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Question 13 Marks
Differentiate between the following:

mRNA and tRNA.

Answer
mRNA and tRNA.
mRNA tRNA
It is called messenger RNA and carries the codes for amino acid sequence. It is called transfer RNA as it carries amino acids to the site of protein synthesis.
It is a linear molecule. It has clover leaf shape.
It is synthesised by RNA polymerase II It is synthesised by RNA polymerase III.
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Question 23 Marks
What is DNA fingerprinting? Mention its application.
Answer
DNA Fingerprinting is a technique to find out variations in individuals of a population at DNA level.
Its applications are as follows:
  1. Used in forensic science to identify suspects.
  2. Used to find out history of an organism.
  3. Used to find out paternity and family relations.
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Question 33 Marks
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer
The property of DNA double helix led Watson and Crick are:
  • Two strands running opposite to each other, wherein bases will always pair with their counterpart-A with T and G with C (specific pairing).
  • If H bonds break and bases of one strand lie exposed, unpaired, they will easily pair up with free nucleotides as well. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. where the two strands separate and act as a template for the synthesis of a new complementary strand.
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Question 43 Marks
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer
Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages. In 1952, they chose a bacteriophage known as T2 for their experimental material. They grew some viruses on a medium that contained radioactive phosphorus (p32) and some others on medium that contained radioactive sulphur (s35). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protem does not Similarly. viruses grown on radioactive sulphur contained radioactive protein but not radio'active DNA because DNA does not contain sulphur. Radioactive phages were allowed to attach to E.coli bacteria. Then, as the infection proceeded, the 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., Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. 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. DNA is therefore the genetic material that is passed from virus to bacteria.
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Question 53 Marks
Differentiate between the following:

Template strand and Coding strand.

Answer
Template strand and Coding strand.
Template strand Coding strand
It is the strand which is transcribed into RNA. It has the same sequence as mRNA.
It is called anti sense strand. It is called sense or non-template strand.
It has 3' → 5' polarity. It has 5' → 3' polarity
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Question 63 Marks
Differentiate between the following:

Repetitive DNA and Satellite DNA.

Answer
Repetitive DNA and Satellite DNA.
Repetitive DNA Satelite DNA
It is the non-coding DNA with multiple copies of identical sequences which may lie in tandem or interspersed. It refers to non-coding tandem repeat sequences.
These can be few base pairs to hundreds or thousands of base pairs. These are generally short sequence repeats (up to 60 base pair long).
It appears as light bands. It appears as small dark bands.
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Question 73 Marks
Unambiguous, universal and degenerate are some of the terms used for the genetic code. Explain the salient features of each one of them.
Answer
Unambiguous -One codon codes for one amino acid, e.g. AUG (Methionine).

Universal -Codon and its corresponding amino acid are the same in all organisms.

Example:- more-Bacteria to human UUU codes for phenylalanine (phe).

Degenerate - Some amino acids are by more than one codon.

Example:- UUU and UUC code for phenylalanine (phe).

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Question 83 Marks
With respect to Messelson and Stahl’s Experiment, answer the following questions:
  1. Identify the method used to distinguish between heavy and light isotopes of nitrogen.
  2. With the help of diagrams, compare the results for the DNA isolated after 20 minutes of experiment with the DNA which was isolated after 40 minutes.
Answer
  1. Centrifugation in a CsCl density gradient.
  2.  

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Question 93 Marks
Write any three goals of human genome project.
Answer
Three goals of human genome project:
  1. To develop new and improved medicines.
  2. To predict and prevent diseases.
  3. To ensure that the diagnosis is accurate.
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Question 103 Marks
AtRNA is charged with the amino acid phenylalanine.
  1. At what end of the tRNA is the amino acid attached?
  2. What is the mRNA codon that codes for phenylalanine?
  3. Name the enzyme responsible for this attachment. 
Answer
  1. It is attached at the 3' end of RNA.
  2. UUU and UUC.
  3. Aminoacyl-tRNA synthetase.
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Question 113 Marks
DNA polymerase and RNA polymerase differ in their requirement while functioning. Explain.
Answer
S. No RNA polymerase DNA polymerase
(i) It cannot carry out proofreading. It carries out proofreading for DNA repairmechanism.
(ii) RNA polymerase does not require RNA primer for synthesis of RNA. DNA polymerase requires RNA primer for synthesis of DNA.
(iii) It uses ribonucleotides for RNA synthesis. It uses deoxyribonucleotides for DNA synthesis.
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Question 123 Marks
The base sequence in one of the strands of DNA is TAGCATGAT
  1. Give the base sequence of its complementary strand.
  2. How are these base pairs held together in a DNA molecule?
  3. Explain the base complementarity rules. Name the scientist who framed this rule.
Answer
  1. ATCGTACTA.
  2. Through Hydrogen bonds, between A and T and C and G on the two strands.
  3. A= T and C$\equiv$ G, Watson and Crick/Chargaff.
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Question 133 Marks
Study the figure given below and answer the following questions.

