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NCERT Solutions · Class 12 Biology Molecular Basis of Inheritance

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Exercises 5.1–5.10 (part 1 of 2)

  1. Exercise 5.1

    Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

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    Nitrogenous bases: Adenine, Thymine, Uracil, Cytosine.
    Nucleosides: Cytidine, Guanosine.
    A nucleoside is a nitrogenous base linked to the OH of the $\displaystyle 1$' C of a pentose sugar through an N-glycosidic linkage — so a nucleoside is base + sugar, with no phosphate.
    The name is the clue: cytosine becomes cytidine and guanine becomes guanosine once the sugar is attached.
    Adenine and Guanine are purines; Cytosine, Uracil and Thymine are pyrimidines (Uracil replaces Thymine in RNA).

  2. Exercise 5.2

    If a double stranded DNA has $\displaystyle 20$ per cent of cytosine, calculate the per cent of adenine in the DNA.

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    Adenine = $\displaystyle 30$ per cent.
    In double-stranded DNA the bases pair A with T and G with C, so Chargaff's rule gives \(\displaystyle A = T\) and \(\displaystyle G = C\).
    Cytosine is $\displaystyle 20$ per cent, therefore Guanine is also $\displaystyle 20$ per cent.
    Together \(\displaystyle C + G = 40\%\), leaving \(\displaystyle A + T = 100 - 40 = 60\%\).
    Since \(\displaystyle A = T\), \(\displaystyle A = 60/2 = 30\%\).

  3. Exercise 5.3

    If the sequence of one strand of DNA is written as follows: $\displaystyle 5$'-ATGCATGCATGCATGCATGCATGCATGC-$\displaystyle 3$' Write down the sequence of complementary strand in $\displaystyle 5$'→$\displaystyle 3$' direction.

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    Complementary strand: $\displaystyle 5$'-GCATGCATGCATGCATGCATGCATGCAT-$\displaystyle 3$'
    Pairing base by base against the given strand (A with T, G with C) gives $\displaystyle 3$'-TACGTACGTACGTACGTACGTACGTACG-$\displaystyle 5$'.
    The two chains have anti-parallel polarity, so writing that same strand in the $\displaystyle 5$'→$\displaystyle 3$' direction means reading it from the opposite end — the sequence reverses to GCAT repeated seven times.
    Both strands are $\displaystyle 28$ nucleotides long, as base pairing demands.

  4. Exercise 5.4

    If the sequence of the coding strand in a transcription unit is written as follows: $\displaystyle 5$'-ATGCATGCATGCATGCATGCATGCATGC-$\displaystyle 3$' Write down the sequence of mRNA.

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    mRNA: $\displaystyle 5$'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-$\displaystyle 3$'
    The RNA is actually copied from the template strand (polarity $\displaystyle 3$'→$\displaystyle 5$'), so it turns out to be identical in sequence and polarity to the coding strand.
    The only change is chemical: uracil takes the place of thymine, so every T of the coding strand is written as U.
    Hence the coding strand can be copied out directly, T→U, without reversing it.

  5. Exercise 5.5

    Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

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    The property is the complementary base pairing between the two strands of the double helix — Adenine with Thymine (two H-bonds), Guanine with Cytosine (three H-bonds).
    Because the strands are complementary, knowing the sequence of one strand lets you predict the sequence of the other exactly.
    Watson and Crick therefore proposed that the two strands separate, and each acts as a template for the synthesis of a new complementary strand.
    The two daughter DNA molecules so formed are identical to the parental DNA molecule.
    Each daughter molecule keeps one parental strand and one newly synthesised strand — this is why the scheme is called semiconservative replication.
    Watson and Crick themselves remarked that the specific pairing they had postulated at once suggested a copying mechanism for the genetic material.

  6. Exercise 5.6

    Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.

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    Four types, one for each template–product combination:
    DNA-dependent DNA polymerase — DNA template, DNA product; catalyses replication (in \(\displaystyle 5' \to 3'\) direction only).
    DNA-dependent RNA polymerase — DNA template, RNA product; catalyses transcription.
    RNA-dependent RNA polymerase — RNA template, RNA product; replicates the genome of RNA viruses such as Tobacco Mosaic Virus and QB bacteriophage.
    RNA-dependent DNA polymerase (reverse transcriptase) — RNA template, DNA product; carries out reverse transcription, the reverse flow of information (RNA → DNA) seen in some viruses.
    The naming convention follows the chapter: the polymerase is named for the template it reads and the nucleic acid it makes.

