungal lactose
Kluyveromyces
Breaking down of lactose
glucose + glactose
Amylases
Aspergillus
Removal of starch in woven cloth production

Answer:

Enzyme
Source
Application
Bacterial protease
Bacillus
Biological detergents
Bacterial glucose isomerase
Bacillus
Fructose- Syrup manufacture
Fungal lactose
Kluyveromyces
Breaking down of lactose
glucose + glactose
Amylases
Aspergillus
Removal of starch in woven cloth production

Q.7Enumerate the properties of Enzyme.v
Solution

The properties of Enzyme:
* Enzymes are globular proteins.
* They act as catalysts and effective even in small quantities.
* They remain unchanged at the end of the reaction.
* They are highly specific.
* They have an active site where the reaction takes place.
* Enzymes lower the activation energy of the reaction they catalyse.

Answer:

The properties of Enzyme:
* Enzymes are globular proteins.
* They act as catalysts and effective even in small quantities.
* They remain unchanged at the end of the reaction.
* They are highly specific.
* They have an active site where the reaction takes place.
* Enzymes lower the activation energy of the reaction they catalyse.

Q.8Explain lock and key mechanism of enzymes.v
Solution

The substrate Enzyme product
The substrate binds to a specific pocket in an enzyme known as the Active site.
Active site = Lock
Substrate = Key
* The substrate binds to the active site of the enzyme
* As enzyme and substrate form ES- Complex, the substrate is raised in Energy —
* This was explained by Fischer
* transition stage break down in to, products, Enzyme remain unchanged.

Answer:

The substrate Enzyme product
The substrate binds to a specific pocket in an enzyme known as the Active site.
Active site = Lock
Substrate = Key
* The substrate binds to the active site of the enzyme
* As enzyme and substrate form ES- Complex, the substrate is raised in Energy —
* This was explained by Fischer
* transition stage break down in to, products, Enzyme remain unchanged.

Q.9What are the various types of inhibitors of enzymes.v
Solution

Definition:
Substances present in the cells may react with enzyme and lower the rate of reactions Inhibitors
I. Competitive Inhibitors:
Substances resemble the shape of substrate & compete to occupy active sites
Eg. 1. Enzyme RUBISCO – is competitively inhibited by oxygen/carbon dioxide in the chloroplast
2. Succinic dehydrogenase – Inhibited by malonate.
II. Non-Competitive Inhibitors
Unlike substrates, blocks by binding on active sites, change its shape so enzyme unable to accept substrate.
Enzyme – pyruvate kinase- inhibited by amino acid Alanine.
III. Non-reversible/ Irreversible Inhibitors
They bind to an enzyme tightly & permanently destroying their catalytic nature Enzyme cytochrome oxidase inhibited by cyanide ions Neurotransmitter – blocked by nerve gas sarin.

Answer:

Definition:
Substances present in the cells may react with enzyme and lower the rate of reactions Inhibitors
I. Competitive Inhibitors:
Substances resemble the shape of substrate & compete to occupy active sites
Eg. 1. Enzyme RUBISCO – is competitively inhibited by oxygen/carbon dioxide in the chloroplast
2. Succinic dehydrogenase – Inhibited by malonate.
II. Non-Competitive Inhibitors
Unlike substrates, blocks by binding on active sites, change its shape so enzyme unable to accept substrate.
Enzyme – pyruvate kinase- inhibited by amino acid Alanine.
III. Non-reversible/ Irreversible Inhibitors
They bind to an enzyme tightly & permanently destroying their catalytic nature Enzyme cytochrome oxidase inhibited by cyanide ions Neurotransmitter – blocked by nerve gas sarin.

