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34 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Explain the role of lactose in ‘Lac Operon’.
Answer
  1. A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
  2. Lactose binds to repressor and inactivates it.
  3. Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
  4. RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
  5. Thus, lactose acts as an inducer.
  6. When the inducer level falls, the operator is blocked again by repressor and structural genes are repressed again. This is negative feedback.
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Question 23 Marks
Give the applications of DNA fingerprinting.###Mention DNA fingerprinting application.###List any two applications of DNA fingerprinting technique.
Answer
Applications of DNA fingerprinting are as follows:
  1. In forensic science to solve rape and murder cases.
  2. Finds out the biological father or mother or both, of the child, in case of disputed parentage.
  3. Used in pedigree analysis in cats, dogs, horses and humans.
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Question 33 Marks
Enlist the characteristics of genetic code.
Answer
The characteristics of genetic code are
  1. Genetic code is triplet, commaless and non-overlapping.
  2. It is degenerate and non-ambiguous.
  3. It is universal
  4. It has polarity.
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Question 43 Marks
Write short note on : DNA packaging in eukaryotic cell.
Answer
1. In eukaryotic cells, DNA ( $2.2$ metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus ( $10^{-16} m$ ).
2. DNA is associated with histone and non-histone proteins.
3. Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
4. Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. $H _2 A, H _2 B, H _3$ and $H _4$ forming histone octamer) and negatively charged DNA $( 146\ bps )$ that wraps around the histone octamer by $1 3/4$ turns.
5. $H _1$ protein binds the DNA thread where it enters and leaves the nucleosome.
6. Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to $114\ bp$, average length of linker DNA is about 54 bp ).
7. Each nucleosome contains $200\ bp$ of DNA.
8. Packaging involves formation of - Beads on string ( $10\ nm$ diameter), Solenoid fibre (looks like coiled telephone wire, $30\ nm$ diameter/ $300 \AA$ ), Chromatin fibre and Chromosome.
9. Non-Histone Chromosomal Proteins (NHC) contribute to the packaging of chromatin at a higher level.
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Question 53 Marks
What is the role of ribosomes in protein synthesis?
Answer
  1. Ribosomes serve as site for protein synthesis.
  2. A ribosome has one binding site for m-RNA and 3 binding sites for t-RNA. They are P-site (peptidylt-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
  3. In Eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
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Question 63 Marks
Explain Genetic code is Non-overlapping.
Answer
  1. Each single base is a part of only one codon.
  2. Adjacent codons do not overlap.
  3. If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
  4. Experimental evidence favours non-overlapping nature of genetic code.
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Question 73 Marks
Explain why it was suggested that codon is a sequence of three consecutive nucleotides on $\text{m-RNA.}$
Answer
$1.$ If each codon has only one nucleotide, then there will be $4^1 = 4$ codons, which can encode for only four different types of amino acids.
$2.$ If each codon has two nucleotides, then there will be $4^2 = 16$ codons, which can encode for only $16$ different types of amino acids.
$3.$ If each codon has three nucleotides, then there will be $4^3 = 64$ codons, which are sufficient to specify $20$ different types of amino acids.
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Question 83 Marks
What are the functions of three types of RNAs in bacteria? Which enzyme is involved in transcription of DNA to form RNAs in bacteria? What is its function?
Answer
  1. In bacteria, m-RNA provides the encoded message for protein synthesis; t-RNA brings specific amino acid to the site of translation; r-RNA plays role in providing binding site to m-RNA and t-RNA.
  2. There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
  3. RNA polymerase binds to promoter and initiates transcription (initiation) and synthesizes RNA.
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Question 93 Marks
Why is HGP important?
Answer
  1. HGP is associated with rapid development of Bioinformatics.
  2. Knowledge gained about the functions of genes and proteins has a major impact in the fields like Medicine. Biotechnology and the Life sciences.
  3. It has helped in identifying the genes that are associated with genetic characteristics.
  4. The genetic basis of many hereditary diseases can be understood.
  5. It has increased the understanding of gene structure and function in other species. As human beings have many of the genes same as those of flies, roundworms and mice, such studies will enhance understanding of human evolution.
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Question 103 Marks
What are the application of genomics?
Answer

Applications of genomics are as follows:

