Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 8 Biomolecules Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules

11th Bio Botany Guide Biomolecules Text Book Back Questions and Answers

Part I

Question 1.
The most basic amino acid is
a. Arginine
b. Histidene
c. Glycine
d. Glutamine
Answer:
c. GIycine

Question 2.
An 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 malonate
Answer:
c. Allosteric inhibition of hexokinase by glucose-6-phosphate

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Enzymes the catalyse interconversion of optical, geometrical or positional isomers are
a. Ligases
b. Lyases
c. Hydrolases
d. Isomerases
Answer:
d. Isomerases

Question 4.
Proteins 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. Hormones
Answer:
d.Hormones

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Given 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.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 1

Category   Compound
I. Cholesterol A. Guanine
II. Amino acid B. IVH2
III. Nucleotide C. Adenine
IV. Nucleoside D. Uracil

Answer:
IV

Question 6.
Distinguish  between Nitrogen base and a base found in inorganic chemistry
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 2
Question 7.
What are the factors affecting the rate of enzyme reactions?
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.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Briefly outline the classification of enzymes?
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 3

Question 9.
Write down the characteristic features of DNA?
Answer:
The characteristic feature of DNA.

  1. 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.
  2. 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.
  3. 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.
  4. 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°).
  5. 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.
  6. Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
  7. The phosphodiester linkages give an inherent polarity to the DNA helix. They form strong covalent bonds, gives strength and stability to the polynucleotide chain.
  8. 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.
  9. Based on the helix and the distance between each turn, the DNA is of three forms – A DNA, B DNA and Z DNA.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 10.
Explain the structure and function of different types of RNA?
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.

Question 1.
Who invented the electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer:
(c) Knoll and Ruska

Question 2.
Polysaccharides also called
a. Polymers
b. Glycans
c. Glycosidic compounds
d. Glycones
Answer:
b. Glycans

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Omnis – cellula – e – cellula was given by ……………. (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer:
(a) Virchow

Question 4.
Nitrocellulose is used in making
a. cellophane
b. drapers
c. explosives
d. pain balms
Answer:
c. explosives

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Genes present in the cytoplasm of eukaryotic cells are found in ……………. (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids inherited via male gametes
Answer:
(a) mitochondria and inherited via egg cytoplasm

Question 6.
Chitin when added with amino acid becomes
a.myeopolysaccharide
b. amylopolysaceharide
c .mucopolysaccharide
d. peptidopolysaccharide
Answer:
c. mucopolysaccharide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 7.
A quantosome is present in …………… . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer:
(b) Chloroplast

Question 8.
Among the following one is not a non-polar solvent
a. benzene
b. sulphuric acid
c. ether
d. chloroform
Answer:
b. sulphuric acid

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 9.
One of the given below is a complex found in the cell membrane of animal cell
a. cholesterol
b. myelin
c. proline
d. Ieeithin
Answer:
a. cholesterol

Question 10.
A major site for the synthesis of lipids ……………. (2013 NEET)
(a) Rough ER
(b) smooth ER
(c) Centriole
(d) Lysosome
Answer:
(b) smooth ER

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 11.
Principle information molecules of the cell are known as
a. Nucleus
b. DNA
c. RNA
d. Nucleic acids
Answer:
d. Nucleic acids

Question 12.
(I) Cellulose – A most abundant organic compound
(II) Morphine – Pain relieving alkaloid
(III) Aldose – reducing sugar & Ketose
(IV) Glycogen – mucopolysaccharide
Answer:
(IV) Glycogen – mucopolysaccharide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 13.
The following is a general formula of
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 4
a. Amino acid
b. Fatty acid
c. Nucleotide
d. Monosaccharide
Answer:
a. Amino acid

Question 14.
Lactose is a disaccharide of
a. Glucose – Glucose
b. Fructose – Fructose
c. Glucose – Galactose
d. Fructose – Galactose
Answer:
c. Glucose – Galactose

