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12.2: Modifications at the 5' and 3' ends of mRNA - Biology

12.2: Modifications at the 5' and 3' ends of mRNA - Biology


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As discussed previously, eukaryotic mRNAs are capped at their 5' end and polyadenylated at their 3' end. In vitro assays for these reactions have been developed, and several of the enzymatic activities have been identified. These will be reviewed in this section. Polyadenylation is not limited to eukaryotes. Several mRNAs in E. coli are polyadenylated as well. This is a fairly new area of study.

Modification at the 5' end: Cap Structure

The "cap" is a methylated 5'‑GMP that is linked via its 5' phosphate to the b‑phosphoryl of the initiating nucleotide (usually A); see Figure 3.3.6. Capping occurs shortly after transcription has begun. It occurs in a series of enzymatic steps (Figure 3.3.7):

  1. Remove the g‑phosphoryl of the initiating nucleotide (RNA triphosphatase)
  2. Link a GMP to the b‑phosphoryl of the initiating nucleotide (mRNA guanylyl transferase). The GMP is derived from GTP, and is linked by its 5' phosphate to the 5' diphosphate of the initiating nucleotide. Pyrophosphate is released.
  3. The N‑7 of the cap GMP is methylated (methyl transferase), donor is S‑adenosyl methionine.
  4. Subsequent methylations occur on the 2' OH of the first two nucleotides of the mRNA.

Capping has been implicated in having a role in efficiency of translation and in mRNA stability.

Several proteins are required for cleavage and polyadenylation at the 3' end:

  1. CPSF is a tetrameric specificity factor; it recognizes and binds to the AAUAAA polyadenylation signal.
  2. CFI and CFII are cleavage factors.
  3. PAP is the polyA polymerase.
  4. CFI, CFII and PAP form a complex that binds to the nascent RNA at the cleavage site, directed by the CPSF specificity factor.
  5. CstF is an additional protein implicated in this process in vitro, but its precise function is currently unknown.

Bookshelf

NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health.

Alberts B, Johnson A, Lewis J, et al. Molecular Biology of the Cell. 4th edition. New York: Garland Science 2002.

  • By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.


Steps of Transcription

Transcription takes place in three steps: initiation, elongation, and termination. The steps are illustrated in Figure 2.

Figure 2. Transcription occurs in the three steps—initiation, elongation, and termination—all shown here.

Step 1: Initiation

Initiation is the beginning of transcription. It occurs when the enzyme RNA polymerase binds to a region of a gene called the promoter. This signals the DNA to unwind so the enzyme can ‘‘read’’ the bases in one of the DNA strands. The enzyme is now ready to make a strand of mRNA with a complementary sequence of bases.

Step 2: Elongation

Elongation is the addition of nucleotides to the mRNA strand. RNA polymerase reads the unwound DNA strand and builds the mRNA molecule, using complementary base pairs. There is a brief time during this process when the newly formed RNA is bound to the unwound DNA. During this process, an adenine (A) in the DNA binds to an uracil (U) in the RNA.

Step 3: Termination

Termination is the ending of transcription, and occurs when RNA polymerase crosses a stop (termination) sequence in the gene. The mRNA strand is complete, and it detaches from DNA.

This video provides a review of these steps. You can stop watching the video at 5:35. (After this point, it discusses translation, which we’ll discuss in the next outcome.)


Important Questions for CBSE Class 12 Biology The DNA and RNA World

1.Over the years after Mendel, the nature of the genetic material was investigated, resulting in the realisation that DNA is the genetic material in majority of organisms.

2.Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA) are the two types of nucleic acid found in living systems. Nucleic acids are polymers of nucleotides.

3DNA acts as a genetic material in most organisms, whereas RNA acts as a genetic material in some viruses.

4.RNA mostly functions as messenger. RNA has other functions as adapter, structural or as a catalytic molecule.

5.Structure of Polynucleotide Chain
(i)A nucleotide has three parts, i.e. a nitrogenous base, a pentose sugar (deoxyribose in DNA and ribose in RNA) and a phosphate group.
(ii)Nitrogenous bases are purines, i.e. adenine, guanine and pyrimidines, i.e. cytosine, uracil and thymine.
(iii)Cytosine is common for both DNA and RNA and thymine is present in DNA. Uracil is present in RNA at the place of thymine.
(iv)A nitrogenous base is linked to the pentose sugar through a N-glycosidic linkage to form a nucleoside, i.e. adenosine and guanosine, etc.
(v)When a phosphate group is linked to 5′ —OH of a nucleoside through phosphodiester linkage, a corresponding nucleotide is formed.
(vi)Two nucleotides are linked through 3′ -> 5′ phosphodiester linkage to form a dinucleotide.
(vii)Several nucleotides can be joined to form a polynucleotide chain.
(viii) The backbone in a polynucleotide chain is formed due to sugar and phosphates.
(ix)The nitrogenous bases linked to sugar moiety project from the backbone.
(x)The base pairs are complementary to each other.

6.In case of RNA, every nucleotide residue has an additional—OH group present at 2-position in the ribose. Also, the uracil is found at the place of thymine (5-methyl uracil)

7.Discoveries Related to Structure of DNA
(i)Friedrich Meischer in 1869, first identified DNA as an acidic substance present in the nucleus and named it as ‘nuclein’.
(ii)James Watson and Francis Crick, proposed a very simple double helix model for the structure of DNA in 1953 based on X-ray diffraction data.
(iii)Erwin Chargaff proposed that for a double-stranded DNA, the ratios between adenine and thymine and guanine and cytosine are constant and equals to one.

8.Salient Features of Double-helix Structure of DNA
(i)DNA is a long polymer of deoxyribonucleotides. It is made up of two polynucleotide chains, where the backbone is constituted by sugar-phosphate and the bases project inside.
(ii)The two chains have anti-parallel polarity, i.e. 5′—- > 3′ for one, 3′– > 5′ for another.
(iii)The bases in two strands are paired through hydrogen bond (H—bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with thymine from opposite strand and vice-versa. Guanine bonds with cytosine by three H—bonds. Due to this, purine always comes opposite to a pyrimidine. This forms a uniform distance between the two strands.
(iv)The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm and there are roughly 10 bp in each turn. Due to this, the distance between a base pair in a helix is about 0.34 nm.
(v)The plane of one base pair stacks over the other in double helix. This confers stability to the helical structure in addition to H—bonds.

9.The length of a DNA double helix is about 2.2 meters
Therefore, it needs special packaging in a cell.
(i)In prokaryotic cells (which do not have a defined nucleus), such as E.coli, DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called as nucleoid. The DNA in nucleoid is organised in large loops held by proteins.
(ii)In eukaryotes, there is a set of positively charged proteins called histones that are rich in basic amino acid residues, lysines and arginines (both positive).Histones are organised to form a unit of eight molecules called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.
(iii)A typical nucleosome contains 200 bp of DNA helix.Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin (thread-like stained structure). Under electron microscope, the nucleosomes in chromatin can be seen as beads-on-string. This structure in chromatin is packaged to form chromatin fibres that further coils and condense to form chromosomes at metaphase stage.
(iv)The packaging of chromatin at higher level requires additional set of proteins which are collectively called Non-Histone Chromosomal (NHC) proteins.
(v)In a nucleus, some regions of chromatin are loosely packed (stains light) called euchromatin (transcriptionally active chromatin). In some regions, chromatin is densely packed (stains dark) called heterochromatin (inactive chromatin).

10.Transfonning Principle
(i)Frederick Griffith (1928) carried out a series of experiments with Streptococcus pneumoniae (bacterium causing pneumonia).
(ii)According to him, when the bacteria are grown on a culture plate, some produce smooth shiny colonies (S), while others produce rough (R) colonies.
(iii)This is because the S-strain bacteria have a mucous (polysaccharide) coat, while R-strain does not.
(iv)Mice infected with S-strain (virulent) die from pneumonia but mice infected with R-strain do not develop pneumonia.

(v)Griffith killed bacteria by heating and observed that heat-killed S-strain bacteria injected into mice did not kill them. On injecting mixture of heat-killed S and live R bacteria, the mice died. He recovered living S-bacteria from dead mice.

(vi)From this experiment, he concluded that the ‘R-strain bacteria’ had been transformed by the heat-killed S-strain bacteria. Some transforming principle transferred from heat-killed S-strain, had enabled the R-strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to transfer of the genetic material.

11.Biochemical Nature of Transforming Principle
(i)Oswald Avery, Colin MacLeod and Maclyn McCarty, worked to determine the biochemical nature of transforming principle in Griffith’s experiment.
(ii)They purified biochemicals (proteins, RNA and DNA, etc) from heat-killed S-cells and discovered that DNA alone from S-bacteria caused R-bacteria to be transformed.
(iii)They also discovered that protease (protein digesting enzyme) and RNAases (RNA-digesting enzymes) did not affect transformation.
(iv)Digestion with DNAse did inhibit transformation, indicating that DNA caused transformation.
(v)They concluded that DNA is the hereditary material. But, still all the biologists were not convinced.

12.DNA is the Genetic Material
(i)Alfred Hershey and Martha Chase (1952) gave unequivocal proof that DNA is the genetic material.
(ii)In their experiments, bacteriophages (viruses that infect bacteria) were used.
(iii)They grew some viruses on a medium that contained radioactive phosphorus and some others on sulphur containing radioactive medium
(iv)Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. In the same way, viruses grown on radioactive sulphur contained radioactive protein, but not radioactive DNA because DNA does not contain sulphur

(v)Radioactive phages were allowed to attach to coli bacteria. As the infection proceeded, viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
(vi) Bacteria which were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
(vii)Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that the proteins did not enter the bacteria from viruses. It proved that DNA is a genetic material that is passed from virus to bacteria.

