Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Multiple Choice Questions

Question 1.
How many of the following characteristics are shown by the R-strain of Streptococcus pneumonia? Avirulent, Smooth, Pathogenic, Capsulated ………………..
(a) One
(b) TWo
(c) Three
(d) Four
Answer:
(a) One

Question 2.
Griffith obtained …………….. from the blood of the dead mice.
(a) dead S-strain bacteria
(b) live R-strain bacteria
(c) dead R-strain bacteria
(d) live S-strain bacteria
Answer:
(d) live S-strain bacteria

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 3.
Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty demonstrated that ………………..
(a) transformation of live S-strain bacteria into R-strain type was because of DNA of bacteria of S-strain.
(b) the transforming substance was either a protein or RNA.
(c) only DNA was able to transform harmless R-strain into virulent S-strain.
(d) when DNA isolated from S-strain bacteria, was digested with DNase, the transformation occurred.
Answer:
(c) only DNA was able to transform harmless R-strain into virulent S-strain

Question 4.
Which of the following was NOT observed in Hershey and Chase experiment?
(a) Viruses grown in the presence of radioactive sulphur, had radioactive protein but not radioactive DNA.
(b) Radioactive ‘P’ remained in suspension.
(c) Only radioactive ‘P’ was found inside the bacterial cells in the pellet.
(d) Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive proteins.
Answer:
(b) Radioactive ‘P’ remained in suspension.

Question 5.
Enzymes like ……………….. and DNA topoisomerase-I, play important role in maintaining super-coiled state in prokaryotic DNA.
(a) DNA ligase
(b) DNA gyrase
(c) RNA polymerase
(d) None of these
Answer:
(b) DNA gyrase

Question 6.
Histone octamer of nucleosome has two molecules, each of ……………….. proteins.
(a) H2A, H2B, H3 and H4
(b) H2A, H2B, H3 and H1
(c) H2A, H2B. H3A and H3B
(d) H1A, H2B, H3A and H4
Answer:
(a) H2A, H2B, H3 and H4

Question 7.
Select the CORRECT statement.
(a) Euchromatin is mainly located near centromere and telomeres.
(b) Heterochromatin replicates at faster rate than euchromatin.
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin.
(d) Heterochromatin is lightly stained region of chromonema.
Answer:
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin

Question 8.
A DNA molecule in which both strands have 14N is allowed to replicate in an environment containing 15N. What will be the exact number of DNA molecules that contain the 14N after three replications?
(a) One
(b) Two
(c) Four
(d) Eight
Answer:
(b) Two

Question 9.
As the base sequence present on one strand of DNA decides the base sequence of other 8 strand, this strand is considered as ………………..
(a) descending strand
(b) leading strand
(c) lagging strand
(d) complementary strand
Answer:
(d) complementary strand

Question 10.
In prokaryotes ……………….. recognizes the promoter sequence.
(a) alpha factor
(b) rho factor
(c) theta factor
(d) sigma factor
Answer:
(d) sigma factor

Question 11.
If the base sequence in DNA is 5′ AAAA 3′, then the base sequence in m-RNA is ………………..
(a) 5′ UUUU 3′
(b) 3′ UUUU 5′
(c) 5′ AAAA 3′
(d) 3′ TTTT 5′
Answer:
(c) 5′ AAAA 3′

Question 12.
During capping, methylated guanosine tri¬phosphate is added to 5′ end of ………………..
(a) m-RNA
(b) t-RNA
(c) hnRNA
(d) r-RNA
Answer:
(c) hnRNA

Question 13.
If each codon has two nucleotides, then there will be ……………….. codons, which can encode for only …………….. different types of amino acids.
(a) 16, 16
(b) 16, 20
(c) 20, 16
(d) 64, 64
Answer:
(a) 16, 16

Question 14.
What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?
(a) A polypeptide of 24 amino acids is formed.
(b) Two polypeptides of 24 and 25 amino acids will be formed.
(c) A polypeptide of 49 amino acids is formed.
(d) A polypeptide of 25 amino acids is formed.
Answer:
(a) A polypeptide of 24 amino acids is formed.

Question 15.
A strand of DNA has following base sequence – 3′ AAAAGTGAATAGTGA 5′. On transcription it produces an m-RNA. Which of the following anticodon of t-RNA recognizes the third codon of this m-RNA?
(a) AAA
(b) CUG
(c) AAG
(d) CTG
Answer:
(c) AAG

Question 16.
Polynucleotide chain consisting of only CUA repeats will give polypeptide chain with only one amino acid ………………..
(a) tryptophan
(b) leucine
(c) serine
(d) methionine
Answer:
(b) leucine

Question 17.
Select the INCORRECT statement.
(a) Dr. Khorana prepared polyribo-nucleotides chains with known repeated sequences of two or three nucleotides by using synthetic DNA.
(b) M. Nirenberg and Matthaei synthesized artificial m-RNA which was a homopolymer of uracil ribonucleotides.
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.
(d) Evidence for triplet nature of geneticcode, was given by Crick (1961) using “frame-shift mutation”.
Answer:
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.

Question 18.
…………… is/are based on complementarity principle.
(a) Replication and translation
(b) Replication and transcription
(c) Translation
(d) Only replication
Answer:
(b) Replication and transcription

Question 19.
Cysteine has codons, while isoleucin has ……………….. codons.
(a) two, three
(b) three, two
(c) two, four
(d) four, two
Answer:
(a) two, three

Question 20.
Out of 64 codons, only 61 code for the 20 different amino acids. This is known as ……………….. of genetic code.
(a) non-ambiguity
(b) overlapping nature
(c) ambiguity
(d) degeneracy
Answer:
(d) degeneracy

Question 21.
Mutation that results in Sickle-cell anaemia is a ………………..
(a) deletion
(b) frame-shiftmutation
(c) point mutation
(d) insertion
Answer:
(c) point mutation

Question 22.
Initiator charged t-RNA occupies the ……………….. of ribosome first.
(a) A-site
(b) P-site
(c) E-site
(d) either A-site or P-site
Answer:
(b) P-site

Question 23.
It takes ……………….. for formation of peptide bond.
(a) 10 seconds
(b) 0.1 second
(c) less than 0.1 second
(d) 60 seconds
Answer:
(c) less than 0.1 second

Question 24.
Anticodon and codon bind by ………………..
(a) glycosidic bond
(b) hydrogen bond
(c) phosphodiester bond
(d) none of these
Answer:
(b) hydrogen bond

Question 25.
The UTRs are present at ………………..
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA
(b) 5′-end, before start codon and at 3′-end, after stop codon of t-RNA
(c) only at 3′-end, after stop codon of m -RNA
(d) only at 5′-end, before start codon of m-RNA
Answer:
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA

Question 26.
The action of structural genes is regulated by …………….. site with the help of a …………….. protein.
(a) operator, inducer
(b) operator, repressor
(c) regulator, repressor
(d) regulator, inducer
Answer:
(b) operator, repressor

Question 27.
Repressor protein is produced by the action of ………………..
(a) gene z
(b) gene y
(c) gene i
(d) gene o
Answer:
(c) gene i

Question 28.
Select the correct pair.
(a) Gene z – Transacetylase
(b) Gene y – Beta-galactocidase
(c) Gene a – Beta-galactoside permease
(d) Gene I – Repressor
Answer:
(d) gene I – repressor

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 29.
Structural genomics involves ……………….. of genome.
(a) mapping
(b) sequencing
(c) analysis
(d) all of these
Answer:
(d) all of these

Question 30.
The technique of transferring DNA fragments separated on agarose gel to a synthetic nitrocellulose membrane is known as ………………..
(a) Southern blotting
(b) Autoradiography
(c) Southern hybridization
(d) None of these
Answer:
(a) Southern blotting

Question 31.
Sequence of various steps in DNA fingerprinting is ………………..
i. Southern blotting.
ii. Restriction digestion
iii. Agarose gel electrophoresis
iv. DNA isolation
v. Photography.
vi. Selection of DNA probe
vii. Hybridization
(a) iv, iii, ii, i, v, vi, vii
(b) iv. v, iii, i, vi, vii, ii
(c) iv, ii, iii, i, vi, vii, v
(d) ii, iii, iv, i, vi, vii,
Answer:
(c) iv, ii, iii, i, vi. vii, v

Match the Columns

Question 1.

Column A Column B
(1) Frederick Griffith (a) Test tube assay
(2) Avery, McCarty and MacLeod (b) Streptococcus pneumoniae
(3) Alfred Hershey and Martha Chase (c) E. coli
(4) Meselson and Stahl (d) Bacteriophages

Answer:

Column A Column B
(1) Frederick Griffith (b) Streptococcus pneumoniae
(2) Avery, McCarty and MacLeod (a) Test tube assay
(3) Alfred Hershey and Martha Chase (d) Bacteriophages
(4) Meselson and Stahl (c) E. coli

Classify the following to form Column B as per the category given in Column A

Question 1.
(i) UUU
(ii) CUA
(iii) UAA
(iv) AUG
(v) UAG
(vi) UGA

Column A Column B
1. Initiator codon ————–
2. Stop codons ————-
3. Codon that codes for Phenyl alanin ————–
4. Codon that codes for leucine ————-

Answer:

Column A Column B
1. Initiator codon (iv) AUG
2. Stop codons (iii) UAA, (v) UAG, (vi) UGA
3. Codon that codes for Phenyl alanin (i) UUU
4. Codon that codes for leucine (ii) CUA

Question 2.
(i) Severo Ochoa
(ii) F. Jacob and J. Monod
(iii) Temin and Baltimore
(iv) H. Winkler
(v) T. H. Roderick
(vi) R Kornberg

Column A Column B
(1) Lac Operon ————–
(2) Central dogma in retroviruses ————-
(3) Coined the term Genome ————–
(4) Coined the term Genomics ————-
(5) Enzymatic synthesis of RNA ————-
(6) DNA is associated with histones and non-histones ————–

Answer:

Column A Column B
(1) Lac Operon (ii) E Jacob and J. Monod
(2) Central dogma in retroviruses (iii) Temin and Baltimore
(3) Coined the term Genome (iv) H. Winkler
(4) Coined the term Genomics (v) T. H. Roderick
(5) Enzymatic synthesis of RNA (i) Severo Ochoa
(6) DNA is associated with histones and non-histones (vi) R. Kornberg

Very Short Answer Questions

Question 1.
What are the two types of bacteria used by F. Griffith and which one out of these is avirulent?
Answer:
S-type and R-type strains of Streptococcus penumoniae were used by F. Griffith and out of these R-type is avirulent.

Question 2.
Enlist the characteristics of S-strain pneumoniae.
Answer:
S-strain pneumoniae are virulent, smooth and encapsulated.

Question 3.
What is the bacteriophage?
Answer:
Bacteriophage is a virus that infects bacterium and injects its genetic material in the bacterium.

Question 4.
What is the length of DNA double helix molecule in a typical mammalian cell?
Answer:
The length of DNA double helix molecule in a typical mammalian cell is approximately 2.2 meters.

Question 5.
What is the approximate size of a typical nucleus ?
Answer:
Approximate size of a typical nucleus is 10-6 m.

Question 6.
What is size of E. coli cell?
Answer:
The size of E. coli cell size is 2-3 µm.

Question 7.
What determines the charge on protein molecules?
Answer:
A protein acquires its charge depending upon the abundance of amino acid residues with charged side chains.

Question 8.
What is nucleosome core?
Answer:
Nucleosome core is a histone octamer.

Question 9.
Where is H1 histone present?
Answer:
H1 histone binds the DNA thread where it enters and leaves the nucleosome.

Question 10.
How are solenoid fibres formed?
Answer:
Six nucleosomes get coiled and then form solenoid that looks like coiled telephone wire of 30 nm diameter (300Å).

Question 11.
How is chromatin fibre formed?
Answer:
Supercoiling of solenoid fibre forms a looped structure called chromatin fibre.

Question 12.
What is NHC?
Answer:
NHC stands for Nonhistone Chromosomal proteins.

Question 13.
List as many different enzyme activities required during DNA synthesis as you can.
Answer:
Phosphorylase, Helicase, DNA polymerase, Primase, DNA ligase, Super helix relaxing enzyme, Topoisomerase (gyrase) are different enzymes required during DNA synthesis.

Question 14.
How many replicons are present in prokaryotes and eukaryotes respectively?
Answer:
Prokaryotes have one replicon. Several replicons in tandem are present in eukaryotes.

Question 15.
What is the function of SSBP?
Answer:
During replication of DNA SSBP proteins remain attached to both the separated strands and prevent them from coiling back.

Question 16.
During which phases of cell cycle, transcription occurs in the nucleus?
Answer:
Transcription occurs in the nucleus during G1 and G2 phases of cell cycle.

Question 17.
Which strand of transcription unit gets transcribed ?
Answer:
DNA strand having 3’ → 5’ polarity acts as template strand and it gets transcribed.

Question 18.
What is a cryptogram?
Answer:
Cryptogram is a genetic code consisting of triplet codons on m-RNA that code for a specific amino acids.

Question 19.
What is meant by the polarity of genetic code?
Answer:
Genetic code is always read in 5′ → 3’ direction. This is called polarity of genetic code.

Question 20.
Which mutation can result in changes in the reading frame?
Answer:
Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.

Question 21.
Which mutation can result in insertion or deletion of amino acids, but reading frame remains unaltered?
Answer:
Insertion or deletion of three or multiples of three bases results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 22.
Which molecule serves as an intermediate molecules between DNA and protein during proteins synthesis?
Answer:
RNA serves as an intermediate molecule between DNA and protein during proteins synthesis.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 23.
What is the function of a groove present between two subunits of ribosome in eukaryotes ?
Answer:
The groove present between two subunits of ribosomes in eukaryotes protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 24.
Enlist different steps of protein synthesis.
Answer:
Steps in protein synthesis are:

  1. Transcription
  2. Activation of amino acids and formation of charged t-RNAs,
  3. Synthesis of polypeptide chain:
  4. initiation
  5. elongation and
  6. termination of polypeptide chain.

Question 25.
What is translocation?
Answer:
During elongation of polypeptide chain, the ribosome moves along the m-RNA in stepwise manner from start codon to stop codon (5′ → 3′), 1 codon ahead each time t, his movement is called translocation and due to this t-RNA carrying a dipeptide at A-site of the ribosome moves to the p-site.

Question 26.
What is meant by inducible enzymes?
Answer:
Bacteria like E.coli adapt to their chemical environment by synthesizing certain enzymes depending upon the substrate present. Such adaptive enzyme is called inducible enzyme.

Question 27.
What is meant by induction and inducer?
Answer:
A set of genes are switched on when a new substrate is to be metabolized. This phenomenon is called induction and small molecule responsible for this is known as inducer.

Question 28.
What is the role of a repressor gene?
Answer:
The role of a repressor gene is to produce repressor protein. Repressor binds with operator gene and this prevents transcription of structural genes in the operon.

Question 29.
Which molecule does act as inducer molecule in lac operon?
Answer:
Allolactose acts as inducer molecule in lac operon.

Question 30.
In which condition, lac operon is switched off?
Answer:
If E.coli bacteria do not have lactose in the surrounding medium as a source of energy, lac operon is switched off.

Question 31.
What lac operon consists of?
Answer:
Lac operon consists of a regulator promoter, operator and three structural genes z, y and a.

Question 32.
Which gene acts as a regulatory gene in lac operon?
Answer:
Repressor protein is produced by the action of gene i (inhibitor). This gene acts as a regulator gene.

Question 33.
When was Human Genome Project started ? When was it completed ?
Answer:
The Human Genome Project was started in 1990 and was completed in 2003.

Question 34.
What is functional genomics?
Answer:
Functional genomics is a branch of genomics that involves the study of functions of all gene sequences and their expressions in organisms.

Question 35.
What is the advantage of sequencing of genomes of non-human organisms?
Answer:
Sequencing of genomes of non-human model organisms allows researchers to study gene functions in these organisms. Since human beings possess many genes which are like those of flies, roundworms and mice, comparative studies will lead to greater understanding of human evolution.

Question 36.
What are VNTRs?
Answer:
Variable Number of Tandem Repeats (VNTRs) are unusual sequences of 20-100 base pairs, which are repeated several times and are arranged tandency.

Question 37.
Do different organisms have the same DNA?
Answer:
Different organisms differ in their DNA sequence.

Question 38.
What is the amino acid sequence encoded by base sequence UCA, UUU, UCC, GGG, AGU of an m-RNA segment?
Answer:
The amino acid sequence: Ser-Phe – Ser – Gly- Ser

Give definitions of the following

Question 1.
Replicon
Answer:
The unit of DNA in which replication occurs is known as replicon.

Question 2.
Transcription
Answer:
Transcription is defined as the process of copying of genetic information from template strand of DNA into a complementary single stranded RNA transcript.

Question 3.
Gene
Answer:
Gene is defined as the DNA sequence coding for m-RNA/ t-RNA or r-RNA.

Question 4.
Cistron
Answer:
Cistron is defined as a segment of DNA coding for a polypeptide.

Question 5.
Monocistronic gene
Answer:
Gene is called monocistronic, when there is a single structural gene in one transcription unit.

Question 6.
Polycistronic gene
Answer:
Gene is called polycistronic, when there is a set of various structural genes in one transcription unit.

Question 7.
Interrupted or Split genes
Answer:
Interrupted or split genes are the structural genes in eukaryotes which have both exons and introns.

Question 8.
Exons
Answer:
Exons are the coding sequences or express sequences in DNA/hnRNA/ m-RNA.

Question 9.
Introns
Answer:
Introns are the non-coding sequences in DNA or hnRNA.

Question 10.
Anticodon
Answer:
Anticodon is a triplet of nucleotides present on the anticodon loop of t-RNA, which is complementary to codon on m-RNA.

Question 11.
Mutation
Answer:
Mutation is a sudden heritable change in the DNA sequence that results in the change of genotype.

Question 12.
Translation
Answer:
Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.

Question 13.
Genomics
Genomics is the study of genomes through analysis, sequencing and mapping of genes along with the study of their functions.

Question 14.
Repressors
Answer:
Repressors are proteins which are able to bind the operator region of operon and prevent the RNA polymerase from transcribing the operon.

Name the following

Question 1.
Enzyme that cleaves DNA.
Answer:
DNase

Question 2.
Enzyme that cleaves proteins.
Answer:
Protease

Question 3.
Enzyme involved in activation of nucleotides.
Answer:
Phosphorylase

Question 4.
Enzyme involved in unwinding of DNA.
Answer:
Helicase

Question 5.
Enzyme involved in synthesis of DNA.
Answer:
DNA polymerase

Question 6.
Enzyme involved in joining of Okazaki fragments.
Answer:
DNA ligase

Question 7.
Enzyme involved in synthesis of RNA primer.
Answer:
Primase

Question 8.
Enzyme involved in removal of RNA primer.
Answer:
DNA polymerase

Question 9.
Enzyme involved in replacement of gaps in prokaryotes.
Answer:
DNA polymerase – I

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 10.
Enzyme involved in replacement of gaps in eukaryotes.
Answer:
DNA polymerase α

Question 11.
Enzyme involved in formation of double helix in daughter DNA molecules.
Answer:
Topoisomerase

Question 12.
Enzyme involved in releasing strain created by unwinding of DNA.
Answer:
Super helix relaxing enzyme

Question 13.
Enzyme involved in synthesis of hnRNA, m-RNA.
Answer:
RNA polymerase – II

Question 14.
Enzyme involved in synthesis of t-RNA, snRNA.
Answer:
RNA polymerase – III

Question 15.
Enzyme involved in synthesis of r-RNA
Answer:
RNA polymerase – I

Question 16.
Enzyme involved in polymerizing RNA in template independent manner.
Answer:
Polynucleotide phosphorylase

Question 17.
Enzyme involved in peptide bond synthesis.
Answer:
Ribozyme

Question 18.
Enzyme involved in Cutting DNA at specific sites.
Answer:
Restriction endonuclease

Question 19.
Name the initiator codon of protein synthesis.
Answer:
AUG is the initiator codon of protein synthesis.

Question 20.
Name three binding sites of ribosome.
Answer:
Three binding sites for t-RNA on ribosomes are P-site (peptidy t-RNA-site), A-site (aminoacyl – t-RNA-site) and E-site (exit site).

Question 21.
Name the different structural genes in sequence of lac operon.
Answer:
There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.

Question 22.
Name the organisms whose genomes have been sequenced?
Answer:
The genomes of several organisms such as bacteria e.g. E.coli, Caenorhabditis elegans (a free living non-pathogenic nematode), Saccharomyces cerevisiae (yeast), Drosophila (fruit fly), plants (rice and Arabidopsis), Mus musculus (mouse), etc. have been sequenced.

Give Significance/Functions of the following

Question 1.
DNA.
Answer:

  1. DNA regulates and controls all the cellular activities.
  2. It replicates and gets distributed equally to the daughter cells when the cell divides.
  3. It is a carrier of genetic information.
  4. Heterocatalytic function : DNA directs the synthesis of chemical molecules other than itself. E.g. Synthesis of RNA (transcription), synthesis of protein (Translation), etc.
  5. Autocatalytic function : DNA directs the synthesis of DNA itself. E.g. Replication.
  6. DNA is a master molecule of a cell that initiates, guides, regulates and controls the process of protein synthesis.

Question 2.
Proteins.
Answer:
Proteins serve as structural components, enzymes and hormones.

Distinguish between the following.

Question 1.
Euchromatin and Heterochromatin.
Answer:

Euchromatin Heterochromatin.
1. Euchromatin is loosely packed region of the chromatain. 1. Heterochromatin is densely packed region of the chromatin.
2. Euchromatin stains lightly. 2. Heterochromatin stains darkly.
3. Euchromatin is transcriptionally active region of the chromatin. 3. Heterochromatin is transcriptionally inactive region of the chromatin.

Question 2.
DNA in prokaryotic and eukaryotic cells.
Answer:

DNA in prokaryotes DNA in eukaryotic
1. It is present in the cytoplasm. 1. It is present in the nucleus.
2. It is not associated with histones. 2. It is associated with histones.
3. It is circular. 3. It is linear.
4. Genes do not contain introns. 4. Genes contain introns along with exons.
5. Genes are polycostronic. 5. Genes are monocistronic.

Question 3.
DNA and RNA
Answer:

DNA RNA
1. DNA is deoxyribonucleic acid. 1. RNA is ribonucleic acid.
2. DNA is double stranded, helical molecule. 2. RNA is single stranded molecule.
3. In DNA, there is deoxyribose sugar. 3. In RNA, there is ribose sugar.
4. The pyrimidine nitrogen bases are cytosine and thymine. 4. The pyrimidine nitrogen bases are cytosine and uracil.
5. DNA is the genetic material in all types of organisms. 5. RNA is genetic material in few viruses only.
6. In eukaryotic cells, DNA is present in nucleus. 6. In eukaryotic cells, RNA is present in nucleus as well as cytoplasm.
7. The number of purine : pyrimidine ratio is always 1 : 1 in DNA molecule. 7. The number of purine : pyrimidine ratio may not be 1 : 1 in RNA molecule.
8. DNA sends the codon for the synthesis of proteins, but otherwise it does not participate in the protein synthesis. 8. RNA takes part in the protein synthesis through transcription and translation.

Question 4.
m-RNA, t-RNA and r-RNA.
Answer:

m-RNA t-RNA r-RNA.
1. m-RNA is a simple molecule which shows linear structure without any folds. 1. t-RNA is a single stranded molecule. There is regular pattern of folding shown by this molecule. 1. r-RNA is a single stranded RNA that is variously folded upon itself. In the folded regions it shows complementary base pairing.
2. It performs the function of transcription during protein synthesis. 2. It performs the function of transferring the amino acids during translation. 2. This RNA remains associated with the ribosomes permanently. It gives the binding site for m-RNA during the process of protein synthesis. It also orients the m-RNA molecule so as to read the message on codons properly.
3. Of the total cellular RNA, m-RNA forms 3-5%. 3. Of the total cellular RNA, t-RNA forms about 10-20%. 3. Of the total cellular RNA, r-RNA forms 80%
4. It has molecular weight of about 5,00,000. 4. t-RNA is the smallest RNA having only 73-93 nucleotides and has molecular weight of about 23,000 to 30,000 daltons. 4. Molecular weight is about 40,000 to 100,000 daltons.

Give Reasons

Question 1.
Nuclein was called as nucleic acid.
Answer:
Nuclein had acidic properties and it was isolated from nucleus. Hence, it was called as nucleic acid.

Question 2.
Initially proteins (and not DNA) were considered as genetic material.
Answer:

  1. Proteins are large, complex molecules and store information required to govern cell metabolism. Hence it was assumed that variations found in species were caused by proteins.
  2. On the other hand, DNA was considered as a small, simple molecule whose composition does not vary much among species.
  3. Variations in the DNA molecules are different than the variation in shape, electrical charge and function shown by proteins.
  4. Hence, initially proteins (and not DNA) were considered as genetic material.

Question 3.
On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.
Answer:

  1. Griffith obtained live S-strain bacteria from the blood of the dead mice.
  2. In a mixture of live R-bacteria and heat killed S-bacteria, live R-strain bacteria picked up something (transforming principle) from the heat-killed S bacterium and got changed into S-type.
  3. Transforming principle allowed R-type bacteria to synthesize capsule and thus they became virulent.
  4. Hence, on injecting a mixture of heat-killed S bacteria and live R bacteria, the mice died.

Question 4.
Viruses obtained by infecting bacteria having radioactive phosphorus contained radioactive DNA (labelled DNA), but not radioactive proteins.
Answer:
Viruses obtained by infecting bacteria having radioactive phosphorus, contained radioactive DNA (labelled DNA). but not radioactive proteins because DNA contains phosphorus (labelled DNA) but proteins do not.

Question 5.
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA.
Answer:
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur and proteins contain sulphur.

Question 6.
In bacteria, m-RNA does not require any processing.
Answer:
In bacteria, m-RNA does not require any processing because it has no introns and it is synthesized in cytoplasm.

Question 7.
Eukaryotic DNA is condensed and supercoiled.
Answer:

  1. In a typical mammalian cell, length of DNA double helix is approximately 2.2 metres.
  2. The size of typical nucleus is approximately 10-6 m
  3. Such a long DNA molecule has to be fitted in small nuclear space.
  4. Therefore, DNA is highly condensed, coiled and supercoiled so that it can be accommodated in the nucleus.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 8.
During translation, complementarity principle is not applicable.
Answer:
During translation, complementarity principle is not applicable as, genetic information is transferred from a polymer of nucleotides to a polymer of amino acids.

Question 9.
Protein synthesis is the most important and essential activity in the living cells.
Answer:

  1. Proteins play a significant role in the metabolism of living cells.
  2. The actual phenotypic expression of living cells is dependent on the biochemical reactions.
  3. Each biochemical reaction needs a specific enzyme for its initiation and completion. All the enzymes are proteins.
  4. In a cell there are many structural proteins too. Thousands of structural and catalytic proteins are constantly required within the cell at all times.
  5. Many functional proteins like hormones are also important for metabolism. Thus, for the synthesis of all such proteins, protein synthesis has become the most important and essential activity of the living cell.

Question 10.
Only 20 amino acids are considered as standard.
Answer:

  1. It was believed that there are total 20 amino acids in the living world. But 21st amino acid called selenocysteine was discovered later.
  2. This amino acid is coded by UGA which is usually a termination codon.
  3. In both prokaryotic and eukaryotic cells polypeptide chains contain 100-300 amino acids and they are formed by specific arrangement of 21 amino acids.
  4. But formation of selenocysteine requires the availability of element selenium in the cells.
  5. Therefore, only 20 amino acids are considered as standard.

Write Short Notes on the following

Question 1.
Friedrich Miescher’s nuclein
Answer:

  1. Nuclein is an acidic substance, having high phosphorus content and it was isolated by Friedrich Miescher in 1869, from the nuclei of pus cells.
  2. As Nuclein had acidic properties and it was isolated from nucleus, it was called as nucleic acid.
  3. E Miescher started working with white blood cells (the major component of pus). He used a salt solution to wash the pus off the bandages. He lysed the cells by adding a weak alkaline solution and isolated nucleic acid from nuclei that precipitated out of the solution.
  4. By the early 1900s, it was known that Miescher’s nuclein was a mixture of proteins and nucleic acids (DNA and RNA).