5III. Match The Following And Find The Correct Answer.5 questions
Q.1(I) Morphine – A. Hectins (II) Concanavalin A – B. Drug (III) Vinblastin – C. Pigment (IV) Anthocyanin – D. Toxinv
Solution

(a) B – A – D – C

Answer:

(a) B – A – D – C

Q.2(I) Lactose – A. Penta saccharide (II) Ramnose – B. Tetra saccharide (III) Stachyose – C. Disaccharide (IV) Verbascose – D. Tri saccharidev
Solution

(b) C-D-B-A

Answer:

(b) C-D-B-A

Q.3(I) Fred Sanger 1st sequenced (II) Di sulphide bridges formed between sulphur & amino acids (III) non-protein enzyme (IV) homo polysaccharide with amino acidv
Solution

(b) D-A-B-C

Answer:

(b) D-A-B-C

Q.4(I) Amino acid chain is twisted into coiled configuration call a helix – A. Tertiary Protein (II) Protein fold into a globular structure called domains – B. Quaternary protein (III) Linear arrangement of amino acids in a Polypeptide chain – C. Secondary protein (IV) more than one polypeptide forms a large multiunit multimer – D. Primary Proteinv
Solution

(c) C – A – D – B

Answer:

(c) C – A – D – B

Q.5(I)) Esters formed between long-chain alcohol another negative. – A. a molecule with two or more & saturated fatty acid function group one +ve and (II) lipids have both hydrophobic & hydrophilic end known for permeability – B. fluid nature & selective (III) The amino acids are both acidic & basic exoskeleton of insects – C. waxy substance coating (IV) Zwitter is also called dipolar – D. amophoteric in naturev
Solution

(c) C-B-D-A
IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.

Answer:

(c) C-B-D-A
IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.

6III. Non-reversible/ Irreversible Inhibitors2 questions
Q.10Distinguish between feedback allosteric inhibition negative feedback (end product) inhibition.v
Solution

Feedback Allosteric Inhibition
Negative feedback Inhibition
Allosteric inhibitors modify enzyme active sites (reversible)
E.g. Glucose Hexokinase G-6 Phosphate
G.6. Phosphate – Inhibit Hexokinase
When end products accumulate they cause negative feedback or end product inhibition
After products get used up the enzyme reaction is switched on once again.

Answer:

Feedback Allosteric Inhibition
Negative feedback Inhibition
Allosteric inhibitors modify enzyme active sites (reversible)
E.g. Glucose Hexokinase G-6 Phosphate
G.6. Phosphate – Inhibit Hexokinase
When end products accumulate they cause negative feedback or end product inhibition
After products get used up the enzyme reaction is switched on once again.

Q.11Tabulate other sugar compounds.v
Solution

Other polysaccharides
Structure
Functions
Inulin
Polymer of fructose
It is not metabolized in the human body and is readily filtered through the kidney
Hyaluronic acid
Heteropolymer of d glucuronic acid and D-N acetyl glucosamine
It accounts for the toughness and flexibility of cartilage and tendon
Agar
Mucopolysaccharide from red algae
Used as a solidifying agent in culture medium in a laboratory
Heparin
Glucosamine glycan contains variably sulphated disaccharide unit present in liver
Used as anticoagulant
Used as anticoagulant
Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid)
Dietary supplement for treatment of osteoarthritis
Keratan sulphate
Sulphated glycosaminoglycan and is a structural carbohydrate
Acts as cushion to absorb mechanical shock

Answer:

Other polysaccharides
Structure
Functions
Inulin
Polymer of fructose
It is not metabolized in the human body and is readily filtered through the kidney
Hyaluronic acid
Heteropolymer of d glucuronic acid and D-N acetyl glucosamine
It accounts for the toughness and flexibility of cartilage and tendon
Agar
Mucopolysaccharide from red algae
Used as a solidifying agent in culture medium in a laboratory
Heparin
Glucosamine glycan contains variably sulphated disaccharide unit present in liver
Used as anticoagulant
Used as anticoagulant
Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid)
Dietary supplement for treatment of osteoarthritis
Keratan sulphate
Sulphated glycosaminoglycan and is a structural carbohydrate
Acts as cushion to absorb mechanical shock

7IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.4 questions
Q.1(I) Esters are formed between long-chain alcohol & saturated fatty acid. (II) Lecithin is a food additive & dietary supplement. ‘ (III) Lipids in their structure have two hydrophilic ends (IV) Solid fats are usually unsaturatedv
Solution

(a) True – True – False – False

Answer:

(a) True – True – False – False

Q.1Label the diagram parts correctly by choosing the right option. A B C D a Deoxyribose sugar Nitrogen base Nucleotide Phosphate b Deoxyribose sugar Phosphate Nucleotide Nitrogen base c Deoxyribose sugar Nitrogen base Nucleotide Phosphate d Deoxyribose sugar Phosphate Nitrogen base Nucleotidev
Solution