13.Properties of Genetic Material
(i)It became establised that DNA is the genetic material from the Hershey-Chase experiment.
(ii)In some viruses, RNA was also reported as genetic material, e.g. Tobacco mosaic viruses, QB bacteriophage, etc.
(iii)Characteristics of a Genetic Material

  • (iv)It should be able to replicate.
  • (v)It should be chemically and structurally stable.
  • (vi)It should provide scope for slow changes (mutation) that are required for evolution.
  • (vii)It should be able to express itself in the form of ‘Mendelian characters’.

(iv) According to the above mentioned rules, both the nucleic acids (DNA and RNA) have the ability to direct duplications.Stability can be explained in DNA as the two strands being complementary if separated by heating come together in appropriate conditions.
(v)The 2′ — OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable, hence it is reactive.
(vi)DNA is chemically less reactive and structurally more stable when compared to RNA. Thymine also confers additional stability to DNA. So, among the two nucleic acids, the DNA is a predominant genetic material.
(vii)Both RNA and DNA are able to mutate. Viruses having RNA genome and having shorter life span mutate and evolve faster.
(viii) DNA is dependent on RNA for protein synthesis, while RNA can directly code for it. The protein synthesising machinery has evolved around RNA. This concluded that the DNA being more stable is suitable for storage of genetic information, while for the transmission of genetic information, RNA is suitable.

14.Francis Crick proposed the central dogma in molecular biology, which states that the genetic information flows from

In some viruses, the flow of information is in reverse direction i.e. from RNA to DNA.

15.Replication Scheme for replication of DNA termed as semiconservative DNA replication was proposed by Watson and Crick (1953). According to it,
(i)The two strands would separate and act as a template for the synthesis of new complementary strands. .
(ii)After replication, each DNA molecule would have one parental and one newly synthesised strand.

16.Experimental proof that DNA replicates semiconservatively, comes first from E. coli and later from higher organisms, such as plants and human cells.
Matthew Meselson and Franklin Stahl performed the following experiments to prove this in 1958.
(i)coli was grown in a medium containing 15 NH4C1 as the only nitrogen source for many generations. 15 N got incorporated into newly synthesised DNA (and other nitrogen containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.
(ii)They then transferred the cells into a medium with normal 14 NH4C1 and took samples at various definite intervals as the cells multiplied and extracted the DNA that remained as double stranded helices. DNA samples were separated independently on CsCl gradients to measure DNA densities.
(iii)The DNA that was extracted from the culture, one generation (after 20 min) after the transfer from 15 N to 14 N medium had a hybrid or intermediate density. DNA extracted from the culture after another generation (after 40 min) was composed of equal amounts of this hybrid DNA and of light DNA.
(iv)Very similar experiments were carried out by Taylor and Colleagues on Vicia faba (faba beans) using radioactive thymidine and the same results, i.e. DNA replicates semiconservatively, were obtained as in earlier experiments.

17.DNA replication machinery and enzymes process of replication requires a set of catalysts (enzymes).
(i)The main enzyme is DNA-dependent DNA polymerase, since it uses a DNA template to catalyse the polymerisation of deoxynucleotides. The average rate of polymerisation by these enzymes is approximately 2000 bp/second.
(ii)These polymerases has to catalyse the reaction with high degree of accuracy because any mistake during replication would result into mutations.
(iii) DNA polymerisation is an energy demanding process, so deoxyribonucleoside triphosphates serve dual purposes, i.e. act as substrates and provide energy for polymerisation reaction.
(iv)Many additional enzymes are also required in addition to DNA-dependent DNA polymerase.
(v)(a) Replication in DNA strand occurs within a small opening of the DNA helix, known as replication fork.
(b) DNA-dependent DNA polymerases catalyse polymerisation only in one direction, i.e. 5′ -> 3. It creates additional complications at the replicating fork. Consequently, on one strand (template 3′-5′), the replication is continuous, while on the other strand (template 5′-3′), it is discontinuous. The discontinuously synthesised fragments called Okazaki fragments are later joined by DNA ligase.

18.Origin of Replication
(i)DNA polymerases cannot initiate the process of replication on their own. Also, replication does not initiate randomly at any place in DNA. So, there is a definite region in coli DNA where the replication originates. The region is termed as origin of replication.
(ii)Due to this requirement, a piece of DNA, if needed to be propagated during recombinant DNA procedures, requires a vector. The vectors provide the origin of replication.

19.RNA world RNA was the first genetic material. There are evidences to prove that essential life processes, such as metabolism, translation, splicing, etc., have evolved around RNA.
(i)There are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein enzyms.
(ii) DNA has evolved from RNA with chemical modifications that make it more stable because RNA being a catalyst was reactive and hence, unstable.

20. There are following three types of RNAs:
(i) mRNA (messenger RNA) provides the template for transcription.
(ii) tRNA (transfer RNA) brings amino acids and reads the genetic code.
(iii) rRNA (ribosomal RNA) plays structural and catalytic role during translation.
All the three RNAs are needed to synthesise a protein in a cell.

21.Trascription is the process of copying genetic information from one strand of the i into RNA. The principle of complementarity governs the process of transcription, except the adenosine now forms base pair with uracil instead of l thymine.
(i) In transcription, only a segment of DNA is duplicated and on Iv one of the strands is . copied into RNA. Both the strands are not copied because

  • If both the strands code for RNA, two different RNA molecules and two different proteins would be formed, hence complicating the genetic information transfer machinery.
  • Since two RNA produced would be complementary to each other, they would form a double-stranded RNA without translation, making the process of transcription futile.

(ii)A transcription unit in DNA is defined by three regions in the DNA which are as follows:
(a)A promoter (b) The structural gene (c) A terminator
(iii) The two strands of DNA have opposite polarity and the DNA-dependent RNA polymerase also catalyse the polymerisation in only one direction that is 5′ -> 3′.
(iv)The strand that has the polarity 3′ -> 5′ acts as a template and is referred to as template strand. The other strand which has the polarity (5′ -> 3′) and the sequence same as RNA (T at the place of U) is displaced during transcription. This strand is called as coding strand.

(v)The promoter and terminator flank the structural gene in a transcription unit.
(vi)The promoter is located towards 5′ end (upstream) of the structural gene.
(vii)It is the DNA sequence that provides binding site for RNA polymerase and the presence of promoter defines the template and coding strands. By switching its position with terminator, the definition of coding and template strands could be reversed.
(viii) The terminator is located towards 3 f -end (downstream) of the coding strand and it usually defines the end of the process of transcription.
(ix) There are additional regulatory sequences that may be present further upstream or downstream to the promoter.
Transcription Unit and the Gene
(i)A gene can be defined as the functional unit of inheritance.
(ii)A cistron is a segment of DNA coding for a polypeptide.
(iii) The structural gene in a transcription unit could be said as monodstronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes).
(iv)The coding sequences or expressed sequences are defined as exons. Exons appear in mature or processed RNA. The exons are interrupted by introns.
(v)Introns or intervening sequences do not appear in mature or processed RNA.
(vi)Sometimes, the regulatory sequences are loosely defined as regulatory genes ,even
though these sequences do not code for any RNA or protein.

22.Transcription in prokaryotes occur in the following steps:
(i)A single DNA-dependent RNA polymerase catalyses the transcription of all types of RNA in bacteria.
(ii)RNA polymerase binds to promotor and initiates transcription (initiation).
(iii) It uses nucleoside triphosphates as substrate and polymerises in a template depended fashion following the rule of complementarity.
(iv) It also facilitates opening of the helix and continues elongation.
(v) Once the polymerase reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription.
(vi) RNA polymerase is only capable of catalysing the process of elongation.
It associates transiently with initiation-factor (a) and terminator factor (p), to initiate and terminate the transcription, respectively. Thus, catalysing all the three steps.
(vii) Since, the mRNA does not require any processing to become active and also since transcription and translation take place in the same compartment, many times the translation can begin much before the mRNA is fully transcribed. As a result,transcription and translation can be coupled in bacteria.

23. Transcription in eukaryotes have additional complexities than prokaryotes.
(i) There are at least three RNA polymerases in the nucleus other than the RNA polymerase in organelles. The RNA polymerase I transcribes rRNAs (28S, 18S and5.8S). RNA polymerase III is responsible for transcription of fRNA, 5srRNA and SnRNAs (small nuclear RNAs). RNA polymerase II transcribes precursor of mRNA, the heterogenous nuclear RNA (hnRNA).
(ii) Another complexity is that, the primary transcripts contain both the exons and the ‘ introns and are non-functional. Hence, subject to a process called splicing. In this process, introns are removed and exons are joined in a definite order.
(iii) hnRNA undergoes additional processing called as capping and tailing. In capping, an unusual nucleotide is added to the 5′-end of hnRNA. In tailing, adenylate residues (200-300) are added at 3’-end in a template. It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation process

Significance of these complexities are:
(i)The split gene arrangements represent an ancient feature of genome.
(ii)The presence of introns is reminescent of antiquity.
(iii) The process of splicing represents the dominance of RNA world.