Question 2.
Packaging in Prokaryotes
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 1

  1. Size of cell in E. coli size is 2-3µ.
  2. The nucleoid is small, circular, highly folded, naked DNA (1100 µm long in perimeter and contains about 4.6 million base pairs).
  3. When the negatively charged DNA becomes circular, the size reduces to 350 µm in diameter.
  4. Folding/looping (40-50 domains (loops) further reduce it to 30 µm in diameter.
  5. RNA connectors assist in loop formation.
  6. Further coiling and super coiling of each domain reduces the size to 2 µm in diameter.
  7. This coiling (packaging) is assisted by positively charged HU (Histone like DNA binding proteins) proteins and enzymes like DNA gyrase and DNA topoisomerase-I, which maintain supercoiled state.

Question 3.
Experimental confirmation of semiconservative replication of DNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 2
(1) Matthew Meselson and Franklin Stahl (1958) used equilibrium – density – gradient – centrifugation technique to experimentally prove semiconservative DNA replication.

(2) They cultured bacteria E.coli in the medium containing 14N (light nitrogen). They obtained equilibrium density gradient band by using 6M CsCl2. The position of this band is recorded.

(3) E. coli cells were then transferred to 15N medium (heavy isotopic nitrogen) and allowed to replicate for several generations. At equilibrium point density gradient band was obtained, by using 6M CsCl2. The position of this band is recorded.

(4) The heavy DNA (15N) molecule can be distinguished from normal DNA by centrifugation in a 6M Cesium chloride (CsCl2) density gradient. At the equilibrium point 15N DNA will form a band. In this both the strands of DNA are labelled with 15N.

(5) Such E. coli cells were then transferred to another medium containing 14N i.e. normal (light) nitrogen. After first generation, the density gradient band for 14N 15N was obtained and its position was recorded. After second generation, two density gradient bands were obtained – one at 14N 15N position and other at 14N position.

(6) The position of bands after two generations clearly proved that DNA replication is semiconservative.

Question 4.
Central Dogma of molecular biology.
Answer:
(1) Central dogma of molecular biology was postulated by EH.C. Crick in 1958.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 3

(2) DNA gets transcribe to form m-RNA, m-RNA acts as a messenger and gets translated to form a polypeptide chain (protein) having specific amino acid sequence.

(3) This unidirectional flow of information from DNA to RNA and from RNA to proteins is referred as central dogma of molecular biology.

(4) Temin (1970) and Baltimore (1970) : Central dogma in retroviruses.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 4

Question 5.
Processing of hnRNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 5

  1. Eukaryotes have split genes.
  2. Primary transcript or hnRNA is non-functional and contains both exons and introns.
  3. Processing of hnRNA results in functional m-RNA.
  4. The fully processed hnRNA is called m-RNA.
  5. m-RNA comes out of the nucleus through nuclear pore for getting translated.

hnRNA undergoes capping, tailing and splicing.

  1. Capping : Methylated guanosine tri¬phosphate is added to 5’ end of hnRNA.
  2. Tailing : Polyadenylation take place at 3′ end.
  3. Splicing : It is removal of introns.
  4. DNA ligase joins exons in a definite sequence (order).

Question 6.
Cracking of genetic code.
Answer:

  1. M. Nirenberg and Matthaei synthesized artificial poly-U m-RNA.
  2. Using this synthetic poly-U m-RNA and cell free system of protein synthesis, a small polypeptide consisting of only amino acid phenylalanine was obtained.
  3. It suggested that UUU codes for phenylalanine.
  4. Dr. Har Gobind Khorana devised a technique for synthesis of artificial m-RNA with repeated sequences of known nucleotides.
  5. He synthesized artificial RNA consisting of known repeated sequences of two or three nucleotides. E.g. CUC, UCU, CUC, UCU, by using synthetic DNA.
  6. This resulted in formation of polypeptide chain consisting of alternate amino acids leucine and serine.
  7. Synthetic RNA consisting of CUA, CUA, CUA, CUA repeats gave polypeptide chain with only one amino acid – leucine.
  8. Severo Ochoa established that the enzyme (polynucleotide phosphorylase) also helps in polymerizing RNA of defined sequences in a template-independent manner (i.e. enzymatic synthesis of RNA).
  9. Thus all the 64 codons in the dictionary of genetic code were deciphered.

Question 7.
Types of mutations.
Answer:

  1. Chromosomal mutations : Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in alteration in the chromosome.
  2. Point mutations : They involve change in a single base pair of DNA. E.g. Mutation that results in Sickle-cell anaemia.
  3. Deletion or insertion of base pairs of DNA : It causes frame-shift mutations or deletion mutation.
  4. Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
  5. Insertion or deletion of three or multiples of three bases (insert or delete) results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 8.
Transfer-RNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 6
(1) As t-RNA can read the codon and also can bind with the amino acid, t-RNA is considered as an adapter molecule.

(2) Clover leaf structure (2 dimensional) of t-RNA:

  • t-RNA has four arms – DHU arm (has amino acyl binding loop), middle arm (has anticodon loop), Tif/C arm (has ribosome binding loop) and variable arm.
  • It has G nucleotide at 5’ end.
  • Amino acid acceptor end (3’ end) having unpaired CCA bases (i.e. amino acid binding site).

(3) For every amino acid, there is specific t-RNA.
(4) Initiator t-RNA is specific for methionine.
(5) There are no t-RNAs for stop codons.
(6) In the actual structure, the t-RNA molecule looks like inverted L (3 dimensional : structure)

Question 9.
UTRs.
Answer:

  1. m-RNA has some additional sequences that are not translated. These sequences are referred as untranslated regions (UTR).
  2. The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon).
  3. They are required for efficient translation process.

Short Answer Questions

Question 1.
Why are Okazaki fragments formed on lagging strand only?
Answer:

  1. The lagging template is the template strand with free 5’ end.
  2. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
  3. Both the strands of the parental DNA are antiparallel and new strands are always formed in 5′ → 3′ direction, i.e. DNA polymerase synthesizes new strand in only one direction i.e. 5′ → 3′ direction.
  4. Hence, the lagging templates becomes available for replication only discontinuously in small patches.
  5. The new lagging strand develops discontinuously away from the replicating fork in the form of small Okazaki fragments.
  6. Hence, Okazaki fragments formed on lagging strand only.

Question 2.
Why t-RNA is called as adapter molecule?
Answer:
t-RNA can read the codon on m-RNA. It also can bind with the amino acid at 3’ end and transport it to m-RNA-ribosome complex during translation. It can bind with specific codon which is complementary to its anticodon. So t-RNA is considered as an adapter molecule.

Question 3.
Why DNA replication is called semiconservative replication?
Answer:

  1. In each of the two daughter DNA molecules thus formed, one strand is parental and the other one is newly synthesized.
  2. 50% is contributed by mother DNA.
  3. Hence, DNA replication is described as semiconservative replication.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 4.
What are the functions of three types of RNAs in bacteria? Which enzyme is involved in transcription of DNA to form RNAs in bacteria? What is its function?
Answer:

  1. In bacteria, m-RNA provides the encoded message for protein synthesis; t-RNA brings specific amino acid to the site of translation; r-RNA plays role in providing binding site to m-RNA and t-RNA.
  2. There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
  3. RNA polymerase binds to promoter and initiates transcription (initiation) and synthesizes RNA.

Question 5.
Explain why it was suggested that codon is a sequence of three consecutive nucleotides on m-RNA.
Answer:

  1. If each codon has only one nucleotide, then there will be 41 = 4 codons, which can encode for only four different types of amino acids.
  2. If each codon has two nucleotides, then there will be 4² = 16 codons, which can encode for only 16 different types of amino acids.
  3. If each codon has three nucleotides, then there will be 4³ = 64 codons, which are sufficient to specify 20 different types of amino acids.

Question 6.
Explain Genetic code is Non-overlapping.
Answer:

  1. Each single base is a part of only one codon.
  2. Adjacent codons do not overlap.
  3. If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
  4. Experimental evidence favours non-overlapping nature of genetic code.

Question 7.
How degeneracy of the code is explained by Wobble hypothesis?
Answer:

  1. In 1966, Crick proposed Wobble hypothesis.
  2. According to this hypothesis, in codon- anticodon base pairing, the third base may not be complementary.
  3. The third base of codon is called wobble base and this position is called wobble position.
  4. This results in economy of t-RNA as anticodon of a t-RNA may bind with codon even when only first two bases are complementary.
  5. For example, GUU, GUC, GUA and GUG codons code for amino acid Valine.
  6. Degeneracy of genetic code means many codons can code for same amino acid.
  7. Thus, the degeneracy of genetic code gets explained by Wobble hypothesis.

Question 8.
a. What is meant by universal genetic code? Give example.
b. Why genetic code is called Non- ambiguous?
Answer:
a. Universal genetic code means that the specific codon codes for same amino acid in all living organisms, e.g. codon AUG always specifies amino acid methionine.

b. Genetic code is called non-ambiguous because, each codon specifies a particular amino acid.

Question 9.
Give examples of termination codons. Why are they known as termination codons?
Answer:

  1. UAA, UAG and UGA are known as termination codons.
  2. They do not code for any amino acid.
  3. They terminate or stop the process of elongation of polypeptide chain.
  4. Hence, they are known as termination codons.

Question 10.
How many amino acids are required for protein synthesis? From where are they obtained?
Answer:

  1. About 20 different types of amino acids are required for protein synthesis.
  2. They are available in the cytoplasm.

Question 11.
How DNA regulates protein synthesis?
Answer:

  1. DNA regulates protein synthesis by coding for the specific sequence of amino acids in a protein.
  2. This control is possible through transcription of m-RNA.
  3. Genetic code is specific for particular amino acid.

Question 12.
What is the role of ribosomes in protein synthesis?
Answer:

  1. Ribosomes serve as site for protein synthesis.
  2. A ribosome has one binding site for m-RNA and 3 binding sites for t-RNA. They are P-site (peptidylt-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
  3. In Eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 13.
Give examples of coordinated regulation or expression, of several sets of genes.
Answer:
Examples of coordinated regulation or expression of several sets of gene are:

  1. If E.colt bacteria do not have lactose in the surrounding medium as a source of energy, then structural genes in Lac operon do not get transcribed and enzyme like β -galactosidase is not synthesized.
  2. The development and differentiation of embryo into an adult organism.

Question 14.
Explain with example what is meant by positive control of gene regulation.
Answer:

  1. A set of genes will be switched on when a substrate is to be metabolized.
  2. This phenomenon is called induction and small molecule responsible for this, is known as inducer.
  3. It is positive control of gene regulation.
  4. For example, lactose acts as an inducer in Lac operon.

Question 15.
If operator gene is deleted due to mutation, how will E.coli metabolise lactose?
Answer:
If operator gene is deleted due to mutation, lac operon cannot be regulated. It will get transcribed continuously and enzymes required for lactose metabolism will get synthesized continuously.

Question 16.
Explain in brief the process of initiation during protein synthesis.
Answer:
Initiation of synthesis of polypeptide chain takes place as follows:

  1. Small subunit of ribosome binds to the m-RNA at 5’ end.
  2. Start codon is positioned properly at P-site.
  3. Initiator t-RNA, (carrying methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA by it’s anti-codon (UAC). Codon-anti-codon pairing involves formation of hydrogen bonds.
  4. The large subunit and smaller subunit of ribosome bind in the presence of Mg++.
  5. Thus, initiator charged t-RNA occupies the P-site and A-site is vacant for next charged t-RNA.

Question 17.
What are the application of genomics?
Answer:
Applications of genomics are as follows:

  1. Structural and functional genomics is used in the improvement of crop plant, human health and livestock.
  2. The knowledge and understanding acquired by genomics research can be applied in medicine, biotechnology and social sciences.
  3. It helps in the treatment of genetic disorders through gene therapy.
  4. It helps in the development of transgenic crops having desirable characters.
  5. Genetic markers have applications in forensic analysis.
  6. Genomics can lead to introduction of new gene in microbes to produce enzymes, therapeutic proteins and biofuels.

Question 18.
Why is HGP important?
Answer:

  1. HGP is associated with rapid development of Bioinformatics.
  2. Knowledge gained about the functions of genes and proteins has a major impact in the fields like Medicine. Biotechnology and the Life sciences.
  3. It has helped in identifying the genes that are associated with genetic characteristics.
  4. The genetic basis of many hereditary diseases can be understood.
  5. It has increased the understanding of gene structure and function in other species. As human beings have many of the genes same as those of flies, roundworms and mice, such studies will enhance understanding of human evolution.

Chart based / Table based Questions

Question 1.
Complete the following table:

Organism Diploid chromosome number
Mouse ————-
Fruitfly ————
Roundworm ————-
Yeast ————–

Answer:

Organism Diploid chromosome number
Mouse 40
Fruitfly 8
Roundworm 12
Yeast 32

Diagram Based Questions

Question 1.
a. Identify A and B in the following diagram.
b. Name the scientist who conducted this experiment.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 7
Answer:
a. A : Smooth strain (III-S)
B : Rough Strain and Heat-killed Smooth Strain

b. F. Griffith conducted the experiment shown in the diagram.

Question 2.
a. Identify A and B in the given diagram.
b. What is the conclusion of the given experiment?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 8
Answer:
a. A : Rough, nonvirulent R-strain B : Heat-killed virulent S-strain
b. When DNA of heat-killed S-strain bacteria is treated with DNase, mouse remains alive as transformation does not take place. This proves that DNA is the genetic material.

Question 3.
Identify A, B, C and name the scientists who carried out the experiment given in the diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 9
Answer:
Answer: A : Infection B : Blending C : Centrifugation Scientists who carried out experiment are Alfred Hershey and Martha Chase.

Question 4.
(1) Identify A, B and C.
(2) What are their sizes?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 10
Answer:
(1) A : Circular, unfolded chromosome
b : Folded chromosome (40 to 50 loops)
c : Supercoiled, folded chromosome

(2) (A) 350 µ (B) 30 µ (C) 2 µ

Question 5.
Draw a labelled diagram of Nucleosome with H1 histone.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 11

Question 6.
Identify A and B in the given diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 12
Answer:
A : Nucleosome
B : Linker DNA

Question 7.
(a) Identify A in the given diagram.
(b) What is the dimension denoted in ‘B’
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 13
Answer:
A : Solenoid B : 300 Å

Question 8.
Sketch and label process of formation of beads on strings/fibres from chain of nucleosomes.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 14

Question 9.
a. Identify A. Which enzyme joined them?
b. Identify B. What is its function ?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 15
Answer:
a. A is Okazaki fragment. These fragments are joined by DNA ligase enzyme
b. B is RNA primer. It provides 3’ OH to which nucleotide gets attached by ester bond.

Question 10.
Draw a labelled diagram showing semi-conservative replication of DNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 16

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 11.
Sketch and label, Meselson and Stahls experiment.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 17

Question 12.
Identify A, B and C in the following diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 18
Answer:
A : Transcription
B : Translation and
C : Reverse Transcription

Question 13.
a. Draw a labelled diagram of transcription unit.
b. What is the sequence of m-RNA and coding strand if sequence of template strand of DNA is
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 19
Answer:
a. Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 20
b. Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 21

Question 14.
a. The following diagram shows processing of ………….
b. What is capping?
c. Identify A, B and C.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 22
Answer:
a. hnRNA
b. Capping is addition of methylated guanosine tri-phosphate at 5′ end of hnRNA.
c. A : Exon and B : Intron C : m-RNA

Question 15.
Draw a labelled diagram of t-RNA carrying Glutamic acid.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 23

Question 16.
Observe the diagrams (a), (b) and (c)

  1. Which step of protein synthesis is shown in the following diagrams?
  2. During initiation, which subunit of ribosome binds with m-RNA?
  3. What are the three binding sites for t-RNA on ribosomes?
  4. On which site of ribosome second and subsequent t-RNA arrives?
  5. Which link is binding amino acids in diagram (b)?
  6. Which chain is being released from ribosome in diagram (c) ?
    Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 24

Answer:

  1. Translation
  2. 30S or 40S
  3. P site, A site and E site
  4. A site
  5. Peptide link
  6. Polypeptide chain

Question 17.
Draw a labelled diagram – Working of lac operon.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 25

Question 18.
What is the process shown in the following diagram? Mention all the steps given in the diagram in a proper sequence.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 26
Answer:
The process shown in the above diagram is DNA fingerprinting. The sequence of steps in the process are:

  1. Isolation of DNA
  2. Restriction digestion
  3. Gel electrophoresis
  4. Southern blotting
  5. Selection of DNA probe
  6. Hybridization
  7. Photography,

Long Answer Questions

Question 1.
Describe Griffith’s transformation experiment.
OR
In the light of Griffith’s experiment, explain the action of two strains of streptococcus pneumoniae and give his conclusion.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 27
(1) In 1928, Frederick Griffith, carried out experiments with bacterium Streptococcus pneumoniae (which causes pneumonia in humans and other mammals).

(2) Griffith used two strains of Streptococcus pneumonia:

  • S-type (Virulent, smooth, pathogenic and encapsulated).
  • R-type (Non-virulent, rough, non- pathogenic and non-capsulated).

(3) Experiments carried out by E Griffith:

  • Mice were injected with R-strain bacteria and they survived (no pneumonia).
  • Mice injected with S-strain bacteria developed pneumonia and died.
  • When heat-killed S-strain bacteria were injected in mice, the mice survived.
  • On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.

(4) Griffith obtained live S-strain bacteria from the blood of the dead mice.

(5) He concluded that the live R-strain bacteria must have picked up something (which he called transforming principle) from the heat killed S bacterium, and got changed into S-type. Transforming principle allowed R-type to synthesize capsule and it became virulent.

(6) Thus, F. Griffith first demonstrated genetic transformations.

Question 2.
Describe Avery, McCarty and MacLeod’s experiments.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 28

  • In 1944, Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty proved that the DNA is a genetic material (transforming principle).
  • They purified DNA, RNA, proteins (enzymes) and other materials from heat killed S-strain cells and mixed them with cells of R-strain bacteria separately to confirm which one could transform living R cells into S cells.
  • Only DNA was able to transform harmless R-strain into virulent S-strain.
  • They also demonstrated that proteases, RNases did not affect transformation. Thus it was proved that the transforming substance was neither a protein nor-RNA.
  • When DNA was digested with DNase, there was no transformation.
  • These experiments proved that the transformation of Live R-strain bacteria into S-strain type was because of DNA of bacteria of S-strain.
  • Thus, they proved that the DNA is transforming principle.

Question 3.
Explain how Hershey – Chase experimentally proved that DNA is the genetic material.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 29
(1) Hershey and Chase worked with bacteriophages (viruses that infect bacteria and which are composed of DNA and protein coat).

(2) They cultured E. coli bacteria in medium containing radioactive phosphorus 32p. By infecting these bacteria with bacteriophages, Hershey and Chase could develop bacteriophages having DNA labelled with 32p. as DNA contains phosphorus and proteins do not.

(3) They also cultured E. coli bacteria in medium containing radioactive sulphur 35s. By infecting these bacteria with bacteriophages, they developed
bacteriophages whose protein coat was labelled with 35s, as proteins contain sulphur and DNA does not.
[Note : Viruses cannot be cultivated in medium.]

(4) Experiment involved three steps.

  • Infection : Both the types of radioactive phages were allowed to infect E.coli bacteria grown on the medium containing normal ‘P‘ and ‘S’.
  • Blending : Then bacterial cultures were agitated in blender to break contact between bacteria and parts of viruses that did not enter bacterial cells.
  • Centrifugation : It was done to separate bacterial cells as a pellet. Parts of viruses which did not enter bacteria remained in the suspension.

(5) Observation:

  • Radioactive ‘S’ remained in suspension.
  • Only radioactive ‘P’ was found inside the bacterial cell in the pellet.

(6) Thus it was proved that DNA is the genetic material which enters bacterial cell and not protein.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 4.
Explain the formation of beads on string, solenoid fibre, chromatin fibre and chromosome.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 30

  • Beads on string (11 nm in diameter) : Under an electron microscope, nucleosomes in thread like chromatin look like ’beads-on- string’.
  • Solenoid fibre (30 nm in diameter) : Six such nucleosomes get coiled and then form solenoid that looks like coiled telephone wire.
  • Chromatin fibre (200 nm in diameter) : Further supercoiling tends to form a looped structure called chromatin fibre.
  • Chromosome (1400 nm in diameter) : Chromatin fibre further coils and condenses at metaphase stage to form the chromosomes. Each chromatid is 700 nm in diameter.
  • Non-Histone Chromosomal Proteins (NHC) are the additional sets of proteins that contribute to the packaging of chromatin at a higher level.

Question 5.
Explain the components of a transcription unit with the help of a diagram?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 31
Transcription unit (Each transcribed segment of DNA) consists of the promoter, the structural gene and the terminator.
(1) The promoter:

  • The promoter is located towards 5′ end of structural gene, i.e. upstream.
  • It is a DNA sequence that provides binding site for enzyme RNA polymerase.
  • In prokaryotes, sigma factor sub unit of the enzyme recognizes the promoter.

(2) Structural genes:

  • Template strand (Antisense strand) : DNA strand having 3’→ 5’ polarity acts as template strand as DNA dependent- RNA polymerase catalyses polymerization in 5’ → 3’ direction.
  • Sense strand : The other strand of DNA having 5’ → 3’ polarity is complementary to template strand. It is called as sense strand. The sequence of bases in this strand, is same as in RNA (where Thymine is replaced by Uracil). It is the actual coding strand.

(3) The terminator:

  • The terminator is located at 3’ end of coding strand, i.e. downstream.
  • It defines the end of the transcription process.

Question 6.
What have we learnt from the Human Genome Project?
Answer:
We have learnt following salient features of human genome from the Human Genome Project.

  1. The human genome contains 3164.7 million nucleotide bases.
  2. The average gene consists of 3000 bases.
  3. Largest known human gene is dystrophin at 2.4 million bases.
  4. Total number of genes is estimated to be 3000.
  5. 99.9% nucleotide bases are exactly the same in all people.
  6. The function of about 50% of the discovered genes are unknown.
  7. Less than 2% of the genome codes for proteins.
  8. Repeated sequences make up a very large portion of the human genome. They can shed light on chromosome structure, dynamics and evolution.
  9. Chromosome 1 has most genes (2968) and the Y has the fewest (231).
  10. Single nucleotide polymorphism have been identified at about 1.4 million locations. It is useful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 7.
Describe the steps involved in DNA fingerprinting ?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 32
Steps involved in DNA fingerprinting are as follows:
1. Isolation of DNA : The DNA can be isolated even from the small amount of tissue like blood, hair roots, skin, etc.

2. Restriction digestion:

  • The isolated DNA is treated with restriction enzymes which cut the DNA at specific sites to form small fragments of variable lengths.
  • Variations in the lengths of restriction fragments are known as Restriction Fragment Length Polymorphism (RFLP).

3. Gel electrophoresis:

  • The DNA samples are loaded on agarose gel and electrophoresis is carried out.
  • Negatively charged DNA fragments move to the positive pole.
  • Separation of fragments depends on their length and it results in formation of bands.
  • dsDNA is then denatured into ssDNA by alkali treatment.

4. Southern blotting : The separated DNA fragments are transferred to a nylon membrane or a nitrocellulose membrane.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

5. Selection of DNA probe:

  • DNA Probe is a known sequence of single- stranded DNA.
  • It is obtained from organisms or prepared by cDNA preparation method.
  • The DNA probe is labelled with radioactive isotopes.

6. Hybridization:

  • In this process, probe is added to the nitrocellulose membrane containing DNA fragments.
  • The single-stranded DNA probe pairs with the complementary base sequence of the DNA strand.
  • As a result DNA-DNA hybrids are formed on the nitrocellulose membrane. Unbound single-stranded DNA probe fragments are washed off.

7. Photography : The nitrocellulose membrane is then kept in contact with X-ray film. DNA bands, due to radioactive probe, give photographic image on X-ray film. This is autoradiography.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 4 Kingdom Animalia Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 4 Kingdom Animalia

Question 1.
What are grades of organization in animals?
Answer:
Cellular, cell- tissue, tissue-organ are the grades of organization in animals.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 2.
How are the animals classified based on body cavity?
Answer:
The animals are classified as acoelomates, pseudocoelomates, and coelomates based on body cavity.

Question 3.
Explain in detail the body plan in animals.
Answer:
Animals show three fundamental body plans as follows:
1. Cell aggregate body plan.
2. Blind sac body plan,
3. Tube within tube body plan.

1. Cell aggregate body plan:
a. In this body plan, cells do not form tissues or organs.
b. Differentiation and division of labour among the cells is minimal.
c. Members of phylum Porifera show cell aggregate body plan.

2. Blind sac body plan:
a. In this body plan, the body is sac-like with a single opening. Digestion is carried out in this sac-like structure.
b. The food is ingested and egested through the same
opening.
c. Members of phylum Cnidaria show a blind sac body plan.

3. Tube within tube body plan:
a. Digestive system is present in tube-like body cavity.
b. Mouth and anus are present at two separate ends of the digestive system.
c. Phylum Annelida onwards all phyla show tube within tube body plan.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 4.
Give the characteristic features of phylum Porifera.
Answer:
Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

  1. Habitat: They are aquatic, mostly marine but few species are found in fresh water.
  2. Forms: They are sedentary animals (attached to substratum or rock).
  3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
    division of labour among cells. Hence, their body is considered as a colony of different types of cells.
  4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.
  5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
  6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
  7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
  8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
  9. Sponges have great power of regeneration.
    e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Question 5.
Given below is a typical sponge body. Identify i, ii, and iii.
Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).
Answer:

  1. Ostium,
  2. Choanocyte,
  3. Osculum

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 6.
Identify the organism and enlist the general characters of its phylum.
Answer:
The given organism is Euplectella.
For characters: Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

  1. Habitat: They are aquatic, mostly marine but few species are found in freshwater.
  2. Forms: They are sedentary animals (attached to substratum or rock).
  3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
    division of labour among cells. Hence, their body is considered as a colony of different types of cells.
  4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘oscu lum’. Beating of flagella creates water current.
  5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
  6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
  7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
  8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
  9. Sponges have great power of regeneration.
    e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Question 7.
State the characteristics of members belonging to phylum Cnidaria.
Answer:
Characteristics of members belonging to phylum Cnidaria:

  1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
  2. Forms: They are sessile or free swimming.
  3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
  4. Body Symmetry: They have radially symmetrical body.
  5. Germ layer: They are diploblastic.
  6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
  7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
  8. Digestion: They have extracellular and intracellular digestion.
  9. Reproduction: Cnidarians reproduce asexually and sexually.

Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 8.
Describe the salient features of phylum Ctenophora.
Answer:
The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

  1. Habitat: They are exclusively marine.
  2. Forms: They are free swimming animals.
  3. Germ layers: Members of this phylum are diploblastic.
  4. Body Symmetry: They are radially symmetrical.
  5. Body plan: The animals of this phylum show blind-sac body plan.
  6. Body organization: They show tissue level organization.
  7. Locomotion: It is earned out by eight rows of ciliated comb plates.
  8. Bioluminescence: It is the characteristic feature of the members of this phylum.
  9. Digestion: It is extracellular and intracellular.
  10. Reproduction: Reproduction is sexual with indirect development.
  11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplana

Question 9.
Draw a neat and labelled diagram:
1. Cnidoblast
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 1

2. Colloblast
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 2

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 10.
Mention the unique features of phylum Platyhelminthes.
Answer:
The members of this phylum are called as flatworms.
Unique features of phylum Platyhelminthes:

  1. Body shape: Body of these animals is dorsoventrally flattened.
  2. Coelom: They are acoelomates.
  3. Germ layers: Platyhelminthes are triploblastic.
  4. Body organization: They show organ-system grade of organization.
  5. Forms: Most of them are endoparasites and few are free-living. Parasitic forms have hooks and suckers for attachment to the hosts’ body. In parasitic forms, body is covered by cuticle and in free-living forms it is covered by cilia.
  6. Digestive system: Parasitic forms generally lack digestive system. In free-living forms, the digestive system is incomplete.
  7. Body plan: They show blind-sac body plan.
  8. Excretion and osmoregulation: It occurs by flame cells or protonephridia.
  9. Reproduction: They are mostly hermaphrodite (bisexual).
  10. Self-fertilization is seen. Few animals show high power of regeneration and show polyembryony.
    e.g. Planaria, Taenia (Tapeworm), Fasciola (Liver fluke) are included in this phylum.

Question 11.
Identify the organisms and label their diagrams,

Question 1.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 3
Answer:
The given organism is Taenia or Tapeworm.

Question 2.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 4
Answer:
The given organism is Fasciola or Liver fluke.

Question 3.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 5
Answer:
The given organism is Planaria.