(d) Deoxyribose sugar – Phosphate Nitrogen base – Nucleotide

Answer:

(d) Deoxyribose sugar – Phosphate Nitrogen base – Nucleotide

Q.2(I) In saturated fatty acids, the hydrocarbon chain is single-bonded (II) Triglycerides are composed of a single molecule of glycerol bound to 2 fatty acids (III) Palmitic acid is an example of saturated fatty acid. (IV) Oleic acid is an example of unsaturated fatty acid.v
Solution

(d) Deoxyribose sugar – Phosphate – Nitrogen base – Nucleotide
V.

Answer:

(d) Deoxyribose sugar – Phosphate – Nitrogen base – Nucleotide
V.

Q.2A B C D a Q arm Centromere Sister Chromatids Q arm b P arm Centromere Sister Chromatids Q arm c Sister Chromatids Centromere Q arm P arm d Q arm Centromere P arm Sister Chromatidsv
Solution

(b) P arm – Centromere – Sister chromatids – Q arm
VI. Assertion & Reason – Find Out The Correct Answer.

Answer:

(b) P arm – Centromere – Sister chromatids – Q arm
VI. Assertion & Reason – Find Out The Correct Answer.

8IV. Nucleoside3 questions
Q.7What are the factors affecting the rate of enzyme reactions?v
Solution

Enzymes being bio-molecules sensitive to environmental condition
(i) Temperature
* Heating increases molecular motion-quicken enzyme reaction
* Optimum temperature is the temperature that promotes maximum activity
(ii) pH –
* Change in the pH – leads to an alteration of enzyme shape (active site)
* Extremes of pH’ denatures enzymes
* Optimum pH’ is that at which the maximum rate of reaction occurs
(iii) Substrate concentration
For a given enzyme concentration, the rate of reaction increase with increasing substrate concentration
(iv) Enzyme concentration
The rate of reaction is directly proportional to enzyme concentration.

Answer:

Enzymes being bio-molecules sensitive to environmental condition
(i) Temperature
* Heating increases molecular motion-quicken enzyme reaction
* Optimum temperature is the temperature that promotes maximum activity
(ii) pH –
* Change in the pH – leads to an alteration of enzyme shape (active site)
* Extremes of pH’ denatures enzymes
* Optimum pH’ is that at which the maximum rate of reaction occurs
(iii) Substrate concentration
For a given enzyme concentration, the rate of reaction increase with increasing substrate concentration
(iv) Enzyme concentration
The rate of reaction is directly proportional to enzyme concentration.

Q.9Write down the characteristic features of DNA?v
Solution

The characteristic feature of DNA.
* If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in the opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
* The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow-angle and 240° for the wide-angle. The narrow-angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
* Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
* DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X-ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
* Thermodynamic stability of the helix and specificity of base pairing includes
* The hydrogen bonds between the complementary bases of the double helix
* stacking interaction between bases tend to stack about each other perpendicular to the direction of the helical axis.
* Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
* The phosphodiester linkages give an inherent polarity to the DNA helix. They form strong covalent bonds, gives strength and stability to the polynucleotide chain.
* Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
* Based on the helix and the distance between each turn, the DNA is of three forms – A DNA, B DNA and Z DNA.

Answer:

The characteristic feature of DNA.
* If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in the opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
* The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow-angle and 240° for the wide-angle. The narrow-angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
* Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
* DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X-ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
* Thermodynamic stability of the helix and specificity of base pairing includes
* The hydrogen bonds between the complementary bases of the double helix
* stacking interaction between bases tend to stack about each other perpendicular to the direction of the helical axis.
* Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
* The phosphodiester linkages give an inherent polarity to the DNA helix. They form strong covalent bonds, gives strength and stability to the polynucleotide chain.
* Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
* Based on the helix and the distance between each turn, the DNA is of three forms – A DNA, B DNA and Z DNA.