Previous Year Examinations Questions

1 Mark Questions

1.What will happen if DNA replication is hot followed by cell division in a eukaryotic cell? [All India 2014 c]
Ans.If cell division is not followed after DNA replication then replicated chromosomes (DNA) would not be distributed to daughter nuclei. A repeated replication of DNA without any cell division results in the accumulation of DNA inside the cell.
This would increase the volume of the cell nucleus, thereby causing cell expansion,

2.Name the specific components and the linkage between them that form deoxyadenosine. [Delhi 2013c]
Ans.Adenine (N-glycosidic Linkage) + Deoxyribose -> Deoxyadenosine

3.Which one out of rho factor and sigma factor act as an initiation factor during transcription in a prokaryote? [Delhi 2013 C]
Ans.Sigma factor acts as an initiation factor during transcription in prokaryotes

4.Name the enzyme involved in continuous replication of DNA strand. Mention the polarity of the template strand. [All India 2012]
Ans.Enzyme involved in continuous replication of DNA strand is DNA polymerase. Template strand has 3′->5′ polarity

5.Name the positively charged protein around which the negatively charged DNA wrapped.[All India 2012 C]
Ans.Histones are the positively charged proteins around which the negatively charged DNA wrapped.

6.A structural gene has two DNA strands X and Y shown below. Identify the template strand.
[HOTS All India 2010 C]

Ans.’X’ is template strand. It is because the template strand has the polarity in 3′ -> 5′ direction.

7.Why is hnRNA required to undergo splicing? [HOTS Delhi 2009]
Ans.hnRNA is required to undergo splicing because of the introns (the non-coding sequences). These are needed to be removed and the exons (the coding sequences) have to be joined in a specific sequence.

8.Mention the two additional processes, which hnRNA needs to undergo after splicing so, as to become functional. [Delhi 2009]
Ans.The additional processes bnRNA needs to undergo after splicing are capping and tailing.

9.When and at what end does the tailing of hnRNA takes place?[All India 2009]
Ans.When hnRNA is processed to make mRNA, tailing takes place at the 3′-end.

10.At which ends do capping and tailing of hnRNA occur respectively? [Foreign 2009]
Ans.Capping – At 5′-end Tailing -At 3′-end

11.How is the length of DNA usually Calculated? [All India 2009 C]
Ans.Length of DNA can be calculated by simply multiplying the total number of base pair with distance between two consecutive bp,
i.e.6.6x 10 9 bpx 0.34x 10 _9 m/bp.
It comes about 2.2 m. (0.34 x 10 _9 m is the distance between two consecutive base pairs)

12.Name the parts A and B of the transcription unvit given below.

Ans.A -Promoter, 8-Coding strand.

13.Name the components A and B in the nucleotide with a purine, given below.

Ans.A – Phosphate B-Any nitrogenous base (e.g. adenine, guanine, cytosine or thymine).

14.Name the type of synthesis A and B occurring in the replication fork of DNA as shown below. [Delhi 2008]

Ans.A – Continuous synthesis (Leading strand) B – Discontinuous synthesis (Lagging strand)

15.Mention the polarity of the DNA strands A-B and C – D shown in the replicating fork given below.
[Ail India 2008]

Ans. A – B = 3′ -> 5′ C – D = 5′ -> 3′

16.Mention the carbon positions to which nitrogenous base and the phosphate molecule are respectively linked in the nucleotide given below. [All India 2008]

Ans.Nitrogenous base at first C Phosphate molecule at 5thC

17.What are A and B in the transcription unit represented below?[Foreign 2008]

Ans.A – Promoter B – Terminator

2 Marks Questions

18.Explain the two factors responsible for conferring stability to double helix structure of DNA. [All India 2014]
Ans.Two factors responsible for conferring stability to double helix structure of DNA are
(i) Stacking of one base pair over other.
(ii) H-bond between nitrogenous base

19.State the difference between the structural genes in a transcription unit of prokaryotes and eukaryotes. [All India 2014]
Ans.Prokaryotic structural genes are found continuously with any non-coding region, while eukaryotic structural genes are divided into exons (coding DNA) and introns (non-coding DNA).

20.Show DNA replication with the help of a diagram only.[All India 2014 C Delhi 2012]
Ans.The replication fork of DNA formed during DNA replication.

21.A template strand is given below. Write down the corresponding coding strand and the m-RNA strand that can be formed, along with their polarity.
3′ ATGCATGCATGCATGCATGCATGC 5′ [Foreign 2014]
Ans.For the given template strand
3’ATGCATGCATGCATGCATGC A T G C 5′
Coding strand is
5’TACGTACGTACGTACGTACG T A C G 3′
And mRNA strand is
5’UACGUACGUACGUACGUA C G UA C G 3′

22.Draw a labelled diagram of a nucleosome. Where is it found in a Cell? [Foreign 2014, All India 2012]
or
How do histones acquire positive charge? [Delhi 2011]
Ans.Structure of a nucleosome

A nucleosome is found in the nucleus of the cell. It contains histone proteins acquiring positive charge depending upon the abundance of amino acid residues, i.e. lysine and arginines, with charged side chains. Both these amino acids carry positive charges in their side chains.

23.Draw a schematic diagram of a part of double stranded dinucleotide DNA chain having all the four nitrogenous bases showing the correct polarity. [Delhi 2012]
Ans.Schematic diagram of a double stranded dinucleotide DNA chain having all the four nitrogenous bases (A,T,G,C) with polarity

24.State the dual role of deoxyribonucleoside triphosphates during DNA replication. [Delhi 2011]

Ans. (i) The deoxyribonucleoside triphosphates are the building blocks for the DNA strand (polynucleotide chain) i.e. they act as substrates.

(ii) These also serve as energy source in the form of ATP and GTP.
25.Answer the questions based on the dinucleotide shown below

(i)Name the type of sugar guanine base is attached to.

(ii)Name the linkage connecting the two nucleotides.

(iii)Identify the 3′ end of the dinucleotide. Give a reason for your answer. [All India 2010c]

Ans.(i)Pentose sugar or deoxyribose sugar.
(ii) Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a dinucleotide.

(iii) The polymer ribose has a free 3′ — OH group which is referred to as 3′ – end of the polynucleotide chain.

26.Make a labelled diagram of an RNA dinucleotide showing its 3′ ->5′ polarity. [All India 2010 c]
Ans.RNA dinucleotide.

27.Study the given portion of double stranded polynucleotide chain carefully. Identify A, B, C and the 5′ end of the chain. [All India 2009]

Ans.A – hydrogen bonds, B – purine base,
C – pentose (deoxyribose) sugar, D – 5′ end.

28.Differentiate between a template strand and a coding strand of DNA.[Foreign 2009]
Ans.Differences between template strand and coding strand of DNA are:

29.Give one function each of histone protein and non-histone chromosomal protein in an eukaryotic nucleus,[All India 2009 C]
Ans.(i)Histone proteins help in packaging of DNA. These are organised to form a unit of eight molecules called as histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to from a structure called nucleosome.
(ii) Non-histone chromosomal proteins helps in packaging of chromatin at higher levels.

30.

Look at the above sequence and mention the event A,B and C
(ii)What does central dogma state in molecular biology? How does it differ in some viruses?[Delhi 2009 c]
Ans.(i) A – Replication of DNA B – Transcription C – Translation
(ii) Central dogma states that the genetic information flows from DNA to RNA to Proteins. In some viruses the flow of information is reverse in direction, i.e. RNA to DNA.

31.Compare the roles of the enzymes DNA polymerase and DNA ligase in the replication fork of DNA.[All India 2008 C]
Ans.Differences between the roles of DNA polymerase and DNA ligase are:

3 Marks Questions

32.With the help of a schematic diagram , explain the location and role of the following in a transcription unit.Promoter structural gene, terminator[All India 2014 c]
Ans. Structure of a transcription unit

The promoter and terminator flank the structural gene in a transcription unit. The promoter is located towards 5′-end (upstream) of the structural gene. The terminator is located toward 3′-end (downstream) of the coding strand and it usually defines the end of the process of transcription

33.(i) What are the transcriptional products of RNA polymerase III?
(ii)Differentiate between’capping’ and ‘tailing’.
(iii)Expand ZmRNA. [All India 2014C]
Ans.(i)RNA polymerase III is responsible for transcription of tRNA, 55 rRNA and snRNAs (small nuclear RNAs).(ii) In capping process, an unusual nucleotide (methyl guanosine triphosphate) is added to 5′-end ofbnRNA.In tailing process, 200-300 adenylate residues are added at 3′-end in a template independent manner. It is now called as mRNA
(iii)hn RNA is heterogenous nuclear RNA.

34.It is established that RNA is the first genetic material. Explain giving three reasons
[Delhi 2012,2008 C]
Ans.RNA is the first genetic material in cells because
(i) RNA is capable of both storing genetic information and catalysing chemical reactions.
(ii) Essential life processes (such as metabolism, translation, splicing, etc.), were evolved around RNA.
(iii) It shows the power of self-replication.

35.(i) Construct a complete transcription unit with promoter and terminator on the basis of hypothetical template strand given below,

(ii)Write the RNA strand transcribed from the above transcription unit along with its polarity. [Delhi 2012]
Ans.