Question 4.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 6
Answer:
The given organism is Wuchereria.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 12.
Describe the characteristics of Aschelminthes.
Answer:
Phylum Aschelminthes (ascus – sac, helminth – worm) is also called as Nemathelminthes (Nema = thread, helmins = worms).
Characteristics of Aschelminthes:

  1. Forms: These are mostly parasitic. However, few forms are free-living.
  2. Body shape: The body is long, cylindrical, thread-like, circular in cross-section, hence they are known as roundworms.
  3. Body symmetry: These are bilaterally symmetrical.
  4. Coelom: They are pseudocoelomate animals.
  5. Germ layers: These animals are triploblastic.
  6. Body plan: They show tube within a tube-type body plan.
  7. Body covering: The body is covered by tough, resistant cuticle.
  8. Muscles: Body wall has longitudinal muscles, but circular muscles are absent.
  9. Digestive system: Alimentary canal is complete with mouth and anus, at opposite ends.
  10. Excretion: Excretion takes place either by canals or gland cells.
  11. Nervous system: Nervous system consists of a nerve ring and nerves.
  12. Reproduction: Animals are unisexual i.e. sexes are separate.
  13. Fertilization is internal. Development may or may not include larval stages. It shows sexual dimorphism.
    e.g. Ascaris (Roundworm), Wuchereria (filarial worm) and Ancylostoma (hookworm).

Question 13.
Explain the sexual dimorphism in Ascaris.
Answer:
Animals like Ascaris show sexual dimorphism. The male Ascaris is shorter and narrower than the female and has a curved posterior end with a pair of penial setae for copulation. The female Ascaris is relatively longer and broader and has a straight posterior end without penial setae.

Question 14.
Draw a neat and labelled diagram of Ascaris.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 7

Question 15.
Enlist the characteristic features of phylum Annelida.
Answer:
Annelids are commonly called as ring worms or segmented worms.
Characteristic feature of phylum Annelida:

  1. Forms: Annelids may be aquatic, ectoparasitic or free – living or burrowing in moist soil.
  2. Body symmetry: They are bilaterally symmetrical.
  3. Body coelom: They are true coelomates.
  4. Segmentation: Body is metamerically segmented and has a special region called clitellum.
  5. Digestive system: Alimentary canal is complete.
  6. Locomotion: Locomotion takes place with the help of setae (earthworm), parapodia (Nereis) or suckers (leech). Well developed longitudinal and circular muscles help in locomotion.
  7. Nervous system: It consists of nerve ring and ventral solid and ganglionated nerve cord.
  8. Reproduction: Mostly are hermaphrodites and few are dioecious (Nereis).
  9. Respiration: Exchange of gases takes place through body wall.
  10. Circulation: Circulatory system is of closed type. Excretion and osmoregulation is carried out with help of nephridia. e.g. Nereis (Aquatic annelid), Pheretima (Earthworm), Hirudinaria (Leech).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 16.
Which phylum onwards all phyla show tube within tube body plan?
Answer:
Phylum Annelida onwards all phyla show tube within tube body plan.

Question 17.
State the unique features of phylum Mollusca.
Answer:
Mollusca (Mollis: Soft) is the second-largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper-containing blue-coloured respiratory pigment called hemocyanin.
  9. Excretion: Excretion occurs by kidney-like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
    e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Question 18.
Give the economic importance of molluscs.
Answer:
Economic importance of molluscs:

  1. Pearl oyster (Pinctada) gives precious pearls.
  2. Many molluscs are edible.
  3. Molluscan shells are rich source of calcium.

Question 19.
Fill in the blanks.

  1. The stinging cells on the tentacles of cnidarians are known as _______.
  2. Laccifer lacca which produces lac, belongs to phylum __________.
  3. Excretion in molluscs occurs by _________.
  4. The annelid with locomotory structures like setae is ________.

Answer:

  1. cnidoblasts
  2. Arthropoda
  3. Organ of Bojanus
  4. earthworm

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 20.
Identify the phylum to which the given organism belongs to and enlist the characteristics of this phylum.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 8
Answer:
The given organism (Balanoglossus) belongs to phylum Hemichordata.
Characteristics of phylum Hemichordata:

  1. Habitat: Hemichordates are exclusively marine animals, usually living at the bottom of the sea in burrows. These are mostly free – living but animals like Rhabdopleura are sedentary.
  2. Body shape and division: Body is soft and vermiform. It is unsegmented and divided into three parts namely – proboscis, collar and trunk.
  3. Digestive system: Alimentary canal is complete, straight or ‘U’ shaped. Buccal cavity gives rise to a rod-like buccal diverticulum.
  4. Respiration: Respiration is brought about by numerous gills arranged in two longitudinal rows present in the pharyngeal region. Gills open by gill slits.
  5. Circulation: Circulatory system is simple and open type.
  6. Excretion: It takes place with help of with the glomerulus.
  7. Nervous system: Nervous tissue is embedded in epidermis on the dorsal as well as the ventral side.
  8. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external and development is indirect through free swimming larva.

Question 21.
Name the various subphyla of phylum chordata.
Answer:
Subphyla of phylum chordata:

  1. Urochordata
  2. Cephalochordata
  3. Vertebrata

Question 22.
Members of which subphyla are called protochordates?
Answer:
The members of subphyla – Urochordata and Cephalochordata are collectively called protochordates.

Question 23.
Which subphylum includes the tunicates or ascidians?
Answer:
Subphylum Urochordata includes tunicates or Ascidians.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 24.
Give the distinguishing features of Tunicata.
Answer:
Distinguishing features of Tunicata or Urochordata:

  1. Habitat: They are exclusively marine.
  2. Body covering: Soft body is covered by ‘test’ or ‘tunic’ which is made up of tunicine.
  3. Notochord: Notochord is present only in the tail of the larva and is lost during metamorphosis. Hence, the name Urochordata.
  4. Respiration: Pharynx with many gill slits for respiration.
  5. Circulation: Closed circulatory system is present.
  6. Reproduction: Development is indirect, e. g. Herdmania, Salpa, Doliolum, Ascidia.

Question 25.
Write a short note on lancelets.
Answer:

  1. Cephalochordates are also known as lancelets and are small fish-like animals that rarely exceed 5 cm in length.
  2. Lancelets are exclusively marine and live partly buried in soft marine sediments.
  3. Notochord extends throughout entire length of the body and persists throughout life.
  4. Myotomes (muscle blocks) are present.
  5. Post anal tail is present.
  6. Circulatory system is closed type. Blood lacks pigment, e.g. Branchiostoma

Question 26.
Classify Branchiostoma.
Answer:
Classification of Branchiostoma:
Kingdom: Animalia Phylum: Chordata Subphylum: Cephalochordata

Question 27.
In some chordates, the notochord is replaced by cartilaginous or bony vertebral column. Name the chordates which possess this character.
Answer:
Vertebrates

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 28.
Explain in brief the divisions of sub-phylum Vertebrata.
Answer:
Sub-phylum vertebrata is divided into two divisions: Agnathostomata (lacks jaw) and Gnathostomata (bears jaw) on the basis of presence/absence of jaws.
1. Division Agnathostomata:
This division consist of the lowest or most primitive vertebrates that lack jaws.
They include only one class of living vertebrates, the Cyclostomata.
2. Division Gnathostomata:
This division includes animals with jaws.
It is divided into two superclasses: Pisces (bear fins) and Tetrapoda (bear four limbs)
[Note: Students can scan the given Q.R code for understanding the characteristics of vertebrates.]

Question 29.
Mention the characteristic feature of class Cyclostomata.
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

  1. Members of class Cyclostomata are jaw-less and eel like organisms.
  2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
  3. Median fms are present but paired fins are absent.
  4. They are ectoparasites on fishes.
  5. They have sucking circular mouth, without jaws.
  6. Cranium and vertebral column are made up of cartilage.
  7. Their digestive system lacks stomach.
  8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
  9. Heart is two chambered with one auricle and one ventricle.
  10. Gonad is single, large and without gonoduct.
  11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
  12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.
    e.g Petromyzon (Lamprey), Myxine (Hagfish).

Question 30.
Mention the important features of superclass Pisces.
Answer:
Important features of superclass Pisces:

  1. Habitat: These are aquatic animals and are present in fresh, marine and brackish waters.
  2. Body temperature: Pisces are poikilothermic animals i.e., cold blooded animals, in which body temperature changes according to the change in the surrounding temperature.
  3. Sensory organs:They have lateral line system which shows the presence of rheoreceptors for the detection of water current.
  4. Locomotion: Locomotion is by body muscles and fins. Caudal fin helps in steering wheel.
  5. Skeleton: Exoskeleton is made up of dermal scales. It is either bony or cartilaginous.
  6. Body shape: Body is streamlined and boat-shaped and helps to overcome resistance during swimming.
  7. Respiration: Respiration is by gills.
  8. Circulation: It shows single and closed circulation. Heart is two-chambered and ventral in position. Heart of fishes is described as venous heart (presence of deoxygenated blood).
  9. Nervous system: They have a well-developed brain with large olfactory lobes.
  10. Reproduction: Sexes are separate. Most of the fishes are oviparous, however some are viviparous.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 31.
Distinguish between Chondrichthyes and Osteichthyes.
Answer:

Chondrichthyes Osteichthyes
1. Endoskeleton is made of cartilage. Endoskeleton is made of bones.
2. Exoskeleton is made of minute scales called placoid scales. Exoskeleton is made of large, flat and overlapping cycloid or ctenoid scales.
3. Mouth is ventral in position. Mouth is mostly terminal in position.
4. 5-7 pairs of gill slits without operculum are present. Four pairs of gills covered by operculum are present.
5. Caudal fin is heterocercal. Caudal fin is homocercal.
6. Males have copulatory organs called claspers located between the pelvic fins. Males lack claspers.
7. Air bladder is absent. Air bladder is present to maintain buoyancy. Thus, these fishes do not need to swim constantly.
8. Fertilization is internal. Fertilization is external.
9. Many of them are viviparous animals. They are oviparous animal
10. Scolidon (Dogfish), Pristis (Sawfish), Electric ray, Common skate, Hammer headed shark, Carcharodon (Great white shark), Trygon (Stingray), Anoxypristis. Exocoetus (Flying fish), Hippocampus (Sea horse), Labeo rohita, Pomphret (Rohu), Catla (Katla), Clarius (Magur), Pterophyllum (Angle fish), Bombay duck, Lung fishes (Protopterus, Lepidosireri), Aquarium fishes like Betta (Fighting fish).

Question 32.
Match the columns:

Column I (Organism) Column II (Characteristic Feature)
1. Euplectella (a) Exoskeleton formed of placoid scales
2. Periplaneta (b) Presence of mantle
3. Sepia (c) Water canal system
4. Scoliodon (d) Jointed appendages
5. Clarias (e) Exoskeleton formed of cycloid scales

Answer:

Column I (Organism) Column II (Characteristic Feature)
1. Euplectella (c) Water canal system
2. Periplaneta (d) Jointed appendages
3. Sepia (b) Presence of mantle
4. Scoliodon (a) Exoskeleton formed of placoid scales
5. Clarias (e) Exoskeleton formed of cycloid scales

Question 33.
What are tetrapods?
Answer:
Tetrapods are group of vertebrates that includes amphibians, reptiles, birds and mammals. It includes animals that bear two pairs of appendages (with some exceptions e.g. Snakes are limbless, etc.)

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 34.
Write a short note on amphibians.
Answer:

  1. These animals live on land as well as in water (freshwater only).
  2. Amphibians are poikilothermic animals.
  3. Body is differentiated into head, and trunk. Neck and tail are absent in many adults with few exceptions.
  4. Two pairs of limbs arise from the pectoral and pelvic girdles respectively, which help in locomotion.
  5. Skin is moist and glandular with mucous glands.
  6. Exoskeleton is absent.
  7. Eyelids are present. Tympanum represents the ear.
  8. Excretory products, digestive products and gametes are released through the common chamber cloaca.
  9. Circulatory system is of closed type. Heart is three-chambered and ventral. RBCs are biconvex and nucleated.
  10. Respiration is by skin, lungs and bucco-pharynx.
  11. Nervous system is well developed.
  12. Sexes are separate. Amphibians are oviparous. Fertilization is external and development is indirect through aquatic larval stage.
  13. They exhibit metamorphosis. e.g. Rana (Frog), Bufo (Toad), Salamandra (Salamander), Ichthyophis, Hyla (Tree frog), etc.

Question 35.
Name the limbless amphibian.
Answer:
Ichthyophis is a limbless amphibian.

Question 36.
Complete the table.

Phylum/Class Excretory organ Circulation Respiratory organ
Arthropoda Lungs/Gills/Tracheal system
Nephridia Closed Skin/Parapodia
Organ of Bojanus Open
Amphibia Closed Lung

Answer:

Phylum/Class Excretory organ Circulation Respirators organ
Arthropoda Malpighian tubule Open Lungs/Gills/Tracheal system
Annelida Nephridia Closed Skin/Parapodia
Mollusca Organ of Bojanus Open Ctenidia
Amphibia Kidneys Closed Lung

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 37.
Give the diagnostic characters of Reptilia.
Answer:
Diagnostic characters of Reptilia:

  1. Habitat: They are crawling animals. They are the first true terrestrial vertebrates. Few may be aquatic or semi- aquatic and are also found in marshy areas.
  2. Locomotion: Locomotion occurs by limbs in most animals. The limbs are pentadactyl with clawed digits, which help the animal to walk, creep or crawl. Snakes are limbless and crawl on their belly.
  3. Body temperature: They are poikilotherms.
  4. Exoskeleton: Skin is dry, non-glandular and covered by an exoskeleton of epidermal scales or scutes, shields or plates. Lizards and snake shed their skin periodically.
  5. Ear: Tympanum is present
  6. Circulatory system: It has two complete auricles but the ventricles are incompletely partitioned. Therefore, the heart of reptiles is not perfectly four chambered (except in crocodile the heart is four chambered).
  7. Nervous system: The brain is well developed. The olfactory lobes and cerebellum are better developed as compared to amphibians.
  8. Reproduction: Sexes are separate and exhibit prominent sexual dimorphism. Fertilization is internal and the animals are oviparous (exception – viper, it is viviparous). They show little parental care.
  9. e.g. Naja naja (Cobra), Hemidactylus (Wall lizard), Chelonia (Turtle), Crocodilus (Crocodile), Testudo (Tortoise), Chameleon (Tree lizard), Bangarus (Krait), Vipera (Viper).

Question 38.
Enlist the salient features of class Aves.
Answer:
The salient features of class Aves:

  1. Habitat: These animals are aerial in habitat.
  2. Locomotion: Forelimbs are modified into wings for flying. Hind limbs are used for walking, clasping tree branches and running. Aquatic birds have webbed toes. This helps in swimming, e.g. Duck.
  3. Body division: Body is differentiated into head, neck, trunk and tail.
  4. Body shape: Body is streamlined (boat-shaped) to reduce resistance during flight.
  5. Body temperature: These are warm-blooded animals (homeotherms) i.e., keep the body temperature constant irrespective of fluctuations in environmental temperature.
  6. Exoskeleton: Exoskeleton is made up of feathers. Scales are present on hind-limbs. Skin is thin, dry and non-glandular except oil gland at the base of tail (uropygial gland).
  7. Endoskeleton: Bones are hollow (pneumatic) with air cavities to reduce body weight.
  8. Digestion: Jaws are modified into beaks. Teeth are absent. Special structures such as crop and gizzard are present.
  9. Circulatory system: They show double circulation. Blood is red in colour due to presence of biconvex and nucleated RBCs. Heart is perfectly four-chambered, with two auricles and two ventricles.
  10. Respiration: Respiration occurs by lungs. Presence of air sacs increases the buoyancy.
  11. Nervous system: Brain is enlarged with a well developed cerebellum for equilibrium.
  12. Reproduction: Sexes are separate and the animals exhibit prominent sexual dimorphism.
  13. The female shows presence of only left ovary and left oviduct.

This helps to reduce body weight during flying. Fertilization is internal. Avians are oviparous. Parental care is very well developed.
e.g. Columba (Pigeon), Psittacula (Parrot), Struthio (Ostrich), Kiwi, Aptenodytes (Penguin), Corvus (Crow), Neophron (Vulture), Passer (Sparrow), Pavo (Peacock), etc.

Question 39.
Name the flightless bird.
Answer:
Ostrich

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 40.
Give important features of class Mammalia.
Answer:
Features of class Mammalia (mammae: breasts, nipple):

  1. Special feature: Presence of mammary glands (milk-producing glands) for the nourishment of young ones. Mammary glands are modified sweat glands.
  2. Habitat: Mammals are omnipresent (present everywhere). These are mostly terrestrial, some are aquatic and few are aerial and arboreal (living on trees).
  3. Locomotion: Limbs are the organs of locomotion and are modified for walking, climbing, burrowing, swimming, etc.
  4. Body division: Body is differentiated into head, neck, trunk and tail. They have external ear (pinna).
  5. Body temperature: Mammals are homeotherms or warm-blooded animals.
  6. Exoskeleton: It is in the form of hair, fur, nails, hooves, horns, etc.
  7. Skin: Skin is glandular and has sweat glands and sebaceous (oil) glands.
  8. Mouth cavity: Mammals show heterodont dentition (various types of teeth like incisors, canines, premolars and molars).
  9. Circulation: Heart is ventral in position, four chambered with two auricles and two ventricles. RBCs are biconcave and enucleated (except camel). Blood is red in colour.
  10. Respiration: Respiration takes place by lungs.
  11. Nervous system: Brain is highly developed. Cerebrum shows a transverse band called corpus callosum.
  12. Reproduction and development: Only few mammals are oviparous, e.g. Duck billed platypus. Some have pouches for development of immature young ones. These are called marsupials, e.g. Kangaroo. Most of the mammals are placental and viviparous.

Question 41.
Give examples of animals belonging to class Mammalia.
Answer:
Bat, Rattus(rat), Macaca (monkey), Camelus (camel), Whale, Human being, Canis (dog), Elephas (elephant), Equus (horse), Pteropus (flying fox), Ornithorhynchus (platypus), Macropus (kangaroo), Trachypithecus.

Question 42.
Distinguish between Reptilia, Amphibia and Aves.
Answer:

Reptilia Amphibia Aves
Members of Reptilia are terrestrial, with few exceptions. Members of Amphibia live on land as well as in water. Aves are terrestrial and aquatic.
They are poikilothermic. They are poikilothermic. They are homeothermic.
All reptiles have three chambered heart, except for crocodiles. They have three chambered heart. They have four-chambered heart.
Olfactory lobes and cerebellum are better developed than those of amphibians. Olfactory lobes and cerebellum are less developed as compared to reptiles. Cerebellum is well developed for equilibrium.
Skin is dry, non-glandular and covered by scales and plates. Skin is moist, glandular with mucous glands. Skin is thin, dry, non-glandular except oil gland at the base of tail.
Digits bear claws. Digits do not bear claws. Digits bear claws.
Exoskeleton bears epidermal scales or scutes, shields or plates. Exoskeleton is absent. Exoskeleton is made up of feathers.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 43.
Name the phyla to which the following animals belong:

Question 1.
Diploria
Answer:
Cnidaria

Question 2.
Ancyclostoma
Answer:
Aschelminthes

Question 3.
Nereis
Answer:
Annelida

Question 4.
Hottentotta
Answer:
Arthropoda

Question 5.
Chaetopleura
Answer:
Mollusca

Question 6.
Ophiothrix
Answer:
Echinodermata

Question 7.
Rhabdopleura
Answer:
Hemichordata

Question 8.
Exocoetus
Answer:
Chordata

Question 9.
Lepidosiren
Answer:
Chordata

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 44.
Apply Your Knowledge:

Question 1.
A college conducted an inter-college quiz competition. During a round in the quiz, the students were asked to identify animals with respect to their characteristic features given below.

  1. A limbless reptile
  2. Gastrovascular cavity in Hydra
  3. The phylum which includes ringworms or segmented worms
  4. An oviparous mammal
  5. The phylum which includes comb jellies

Answer:

  1. Snake
  2. Coelenteron
  3. Phylum Annelida
  4. Duck-billed platypus
  5. Phylum Ctenophora

Question 45.
An organism has long cylindrical thread-like body. Its body wall has longitudinal muscles but no circular muscles. It is a pseudocoelomate. Identify the phylum to which it belongs.
Answer:
Aschelminthes

Question 46.
Classify the given animals in their respective groups.
Macropus, Struthio, Equus, Bufo, Anura, Salamander, Naja naja, Hippocampus, Bombay duck, Lamprey, Hagfish, Doliolum, Aplysia, Wuchereria, Physalia, Euplectella, Krait, Scypha, Ctenoplana, Brain coral, Obelia, Loligo.
Answer:

Phylum / Subphylum / Class Animals
1. Porifera Euplectella, Scypha
2. Cnidaria Physalia, Brain Coral, Obelia
3. Ctenophora Ctenoplana
4. Mollusca Aplysia, Loligo
5. Aschelminthes Wuchereria
6. Osteichthyes Hippocampus, Bombay duck
7. Cyclostomata Lamprey, Hagfish
8. Urochordata Doliolum
9. Amphibia Bufo, Anura, Salamander
10. Reptilia Naja naja, Krait
11. Aves Struthio
12. Mammalia Macropus, Equus

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 47.
Match the Column.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 9
Answer:
1 – b,
2 – c,
3 – d,
4 – a

Quick Review:

Classification of animals at a glance:

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 10

Question 48.
Exercise:

Question 1.
Give the difference between diploblastic and triploblastic animals?
Answer:
Number of germ layers:
(a) When an organism shows only two germ layers, they are called diploblastic animals. In this case, the outer ectoderm is separated from the inner endoderm by a non-living substance called mesoglea.
(b) When an organism shows three germinal layers, they are called triploblastic animals. The three layers are namely – outer ectoderm, middle mesoderm and inner endoderm.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 2.
Name the superclass under the division Gnathostomata.
Answer:
Division Gnathostomata:
This division includes animals with jaws.
It is divided into two superclasses: Pisces (bear fins) and Tetrapoda (bear four limbs)
[Note: Students can scan the given Q.R code for understanding the characteristics of vertebrates.]

Question 3.
In which group, notochord is present only in the tail of larva.
Answer:
Notochord: Notochord is present only in the tail of the larva and is lost during metamorphosis. Hence, the name Urochordata.

Question 4.
Comment on respiration in Aves.
Answer:
Respiration: Respiration occurs by lungs. Presence of air sacs increases the buoyancy.

Question 5.
Aves and Pisces have stream-lined body.
What is the significance of this type of body.
Answer:
Exoskeleton: Exoskeleton is made up of feathers. Scales are present on hind-limbs. Skin is thin, dry and non-glandular except oil gland at the base of tail (uropygial gland).

Question 6.
Which type of circulation occurs in Aves?
Answer:
Circulatory system: They show double circulation. Blood is red in colour due to presence of biconvex and nucleated RBCs. Heart is perfectly four chambered, with two auricles and two ventricles.

Question 7.
Which group of chordates possess sucking and circular mouth without jaws?
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

  1. Members of class Cyclostomata are jaw-less and eel like organisms.
  2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
  3. Median fms are present but paired fins are absent.
  4. They are ectoparasites on fishes.
  5. They have sucking circular mouth, without jaws.
  6. Cranium and vertebral column are made up of cartilage.
  7. Their digestive system lacks stomach.
  8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
  9. Heart is two chambered with one auricle and one ventricle.
  10. Gonad is single, large and without gonoduct.
  11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
  12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.e.g Petromyzon (Lamprey), Myxine (Hagfish).

Question 8.
Give any four characteristic features of sponges.
Answer:
Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

  1. Habitat: They are aquatic, mostly marine but few species are found in freshwater.
  2. Forms: They are sedentary animals (attached to substratum or rock).
  3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
    division of labour among cells. Hence, their body is considered as a colony of different types of cells.
  4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.
  5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
  6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
  7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
  8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
  9. Sponges have great power of regeneration.
    e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 9.
Mention the role of cnidoblasts.
Answer:
Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.

Question 10.
Name the oil gland present at the base of tail in Aves.
Answer:
Exoskeleton: Exoskeleton is made up of feathers. Scales are present on hind-limbs. Skin is thin, dry and non-glandular except oil gland at the base of tail (uropygial gland).

Question 11.
Why are cyclostomes termed as anadromous?
Answer:
Nervous system: Brain is enlarged with a well developed cerebellum for equilibrium.

Question 12.
Write a short note on Urochordates.
Answer:
Distinguishing features of Tunicata or Urochordata:

  1. Habitat: They are exclusively marine.
  2. Body covering: Soft body is covered by ‘test’ or ‘tunic’ which is made up of tunicine.
  3. Notochord: Notochord is present only in the tail of the larva and is lost during metamorphosis. Hence, the name Urochordata.
  4. Respiration: Pharynx with many gill slits for respiration.
  5. Circulation: Closed circulatory system is present.
  6. Reproduction: Development is indirect, e. g. Herdmania, Salpa, Doliolum, Ascidia.

Question 13.
Enlist the characters of second largest phylum of animal kingdom.
Answer:
Mollusca (Mollis: Soft) is the second largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese t&e,
    interconnected by commissures and connectives.
  11. Sense orgAnswer:Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Question 14.
Members of which phylum are known as segmented or ring worms?
Answer:
Annelids are commonly called as ring worms or segmented worms.
Characteristic feature of phylum Annelida:

  1. Forms: Annelids may be aquatic, ectoparasitic or free – living or burrowing in moist soil.
  2. Body symmetry: They are bilaterally symmetrical.
  3. Body coelom: They are true coelomates.
  4. Segmentation: Body is metamerically segmented and has a special region called clitellum.
  5. Digestive system: Alimentary canal is complete.
  6. Locomotion: Locomotion takes place with the help of setae (earthworm), parapodia (Nereis) or suckers (leech). Well developed longitudinal and circular muscles help in locomotion.
  7. Nervous system: It consists of nerve ring and ventral solid and ganglionated nerve cord.
  8. Reproduction: Mostly are hermaphrodites and few are dioecious (Nereis).
  9. Respiration: Exchange of gases takes place through body wall.
  10. Circulation: Circulatory system is of closed type. Excretion and osmoregulation is carried out with help of nephridia. e.g. Nereis (Aquatic annelid), Pheretima (Earthworm), Hirudinaria (Leech).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 15.
Mention the unique features of ctenophores.
Answer:
The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

  1. Habitat: They are exclusively marine.
  2. Forms: They are free swimming animals.
  3. Germ layers: Members of this phylum are diploblastic.
  4. Body Symmetry: They are radially symmetrical.
  5. Body plan: The animals of this phylum show blind-sac body plan.
  6. Body organization: They show tissue level organization.
  7. Locomotion: It is earned out by eight rows of ciliated comb plates.
  8. Bioluminescence: It is the characteristic feature of the members of this phylum.
  9. Digestion: It is extracellular and intracellular.
  10. Reproduction: Reproduction is sexual with indirect development.
  11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplan

Question 16.
What is ecdysis?
Answer:
Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.

Question 17.
Give the characteristic features of class Cephalochordata.
Answer:

  1. Cephalochordates are also known as lancelets and are small fish-like animals that rarely exceed 5 cm in length.
  2. Lancelets are exclusively marine and live partly buried in soft marine sediments.
  3. Notochord extends throughout entire length of the body and persists throughout life.
  4. Myotomes (muscle blocks) are present.
  5. Post anal tail is present.
  6. Circulatory system is closed type. Blood lacks pigment, e.g. Branchiostoma

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 18.
Give an example of:

Question 1.
Animals whose body is covered by shell.
Answer:
Mollusca (Mollis: Soft) is the second largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Question 2.
Animals with organs of Bojanus for excretion.
Answer:
Mollusca (Mollis: Soft) is the second largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper-containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 19.
Distinguish between Arthropoda and Mollusca
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs:Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.
Mollusca (Mollis: Soft) is the second largest phylum.

Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 20.
Name the phylum that forms connecting link between Chordates and Non-chordates.
Answer:

  1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered as notochord. It is now placed as a separate phylum under Non-Chordata.
  2. It possesses certain characteristics of both Chordates and Non-chordates.
  3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
  4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata.
  5. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Question 21.
Comment on the reproduction of the members of the phylum Platyhelminthes.
Answer:
The members of this phylum are called as flatworms.
Unique features of phylum Platyhelminthes:

  1. Body shape: Body of these animals is dorsoventrally flattened.
  2. Coelom: They are acoelomates.
  3. Germ layers: Platyhelminthes are triploblastic.
  4. Body organization: They show organ-system grade of organization.
  5. Forms: Most of them are endoparasites and few are free-living. Parasitic forms have hooks and suckers for attachment to the hosts’ body. In parasitic forms, body is covered by cuticle and in free-living forms it is covered by cilia.
  6. Digestive system: Parasitic forms generally lack digestive system. In free-living forms, the digestive system is incomplete.
  7. Body plan: They show blind-sac body plan.
  8. Excretion and osmoregulation: It occurs by flame cells or protonephridia.
  9. Reproduction: They are mostly hermaphrodite (bisexual). Self fertilization is seen.
  10. Few animals show high power of regeneration and show polyembryony. e.g. Planaria, Taenia (Tapeworm), Fasciola (Liver fluke) are included in this phylum.