  1. Name the molecule 'X' synthesised by i gene. How does this molecule get inactivated?
  2. Which one of the structural genes codes for B-galactosidase?
  3. When will the transcription of this gene stop?
Answer
  1. Molecule 'X' is a repressor protein. When an inducer combines with it, it is inactivated.
  2. 'z' gene.
  3. Transcription of gene stops.
  1. Substrate (lactose) is not available.
  2. Energy source (glucose) is available to the cells.
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Question 143 Marks

  1. Identify the polarity from a to a', in the above diagram and mention how many more amino acids are expected to be added to this polypeptide chain.
  2. Mention the DNA sequence coding for serine and the anticodon of tRNA for the same amino acid.
  3. Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA?
Answer
  1. a to a' is 5' → 3'. No more amino acid will be added.
  2. TCA; anticodon is UCA.
  3.  
  1. The untranslated regions are required for efficient translation process.
  2. They are present before the initiation codon at the 5'-end and after the stop/ termination codon, at the 3'-end.
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Question 153 Marks

Why do you see two different types of replicating strands in the given DNA replication fork? Explain. Name these strands.
Answer
  1. Both the strands of parent DNA function as template strands.
  2. On the template strand with 3' → 5' polarity, the new strand is synthesised as a continuous stretch as the 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.
  3. 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.
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Question 163 Marks
The following is the flow chart highlighting the steps in DNA fingerprinting technique. Identify a, b, c, d, e and f.

Answer
  1. Restriction endonuclease.
  2. Agarose gel.
  3. Nitrocellulose/ nylon/ synthetic membrane.
  4. Variable Number Tandem Repeats (VNTR).
  5. Hybridisation with VNTR probe.
  6. Autoradiography.
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Question 173 Marks
Describe the discontinuous synthesis of DNA.
Answer
The discontinuous synthesis of DNA is as follows:
  1. At the replication site, unwinding of double stranded DNA takes place by DNA gyrase and helicase.
  2. ssBPs (single-stranded binding proteins) bind to the separated strands to avoid recoiling or to provide stability.
  3. Since DNA polymerase can synthesise DNA only in 5′ → 3′ direction, DNA synthesis occurs discontinuously on the lagging strand.
  4. These small fragments of DNA are called Okazaki fragments.
  5. The enzyme primase adds primers after every fragment is formed.
  6. These Okazaki fragments are then joined by DNA ligase.
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Question 183 Marks
Explain the post-transcriptional modifications, the hnRNA undergoes in eukaryotic cells.
Answer
  1. Since eukaryotes have split gene arrangement, the hRNA has both coding sequences (exons) and non-coding sequences (introns) and is non-functional; so it has to undergo splicing, the process, in which introns are removed and exons are joined.
  2. It has to undergo two other processes, namely capping and tailing to become functional.
  3. In capping, an unusual nucleotide, called methyl guanosine triposphate residues are added at the 5' end of hnRNA.
  4. In-tailing, about 200-300 adenylate residues are added at the 3' end, in a template-independent manner.
  5. These changes take place in the nucleus of the cell.
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Question 193 Marks
Describe the initiation process of transcription in bacteria.
Answer

(DNA dependent) RNA polymerase, binds to the promoter, at 5’ end, associates transiently with initiation factor/sigma factor, using nucleoside triphosphates as substrate, and energy initiates transcription.