36.List the salient features of double helix structure of DNA.[All India 2012]
Ans.Salient features of DNA double helix
(i) It is made up of two polynucleotide chains containing the backbone of sugar- phosphate and the bases project inside.
(ii) The two chains have anti-parallel polarity one of them is 5′-»3′, the other has 3′-* 5′ polarity.
(iii) The bases in two strands are paired through hydrogen bond (H—bonds) forming base pairs (bp). Adenine pairs through two hydrogen bonds with thymine from opposite strand and vice-versa. In the same way, guanine is bonded with cytosine through three H—bonds. Due to which, purine always comes opposite to a pyrimidine.
(iv)The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm and there are roughly 10 bp in each turn. Consequently, the distance between base pair in a helix is about 0.34 nm.
(v)The plane of one base pair stacks over the other in double helix. This confers stability to the helical structure

37.How is hit RNA processed to form mRNA? [Foreign 2012,2008]
Ans.The precursor of mRNA transcribed by RNA polymerase II is called heterogenous nuclear RNA (hnRNA). It undergoes following changes:
(i) Splicing In this process, the non-coding introns are removed and coding sequences called exons are joined in a definite order. This is required because primary transcript contain introns and exons. (ii) Capping RNA polymerase III is responsible for transcription of tRNA, 55 rRNA and snRNAs (small nuclear RNAs).
(a) In capping process, an unusual nucleotide (methyl guanosine triphosphate) is added to 5′-end ofbnRNA.In tailing process, 200-300 adenylate residues are added at 3′-end in a template independent manner. It is now called as mRNA
(b)hn RNA is heterogenous nuclear RNA.
(iii) Tailing RNA polymerase III is responsible for transcription of tRNA, 55 rRNA and snRNAs (small nuclear RNAs).
(a) In capping process, an unusual nucleotide (methyl guanosine triphosphate) is added to 5′-end ofbnRNA.In tailing process, 200-300 adenylate residues are added at 3′-end in a template independent manner. It is now called as mRNA
(b)hn RNA is heterogenous nuclear RNA.
(iv) The fully processed mRNA is released from the nucleus into cytoplasm for translation.

38.Why is DNA considered a better hereditary material than RNA?[Foreign 2012]
Ans.DNA is considered as a better genetic material because it is stable and does not change with age or change in physiology due to its double-stranded nature and presence of thymine.RNA is not considered as a better genetic material because
(i)2—-OH group of RNA nucleotide is a reactive group that make RNA labile and easily degradable.
(ii) RNA (23S r-RNA) is catalytic, i.e. it is reactive.

39.The base sequence in one of the strands of DNA is TAGCATGAT.
(i) Give the base sequence of the complementary strand.
(ii)How are these base pairs held together in a DNA molecule?
(iii)Explain the base complementarity rule. Name the scientist who framed this rule[HotsDelhi 2011]
Ans.(i)ATCGTACTA
(ii) Base pairs are held together by weak hydrogen bonds, adenine pairs with thymine by two H—bonds and guanine pairs with cytosine forming three H—bonds.
(iii) Base complementarity rule For a double-stranded DNA, the ratios between adenine and thymine and guanine and cytosine are constant and equal to one. Erwin Chargaff framed this rule.

40.Why do you see two different types of replicating strands in the givenDNA replication fork? Explain.Name these strands, [hots Delhi 2011]

Ans.Two different types of parent strands function as template strands.
On the template strand with 3′ ->5′ polarity, the new strand is synthesised as a continuous strand. The enzyme DNA polymerase can carry out polymerisation of the nucleotides only in 5′ -» 3′ direction. This is called continuous synthesis and the strand is called leading strand.
On the other template strand with 5′ -> 3′ polarity, the new strand is synthesised from the point of replication fork, also in 5′ -> 3′ direction. But, in short stretches, they are later joined by DNA ligases to form a strand, called lagging strand.

41.(i)Name the enzyme that catalyses the transcription of hnRNA.
(ii) Why does the hnRNA needs to undergo changes? List the changes hnRNA undergoes and 1 where in the cell such changes take place. [HOTS All India 2011]
Ans.(i) RNA polymerase II catalyses the transcription of hnRNA.
(ii) hnRNA undergoes changes because it contains introns and exons and is non-functional. Changes in hnRNA are:
The precursor of mRNA transcribed by RNA polymerase II is called heterogenous nuclear RNA (hnRNA). It undergoes following changes:
(i) Splicing In this process, the non-coding introns are removed and coding sequences called exons are joined in a definite order. This is required because primary transcript contain introns and exons. (ii) Capping RNA polymerase III is responsible for transcription of tRNA, 55 rRNA and snRNAs (small nuclear RNAs).
(a) In capping process, an unusual nucleotide (methyl guanosine triphosphate) is added to 5′-end ofbnRNA.In tailing process, 200-300 adenylate residues are added at 3′-end in a template independent manner. It is now called as mRNA
(b)hn RNA is heterogenous nuclear RNA.
(iii) Tailing RNA polymerase III is responsible for transcription of tRNA, 55 rRNA and snRNAs (small nuclear RNAs).
(a) In capping process, an unusual nucleotide (methyl guanosine triphosphate) is added to 5′-end ofbnRNA.In tailing process, 200-300 adenylate residues are added at 3′-end in a template independent manner. It is now called as mRNA
(b)hn RNA is heterogenous nuclear RNA.
(iv) The fully processed mRNA is released from the nucleus into cytoplasm for translation.

42.Answer the following questions based on Meselson and Stahl’s experiment.
(i) Write the name of the chemical substance used as a source of nitrogen in the experiment by them.
(ii) Why did the scientists synthesise the light and the heavy DNA molecules in the organism used in the experiment?
(iii) How did the scientists make it possible to distinguish the heavy DNA molecule from the light DNA molecule? Explain.
(iv) Write the conclusion the scientists arrived at, after completing the experiment. [All India 2011]
Ans.(i)NH4CI (Ammonium chloride).
(ii) It is to show that after one generation E.coli with 15 N-DNA in a medium of 14 N, has DNA of intermediate density between the light and heavy DNAs. It shows that of the two strands, only one strand is synthesised newly, using the 14 N-nitrogen source in the medium.
(iii)The heavy and light DMA molecules can be differentiated by centrifugation in a cesium chloride (CsCI) density gradient. The 15 N-DNA was heavier than 14 N -DNA and the hybrid 15 N – 14 N -DNA was intermediate between the two.
(iv)Scientists concluded that the DNA replication is semiconservative, i.e. of the two strands of DNA, one is the parental strand while the other is newly synthesised.

43. Describe the initiation process of transcription in bacteria. [Delhi 2010]
Ans.Initiation process of transcription in bacteria RNA polymerase becomes associated transiently to an initiation factor (o) and binds to specific sequence on DNA called promoter to initiate transcription (initiation).

44. Describe the elongation process of transcription in bacteria. [Delhi 2010]
Ans.Elongation process of transcription in bacteria RNA polymerase facilitates opening of the DNA helix after binding to promoter it uses nucleoside triphosphates as substrate and polymerises the nucleotides in a template dependent fashion following complementarity.
The process continues till RNA polymerase reaches the terminator region on the DNA strand.

45.Describe the termination process of transcription in bacteria. [Delhi 2010]
Ans.Termination occurs when RNA polymerase reaches the terminator region and the nascent RN A falls off. The RNA polymerase becomes transiently associated with termination factor (p) and falls off the transcription unit

46.In a series of experiments with Streptococcus and mice, F Griffith concluded that R-strain bacteria had been transformed. Explain.[All India 2010]
Ans.F Griffith’s Experiment
(i) The two strains of bacterium Streptococcus pneumoniae (causing pneumonia) one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type) were taken for the experiment.
(ii) The S-type cells were virulent and R-types were not virulent.
(iii) When live S-type cells were injected into the mice, they died.
(iv) When live R-type cells were injected into mice, they did not show pneumonia.
S-strain —– > Injected into mice ———- >Mice died
R-strain —– > Injected into mice ———- >Mice lived
(v)When S-strain bacteria were killed by heating and injected into the mice, they did not develop disease.
S-strain —– > (heat-killed) —– > Injected into mice—-> Mice lived
(vi)When a mixture of heat-killed S-type cells and live R-cells were injected into the mice, the mice died of pneumonia.
(vii)Griffith recovered living S-strain cells from the dead mice..
(viii) According to him, R-strain bacteria had somehow been transformed by the heat-killed S-strain bacteria. This may be due to some transforming principle. A factor may be transferred from the heat-killed S-strain, which enabled the R-strain to synthesise a smooth capsule and become virulent.
(ix) This transforming principle must be the genetic material.

47.Draw a schematic representation of a dinucleotide. Label the following.
(i)The component of a nucleotide
(ii)5′ end
(iii)N-glycosidic linkage
(iv)Phosphodiester linkage[Foreign 2010]
Ans. Schematic representation of a dinucleotide.

48.(i) Draw a schematic representation of transcription unit showing the polarity of both the strands. Label the promoter gene and the template strand.
(ii)Mention the condition when template strand becomes coding strand.
(iii)Give the function of the promoter gene. [All India 2009 C]
Ans.(i) Structure of a transcription unit

The promoter and terminator flank the structural gene in a transcription unit. The promoter is located towards 5′-end (upstream) of the structural gene. The terminator is located toward 3′-end (downstream) of the coding strand and it usually defines the end of the process of transcription
(ii) The two strands in DNA have opposite polarity and the DNA-dependent RNA polymerase catalyses the polymerisationin only one direction, i.e. 5′-> 3′. The strand that has the polarity 3′-> 5′ acts as a template, called as template strand. The other strand which has the polarity (5′-» 3′) and the sequence same as RNA I (except thymine in place of uracil), is displaced during transcription. This strand which does not code for anything) is . called coding strand.
(iii) The promoter gene defines the template and coding strands. By switching its position with terminator, the definition of coding and template strands can be reversed

49.(i) Why does DNA replication occur in small replication fork and not in its entire length?
(ii)Why is DNA replication continuous and discontinuous in a replication fork?
(iii)Explain the importance of origin of replication in a replication fork.[HOTS All India 2009 C]
Ans.(i)Because DNA molecule is very long, so two strands cannot be separated in its entire length, as it requires very high energy. The replication occurs within a small opening of I the helix called as replication fork.
(ii) DNA polymerase can catalyse the polymerisation of nucleotides only in 5′ —> 3′ direction. So, on the template strand with 3′ ->5′ polarity, DNA replication is continuous .On the template strand with 5′-> 3′ polarity, DNA synthesis occurs in short stretches as the opening of replication for,continues. Later, these short stretches are joined by the action of DNA ligases
(iii) Replication of DNA does not initiate randomly, and DNA polymerases on their own cannot initiate replication.So,there is a need of specific sequence can DNA, called origin of replication. DNA polymerase bind to it and continues the process.

