Answer the following questions (Pure Science Group)
Botany : Chromosomal Basis of Inheritance
32. When two different genes came from same parent they tend to remain together.
i) What is the name of this phenomenon?
ii) Draw the cross with suitable example.
iii) Write the observed phenotypic ratio.
Answer: (i) The phenomenon is called linkage.
(ii) William Bateson and Reginald C. Punnet crossed one homozygous strain of sweet peas having purple flowers and long pollen grains with another homozygous strain having red flowers and round pollen grains. Results of Test Cross showed that a greater number of F2 plants had purple flowers and long pollen or red flowers and round pollen. So they concluded that genes for purple colour and long pollen grain and the genes for red colour and round pollen grain were found close together in the same homologous pair of chromosomes. This type of tendency of genes to stay together during separation of chromosomes is called Linkage.
Alleles in coupling or cis configuration
(iii) Phenotypic ratio 7 : 1 : 1 : 7
33. If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain F1 hybrid. Now you cross F1 male with double recessive female.
i) What type of linkage is seen?
ii) Draw the cross with correct genotype.
iii) What is the possible genotype in F2 generation?
Answer: (i) Complete linkage
(The tendency of genes to stay together during separation of chromosomes)
(iii) PV/pv and pv/pv in the ratio 1:1.
i) What is the name of this test cross?
ii) How will you construct gene mapping from the above given data?
iii) Find out the correct order of genes.
Answer: (i) Three point test cross - gene mapping.
between A and B:
RF = [ Total no. of recombinants / Total no. of progenies ] x 100
= (239 / 1200) x 100 = 19.9 %
Recombinant frequency between A and C
RF = (482 / 1200) x 100 = 40.1%
Recombinant frequency between B and C
RF = (261 / 1200) x 100 = 21.7%
RF is higher between A and C, so they must be farthest apart. The locus B must lie between them.
(iii) ABC / abc
35. What is the difference between missense and nonsense mutation?
Answer: The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations. The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.
From the above figure identify the type of mutation and explain it.
Answer: The figure shows structural chemical aberration- Duplication-Reverse tandem type.
Reverse tandem duplication:
The duplicated segment is located immediately after the normal segment but the gene sequence order will be reversed.
37. Write the salient features of Sutton and Boveri concept.
Answer: Sutton and Boveri (1903) independently proposed the chromosome theory of inheritance. Sutton united the knowledge of chromosomal segregation with Mendelian principles and called it chromosomal theory of inheritance.
Salient features of the Chromosomal theory of inheritance:
Answer: (i) Somatic cells of organisms are derived from the zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
(ii) Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
(iii) Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.
(iv) The behaviour of chromosomes during the gamete formation (meiosis) provides evidence to the fact that genes or factors are located on chromosomes.
38. Explain the mechanism of crossing over.
Answer: Mechanism of Crossing Over
Crossing over is a precise process that includes stages like synapsis, tetrad formation, cross over and terminalization.
Intimate pairing between two homologous chromosomes is initiated during zygotene stage of prophase I of meiosis I. Homologous chromosomes are aligned side by side resulting in a pair of homologous chromosomes called bivalents. This pairing phenomenon is called synapsis or syndesis. It is of three types,
1. Procentric synapsis: Pairing starts from middle of the chromosome.
2. Proterminal synapsis: Pairing starts from the telomeres.
3. Random synapsis: Pairing may start from anywhere.
(ii) Tetrad Formation:
Each homologous chromosome of a bivalent begin to form two identical sister chromatids, which remain held together by a centromere. At this stage each bivalent has four chromatids. This stage is called tetrad stage.
(iii) Cross Over:
After tetrad formation, crossing over occurs in pachytene stage. The non-sister chromatids of homologous pair make a contact at one or more points. These points of contact between non-sister chromatids of homologous chromosomes are called Chiasmata (singular-Chiasma). At chiasma, cross-shaped or X-shaped structures are formed, where breaking and rejoining of two chromatids occur. This results in reciprocal exchange of equal and corresponding segments between them. Synapsis and chiasma formation are facilitated by a highly organised structure of filaments called Synaptonemal Complex (SC). This synaptonemal complex formation is absent in some species of male Drosophila. Hence crossing over does not takes place.