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Question 203 Marks
Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer
  1. They do not code for any proteins,
  2. They form large part of the human genome,
  3. They show high degree of polymorphism/Specific to each individual.
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Question 213 Marks
Explain the mechanism of DNA replication with the help of a replication fork. What role does the enzyme DNA-ligase play in a DNA replication fork?
Answer
  1. DNA replication occurs in small replication forks. It does not occur in its entire length in one time as DNA is a very large molecule and only that part of DNA opens up which is being replicated. The opening of the whole DNA molecule would be an energetically more expensive process.
  2. The main enzyme involved in DNA replication is the DNA-dependent DNA polymerase. This enzyme catalyzes the polymerization of deoxynucleotides along the 5′ → 3′ direction, and hence, replication is continuous along the 3′ → 5′ strand (leading strand) and discontinuous along the template, i.e., the 5′ → 3′ direction (lagging strand).
  3. Okazaki fragments are short DNA segments on the lagging strand, formed in the 5’ – 3’ direction, starting from RNA primers. A separate RNA primer is needed for the synthesis of each Okazaki fragment. These discontinuously synthesized fragments are later joined by the enzyme DNA ligase.
  4. Ori stands for Origin of replication. This site has the highly conserved sequence of DNA among various species. The replication of DNA starts here because this site attracts some proteins which help in the opening and unwinding of DNA and this leads to the initiation of replication.

The function of DNA Ligase is to join the two nucleotides. During the DNA replication process, it joins the Okazaki fragments of the daughter DNA to form the complete DNA molecule on the lagging strand.

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Question 223 Marks
A typical mammalian cell has 2.2m long DNA molecule, whereas the nucleus in which it is packed measures about 10m. Explain how such a long DNA molecule is packed within a tiny nucleus in the cell.
Answer
  1. In the mammalian cells (or eukaryotes) there is a set of positively-charged basic proteins, called histones.
  2. Histones are organised to form a unit of eight molecules, called histone octamer.
  3. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure, called nucleosome.
  4. A typical nucleosome contains 200 bp of DNA helix.
  5. The nucleosomes constitute the repeating units of chromatin, which appear as beads-on-string structure under an electron microscope.
  6. These are further packaged to form the chromatin fibres, which condense to form chromosomes.
  7. The packaging of chromatin at higher levels requires additional set of proteins called non-histone chromosomal (NHC) proteins.
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Question 233 Marks
AtRNA is charged with the amino acid methionine.
  1. Name the process involved in attachment.
  2. Point out the mRNA codon and anticodon on tRNA for this amino acid.
  3. Name the enzyme responsible for this attachment.
Answer
  1. Aminoacylation of RNA.
  2. mRNA codon-AUG, RNA anticodon-UAC.
  3. Enzyme-aminoacyl RNA synthetase.
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Question 243 Marks
Explain the following salient features of a DNA molecule:
  1. Complimentarity of base pairs of its two chains.
  2. Antiparallel polarity of the two chains.
Answer
  1. The two strands of a DNA are said to be complementary to each other, i.e., a purine of one strand always pair with a pyrimidine.
  1. Adenine (a purine) pairs with thymine (a pyrimidine) by forming two hydrogen bonds.
  2. Guanine (a pyrimidine) by forming three hydrogen bonds.
  1. The two chains of a DNA shows antiparallel polarity, which means that if one strand has 5' 3' polarity, the other strand has 3' → 5' polarity.
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Question 253 Marks
Explain the role of regulatory gene in lac operon. Why is the regulation of the operon called negative regulation?
Answer
  1. The regulatory gene in lac operon (also called i gene or inhibitor gene) codes for a protein, called repressor; the repressor is synthesised all the time constitutively in the cell.
  2. The repressor has high affinity to the operator; it binds to the operator region and prevents the RNA polymerase from transcribing the structural genes of the operon.
  3. It is called negative regulation because the operon is switched off and transcription is prevented.
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Question 263 Marks
Name the specific enzyme responsible for nucleotide polymerisation in DNA replication. Write two characteristic features of this enzyme. Name the region on E.coli DNA, where this enzyme can initiate replication.
Answer
  1. DNA-dependent DNA polymerase is the enzyme.
  2. Two characteristic features of this enzyme include:
  1. It catalyses the polymerisation of nucleotides in 5' → 3' direction only.
  2. It polymerises approximately 2000bp per second, with high degree of accuracy.
  1. Origin of replication is the region on E.coli DNA, where this enzyme can initiate replication.
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Question 273 Marks
Differentiate between a template strand and coding strand of DNA.
Answer
S. No Role/ Strand Template strand Coding strand
(i) Function Codes for the protein molecule. Does not code for anything.
(ii) Polarity 3′ → 5′ 3′ → 5′
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Question 283 Marks

  1. Identify the polarity of strand in the diagram from a to a', and mention how many more amino acids can be expected to be added to this polypeptide chain.
  2. Mention the DNA sequence coding for serine and the anticodon of tRNA for the same amino acid.
  3. Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide?