  1. Structural and functional genomics is used in the improvement of crop plant, human health and livestock.
  2. The knowledge and understanding acquired by genomics research can be applied in medicine, biotechnology and social sciences.
  3. It helps in the treatment of genetic disorders through gene therapy.
  4. It helps in the development of transgenic crops having desirable characters.
  5. Genetic markers have applications in forensic analysis.
  6. Genomics can lead to introduction of new gene in microbes to produce enzymes, therapeutic proteins and biofuels.
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Question 113 Marks
Explain in brief the process of initiation during protein synthesis.
Answer
Initiation of synthesis of polypeptide chain takes place as follows:
  1. Small subunit of ribosome binds to the $ \text{m-RNA}$ at $5\ ’$ end.
  2. Start codon is positioned properly at $P-$site.
  3. Initiator $ \text{t-RNA}, ($carrying methionine in eukaryotes or formyl methionine in prokaryotes$)$ binds with initiation codon $ \text{(AUG)}$ of $ \text{m-RNA}$ by it’s anti$-$codon $ \text{(UAC)}$. Codon$-$anti$-$codon pairing involves formation of hydrogen bonds.
  4. The large subunit and smaller subunit of ribosome bind in the presence of $Mg^{++}.$
  5. Thus, initiator charged $ \text{t-RNA}$ occupies the $P-$site and $A-$site is vacant for next charged $ \text{t-RNA}.$
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Question 123 Marks
If operator gene is deleted due to mutation, how will E.coli metabolise lactose?
Answer
If operator gene is deleted due to mutation, lac operon cannot be regulated. It will get transcribed continuously and enzymes required for lactose metabolism will get synthesized continuously.
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Question 133 Marks
Explain with example what is meant by positive control of gene regulation.
Answer
  1. A set of genes will be switched on when a substrate is to be metabolized.
  2. This phenomenon is called induction and small molecule responsible for this, is known as inducer.
  3. It is positive control of gene regulation.
  4. For example, lactose acts as an inducer in Lac operon.
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Question 143 Marks
Give examples of coordinated regulation or expression, of several sets of genes.
Answer

Examples of coordinated regulation or expression of several sets of gene are:

  1. If E.colt bacteria do not have lactose in the surrounding medium as a source of energy, then structural genes in Lac operon do not get transcribed and enzyme like β -galactosidase is not synthesized.
  2. The development and differentiation of embryo into an adult organism.
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Question 153 Marks
What is the role of ribosomes in protein synthesis?
Answer
  1. Ribosomes serve as site for protein synthesis.
  2. A ribosome has one binding site for m-RNA and 3 binding sites for t-RNA. They are P-site (peptidylt-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
  3. In Eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
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Question 163 Marks
How DNA regulates protein synthesis?
Answer
  1. DNA regulates protein synthesis by coding for the specific sequence of amino acids in a protein.
  2. This control is possible through transcription of m-RNA.
  3. Genetic code is specific for particular amino acid.
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Question 173 Marks
How many amino acids are required for protein synthesis? From where are they obtained?
Answer
  1. About 20 different types of amino acids are required for protein synthesis.
  2. They are available in the cytoplasm.
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Question 183 Marks
Give examples of termination codons. Why are they known as termination codons?
Answer
  1. UAA, UAG and UGA are known as termination codons.
  2. They do not code for any amino acid.
  3. They terminate or stop the process of elongation of polypeptide chain.
  4. Hence, they are known as termination codons.
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Question 193 Marks

a. What is meant by universal genetic code? Give example.
b. Why genetic code is called Non- ambiguous?

Answer

a. Universal genetic code means that the specific codon codes for same amino acid in all living organisms, e.g. codon AUG always specifies amino acid methionine.