50.The length of a DNA molecule in a typical mammalian cell is calculated to be approximately 2.2 m. How is the packaging of this long molecule done to accomodate it within the nucleus of the cell?[Delhi 2009,2008]
Ans.Eukaryotic cells have a set of positively charged basic proteins called histones. They are rich in lysine and arginine. The histones are organised to form a unit of eight molecules called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a nucleosome. The nucleosome contains 200 bp of the DNA helix and nucleosomes form the repeating units of a structure of the nucleolus, called chromatin.

5 Marks Questions

51.’DNA replication is semiconservative’. Name the scientists who proposed it and who proved it. How was it proved experimentally?[All India 2014C Delhi 2008 Foreign 2008]
or
Who proposed that DNA replication is semiconservative? How did Meselson and Stahl prove it.
[Delhi 2008C]
or
Describe Meselson and Stahl’s experiment and write the conclusion they arrived at. [Foreign 2014 Delhi 2012]
Ans.Watson and Crick proposed that DNA replication is semiconservative. Later in the year 1958, Meselson and Stahl proved this. The semiconservative nature of DNA suggests that, after the completion of replication, each DNA molecule will have one parental and one newly-synthesised strand.
Experimental Proof
(i) E. coli was grown in a medium containing 15 NH4CI ( 15 N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. As a result, 15 N was incorporated into the newly-synthesised DNA. This heavy DNA could be distinguished by centrifugation in CsSI density gradient.
(ii) Then, these E. coli cells were transferred to a medium with normal 14 NH4CI and the DNA was extracted as double stranded helix. The various samples were separated on CsCI gradients for measuring the density of DNA (after 20 min).
The hybrid had intermediate density.

(iii)After 40 min, the DNA of the second generation was extracted from the 14 NH4CI medium and was found to have equal amounts of hybrid and light DNA.
(iv) This proves that after replication, each DNA molecule has one parental strand and one newly synthesised strand.

52.(i) Describe the various steps of Griffith’s experiment that led to the conclusion of the ‘transforming principle’.
(ii)How did the chemical nature of the ‘transforming principle’ get established? [All India 2014]
or
(i)Write the conclusion drawn by Griffith at the end of his experiment with Streptococcus pneumoniae.
(ii)How did O Avery, C MacLeod and M McCarty prove that DNA was the genetic material? Explain.
[All India 2013,2009]
or
Describe Frederick Griffith’s experiment on Streptococcus pneumoniae. Discuss the conclusion he arrived at. [All India 2012]
or
(i)Write the scientific name of the bacterium used by Frederick Griffith in his experiment.
(ii)How did he prove that some transforming principle is responsible for transformation of the non-virulent strains of bacteria into the virulent form?
(iii)State the biochemical nature of the transforming principle.
(iv)Name the scientists who proved it [Foreign 2011,2009,2008Delhi 2009 C, 2008 C]
Ans.(i)RNA dinucleotide.

(ii)Biochemical nature of transforming principle of Griffith’s experiment.

  • Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44) worked to determine the biochemical nature of transforming principle in Griffith’s experiment.
  • They purified biochemicals (proteins, DNA, RNA, etc) from the heat-killed S-cel Is to see which ones could transform live R-cells into S-cel Is.
  • They discovered that DNA alone from S-bacteria caused R-bacteria to become transformed.
  • They also discovered that protein digesting enzymes (proteases) and RNA digesting enzyme (RNAse) did not affect transformation.
  • Digestion with DNAase did inhibit transformation. It suggested that the DNA causes the transformation.
  • They thus, finally concluded that DNA is the genetic material

53.Describe the Hershey and Chase’s experiment. Write the conclusion drawn by the scientists after their experiment. [ All India 2014]
or
Name the scientists, who proved experimentally that DNA is the genetic material. Describe their experiment.
or
(i)Describe Hershey and Chase’s experiment.
(ii)Write the aim of the experiment. [Delhi 2010 C All India 2010,2008 C]
Ans.Hershey and Chase’s experiment Their experiment is to prove unequivocally that DNA is the genetic material and not the protein .They worked with T2 bacteriophage, which attacks bacterium E. coli. They grew some viruses on a medium that contained radioactive phosphorus ( 32 P) and some others on medium that contained radioactive sulphur ( 32 5).
(i) Radioactive phages were allowed to attack E.coli bacteria. The infection proceeded, the viral coats were removed form the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
(ii)Bacteria which were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
(iii) Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. Hence, DNA is genetic material that is passed from virus to bacteria.

54.(i) Explain the process of DNA replication with the help of a schematic diagram.
(ii) In which phase of the cell cycle does replication occur in eukaryotes? What would happen if cell division is not followed after DNA replication. [Delhi 2014]
Ans.(i)The replication fork of DNA formed during DNA replication.

(ii)DNA replication occurs in S-phase of cell cycle in eukaryotes. Refer to answer.

55.Name the major types of RNAs and explain their role in the process of protein synthesis in a prokaryote. [Foreign 2014]
Ans.There are major three types of RNAs in prokaryotes which helps in protein synthesis as follows:
(i) Messenger RNA (mRNA) It is formed as a complementary strand on one of the two strands of DNA inside nucleus. Soon after its formation, mRNA comes out in cytoplasm. Formation of mRNA from DNA is called transcription. Function of mRNA is to carry the genetic information present in DNA (inside nucleus) to cytoplasm for protein synthesis.
(ii) Ribosomal RNA (rRNA) It is formed in nucleolus and it forms 80% of total RNA present inside the cell. It is also the moststable type of RNA. rRNA is associated with structural organisation of ribosomes (rRNA forms about 60% of weight of ribosomes), which are seats of protein synthesis.
(iii) Transfer RNA (tRNA) It is also called soluble RNA (sRNA) or adapter RNA or adaptive RNA. tRNA forms 10-15% of total RNA present in the cell. It acts as adapter molecule which carries amino acids to the site of protein synthesis i (i.e. ribosomes).

56.Describe the process of transcription in bacterium. [All India 2014 C]
Ans.Initiation process of transcription in bacteria RNA polymerase becomes associated transiently to an initiation factor (o) and binds to specific sequence on DNA called promoter to initiate transcription (initiation).

Elongation process of transcription in bacteria RNA polymerase facilitates opening of the DNA helix after binding to promoter it uses nucleoside triphosphates as substrate and polymerises the nucleotides in a template dependent fashion following complementarity.
The process continues till RNA polymerase reaches the terminator region on the DNA strand.

Termination occurs when RNA polymerase reaches the terminator region and the nascent RN A falls off. The RNA polymerase becomes transiently associated with termination factor (p) and falls off the transcription unit

57.(i)Explain the role of DNA dependent RNA polymerase in initiation, elongation and termination during transcription in bacterial cell.
(ii) How is transcription a more complex process in eukaryotic cells? Explain. [Foreign 2011]
Ans.(i)Role of DNA dependent RNA polymerase.
(a) RNA polymerase becomes associated transiently with initiation factor and binds to the promoter site on DNA and initiates transcription.
(b) It uses the nucleoside triphosphate as substrates and polymerises them in a template-dependent fashion following the base complementarity rule in the 5′-> 3’direction.
(c) It also facilitates the opening of the DNA helix and continues the elongation process.
(d) When the polymerase falls off a terminator region on the DNA, the nascent RNA separates. This results in termination.
(ii) Reasons that transcription is more complex in eukaryotes are:
(a)The three types of RNA polymerases in r the nucleus show division of labour

  • RNA polymerase I transcribes rRNAs (28S, 18S and 5.8S).
  • RNA polymerase II transcribes the precursor of mRNA, called hnRNA
  • RNA polymerase III transcribes tRNA, 5 srRNA and snRNAse.

(b)hnRNA contains both coding sequences called exons and non-coding sequences called introns. So, it undergoes a process called splicing, in which the non-coding sequences (introns) are removed and the coding sequences (exons) are joined together in a defined order.
(c)In capping, unusual nucleotide,methyl guanosine triphosphate residues are added at the 5-end of the hnRNA.
(d)In tailing, 200-300 adenylate residues are added at the 3-end of the hnRNA.

58.Study the flow chart given below and answer the questions that follow :

(a)Name the organism and differentiate between, its two strains R and S respectively.
(b)Write the result A and B obtained in step (iii) and (iv) respectively.
(c)Name the scientist who performed the steps (i), (ii) and (iii)
(d)Write the specific conclusion drawn from the step (iv).[Ail India 2010 C]
Ans.(a)The organism is bacterium Streptococcus pneumoniae. Differences between S-type cells and
R-type cells are:

(b) A – Mice died B – Mice lived.
(c)Frederick Griffith performed these steps.
(d)This indicates that DNA is the transforming principle. When DNase is added to the medium, the DNA of the heat killed cells get denatured and is unable to carry transformation.