  7. Exercise 5.7

    How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

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    They used radioactive labels that can enter only one of the two molecules: DNA contains phosphorus but no sulfur, while protein contains sulfur but no phosphorus.
    One batch of bacteriophages was grown on a medium containing radioactive phosphorus — these viruses had radioactive DNA but non-radioactive protein.
    Another batch was grown on a medium containing radioactive sulfur — these had a radioactive protein coat but non-radioactive DNA.
    Both sets of labelled phages were allowed to attach to and infect E. coli.
    The empty viral coats were then stripped off the bacteria by agitating them in a blender, and the virus particles were separated from the bacteria by centrifugation.
    Bacteria infected by the phosphorus-labelled phages were radioactive; bacteria infected by the sulfur-labelled phages were not.
    Conclusion: only DNA passed from the virus into the bacterium, so DNA — not protein — is the genetic material.

  8. Exercise 5.8

    Differentiate between the followings:
    (a)
    Repetitive DNA and Satellite DNA
    (b)
    mRNA and tRNA
    (c)
    Template strand and Coding strand

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    (a) Repetitive DNA and Satellite DNA
    Repetitive DNA is any stretch of DNA sequence repeated many times — sometimes hundred to thousand times; satellite DNA is that fraction of repetitive DNA which separates out as small peaks away from the bulk DNA during density gradient centrifugation.
    Repetitive DNA makes up a very large portion of the human genome and is thought to have no direct coding function, though it sheds light on chromosome structure, dynamics and evolution.
    Satellite DNA is classified by base composition (A:T rich or G:C rich), length of segment and number of repeat units into micro-satellites, mini-satellites, etc.
    Satellite DNA shows a very high degree of polymorphism and forms the basis of DNA fingerprinting; repetitive DNA as a whole is the broader category to which it belongs.
    (b) mRNA and tRNA
    mRNA provides the template for protein synthesis; tRNA brings amino acids and reads the genetic code.
    mRNA carries the message as triplet codons read contiguously, flanked by a start codon (AUG) and a stop codon, with untranslated regions (UTRs) at both the $\displaystyle 5$'- and $\displaystyle 3$'-ends.
    tRNA is the adapter molecule: it has an anticodon loop whose bases are complementary to the codon, and an amino acid acceptor end.
    There are specific tRNAs for each amino acid plus a special initiator tRNA, and no tRNA for stop codons.
    tRNA looks like a clover-leaf in secondary structure but is a compact inverted L in its actual shape.
    (c) Template strand and Coding strand
    The template strand has polarity $\displaystyle 3$'→$\displaystyle 5$' and is the one actually copied into RNA; the coding strand has polarity $\displaystyle 5$'→$\displaystyle 3$' and is displaced during transcription.
    The RNA is complementary to the template strand, but has the same sequence as the coding strand except that uracil replaces thymine.
    The coding strand does not code for anything, yet all reference points while defining a transcription unit are made with respect to it.
    Which strand is which is fixed by the position of the promoter: by switching the promoter's position with the terminator, the definitions of template and coding strand would be reversed.

  9. Exercise 5.9

    List two essential roles of ribosome during translation.

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    It provides the platform for joining amino acids. The large subunit has two sites at which successive aminoacyl-tRNAs bind, holding the two amino acids close enough for a peptide bond to form; the small subunit binds the mRNA and the ribosome then moves codon to codon along it.
    It acts as a catalyst for peptide bond formation. In bacteria the 23S rRNA of the large subunit is the enzyme — a ribozyme — that catalyses the reaction.
    (The ribosome is the cellular factory for protein synthesis: structural RNAs plus about $\displaystyle 80$ proteins, existing as a large and a small subunit that come together on mRNA.)

  10. Exercise 5.10

    In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?

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    Because the inducer itself gets used up.
    Once the operon is switched on, the z gene makes beta-galactosidase, which hydrolyses lactose into galactose and glucose for use as an energy source.
    As lactose is consumed, its concentration falls, so there is no longer enough inducer (lactose/allolactose) to bind the repressor.
    The repressor protein is made all the time (constitutively) from the i gene; it is inactive only while it is bound to the inducer.
    With the inducer gone, free active repressor binds the operator region again and prevents RNA polymerase from transcribing the structural genes z, y and a.
    The operon is therefore shut down — an example of negative regulation, and of regulation of enzyme synthesis by its own substrate.