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 15.
Number of fatty acids in triglyceride is …………… .
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
Heparin the anti-coagulant is got from
a. D – glucuronic acid
b. Polymer of fructose
c. Mucopolysaccharide from red algae
d. Glucosaminoglycan
Answer:
d. Glycosaminoglycan

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 17.
The pH at which Zwitterion is formed is known as
a. Iso ionic balance
b. Isoelectric potential
c. Isoelectric point
d. Iso ionic point
Answer:
c. Isoelectric point

Question 18.
Aspartate 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’ groups
Answer:
a. Negatively charged ‘R’ groups

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 19.
The test for protein ¡s
a. iodine test
b. Biuret test
c. Benedict’s test
d. Hydrolysis test
Answer:
b. Biuret test

Question 20.
The competitive inhibitor is …………… for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer:
(a) malonate

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 21.
Formation of new chemical bonds using ATP as a source of energy
a. Lyase
b. Hydrolase
c. Telomerase
d. Ligase
Answer:
d. Ligase

Question 22.
Uridylic acid is an
a. Dinucleotide
b. Nucleoside
c. Nucleotide
d. Ribo nucleotide
Answer:
d. Ribonucleotide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 23.
Phosphate forming linkage with sugar is known as
a. diester linkage
b. peptide linkage
c. phosphodiester linkage
d. Ionic linkage
Answer:
c. phosphodiester linkage

Question 24.
…………… is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer:
(b) Ribozyme

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 25.
A class of lipid that serves as a major component of the cell membrane is
a. triglyceride
b. glycerol
c. phospho lipid
d. lipoprotein
Answer:
c. phospholipid

Question 26.
One 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 fructose
Answer:
b. 1 molecule glucose & 1 molecule fructose

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 27.
In fibrous proteins polypeptide chai are held together by
a. Vander Waals forces
b. disulphide linkage
c. electrostatic forces
d. hydrogen bonds
Answer:
a. Vander Waals forces

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

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 29.
Which polymer is stored in liver
a. Amylose
b. Amylo pectin
c. Cellulose
d. Glycogen
Answer:
d. Glycogen

Question 30.
The bond that is not needed for protein formation is
a. Hydrogen bond
b. Peptide bond
c. Ionic bond
d. glucosidic bond
Answer:
d. glucosidic bond

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 31.
A complete turn of the helix comprises …………….
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer:
(b) 3.4 nm

Question 32.
The acid is also known as vitamin C
a. Aspartic acid
b. Tartaric acid
c. Ascorbic acid
d. Adipic acid
Answer:
c. Ascorbic acid

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 33.
Which is the left-handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer:
(c) Z – DNA

II. Choose the wrong answer.

Question 1.
(a) Hydrolase – Amylase
(b) Oxidoreductase – Dehydrogenase
(c) Transferase – Transaminase
(d) Isomerase – Hexokinase
Answer:
(d) Isomerase – Hexokinase

Question 2.
This has nothing to do with the structure of
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 5
a. Cytosine
b. Pyrimidine
c. Adenine
d. Thyamine
Answer:
c. Adenine

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Choose 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 group
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.

Question 1.
(I) Morphine – A. Hectins
(II) Concanavalin A – B. Drug
(III) Vinblastin – C. Pigment
(IV) Anthocyanin – D. Toxin
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 6
Answer:
(a) B – A – D – C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 2.
(I) Lactose – A. Penta saccharide
(II) Ramnose – B. Tetra saccharide
(III) Stachyose – C. Disaccharide
(IV) Verbascose – D. Tri saccharide
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 7
Answer:
(b) C-D-B-A

Question 3.
(I) Fred Sanger 1st sequenced
(II) Di sulphide bridges formed between sulphur & amino acids
(III) non-protein enzyme
(IV) homo polysaccharide with amino acid
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 8
Answer:
(b) D-A-B-C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 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 Protein
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 9
Answer:
(c) C – A – D – B

Question 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 nature
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 10
Answer:
(c) C-B-D-A

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.

Question 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 unsaturated
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 11
Answer:
(a) True – True – False – False

Question 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.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 12
Answer:
(d) Deoxyribose sugar – Phosphate – Nitrogen base – Nucleotide

V.