Where exactly are they present on mRNA?

Answer
  1. a to d' is 5 → 3'.

No more amino acid will be added.

  1. TCA; anticodon is UCA.
  2. The untranslated regions are required for efficient translation process.

They are present before the initiation codon at the 5' end and after the stop/termination codon, at the 3' end.

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Question 293 Marks
ATGGAGTACTTCGTGTGA is the coding strand of DNA in a transcription unit.
  1. Write the mRNA transcribed from this DNA segment.
  2. How many amino acids does it code for? Why?
Answer
  1. AUG GAG UAC UUC GUG UGA
  2. It will code for five amino acids.

The last codon, UGA is a termination codon. that does not code for any amino acid.

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Question 303 Marks
Identify by giving reasons, the salient features of genetic code by studying the following nucleotide sequence of mRNA strand and the polypeptide translated from it.
(AUG UUU UCU UUU UUU UCU UAG) (Met - Phe - Ser - Phe - Phe - Ser)
Answer
S. No.
Salient features of genetic code
Reason
1. The condin is triplet. AUG, UUU, etc., are triplets.
2.
One codon codes for only one amino acid, so it is unambigus and specific.
UUU codes for serine, AUG, codes for methionine, etc.
3.
AUG has dual function as it codes for methionine and it also acts as initator codon.
AUG is seen at the beginning of the polypeptide chain.
4.
UAG act as a stop codon.
No amino acid is coded by UAG in the polypeptide chain given.
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Question 313 Marks
The base sequence on one of the strands of DNA is ATGTCTATA.
  1. Give the base sequence of its complementary strand.
  2. If an RNA strand is transcribed by this strand, what would be the base sequence of RNA?
  3. In what other respect, an RNA molecule differs from a DNA molecule?
Answer
  1. TACAGATAT
  2. UACAGAUAU
  3.  
S. No DNA RNA
(i) It is a double-stranded structure. It is a single-stranded structure.
(ii) It contains deoxyribose sugar. It contains ribose sugar.
(iii) Bases are A, T, G, C Bases are A, G, C, U
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Question 323 Marks
In a typical nucleus some regions of chromatin are stained light and others dark. Explain why it is so and what is its significance.
Answer
  1. The regions of a chromatin, which are stained light, are the regions where chromatin is loosely packed; they are called euchromatin.
  2. The regions of a chromatin, which are stained dark, are the regions where chromatin is tightly packed; they are called heterochromatin.
  3. Euchromatinis transcriptionally more active, while heterochromatin is inactive.
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Question 333 Marks
Where is an operator' located in a prokaryotic DNA? How does an operator regulate gene expression at transcription level in a prokaryote? Explain.
Answer
  1. The operator' region is adjacent to the promoter region.
  2. The operator regulates the accessibility of promoter region to the RNA polymerase for transcription.
  3. The operator binds to the repressor protein coded by the regulatory gene and prevents the RNA polymerase from binding to the promoter; the operon is switched "off".
  4. When an inducer binds to the repressor and inactivates it, it cannot bind to the operator; this allows RNA polymerase access to the promoter and transcription continues, i.e., the operon is switched 'on'.
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Question 343 Marks
Describe the experiments that established the identity of 'transforming principles' of Griffith.
Answer
  • Purification of biochemicals like Proteins, RNA & DNA from S cells. (heat killed)
  • Presence of Protein & RNA in medium did not affect transformation.
  • DNA alone from S Bacteria caused R Bacteria to transform.
  • Digestion with DNAase did inhibit transformation,

Conclusion: DNA is the transforming chemical/biochemical.