Question 22.
Give a list of aerial adaptations shown by birds.
Answer:

  1. In birds, the forelimbs are modified into wings for flying.
  2. They possess stream-lined body to reduce resistance during flight.
  3. Bones are hollow or pneumatic to reduce body weight.
  4. In order to reduce body weight, urinary bladder is absent. Also, females possess only left ovary and oviduct.
  5. Body is covered by feathers to facilitate flying.

Question 23.
Write a short note on Superclass Pisces. Give one example.
Answer:
Important features of superclass Pisces:

  1.  Habitat: These are aquatic animals and are present in fresh, marine and brackish waters.
  2. Body temperature: Pisces are poikilothermic animals i.e., cold blooded animals, in which body temperature changes according to the change in the surrounding temperature.
  3. Sensory orgAnswer:They have lateral line system which shows the presence of rheoreceptors for the detection of water current.
  4. Locomotion: Locomotion is by body muscles and fins. Caudal fin helps in steering wheel.
  5. Skeleton: Exoskeleton is made up of dermal scales. It is either bony or cartilaginous.
  6. Body shape: Body is streamlined and boat-shaped and helps to overcome resistance during swimming.
  7. Respiration: Respiration is by gills.
  8. Circulation: It shows single and closed circulation. Heart is two-chambered and ventral in position. Heart of fishes is described as venous heart (presence of deoxygenated blood).
  9. Nervous system: They have a well-developed brain with large olfactory lobes.
  10. Reproduction: Sexes are separate. Most of the fishes are oviparous, however some are viviparous.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 24.
Give any six salient features of class Cyclostomata.
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

  1. Members of class Cyclostomata are jaw-less and eel like organisms.
  2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
  3. Median fms are present but paired fins are absent.
  4. They are ectoparasites on fishes.
  5. They have sucking circular mouth, without jaws.
  6. Cranium and vertebral column are made up of cartilage.
  7. Their digestive system lacks stomach.
  8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
  9. Heart is two chambered with one auricle and one ventricle.
  10. Gonad is single, large and without gonoduct.
  11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
  12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.
    e.g Petromyzon (Lamprey), Myxine (Hagfish).

Question 25.
Describe salient features of Phylum Echinodermata.
Answer:
Salient features of phylum Echinodermata (Echinus – spines, derma – skin)

  1. Habitat: These are exclusively marine.
  2. Forms: Members of this phylum are solitary, sedentary or free-living and gregarious, benthic.
  3. Body symmetry: These animals are radially symmetrical with pentamerous symmetry.
  4. Shape: Members of Echinodermata are spherical, elongated or star – shaped.
  5. Body: The endoskeleton is made up of calcareous ossicles. Spines are formed on the body. Hence, they are known as echinoderms. The body has two sides oral and aboral and lacks definite divisions. Mouth is ventrally present on oral surface and anus on aboral surface.
  6. Water vascular system: Presence of water vascular system is the peculiar character of echinoderms.
  7. MadrepOrite is the opening of water vascular system through which water enters. Water vascular system is useful in locomotion, food capturing, respiration.
  8. Digestion: Digestive system is complete.
  9. Respiration: Peristomial gills, papillae, respiratory tree, etc. are used for respiration.
  10. Circulatory and excretory systems: Absent in echinoderms.
  11. Nervous system: Nervous system is simple with a nerve ring around the mouth and radial nerves in arms.
  12. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external.
  13. Development is indirect, i.e. through larval stages. They show high power of regeneration.
    e.g. Sea lily (Antedon), Sea star (Asterias), Sea cucumber (Cucumaria), Brittle star (Ophiothrix), Sea urchin (Echinus).

Question 26.
Explain in brief the characteristic features of Phylum Hemichordata.
Answer:
The given organism (Balanoglossus) belongs to phylum Hemichordata.
Characteristics of phylum Hemichordata:

  1. Habitat: Hemichordates are exclusively marine animals, usually living at the bottom of the sea in burrows. These are mostly free – living but animals like Rhabdopleura are sedentary.
  2. Body shape and division: Body is soft and vermiform. It is unsegmented and divided into three parts namely – proboscis, collar and trunk.
  3. Digestive system: Alimentary canal is complete, straight or ‘U’ shaped. Buccal cavity gives rise to a rod-like buccal diverticulum.
  4. Respiration: Respiration is brought about by numerous gills arranged in two longitudinal rows present in the pharyngeal region. Gills open by gill slits.
  5. Circulation: Circulatory system is simple and open type.
  6. Excretion: It takes place with help of with the glomerulus.
  7. Nervous system: Nervous tissue is embedded in epidermis on the dorsal as well as the ventral side.
  8. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external and development is indirect through free swimming larva.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 27.
Write the distinguishing features of class Reptilia.
Answer:
Diagnostic characters of Reptilia:

  1. Habitat: They are crawling animals. They are the first true terrestrial vertebrates. Few may be aquatic or semi- aquatic and are also found in marshy areas.
  2. Locomotion: Locomotion occurs by limbs in most animals. The limbs are pentadactyl with clawed digits, which help the animal to walk, creep or crawl. Snakes are limbless and crawl on their belly.
  3. Body temperature: They are poikilotherms.
  4. Exoskeleton: Skin is dry, non-glandular and covered by an exoskeleton of epidermal scales or scutes, shields or plates. Lizards and snake shed their skin periodically.
  5. Ear: Tympanum is present
  6. Circulatory system: It has two complete auricles but the ventricles are incompletely partitioned. Therefore, the heart of reptiles is not perfectly four chambered (except in crocodile the heart is four chambered).
  7. Nervous system: The brain is well developed. The olfactory lobes and cerebellum are better developed as compared to amphibians.
  8. Reproduction: Sexes are separate and exhibit prominent sexual dimorphism. Fertilization is internal and the animals are oviparous (exception – viper, it is viviparous). They show little parental care.
  9. e.g. Naja naja (Cobra), Hemidactylus (Wall lizard), Chelonia (Turtle), Crocodilus (Crocodile), Testudo (Tortoise), Chameleon (Tree lizard), Bangarus (Krait), Vipera (Viper).

Question 28.
Mention the unique features of Phylum Cnidaria.
Answer:
Characteristics of members belonging to phylum Cnidaria:

  1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
  2. Forms: They are sessile or free swimming.
  3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
  4. Body Symmetry: They have radially symmetrical body.
  5. Germ layer: They are diploblastic.
  6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
  7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
  8. Digestion: They have extracellular and intracellular digestion.
  9. Reproduction: Cnidarians reproduce asexually and sexually. Asexual reproduction takes place by budding and regeneration.
  10. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
  11. e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

Question 29.
Describe salient features of phylum Arthropoda.
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs: Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 30.
Name the following.

Question 1.
Pores on the body of sponges through which the water enters.
Answer:
Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.

Question 2.
Brain coral belongs to this phylum.
Answer:
Characteristics of members belonging to phylum Cnidaria:

  1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
  2. Forms: They are sessile or free swimming.
  3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
  4. Body Symmetry: They have radially symmetrical body.
  5. Germ layer: They are diploblastic.
  6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
  7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
  8. Digestion: They have extracellular and intracellular digestion.
  9. Reproduction: Cnidarians reproduce asexually and sexually. Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation.
  10. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
  11. e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

Question 3.
Annelid with parapodia.
Answer:
Locomotion: Locomotion takes place with the help of setae (earthworm), parapodia (Nereis) or suckers (leech). Well developed longitudinal and circular muscles help in locomotion.

Question 4.
Groups under phylum Chordata which include poikilotherms?
Answer:
Important features of superclass Pisces:

  1. Habitat: These are aquatic animals and are present in fresh, marine and brackish waters.
  2. Body temperature: Pisces are poikilothermic animals i.e., cold blooded animals, in which body temperature changes according to the change in the surrounding temperature.
  3. Sensory organs: They have lateral line system which shows the presence of rheoreceptors for the detection of water current.
  4. Locomotion: Locomotion is by body muscles and fins. Caudal fin helps in steering wheel.
  5. Skeleton: Exoskeleton is made up of dermal scales. It is either bony or cartilaginous.
  6. Body shape: Body is streamlined and boat-shaped and helps to overcome resistance during swimming.
  7. Respiration: Respiration is by gills.
  8. Circulation: It shows single and closed circulation. Heart is two-chambered and ventral in position. Heart of fishes is described as venous heart (presence of deoxygenated blood).
  9. Nervous system: They have a well-developed brain with large olfactory lobes.
  10. Reproduction: Sexes are separate. Most of the fishes are oviparous, however some are viviparous.

Question 5.
The phenomenon of alternation of generation between asexual and sexual reproduction in cnidarians.
Answer:
Reproduction: Cnidarians reproduce asexually and sexually. Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 31.
Why was phylum Elemichordata earlier considered as a sub phylum of Chordata?
Answer:

  1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered a notochord. It is now placed as a separate phylum under Non-Chordata.
  2. It possesses certain characteristics of both Chordates and Non-chordates.
  3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
  4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Question 32.
Distinguish between Platyhelminthes and Nemathelminthes.
Answer:
The members of this phylum are called as flatworms.
Unique features of phylum Platyhelminthes:

  1. Body shape: Body of these animals is dorsoventrally flattened.
  2. Coelom: They are acoelomates.
  3. Germ layers: Platyhelminthes are triploblastic.
  4. Body organization: They show organ-system grade of organization.
  5. Forms: Most of them are endoparasites and few are free-living. Parasitic forms have hooks and suckers for attachment to the hosts’ body. In parasitic forms, body is covered by cuticle and in free-living forms it is covered by cilia.
  6. Digestive system: Parasitic forms generally lack digestive system. In free-living forms, the digestive system is incomplete.
  7. Body plan: They show blind-sac body plan.
  8. Excretion and osmoregulation: It occurs by flame cells or protonephridia.
  9. Reproduction: They are mostly hermaphrodite (bisexual). Self fertilization is seen. Few animals show high power of regeneration and show polyembryony.
  10. e.g. Planaria, Taenia (Tapeworm), Fasciola (Liver fluke) are included in this phylum.
    Phylum Aschelminthes (ascus – sac, helminth – worm) is also called as Nemathelminthes (Nema = thread, helmins = worms).

Characteristics of Aschelminthes:

  1. Forms: These are mostly parasitic. However, few forms are free-living.
  2. Body shape: The body is long, cylindrical, thread-like, circular in cross-section, hence they are known as roundworms.
  3. Body symmetry: These are bilaterally symmetrical.
  4. Coelom: They are pseudocoelomate animals.
  5. Germ layers: These animals are triploblastic.
  6. Body plan: They show tube within a tube type body plan.
  7. Body covering: The body is covered by tough, resistant cuticle.
  8. Muscles: Body wall has longitudinal muscles, but circular muscles are absent.
  9. Digestive system: Alimentary canal is complete with mouth and anus, at opposite ends.
  10. Excretion: Excretion takes place either by canals or gland cells.
  11. Nervous system: Nervous system consists of a nerve ring and nerves.
  12. Reproduction: Animals are unisexual i.e. sexes are separate. Fertilization is internal. Development may or may not include larval stages. It shows sexual dimorphism.
  13. e.g. Ascaris (Roundworm), Wuchereria (filarial worm) and Ancylostoma (hookworm).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 33.
Give one word for the following.

Question 1.
The gastrovascular cavity in cnidarians
Answer:
Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.

Question 2.
Animals known as acnidarians.
Answer:
The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

  1. Habitat: They are exclusively marine.
  2. Forms: They are free-swimming
  3. Germ layers: Members of this phylum are diploblastic.
  4. Body Symmetry: They are radially symmetrical.
  5. Body plan: The animals of this phylum show blind-sac body plan.
  6. Body organization: They show tissue-level organization.
  7. Locomotion: It is earned out by eight rows of ciliated comb plates.
  8. Bioluminescence: It is the characteristic feature of the members of this phylum.
  9. Digestion: It is extracellular and intracellular.
  10. Reproduction: Reproduction is sexual with indirect development.
  11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplana

Question 3.
The largest phylum of kingdom Animalia.
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs:Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 4.
Copper containing respiratory pigment in blood of molluscs.
Answer:
Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper-containing blue-coloured respiratory pigment called haemocyanin.

Question 39.
What is the role of radula in mollusca?
Answer:
Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.

Question 40.
What are choanocytes?
Answer:
Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.

Question 41.
Name the worm which causes filariasis.
Answer:
e.g. Nereis (3 annelid), Pheretima (Earthworm), Hirudinaria (Leech).

Question 42.
Multiple-choice Questions

Question 1.
Blind sac body plan occurs in
(a) Cnidaria
(b) Arthropoda
(c) Echinodermata
(d) Hemichordata
Answer:
(a) Cnidaria

Question 2.
Physalia belongs to phylum
(a) Platyhelminthes
(b) Cnidaria
(c) Nemathelminthes
(d) Arthropoda
Answer:
(b) Cnidaria

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 3.
Flame cells are found in phylum
(a) Porifera
(b) Coelenterata
(c) Platyhelminthes
(d) Arthropoda
Answer:
(c) Platyhelminthes

Question 4.
___________ is commonly known as hookworm.
(a) Wuchereria
(b) Ancyclostoma
(c) Ascaris
(d) Nereis
Answer:
(b) Ancyclostoma

Question 5.
Which of the following is bilaterally symmetrical?
(a) Pleurobrachia
(b) Cucumaria
(c) Aurelia
(d) Pheretima
Answer:
(d) Pheretima

Question 6.
Malpighian tubules or coxal glands are organs of excretion found in
(a) molluscs
(b) arthropods
(c) hemichordates
(d) platyhelminthes
Answer:
(b) arthropods

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 7.
________ is known as living fossil.
(a) Limulus
(b) Locusta
(c) Laccifer
(d) Loligo
Answer:
(a) Limulus

Question 8.
The member of second largest phylum is
(a) Lobster
(b) Squid
(c) Saccoglossus
(d) Antedon
Answer:
(b) Squid

Question 9.
Excretory system of these molluscs is of open type except,
(a) Sea hare
(b) Pila
(c) Octopus
(d) Sepia
Answer:
(d) Sepia

Question 10.
_________ are exclusively marine animals.
(a) Cnidarians
(b) Echinoderms
(c) Molluses
(d) Arthropoda
Answer:
(b) Echinoderms

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 11.
The peculiar character of echinoderms is
(a) Presence of mantle cavity
(b) Presence of water vascular system
(c) Presence of jointed appendages
(d) Presence of ostia and osculum
Answer:
(b) Presence of water vascular system

Question 12.
Which one of the following belongs to Subphylum Cephalochordata?
(a) Amphioxus
(b) Herdmania
(c) Petromyzon
(d) Ascidia
Answer:
(a) Amphioxus

Question 13.
Complete the analogy:
Salpa: Tunicata : : Myxine : ________ .
(a) Cyclostomata
(b) Chondrichthyes
(c) Cephalochordata
(d) Amphibia
Answer:
(a) Cyclostomata

Question 14.
Members of class Reptilia
(a) are limbless except for Salamander
(b) have moist, glandular skin
(c) have better developed olfactory lobes and cerebellum than amphibians
(d) have four chambered heart except for crocodile
Answer:
(c) have better developed olfactory lobes and cerebellum than amphibians

Question 15.
Which of the following are the first true terrestrial vertebrates?
(a) Mammals
(b) Amphibians
(c) Reptiles
(d) Both (b) and (c)
Answer:
(c) Reptiles

Question 16.
________ is an oviparous mammal.
(a) Macaca
(b) Pteropus
(c) Macropus
(d) Duck billed platypus
Answer:
(d) Duck billed platypus

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 43.
Competitive Corner:

Question 1.
Which of the following animals are TRUE coelomates with bilateral symmetry?
(a) Annelids
(b) Adult Echinoderms
(c) Aschelminthes
(d) Platyhelminthes
Answer:
(a) Annelids

Question 2.
Consider following features:
1. Organ system level of organization
2. Bilateral symmetry
3. True coelomates with segmentation of body Select the correct option of animal groups which possess all the above characteristics.
(a) Arthropoda, Mollusca and Chordata
(b) Annelida, Mollusca and Chordata
(c) Annelida, Arthropoda and Chordata
(d) Annelida, Arthropoda and Mollusca
Hint: In Annelida, Arthropoda and Chordata true segmentation occurs.
Answer:
(c) Annelida, Arthropoda and Chordata

Question 3.
Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system.
(a) Aves
(b) Reptilia
(c) Amphibia
(d) Osteichthyes
Hint: In Aves, crop is associated with storage of food grains and gizzard is used to crush food grain.
Answer:
(a) Aves

Question 4.
Match the following organisms with their respective Characteristics.

1. Pila (P) Flame cells
2. Bomby (q) Comb plates
3. Pleurobrachia (r) Radula
4. Taenia (s) Malpighian tubules

Select the correct option from the following:
(a) i – q, ii – s, iii – r, iv – p
(b) i – r, ii – q, iii – s, iv – p
(c) i – r, ii – q, iii – p, iv – s
(d) i – r, ii – s, iii – q, iv – p
Answer:
(d) i – r, ii – s, iii – q, iv – p

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 5.
An important characteristic that Hemichordates share with Chordates is
(a) absence of notochord
(b) ventral tubular nerve cord
(c) pharynx with gill slits
(d) pharynx without gill slits
Answer:
(c) pharynx with gill slits

Question 6.
Which among these is the CORRECT combination of aquatic mammals?
(a) Seals, Dolphins, Sharks
(b) Dolphins, Seals, Trygon
(c) Whales, Dolphins, Seals
(d) Trygon, Whales, Seals
Hint: Shark and Trygon (sting ray) are cartilaginous fishes. They belong to class Chondrichthyes. While Dolphins, Seals and Whales are aquatic mammals.
Answer:
(c) Whales, Dolphins, Seals

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

I. Objective Questions:

A. Select the most appropriate alternatives from the following & rewrite the sentences:

Question 1.
When dates of drawings are not given, interest on drawings is charged for _____________ months.
(a) three
(b) six
(c) nine
(d) twelve
Answer:
(b) six

Question 2.
A debit balance of the partner’s current account will appear on the _____________ side of the Balance Sheet.
(a) assets
(b) liabilities
(c) debit
(d) credit
Answer:
(a) assets

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 3.
The interest on partner’s capital is credited to _____________ Account.
(a) Trading
(b) Profit and Loss
(c) Capital
(d) Cash
Answer:
(c) Capital

Question 4.
Under fixed capital method, salary or commission to partner is credited to _____________ Account.
(a) Partner’s Capital
(b) Partner’s Current
(c) Partner’s Drawings
(d) Partner’s Salary
Answer:
(b) Partner’s Current

Question 5.
If fixed capital method is adopted, net divisible profit is transferred to _____________ Account.
(a) Partner’s Current
(b) Partner’s Capital
(c) Profit and Loss
(d) Trading
Answer:
(a) Partner’s Current

Question 6.
A statement showing financial position of a business is called a _____________
(a) Balance Sheet
(b) Trial Balance
(c) Capital A/c
(d) Trading A/c
Answer:
(a) Balance Sheet

Question 7.
Wages paid for Installation of machinery should be debited to _____________ Account.
(a) Machinery
(b) Wages
(c) Trading
(d) Profit and Loss
Answer:
(a) Machinery

Question 8.
All indirect expenses are debited to _____________ Account.
(a) Trading
(b) Capital
(c) Profit and Loss
(d) Current
Answer:
(c) Profit and Loss

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 9.
Return outwards are deducted from _____________
(a) Purchase
(b) Sales
(c) Capital
(d) Debtors
Answer:
(a) Purchase

Question 10.
Debit balance of Trading Account indicates _____________
(a) Gross Profit
(b) Gross Loss
(c) Net Profit
(d) Net Loss
Answer:
(b) Gross Loss

Question 11.
Credit balance of Profit and Loss Account indicates _____________
(a) Gross Profit
(b) Gross Loss
(c) Net Profit
(d) Net Loss
Answer:
(c) Net Profit

Question 12.
Income received in advance is shown on _____________ side of the Balance Sheet.
(a) Debit
(b) Credit
(c) Assets
(d) Liabilities
Answer:
(d) Liabilities

Question 13.
Amount irrecoverable from debtors is known as _____________
(a) discount
(b) bad debts
(c) allowance
(d) none of these
Answer:
(b) bad debts

Question 14.
Trading Account is prepared on the basis of _____________ expenses.
(a) indirect
(b) direct
(c) revenue
(d) capital
Answer:
(b) direct

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 15.
Royalty paid on production is shown in the _____________
(a) Balance Sheet
(b) Trading A/c
(c) Profit and Loss A/c
(d) Partner’s Current Account
Answer:
(b) Trading A/c

Question 16.
Prepaid expenses are shown on the _____________ side of the Balance Sheet.
(a) Assets
(b) Liabilities
(c) Debit
(d) Credit
Answer:
(a) Assets

Question 17.
Advertisement expenditure to be written off yet will appear on the _____________ side of Balance Sheet.
(a) Debit
(b) Liabilities
(c) Assets
(d) Credit
Answer:
(c) Assets

Question 18.
_____________ is the list of all ledger balances.
(a) Balance Sheet
(b) Trial Balance
(c) Trading A/c
(d) Profit and Loss A/c
Answer:
(b) Trial balance

Question 19.
Final accounts are prepared on the basis of _____________ and adjustments.
(a) Trial balance
(b) Trading A/c
(c) Profit and Loss A/c
(d) Capital A/c
Answer:
(a) Trial balance

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 20.
The personal medical bill of a partner paid from the business is known as _____________ of the partner.
(a) capital
(b) profit
(c) cash
(d) drawings
Answer:
(d) drawings

B. Write the word/phrase/term, which can substitute each of the following sentences.

Question 1.
The capital method in which the partner’s Current Account is opened.
Answer:
Fixed Capital Method

Question 2.
The capital method in which the partner’s Current Account is not opened.
Answer:
Fluctuating Capital Method

Question 3.
Method of Capital Account in which capital balances of partners change every year.
Answer:
Fluctuating Capital Method

Question 4.
Expenses that are due but not paid at the end of the year.
Answer:
Outstanding/Unpaid expenses

Question 5.
A provision that is created on sundry debtors for likely bad debts.
Answer:
Reserve for Doubtful Debts

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 6.
Income is received before it is due.
Answer:
Income received in advance

Question 7.
The stock is valued at cost price or market price whichever is less.
Answer:
Closing stock

Question 8.
Reduction in the value of fixed assets due to its continuous use.
Answer:
Depreciation

Question 9.
The transport expenses incurred to carry the goods purchased by the firm.
Answer:
Carriage Inward

Question 10.
Income due but not received.
Answer:
Accrued income

Question 11.
Account prepared on the basis of direct expenses and incomes.
Answer:
Trading Account

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 12.
Account prepared on the basis of indirect expenses and incomes.
Answer:
Profit and Loss Account

Question 13.
The transport expenses are paid to Railway, the Airways company, or the Shipping company.
Answer:
Freight

C. State whether the following statements are True or False with reasons:

Question 1.
The partnership agreement must be in written form.
Answer:
This statement is Raise.
A partnership agreement can be in oral or written form. It is advisable to have a partnership agreement in written form, to avoid future conflicts and disputes among the partners. However, it is not compulsory.

Question 2.
There is no limit to a maximum number of partners in a firm.
Answer:
This statement is Raise.
Minimum two persons are required to form the partnership firm. As per the provisions made under the Companies Act 2013 (amended in 2014) the maximum number of partners in a firm is restricted to 50.

Question 3.
Partners are entitled to get a salary or commission.
Answer:
This statement is False.
In Partnership Deed when it is clearly mentioned that all partners or specific partners are entitled to salary or commission then only partners are entitled to get salary or commission. When partnership deed remains silent on salary or commission, then partners are not able to get any salary or commission.

Question 4.
Closing stock is always valued at market price.
Answer:
This statement is False.
As per the conservatism concept, the closing stock is always valued at cost price or market price whichever is less. If the market price of closing stock is greater than its cost then closing stock is recorded at cost.

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 5.
The trial balance is the basis of the Final Account.
Answer:
This statement is True.
Based on the Trial balance and other adjustments, one can prepare Final Accounts. So, the Trial balance is the basis of Final Accounts.

Question 6.
Return inward is deducted from purchases.
Answer:
This statement is False.
Return inward means Sales return and it is to be deducted from sales, not from purchase. Return outward is deducted from purchases.

Question 7.
Discount allowed to Debtors is called as Bad debts.
Answer:
This statement is False.
Discount allowed to Debtors is an expense for the business while Bad debts mean irrecoverable amount from debtors and is a loss to the business. Thus, both have different meanings so, we cannot say that discount allowed to debtors is called as Bad debt.

D. Complete the Sentences.

Question 1.
Documentation charges paid for purchasing a building is debited to _____________ Account.
Answer:
Building

Question 2.
Credit balance of Trading Account indicates _____________
Answer:
Gross Profit

Question 3.
Receivable income is shown on _____________ side of the Balance Sheet.
Answer:
Assets

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 4.
Trademark, Copyright, Patents are the examples of _____________ asset.
Answer:
intangible

Question 5.
Balance Sheet is a _____________ but it is not an _____________
Answer:
Statement, Account

Question 6.
Profit and Loss Account is a _____________ Account.
Answer:
Nominal

Question 7.
The income which is due but not yet received is called _____________ income.
Answer:
accrued/receivable

Question 8.
The statement showing list of all ledger balances is known as _____________
Answer:
Trial balance

Question 9.
Debit balance of Profit and Loss Account means _____________
Answer:
Net Loss

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 10.
Interest on capital is an _____________ for the partner.
Answer:
income

Question 11.
Income accured but not received is _____________ for firm.
Answer:
an asset

Question 12.
Sale of scrap is recorded on _____________ side of _____________ Account.
Answer:
Credit, Profit and Loss

Question 13.
General reserve is recorded in _____________ side of _____________
Answer:
Liability, Balance Sheet

Question 14.
Provision for doubtful debts recorded in _____________ side of _____________ when it is given in the Trial balance only.
Answer:
Liability, Balance Sheet

Question 15.
Provident fund amount is a _____________ for the firm.
Answer:
Liability

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

E. Answer in one sentence only:

Question 1.
Define: ‘Partnership’ as per the Indian Partnership Act 1932.
Answer:
As per the Indian Partnership Act 1932, “partnership is the relation between persons who have agreed to share the profits of a business carried on by all or any one of them acting for all”.

Question 2.
State the two-fold capacities of each partner who works in a business.
Answer:
Each partner works in two-fold capacities viz. Principal and Agent in a business.

Question 3.
As per Income Tax Act 1961, write the dates for Financial or Accounting Year.
Answer:
As per Income Tax Act, 1961, the Financial or Accounting year starts from 1st April of the current year to 31st March of next year. [e.g. 01/04/2019 to 31/03/2020].

Question 4.
How many effects for the hidden adjustment given in the Trial balance are to be passed?
Answer:
Two effects for every hidden adjustment, given in the Trial balance are to be passed, though no special instruction is given in the problem.

F. Do you agree/disagree with the following statements.

Question 1.
A profit and Loss Account is a Real Account.
Answer:
Disagree

Question 2.
Carriage outward means carriage on sales.
Answer:
Agree

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 3.
Adjustments are recorded in Partners Current Account in Fluctuating Capital method.
Answer:
Disagree

Question 4.
Outstanding incomes are treated as an asset.
Answer:
Agree

Question 5.
The balance Sheet is an account.
Answer:
Disagree

Question 6.
R.D.C. is created on creditors.
Answer:
Agree

Question 7.
Depreciation is calculated on fixed assets.
Answer:
Agree

Question 8.
Copyright is a visible asset.
Answer:
Disagree

Question 9.
Interest on drawings is an income for the firm.
Answer:
Agree

Question 10.
Interest on a Partner’s Loan to the firm is always to be allowed.
Answer:
Agree

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 11.
All indirect expenses are debited to Trading Account.
Answer:
Disagree

Question 12.
Capital Account always shows a credit balance.
Answer:
Disagree

Question 13.
Trading Account is prepared to know the profit or loss of the firm.
Answer:
Disagree

Question 14.
Final Accounts are prepared on the basis of Trial Balance and adjustments given.
Answer:
Agree

Question 15.
Royalty paid on production is shown in the Trading Account.
Answer:
Agree

Solved Problems

Question 1.
From the following Trial Balance and Adjustments prepare Trading and Profit and Loss Account and Balance Sheet as on 31st March 2019 for Mr. A and B.
Trial Balance as of 31st March, 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1
Adjustments:
1. Closing stock: Cost price ₹ 60,000, Market price ₹ 52,500.
2. Interest on fixed deposit ₹ 1800 is still outstanding.
3. Provide R.D.D. at 2.5 % on sundry debtors.
4. Depreciate furniture by 5 %.
5. Goods of ₹ 12,000 were destroyed by fire and the insurance company accepted the claim of ₹ 9,000 only.
Solution:
In the books of Mr. A and Mr. B
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1.1
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1.2
Balance Sheet as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1.3
Working Notes:
1. Closing stock is to be considered at a cost price or market price whichever is less. It is valued at ₹ 52,500.