39. Write the steps involved in molecular mechanism of DNA recombination with diagram.
The widely accepted model of DNA recombination during crossing over is Holliday’s hybrid DNA model. It was first proposed by Robin Holliday in 1964. It involves several steps. ( Figure )
(i) Homologous DNA molecules are paired side by side with their duplicated copies of DNAs.
(ii) One strand of both DNAs cut in one place by the enzyme endonuclease.
(iii) The cut strands cross and join the homologous strands forming the Holliday strcture or Holliday junction.
(iv) The Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
(v) DNA strands may cut along through the vertical (V) line or horizontal (H) line.
(vi) The vertical cut will result in heteroduplexes with recombinants.
(vii) The horizontal cut will result in heteroduplex with non recombinants.
40. How is Nicotiana exhibit self-incompatibility. Explain its mechanism.
Answer: (i) In plants, multiple alleles have been reported in association with self-sterility or self-incompatibility. Self - sterility means that the pollen from a plant is unable to germinate on its own stigma and will not be able to bring about fertilization in the ovules of the same plant.
The self-incompatibility in relation to its genotype in tobacco
(ii) East (1925) observed multiple alleles in Nicotiana which are responsible for self-incompatibility or self-sterility. The gene for self-incompatibility can be designated as S, which has allelic series S1, S2, S3, S4 and S5 .
(iii) The cross-fertilizing tobacco plants were not always homozygous as S1S1 or S2S2, but all plants were heterozygous as S1S2, S3S4, S5S6.
(iv) When crosses were made between different S1S2 plants, the pollen tube did not develop normally. But effective pollen tube development was observed when crossing was made with other than S1S2 for example S3S4.
of progeny in self-incompatibility
41. How sex is determined in monoecious plants. write their genes involved in it.
Answer: (i) Zea mays (maize) is an example for monoecious, which means male and female flowers are present on the same plant.
(ii) There are two types of inflorescence. The terminal inflorescence which bears staminate florets develops from shoot apical meristem called tassel. The lateral inflorescence which develop pistillate florets develops from axillary bud and is called ear or cob.
Sex determination in Maize (Superscript (+) denotes
(iii) Unisexuality in maize occurs through the selective abortion of stamens in ear florets and pistils in tassel florets.
(iv) A substitution of two single gene pairs 'ba' for barren plant and 'ts' for tassel seed makes the difference between monoecious and dioecious (rare) maize plants. The allele for barren plant (ba) when homozygous makes the stalk staminate by eliminating silk and ears. The allele for tassel seed (ts) transforms tassel into a pistillate structure that produce no pollen. Most of these mutations are shown to be defects in gibberellin biosynthesis. Gibberellins play an important role in the suppression of stamens in florets on the ears.
42. What is gene mapping? Write its uses.
Answer: The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map. The concept of gene mapping was first developed Alfred H Sturtevant.
Uses of genetic mapping:
(i) It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
(ii) They are useful in predicting results of dihybrid and trihybrid crosses.
(iii) It allows the geneticists to understand the overall genetic complexity of particular organism.
43. Draw the diagram of different types of aneuploidy.
Answer: It is a condition in which diploid number is altered either by addition or deletion of one or more chromosomes. Different types of Aneuploidy is show below. It is classified into two main types - Hyperploidy and Hypoploidy.
Types of Aneuploidy
44. Mention the name of man-made cereal. How it is formed?
Answer: Triticale is the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups.
(i) Tetraploidy: Crosses between diploid wheat and rye.
(ii) Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye
(iii) Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye.
Hexaploidy: Triticale hybrid plants demonstrate characteristics of both macaroni wheat and rye. For example, they combine the high-protein content of wheat with rye’s high content of the amino acid lysine, which is low in wheat.
Colchicine, an alkaloid when applied in low concentration to the growing tips of the plants induce polyploidy.
45. What is DNA repair?
Answer: (i) Genomic stability is maintained in all living organisms which is essential for their survival.
(ii) DNA is unique because it is the only macromolecule where the repair system exists, which recognises and removes mutations.