59.(i)What did Meselson and Stahl observed? When
(a)They cultured coli in a medium containing 15 NH4 Cl for a few generations and centrifuged the content?
(b)They transferred one such bacterium to the normal medium of NH4 Cl and cultured for two generations.
(ii)What did Meselson and Stahl conclude from his experiment? Explain with the help of diagrams.
(iii)Which is the first genetic material? Give reasons in support of your answer.
[Delhi 2009 Foreign 2009 Delhi 2008 C]
Ans.(i) (a)Meselson and Stahl observed that the 15 N was incorporated into the newly synthesised strand of DNA and also other nitrogen containing compounds. This heavy DNA could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCI) density gradient.
(b) DNA from such bacterium had a hybrid or intermediate density, one generation after the transfer from 15 N to 14 N. After another generation, it is composed of equal amount of this hybrid DNA and of light DNA
(ii)Meselson and Stahl concluded that replication of DNA is semiconservative.
Watson and Crick proposed that DNA replication is semiconservative. Later in the year 1958, Meselson and Stahl proved this. The semiconservative nature of DNA suggests that, after the completion of replication, each DNA molecule will have one parental and one newly-synthesised strand.
Experimental Proof
(i) E. coli was grown in a medium containing 15 NH4CI ( 15 N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. As a result, 15 N was incorporated into the newly-synthesised DNA. This heavy DNA could be distinguished by centrifugation in CsSI density gradient.
(ii) Then, these E. coli cells were transferred to a medium with normal 14 NH4CI and the DNA was extracted as double stranded helix. The various samples were separated on CsCI gradients for measuring the density of DNA (after 20 min).
The hybrid had intermediate density.

(iii)After 40 min, the DNA of the second generation was extracted from the 14 NH4CI medium and was found to have equal amounts of hybrid and light DNA.
(iv) This proves that after replication, each DNA molecule has one parental strand and one newly synthesised strand.
(iii)RNA is the first genetic material in cells because
(a) RNA is capable of both storing genetic information and catalysing chemical reactions.
(b) Essential life processes (such as metabolism, translation, splicing, etc.), were evolved around RNA.
(c) It shows the power of self-replication.

60.Why is DNA molecule more stable genetic material than RNA? Explain. [All India 2008]
Ans.DNA is more stable genetic material because
(i) The 2’— OH group in the nucleotides of RNA is a reactive group and makes RNA labile and easily degradable. But, DNA is chemically less reactive and structurally more stable.
(ii) The presence of thymine in place of uracil also confers more stability to DNA.
(iii)Two strands of DNA are complementary to each other and even if separated by heat, come together, when suitable conditions are created on the other hand, RNA is usually single stranded

61.Draw the labelled schematic structure of a transcription unit. Explain the function of each component of the unit in the process of transcription. [All India 2008]
Ans. Structure of a transcription unit

The promoter and terminator flank the structural gene in a transcription unit. The promoter is located towards 5′-end (upstream) of the structural gene. The terminator is located toward 3′-end (downstream) of the coding strand and it usually defines the end of the process of transcription
Functions of components of transcription unit
(i) Promoter (DNA sequence) Provides binding site for the RNA polymerase.
(ii) Structural genes Code for enzymes/ proteins and transcribe the mRNA for the same.
(iii)Terminator (sequence of bases) Defines the end of transcription process
(iv)DNA strand with 3′ – 5′ polarity – Acts as the template for transcription of mRNA.
(v)DNA strand with 5′ – 3′ polarity – Coding strand it does not code for RNA, but all reference points regarding transcription are made with this strand

62.(i) State the central dogma in molecular biology. Who proposed it? Is it universally applicable? Explain.
(ii)List any four properties of a molecule to be able to act as a genetic material. [All India 2008 C]
Ans.(i) Francis Crick proposed the central dogma in molecular biology, which states that the genetic information flows from


It is not universally applicable. In some viruses, the flow of information is in reverse direction, that is from RNA to DNA.
(ii) Properties of a molecule to act as a genetic material

  • It should be able to generate its replica.
  • It should chemically and structurally be stable.
  • It should provide scope for slow changes (mutation) that are required for evolution.
  • It should be able to express itself in the form of Mendel ian characters

63.Diagrammatically represent a portion of the double stranded polynucleotide chain sequence in a DNA molecule involving all the four nitrogenous bases.[All India 2008 C]
Ans.Double stranded polynucleotide chain sequence in a DNA molecule involving all the four nitrogenous bases, i.e. A, T, G, C is represented below


RNA Types: 3 Main Types of RNA (With Diagram)

It delivers amino acids to ribosome and decodes the information of mRNA. Each nucleotide triplet codon on mRNA represents an amino acid. The tRNA plays the role of an adaptor and matches each codon to its particular amino acid in the cytopolasmic pool.

The tRNA has two properties:

(a) It represents a single amino acid to which it binds covalently.

(b) It has two sites. One is a trinucleotide sequence called anticodon, which is complementary to the codon of mRNA. The codon and anticodon form base pairs with each other. The other is amino acid binding site.

There are many different kinds of tRNA molecules in a cell. Each tRNA is named after the amino acid it carries. For example if tRNA carries amino acid tyrosine it is written as tRNA Tyr .Sometimes there are more than one tRNA for an amino acid, then it is denoted as tRNA1 Try and tRNA2 Try . A minimum of 32 tRNAs are required to translate all 61 codons.

The tRNA charged with an amino acid is called amino acyl tRNA.

Clover Leaf Structure of tRNA:

The primary structure of all tRNA molecules is small, linear, single stranded nucleic acid ranging in size from 73 to 93 nucleotides. The tRNA due to its property of having stretches of complementary base pairs forms secondary structure, which is in the form of a cloverleaf.

Several regions of the single stranded molecule form double stranded stems or arms and single stranded loops due to folding of various regions of the molecule. These double stranded stems have complementary base pairs. A typical tRNA has bases numbering from 1-76, using the standard numbering convention where position 1 is the 5′ end and 76 is the 3′ end.

The various regions of the clover leaf model of tRNA are as follows:

It has a seven base pairs stem formed by base pairing between 5′ and 3′ ends of tRNA. At 3′ end a sequence of 5′-CCA-3′ is added. This is called CCA arm or amino acid acceptor arm. Amino acid binds to this arm during protein synthesis.

Going from 5′ to 3′ direction or anticlockwise direction, next arm is D-arm. It has a 3 to 4 base pair stem and a loop called D-loop or DHU-loop. It contains a modified base dihydrouracil.

Next is the arm which lies opposite to the acceptor arm. It has a five base pair stem and a loop in which there are three adjacent nucleotides called anticodon which are complementary to the codon of mRNA.

Next lies an extra arm which consists of 3-21 bases. Depending upon the length, extra arms are of two types, small extra arm with 3-5 bases and other a large arm having 13-21 bases.

It has a modified base pseudouridine ψ. It has a five base pair stem with a loop.

There are about 50 different types of modified bases in different tRNAs, but four bases are more common. One is ribothymidine which contains thymine which is not found in RNA. Other modified bases are pseudouridine ψ, dihyrouridine and inosine.

Three Dimensional Structure of tRNA:

X-ray crystallographic analysis of tRNA shows three dimensional structure called tertiary structure. The molecule is folded and has two helical double stranded branches. One branch consists of acceptor arm and T ψ C arm. The other arm consists of DHU loop and anticodon arm with loop.

The tRNA molecule is L- shaped. The tertiary structure creates two double helices at right angle to each other. The amino acid binding site is opposite to the anticodon arm. This facilitates protein synthesis.

The tRNA constitutes about 10% of the total cellular RNA.

RNA Type # 2. Messenger RNA (mRNA):

Messenger RNA is a linear molecule transcribed from one strand of DNA. It carries the base sequence complementary to DNA template strand. The base sequence of mRNA is in the form of consecutive triplet codons. Ribosomes translate these triplet codons into amino acid sequence of polypeptide chain.

Length of mRNA:

Length of mRNA depends upon the length of polypeptide chain it Codes for. Polypeptide length varies from a chain of a few amino acids to thousands of amino acids. For example, a sequence of 600 nucleotides will code for a polypeptide having a chain of 200 amino acids. The message is read in the groups of three consecutive bases from a fixed starting point.

Life Span of mRNA:

In bacteria, mRNA is transcribed and translated in a single cellular compartment and the two processes are so closely linked that they occur simultaneously. Transcription begins when the enzyme RNA polymerase binds to DNA and then moves along making a copy of one strand. As soon as the transcription begins, the ribosomes attach to the 5′ end (free end) of the mRNA and start translation while the other end of mRNA is still under synthesis.

This is known as coupled transcription and translation in prokaryotes. After the translation of whole of mRNA is completed, the mRNA is then degraded in 5′ → 3′ direction. The mRAN is synthesized, translated and degraded all in rapid succession and all in 5′ → 3′ direction. An individual mRNA molecule survives only for a minute or less.

In eukaryotes, transcription occurs in the nucleus while translation takes place in cytoplasm. Eukaryotic mRNA is quite stable and survives from a few minutes to more than a day. In mammalian RBC, through the nucleus is lost, mRNA continues to produce haemoglobin for many days.

Eukaryotic mRNA constitutes only a small proportion of the total cellular RNA. It is only about 3% of the total RNA.

Coding and Non-coding Regions:

All mRNAs have two types of regions. The coding region consists of series of codons. But the mRNA is longer than the coding regions. Length of newly synthesized mRNA is much larger than the length of mRNA used for translation. The coding regions are called exons. Between the coding regions lie various non-coding regions called introns. Genes with these intervening sequences are called Split genes or Interrupted genes.