2. Goods of ₹ 12,000 were destroyed and the insurance company accepted a claim of ₹ 9,000, which means ₹ 3,000 is a loss for the firm. The insurance company accepted the claim, (not yet paid the amount) therefore, it is recorded on the Asset side of the Balance Sheet.

3. Advertisement expense ₹ 45,000 is given for 3 years means for one year, we have to take 45000/3 = ₹ 15,000 as advt. exp. and ₹ 30,000 (45,000 – 15,000) is to be taken as prepaid advt. exp. (Asset side)

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 2.
From the following Trial Balance of Parth and Zalak and given Adjustments, prepare Final Accounts for the year ending on 31st March 2019.
Trial Balance as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2
Adjustments:
1. Closing stock is valued at ₹ 99,000.
2. Write off ₹ 3,000 as further bad debts and maintain 5% R.D.D. on debtors.
3. Depreciate Plant and Machinery by 10%, Motor car by 15%, Patents by 20%.
4. Furniture costing ₹ 12,000 sold for ₹ 7,500 was wrongly included in sales and the remaining furniture is valued at ₹ 33,000.
5. Outstanding expenses are Wages ₹ 8,100, Salaries ₹ 6,750. The insurance premium is paid for the year ended 31st December 2019.
6. Goods worth ₹ 67,500 were destroyed by fire and the insurance company accepted the claim for only ₹ 57,000.
7. Sale of goods of ₹ 15,000 was wrongly considered as the sale of machinery.
Solution:
In the books of Parth and Zalak
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2.1
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2.2
Balance Sheet as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2.3
Working Notes:
1. Interest on loan @18 % is calculated for 3 months, (i.e. 1/1/2019 to 31/3/2019)
I = \(\frac{\text { PRN }}{100}\)
= 30,000 × \(\frac{18}{100} \times \frac{3}{12}\)
= ₹ 1,350

2. Net loss by fire = 67,500 – 57,000 = ₹ 10,500

3. Depreciation on furniture = Book value – Value given in adj.
= 45,000 – 33,000
= ₹ 12,000

4. Rent is paid for 10 months i.e. 2 months rent is outstanding.

5. Insurance premium is paid upto 31st Dec., 2019. i.e. 9 months insurance premium is prepaid.
= 5,400 × \(\frac{9}{12}\)
= ₹ 4,050

6. Loss on sale of furniture = Cost of furniture sold – Sale proceeds of furniture
= 12,000 – 7,500
= ₹ 4,500

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 3.
From the following trial balance and adjustments of Rushabh and Yesha, you are required to prepare final accounts as of 31st March 2019. The profit and Loss sharing ratio of partners is their capital ratio.
Trial Balance as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q3
Adjustments:
1. Closing stock is ₹ 1,10,000. Its market value is 20% more than its value.
2. Calculate interest on capital @ 6% p.a.
3. Interest on drawings to be charged from partners: Rushabh ₹ 900, Yesha ₹ 600
4. Provision for doubtful debts is to be kept at 5%.
5. Outstanding expenses at the end of the year: Rent ₹ 300, Salary ₹ 950.
6. Provide depreciation at 10% on machinery and at 5% on furniture.
7. Write off ₹ 4,000 from leasehold building.
Solution:
In the books of Rushabh and Yesha
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q3.1
Balance Sheet as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q3.2
Working Notes:
1. Consider closing stock value ₹ 1,10,000 as its market value is 20% more.

2. Interest in drawings: Record it on the Cr. side of P & L A/c subtract it from the capital. (As shown)

3. Interest on an 8 % loan is calculated for 5 months (i.e. 1/11/18 to 31/3/19)
I = \(\frac{\mathrm{PRN}}{100}\)
= 30,000 × \(\frac{8}{100} \times \frac{5}{12}\)
= ₹ 1,000

4. In the Trial Balance, the following balances have credit balance:
Bills payable, Bank Overdraft, Provision for doubtful debts, 8 % Loan, etc.

5. O/s wages – Cr. bal. – write it on the Liability side of the Balance Sheet.

6. Prepaid insurance – Dr. bal. write it on the Asset side of the Balance Sheet.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation

Multiple Choice Questions

Question 1.
Which one of the following characters is recessive in the case of the pea plants?
(a) Axial flower
(b) Green pod
(c) Green seed
(d) Inflated pod
Answer:
(c) Green seed

Question 2.
Which of the following trait is dominant in Pisum sativum?
(a) White flowers
(b) Green seeds
(c) Yellow pods
(d) Inflated pods
Answer:
(d) Inflated pods

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 3.
When phenotypic and genotypic ratio is the same, then it is an example of ……………….
(a) incomplete dominance
(b) cytoplasmic inheritance
(c) quantitative inheritance
(d) incomplete dominance and co-dominance
Answer:
(a) incomplete dominance

Question 4.
A pea plant with yellow and round seeds is crossed with another pea plant with green and wrinkled seeds produce 51 yellow round seeds and 49 yellow wrinkled seeds, the genotype of plant with yellow round seeds must be ……………….
(a) YYRr
(b) YyRr
(c) YyRR
(d) YYRR
Answer:
(a) YYRr

Question 5.
When a single gene produces two effects and one of it is lethal, then the ratio is ……………….
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(a) 2 : 1

Question 6.
When two genes control a single character and have cumulative effect, the ratio is ……………….
(a) 1 : 1 : 1 : 1
(b) 1 : 4 : 6 : 4 : 1
(c) 1 : 2 : 1
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1
Answer:
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1

Question 7.
Genes located on the same locus but show more than two different phenotypes are called ……………….
(a) polygenes
(b) multiple alleles
(c) co-dominants
(d) pleiotropic genes
Answer:
(b) multiple alleles

Question 8.
Genotype refers to the genetic composition of ……………….
(a) an organism
(b) an organ
(c) chromosomes
(d) germ cells
Answer:
(a) an organism

Question 9.
Individuals having identical alleles of a gene are known as ……………….
(a) homozygous
(b) heterozygous
(c) hybrids
(d) dominants
Answer:
(a) homozygous

Question 10.
If a heterozygous tall plant is crossed with a homozygous dwarf plant, the proportion of dwarf progeny will be ……………….
(a) 100 per cent
(b) 75 per cent
(c) 50 per cent
(d) 25 per cent
Answer:
(c) 50 percent

Question 11.
Inheritance of AB blood group is due to ……………….
(a) incomplete dominance
(b) polyploidy
(c) polygeny
(d) co-dominance
Answer:
(d) co-dominance

Question 12.
The recombination of characters in a dihybrid cross is related to ……………….
(a) law of dominance
(b) incomplete dominance
(c) co-dominance
(d) independent assortment
Answer:
(d) independent assortment

Question 13.
Which one of the following is a true pleiotropic gene?
(a) HbA
(b) Hbs
(c) HbD
(d) HbP
Answer:
(b) Hbs

Question 14.
For demonstrating the law of independent assortment, one should carry out ……………….
(a) back cross
(b) test cross
(c) dihybrid cross
(d) monohybrid cross
Answer:
(c) dihybrid cross

Question 15.
Which one of the following is an example of multiple alleles?
(a) Height in pea plant
(b) Hair colour in cattle
(c) Petal colour in four o’clock plant
(d) Wing-size in Drosophila
Answer:
(d) Wing-size in Drosophila

Question 16.
For the formation of 50 seeds, how many minimum meiotic divisions are necessary?
(a) 25
(b) 50
(c) 75
(d) 63
Answer:
(d) 63

Question 17.
A cross used to verify the unknown genotype of F1 hybrid is a ………………. cross.
(a) test
(b) back
(c) dihybrid
(d) monohybrid
Answer:
(a) test

Question 18.
Appearance of new combinations in F2 generation in a dihybrid cross proves the law of ……………….
(a) dominance
(b) segregation
(c) independent assortment
(d) purity of gametes
Answer:
(c) independent assortment

Question 19.
Genotype of human blood group ‘O’ will be ……………….
(a) IAIA
(b) IAIB
(c) ii
(d) IAi
Answer:
(c) ii

Question 20.
The genotype of human blood group B is ……………….
(a) IAIA
(b) IBi
(c) IAIB
(d) ii
Answer:
(b) IBi

Question 21.
……………… chromosome appears ‘V’-shaped during anaphase.
(a) Metacentric
(b) Acrocentric
(c) Telocentric
(d) Sub-Metacentric
Answer:
(a) Metacentric

Question 22.
The sister chromatids are held together by ……………….
(a) centrioles
(b) chromonemata
(c) chromomere
(d) centromere
Answer:
(d) centromere

Question 23.
Which of the following is not X-linked disorder ?
(a) Haemophilia
(b) Night-blindness
(c) Hypertrichosis
(d) Myopia
Answer:
(c) Hypertrichosis

Question 24.
Which of the following is also called bleeder’s disease ?
(a) Anaemia
(b) Thrombocytopenia
(c) Polycythemia
(d) Haemophilia
Answer:
(d) Haemophilia

Question 25.
The person with Turner’s syndrome has ……………….
(a) 45 autosomes and X sex chromosome
(b) 44 autosomes and XYY sex chromosome
(c) 45 autosomes and Y chromosome
(d) 44 autosomes and X chromosome
Answer:
(d) 44 autosomes and X chromosome

Question 26.
Which of the following is sex chromosomal disorder ?
(a) Colour blindness
(b) Turner’s syndrome
(c) Thalassemia
(d) Down’s syndrome
Answer:
(b) Turner’s syndrome

Question 27.
The word chroma means ……………….
(a) a part of nucleus
(b) a part of chromosome
(c) colour
(d) filamentous body
Answer:
(c) colour

Question 28.
Presence of whole sets of chromosomes is called ……………….
(a) aneuploidy
(b) euploidy
(c) ploidy
(d) chromatography
Answer:
(b) euploidy

Question 29.
The synonymous term for centromere is ……………….
(a) primary constriction
(b) secondary constriction
(c) telomere
(d) satellite
Answer:
(a) primary constriction

Question 30.
Small swellings on the surface of the chromosome are called ……………….
(a) centromeres
(b) chromonemata
(c) chromomeres
(d) telomeres
Answer:
(c) chromomeres

Question 31.
On what basis are the chromosomes usually classified?
(a) On the basis of their function
(b) On the basis of their length
(c) On the basis of the position of the centromere
(d) On the basis of their number
Answer:
(c) On the basis of the position of the centromere

Question 32.
Find the mismatched pair :
(a) Metacentric – V-shaped
(b) Sub-Metacentric – L-shaped
(c) Acrocentric – J-shaped
(d) Telocentric – S-shaped
Answer:
(d) Telocentric – S-shaped

Question 33.
Out of the following combinations which individual will have maximum genetically active DNA?
(a) 44 + XX
(b) 44 + XY
(c) 44 + XYY
(d) Down’s syndrome
Answer:
(a) 44 +XX

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 34.
Crossing over occurs at the time of ……………….
(a) diplotene
(b) pachytene
(c) leptotene
(d) zygotene
Answer:
(b) pachytene

Question 35.
A mature woman has ………………. linkage groups.
(a) 44
(b) 22
(c) 46
(d) 23
Answer:
(d) 23

Question 36.
The pairing of homologous chromosomes is called ……………….
(a) crossing over
(b) terminalisation
(c) synapsis
(d) bivalent
Answer:
(c) synapsis

Question 37.
If only one ‘X’ chromosome is found in a female person, which of the following symptoms will she show?
(a) epicanthal skin fold
(b) webbing of neck
(c) small testis and absence of spermatogenesis
(d) presence of simian crease on the palm
Answer:
(b) webbing of neck

Question 38.
If centromere is situated in the middle of the chromosome, it is called ……………….
(a) metacentric
(b) acrocentric
(c) submetacentric
(d) telocentric
Answer:
(a) metacentric

Question 39.
In which of the following disorders the number of chromosomes present is (extra) 47?
(a) Turner’s syndrome
(b) Cushing’s syndrome
(c) Acquired immuno-deficiency syndrome
(d) Down’s syndrome
Answer:
(d) Down’s syndrome

Question 40.
Myopia is an example of ……………….
(a) complete sex linkage
(b) incomplete sex linkage
(c) recombination
(d) crossing over
Answer:
(a) complete sex linkage

Question 41.
Down’s syndrome is represented by ……………….
(a) n + 1
(b) 2n + 1
(c) 3n + 1
(d) n – 1
Answer:
(b) 2n + 1

Classify the following to form Column B as per the category given in Column A

Question 1.
Types of traits:
[Sickle-cell anaemia, Flower colour of Mirabelis jalapa, Coat colour of cattle, Human blood groups, Widow’s peak, Height in human beings.]

Column A Column B
(1) Co-dominance —————–
(2) Incomplete dominance —————–
(3) Multiple allelism —————-
(4) Pleiotropy —————–
(5) Polygenes —————-
(6) Autosomal dominance ——————

Answer:

Column A Column B
(1) Co-dominance Coat colour of cattle
(2) Incomplete dominance Flower colour of Mirabelis jalapa
(3) Multiple allelism Human blood groups
(4) Pleiotropy Sickle-cell anaemia
(5) Polygenes Height in human beings
(6) Autosomal dominance Widow’s peak

Question 2.
Types of sex-linked genes:
[Haemophilia, Ichthyosis, Nephritis, Myopia, Hypertrichosis, Retinitis pigmentosa]

Column A Column B
(1) Completely X-linked genes —————–
(2) Completely Y-linked genes —————–
(3) Incompletely sex-linked genes —————-

Answer:

Column A Column B
(1) Completely X-linked genes Haemophilia, Myopia
(2) Completely Y-linked genes Ichthyosis,Hypertrichosis
(3) Incompletely sex-linked genes Nephritis, Retinitis pigmentosa

Question 3.
Genetic Disorders
[Turner’s syndrome, Sickle-cell anaemia, Thalassemia, Edward’s syndrome, Klinefelter’s syndrome, Down’s syndrome]

Column A Column B
(A) Autosomal disorder —————–
(B) Sex chromosomal disorder —————–
(C) Mendelian disorder —————-

Answer:

Column A Column B
(A) Autosomal disorder Edward’s syndrome, Down’s syndrome
(B) Sex chromosomal disorder Turner’s syndrome, Klinefelter’s syndrome
(C) Mendelian disorder Sickle-cell anemia, Thalassemia

Very Short Answer Questions

Question 1.
What is hybrid?
Answer:
Hybrid is a heterozygous individual produced from a cross involving two parents differing in one or more contrasting characters.

Question 2.
What are homologues?
Answer:
Homologues are homologous chromosomes which are morphologically similar to each other.

Question 3.
Which law of Mendelian genetics is universally applicable?
Answer:
The law of segregation of Mendelian genetics is universally applicable.

Question 4.
Which law of Mendelian genetics is not universally applicable?
Answer:
The law of independent assortment of Mendelian genetics is not universally applicable.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Give the alternative term for checker board.
Answer:
Punnett’s square is the alternative term for the checker board.

Question 6.
Give the genotypic dihybrid ratio.
Answer:
1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 is the genotypic dihybrid ratio.

Question 7.
What is a lethal gene?
Answer:
The gene which causes the death of the bearer is called lethal gene.

Question 8.
A pea plant pure for yellow seed colour is crossed with a pea plant pure for green seed colour. In F1 generation, all pea plants were with yellow seed. Which law of Mendel is applicable?
Answer:
Mendel’s law of dominance is applicable in this example.

Question 9.
Identify which one of the following is a test cross.

  1. Tt × Tt
  2. TT × tt
  3. Tt × tt

Answer:
3. Tt × tt is a test cross.

Question 10.
What colouration do roans possess? Why?
Answer:
Roans possess the mixture of red and white colour side by side due to codominant alleles for red and white traits.

Question 11.
What are polygenes?
Answer:
When a character is controlled by two or more than two pairs of genes, the genes are called polygenes.

Question 12.
In which region of chromosomes does crossing over take place?
Answer:
Crossing over takes place in the homologous region of the chromosomes.

Question 13.
What are the four sequential steps of crossing over?
Answer:
There are four sequential steps such as synapsis, tetrad formation, crossing over and terminalisation.

Question 14.
Give one example of complete linkage.
Answer:
X chromosome of Drosophila males show complete linkage.

Question 15.
What is the number of linkage groups found in honey bee?
Answer:
The number of linkage group corresponds to the haploid number of chromosomes. Honey bee’s haploid chromosomes number is 16 and thus it has 16 linkage groups.

Question 16.
Name the term for genes located on non-homologous region of Y chromosomes.
Answer:
The genes located on non-homologous region of Y chromosomes are known as holandric genes or Y-linked genes.

Question 17.
What are linkage groups?
Answer:
The genes present on the same chromosome and inherited together are called linkage group.

Question 18.
How are RBCs changed due to sickle-cell anaemia ?
Answer:
RBCs undergo change in their shape and look like a sickle, resulting in reduced capacity to carry haemoglobin.

Question 19.
Which part of a chromosome is called nucleolar organizer?
Answer:
The secondary constriction present on the chromatid arms of a chromosome is called nucleolar organizer.

Question 20.
Why is Y chromosome genetically less active?
Answer:
Since Y-chromosome possesses small amount of euchromatin that contains DNA or genes, therefore it is genetically less active.

Question 21.
Why hypertrichosis is called holandric gene?
Answer:
Hypertrichosis is Y linked gene which can be seen only in males, therefore it is called holandric gene.

Question 22.
What is the genetic difference between total colour blindness and red-green colour blindness ?
Answer:
Total colour blindness is due to incomplete sex-linked genes while red-green colour blindness is due to complete sex linkage.

Question 23.
What happens if the gene for production of factor VIII and IX becomes recessive?
Answer:
The person having recessive gene for haemophilia is deficient in clotting factors (VIII or IX) in blood, such person’s blood does not clot and he thus becomes a patient of haemophilia.

Question 24.
What is the cause of Thalassemia?
Answer:
Thalassemia is caused due to deletion or mutation of gene which codes for alpha (α) and beta (β) globin chains, causing abnormal synthesis of haemoglobin. Thus it is a quantitative abnormality of polypeptide globin chain synthesis.

Question 25.
What is monosomy? Give one example of the same.
Answer:
Monosomy is lack of one chromosome from the usual chromosomal complement. Turner’s syndrome is the example of monosomy.

Give Definitions

Question 1.
Factor
Answer:
The unit of heredity which is responsible for the inheritance and expression of a character and which is responsible for the genetic character is called a factor.

Question 2.
Gene
Answer:
The specific segment of DNA or sequence of nucleotides which is responsible for the inheritance and expression of that character is called a gene.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 3.
Alleles or Allelomorphs
Answer:
The two or more alternative forms of a given gene which are present on the identical loci on the homologous chromosomes are called alleles of each other.

Question 4.
Phenotype
Answer:
The external appearance of an individual for any trait is called phenotype for that trait.

Question 5.
Genotype
Answer:
Genetic constitution of an organism with respect to a particular trait is called genotype.

Question 6.
Homologous Chromosomes?
Answer:
The morphologically, genetically and structurally essentially identical chromosomes present in a diploid cell are called homologous chromosomes.

Question 7.
Back cross
Answer:
The cross of Fx progeny with any of the parents, irrespective of being dominant or recessive, is called back cross.

Question 8.
Linkage
Answer:
Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome.

Question 9.
Non-disjunction
Answer:
Non-disjunction is the phenomenon in which chromosomes fail to separate at the time of cell division, resulting in abnormal chromosomal combinations.

Question 10.
Syndrome
Answer:
The appearance of different types of symptoms at the same time in an individual is called a syndrome.

Question 11.
Aneuploidy
Answer:
Addition or deletion of one or two chromosomes in a diploid chromosome set is called aneuploidy.

Distinguish Between

Question 1.
Homozygous and Heterozygous.
Answer:

Homozygous Heterozygous
1. Individuals with similar gene pairs are called homozygous. 1. Individuals with different gene pairs are called heterozygous.
2. Homozygous individuals form only one type of gametes. 2. Heterozygous individuals form more than one type of gametes.
3. Individuals with similar gene pairs TT, tt, RR and rr are homozygous. 3. Individuals with dissimilar gene pairs Tt and Rr are heterozygous.
4. Homozygous are also called pure breed. 4. Heterozygous are referred to as hybrids.

Question 2.
Monohybrid cross and Dihybrid cross.
Answer:

Monohybrid cross Dihybrid cross
1. Crosses involving a single pair of alleles are called monohybrid crosses. 1. Crosses involving two pairs of alleles are called dihybrid crosses.
2. Monohybrid crosses yield a phenotypic ratio of 3 : 1 in the F2 generation. 2. Dihybrid crosses yield a 9 : 3 : 3 : 1 ratio in F2 generation.
3. Monohybrid crosses yield 1 : 2 : 1 genotypic ratio in F2 generation. 3. Dihybrid crosses yield 1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 genotypic ratio in F2 generation.
4. Application of the law of independent assortment is not applicable in monohybrid crosses. 4. Application of the law of independent assortment is applicable in dihybrid crosses.

Question 3.
Dominant characters and Recessive characters.
Answer:

Dominant characters Recessive characters
1. The characters that are expressed in the F1 generation are called dominant characters. 1. The characters that are not expressed in the F1 generation are called recessive characters. They are prevented from expressing themselves, due to presence of dominant allele.
2. Dominant character is expressed either in homozygous or heterozygous combination. 2. Recessive characters are expressed only when they are in homozygous combination.
3. Dominant characters cannot be masked by recessive characters.
E.g. Round seed and yellow seed are dominant characters in pea plant.
3. Recessive characters are masked by dominant characters.

E.g. Wrinkled seed and green seed are recessive characters in pea plant.

Question 4.
Phenotype and Genotype.
Answer:

Phenotype Genotype
1. Phenotype refers to the outward appearance of an individual such as shape, colour, sex, etc. 1. Genotype refers to the genetic composition of an individual.
2. Phenotype can be observed directly in an individual. 2. Genotype cannot be seen, but can be found out by modern techniques like DNA fingerprinting.
3. Individuals resembling each other may or may not have the same genotype. 3. Individuals possessing the same genotype usually have the same phenotypic expression.
4. The phenotypic ratio obtained in the F2 generation of a monohybrid cross is 3 : 1. 4. The genotypic ratio obtained in the F2 generation of a monohybrid cross is 1 : 2 : 1.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Incomplete dominance and Co-dominance.
Answer:

Incomplete dominance Co-dominance
1. Incomplete dominance is seen when the phenotypes of the two parents blend together to create a new phenotype for their offspring. 1. Co-dominance is seen when the two parent phenotypes are expressed together in the offspring.
2. Both the genes of an allelomorphic pair express themselves partially in F1 hybrids. 2. Both the genes of an allelomorphic pair express themselves equally in F1 hybrids.
3. In incomplete dominance, a mixture of the alleles in the genotype is seen in the phenotype. 3. In co-dominance, both alleles in the genotype are seen in the phenotype.
4. The phenotypic effect of one allele is more prominent than the other. 4. The phenotypic effect of both the alleles is equally prominent.
5. Blending or intermixing of two alleles can be observed. A white flower and a red flower alleles mix and produce pink flowers.

Example : Pink flowers in Mirabilis jalapa.

5. No intermixing or blending effect of two alleles is observed. The colours don’t mix but are seen in patches.

Example : Roan colour in cattle.

Question 6.
Turner’s syndrome and Klinefelter’s syndrome.
Answer:

Turner’s syndrome Klinefelter’s syndrome
1. Individual with Turner’s syndrome has total 45 chromosomes in each of her cell. 1. Individual with Klinefelter’s syndrome has total 47 chromosomes in each of his cell.
2. Turner’s syndrome is XO female, caused due to monosomy of X-chromosome. 2. Klinefelter’s syndrome is XXY male, caused due to trisomy of X chromosome.
3. The external phenotype is of female. 3. The external phenotype is of male.
4. The stature is short. 4. The stature is tall and thin.
5. Secondary sexual characteristics are not developed in Turner’s syndrome. 5. Secondary sexual characteristics are poorly developed in Klinefelter’s syndrome.

Give Reasons or Explain the Statements

Question 1.
Law of segregation is universally applicable.
Answer:

  1. According to the law of segregation, the members of the allelic pair remain together without mixing with each other.
  2. They segregate or separate when the gametes are formed.
  3. Thus the gametes that are formed receive only one of the two factors.
  4. Now it is known that the organisms are diploid and the gametes produced by them are haploid.
  5. The law of segregation therefore is universally applicable.

Question 2.
Mendel selected garden pea for his breeding experiments.
Answer:
Mendel selected garden pea for his breeding experiments, because:

  1. The pea plants were true breeding varieties.
  2. The pea plants being annual, it was possible to cross and study many generations within a short period.
  3. The pea plants had a number of distinguishable, contrasting characters such as tall habit and dwarf habit, round seed and wrinkled seed.
  4. The pea plants were easy to handle for breeding experiments.

Question 3.
When Mendel crossed a tall pea plant with a dwarf pea plant the offspring obtained from this cross were all tall.
Answer:

  1. The tall habit of the pea plant is dominant over the dwarf habit of the pea plant.
  2. Hence, when Mendel crossed a tall pea plant with a dwarf pea plant, the offspring obtained from this cross were all tall.

Question 4.
A cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F2 generation.
Answer:

  1. A cross between a homozygous tall (TT) and a homozygous dwarf (tt) gives rise to a heterozygous tall (Tt) plant in the F1 generation.
  2. When the F1 plant is selfed (Tt × Tt), it gives rise to three tall plants of which two- are heterozygous (Tt) tall and one is homozygous (TT) tall.
  3. Hence a cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F2 generation.

Question 5.
Possibility of female becoming a haemophilic is extremely rare.
Answer:

  1. Haemophilia is caused due to X-linked recessive gene. Females have double X chromosomes.
  2. Even if she has haemophilic gene on one of her X-chromosome, the dominant gene on other X-chromosome, suppresses its expression. Female therefore, does not become haemophilic.
  3. If she inherits haemophilic gene on both of her X-chromosomes, this combination becomes lethal. Such embryo is aborted. If born, she dies soon. This makes the possibility of female becoming a haemophilic extremely rare.

Question 6.
Human female is referred to as carrier of colour blindness.
Answer:
Human female is referred to as carrier of colour blindness because of the following reasons:

  1. Females possess double X-chromosomes in her gametes.
  2. If one X-chromosome is carrying recessive gene for colour blindness, her other dominant X hides the expression of colour blindness and hence she does not become a patient.
  3. But such female can carry the defective gene to her progeny. Thus she is called carrier of colour-blindness.
  4. A female having one recessive gene on X-chromosome is a carrier female, while a female possessing both recessive genes on both the X-chromosomes will be colour blind which is very rare.

Write Short Notes

Question 1.
Linkage.
Answer:

  1. Linkage is the tendency of genes to be inherited together because they are present in the same chromosome.
  2. All the genes on a chromosome are linked to one another. In the linkage group some of the genes are included.
  3. The number of linkage groups of a particular species corresponds to its haploid number of chromosomes present in the organism.
  4. In human beings, there are 23 linkage groups which correspond to the pairs of chromosomes found in each cell.
  5. Linkage groups can be separated only at the time of crossing over during meiosis. The linkage group can form a new combination of genes after crossing over.
  6. Linkages Eire of two types, viz, complete linkage and incomplete linkage.
  7. Morgan discovered linkage in animals while Bateson and Punnett discovered it in plants.