(iii) DNA is subjected to various types of damaging reactions such as spontaneous or environmental agents or natural endogenous threats. Such damages are corrected by repair enzymes and proteins, immediately after the damage has taken place.
(iv) DNA repair system plays a major role in maintaining the genomic / genetic integrity of the organism. DNA repair systems protect the integrity of genomes from genotoxic stresses.
46. What is replication fork?
Answer: (i) During DNA replication, replication fork is the site (point of unwinding) of separation of parental DNA strands where new daughter strands are formed. Multiple replication forks are found in eukaryotes.
(ii) The enzyme helicases are involved in unwinding of DNA by breaking hydrogen bonds holding the two strands of DNA and replication protein A (RPA) prevents the separated polynucleotide strand from getting reattached.
47. Write about the energetics of DNA replication.
Answer: The Energetics of DNA replication - Deoxyribonucleotides such as deoxyadenosine triphosphate dATP, dGTP, dCTP and dTTP provide energy for the synthesis of DNA. Purpose of Deoxyribonucelotides (1) acts as a substrate (2) provide energy for polymerisation.
48. What is TATA box?
Answer: In protein synthesis during transcription, Eukaryotic structural gene has 3 features in promoter
(i) Regulatory elements
(ii) TATA box
(iii) A transcriptional start site
The transcription start site contains about 25 bp (basepairs) upstream, the sequence is TATAAT known as TATA or Hogness box which is present in core promoter.
49. What is alternative splicing?
Answer: It is very useful in regulating gene expression to overcome the environmental stress in plants.
(i) Alternative splicing is an important mechanism / process by which multiple mRNA’s and multiple proteins products can be generated from a single gene. The different proteins generated are called isoforms.
(ii) There are various modes of alternative splicing. When multiple introns are present in a gene, they are removed separately or as a unit. In certain cases one or more exons which is present between the introns are also removed.
Significance of alternative splicing
(i) The proteins transcribed from alternatively spliced mRNA containing different amino acid sequence lead to the generation of protein diversity and biological functions.
(ii) Multiple protein isoforms are formed.
(iii) It creates multiple mRNA transcripts from a single gene. A process of producing related proteins from a single gene thereby the number of gene products are increased.
(iv) It plays an important role in plant functions such as stress response and trait selection. The plant adapts or regulates itself to the changing environment.
50. What is coding strand?
Answer: During protein synthesis, in transcription, the strand of DNA which is not transcribed is called the coding strand.
51. What are the enzymes involved in DNA replication in eukaryotes?
Answer: (i) The enzyme helicases are involved in unwinding of DNA by breaking hydrogen bonds holding the two strands of DNA.
(ii) Topoisomerase is an enzyme which breaks DNAs covalent bonds and removes positive supercoiling ahead of replication fork.
(iii) DNA replication is initiated by an enzyme DNA polymerase α / primase which synthesizes short stretch of RNA primers on both leading strand (continuous DNA strand) and lagging strands (discontinuous DNA strand).
(iv) DNA Pol α (alpha), DNA Pol δ (delta) and DNA Pol ε (Epsilon) are the 3 enzymes involved in nuclear DNA replication.
(v) DNA Pol α - Synthesizes short primers of RNA
(vi) DNA Pol δ - Main Replicating enzyme of cell nucleus
(vii)DNA Pol ε - Extend the DNA Strands in replication fork
(viii) DNA ligase joins any nicks in the DNA by forming a phosphodiester bond between 3’ hydroxyl and 5’ phosphate group.
52. Differentiate coding and non coding strand.
Answer: (i) Coding strand / Non-template strand / Sense Strand
During protein synthesis in transcription, the strand of DNA which is not transcribed is called the Coding Strand.
(ii) Template Strand / Non-Coding Strand / Antisense Strand
During trascription in protein synthesis, The strand of DNA which is oriented in 3’ → 5’ direction that serves as a template for the synthesis of mRNA is called template strand.
53. What are splicesomes?
Answer: (i) RNA splicing is a process which involves the cutting or removing out of introns and knitting of exons. This process takes place in spherical particles which is a multiprotein complex called SPLICISOMES. It is approximately 40 - 60 nm in diameter.