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Question 353 Marks
Study the diagram given below:

Name the linkages X, Y, Z and the respective molecules formed by them.

Answer
  1. X is N-glycosidic linkage; it forms a nucleoside.
  2. Y is phosphoester linkage; it forms a nucleotide.
  3. Z is a phosphodiester linkage; it forms a dinucleotide.
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Question 363 Marks
Given below are the sequences of nucleotides in a particular mRNA and amino acids coded by it.
UUU AUG UUC GAG UUA GUG UAA
Phe - Met - Phe - Glu - Leu – Val
Write the properties of genetic code that can be and that cannot be correlated from the above given data.
Answer
  1. The properties that can be correlated:
  1. UAA does not code for any amino acid; it is a termination codon.
  2. Genetic code is specific and unambiguous, i.e. one codon codes for a particular amino acid only.
  3. Each codon is a triplet.
  4. Genetic code is degenerate, as one amino acid is coded by more than codon, e.g. UUU and UUC code for phenylalanine.
  5. Genetic code is read in a contiguous manner without any punctuation.
  1. The property that cannot be correlated:

AUG has a dual function; it is initiation codon as well as codes for methionine.

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Question 373 Marks
  1. Identify the polarity at A and B respectively in the figure given below.
  2. Explain the mechanism the figure represents.

Answer
  1. A-5', B-3
  2.  
  1. The figure represents the continuous and discontinuous synthesis of DNA strands at the replication fork, during replication of DNA.
  2. Replication fork is the Y-shaped structure formed with small opening of DNA double helix.
  3. DNA polymerase catalyses polymerisation of nucleotides only in 5' → 3' direction.
  4. Both the strands of parental) DNA act as templates for the synthesis of new strands.
  5. On the template strand with 3' → 5' polarity, the new strand is synthesised as a continuous stretch; it is called continuous synthesis.
  6. On the other strand with 5' → 3' polarity, DNA is synthesised as short stretches; it is called discontinuous synthesis.
  7. Later the short stretches of DNA are joined by DNA-ligases into a continuous strand.
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Question 383 Marks
Draw a labelled schematic sketch of replication fork of DNA. Explain the role of the enzymes involved in DNA replication.
Answer
A. 

B. DNA polymerase catalyses the polymerisation of nucleotides, Ligase joins the fragments of discontinuous synthesis.

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Question 393 Marks
What is ‘semi-conservative’ DNA replication? How was it experimentally proved and by whom?
Answer
After the completion of replication (of the DNA molecule into two)) each DNA molecule will have one parental strand and newly synthesised strand.

 Meselson and Stahl

  1. E.coli grown in medium containing 15NH4CL (15N - heavy Nitrogen) for many generations to ensure that all DNA in the bacteria were heavy,
  2. Heavy E.coli transferred to a medium with normal 14NH4CL,(after 20 minutes) DNA of generation I extracted to measure their densities, they were of intermediate density,
  3. After 40 minutes DNA of II generation were extracted and tested for their densities, they were of equal amounts of (hybrid) intermediate DNA, and light DNA/14NH4Cl.
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Question 403 Marks
Discuss the significance of heavy isotope of nitrogen in the Meselson and Stahl’s experiment.
Answer
The heavy isotope of nitrogen was used in Meselson and Stahl’s experiment for various reasons. The heavy DNA molecule (containing 15N) could be easily distinguished from the normal DNA by centrifugation in a Cesium chloride (CsCl) density gradient. It could be easily separated from lighter nitrogen (14N) on the basis of density. Use of lighter and heavier nitrogen made the task of identifying transfer of DNAs through subsequent generation quite easier.
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Question 413 Marks
Given below is a part of the template strand of a structural gene:

TAC CAT TAG GAT

  1. Write its transcribed mRNA strand with its polarity.
  2. Explain the mechanism involved in initiation of transcription of this strand.
Answer
  1. 5’ AUG GUA AUC CUA 3’ (Polarity),
  2. RNA polymerase, blinds at 5 end, in the presence of (initiation factor) $\sigma$ factor, at the promoter site.
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Question 423 Marks
A tRNA is charged with amino acid phenylalanine:
  1. At what end of the tRNA is the amino acid attached?
  2. What is the mRNA codon that codes for phenylalanine?
  3. Name the enzyme responsible for this attachment.
Answer
  1. At the 3' end.
  2. UUU or UUC.
  3. Aminoacyl tRNA synthetase.
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Question 433 Marks
Explain Hershey - Chase experiment. What was proved through this experiments?
Answer