Question 2.
Multiple alleles.
Answer:

  1. Multiple alleles are more than two alternative alleles of a gene in a population situated on the same locus on a chromosome or its homologue.
  2. Multiple alleles arise by mutations of the wild type of gene. Series of multiple alleles are formed due to several mutations that take place in the wild type of allele. This series show alternative expression.
  3. Different alleles in a series show dominant-recessive relation or may show co-dominance or incomplete dominance among themselves. Among all the wild type is the most dominant one over all other mutant alleles.
  4. In Drosophila, a large number of multiple alleles are known. E.g. The size of wings from normal wings to vestigial wings is due to one allele (vg) in homozygous condition. The normal wing is dominant and wild type while vestigial wing is recessive type.
  5. Human blood groups A, B, AB and O Eire also due to series of multiple alleles.

Question 3.
Autosomal inheritance.
Answer:

  1. Transmission of body characters occurs due to autosomes. They are not concerned with sex determination or sex linkage.
  2. All the body characters from parents are passed on to their offspring through autosomes. This is called autosomal inheritance.
  3. Some autosomal characters are due to dominant genes while some other are due to recessive genes. E.g. Widow’s peak and Huntington’s disease is also autosomal dominant character, etc.
  4. Phenyl ketonuria (PKU), Cystic fibrosis and Sickle-cell anaemia are autosomal recessive traits.

Question 4.
Widow’s peak.
Answer:

  1. Widow’s peak is a prominent ‘V’ shaped hairline on forehead.
  2. It is due to autosomal dominant gene.
  3. Widow’s peak occurs in homozygous dominant (WW) and also heterozygous (Ww) individuals.
  4. Individuals with homozygous recessive (ww) genotype do not have widow’s peak but have a straight hair line.
  5. Both males and females have equal chance of inheritance.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Environmental sex determination
Answer:

  1. Environmental sex determination is shown by lower organisms such as Bonellia viridis.
  2. In this animal the environmental factors decide the sex of an offspring.
  3. There is extreme sexual dimorphism in this worm. Female is about 10 cm long while male is tiny and parasitic in the reproductive parts of mature female.
  4. If larva is reared in vicinity of mature female then it becomes a male. By settling on the proboscis of mature female, larva becomes parasitic, enters the female’s mouth and then takes permanent shelter in the female uterus. Such males then produce gametes and fertilize the eggs.
  5. If larvae are drifted away from mature female or if they settle on the sea bottom, they develop into females. Thus determination of sex is due to environmental factors.

Question 6.
Y-linked or Holandric genes.
Answer:

  1. Holandric means entirely of male sex. Y-linked genes are called holandric genes because they are located on non-homologous region of Y chromosome.
  2. The Y-linked genes are inherited directly from male to male.
  3. These genes are never seen in females due to lack of Y chromosome in them.
  4. Hyper Mchosis and ichthyosis are examples of holandric genes.
  5. Hypertrichosis means excessive development of hair on pinna of ear. This character is transmitted directly from father to son.
  6. Ichthyosis person with rough skin.

Question 7.
X-body.
Answer:

  1. German biologist, Henking in 1891, was studying spermatogenesis of the squash bug (Anasa tristis).
  2. He noted that 50% of sperms receive the unpaired chromosomes while other 50% sperms do not receive it.
  3. Henking gave a name to this structure as the X-body. He was unable to explain its role in sex determination.
  4. Further investigations by other scientists led to conclusion that the ‘X-body’ of Henking was a chromosome and gave the name as X-Chromosome to X-body.

Question 8.
Thalassemia
Answer:
(1) Thalassemia is an autosomal-eeessive disorder. The synthesis of alpha ciiains are controlled by two genes, (HBA1 and HBA2) on chromosome 16. Beta chain synthesis is controlled by gene HBB located on chromosome 11. Two alpha chains and two beta chains together form four polypeptide chains that make heterotetrameric haemoglobin molecule. But when there is defective gene on either of chromosome 16 or 11, there is quantitative abnormality of polypeptide globin chain synthesis. This results into thalassemia.

(2) Depending upon which chain is affected, thalassemia is classified as, alpha (α) thalassemia and beta (β) thalassemia.

(3) The clinical symptoms of thalassemia are as follows:

  • Pale yellow skin.
  • Anaemia due to inability to synthesize haemoglobin.
  • Slow growth and development.
  • Variation in the shape and size of RBCs.

(4) Patients need regular blood transfusions to cope with the disorder.

Short Answer Questions

Question 1.
Write the statements of three laws of inheritance given by Mendel.
Answer:
(1) Statement of Law of Dominance : When two homozygous individuals with one or more sets of contrasting characters are crossed, the alleles that appear in F1 are dominant and those which do not appear in F1 are recessive.

(2) Statement of Law of Segregation or Law of purity of gametes : When F1 hybrid forms gametes, the alleles segregate from each other and enter in different gametes. The gametes formed are pure because they carry only one either dominant allele or recessive allele each. Due to this the law is also called “Law of purity of gametes”.

(3) Statement of Law of Independent Assortment : When hybrid possessing two (or more) pairs of contrasting alleles forms gametes, these alleles in each pair segregate independently of the other pair.

Question 2.
Why are farmers and gardeners advised to buy new F1 hybrid seeds every year?
Answer:

  1. Farmers use hybrid seeds for agriculture or horticulture. Hybrid seeds are produced by crossing two unrelated parent plants.
  2. Hybrid seed varieties give improved yields and crop vigour to the farmer.
  3. Hybrids are made by crossing two highly inbred ‘parent’ plants. First generation hybrids, however, do not breed true to type, meaning that the seed they set may not grow into crops that are identical to the ‘parent’ plants.
  4. This can result in variations in yield and quality therefore many farmers prefer to buy new hybrid seed each year to ensure consistency in their final product.

Question 3.
What are the main generalizations given after Mendel’s experiments on the pea plant?
Answer:
After the Mendel’s laws of inheritance and his experiments, following generalizations were made:

  1. Single trait is shown due to single gene. Every single gene has two contrasting alleles.
  2. Two alleles are always in interaction in which one is completely dominant while other is completely recessive.
  3. Factors which were later called genes for different traits are always present on different chromosomes. These traits can assort independently of each other.

Question 4.
Mention the types of deviations from Mendel’s finding.
OR
Describe Neo-Mendelism in short.
Answer:
As the science of genetics progressed, many changes were seen from Mendel’s generalizations. These are called as Neo- Mendelism.
The deviations from Mendel’s findings can be categorised under following heads:

  1. Intragenic interactions : These interactions : are seen between the alleles of same gene. e.g. incomplete dominance and co-dominance. They are also seen in multiple allele series of a gene.
  2. Intergenic interactions : Intergenic interactions are between the alleles of different genes present on the same or different chromosomes, e.g. pleiotropy, polygenes, epistasis, supplementary and complementary genes, etc.

Question 5.
Why Drosophila is most suitable organism for genetics experiments?
Answer:
Drosophila is most suitable organism because of the following reasons:

  1. Drosophila cam easily be cultured under laboratory conditions.
  2. Life span of Drosophila is short for about two weeks.
  3. Drosophila has high rate of reproduction and hence newer organisms can be obtained rapidly.

Question 6.
What are the causes of Down’s syndrome?
Answer:

  1. Down’s syndrome is caused due to aneuploidy.
  2. Aneuploidy is due to non-disjunction of chromosome at the time of gamete formation during meiosis. Due to non-disjunction, chromosomes fail to separate.
  3. In addition to a homologous pair of 21st chromosome there is an extra 21st, therefore it is called trisomy (2n + 1) of 21st chromosome.

Question 7.
What are the characteristic symptoms of Down’s syndrome?
Answer:
Symptoms of Down’s syndrome:

  1. Typical facial features.
  2. An epicanthal skin fold, over the inner corner of eyes causing downward slanting eyes.
  3. Typical flat face, rounded flat nose, mouth always open with protruding tongue.
  4. Mental retardation.
  5. Poor skeletal development.
  6. Short stature, relatively small skull and arched palate.
  7. Flat hand with simian crease that runs across the palm.

Question 8.
Write a brief account of Turner’s syndrome.
Answer:

  1. Turner’s syndrome is a genetic disorder caused due to monosomy of X chromosome.
  2. It was first described by H. H. Turner.
  3. Turner’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
  4. Chromosomal complement of Turner’s syndrome is 44 + XO, having a total of 45 chromosomes.

Symptoms of Turner’s syndrome are as follows:

  1. Female phenotype.
  2. Short stature with webbing of neck.
  3. Low posterior hair line.
  4. Secondary sexual characters fail to develop.
  5. Mental retardation.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 9.
Write a brief account of Klinefelter’s syndrome.
Answer:

  1. Klinefelter’s syndrome is a genetic disorder caused due to trisomy of X chromosome.
  2. It was first described by Harry Klinefelter.
  3. Klinefelter’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
  4. Chromosomal complement of Klinefelter’s syndrome is 44+XXY, having a total of 47 chromosomes.

Symptoms of Klinefelter’s syndrome are as follows:

  1. The Klinefelter’s syndrome individuals are tall, thin and eunuchoid.
  2. They are sterile with poorly developed sexual characteristics.
  3. Testes are underdeveloped and small. Spermatogenesis does not take place.
  4. They have subnormal intelligence and show partial mental retardation.

Diagram Based Questions

Question 1.
Give the graphical representation of test cross and back cross.
Answer:
(1) Test cross
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 1
The F2 generation of test cross consists of 50% heterozygous tall plants and 50% homozygous dwarf plants.

(2) Back cross
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 2
F1 crossed back with its dominant parent
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 3

Question 2.
Give a cross for incomplete dominance using a suitable example.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 4
Result:
Genotypic ratio – 1RR : 2 Rr : 11rr
Phenotypic ratio – 1Red : 2 Pink : 1 White

Question 3.
Give a cross of co-dominance using a suitable example.
Answer:
Coat colour in cattle
Red female RR × White male WW
P1 generation : RR × WW
Gametes : R and W
F1 generation all RW Roan coloured
P2 generation RW × RW

R W
R RR RW
W RW WW

Genotypic ratio : 1 RR : 2 RW : 1 WW
Phenotypic ratio : 1 Red : 2 Roan : 1 White

Question 4.
Draw two crosses to show inheritance of colour blindness, (i) A cross between normal female and colour blind male, (ii) A cross of carrier woman with normal man.
Answer:
(i) A cross between normal female and colour-blind male.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 5

(ii) A cross of carrier female with normal male.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 6

Question 5.
Draw the following crosses to show inheritance of haemophilia : Normal female with haemophilic male. Show their progeny. If one of their carrier daughters marries a normal male what would be possible genotypes of this generation.
Answer:
(1) Haemophilic male crossed with normal female:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 7

(2) Carrier female crossed with normal male :
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 8

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 6.
Sketch and label structure of X and Y chromosomes.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 9

Question 7.
Give the graphical representation of pleiotropy to show inheritance of Sickle-cell anaemia.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 10

Long Answer Questions

Question 1.
There are 16 possible individuals in F2 generation. Try to find out the phenotypes as well as the genotypic and phenotypic ratios.
Answer:
In the above dihybrid cross there are 4 phenotypes such as yellow round, yellow wrinkled, green round, green wrinkled.
There are 9 different genotypes as follows:
1 : RRYY / 2 : RRYy / 1 : RRyy / 2 : RrYY / 4 : RrYy / 2 : Rryy / 1 : rrYY /2 : rrYy / 1 : rryy

Question 2.
Explain with suitable diagram how test cross is used to find out genotype of dominant plant.
Answer:
Test cross is used to find out the exact genotype by crossing the F1 individual with homozygous recessive one.
E.g. To find out the genotype of unknown violet flower obtained in F1 generation, one can conduct two crosses as follows:
I. Unknown flower considering as RR (homozygous dominant)
RR × rr Homozygous dominant

R R
R Rr Rr
R Rr Rr

II. In such cross, all the flowers will be violet. II. Unknown flower considering as Rr (heterozygous)
Rr × rr Homozygous recessive with heterozygous

R R
R Rr Rr
R Rr Rr

In such cross half the flowers will be violet and half will be white.

Question 3.
Describe briefly Morgan’s Experiments showing linkage and crossing over. (Diagram is not needed)
Answer:
(1) Morgan used Drosophila melanogaster for his experiments.

(2) He carried out several dihybrid cross experiments to study sex-linked genes of Drosophila.

(3) Crosses between yellow-bodied, white-eyed female and brown-bodied, red-eyed males were done in P1 generation. Brown-bodied and red-eyed forms were wild.

(4) Morgan intercrossed their F1 progeny and noted that two genes did not segregate independently of each other and F2 ratio deviated very significantly from Mendelian 9 : 3 : 3 : 1 ratio.

(5) When genes are grouped on the same chromosome, some genes are strongly linked. They show very few recombinations (1.3%).

(6) When genes are loosely linked, i.e. located away from each other on chromosome, they show more (higher) recombinations (37.2%).

(7) For example, the genes for yellow body and white eye were strongly linked and showed only 1.3 per cent recombination (in cross-I).

(8) White-bodied and miniature wings showed 37.2 per cent recombination (in cross-II). Cross I shows crossing over between genes y and w.

(9) Cross II shows crossing over between genes white (w) and miniature wing (m). Here dominant wild type alleles are represented with (+) sign.

(10) Parental combinations occur more due to linkage and new combinations less due to crossing over.

Question 4.
Describe the mechanism of sex determination in human beings with a suitable cross.
Answer:
1. Sex determination in human beings:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 11
(1) In human beings, the sex is determined with the help of X and Y chromosomes. This chromosomal mechanism of sex determination is of XX-XY type.

(2) In male, the nucleus of each cell contains 46 chromosomes or 23 pairs of chromosomes. Of these 22 pairs are autosomes and one pair of sex chromosomes. Males are thus heteromorphic as they have two different types of sex chromosomes.

(3) Autosomes or somatic chromosomes are responsible for determination of other characters of the body, but not the sex.

(4) In female cells, there are 22 pairs of autosomes and one pair of X chromosomes. Females are thus homomorphic as they have similar sex chromosomes.

(5) Thus the genotypes of female and male are as follows:
Female : 46 chromosomes = 44 autosomes + XX sex chromosomes
Male : 46 chromosomes = 44 autosomes + XY sex chromosomes

(6) During gamete formation, the diploid germ cells in the testes and ovaries undergo meiosis to produce haploid gametes (sperms and eggs). The homologous chromosomes separate and enter into different gametes during this process.

(7) The human male produces two different types of sperms, one containing 22 autosomes and one X chromosome and the other containing 22 autosomes and one Y chromosome. Human males are therefore called heterogametic, i.e. they produce different types of gametes.

(8) The human female produces only one type of eggs containing 22 autosomes and one X chromosome and therefore she is homogametic.

(9) During fertilization, if X containing sperm fertilizes the egg having X chromosome, then a female child with XX chromosomes is conceived.

(10) If Y containing sperm fertilizes the egg having X chromosome then a male child with XY chromosomes is conceived.

(11) The sex of the child thus depends upon the type of sperm fertilizing the egg. The heterogametic parent determines the sex of the child and thus the father is responsible for the determination of the sex of the child and not the mother.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 12

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Explain the mechanism of sex determination in case of birds.
Answer:

  1. Sex determination in birds is by ZW-ZZ mechanism.
  2. In birds, males are homogametic while females are heterogametic.
  3. Males produce all similar types of sperms, containing 8 autosomes and ‘Z’ sex chromosome.
  4. Females produce two different types of eggs, one containing 8 autosomes and Z chromosome and the other containing 8 autosomes and W chromosome.
  5. When Z bearing egg is fertilized by a sperm a male offspring is produced. If W bearing egg is fertilized then female offspring is produced.
    Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 13

[Note : In domestic fowl chromosome number is 18, with 16 autosomes and two sex chromosomes.]

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Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Balbharti Maharashtra State Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India Important Questions and Answers.

Maharashtra State Board 12th Sociology Important Questions Chapter 4 Processes of Social Change in India

1A. Complete the following statements by choosing the correct alternative given in the brackets and rewrite it.

Question 1.
___________ is the process of the use of unbiotic power for the mass production of goods. (digitalisation, urbanisation, industrialisation)
Answer:
industrialization

Question 2.
Process of industrialization spread from ___________ to other regions of the world. (Asia, Europe, Australia)
Answer:
Europe

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 3.
In the process of ___________ human society is transformed from a state pre-industrial to an industrial. (urbanisation, modernisation, industrialisation)
Answer:
industrialization

Question 4.
The development of industries led to the of workplaces. (urbanisation, digitalisation, mechanisation)
Answer:
mechanization

Question 5.
The use of precision techniques and accuracy in production is required in ___________ (mechanization, computerisation, capital)
Answer:
mechanization

Question 6.
In the process of mechanization workers led to feel ___________ from the process of production. (alienated, integrated, neutral)
Answer:
alienated

Question 7.
The high mechanization and automation of industrial processes naturally depend on ___________ resources available. (social, financial, natural)
Answer:
financial

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 8.
Industries required skilled workforce and ___________ of apprentices at the workplace. (specific training, unskill, eatables)
Answer:
specific training

Question 9.
Early industries required skilled and unskilled ___________ workforce to complete various tasks at all levels. (animal, human, machines)
Answer:
human

Question 10.
Special institutes like ___________ are established to impart technical education and also for professional education. (management training, urbanisation, modernisation)
Answer:
management training

Question 11.
Industrialisation led to ___________ on the basis of specialisation and expertise. (capital, labour, division of labour)
Answer:
division of labour

Question 12.
Modern ___________ was a result of industrialisation. (education, facilities, urbanisation)
Answer:
urbanisation

Question 13.
___________ is a typical feature of urban living. (urbanism, rural, tribal)
Answer:
urbanism

Question 14.
Urbanisation is a ___________ way process. (two, three, four)
Answer:
two

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 15.
Urbanisation consists of large ___________ number of people from rural to urban areas. (outward flow, overflow, inward flow)
Answer:
inward flow

Question 16.
The gradual emergence of factories led to the ___________ of people from rural and tribal areas, to the factory locations. (communication, migration, specialisation)
Answer:
migration

Question 17.
The flux of people for the purpose of employment has resulted in cities getting ___________ (overpopulated, less dense, shut down)
Answer:
overpopulated

Question 18.
Overpopulated cities are expanding and turning into ___________ such as Mumbai, Pune. (rural, metropolises, tribal)
Answer:
metropolises

Question 19.
The place of residence and one’s place of which work drift apart with the passage of time which means ___________ (a division of labour, spatial segregation, capital intensive)
Answer:
spatial segregation

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 20.
Urbanisation led to a ___________ gathering of people of different gender, sexuality, caste, creed, class, language and so forth. (homogeneous, barriers, heterogeneous)
Answer:
heterogeneous

Question 21.
Urbanism as a ___________ of life. (solution, way, tradition)
Answer:
way

Question 22.
Secondary modes of security control in urban areas are ___________ (law, belief, morals)
Answer:
law

Question 23.
Division of labour is based on one’s ___________ and ___________ (skills and expertise, knowledge and experience, limitations and unskilled)
Answer:
skills and expertise

Question 24.
The term modernisation was coined by ___________ (Anderson, Daniel Lerner, Giddens)
Answer:
Daniel Lerner

Question 25.
___________ is the application of modern science to human affairs. (Globalisation, Digitalisation, Modernisation)
Answer:
Modernisation

Question 26.
Self-criticism, willingness to introspect critically, is also an aspect of ___________ thinking. (logical, critical, analytical)
Answer:
critical

Question 27.
___________ means the approach and ability to provide logical explanations for any phenomenon. (rationalism, science, art)
Answer:
rationalism

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 28.
Scientific reasoning explains ___________ relationships between factors. (mortal, casual, immortal)
Answer:
Causal

Question 29.
Being ‘modern’ cannot be limited to only using modern devices or gadgets but there should be a willingness to receive ___________ ideas. (new, old, superstitious)
Answer:
new

Question 30.
The new economic policy means ___________ policy. (PNG, LPG, CNG)
Answer:
LPG

Question 31.
LPG stands for Liberalisation, Privatisation and ___________ (Googlisation, Globalisation, Geometrisation)
Answer:
Globalisation

Question 32.
Process of ___________ opened up the skies for Indian economy. (modernisation, digitalisation, globalisation)
Answer:
globalisation

Question 33.
This new economic policy brought in much ___________ and criticism. (superstitions, beliefs, skepticism)
Answer:
skepticism

Question 34.
The principle of ___________ is an integral part of globalisation as a process of change. (laissez faire, loss, profit)
Answer:
laissezfaire

Question 35.
___________ is a process where government control services are opened up for private service providers. (privatisation, marketisation, globalisation)
Answer:
privatisation

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 36.
Privatisation has encouraged many service providers to indulge in ___________ (donation, profiteering, social service)
Answer:
profiteering

Question 37.
Privatisation is mainly oriented to ___________ (production, marketing, profits)
Answer:
profits

Question 38.
Globalisation led to increase in production, this in turn has led to large-scale ___________ (marketisation, privatisation, liberalisation)
Answer:
marketisation

Question 39.
The term ___________ was coined by Capgemini and MIT. (bhakti movement, digital transformation, laissez-faire)
Answer:
digital transformation

Question 40.
___________ has led to frequent changes in business models. (consumerism, digitalisation, materialistic)
Answer:
digitalisation

Question 41.
Now a days we use ___________ for various purposes such as production, surgery, robotics etc. (artificial intelligence (AI), machines, capital)
Answer:
artificial intelligence (AI)

Question 42.
___________ has escalated the speed of the processes with a far greater extent of accuracy. (industrialisation, modernisation, digitalisation)
Answer:
digitalisation

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 43.
___________ is based on technology, innovation, research and development. (modernisation, urbanisation, digitalisation)
Answer:
digitalisation

Question 44.
In the field of education, we are working towards ___________ technology for the purpose of education in the 21st century. (integrating, managing, old)
Answer:
integrating

Question 45.
Industrialization with the growth of cities has caused the breakdown of the ___________ (nuclear families, joint families, marriage)
Answer:
joint families

Question 46.
India is now an integral part of the ___________ economy. (personal, social, global,)
Answer:
global

Question 47.
___________ and ___________ has opened up a range of options to the user with a click of button. (computerisation and digitalisation, modernisation and urbanisation, transportation and communication)
Answer:
computerisation and digitalisation

Question 48.
Digitization has increased ___________ networking. (cultural, economic, social)
Answer:
social

1B. Correct the incorrect pair and rewrite it.

Question 1.
(a) Growth Of industries – Industrialisation
(b) Alienated from the process of production – Mechanisation
(c) Extent of mechanization and automation depend on – Finances
(d) Skilled workforce – Capital
Answer:
(d) Skilled workforce – Specific Training

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 2.
(a) Formation of economic classes – Industrialisation
(b) Industrial expansion – Modernisation
(c) Tasks assigned on the basis of – Division of labour
(d) Emergence of metropolis – Urbanisation
Answer:
(b) Industrial expansion – Spatial Segregation

Question 3.
(a) Metropolises – Mumbai, Pune
(b) Heterogenous gathering of people – Urbanisation
(c) Secondary modes of security control – Family
(d) A way of life – Urbanism
Answer:
(c) Secondary modes of security control – Law, city traffic signal, police, etc

Question 4.
(a) Based on one’s skills of expertise – Division of labour
(b) Modernisation – Daniel Lerner
(c) Scientific reasoning – Causal Relationships
(d) Shift to secular and rational values from spiritual values – Nationalism
Answer:
(d) Shift to secular and rational values from spiritual values – Rationalism

Question 5.
(a) Scepticism and criticism – Self-criticism
(b) Technological advancement – Industrialisation
(c) Ability to explain the constructive and destructive aspect – Critical thinking
(d) New economic policy – LPG
Answer:
(a) Scepticism and criticism – New economic policy

Question 6.
(a) Free trade and free competition – Laissez Faire
(b) Private service provider – Insurance, radio, etc.
(c) Increase production need – Distribution
(d) To make max profit – Privatisation
Answer:
(c) Increase production need – Marketisation

Question 7.
(a) Increased consumerism – Large production
(b) Sharing of resources – Nationalist movement
(c) Materialistic – Globalisation
(d) All nations interdependent – Global economy
Answer:
(b) Sharing of resources – Technological outsourcing

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 8.
(a) Use of digital technology – Digitalisation
(b) Digital transformation – Capgemini and MIT
(c) Use of computers – Computerisation
(d) Technology-driven – Rural
Answer:
(d) Technology-driven – Digitalisation

Question 9.
(a) Migration of people from rural to urban – Westernisation
(b) Developed scientific temperament – Modernisation
(c) India is part of the global economy – Globalisation
(d) Impact of computers on various aspects of life – Digitalisation
Answer:
(a) Migration of people from rural to urban – Urbanisation

1C. Identify the appropriate term from the given options in the box and rewrite it against the given statement.

Rational Outlook, Factory System, Division of Labour, Technological Advancement, Digitalisation, Mechanisation, Labour Intensive System, Global Economy, Marketisation, Industrialisation, Capital Intensive, Industrial Growth, Law and City Police, etc., Liberal Principle, Urbanisation, Modernisation, Scientific Temperament, Critical Thinking, Laissez-Faire, Profiteering, Interdependence, Outsourcing, Capgemini of MIT, Digitalisation, Computerisation.

Question 1.
Establishment of large factories for the purpose of production.
Answer:
Factory System

Question 2.
Use of heavy machines and techniques for the production of goods and services.
Answer:
Mechanisation

Question 3.
In industrialisation extent of automation and mechanisation.
Answer:
Capital Intensive

Question 4.
The need for skilled force at the workplace.
Answer:
Labour Intensive System

Question 5.
Individuals’ qualities, skills, efficiency, education, and training are the determinants.
Answer:
Division of Labour

Question 6.
A process of migration of rural population to urban areas.
Answer:
Urbanisation

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 7.
Advanced means of commutation are the pre-requisites.
Answer:
Industrial Growth

Question 8.
Means of secondary control.
Answer:
Law and City Police etc.

Question 9.
Daniel Lerner coined the term.
Answer:
Modernisation

Question 10.
The development of a scientific way of understanding and explaining any phenomenon.
Answer:
Scientific Temperament

Question 11.
The approach and ability to provide logical explanations for any phenomenon.
Answer:
Rational Outlook

Question 12.
Use of advanced technology in industries.
Answer:
Technological Advancement

Question 13.
Ability to explain the constructive and destructive aspects of a phenomenon.
Answer:
Critical Thinking

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 14.
Opening up of the economy to private players.
Answer:
Liberal Principle

Question 15.
Free trade and free competition.
Answer:
Laissez-Faire

Question 16.
Privatisation has encouraged many service providers to indulge in it.
Answer:
Profiteering

Question 17.
Increase in production results in large scale of.
Answer:
Marketization

Question 18.
Parts of a product being manufactured in one country and assembled in faraway places is an example of.
Answer:
Interdependence

Question 19.
It has made all people and nations interdependent.
Answer:
Global Economy

Question 20.
People go beyond geographical borders, to perform specific tasks without moving out from their location.
Answer:
Outsourcing

Question 21.
The integration of digital technologies into everyday life.
Answer:
Digitalisation

Question 22.
The term ‘digital transformation’ was coined by.
Answer:
Capgemini of MIT

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 23.
Expansion of the use of computers in all walks of life.
Answer:
Computerisation

1D. Correct the underlined words and complete the statement.

Question 1.
Social change means a change in social life.
Answer:
Social change means a change in social structure.

Question 2.
The domestic production system is replaced by traders.
Answer:
The domestic production system is replaced by a factory system.

Question 3.
The process of industrialisation was started in India.
Answer:
The process of industrialisation was started in Europe.

Question 4.
Industrialisation leads to agrarianism.
Answer:
Industrialisation leads to urbanisation.

Question 5.
Urbanisation is a process of migration of rural populations to tribal areas.
Answer:
Urbanisation is a process of migration of rural population to urban areas.

Question 6.
Daniel Lerner coined the term industrialisation.
Answer:
Daniel Lerner coined the term modernisation.

Question 7.
Urbanism is a way of society.
Answer:
Urbanism is a way of life.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 8.
Urbanisation implies controls and obligations by traditional bodies.
Answer:
Urbanisation implied controls and obligations by civil administrations.

Question 9.
Industrialisation leads to unity.
Answer:
Industrialisation leads to urbanisation.

Question 10.
Urbanisation leads to homogeneity.
Answer:
Urbanisation leads to heterogeneity.

Question 11.
Industrialisation is a process whereby human energy to produce was replaced by the social process for higher production.
Answer:
Industrialisation is a process whereby human energy to produce was replaced by mechanical processes for higher production.

Question 12.
This is the era of computerisation and modernisation.
Answer:
This is the era of computerisation and digitalisation.

Question 13.
Digitalisation is based on belief.
Answer:
Digitalisation is based on technology.