(ii) The spliceosomes have many small nuclear ribonucleic acids (snRNAs) and small nuclear ribonuclear protein particles (snRNPs) which identify and helps in the removal of introns.
(iii) A spliceosome removes the introns with an enzyme ribozyme. Now the mature mRNA comes away from the spliceosomes through the nuclear pore and is transported out from the nucleus into cytoplasm, and gets attached to ribosome to carry out translation.
54. What is meant by capping and tailing?
Answer: hnRNA or heterogenous nuclear RNA is the precursor of mRNA modification at the 5’ end of the primary RNA transcript (hnRNA) with methylguanosine triphosphate is called capping.
Purpose of capping
(i) Protects RNA from degradation
(ii) Capping plays an important role in removal of first intron in pre mRNA.
(iii) It regulates the mRNA export from the nucleus into the cytoplasm.
(iv) It helps in binding of mRNA to theribosome.
The 3’ end of hnRNA is cleaved by an endonuclease and a string of adenine nucleotides is added to the 3’ end of hnRNA (pre mRNA) is known as Poly (A) tail - Polyadenylation. This process is called tailing or polyadenylation.
Purpose of tailing
(i) Translation of RNA transcript is facilitated.
(ii) Helps in the synthesis of polypeptides.
(iii) It enhances the mRNA stability in the cytoplasm.
55. What is RNA editing?
Answer: Chemical modification such as base modification, nucleotide insertion or deletions and nucleotide replacements of mRNA results in the alteration of amino acid sequence of protein that is specified is called RNA editing. This results in the change in the protein coding sequence of RNA following transcription. The coding properties of the RNA transcript is changed.
56. Explain the DNA replication in eukaryotes.
Answer: (i) Replication starts at a specific site on a DNA sequence known as the origin of replication. There are more than one origin of replication in eukaryotes.
(ii) DNA replication in eukaryotes starts with the assembly of a prereplication complex (preRC) consisting of 14 different proteins.)
(iii) The origin of replication in yeast is called as ARS sites (Autonomously Replicating Sequences )
Eukaryotic replication fork
(iv) Replication fork is the site (point of unwinding) of separation of parental DNA strands where new daughter strands are formed. Multiple replication forks are found in eukaryotes.
(v) The enzyme helicases are involved in unwinding of DNA by breaking hydrogen bonds holding the two strands of DNA and replication protein A (RPA) prevents the separated polynucleotide strand from getting reattached.
(vi) Topoisomerase is an enzyme which breaks DNAs covalent bonds and removes positive supercoiling ahead of replication fork.
(vii) DNA replication is initiated by an enzyme DNA polymerase α / primase which synthesizes short stretch of RNA primers on both leading strand (continuous DNA strand) and lagging strands (discontinuous DNA strand).
(viii) Primers are needed because DNA polymerase requires a free 3’ OH to initiate synthesis. DNA polymerase covalently connects the nucleotides at the growing end of the new DNA strand.
(ix) DNA Pol α (alpha), DNA Pol δ (delta) and DNA Pol ε (Epsilon) are the 3 enzymes involved in nuclear DNA replication.
(x) DNA Pol α - Synthesizes short primers of RNA.
(xi) DNA Pol δ - Main replicating enzyme of cell nucleus.
(xii) DNA Pol ε - Extend the DNA Strands in replication fork.
(xiii) DNA synthesis takes place in 5’ → 3’ direction and it is semidiscontinuous. When DNA is synthesized in 5’ → 3’ direction, only in the free 3’ end (OH end) DNA is elongated.
(xiv) In discontinuous strand the new DNA strand is synthesized in short pieces called Okazaki fragments. The Okazaki fragments are united by ligase and is also called Lagging strand where the replication direction is 5’ → 3’ which is opposite to the direction of fork movement.
(xv) The continuous strand is called Leading strand where the replication direction is 5’ → 3’ which is same to the direction to that of the replication fork movement.
(xvi) DNA ligase joins any nicks in the DNA by forming a phosphodiester bond between 3’ hydroxyl and 5’ phosphate group.