RNA Type # 3. Ribosomal RNA (rRNA):

Most of the RNA of the cell is in the form of ribosomal RNA which constitutes about 85% of the total RNA. Ribosomes consist of many types of rRNA. The 70S ribosome of prokaryotes, in its smaller subunit of 30S has 16S rRNA. The 50S larger subunit consists of 23S and 5S rRNA. Similarly 80S ribosome has 18S rRNA in its smaller subunit of 40S. The 60S larger subunit has 28S, 5.8S and 5S rRNA.

The rRNA molecules form secondary structure of double stranded stems and single stranded loops by extensive complementary base pairings. The rRNA plays major role in protein synthesis. They interact with mRNA and tRNA at each step of translation or protein synthesis.

The 3′ terminus of rRNA of 16S rRNA interacts with initiation site on mRNA which is called Shine-Dalgarno sequence and lies just before the start codon AUG.

The 23S rRNA plays an active role in peptidyl transferase activity. Movement of tRNA between A and P site on ribosome is aided by 23 S rRNA.

The rRNA molecules form complexes with specific proteins in ribosomes. The RNA- protein complexes are called ribonucleoproteins (RNP).

Some RNAs are Enzymes:

Recently it has been discovered that some RNAs play the role of enzymes, they are called ribozymes. Like typical enzymes, a ribozyme has an active site, a binding site for substrate and a binding site for a co-factor. Ribozymes are mainly involved in splicing of introns present on RNA molecules.

Small Nuclear RNAs:

RNA polymerase enzyme transcribes several small RNAs in the nucleus of eukaryotes. These are called small nuclear RNAs (snRNA). These form complex with specific proteins and are called small nuclear ribonuclear proteins (snRNP) also known as snurps. The proteins associated with snRNAs are called S proteins.

These sn RNAs are rich in uracil and are of several types i.e., U1, U2, U4, U5 and U6. Each of these RNAs is between 200-300 nucleotide long. They are involved in the splicing of group II introns. They form a complex with intron. This complex is called spliceosome which is involved in splicing of the intron.

These snRNP molecules contain small RNA sequences which are complementary to the introns of mRNA and form RNA-RNA base pairs at 5′ and 3′ splice sites where actual splice reaction occurs.

Small Nucleolar RNA (sno RNA):

This sno RNA is required for the processing of eukaryotic rRNA molecules. The snoRNAs are associated with proteins. These snoRNAs are present in the nucleolus where processing of rRNA takes place. Ribosome are also assembled in nucleolus.

Many small RNAs like micro RNAs (miRNAs), small interfering RNAs (siRNAs) play their role in silencing of genes. They act on mRNA resulting in the disruption of translation.


mRNA is produced by synthesising a ribonucleic acid (RNA) strand from nucleotide building blocks according to a deoxyribonucleic acid (DNA) template, a process that is called transcription. [2] When the building blocks provided to the RNA polymerase include non-standard nucleosides such as pseudouridine — instead of the standard adenosine, cytidine, guanosine, and uridine nucleosides — the resulting mRNA is described as nucleoside-modified. [3]

Production of protein begins with assembly of ribosomes on the mRNA, the latter then serving as a blueprint for the synthesis of proteins by specifying their amino acid sequence based on the genetic code in the process of protein biosynthesis called translation. [4]

To induce cells to make proteins that they do not normally produce, it is possible to introduce heterologous mRNA into the cytoplasm of the cell, bypassing the need for transcription. In other words, a blueprint for foreign proteins is "smuggled" into the cells. To achieve this goal, however, one must bypass cellular systems that prevent the penetration and translation of foreign mRNA. There are nearly-ubiquitous enzymes called ribonucleases (also called RNAses) that break down unprotected mRNA. [5] There are also intracellular barriers against foreign mRNA, such as innate immune system receptors toll-like receptor (TLR) 7 and TLR8, located in endosomal membranes. RNA sensors like TLR7 and TLR8 can dramatically reduce protein synthesis in the cell, trigger release of cytokines such as interferon and TNF-alpha, and when sufficiently intense lead to programmed cell death. [6]

The inflammatory nature of exogenous RNA can be masked by modifying the nucleosides in mRNA. [7] For example, uridine can be replaced with a similar nucleoside such as pseudouridine (Ψ) or N1-methyl-pseudouridine (m1Ψ), [8] and cytosine can be replaced by 5-methylcytosine. [9] Some of these, such as pseudouridine and 5-methylcytosine, occur naturally in eukaryotes. [10] Inclusion of these modified nucleosides alters the secondary structure of the mRNA, which can reduce recognition by the innate immune system while still allowing effective translation. [9]

A normal mRNA starts and ends with sections that do not code for amino acids of the actual protein. These sequences at the 5′ and 3′ ends of an mRNA strand are called untranslated regions (UTRs). The two UTRs at their strand ends are essential for the stability of an mRNA and also of a modRNA as well as for the efficiency of translation, i.e. for the amount of protein produced. By selecting suitable UTRs during the synthesis of a modRNA, the production of the target protein in the target cells can be optimised. [5] [11]

Various difficulties are involved in the introduction of modRNA into certain target cells. First, the modRNA must be protected from ribonucleases. [5] This can be accomplished, for example, by wrapping it in liposomes. Such "packaging" can also help to ensure that the modRNA is absorbed into the target cells. This is useful, for example, when used in vaccines, as nanoparticles are taken up by dendritic cells and macrophages, both of which play an important role in activating the immune system. [12]

Furthermore, it may be desirable that the modRNA applied is introduced into specific body cells. This is the case, for example, if heart muscle cells are to be stimulated to multiply. In this case, the packaged modRNA can be injected directly into an artery such as a coronary artery. [13]

An important field of application are mRNA vaccines, of which the first authorized for use in humans were COVID-19 vaccines to address SARS-CoV-2. [14] [15] [16] [17] [18] [19] [20] Examples of COVID-19 vaccines using modRNA include those developed by the cooperation of BioNTech/Pfizer/Fosun International (BNT162b2), and by Moderna (mRNA-1273). [21] [22] [23] The zorecimeran vaccine developed by Curevac, however, uses unmodified RNA. [24]

Other possible uses of modRNA include the regeneration of damaged heart muscle tissue [25] [26] and cancer therapy. [27] [28]


NMD-degradome sequencing reveals ribosome-bound intermediates with 3'-end non-templated nucleotides

Nonsense-mediated messenger RNA decay (NMD) controls mRNA quality and degrades physiologic mRNAs to fine-tune gene expression in changing developmental or environmental milieus. NMD requires that its targets are removed from the translating pool of mRNAs. Since the decay steps of mammalian NMD remain unknown, we developed assays to isolate and sequence direct NMD decay intermediates transcriptome-wide based on their co-immunoprecipitation with phosphorylated UPF1, which is the active form of this essential NMD factor. We show that, unlike steady-state UPF1, phosphorylated UPF1 binds predominantly deadenylated mRNA decay intermediates and activates NMD cooperatively from 5'- and 3'-ends. We leverage method modifications to characterize the 3'-ends of NMD decay intermediates, show that they are ribosome-bound, and reveal that some are subject to the addition of non-templated nucleotide. Uridines are added by TUT4 and TUT7 terminal uridylyl transferases and removed by the Perlman syndrome-associated exonuclease DIS3L2. The addition of other non-templated nucleotides appears to inhibit decay.


Biology 171

By the end of this section, you will be able to do the following:

  • Describe the different steps in RNA processing
  • Understand the significance of exons, introns, and splicing for mRNAs
  • Explain how tRNAs and rRNAs are processed

After transcription, eukaryotic pre-mRNAs must undergo several processing steps before they can be translated. Eukaryotic (and prokaryotic) tRNAs and rRNAs also undergo processing before they can function as components in the protein-synthesis machinery.

MRNA Processing

The eukaryotic pre-mRNA undergoes extensive processing before it is ready to be translated. Eukaryotic protein-coding sequences are not continuous, as they are in prokaryotes. The coding sequences (exons) are interrupted by noncoding introns, which must be removed to make a translatable mRNA. The additional steps involved in eukaryotic mRNA maturation also create a molecule with a much longer half-life than a prokaryotic mRNA. Eukaryotic mRNAs last for several hours, whereas the typical E. coli mRNA lasts no more than five seconds.

Pre-mRNAs are first coated in RNA-stabilizing proteins these protect the pre-mRNA from degradation while it is processed and exported out of the nucleus. The three most important steps of pre-mRNA processing are the addition of stabilizing and signaling factors at the 5′ and 3′ ends of the molecule, and the removal of the introns ((Figure)). In rare cases, the mRNA transcript can be “edited” after it is transcribed.


The trypanosomes are a group of protozoa that include the pathogen Trypanosoma brucei, which causes nagana in cattle and sleeping sickness in humans throughout great areas of Africa ((Figure)). The trypanosome is carried by biting flies in the genus Glossina (commonly called tsetse flies). Trypanosomes, and virtually all other eukaryotes, have organelles called mitochondria that supply the cell with chemical energy. Mitochondria are organelles that express their own DNA and are believed to be the remnants of a symbiotic relationship between a eukaryote and an engulfed prokaryote. The mitochondrial DNA of trypanosomes exhibit an interesting exception to the central dogma: their pre-mRNAs do not have the correct information to specify a functional protein. Usually, this is because the mRNA is missing several U nucleotides. The cell performs an additional RNA processing step called RNA editing to remedy this.


Other genes in the mitochondrial genome encode 40- to 80-nucleotide guide RNAs. One or more of these molecules interacts by complementary base pairing with some of the nucleotides in the pre-mRNA transcript. However, the guide RNA has more A nucleotides than the pre-mRNA has U nucleotides with which to bind. In these regions, the guide RNA loops out. The 3′ ends of guide RNAs have a long poly-U tail, and these U bases are inserted in regions of the pre-mRNA transcript at which the guide RNAs are looped. This process is entirely mediated by RNA molecules. That is, guide RNAs—rather than proteins—serve as the catalysts in RNA editing.