Q.10Explain the structure and function of different types of RNA?v
Solution

I. mRNA (messenger RNA)
* single-stranded
* carries a copy of instructions to carry out amino acid assembling &protein synthesis
* unstable
* 5% of total RNA
* In Prokaryotes – it is (polycistronic carrying coding sequence for many polypeptides
* Eukaryotes – (monocistronic) contain information for only one polypeptide
II. tRNA (transfer RNA)
* single-stranded clover-shaped with 4 arms highly folded -3 D structure
* translates the code from mRNA and transfers amino acid to ribosomes (to built proteins)
* unstable (also known as soluble RNA)
* 15% of total RNA
III. rRNA (ribosomal RNA)
* single-stranded
* make up the 2 subunits of ribosomes
* metabolically stable
* 80% total RNA
* A polymer with varied length from 120 – 3000 nucleotides & give ribosomes their shape
* Genes of rRNA employed for phylogenetic studies
Part II
11th Bio Botany Guide Biomolecules Additional Important Questions and Answers
I Choose the right answer.

Answer:

I. mRNA (messenger RNA)
* single-stranded
* carries a copy of instructions to carry out amino acid assembling &protein synthesis
* unstable
* 5% of total RNA
* In Prokaryotes – it is (polycistronic carrying coding sequence for many polypeptides
* Eukaryotes – (monocistronic) contain information for only one polypeptide
II. tRNA (transfer RNA)
* single-stranded clover-shaped with 4 arms highly folded -3 D structure
* translates the code from mRNA and transfers amino acid to ribosomes (to built proteins)
* unstable (also known as soluble RNA)
* 15% of total RNA
III. rRNA (ribosomal RNA)
* single-stranded
* make up the 2 subunits of ribosomes
* metabolically stable
* 80% total RNA
* A polymer with varied length from 120 – 3000 nucleotides & give ribosomes their shape
* Genes of rRNA employed for phylogenetic studies
Part II
11th Bio Botany Guide Biomolecules Additional Important Questions and Answers
I Choose the right answer.

9Part I5 questions
Q.1The most basic amino acid is a. Arginine b. Histidene c. Glycine d. Glutaminev
Solution

c. GIycine

Answer:

c. GIycine

Q.2An example of feed back inhibition is a. cyanide action on cytochrome b. Sulpha drug on folic acid c. Allosteric inhibition of hexokinase by glucose- 6- phosphate d. The inhibition of succinic dehydrogenase by malonatev
Solution

c. Allosteric inhibition of hexokinase by glucose-6-phosphate

Answer:

c. Allosteric inhibition of hexokinase by glucose-6-phosphate

Q.3Enzymes the catalyse interconversion of optical, geometrical or positional isomers are a. Ligases b. Lyases c. Hydrolases d. Isomerasesv
Solution

d. Isomerases

Answer:

d. Isomerases

Q.4Proteins perform many physiological functions, for example, some functions as enzymes one of the following represents an additional function that some proteins discharge a. Antibiotics b. Pigment conferring colour to skin c. Pigments making colours of flowers d. Hormonesv
Solution

d.Hormones

Answer:

d.Hormones

Q.5Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown &one Blank component ‘X’ in it. Category Compound I. Cholesterol A. Guanine II. Amino acid B. IVH 2 III. Nucleotide C. Adenine IV. Nucleoside D. Uracilv
Solution

IV

Answer:

IV

10Part II33 questions
Q.1Who invented the electron microscope? (2010 AIIMS, 2008 JIPMER)v
  1. (a) Janssen
  2. (b) Edison
  3. (c) Knoll and Ruska
  4. (d) Landsteiner
Solution

(c) Knoll and Ruska

Answer:

(c) Knoll and Ruska

Q.2Polysaccharides also called a. Polymers b. Glycans c. Glycosidic compounds d. Glyconesv
Solution

b. Glycans

Answer:

b. Glycans

Q.3Omnis – cellula – e – cellula was given by ……………. (2007 AIIMS)v
  1. (a) Virchow
  2. (b) Hooke
  3. (c) Leeuwenhoek
  4. (d) Robert Brown
Solution

(a) Virchow

Answer:

(a) Virchow

Q.4Nitrocellulose is used in making a. cellophane b. drapers c. explosives d. pain balmsv
Solution

c. explosives

Answer:

c. explosives

Q.5Genes present in the cytoplasm of eukaryotic cells are found in ……………. (2006 AIIMS)v
  1. (a) mitochondria and inherited via egg cytoplasm
  2. (b) lysosomes and peroxisomes
  3. (c) Golgi bodies and smooth endoplasmic reticulum
  4. (d) Plastids inherited via male gametes
Solution