57. With reference to the given diagram correctly match the following pairs.
Column I - Column II
A - Gene
B - Transcribed region
C - Termination of transcription
D - Regulation of initiation of transcription
E - Protein-encoding sequence
58. What attributes make Arabidopsis a suitable model plant for molecular genetic research?
Answer: Arabidopsis thaliana - Thale cress, Mouse ear cress
(i) It is a model plant for the study of genetic and molecular aspects of plant development.
(ii) It belongs to mustard family and it is the first flowering plant, where its entire genome is sequenced.
(iii) The two regions of the nucleolar organiser ribosomal DNA which codes for the ribosomal RNA are present at the extremity of chromosomes 2 and 4.
(iv) It is diploid plant having small genome with 2n = 10 chromosomes. Several generations can be produced in one year. So it facilitates rapid genetic analysis. The genome has low repetitive DNA, over 60% of the nuclear DNA have protein coding functions.
(v) The plant is small, self fertilizes, annual long-day plant with short-life cycle. Large numbers of seeds are produced and they are easy to be grown in laboratory. It is easy to induce mutations. It has many genomic resources and the transformation can be done easily.
(vi) In 1982, Arabidopsis has successfully completed its life cycle in microgravity i.e. space. This shows that Human Space Missions with plant companions may be possible.
59. Describe the molecular mechanism of RNA modification.
Answer: The processing of pre-mRNA to mature mRNA / Molecular mechanism of RNA modification.
In eukaryotes three major types of RNA, mRNA, tRNA and rRNA are produced from a precursor RNA molecule termed as the primary transcript or preRNA. The RNA polymerase II transcribes the precursor of mRNA, which are also called the heterogenous nuclear RNA or hnRNA which are processed in the nucleus before they are transported into the cytoplasm.
Modification at the 5’ end of the primary RNA transcript (hn RNA) with methylguanosine triphosphate is called capping.
Purpose of capping
(a) Protects RNA from degradation.
(b) Capping plays an important role in removal of first intron in pre mRNA.
(c) It regulates the mRNA export from the nucleus into the cytoplasm.
(d) It helps in binding of mRNA to the ribosome.
(ii) Tailing / Polyadenylation
The 3’ end of hnRNA is cleaved by an endonuclease and a string of adenine nucleotides is added to the 3’ end of hnRNA (pre mRNA) is known as Poly (A) tail - Polyadenylation. This process is called tailing or polyadenylation.
Purpose of tailing:
(a) Translation of RNA transcript is facilitated.
(b) Helps in the synthesis of polypeptides.
(c) It enhances the mRNA stability in the cytoplasm.
The protein coding regions are not continuous in eukaryotes. Exons are the coding sequences or expressed sequences contain biological informations in the matured processed mRNA. Introns are intervening sequences, which are non-coding sequences (non-amino acid- coding sequences) that should be removed from a gene before the mRNA product is made. These exons and introns are known as Split Genes.
(iii) RNA splicing in plants
(a) RNA splicing is a process which involves the cutting or removing out of introns and knitting of exons. This process takes place in spherical particles which is a multiprotein complex called SPLICISOMES. It is approximately 40 - 60 nm in diameter.
(b) A spliceosome removes the introns with an enzyme ribozyme. Now the mature mRNA comes away from the spliceosomes through the nuclear pore and is transported out from the nucleus into cytoplasm, and gets attached to ribosome to carry out translation.
60. Explain ribosomal translocation in protein synthesis.
Answer: (i) The translation begins with the AUG codon (start codon) of mRNA.
(ii) Translation occurs on the surface of the macromolecular arena called the ribosome.
(iii) It is a non-membranous organelle.
(iv) During the process of translation the two sub-units of ribosomes unite (combine) together and hold mRNA between them.
(v) The protein synthesis begins with the reading of codons of mRNA.
(vi) The tRNA brings amino acid to the ribosome, a molecular machine which unites amino acids into a chain according to the information given by mRNA.
(vii) rRNA plays the structural and catalytic role during translation.
A ribosome has one binding site for mRNA and two for tRNA. The two binding sites of tRNA are
(i) P-Site - The peptidyl - tRNA binding site is one of the tRNA binding site. At this site tRNA is held and linked to the growing end of the polypeptide chain.
(ii) A-Site - The Aminoacyl - tRNA binding site. This is another tRNA binding site which holds the incoming amino acids called aminoacyl tRNA. The anticodons of tRNA pair with the codons of the mRNA in these sites.