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

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Question 203 Marks
How degeneracy of the code is explained by Wobble hypothesis?
Answer
  1. In 1966, Crick proposed Wobble hypothesis.
  2. According to this hypothesis, in codon- anticodon base pairing, the third base may not be complementary.
  3. The third base of codon is called wobble base and this position is called wobble position.
  4. This results in economy of t-RNA as anticodon of a t-RNA may bind with codon even when only first two bases are complementary.
  5. For example, GUU, GUC, GUA and GUG codons code for amino acid Valine.
  6. Degeneracy of genetic code means many codons can code for same amino acid.
  7. Thus, the degeneracy of genetic code gets explained by Wobble hypothesis.
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Question 213 Marks
Explain Genetic code is Non-overlapping.
Answer
  1. Each single base is a part of only one codon.
  2. Adjacent codons do not overlap.
  3. If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
  4. Experimental evidence favours non-overlapping nature of genetic code.
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Question 223 Marks
Explain why it was suggested that codon is a sequence of three consecutive nucleotides on $ \text{m-RNA.}$
Answer
If each codon has only one nucleotide, then there will be $4^1 = 4$ codons, which can encode for only four different types of amino acids.
If each codon has two nucleotides, then there will be $4^2= 16$ codons, which can encode for only $16$ different types of amino acids.
If each codon has three nucleotides, then there will be $4^3 = 64$ codons, which are sufficient to specify $20$ different types of amino acids.
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Question 233 Marks
What are the functions of three types of RNAs in bacteria? Which enzyme is involved in transcription of DNA to form RNAs in bacteria? What is its function?
Answer
  1. In bacteria, m-RNA provides the encoded message for protein synthesis; t-RNA brings specific amino acid to the site of translation; r-RNA plays role in providing binding site to m-RNA and t-RNA.
  2. There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
  3. RNA polymerase binds to promoter and initiates transcription (initiation) and synthesizes RNA.
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Question 243 Marks
Why DNA replication is called semiconservative replication?
Answer
  1. In each of the two daughter DNA molecules thus formed, one strand is parental and the other one is newly synthesized.
  2. 50% is contributed by mother DNA.
  3. Hence, DNA replication is described as semiconservative replication.
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Question 253 Marks
Why t-RNA is called as adapter molecule?
Answer
t-RNA can read the codon on m-RNA. It also can bind with the amino acid at 3’ end and transport it to m-RNA-ribosome complex during translation. It can bind with specific codon which is complementary to its anticodon. So t-RNA is considered as an adapter molecule.
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Question 263 Marks
Why are Okazaki fragments formed on lagging strand only?
Answer
  1. The lagging template is the template strand with free 5’ end.
  2. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
  3. Both the strands of the parental DNA are antiparallel and new strands are always formed in 5′ → 3′ direction, i.e. DNA polymerase synthesizes new strand in only one direction i.e. 5′ → 3′ direction.
  4. Hence, the lagging templates becomes available for replication only discontinuously in small patches.
  5. The new lagging strand develops discontinuously away from the replicating fork in the form of small Okazaki fragments.
  6. Hence, Okazaki fragments formed on lagging strand only.
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Question 273 Marks
Give Significance: DNA.
Answer
  1. DNA regulates and controls all the cellular activities.
  2. It replicates and gets distributed equally to the daughter cells when the cell divides.
  3. It is a carrier of genetic information.
  4. Heterocatalytic function : DNA directs the synthesis of chemical molecules other than itself. E.g. Synthesis of RNA (transcription), synthesis of protein (Translation), etc.
  5. Autocatalytic function : DNA directs the synthesis of DNA itself. E.g. Replication.
  6. DNA is a master molecule of a cell that initiates, guides, regulates and controls the process of protein synthesis.
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Question 283 Marks
Give Reasons: Protein synthesis is the most important and essential activity in the living cells.
Answer
  1. Proteins play a significant role in the metabolism of living cells.
  2. The actual phenotypic expression of living cells is dependent on the biochemical reactions.
  3. Each biochemical reaction needs a specific enzyme for its initiation and completion. All the enzymes are proteins.
  4. In a cell there are many structural proteins too. Thousands of structural and catalytic proteins are constantly required within the cell at all times.
  5. Many functional proteins like hormones are also important for metabolism. Thus, for the synthesis of all such proteins, protein synthesis has become the most important and essential activity of the living cell.
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Question 293 Marks
Give Reasons: Only 20 amino acids are considered as standard.
Answer
  1. It was believed that there are total 20 amino acids in the living world. But 21st amino acid called selenocysteine was discovered later.
  2. This amino acid is coded by UGA which is usually a termination codon.
  3. In both prokaryotic and eukaryotic cells polypeptide chains contain 100-300 amino acids and they are formed by specific arrangement of 21 amino acids.
  4. But formation of selenocysteine requires the availability of element selenium in the cells.
  5. Therefore, only 20 amino acids are considered as standard.
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Question 303 Marks
Give Reasons: On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.
Answer
  1. Griffith obtained live S-strain bacteria from the blood of the dead mice.
  2. In a mixture of live R-bacteria and heat killed S-bacteria, live R-strain bacteria picked up something (transforming principle) from the heat-killed S bacterium and got changed into S-type.
  3. Transforming principle allowed R-type bacteria to synthesize capsule and thus they became virulent.
  4. Hence, on injecting a mixture of heat-killed S bacteria and live R bacteria, the mice died.
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Question 313 Marks
Give Reasons: Initially proteins (and not DNA) were considered as genetic material.
Answer
  1. Proteins are large, complex molecules and store information required to govern cell metabolism. Hence it was assumed that variations found in species were caused by proteins.
  2. On the other hand, DNA was considered as a small, simple molecule whose composition does not vary much among species.
  3. Variations in the DNA molecules are different than the variation in shape, electrical charge and function shown by proteins.
  4. Hence, initially proteins (and not DNA) were considered as genetic material.
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Question 323 Marks
Eukaryotic DNA is condensed and supercoiled.
Answer
  1. In a typical mammalian cell, length of DNA double helix is approximately 2.2 metres.
  2. The size of typical nucleus is approximately 10-6 m
  3. Such a long DNA molecule has to be fitted in small nuclear space.
  4. Therefore, DNA is highly condensed, coiled and supercoiled so that it can be accommodated in the nucleus.
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Question 333 Marks
DNA in prokaryotic and eukaryotic cells.
Answer

DNA in prokaryotes

DNA in eukaryotic

1. It is present in the cytoplasm.

1. It is present in the nucleus.

2. It is not associated with histones.

2. It is associated with histones.

3. It is circular.

3. It is linear.

4. Genes do not contain introns.

4. Genes contain introns along with exons.

5. Genes are polycostronic.

5. Genes are monocistronic.

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Question 343 Marks
Euchromatin and Heterochromatin.
Answer

Euchromatin

Heterochromatin.

1. Euchromatin is loosely packed region of the chromatain.

1. Heterochromatin is densely packed region of the chromatin.

2. Euchromatin stains lightly.

2. Heterochromatin stains darkly.

3. Euchromatin is transcriptionally active region of the chromatin.

3. Heterochromatin is transcriptionally inactive region of the chromatin.

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