(a) mitochondria and inherited via egg cytoplasm

Answer:

(a) mitochondria and inherited via egg cytoplasm

Q.6Chitin when added with amino acid becomes a.myeopolysaccharide b. amylopolysaceharide c.mucopolysaccharide d. peptidopolysaccharidev
Solution

c. mucopolysaccharide

Answer:

c. mucopolysaccharide

Q.7A quantosome is present in ……………. (JIPMER 2012)v
  1. (a) Mitochondria
  2. (b) Chloroplast
  3. (c) Golgi bodies
  4. (d) ER
Solution

(b) Chloroplast

Answer:

(b) Chloroplast

Q.8Among the following one is not a non-polar solvent a. benzene b. sulphuric acid c. ether d. chloroformv
Solution

b. sulphuric acid

Answer:

b. sulphuric acid

Q.9One of the given below is a complex found in the cell membrane of animal cell a. cholesterol b. myelin c. proline d. Ieeithinv
Solution

a. cholesterol

Answer:

a. cholesterol

Q.10A major site for the synthesis of lipids ……………. (2013 NEET)v
  1. (a) Rough ER
  2. (b) smooth ER
  3. (c) Centriole
  4. (d) Lysosome
Solution

(b) smooth ER

Answer:

(b) smooth ER

Q.11Principle information molecules of the cell are known as a. Nucleus b. DNA c. RNA d. Nucleic acidsv
Solution

d. Nucleic acids

Answer:

d. Nucleic acids

Q.12(I) Cellulose – A most abundant organic compound (II) Morphine – Pain relieving alkaloid (III) Aldose – reducing sugar & Ketose (IV) Glycogen – mucopolysaccharidev
Solution

(IV) Glycogen – mucopolysaccharide

Answer:

(IV) Glycogen – mucopolysaccharide

Q.13The following is a general formula of a. Amino acid b. Fatty acid c. Nucleotide d. Monosaccharidev
Solution

a. Amino acid

Answer:

a. Amino acid

Q.14Lactose is a disaccharide of a. Glucose – Glucose b. Fructose – Fructose c. Glucose – Galactose d. Fructose – Galactosev
Solution

c. Glucose – Galactose

Answer:

c. Glucose – Galactose

Q.15Number of fatty acids in triglyceride is …………….v
  1. (a) 1
  2. (b) 2
  3. (c) 3
  4. (d) 4
Solution

(c) 3

Answer:

(c) 3

Q.16Heparin the anti-coagulant is got from a. D – glucuronic acid b. Polymer of fructose c. Mucopolysaccharide from red algae d. Glucosaminoglycanv
Solution

d. Glycosaminoglycan

Answer:

d. Glycosaminoglycan

Q.17The p H at which Zwitterion is formed is known as a. Iso ionic balance b. Isoelectric potential c. Isoelectric point d. Iso ionic pointv
Solution

c. Isoelectric point

Answer:

c. Isoelectric point

Q.18Aspartate and Glutamate are amino acids of a. Negatively charged ‘R’ groups b. Positively charged ‘R’ groups c. Non-polar aliphatic ‘W groups d. Non-polar aromatic ‘R’ groupsv
Solution

a. Negatively charged ‘R’ groups

Answer:

a. Negatively charged ‘R’ groups

Q.19The test for protein ¡s a. iodine test b. Biuret test c. Benedict’s test d. Hydrolysis testv
Solution

b. Biuret test

Answer:

b. Biuret test

Q.20The competitive inhibitor is …………… for succinic dehydrogenase.v
  1. (a) malonate
  2. (b) succinate
  3. (c) oxalate
  4. (d) citrate
Solution

(a) malonate

Answer:

(a) malonate

Q.21Formation of new chemical bonds using ATP as a source of energy a. Lyase b. Hydrolase c. Telomerase d. Ligasev
Solution

d. Ligase

Answer:

d. Ligase

Q.22Uridylic acid is an a. Dinucleotide b. Nucleoside c. Nucleotide d. Ribo nucleotidev
Solution

d. Ribonucleotide

Answer:

d. Ribonucleotide

Q.23Phosphate forming linkage with sugar is known as a. diester linkage b. peptide linkage c. phosphodiester linkage d. Ionic linkagev
Solution

c. phosphodiester linkage

Answer:

c. phosphodiester linkage

Q.24…………… is a catalytic RNA.v
  1. (a) mRNA
  2. (b) Ribozyme
  3. (c) Ribonuclease
  4. (d) rRNA
Solution