Question 14.
The main aim of digitalisation is the importance of the material.
Answer:
The main aim of digitalisation is important to customers.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 15.
The principle of‘Laissez-Faire’ is an integral aspect of medicines.
Answer:
The principle of ‘Laissez-Faire’ is an integral aspect of globalisation.

Question 16.
The principle of ‘Laissez-Faire’ is originally a Greek term.
Answer:
The principle of ‘Laissez-Faire’ is originally a French term.

Question 17.
Being ‘modern’ means openness to traditional ideas.
Answer:
Being ‘modern’ means openness to new ideas.

Question 18.
Self- criticism, willingness to introspect critically is also an aspect of spiritual thinking.
Answer:
Self-criticism, willingness to introspect critically is also an aspect of critical thinking.

Question 19.
Globalisation involves two processes like liberalisation and generalisation.
Answer:
Globalisation involves two processes like liberalisation and privatisation.

Question 20.
Globalisation is the process of the creation of a global city.
Answer:
Globalisation is the process of the creation of a global economy.

Question 21.
Globalisation is a process that ‘opened up the skies’ for the Japanese economy.
Answer:
Globalisation is a process that ‘opened up the skies’ for the Indian economy.

Question 22.
The flux of people from all over the country to urban areas has resulted in cities getting popular.
Answer:
The flux of people from all over the country to urban areas has resulted in cities getting overpopulated.

Question 23.
Industrialisation is the way by which people go beyond geographical borders, without moving out from their location.
Answer:
Outsourcing is the way by which people go beyond geographical borders without moving out from their location.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 24.
Digital transformation means radically improving knowledge.
Answer:
Digital transformation means radically improving performance.

Question 25.
Maintain individual privacy in the web world is a great challenge in urbanisation.
Answer:
Maintaining individual privacy in the web world is a great challenge in digitalisation.

Question 26.
Due to growth in newer technologies business modules get frequent repetitions.
Answer:
Due to growth in newer technologies business modules get frequent changes.

Question 27.
Cap Gemini is a Spanish data processing company.
Answer:
Cap Gemini is a French data processing company.

2. Write short notes.

Question 1.
Privatisation/Private enterprise
Answer:

  • Privatisation signifies the process wherein, the government transfers ownership, management, and control of the public sector enterprises to the private sector entities.
  • Allows the private sector to set up industries in the field earlier reserved for the public sector.
  • Privatisation has substantially reduced the role of the government in economic activities.
  • Privatisation is an allied process that accompanies globalisation.
  • Under the process of privatisation, reducing the involvement of the state in economic activities and increasing the involvement of the private sector are expected.
  • The government has been thereby trying to mobilise resources for improving the efficiency of the remaining public sector units.
  • Under the policy of globalisation, various countries have adopted the process of privatisation.

Question 2.
Liberalisation/Liberal Principle
Answer:

  • Liberalisation implies the withdrawal of restrictions on industry and business.
  • It also means opening up the economy to private players.
  • It has more reliance on ‘Laissez-Faire’ means free trade and free competition.
  • It encourages foreign trade by a reduction in the tariff rates.
  • Adopting uniform exchange rate.
  • Removal of import/export duties.
  • Duty-free access to foreign goods and services.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 3.
Negative Impact of globalisation
Answer:

  • There is a considerable increase in the immigration of the young technocrats to the developed countries, leaving behind aging parents.
  • There is an increase in the family arguments and break up of more and more marriages and families.
  • The role of the state has been greatly curtailed in economic activities.
  • The role of the state-owned, managed, and controlled public sector has also been curtailed by privatisation and disinvestment.
  • The state also withdrawing from essential social services like health insurance and education by means of privatisation.
  • There is the cultural invasion of western culture. The traditional value and norms of behaviour have slackened up.
  • Consumerism and the pursuit of materialist culture are increasing.
  • The conflict of traditional and modern values has threatened the traditional culture of India.

Question 4.
Modernisation
Answer:

  • It is a process of social transformation.
  • It develops new attitudes, new values, and social relationships. There is a shift to secular and rational values from spiritual-religious values.
  • It has paved a way for developing a scientific temperament. Emphasis has been set on the need for empirical evidence in support of given arguments.
  • It encompasses social, economic, political, religious, and intellectual changes. There is a willingness to receive new ideas, to examine daily events, literature, culture, art, customs, beliefs from a critical point of view.
  • Modernisation is a current term of an old process of social change.

3. Write differences.

Question 1.
Liberalisation and Globalisation
Answer:

Liberalisation Globalisation
(i) Meaning: Liberalisation refers to the removal of undue restrictions and eliminations of bureaucratic controls on productive activities and paves the way for economic development. (i) Meaning: Globalisation refers to the integration of the domestic economy with the world economy.
(ii) Liberalisation is a means to achieve globalisation.

e.g. reduction in tariff is a liberal measure.

(ii) Globalisation can be realised through external liberalisation.
e.g. if the tariff is reduced, transfer of resources, goods, etc., can be made easier.
(iii) Effect: Liberalisation has given a boost to foreign trades. (iii) Effect: Globalisation has led to the increase in foreign direct investment and technical collaboration.
(iv) Manifestations: Abolishing industrial licenses, scrapping the MRTP limit, etc., are the measures adopted to liberate the Indian economy. (iv) Manifestations: Relaxing the FEAR/FEMA regulations, eliminating the trade barriers, providing tax concessions and other incentives to the foreign investors.

Question 2.
Globalisation and Privatisation
Answer:

Globalisation Privatisation
(i) Globalisation refers to “all those processes by which the people of the world are incorporated, into a single world society.” (i) Privatisation means “transferring of ownership rights from public sector to the private sector”.
(ii) Eliminating the trade barriers, relaxing the FERA/FEMA regulations, and other incentives to the foreign investors. (ii) The policy of disinvestment is adopted to privatise the public sector enterprises.
(iii) Globalisation leads to sharing of resources, goods, and capital across the country. (iii) The public sector enterprises are taken over by the private sector. It enables the country to improve the efficiency of these enterprises.
(iv) It has adversely affected agriculture and is a cause of misery in the rural area. (iv) Their policies lead to an increase in unemployment.

Question 3.
Privatisation and Liberalisation
Answer:

Privatisation Liberalisation
(i) Privatisation means reducing the involvement of the public sector and increasing the involvement of the private sector in the country’s economic activities. (i) Liberalisation means reducing or relaxing unnecessary restrictions over economic activities.
(ii) The policy of reduction investment is adopted to privatise the public sector. (ii) Automatic approval to the foreign technology, providing tax concessions and other incentives to the foreign investors, etc.
(iii) The public sector enterprises are taken over by the private sector. It enables the country to improve the efficiency of these enterprises. (iii) Liberalisation has given a boost to the industries in the private sector and given momentum to the industrial development of India.
(iv) Their policies lead to an increase in unemployment. (iv) It has encouraged the culture of consumerism.

4. Explain the following concepts with examples.

Question 1.
Industrialisation
Answer:

  • The process of industrialisation was started in England in the 17th and 18th Centuries.
  • Industrialisation means the process whereby human energy to produce was replaced by mechanical processes and machines to enable higher production.
  • The village population migrated to the industrial centres in the search of employment.
  • Industrialisation signifies the mechanisation of the production process, growth of industries, division of labour and specialisation, capital and labour intensive, etc.
  • Due to industrialisation decline in the Baluta system, the emergence of industry-oriented economy, the emergence of the class system, etc.
  • Examples:
    • Industrial hubs like Mumbai, Pune, Bangalore.
    • The transformation of an urban area into an IT hub is an example of industrialisation.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 2.
Modernisation
Answer:

  • The term modernisation was coined by Daniel Lerner.
  • Modernisation is the process where there is the use of scientific and rational thinking that is deep-seated.
  • Modernisation has led to changes in values, beliefs, and norms.
  • Modernisation is a process of transformation from traditional society to technological modern society.
  • It is a notion of rationalism and the ability to provide scientific and logical explanations for any phenomenon.
  • Being ‘Modern’ means not only using modern devices or gadgets but also openness to receive new ideas, examine alternatives, find new pathways, use creative ways to solve problems, do critical thinking, etc.
  • Examples:
    • Acceptance of rational scientific outlook
    • Individualism
    • Secularism.

Question 3.
Globalisation
Answer:

  • The process of globalisation in the Indian context received an impetus in 1991.
  • In India, New Economic Policy was declared in 1991 by Finance Minister Dr. Manmohan Singh.
  • Globalisation is basically an economic process. It proposes to integrate the national economy with the global economy.
  • There is free flow not only of capital, goods, and technology but also of skilled human resources and ideas across the globe.
  • It proposes to remove barriers in the free movement of human beings and broaden their mental horizons.
  • Due to globalisation, the impact of the western culture and decline of the traditional values as well as changes in the social institutions like family, marriage, etc.
  • We are beset with the positive and negative impacts of globalisation.
  • Examples: International companies such as Pepsi, Coca-Cola, McDonald’s.

Question 4.
Rational outlook
Answer:
Rational outlook refers to an approach that is based on reasoning, and the ability to provide logical explanations for any phenomenon.

Explanation:
A rational outlook is based on the idea of rationalism or reasoning. This means Here, one seeks to establish laws that link facts and which govern social life. To have a rational outlook means that one’s view of a situation or a problem or an event is understood on the basis of reason. This refers to the development of a scientific way of understanding and explaining any phenomenon. Scientific reasoning explains causal relationships between factors. There is a shift to secular and rational values from spiritual-religious values.

For example, non-rational explanation of all-natural calamities are the expressions of God’s anger.
Rational explanation: Scientific explanation for natural calamities.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

5A. Complete the concept map.

Question 1.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q2
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q2.1

Question 2.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q3
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q3.1

Question 3.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q4
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q4.1

Question 4.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q5
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q5.1

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 5.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q6
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q6.1

Question 6.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q7
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q7.1

Question 7.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q8
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q8.1

Question 8.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q9
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q9.1

Question 9.
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q10
Answer:
Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India 5A Q10.1

5B. State whether the following statements are True or False with reasons.

Question 1.
In industrialisation workers feel alienated from the process of production.
Answer:
This statement is True.

  • In industries for large-scale production, purpose capitalists started automation and mechanisation of workplaces.
  • It led to mass production due to which machine-made goods were much cheaper than hand-made goods.
  • The use of machines gives more precision, techniques, and accuracy in production.
  • This leads to workers feeling less important in the work of production and they get alienated from the process of production.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 2.
Industrialisation leads to Urbanisation.
Answer:
This statement is True.
The process of industrial growth led to the large-scale emergence of factories. This in turn resulted in migration to places nearer the factory sites, leading to the growth of towns, which soon became cities and then metropolitan cities.
Since cities become the centres of trade, commerce, industry, education, etc, People from rural areas migrate towards urban areas for getting employment.
The process of urbanisation is the consequence of industrialisation as urbanisation is characteristic of industrialisation.

Question 3.
Urbanisation implies primary means of social control.
Answer:
This statement is False.

  • Urbanisation means a flux of people migrated from all over the country to urban areas.
  • Due to that urban centres are overpopulated. The hold of customs, traditions, religion on people’s behavior, has diminished.
  • The urban environment and way of life are more materialist, radical, commercial, individualist, and non-conforming.
  • Urbanisation implies controls and obligations that are not administered by traditional bodies such as panchayats There are secondary modes of security control, e.g., law enforcement systems such as traffic signals, city police, etc.
  • So, in urban areas, secondary means of social control are useful such as law, city police, etc.

Question 4.
Raj stays in Mumbai and has never faced any problems with commutation.
Answer:
This statement is False.

  • As per the above statement, Raj stays in Mumbai i.e., a Metropolitan city.
  • Which is overpopulated and has a shortage of means of transportation.
  • In metropolises, it is not uncommon to find people spending 3-4 hours commuting to and from the workplace.
  • So, Raj stays in Mumbai means every day he must face problems in commutation.

Question 5.
Modernisation is rational and scientific change.
Answer:
This statement is True.

  • Modernisation is the application of modern science to human affairs.
  • It is linked to the notion of rationalism; the approach and ability to provide logical explanations for any phenomenon.
  • Scientific reasoning explains causal relationships between factors. There N is a shift to secular and rational values from spiritual-religious values.
  • Persons who claim to be modern are willing to examine daily events, literature, culture, art, customs, beliefs from a critical point of view and be able to explain the constructive and destructive aspects of a phenomenon.
  • The ultimate aim of this rational and scientific perspective in modernisation is to make human life better and satisfactory.

Question 6.
In digitalisation, with one click of a button, one can open up a web world.
Answer:
This statement is True.

  • The impact of changes resulting from computerisation and digitisation. Processes have had far-reaching changes in
  • Indian society in terms of access to knowledge artificial intelligence, e-governance, e-commerce, e-learning, e-trade, e-shopping, etc., the list is endless.
  • Digitalisation has sped up the processes of data mining and data management and has made this world a global village.
  • The click of a button can open up a range of options to the user through a very simple procedure.

Question 7.
Digitisation is the use of digital technologies for handling data of various nature
Answer:
This statement is True.

  • Programming, information technology, and computer science have aided the process of computerisation, which in turn has digitised processes for several sectors, e.g., education, banking, revenue, taxation, marketing, etc.
  • Digitisation is based on technology, innovation, research, and development.
  • Digitisation had led to frequent changes in business models due to growth in newer technologies, e.g., Artificial
  • Intelligence is used for various purposes such as production, manufacturing, surgery, etc.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

Question 8.
Globalisation is an eco-friendly process.
Answer:
This statement is False.

  • Due to globalisation, the environmental problem has become more serious.
  • To gain more and more profit, industrialists and multinational companies are excessively using natural resources.
  • Because of this, the intensity of environmental problems such as deforestation, pollution, rise in atmospheric temperature, etc.

6. Give your personal response.

Question 1.
Do you think, during the pandemic situation of COVID-19, people in India will accept digitalised ways for work from home? Why?
Answer:
Being a student of HSC, I feel yes most Indians will accept digitalised ways for work. Because being Indian this pandemic situation is such an eyewash situation for all. As our nation is a developing nation and if we want economic survival then most Indians will start to learn E-Content, even though this will be our first step towards standing in the market, but most Indians will accept digitalized way as this is need of the hour.

Question 2.
Being vigilant student how you will do an online awareness campaign on COVID-19.
Answer:
As COVID-19 is a pandemic and a contagious disease. I will create an online group of my friends and we will remain connected online while staying at home. United through social media groups we will jointly put posters and graffiti on social media to make the general public aware of the deadly disease.

Question 3.
Do you think in the 21st Century people prefer living in a nuclear family? Why?
Answer:
As we are in 2020, in the modern era, so I think yes people will prefer living in a nuclear family. Since the nuclear family provides them with extra space, quality of life and mainly in the nuclear family status of women are high, gender equality, the standard of living is high as the couple is working, less impact of traditional beliefs and superstitions. So, one will prefer individuality i.e. nuclear family.

7. Answer the following question in detail. (About 150-200 words)

Question 1.
You belong to a generation that has been by and large part of digital India. Discuss how digitalisation has brought about positive and negative effects.
Answer:
(i) Positive effects of digitalisation

  • Due to digitalisation, closer cross-border ties.
  • It makes our lives better and easy. It is a most essential technology.
  • It makes people become more efficient and this leads to increased productivity.
  • Digitalisation saves time, cost and gives quick and precise service,
  • Digitalisation boosts the country to do cashless transactions.
  • For example, for filing Income Tax returns, obtaining Birth and Death Certificates from the Municipal Corporation, for online admission, for declaration of election results, etc. Nowadays all these processes make use of digitalisation and it has radically transformed the processes, compared to those used just a few decades ago.

Maharashtra Board Class 12 Sociology Important Questions Chapter 4 Processes of Social Change in India

(ii) Negative effects of digitalisation

  • The excessive use of digital technology has health hazards like affecting sleep.
  • Preventing children from engaging in physical activity that resulting in cases of obesity among children.
  • In extreme cases, there can be mental illnesses like social isolation, aggression, etc.
  • In short, it affects social, physical, and mental health.
  • Due to the acceptance of digitalisation big industries take over the smaller ones.
  • One of the major problems is individual privacy in the web world. This is always under threat.

Maharashtra State Board 12th Book Keeping & Accountancy Important Questions and Answers

Maharashtra State Board HSC 12th Commerce Book Keeping & Accountancy Important Questions and Answers

Maharashtra Board 12th HSC Important Questions

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

1A. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Headquarters of Ministry of Corporate Affairs.
Answer:
New Delhi

Question 2.
The number of Regional Directors.
Answer:
Seven

Question 3.
Total number of Registrar of Companies.
Answer:
Twenty-two

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

Question 4.
The number of ROC’s – cum Official liquidator.
Answer:
Nine

Question 5.
Officer appointed by High Court to look into the matter of winding up of a company.
Answer:
Official Liquidator

Question 6.
Principal Bench of NCLT.
Answer:
New Delhi

1B. State whether the following statements are True or False.

Question 1.
MCA operates with the help of 10 Regional Directors.
Answer:
False

Question 2.
NCLT works through 25 benches.
Answer:
False

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

Question 3.
MCA supervises the working of professional bodies like ICAI, and ICSI.
Answer:
True

1C. Complete the sentences.

Question 1.
To issue an order for seizure of books and papers, ROC has to seek _____________
Answer:
Special Court

Question 2.
MCA is concerned with administration of _____________
Answer:
The Companies Act, 2013

Question 3.
MCA supervises professional body like _____________
Answer:
Institute of Secretaries of India (ICSI)

Question 4.
The headquarter of MCA is at _____________
Answer:
New Delhi

Question 5.
An organization who act as a middleman between capital provider and capital seekers is called as _____________
Answer:
Market Intermediaries

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

Question 6.
Informing about difficulty in uploading e-form means _____________
Answer:
Raising a ticket

1D. Select the correct option from the bracket.

Question 1.

Group ‘A’ Group ‘B’
(1) Public Deposits ……………………….
(2) ……………………… The agent between capital provider and seeker

(Market Intermediaries, 36 Months)
Answer:

Group ‘A’ Group ‘B’
(1) Public Deposits 36 Months
(2) Market Intermediaries Agents between capital provider and seller

1E. Answer in one sentence.

Question 1.
Name the authority who supervises the working of ROC and the Official Liquidator.
Answer:
The authority who supervises the working of ROC’s and the Official Liquidator is Regional Director (RD).

Question 2.
Who can extend the period of helding the Annual General Meeting?
Answer:
Registrar of Company can extend the period of helding the Annual General Meeting.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

Question 3.
Where can an appeal be made against the order of NCLAT?
Answer:
An appeal against the order of NCLAT can be made to Supreme Court within 60 days of receipt of the order of NCLAT.

Question 4.
What do you mean by Market Intermediaries?
Answer:
People or organizations who act as a middleman between the capital provider and capital seeker are called as Market Intermediaries. E.g. Stock Brokers, bankers, underwriters, etc.

1F. Correct the underlined word and rewrite the following sentences.

Question 1.
An appeal can be made against the order issued by NCLAT within 90 days.
Answer:
An appeal can be made against the order issued by NCLAT within 60 days.

Question 2.
An appeal against the order issued by NCLAT can be made to High Court.
Answer:
An appeal against the order issued by NCLAT can be made to Supreme Court.

Question 3.
MCA conducts its operations through 11 Regional Directors.
Answer:
MCA conducts its operations through 7 Regional Directors.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

2. Answer in brief.

Question 1.
Give a flow chart to explain the organizational setup to administer the Companies Act, 2013.
Answer:
Organizational set up to administer the Companies Act, 2013.
Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities 2 Q1

3. Justify the following statements.

Question 1.
Explain the reasons for not holding Annual General Meeting by a company on time.
Answer:
Reasons for not holding Annual General Meeting on time:

  • Directors traveling abroad, so cannot attend Annual General Meeting.
  • Employees Strike in a company.
  • A raid by Income Tax Department.
  • Annual Financial statement not approved or not audited or incomplete audit or loss of financial data due to natural calamity.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

Question 2.
Under what circumstances, Secretary has to correspond with Statutory authorities.
Answer:
Under the following circumstances, Secretary has to correspond with a few of the Statutory authorities:

  • To correspond with ROC for seeking extension of time for holding Annual General Meeting.
  • To correspond with MCA for ‘Ticket Raising’ or other service-related technical complaints.
  • To correspond with SEBI in reply to the complaint by the investor.
  • To correspond with NCLT seeking extension of time to repay Public deposits.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

Question 1.
What are polymers?
Answer:
Polymers are high molecular mass macromolecules (103 – 107 u) and consist of repeating units of monomers.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 2.
Write the classification of polymers based on source. Give examples.
OR
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Based on the source polymers are classified as natural, semisynthetic and synthetic polymers.

  1. Natural polymers : These polymers are obtained either from plants or animals, e.g., cellulose, jute, linene, rubber, silk.
  2. Semisynthetic polymers : The fibres obtained by giving special chemical treatment to natural fibres (cellulose) and further regenerated are called semisynthetic polymers e.g., acetate rayon, viscose rayon, cuprammonium silk.
  3. Synthetic polymers : The man made fibres prepared by polymerization of one monomer or copolymerization of
    two or more monomers are called synthetic polymers, e.g., nylon, terylene, polythene, etc.

Question 3.
Write the classification of polymers based on structure. Give examples
OR
Write the reactions involved in the preparation of (1) Polyvinyl chloride (PVC) (2) Polypropylene.
Answer:
Based on structure polymers are classified as linear chain polymers, branched chain polymers and network or cross linked polymers.

(1) Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 4
They have high melting points; high densities and high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 5

(2) Branched-chain polymers : These polymers consist of long and straight chain with smaller side chains give rise to branched-chain polymers. They have low density. They have lower melting points and tensile strength. Polypropylene having methyl groups as branches.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 6

(3) Network or cross-linked polymers : These polymers consist of cross-linking of chains by strong covalent bonds leading to a network-like structure. Cross-linking results from polyfunctional monomers, e.g., melamine, bakelite, vulcanization of rubber. These polymers are hard rigid and brittle.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 7

Question 3.
How are polymers classified on the basis of mode of polymerization?
OR
Write the classification of polymers based on mode of polymerization.
Answer:
There are three modes of polymerization depending upon the types of reactions taking place between the monomers.

  1. Addition polymerization (or chain growth polymerization)
  2. Condensation polymerization (or step growth polymerization)
  3. Ring opening polymerization
  4. Addition polymerization or chain growth polymerization : It is a process of polymers by the repeated addition of a large number of monomers is called addition polymerization (like alkenes) without loss of any small molecules.
    Example : Formation of polyethylene from ethylene is well known example of addition polymerization. It is a chain growth polymerization.
  5. Condensation polymerization or step growth polymerization : The process of formation of polymers from polyfunctional monomers with the elimination of some small molecules such as water, hydrochloric acid, methanol, ammonia is called condensation polymerization.
    Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 9
    Example : The formation of terylene, a polyester polymer, from ethylene glycol and terephthalic acid. Condensation polymerization is a step growth polymerization.
  6. Ring opening polymerization : The process of formation of polymers from cyclic compounds (like lactams, cyclic ethers, etc.) by ring opening is called ring opening polymerization.
    Example : Polymerization of e-caprolactam.
    Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 10Ring opening polymerization proceeds by addition of a single monomer unit to the growing chain molecules. It is a step growth polymerization.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 4.
Classify the polymers given below as addition, condensation and ring opening polymers:
PVC, polythene, cyclic ethers, polyester, polyacrylonftrile. polystyrene.
Answer:

  • Addition polymers: PVC, polythene. polyacrylonitrile. polystyrene.
  • Condensation polymers: Polyester.
  • Ring opening polymers : Cyclic ethers

Question 5.
Write the classification of polymers based on intermolecular forces. Give examples.
OR
In which dasses, are the polymers classified on the basis of Inter molecular forces?
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question 6.
What is meant by the term homopolymer?
Answer:
A polymer made from only one type of repeating units is called homopolymer. in some cases the repeating unit is formed by condensation of two distinct monomers. Examples. Polythene, PVC. Nylon-6.

Question 7.
What is meant by the term copolymer?
Answer:
A polymer made from two or more types of repeating units is called a copolymer. The different monomer units are randomly sequenced in the copolymer, e.g., Terylene, Nylon-6, 6, Buna-S, Buna-N.

Question 8.
Write the classification of polymers on the basis of biodegradability?
OR
(1) What are biodegradable polymers?
(2) What are nonbiodegradable polymers?
Answer:
Based un biodegradability, polymers are classified as biodegradable polymer and nonbiodegradable polymers.

(1) Biodegradable polymers: Polymers that are affected by microbes or disintegrate by themselves afler a certain period of time due to environmental degradation are called biodegradable polymers.

Examples: PHBV i.e., Polyhydroxy butyrate-CO-β-hydroxy valerate Dextron. Nylon-2-nylon-6.

(2) Non biodegradable polymers: Synthetic polymers do not disintegrate by themselves after a certain period or not affected by microbes, are called nonbiodegradhle polymers.

Examples: Bakelite, Nylon, Terylene.

Question 9.
Explain the following terms :
Answer:

  1. Branched chain polymer : The polymer consists of long and straight chain with smaller side chains give rise to branched chain polymers, e.g. Polypropylene
  2. Addition polymer : The polymer formed by the repeated addition of a large number of monomers (like alkenes) without loss of any small molecules are called addition polymers, e.g. polythene -[-CH2 – CH2-]n. It is a chain growth polymerization.
  3. Condensation polymer : The polymers formed by the repeated condensation reaction between polyfunctional monomers with the elimination of some molecules such as water, hydrochloric acid, methanol, ammonia are called condensation polymers, e.g. Nylon-6, 6.
  4. Elastomers : Polymers in which the intermolecular forces of attraction between the polymer chains are the weakest. When polymer is stretched, it has ability to stretch and when the strain is relieved it returns to its original position. Thus, polymer shows elasticity and is called elastomers, e.g. natural rubber, neoprene, vulcanized rubber.
  5. Homopolymer : A polymer made from only one type of repeating unit of one monomer is called homopolymer e.g. Polythene, PVC, etc.
  6. Biodegradable polymer : Polymers which are affected by microbes or disintegrate by themselves after a certain period of time due to environmental degradation are called biodegradable polymers.
    Example : PHSV i.e. Polyhydroxy butyrate -CO-β-hydroxy valerate Dextron.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 10.
What is natural rubber?
Answer:
Natural rubber is a high molecular mass linear polymer of isoprene (2-methyl-1, 3-butadiene).
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 12

Question 11.
Write a note on natural rubber.
Answer:
Natural rubber is manufactured from rubber latex obtained from the rubber tree in the form of colloidal suspension. Reaction involved in the formation of natural rubber by the process of addition polymerization is as follows :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 13

Natural rubber is -1, 4- polyisoprene. It exhibits elastic properties.

Question 12.
State properties of natural rubber.
Answer:

  1. Polyisoprene molecule has cis configuration of the C = C double bond. It consists of various chains held together by weak van der Waals forces and has coiled structure.
  2. It can be stretched like a spring and exhibits elastic property.
  3. Its molecular mass varies from 130,000 u to 340,000 u.

Question 13.
Write a note on vulcanization of rubber. OR Discuss the main purpose of vulcanization of rubber.
Answer:
The tensile strength, toughness and elasticity of natural rubber can be increased by adding 3 to 5% sulphur and heating at 100-150°C, resulting in cross linking of cis-1, 4-polypropene chains through disulphide bonds, (-S-S-). This process is known as vulcanization of rubber. The physical properties of rubber can be changed by controlling the amount of sulphur in the vulcanization process. Rubber made with 20-30% sulphur is hard, 3 to 10% sulphur is little harder and is used in making tyres and 1 to 3% sulphur is used in making rubber bands.

Question 14.
Write the name and structure of one of the initiators used in free radical polymerisation.
Answer:
The initiator used in free radical polymerisation is acetyl peroxide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 19

Question 15.
What is LDP? How is it prepared? Give its properties and uses.
Answer:
LDP means low-density polyethene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O2 or peroxide as initiator.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 20

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

  • LDP films are extremely flexible, but tough chemically inert and moisture resistant.
  • It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

  • LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
  • It is also used in submarine cable insulation.
  • It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 16.
Whatis HDP ? How is it prepared ? Give its properties and uses ?
Answer:
HDP means high density polyethylene. It is a linear polymer with high density due to close packing.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 21

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

  • HDP is crystalline, melting point in the range of 144 – 150 °C.
  • It is much stiffer than LDP and has high tensile strength and hardness.
  • It is more resistant to chemicals than LDP.

Uses of HDP :

  • HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
  • It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question 17.
How is polyacrylonitrile (PAN) prepared? Give its uses.
Answer:
Acrylonitrile (monomer) on polymerization (addition polymerization) in the presence of peroxide initiator gives polyacrylonitrile.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 23

Uses : Polyacrylonitrile resembles wool and is used as wool substitute and for making orlon or acrilan.