Elongation of polypeptide chain
The P and A sites are near by, so that two tRNA form base pairs with adjacent codon.
The tRNA translates the genetic code from the nucleic acid sequence to the amino acid sequence i.e from gene - Polypeptide. When an amino acid is attached to tRNA it is called aminoacylated or charged.
The translation begins with the AUG codon (start codon) of mRNA. The tRNA which carries first amino acid methionine attach itself to P-site of ribosome. The ribosome adds new amino acids to the growing polypeptides. The second tRNA molecules has anticodons which carries second amino acid and pairs with the mRNA codon in the A-site of the ribosome. The first and second amino acids are close enough so that a peptide bond is formed between them.
The bond between the first tRNA and methionine now breaks. The first tRNA leaves the ribosome and the P-site is vacant. The ribosome now moves one codon along the mRNA strand. The second t-RNA molecule now occupies the P-site. The third t-RNA comes and fills the A site.
Now a peptide bond is formed between second and third amino acids. The mRNA then moves through the ribosome by three bases. This expels deacylated / uncharged tRNA from P-site and moves peptidyl tRNA into the P-site and empties the A-site. This movement of tRNA from A-site to P-site is said to be translocation. The translocation requires thehydrolysis of GTP.
The ribosome (ribozyme - peptidyl transferase) catalyses the formation of peptide bond by adding amino acid to the growing polypeptide chain.
The ribosome moves from codon to codon along the mRNA in the 5’ to 3’ direction. Amino acids are added one by one, translated into polypeptide as dictated by the mRNA. Translation is an energy intensive process. A cluster of ribosomes are linked together by a molecule of mRNA and forming the site of protein synthesis is called as polysomes or polyribosomes.
61. Describe transposons.
Answer: (i) Transposons are the DNA sequences which can move from one position to another position in a genome. This was first reported by an American Geneticist Barbara McClintock as “mobile controlling element” in Maize.
(ii) Barbara McClintock when studying aleurone of single maize kernels, noted the unstable inheritance of the mosaic pattern of blue, brown and red spots due to the differential production of vacuolar anthocyanins.
(iii) In maize plant genome has AC / Ds transposon (AC = Activator, Ds = Dissociation). The activity of AC element is very distinct in maize plant. The transposition in somatic cells results in the changes in gene expression such as variegated pigmentation in maize kernels. Maize genome has transposable elements which regulated the different colour pattern of kernels.
(iv) McClintock’s findings concluded that Ds and AC genes were mobile controlling elements. These are called transposable elements.
(v) McClintock gave the first direct experimental evidence that genomes are not static but are highly plastic entities.
Significance of transposons
(i) They contribute to many visible mutations and mutation rate in an Organism.
(ii) In evolution, they contribute to genetic diversity.
(iii) In genetic research transposons are valuable tools which are used as mutagens, as cloning tags, vehicles for inserting foreign DNA into model organism.
62. Describe RNA editing in plants.
RNA Editing - Post Transcriptional RNA Processing in plants
(i) Chemical modification such as base modification, nucleotide insertion or deletions and nucleotide replacements of mRNA results in the alteration of amino acid sequence of protein that is specified is called RNA editing. This results in the change in the protein coding sequence of RNA following transcription. The coding properties of the RNA transcript is changed. The genetic information encoded in the chloroplast genome is altered by post transcriptional phenomenon which is site - specific (C → U) in chloroplast of higher plants - RNA editing occurs in plant mitochondria and chloroplast.
(ii) In plant cells RNA editing by pyrimidine transitions occurs in mitochondria and plastids (chloroplast). There are two main types of RNA editing.
1. Substitution editing - Alteration of individual nucleotide bases. Mitochondria and chloroplast RNA in plants.
2. Insertion / Deletion editing - Nucleotides are added or deleted from the total number of bases.
Significance of RNA editing
(a) In higher plant chloroplast, it helps to restore the codons for conserved aminoacids which include initiation and termination codon.
(b) It regulates organellar gene expression in plants.
(c) RNA editing results in the restoration of codons for phylogenetically conserved amino acid residues.