(b) Ribozyme

Answer:

(b) Ribozyme

Q.25A class of lipid that serves as a major component of the cell membrane is a. triglyceride b. glycerol c. phospho lipid d. lipoproteinv
Solution

c. phospholipid

Answer:

c. phospholipid

Q.26One molecule of sucrose on hydrolysis give a. 2 molecules of glucose b. 1 molecule glucose & 1 molecule fructose c. 2 molecules of glucose & 1 molecule of fructose d. 2 molecules of fructosev
Solution

b. 1 molecule glucose & 1 molecule fructose

Answer:

b. 1 molecule glucose & 1 molecule fructose

Q.27In fibrous proteins polypeptide chai are held together by a. Vander Waals forces b. disulphide linkage c. electrostatic forces d. hydrogen bondsv
Solution

a. Vander Waals forces

Answer:

a. Vander Waals forces

Q.28According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………….v
  1. (a) 2
  2. (b) 3
  3. (c) 4
  4. (d) Nil
Solution

(a) 2

Answer:

(a) 2

Q.29Which polymer is stored in liver a. Amylose b. Amylo pectin c. Cellulose d. Glycogenv
Solution

d. Glycogen

Answer:

d. Glycogen

Q.30The bond that is not needed for protein formation is a. Hydrogen bond b. Peptide bond c. Ionic bond d. glucosidic bondv
Solution

d. glucosidic bond

Answer:

d. glucosidic bond

Q.31A complete turn of the helix comprises …………….v
  1. (a) 34 nm
  2. (b) 3.4 nm
  3. (c) 20 nm
  4. (d) 2nm
Solution

(b) 3.4 nm

Answer:

(b) 3.4 nm

Q.32The acid is also known as vitamin C a. Aspartic acid b. Tartaric acid c. Ascorbic acid d. Adipic acidv
Solution

c. Ascorbic acid

Answer:

c. Ascorbic acid

Q.33Which is the left-handed DNA?v
  1. (a) B – DNA
  2. (b) A – DNA
  3. (c) Z – DNA
  4. (d) dsDNA
Solution

(c) Z – DNA
II. Choose the wrong answer.

Answer:

(c) Z – DNA
II. Choose the wrong answer.

11VI. Assertion & Reason – Find Out The Correct Answer.12 questions
Q.1Assertion (A): Adhesion refers to the tendency of water molecules to cling together Reason (R): Because of hydrogen bonding, water molecules interact with one another continuous column of water is raised in xylem vessels. (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertion.v
Solution

(a) Assertion & Reason correct Reason Explaining Assertion.

Answer:

(a) Assertion & Reason correct Reason Explaining Assertion.

Q.1Define Micronutrients.v
Solution

Nutrients which are required in trace amounts is called micronutrients.
Cobalt, zinc, boron, copper, molybdenum, and manganese – essential for enzyme action.
Eg – Molybdenum is necessary for the fixation of nitrogen by enzyme nitrogenase.

Answer:

Nutrients which are required in trace amounts is called micronutrients.
Cobalt, zinc, boron, copper, molybdenum, and manganese – essential for enzyme action.
Eg – Molybdenum is necessary for the fixation of nitrogen by enzyme nitrogenase.

Q.2Assertion (A): Glycine is a non-essential amino acid Reason (R): It must be taken through diet (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertion.v
Solution

(c) Assertion is true but Reason is wrong.

Answer:

(c) Assertion is true but Reason is wrong.

Q.2Write down the properties of Water.v
Solution
  • It has Adhesion & cohesion property
  • High latent heat of vaporisation
  • High melting and boiling point
  • Universal solvent
  • Has specific heat capacity.
Answer:
  • It has Adhesion & cohesion property
  • High latent heat of vaporisation
  • High melting and boiling point
  • Universal solvent
  • Has specific heat capacity.
Q.3Assertion (A): In the presence of enzyme substance molecules can be attached by the reagent. Reason (R): Active sites of enzymes hold the substance in a suitable position. (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertionv
Solution

(a) Assertion & Reason correct Reason Explaining Assertion.