DNA is genetic material.
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Question 443 Marks
  1. Why does DNA replication occur in small replication forks and not in its entire length?
  2. Why is DNA replication continuous and discontinuous in a replication fork?
  3. Explain the importance of origin of replication' in a replication fork.
Answer
  1. Replication of DNA occurs in small replication forks, because DNA is such a long molecule that the separation of the two strands along its entirelength requires a very high amount of energy.
  2. DNA polymerase can catalyse the polymerization of nucleotides only in 5' → 3' direction.
  1. So on the template strand with 3' → 5' polarity, DNA replication is continuous.
  2. On the template strand with 5' → 3' polarity, DNA synthesis occurs in short stretches as the opening of replication fork continues; later these short stretches are joined by the action of DNA ligases .
  1. Replication of DNA does not initiate randomly, and DNA polymerases on their own cannot initiate replication.
  1. So, there is a specific sequence on DNA, called origin of replication; DNA polymerase bindsto it and continues the process.
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Question 453 Marks
  1. Draw the structure of the initiator tRNA adaptor molecule.
  2. Why is tRNA called an adaptor molecule?
Answer
  1.  

  1. On one hand, it reads the code.

On the other hand, it binds to specific amino acid.

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Question 463 Marks
Now, sequencing of total genomes getting is getting less expensive day by the day. Soon it may be affordable for a common man to get his genome sequenced. What in your opinion could be the advantage and disadvantage of this development?
Answer
Advantages of Affordable Genome Sequencing: It can help in settling disputes which may arise in case of parentage of a child. This can also help in disputes of property inheritance by finding the bonafide beneficiary. Human genome can also help in preparing a database on people with criminal record. It can help in identifying the chances of genetic disorders in a family.
Disadvantages: Genome sequencing can have serious issues of privacy. Some employers may misuse the data to blackmail their employees. Many private matters may leak into public domain; creating embarrassment for the affected person.
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Question 473 Marks

Study the mRNA Segment given above which is complete to be translated into a polypeptide chain.

  1. Write the codons ‘a’ and ‘b.'
  2. What do they code for?
  3. How is peptide bond formed between two amino acids in the ribosome?
Answer
  1.  

a - AUG.

b - UAA/UAG/UGA.

  1.  

AUG code for Methionine.

UAA/UAG/UGA - Stop codon/Nonsense codon/Does not code for any amino acid.

  1.  

Charged tRNA are brought closer together on mRNA in the ribosomes, and Ribosomes acts as a catalyst (ribozyme) forming peptide bond.

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Question 483 Marks
  1. Expand VNTR and describe its role in DNA fingerprinting.
  2. List any two applications of DNA fingerprinting technique.
Answer
  1. VNTR stands for “Variable Number of Tandem Repeats”.

The VNTR belongs to a class of satellite DNA referred to as mini-satellite. A small DNA sequence is arranged tandemly in many copy numbers. The copy number varies from chromosome to chromosome in an individual. The numbers of repeat show very high degree of polymorphism. As a result th size of VNTR varies in size from 0.1 to 20kb. Consequently, after hybridization with VNTR probe, the autoradiogram gives many bands of differing sizes. These bands give characteristic pattern for an individual DNA which is used to identify individuals.

  1. Since DNA from every tissue (such as blood, hair-follicle, skin, bone, saliva, sperm etc.), from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications to identify criminals. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basic of paternity testing, in case of disputes.
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Question 493 Marks
  1. Why is RNA called an 'adapter'?
  2. Draw and label a secondary structure of RNA. How does the actual structure of RNA look like?
Answer
  1. Since 'RNA on one hand binds to a specific amino acid and on the other hand reads the codon of the amino acid bound to it through its anticodon, it is called an 'adapter'.
  2.  

It actually looks like an inverted L.

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Question 503 Marks
  1. Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical coding strand given below.

  1. Write the RNA strand transcribed from the above transcription unit along with its polarity.
Answer
  1.  

  1. The RNA transcribed is 5' AUGCAUGCAUAG 3'.
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