RNA editing is not just a phenomenon of trypanosomes. In the mitochondria of some plants, almost all pre-mRNAs are edited. RNA editing has also been identified in mammals such as rats, rabbits, and even humans. What could be the evolutionary reason for this additional step in pre-mRNA processing? One possibility is that the mitochondria, being remnants of ancient prokaryotes, have an equally ancient RNA-based method for regulating gene expression. In support of this hypothesis, edits made to pre-mRNAs differ depending on cellular conditions. Although speculative, the process of RNA editing may be a holdover from a primordial time when RNA molecules, instead of proteins, were responsible for catalyzing reactions.

5′ Capping

While the pre-mRNA is still being synthesized, a 7-methylguanosine cap is added to the 5′ end of the growing transcript by a phosphate linkage. This functional group protects the nascent mRNA from degradation. In addition, factors involved in protein synthesis recognize the cap to help initiate translation by ribosomes.

3′ Poly-A Tail

Once elongation is complete, the pre-mRNA is cleaved by an endonuclease between an AAUAAA consensus sequence and a GU-rich sequence, leaving the AAUAAA sequence on the pre-mRNA. An enzyme called poly-A polymerase then adds a string of approximately 200 A residues, called the poly-A tail . This modification further protects the pre-mRNA from degradation and is also the binding site for a protein necessary for exporting the processed mRNA to the cytoplasm.

Pre-mRNA Splicing

Eukaryotic genes are composed of exons , which correspond to protein-coding sequences (ex-on signifies that they are expressed), and intervening sequences called introns (int-ron denotes their intervening role), which may be involved in gene regulation but are removed from the pre-mRNA during processing. Intron sequences in mRNA do not encode functional proteins.

The discovery of introns came as a surprise to researchers in the 1970s who expected that pre-mRNAs would specify protein sequences without further processing, as they had observed in prokaryotes. The genes of higher eukaryotes very often contain one or more introns. These regions may correspond to regulatory sequences however, the biological significance of having many introns or having very long introns in a gene is unclear. It is possible that introns slow down gene expression because it takes longer to transcribe pre-mRNAs with lots of introns. Alternatively, introns may be nonfunctional sequence remnants left over from the fusion of ancient genes throughout the course of evolution. This is supported by the fact that separate exons often encode separate protein subunits or domains. For the most part, the sequences of introns can be mutated without ultimately affecting the protein product.

All of a pre-mRNA’s introns must be completely and precisely removed before protein synthesis. If the process errs by even a single nucleotide, the reading frame of the rejoined exons would shift, and the resulting protein would be dysfunctional. The process of removing introns and reconnecting exons is called splicing ((Figure)). Introns are removed and degraded while the pre-mRNA is still in the nucleus. Splicing occurs by a sequence-specific mechanism that ensures introns will be removed and exons rejoined with the accuracy and precision of a single nucleotide. Although the intron itself is noncoding, the beginning and end of each intron is marked with specific nucleotides: GU at the 5′ end and AG at the 3′ end of the intron. The splicing of pre-mRNAs is conducted by complexes of proteins and RNA molecules called spliceosomes.


Errors in splicing are implicated in cancers and other human diseases. What kinds of mutations might lead to splicing errors? Think of different possible outcomes if splicing errors occur.

Note that more than 70 individual introns can be present, and each has to undergo the process of splicing—in addition to 5′ capping and the addition of a poly-A tail—just to generate a single, translatable mRNA molecule.

View RNA Splicing (video) to see how introns are removed during RNA splicing.

Processing of tRNAs and rRNAs

The tRNAs and rRNAs are structural molecules that have roles in protein synthesis however, these RNAs are not themselves translated. Pre-rRNAs are transcribed, processed, and assembled into ribosomes in the nucleolus. Pre-tRNAs are transcribed and processed in the nucleus and then released into the cytoplasm where they are linked to free amino acids for protein synthesis.

Most of the tRNAs and rRNAs in eukaryotes and prokaryotes are first transcribed as a long precursor molecule that spans multiple rRNAs or tRNAs. Enzymes then cleave the precursors into subunits corresponding to each structural RNA. Some of the bases of pre-rRNAs are methylated that is, a –CH3 methyl functional group is added for stability. Pre-tRNA molecules also undergo methylation. As with pre-mRNAs, subunit excision occurs in eukaryotic pre-RNAs destined to become tRNAs or rRNAs.

Mature rRNAs make up approximately 50 percent of each ribosome. Some of a ribosome’s RNA molecules are purely structural, whereas others have catalytic or binding activities. Mature tRNAs take on a three-dimensional structure through local regions of base pairing stabilized by intramolecular hydrogen bonding. The tRNA folds to position the amino acid binding site at one end and the anticodon at the other end ((Figure)). The anticodon is a three-nucleotide sequence in a tRNA that interacts with an mRNA codon through complementary base pairing.


Section Summary

Eukaryotic pre-mRNAs are modified with a 5′ methylguanosine cap and a poly-A tail. These structures protect the mature mRNA from degradation and help export it from the nucleus. Pre-mRNAs also undergo splicing, in which introns are removed and exons are reconnected with single-nucleotide accuracy. Only finished mRNAs that have undergone 5′ capping, 3′ polyadenylation, and intron splicing are exported from the nucleus to the cytoplasm. Pre-rRNAs and pre-tRNAs may be processed by intramolecular cleavage, splicing, methylation, and chemical conversion of nucleotides. Rarely, RNA editing is also performed to insert missing bases after an mRNA has been synthesized.

Art Connections

(Figure) Errors in splicing are implicated in cancers and other human diseases. What kinds of mutations might lead to splicing errors? Think of different possible outcomes if splicing errors occur.

(Figure) Mutations in the spliceosome recognition sequence at each end of the intron, or in the proteins and RNAs that make up the spliceosome, may impair splicing. Mutations may also add new spliceosome recognition sites. Splicing errors could lead to introns being retained in spliced RNA, exons being excised, or changes in the location of the splice site.

Free Response

Chronic lymphocytic leukemia patients often harbor nonsense mutations in their spliceosome machinery. Describe how this mutation of the spliceosome would change the final location and sequence of a pre-mRNA.

Nonsense spliceosome mutations would eliminate the splicing step of mRNA processing, so the mature mRNAs would retain their introns and be perfectly complementary to the entire DNA template sequence. However, the mRNAs would still undergo addition of the 5’ cap and poly-A tail, and therefore each has the potential to be exported to the cytoplasm for translation.

Glossary


12.2: Modifications at the 5' and 3' ends of mRNA - Biology

Over 100 types of chemical modifications have been identified in cellular RNAs. While the 5′ cap modification and the poly(A) tail of eukaryotic mRNA play key roles in regulation, internal modifications are gaining attention for their roles in mRNA metabolism. The most abundant internal mRNA modification is N 6 -methyladenosine (m 6 A), and identification of proteins that install, recognize, and remove this and other marks have revealed roles for mRNA modification in nearly every aspect of the mRNA life cycle, as well as in various cellular, developmental, and disease processes. Abundant noncoding RNAs such as tRNAs, rRNAs, and spliceosomal RNAs are also heavily modified and depend on the modifications for their biogenesis and function. Our understanding of the biological contributions of these different chemical modifications is beginning to take shape, but it’s clear that in both coding and noncoding RNAs, dynamic modifications represent a new layer of control of genetic information.


The complex enzymology of mRNA decapping: Enzymes of four classes cleave pyrophosphate bonds

The 5' ends of most RNAs are chemically modified to enable protection from nucleases. In bacteria, this is often achieved by keeping the triphosphate terminus originating from transcriptional initiation, while most eukaryotic mRNAs and small nuclear RNAs have a 5'→5' linked N 7 -methyl guanosine (m 7 G) cap added. Several other chemical modifications have been described at RNA 5' ends. Common to all modifications is the presence of at least one pyrophosphate bond. To enable RNA turnover, these chemical modifications at the RNA 5' end need to be reversible. Dependent on the direction of the RNA decay pathway (5'→3' or 3'→5'), some enzymes cleave the 5'→5' cap linkage of intact RNAs to initiate decay, while others act as scavengers and hydrolyse the cap element of the remnants of the 3'→5' decay pathway. In eukaryotes, there is also a cap quality control pathway. Most enzymes involved in the cleavage of the RNA 5' ends are pyrophosphohydrolases, with only a few having (additional) 5' triphosphonucleotide hydrolase activities. Despite the identity of their enzyme activities, the enzymes belong to four different enzyme classes. Nudix hydrolases decap intact RNAs as part of the 5'→3' decay pathway, DXO family members mainly degrade faulty RNAs, members of the histidine triad (HIT) family are scavenger proteins, while an ApaH-like phosphatase is the major mRNA decay enzyme of trypanosomes, whose RNAs have a unique cap structure. Many novel cap structures and decapping enzymes have only recently been discovered, indicating that we are only beginning to understand the mechanisms of RNA decapping. This article is categorized under: RNA Turnover and Surveillance > Turnover/Surveillance Mechanisms RNA Turnover and Surveillance > Regulation of RNA Stability RNA Processing > Capping and 5' End Modifications.

Keywords: ApaH HIT proteins decapping nudix trypanosomes.


Watch the video: Post Transcriptional Modifications of mRNA. 5 Cap. 3 Tail. Splicing of Introns (October 2022).