Answer:

(a) Assertion & Reason correct Reason Explaining Assertion.

Q.3Differentiate between Primary and Secondary Metabolites.v
Solution

Primary metabolites
Secondary Metabolites
Required for the basic metabolic processes, like Photosynthesis, Respiration, Protein & lipid metabolism.
No direct function in the growth and development of organisms.

Answer:

Primary metabolites
Secondary Metabolites
Required for the basic metabolic processes, like Photosynthesis, Respiration, Protein & lipid metabolism.
No direct function in the growth and development of organisms.

Q.4Assertion (A): Aminoacids behave like salt rather than simple amines or carboxylic acid. Reason (R): In aqueous solution, the COOH group of amino acid loses a protein and the NH2 group accepts a proton to form zwitterion (salt). (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertionv
Solution

(a) Assertion & Reason correct Reason Explaining Assertion.
2 Marks

Answer:

(a) Assertion & Reason correct Reason Explaining Assertion.
2 Marks

Q.4Define Polymerisation.v
Solution

A process in which repeating subunits termed monomers is bound into chains of different lengths called polymers.
Eg – Starch – Polynucleotide.

Answer:

A process in which repeating subunits termed monomers is bound into chains of different lengths called polymers.
Eg – Starch – Polynucleotide.

Q.5Distinguish between Glycogen and Cellulose.v
Solution

Glycogen
Cellulose
Storage polysaccharide
Structural polysaccharide
Animal starch
Plant starch
made up of glucose with ( α- 1-6) linked branches
Made up of 1000s of glucose units held by β glucose units held by 1,4 glucosidic linkage
Seen in liver cells skeletal muscle fibre throughout the human body except brain.
Occur in cotton. In the form of nitrocellulose used as explosives.

Answer:

Glycogen
Cellulose
Storage polysaccharide
Structural polysaccharide
Animal starch
Plant starch
made up of glucose with ( α- 1-6) linked branches
Made up of 1000s of glucose units held by β glucose units held by 1,4 glucosidic linkage
Seen in liver cells skeletal muscle fibre throughout the human body except brain.
Occur in cotton. In the form of nitrocellulose used as explosives.

Q.6Distinguish between Dinucleotide & Polynucleotide.v
Solution

Dinucleotide
Polynucleotide
2 nucleotide joined to form Dinucleotide
They are linked through 3′- 5′- Phospho – diester linkage by condensation between phosphate group of one with the sugar of other.
Like dinucleotide,when many nucleotides then it leads to the formation of polynucleotides Eg-DNA-RNA

Answer:

Dinucleotide
Polynucleotide
2 nucleotide joined to form Dinucleotide
They are linked through 3′- 5′- Phospho – diester linkage by condensation between phosphate group of one with the sugar of other.
Like dinucleotide,when many nucleotides then it leads to the formation of polynucleotides Eg-DNA-RNA

Q.7Differentiate between Nucleoside & Nucleotide.v
Solution

Nucleoside
Nucleotide
Nitrogen + Sugar → Nucleoside
Eg – Adenine + Ribose → Adenosine
Nucleoside +→ Nucleotide Phosphoric acid (N + S) + P
Adenine + Ribose → Adenosine
Adenosine + Phosphoric Acid → Adenylic acid

Answer:

Nucleoside
Nucleotide
Nitrogen + Sugar → Nucleoside
Eg – Adenine + Ribose → Adenosine
Nucleoside +→ Nucleotide Phosphoric acid (N + S) + P
Adenine + Ribose → Adenosine
Adenosine + Phosphoric Acid → Adenylic acid

Q.8State Chargaff’s Law.v
Solution

Chargaffs Law in 1949
I. A = T & G = C
Between A & T double bond Between G & C triple bond
II. A + G number equal to T + C
III. But A: T, need not be equal to G: C

Answer:

Chargaffs Law in 1949
I. A = T & G = C
Between A & T double bond Between G & C triple bond
II. A + G number equal to T + C
III. But A: T, need not be equal to G: C