Question 18.
How is nylon-6 prepared?
Write the reaction for the preparation of nylon 6.
Answer:
When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 25

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

Question 19.
Draw the structures of polymers formed using the following monomers.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 28

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 29

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 30
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 31

Question 20.
How is Novolac prepared?
OR
Write the reaction to prepare Novolac polymer.
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 33
Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 21.
How is bakelite prepared?
Answer:
In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 34
Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 22.
How is a melamine-formaldehyde polymer (melamine) prepared?
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 36

Question 23.
Write the preparation of the following synthetic rubbers and give their uses :
(1) Buna-S or styrene-butadiene rubber (SBR) (2) Neoprene rubber
Answer:
(1) Buna-S rubber : Its trade name is SBR (Styrene-butadiene rubber) Buna-S is a copolymer of styrene and 1, 3-butadiene. When 75 parts of butadiene and 25 parts of styrene subjected to addition polymerization by the action of sodium.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 37

Uses : Buna-S is superior to natural rubber with regard to mechanical strength and has abrasion resistance. Hence, it is used in tyre industry.

(2) Neoprene : Neoprene, a synthetic rubber, is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene), Chloroprene polymerizes rapidly in presence of oxygen.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 38
Vulcanization of neoprene takes place in presence of magnesium oxide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 39
Uses : Neoprene is resistant to petroleum, vegetable oils, light as well as heat. It is used in making hose pipes for transport of gasoline and making gaskets. It is used for manufacturing insulator cable, jackets, belts for power transmission and conveying.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 24.
Write structure of natural rubber and neoprene rubber along with the name and structure of their monomers.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 40

Question 25.
Write the preparation of viscose rayon.
Answer:
Viscose rayon is a semisynthetic fibre. It is a regenerated cellulose. The molecular formula of cellulose is (C6H10O5)n. A modified representation of the molecular formula of cellulose Cell-OH is used in the reactions involved in viscose formation.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 42

Cellulose (wood pulp) is treated with the concentrated NaOH to form alkali cellulose. It is then converted to xanthate by treating with CS2. Further xanthate is mixed with dilute NaOH to form viscose solution which is extruted through spinnerates of spinning machine into acid bath, when regenerated cellulose fibres precipitate, i.e. viscose rayon.

Question 26.
How is PHBV polymer prepared?
Answer:
It is a copolymer. The monomers β-hydroxy butyric acid (3-hydroxy butanoic acid) and β-hydroxy valeric acid (3-hydroxy pentanoic acid) undergo polymerization to form PHBV polymer. It has an ester linkage. Hydroxyl group of one monomer forms ester link by reacting with carboxyl group of the other. Thus PHBV is an aliphatic polyester i.e. poly β-hydroxybutyrate-co-β-hydroxy valerate (PHBV). It is a biodegradable polymer.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 45

Question 27.
Write the name/s of monomer/s, polymer structure and one use of each of the following polymers (trade name) :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 47
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 48

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 28.
Write the names of monomers used in preparing following polymers :
(1) Dacron.
Answer:
Monomers : Ethylene glycol and Dmiethyl terephthalate (DMT)
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 49

(2) Bakelite.
Answer:
Monomers : o-hydroxy methyl phenol and formaldehyde.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 50

(3) Nylon-6, 8.
Answer:
Monomers : Hexamethylene diamine and Hexamethylene dicarboxylic acid.
H2N-(CH2)6-NH2 HOOC-(CH2)6-COOH

(4) Melamine.
Answer:
Monomers : Melamine and Formaldehyde
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 51

(5) Buna-S.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 52

(6) Buna-N.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 53

(7) Butyl rubber.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 54

(8) Teflon.
Answer:
Monomers : F2C = CF2 Tetrafluoroethene

(9) Natural rubber.
Answer:
Monomers : 1,3-Butadiene
CH2 = CH – CH = CH2

(10) Neoprene.
Answer:
Monomers : Chloroprene or 2-Chloro-l,3-butadiene
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 55

Question 29.
Write the structures of monomers used in the preparation of following polymers.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 56
Answer:
The monomers used in the preparation of above polymer are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 57

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 58
Answer:
The monomers used in the preparation of above polymer are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 59
Answer:
The monomer used in the preparation of above polymer are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 60
Answer:
The monomer used in the preparation of above polymer is
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 61

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 30.
Following monomers are used to prepare polymers. Predict the structures of polymers:

(1) Ethylene glycol.
Answer:
Ethylene glycol Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 62 is used in the preparation of polyester (terylene or dacron).
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 63

Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 26

Properties :

  • Terylene has relatively high melting point (265 °C)
  • It is resistant to chemicals and water.

Uses :

  • It is used for making wrinkle-free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
  • PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
  • It is used for making many articles like bottles, jams, packaging containers.

(2) ε-Caprolactam.
Answer:
ε-caprolactam is used in the preparation of nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 64

When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 25

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

(3) Ethene.
Answer:
Ethene is used in the preparation of polythene
Polymer: Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 65

Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 4
They have high melting points; high densities and high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 5

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

(4) Formaldehyde.
Answer:
Formaldehye is used in the preparation of bakelite.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 66

The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 33
Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 34
Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 31.
Classify the following polymers as step growth or chain growth polymers :
(1) Nylon-6,6
(2) Terylene
(3) Polyethene (4) PVC.
Answer:
Step growth polymers : Nylon-6,6, terylene
Chain growth polymers : Polythene, PVC.

Question 32.
Classify the following as linear, branched or cross linked polymers :
(1) Bakelite
(2) Starch
(3) Polythene
(4) Nylon.
Answer:
Linear polymers : Polythene, nylon.
Cross-linked polymers : Bakelite, starch.

Question 33.
Classify the following as addition and condensation polymers :
(1) Bakelite
(2) polyvinyl chloride
(3) polythene
(4) terylene.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question 34.
Arrange the polymers in increasing order of their intermolecular forces :
Nylon-6,6, Polythene, Buna-S.
Answer:
The increasing order of their intermolecular forces of attraction follows the order :
Buna-S, Polythene, Nylon-6,6.

Question 35.
Classify the following as natural, synthetic and semisynthetic polymers :
Terylene, cuprammonium silk, jute, melamine
Answer:
Natural polymers : Jute
Synthetic polymers : Terylene, melamine
Semisynthetic polymers : Cuprammonium silk

Question 36.
Complete the following statements :
(1) Chemically teflon is …………………………. .
(2) …………………………. is the catalyst used to obtain HDP by polymerisation of ethene.
(3) Viscose rayon is a …………………………. .
Answer:
(1) polytetrafluoroethylene
(2) Zieglar-Natta
(3) semisynthetic fibre (regenerated fibre).

Question 37.
Answer the following in one sentence each.

(1) Name a polymer used for making LCD screen?
Answer:
The polymer used for making LCD screen is Perspex.

(2) Which of the two is a condensation polymer? Bakelite or Polythene?
Answer:
The condensation polymer is bakelite.

(3) Which of the two is a linear polymer? Nylon or Starch.
Answer:
The linear polymer is nylon.

(4) Which of the two is a step growth polymer? Nylon-6,6 or PVC.
Answer:
The step growth polymer is Nylon-6,6.

(5) Write the use of polyacrylamide gel.
Answer:
Polyacrylamide gel is used in electrophoresis.

(6) Write the use of urea formaldehyde resin.
Answer:
Urea formaldehyde resin is used in making unbreakable dinnerware and decorative laminates.

(7) Give an example of semisynthetic polymer.
Answer:
Semisynthetic polymer : Viscose rayon, cuprammonium silk.

(8) Write the monomer unit of teflon.
Answer:
Monomer unit of teflon : Tetrafluoroethene (F2C = CF2).

(9) Write the equation for the preparation of polypropylene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 72

(10) Name a synthetic polymer which contains amide linkage.
Answer:
Polymer that contains amide linkage : Nylon-6; Nylon 6,6.

(11) Name a synthetic polymer which contains ester linkage.
Answer:
Polymer that contains ester linkage : Terylene or Dacron.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

(12) Name one thermosetting and one thermoplastic polymer.
Answer:
Thermosetting polymer : Bakelite.
Thermoplastic polymer : Polythene, polystyrene.

(13) State the uses of biodegradable polymers.
Answer:
Biodegradable polymers are used as orthopaedic devices, implants, sutures and drug release matrices.

(14) Name a copolymer which is used for making nonbreakable crockeries.
Answer:
The polymer used in making nonbreakable crockeries : Melamine formaldehyde polymer.

(15) Write the structure of monomer used in the preparation of Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 75
Answer:
The structure of monomer : Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 76

(16) Write the structure of melamine.
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 36

(17) What does SBR stand for?
Answer:
SBR stand for styrene(S)butadiene (B) rubber (R).

(18) Draw the structure of repeating unit in nylon-6,10.
Answer:
The structure of repeating unit in nylon-6,10 is
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 77

(19) What are the monomers used to prepare nylon given below?
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 78
Answer:
Monomers used in the preparation of nylon are
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 79

(20) Write the monomers used to prepare nylon given below :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 80
Answer:
Monomers used in the preparation of nylon are
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 81

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

(21) Name the catalyst which is formed from titanium chloride and triethyl aluminium.
Answer:
The catalyst Zieglar Natta is formed from titanium chloride and triethyl aluminium.

(22) Define molecular mass of polymer.
Answer:
Molecular mass of a polymer is an average of the molecular masses of the constituent molecules.

(23) Which factor determines the molecular mass of polymer.
Answer:
Molecular mass of polymer depends upon the degree of polymerization (DP). DP is the number of monomer units in a polymer molecule.

Multiple Choice Questions

Question 38.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which one is the natural polyamide polymer?
(a) Cuprammonium silk
(b) Wool
(c) Perlon-L
(d) Jute
Answer:
(b) Wool

2. The synthetic fibres are prepared from
(a) cellulose
(b) starch
(c) chemical compounds
(d) polymers
Answer:
(c) chemical compounds

3. Which of the following is NOT a vegetable fibre?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

4. Which of the following fibres are made up of polyamides?
(a) Dacron
(b) Rayon
(c) Nylon
(d) Terylene
Answer:
(c) Nylon

5. An example of natural cellulose fibre is
(a) cotton
(b) wool
(c) silk
(d) rayon
Answer:
(a) cotton

6. Cellulose is the main constituent of
(a) nylon-6
(b) cotton
(c) terylene
(d) wool
Answer:
(b) cotton

7. Of the following, which group contains two cellulosic fibres and one protein fibre?
(a) Cotton, keratin, wool
(b) Linen, keratin, wool
(c) Cotton, linen, rayon
(d) Cotton, keratin, linen
Answer:
(d) Cotton, keratin, linen

8. Which of the following is not a vegetable fibres?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

9. Which of the following is polyamide?
(a) Teflon
(b) Nylon 6, 6
(c) Terylene
(d) Bakelite
Answer:
(b) Nylon 6, 6

10. The monomer of e-caprolactam is
(a) styrene
(b) amino acid
(c) aminocaproic acid
(d) adipic acid O
Answer:
(c) aminocaproic acid

11. Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 83is the formula of _ Jn
(a) Nylon-66 salt
(b) Nylon-66
(c) DMT
(d) Nylon-6
Answer:
(d) Nylon-6

12. Nylon-6 is a
(a) polyester fibre
(b) protein fibre
(c) poly caprolactum fibre
(d) poly amine fibre
Answer:
(c) poly caprolactum fibre

13. The condensation product of e-caprolactum is
(a) teflon
(b) nylon-6
(c) nylon-66
(d) bakelite
Answer:
(b) nylon-6

14. \(\left[\overline{\mathrm{O}} \mathrm{OC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COO}-\mathrm{N} \mathrm{H}_{3}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{N} \mathrm{H}_{3}\right]\) is the formula of
(a) nylon-6
(b) nylon-6 salt
(c) nylon-66 salt
(d) nylon-66
Answer:
(b) nylon-6 salt

15. The starting material required for the preparation of Nylon-66 is
(a) glycol
(b) α-amino acid
(c) adipic acid and hexamethylene diamine
(d) dimethyl terephthalate and ethylene glycol
Answer:
(c) adipic acid and hexamethylene diamine

16. Terylene is also known as
(a) styrene
(b) butadiene
(c) dacron
(d) teflon
Answer:
(c) dacron

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

17. Terylene is
(a) vegetable fibre
(b) protein fibre
(c) polyester fibre
(d) polyamide fibre
Answer:
(c) polyester fibre

18. Terylene polymer is formed in
(a) the presence of nitrogen
(b) the presence of hydrogen
(c) the presence of oxygen
(d) the vacuum
Answer:
(d) the vacuum

19. The by-product formed during the preparation of terylene fibre is
(a) glycerol
(b) propylene glycol
(c) ethylene glycol
(d) ethyl alcohol
Answer:
(c) ethylene glycol

20. Nylon polymer cannot be used for making
(a) tyre cords
(b) films
(c) dress materials
(d) fishing nets
Answer:
(b) films

21. Glycol is an important constituent of
(a) nylon-6
(b) nylon-66
(c) terylene
(d) hexamethylene diammonium adipate
Answer:
(c) terylene

22. Terylene is prepared by the process of
(a) halogenation
(b) condensation
(c) esterification
(d) hydrogenation
Answer:
(b) condensation

23. What are the steps during polymerisation to form terylene?
(a) Terephthalic acid is condensed with ethylene glycol at 420 K-460 K.
(b) Ethylene glycol displaces methanol to form dihydroxy diethyl terephthalic acid
(c) Zinc acetate – antimony trioxide is used as catalyst
(d) All of these
Answer:
(d) All of these

24. During polymerisation of nylon salt to nylon-66, the conditions are
(a) room temperature and pressure
(b) temperature 503 K
(c) temperature 553 K in presence of Nitrogen
(d) heating in an autoclave at 373 K
Answer:
(c) temperature 553 K in presence of Nitrogen

25. Which one of the following is a condensation polymer?
(a) Nylon
(b) Polythene
(c) PVC
(d) Teflon
Answer:
(a) Nylon

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

26. Which one of the following is an addition polymer?
(a) Bakelite
(b) Nylon-6,6
(c) Polystyrene
(d) Terylene
Answer:
(c) Polystyrene

27. A polymer of butadiene and Acrylonitrile is called
(a) Buna-S
(b) Buna-N
(c) Buna-B
(d) Buna-A
Answer:
(b) Buna-N

28. Natural rubber is a polymer of
(a) Styrene
(b) Butadiene
(c) Vinyl chloride
(d) Isoprene
Answer:
(d) Isoprene

29. In which of the following pairs both are copolymers?
(a) PHBV, bakelite
(b) Polythene, terylene
(c) Polyacrylonitrile, nylon-6,6
(d) Polystyrene, melamine
Answer:
(a) PHBV, bakelite

30. The polymer used in paints is
(a) Nylon
(b) Glyptal
(c) Neoprene
(d) Terylene
Answer:
(b) Glyptal

31. Which of the following contains biodegradable polymers only?
(a) Cellulose, dextron, PHBV
(b) Starch, PHBV, PVC
(c) Bakelite, nylon-2-nylon-6, nylon-6,6
(d) Cellulose, starch, terylene
Answer:
(a) Cellulose, dextron, PHBV

32. Thermosetting polymer is
(a) Nylon-6
(b) Nylon-6,6
(c) Bakelite
(d) SBR
Answer:
(c) Bakelite

33. Nylon thread contains the polymer
(a) Polyamide
(b) Polyvinyl
(c) Polyester
(d) Polyethylene
Answer:
(a) Polyamide

34. Polythene, PVC, teflon and neoprene are all
(a) Monomers
(b) Homopolymers
(c) Copolymers
(d) Condensation polymers
Answer:
(b) Homopolymers

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

35. Which one of the following is NOT a biodegrad¬able polymer?
(a) Starch
(b) Cellulose
(c) Dextron
(d) Decron
Answer:
(d) Decron

36. The polymer used in making blankets (artificial wool) is
(a) Polyester
(b) Polyacrylonitrile
(c) Polythene
(d) Polystyrene
Answer:
(b) Polyacrylonitrile

37. Which one of the following is a linear polymer?
(a) Bakelite
(b) LDP
(c) Nylon
(d) Formaldehyde melamine polymer
Answer:
(c) Nylon

38. Which one of the following is a branched polymer?
(a) PVC
(b) Polyester
(c) Nylon
(d) Polypropylene
Answer:
(d) Polypropylene

39. The polymer used to make non-stick cookware is
(a) Polyethene
(b) Polystyrene
(c) Polytetrafluoroethylene
(d) Polyvinyl chloride
Answer:
(c) Polytetrafluoroethylene

40. The monomer used to prepare orlon is
(a) CH2 = CH-CN
(b) CH2 = CHCl
(c) CH2 = CH-F
(d) CH2 = CF2
Answer:
(a) CH2 = CH-CN

41. Buna-N rubber is a copolymer of
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 86
Answer:
(b)

42. Bakelite is a polymer of
(a) Formaldehyde and phenol
(b) Benzaldehyde and phenol
(c) Formaldehyde and benzyl alcohol
(d) Acetaldehyde and phenol
Answer:
(a) Formaldehyde and phenol

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

43. The process involving heating of natural rubber with sulphur is known as
(a) Sulphonation
(b) Vulcanisation
(c) Galvanisation
(d) Calcination
Answer:
(b) Vulcanisation

44. A polymer of bisphenol and phosgene is called
(a) Polyamide
(b) Glyptal
(c) Polycarbonate
(d) Polystyrene
Answer:
(c) Polycarbonate

45. Thermocol is a homopolymer of
(a) terephthalic acid
(b) acrylonitrile
(c) methyl a-cyanoacrylate
(d) styrene
Answer:
(d) styrene

46. The polymer is used to prepare shatter resistant glass is called
(a) Perspex/acrylic glass
(b) Soda glass
(c) Buna N
(d) Polyacrylamide
Answer:
(a) Perspex/acrylic glass

47. A polymer used in making shoe soles is
(a) Glyptal
(b) Buna-N
(c) Buna-S
(d) Poly carbonate
Answer:
(b) Buna-N

48. The Zieglar-Natta catalyst is used in the preparation of
(a) LDPE
(b) PHBV
(c) PAN
(d) HDPE
Answer:
(d) HDPE

49. Which of the following is natural rubber?
(a) cis-1, 4-polyisoprene
(b) neoprene
(c) Trans-1. 4-polyisoprene
(d) Butyl rubber
Answer:
(a) cis-1, 4-polyisoprene

50. Which one from the following is the Terylene polymer?
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 84
Answer:
(b)

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

51. Equivalent amount of Hexamethylene diamine and adipic acid on complete neutralization produces :
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 85
Answer:
(a)

52. Polyhydroxy butyrate-CO-β-hydroxy valerate represents
(a) Dextron
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) PHBV
Answer:
(d) PHBV

53. Among the following, the biodegradable polymer is
(a) PVC
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) Neoprene
Answer:
(c) Nylon-2-nylon-6

54. The monomers of Nylon-2-nylon-6 are
(a) glycine and ω-amino caproic acid
(b) lactic acid and glycolic acid
(c) glycolic acid and co-amino caproic acid
(d) isobutylene and isoprene
Answer:
(a) glycine and ω-amino caproic acid

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Negotiable Instrument which can be discounted with the bank _____________
(a) Cheque
(b) Bills of Exchange
(c) Bank Draft
Answer:
(b) Bills of Exchange

Question 2.
Bank overdraft is a _____________ term facility provided by bank.
(a) medium
(b) short
(c) long
Answer:
(b) short

Question 3.
A loan for a period of more than 5 years is called as _____________ loans.
(a) short term
(b) long term
(c) medium-term
Answer:
(b) long term

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 4.
Purchase and sale of securities is _____________ function of Bank.
(a) primary
(b) utility
(c) agency
Answer:
(c) agency

1B. Match the pairs.

Question 1.

Group ‘A’ Group ‘B’
(a) Current Account (1) Time Deposits
(b) Financial Arrangement (2) ATM, Credit Card and Debit Card
(c) Agency Function (3) Demand Deposits
(d) Utility Function (4) Collection of Dividend and Interest
(e) Cash Credit (5) Bank Overdraft
(f) Bill of Exchange (6) Advance against the stock of raw material
(g) Letter of Credit (7) Non-negotiable instrument
(8) Bank Draft
(9) Negotiable Instrument
(10) International trade transaction

Answer:

Group ‘A’ Group ‘B’
(a) Current Account (3) Demand Deposits
(b) Financial Arrangement (5) Bank Overdraft
(c) Agency Function (4) Collection of Dividend and Interest
(d) Utility Function (2) ATM, Credit Card, and Debit Card
(e) Cash Credit (6) Advance against the stock of raw material
(f) Bill of Exchange (9) Negotiable Instrument
(g) Letter of Credit (10) International trade transaction

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Deposits generating saving habits among the people.
Answer:
Saving Deposits

Question 2.
Deposits are not repayable on demand.
Answer:
Time Deposits

Question 3.
Deposits repayable on demand.
Answer:
Demand Deposits

Question 4.
The loan was provided for a period of less than one year.
Answer:
Short Term Loan

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 5.
The loan is provided for a period between 1 to 5 years.
Answer:
Medium-Term Loan

Question 6.
The loan was provided for a period of more than 5 years.
Answer:
Long Term Loan

1D. State whether the following statements are True or False

Question 1.
Letter of credit is issued by banks for domestic trade transactions.
Answer:
False

Question 2.
Bill of Exchange is a negotiable instrument.
Answer:
True

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 3.
The bank acts as a trustee, attorney, and executor of the will.
Answer:
True

Question 4.
RTGS is an agency function of the Bank.
Answer:
False

Question 5.
The bank plays the role of Depository Participant (DP).
Answer:
True

1E. Find the odd one.

Question 1.
Loan, Deposit, Cash credit, Overdraft facility.
Answer:
Deposit

Question 2.
Cheque, Withdrawal slip, Pay in slip.
Answer:
Pay in slip

Question 3.
Travellers Cheque, Safe Deposit Vault, NEFT, Transfer of Money.
Answer:
Transfer of Money

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 4.
Collection of Dividend, Collection of cheque, Buying, and Selling of Securities, Payment of Electricity Bill.
Answer:
Buying and Selling of Securities

1F. Complete the sentences.

Question 1.
The appointment of bankers of a company is made by the _____________
Answer:
Board of Directors

Question 2.
Banker is a dealer in _____________
Answer:
Money

Question 3.
Bank account in which money is deposited at regular interval is a _____________
Answer:
Recurring Deposit Account

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 4.
The instrument which can be discounted with the bank is a _____________
Answer:
Bills of Exchange

Question 5.
Businessman, Companies etc, usually open a _____________ account with a bank.
Answer:
Current

Question 6.
There is no restriction on the withdrawals from _____________ account.
Answer:
Current

Question 7.
Interest is not paid on the _____________ account.
Answer:
Current

Question 8.
_____________ account is suitable for salaried people.
Answer:
Saving

Question 9.
Stop payment letter is sent, when the cheque is _____________
Answer:
lost/misplaced

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 10.
Withdrawals are not permitted from the _____________ accounts.
Answer:
Fixed Deposits

Question 11.
The facility given by the bank to draw more money than the actual balance in the credit is called _____________
Answer:
Overdraft facility

Question 12.
A deposit which is kept for a fixed period in a bank is a _____________
Answer:
Fixed Deposit

Question 13.
A type of bank account which is generally opened by a businessman for his business transactions is a _____________
Answer:
Current account

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 14.
A type of account in which the interest is paid at higher rate is a _____________
Answer:
Fixed Deposit Account

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’ Group ‘B’
(1) Loans not more than one year ……………………….
(2) …………………….. Cash Credit
(3) Collection of cheques ………………………..
(4) …………………….. Letter of Credit

(Interest charged on, actual amount withdrawn, Agency function, Utility function, Short Term loans)
Answer:

Group ‘A’ Group ‘B’
(1) Loans not more than one year Short Term loans
(2) Interest charged on actual amount withdrawn Cash Credit
(3) Collection of cheques Agency function
(4) Utility function Letter of Credit

1H. Answer in one sentence.

Question 1.
What is a demand deposit?
Answer:
Deposits that are repayable on demand are called demand deposits. There are 2 types of demand deposits i.e Saving Deposits and Current Deposits.

Question 2.
What is a time deposit?
Answer:
Deposits that are repayable after a specific period of time are called time deposits. There are 2 types of time deposits i.e. Fixed Deposits and Recurring Deposits.

Question 3.
What do you mean by loan?
Answer:
Any amount granted or lent for a specific period of time against personal security, gold or silver, or other movable or immovable assets is called a loan.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 4.
What is the Bill of Exchange?
Answer:
It is a written unconditional order by the seller to the buyer, to pay a certain sum of money on a future fixed date for payment of goods or services received.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Short-term loans are of more than 5 years.
Answer:
Long-term loans are of more than 5 years.

Question 2.
Bills of Exchange are the non-negotiable instrument.
Answer:
Bills of exchange is a negotiable instrument.

Question 3.
The collection of cheques and bills is a utility function of the bank.
Answer:
The collection of cheques and bills is the agency function of the bank.

Question 4.
The rate of interest is high for saving deposits.
Answer:
The rate of interest is high for fixed deposits.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 5.
Deposits that are repayable after the specific period is Demand Deposit.
Answer:
Deposits that are repayable after the specific period are Time Deposits.

2. Answer in brief.

Question 1.
Explain the types of term loans.
Answer:
A loan granted for a specific time period against personal security, gold or silver, and movable and immovable assets is called a term loan.
Types of term loans:

  • Short-term loan: A loan repayable within 1 year is called a short-term loan. It is generally taken by businessmen to meet their working capital requirements.
  • Medium-term loan: A loan repayable within 2 years to 5 years period is called a medium-term loan.
  • Long-term loan: A loan repayable after 5 years is called a long-term loan. It is taken by businessmen for the growth and development of the business.

Question 2.
Explain the types of Advances.
Answer:
Advance is a credit facility provided by the bank to its customers. Advances are for a shorter period than loans.
Type of Advances:

  • Overdraft: It is an arrangement in which a customer i.e. Current A/c holder, is allowed to overdrew money in excess of the credit balance in his account. It is allowed against collateral securities like – shares, F.D.R., Government securities, etc.
  • Cash credit: In this mode of advance, a separate bank account called ‘Cash Credit Account is opened in the name of the borrower. He can withdraw from this account as and when required. It is allowed against security like – stock of raw materials, finished goods, etc.
  • Discounting of Bills: Discounting of bills means encashing the Bills of Exchange before its due date with the banker. The bank charges a certain amount of interest for this service which is called a discounting rate.

3. Justify the following statements.

Question 1.
In cash credit, the customer’s account is credited by a bank with the sanctioned amount.
Answer:
Cash credit is another kind of credit facility given by the bank to its customers including businessmen, companies, etc.

  • A separate bank account known as a “Cash Credit Account” is required to be opened in the name of the borrower.
  • The bank credits the account as per the sanctioned cash limit from which the customer can utilize funds whenever required for cash credit.
  • Generally, the security of tangible assets like goods, finished stock is required to be kept.
  • The interest is charged only on the actual amount utilized by the customer.
  • Cash credit arrangement is for a longer period as compared to overdraft.
  • This system of lending is prevalent in India only.
  • Thus, in cash credit, the customer’s account is credited by a bank with the sanctioned amount.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

Question 2.
Bank correspondence should be brief and to the point.
Answer:

  • The bank is a financial institution.
  • A company secretary has to conduct bank correspondence as per the instruction of the Board.
  • Bank correspondence needs careful and cautious drafting.
  • The company secretary has to use his knowledge, skill, and experience while conducting bank correspondence,
  • The company secretary has to conduct bank correspondence promptly and accurately.
  • Mistakes and delays in bank correspondence may bring financial loss to the company.
  • Bank correspondence should be always brief, compact, and precise.
  • Unnecessary or irrelevant information should be avoided in bank correspondence.
  • Thus, bank correspondence should be brief and to the point.

Question 3.
No interest is paid by the bank on the current account.
Answer:

  • The Current account is normally opened by businessmen, firms, or companies.
  • A current account is a running account and in practice it never becomes time-barred.
  • This account is opened with a minimum deposit.
  • There is no limit on the amount or number of withdrawals.
  • Interest is not payable on this account.
  • Overdraft facility is given only to current depositors after following the prescribed bank procedure.
  • Hence, interest is paid only in the case of recurring, fixed, and saving accounts and not in the case of the current account.
  • Thus, no interest is paid by the bank on the current account.