Question 1.
Label the diagram parts correctly by choosing the right option.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 13

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 Nucleotide

Answer:
(d) Deoxyribose sugar – Phosphate Nitrogen base – Nucleotide

Question 2.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 14

A

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 Chromatids

Answer:
(b) P arm – Centromere – Sister chromatids – Q arm

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

VI. Assertion & Reason – Find Out The Correct Answer.

Question 1.
Assertion (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.
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

Question 2.
Assertion (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.
Answer:
(c) Assertion is true but Reason is wrong.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Assertion (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 Assertion
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

Question 4.
Assertion (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 Assertion
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

2 Marks

Question 1.
Define Micronutrients.
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.

Question 2.
Write down the properties of Water.
Answer:

  • It has Adhesion & cohesion property
  • High latent heat of vaporisation
  • High melting and boiling point
  • Universal solvent
  • Has specific heat capacity.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Differentiate between Primary and Secondary Metabolites.
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.

Question 4.
Define Polymerisation.
Answer:
A process in which repeating subunits termed monomers is bound into chains of different lengths called polymers.
Eg – Starch – Polynucleotide.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Distinguish between Glycogen and Cellulose.
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.

Question 6.
Distinguish between Dinucleotide & Polynucleotide.
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

Question 7.
Differentiate between Nucleoside & Nucleotide.
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

Question 8.
State Chargaff’s Law.
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

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 9.
Differentiate Between DNA & RNA.
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

Question 10.
Distinguish between Cation, Zwitterion & Anion.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 15

Question 11.
How will you test reducing sugar?
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.

Question 12.
Draw the structure of a basic amino acid.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 16

3 Marks

Question 1.
Classify Polysaccharides.
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

Question 2.
Why do we call Glucose and Fructose Isomers -Discuss.
Answer:
Glucose and Fructose have a same molecular formula, C6H1206 but different structural formulas- so they are known as Isomers.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 17

Question 3.
Explain the formation of Disaccharide Lactose.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 18

Question 4.
Draw the structure of Fatty acid.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 19

Question 5.
Distinguish between Waxes & Steroids.
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.

Question 6.
Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.
Answer:
The non-polar aliphatic R group has 6 amino acids Glycine, Alanine, Valine, Leucine, Methionine & Isoleucine.
Structure of Glvcine Alanine Valine
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 20

Question 7.
Distinguish between Macronutrients & Micronutrients.
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

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Define Activation energy?
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.

Question 9.
Distinguish between Primary metabolite & Secondary metabolite.
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.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 10.
Give examples for Secondary metabolites.
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

Question 11.
Draw the various structures of Protein.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 21

Question 12.
Draw the structure of Purine – Adenine & Guanine Nucleotides.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 22

Question 13.
Why do some people have curly hair?
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.

Question 14.
Deoxyribose (C5H10O4) is not a carbohydrate – Discuss.
Answer:
Carbohydrates are hydrates of carbon Deoxy ribose is a carbohydrate, but its formula C5H10O4 -does not apply the general formula of Carbohydrate (C2H20)X formula has Carbohydrates formula.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

5 Marks Questions

Question 1.
How will you identify the presence of glucose in a given food sample?
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 Cu2+ 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.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 2.
What is protein denaturation.
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.

Question 3.
Tabulate other sugar compounds
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

Question 4.
Classify enzyme reactions:
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 23
II. a. Extracellular enzymes
Enzymes secreted outside & work externally Eg digestive enzymes. b. Intracellular Enzymes
Remain within cells & work there Eg Insulin.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Explain the three types of Co-Factors.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 24

II. Prosthetic groups:
Eg. Vit B2(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.

Question 6.
Tabulate the uses of enzymes
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

Question 7.
Enumerate the properties of Enzyme.
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.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Explain lock and key mechanism of enzymes.
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
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 25

  • 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.

Question 9.
What are the various types of inhibitors of enzymes.
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.

Question 10.
Distinguish between feedback allosteric inhibition negative feedback (end product) inhibition.
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.

Question 11.
Tabulate other sugar compounds.
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

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules