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Is it possible to distinguish between coding and template strands from the sequence?

Is it possible to distinguish between coding and template strands from the sequence?


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Let's say you have the following DNA sequence fragment:

5'-ACCAGTACTTCGT-3' 3'-TGGTCATGAAGCA-5'

Is there any way to determine which strand is the template strand and which is the coding strand for RNA synthesis solely from the above information?


I don't know about your specific example (perhaps look for triplets that code start codons?), but machine learning has been used to predict promoters from sequence information (from promoter regions, you get downstream initiation of transcription of DNA to RNA). You would probably need more sequence to predict a promoter with more certainty. Cites: https://en.m.wikipedia.org/wiki/Promoter_(genetics) and https://academic.oup.com/bioinformatics/article-abstract/35/16/2730/5270663


Given a DNA sequence alone, you can annotate open reading frames (ORFs) in order to identify the coding strand, with the caveat that not all ORFs are genes. ORFs are sequence segments that begin with a start codon (ATG) and end with a stop codon (TAA, TAG, TGA) when read from 5' to 3' in 3-base codons. There are no start or stop codons in either strand of the short sequence you provide, so I've appended a new example sequence:

DNA : 5' - AGGATGCAGGAGTGGTACGATTTATCCTAGGAACCT - 3' <-- Coding strand ^^^ ^^^ Start Stop 3' - TCCTACGTCCTCACCATGCTAAATAGGATCCTTGGA - 5' <-- Template strand RNA : 5' - AUGCAGGAGUGGUACGAUUUAUCC - 3'

Note that the start and stop codons given are part of the standard nuclear genetic code shared by eukaryotes, and there are alternative genetic codes used in prokaryotic and mitochondrial DNA transcription.


The answer is simple : promoter

the promoter is a non-coding sequence of DNA found on one strand of the DNA when reading from (5'->3') which is to be the template strand .

For example the (TATA) box is found in prokaryotic cells DNA (also named prinbow) and in eukaryotic cells DNA .In eukaryotes, It has the following sequence (5'… TATAAA… 3') and is centered (-20-30 bp/20-30 bp upstreaming away from the coding sequence of the gene) it is the sequence to which the transcription factors and RNA polymerase II bind initiating the transcription process.


Is it possible to distinguish between coding and template strands from the sequence? - Biology

It is often useful to distinguish the two strands of DNA -- the strand that is copied into mRNA and subsequently translated has the complementary sequence to the mRNA, while the base sequence of the opposite strand directly corresponds to the codons in the mRNA.

The terms template strand, sense strand, and coding strand are commonly used to describe one of the two strands of DNA, however the nomenclature is quite confusing because different authors have used these terms to describe both strands -- one school argues that the strand copied into mRNA should be considered the template strand, but the other school argues that the opposite strand which reflects the sequence in the mRNA should be considered the template because the corresponding codons are copied into protein. The first definition is used in the figures below, however, to avoid confusion, when using the words template, sense, or coding, it is essential to explicitly define how you are using the terms. I believe that these terms are best defined as described below.

The term template strand refers to the sequence of DNA that is copied during the synthesis of mRNA.

The opposite strand (that is, the strand with a base sequence directly corresponding to the mRNA sequence) is called the coding strand or the mRNA-like strand because the sequence corresponds to the codons that are translated into protein.

Although RNA polymerase must recognize sequences on the template strand, by convention we draw the DNA sequence and regulatory signals on the "mRNA-like" strand. (This makes it simpler to directly determine the sequence of the resulting RNA.) The following cartoon shows this concept for a hypothetical gene.

It may be useful to consider a real gene as well. The DNA sequence of the phage P22 arc gene and some important regulatory sites is shown below. The upper strand of DNA is the "mRNA-like" strand. The lower strand is the strand that is complementary to the mRNA. The -35 region (TTGACA) and -10 region (TATATT) of the promoter sequence and the transcriptional start site (the A ) is indicated on the coding strand. Also note that the DNA sequence of the coding strand corresponding to the RNA codons is shown in bold (of course, the T is a U in the RNA) -- the first codon is ATG the translational start site (fMet) and the last codon is TAA (Ochre) the translational stop codon.


Difference between Sense and Antisense strands of DNA

Sense strand vs antisense strand of DNA
A DNA molecule has a double stranded structure. It consists of two strands. Based on the strand that serves as template for mRNA formation or transcription, one strand is called the sense strand and the other is called the antisense strand.

Sense strand
1. This strand is also called as coding strand, plus strand or non-template strand.
2. Coding strand is same as mRNA except that thymine in DNA is replaced by Uracil in RNA.
3. The coding strand contains codons.
8
Antisense strand
1. This strand is also called as non-coding strand, minus strand or template strand.
2. This strand acts as template for the synthesis of mRNA. Therefore antisense strand is complementary to the sense strand and mRNA (U in RNA in place of T).
3. The non-coding strand contains anti-codons.

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On the anti-sense strand, the 13th nucleotide should be a Cytosine. Overall, excellent summary of the entire process!


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


References

Burge, C. B., Tuschl, T. & Sharp, P. A. in The RNA World 2nd edn (eds Gesteland, R. F., Cech, T. R. & Atkins, J. F.) 525–560 (Cold Spring Harbor Laboratory Press, New York, 1999).

Gerasimova, T. I., Gdula, D. A., Gerasimov, D. V., Simonova, O. & Corces, V. G. Cell 82, 587–597 (1995).

Gerasimova, T. I. & Corces, V. G. Cell 92, 511–521 (1998).

Buchner, K. et al. Genetics 155, 141–157 (2000).

Adams, M. D. et al. Science 287, 2185–2195 (2000).

Agabian, N. Cell 61, 1157–1160 (1990).

Caudevilla, C. et al. Proc. Natl Acad. Sci. USA 95, 12185–12190 (1998).

Harvey, A. J., Bidwai, A. P. & Miller, L. K. Mol. Cell. Biol. 17, 2835–2843 (1997).

Burge, C. B. & Karlin, S. Curr. Opin. Struct. Biol. 8, 346–354 (1998).


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Transcription

For protein synthesis, DNA must first be copied to messenger ribonucleic acid, or mRNA. This process is called transcription. The mRNA holds the coding information to make proteins. Unlike DNA, RNA is single-stranded and not helical in shape. It contains ribose instead of deoxyribose, and its nucleotide bases differ by having uracil (U) instead of thymine (T).

Initially, the enzyme RNA polymerase must assemble the pre-mRNA molecule that complements a section of one DNA’s two strands. Since the goal is not replication but protein synthesis, only one strand of DNA needs copying. The RNA polymerase first attaches to the double helix of DNA and works with proteins called transcription factors to determine what information needs transcribing. The RNA polymerase and transcription factors bind to this DNA strand, called the template strand.

The unit of RNA polymerase and transcription factors moves along the strand in a 3’ to 5’ (3 prime to 5 prime) direction and makes a new strand of mRNA with complementary base pairs. RNA polymerase builds the mRNA with additional nucleotides in elongation. The complementary nucleotides in mRNA, however, differ from DNA in that uracil replaces thymine. The mRNA runs in a 5’ to 3’ (5 prime to 3 prime) direction. After elongation ceases, mRNA separates from the DNA template strand in termination. Then mRNA serves either in a role of messenger in the cell, or it is used in protein formation, or translation.


Since ori is found where the reverse half-strand meets the forward half-strand, we can find ori by seeing when skew switches from decreasing to increasing. AKA, minimum skew.

Take a string Genome as an argument. We start by creating a list Skew where the 0th indexed value is 0. Each element in Skew is a running count of #G — #C at that point along the genome. At the 0th nucleotide (before the first nucleotide), Skew = 0 .

Next we range through the genome and compare the ith character in Genome to “G” or “C”:

  • If it’s the former, we +1 to running count for skew and add it to the list Skew .
  • If it’s the latter, we -1 and do the same.
  • If it’s neither, skew stays the same.

Here’s a diagram of how it plays out:

A different version of this function (credit to Cristian Alejandro Fazio Belan) is more intuitive:

Here, we update the running total score and append that value to the list.

We can take this list and plot it out:

We find the point of minimum skew by writing a quick function to return all values where Skew is at its minimum:

For E. coli, it seems ori is around the 3923620th nucleotide.


Section Summary

In prokaryotes, mRNA synthesis is initiated at a promoter sequence on the DNA template. Elongation synthesizes new mRNA. Termination liberates the mRNA and occurs by mechanisms that stall the RNA polymerase and cause it to fall off the DNA template. Newly transcribed eukaryotic mRNAs are modified with a cap and a poly-A tail. These structures protect the mature mRNA from degradation and help export it from the nucleus. Eukaryotic mRNAs also undergo splicing, in which introns are removed and exons are reconnected with single-nucleotide accuracy. Only finished mRNAs are exported from the nucleus to the cytoplasm.

Exercises

Glossary

exon: a sequence present in protein-coding mRNA after completion of pre-mRNA splicing

intron: non–protein-coding intervening sequences that are spliced from mRNA during processing

mRNA: messenger RNA a form of RNA that carries the nucleotide sequence code for a protein sequence that is translated into a polypeptide sequence

nontemplate strand: the strand of DNA that is not used to transcribe mRNA this strand is identical to the mRNA except that T nucleotides in the DNA are replaced by U nucleotides in the mRNA

promoter: a sequence on DNA to which RNA polymerase and associated factors bind and initiate transcription

RNA polymerase: an enzyme that synthesizes an RNA strand from a DNA template strand

splicing: the process of removing introns and reconnecting exons in a pre-mRNA

template strand: the strand of DNA that specifies the complementary mRNA molecule

transcription bubble: the region of locally unwound DNA that allows for transcription of mRNA


What is a Forward Primer?

Forward orientation is the synthesis of the coding strand or the sense strand of a gene. Taq polymerase catalyzes the synthesis of a new strand in 5’ to 3’ direction. The synthesis of coding strand occurs when the primer anneals with the noncoding or the antisense strand and elongates in 5’ to 3’ direction.

Figure 01: Forward and Reverse Primers

The primer that anneals with the antisense strand or the noncoding strand or the template strand is known as forward primer since forward primer acts as a starting point to the synthesis of coding or the positive strand of the gene. Forward primer has a short nucleotide sequence that is complementary to the 3’ flanking end of the antisense strand. It hybridizes with the antisense strand and facilitates the Taq polymerase to add nucleotides that are complementary to the template strand.


Contents: Difference between Sense Strand and Antisense Strand of DNA

What is Sense Strand of DNA?

The sense strand of DNA has the same base sequence as the mRNA. But it has thymine instead of uracil. This strand is called coding strand, plus strand or non-template strand. Uracil is present in RNA instead of thymine which is present in DNA. Moreover, it has same base sequence as tRNA. It actually runs from 5 primer to 3 primer and is complementary to the antisense strand. This strand undergoes translation and the immediate result of this process is RNA transcript. The translated protein can be inherited by this strand and that is one of the reasons why mRNA makes sense with the genetic code. Most of the times eukaryotic RNA undergoes additional editing before translating to proteins, in this process introns are removed and methylated guanine is added at one end. Additional poly-A tail is added to the other end and we term this process as splicing.

What is Antisense Strand of DNA?

The template strand which is transcribed is known as antisense strand of DNA. This strand is also termed as minus strand, non-coding strand or template strand. It is complementary strand to the sense strand and mRNA. Uracil is present in RNA instead of thymine. This strand carries information that is important for the manufacture of proteins by binding to corresponding mRNA. These strand are alike and only these strand are capable of giving information for the synthesis of protein. It does get transcribed unlike sense strand and it does not have the same base sequence as tRNA.


Watch the video: 1 Next Generation Sequencing NGS - An Introduction (July 2022).


Comments:

  1. Hudson

    It's straight to the point !!! In other words, you can't say it!

  2. Mccloud

    and I vazma probably. come in handy

  3. Tristen

    Very valuable message

  4. Kajishakar

    Got it, thanks for the explanation.



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