EXERCISES

Q1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:

Q2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer: This calculation is based on Chargaff’s rule, which states that for double-stranded DNA, the ratio between Adenine (A) and Thymine (T) and Guanine (G) and Cytosine (C) is constant and equals one.

Given: Cytosine (C) = 20%
Therefore, Guanine (G) = 20%
Total G + C = 40%
The remaining percentage (A + T) = 100% – 40% = 60%
Since A = T, the percentage of Adenine (A) = 60% / 2 = 30%

Hence, the percentage of Adenine in the DNA is 30%.

Q3. If the sequence of one strand of DNA is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′, write down the sequence of complementary strand in 5′ → 3′ direction.
Answer: The complementary strand must be anti-parallel (3′ → 5′) and follow the base pairing rule (A with T, G with C).

Given Strand: 5′-A T G C A T G C A T G C A T G C A T G C A T G C A T G C-3′
Complementary Strand (Natural Polarity): 3′-T A C G T A C G T A C G T A C G T A C G T A C G T A C G-5′
Result (in 5′ → 3′ direction):
5′-G C A T G C A T G C A T G C A T G C A T G C A T G C A T-3′

Q4. If the sequence of the coding strand in a transcription unit is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′, write down the sequence of mRNA.
Answer: The coding strand has the same sequence as RNA (because it has the 5′ → 3′ polarity), except that Uracil (U) is present in RNA at the place of Thymine (T).

The sequence of the mRNA is:
5′-A U G C A U G C A U G C A U G C A U G C A U G C A U G C-3′

Q5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer: The property of the DNA double helix that led Watson and Crick to propose the semi-conservative mode of replication was base pairing and the resulting complementarity between the two polynucleotide strands.

Because the two strands are complementary, knowing the base sequence of one automatically determines that of the other. Thus, each parental strand can serve as a template for the formation of a new complementary strand. After replication, each daughter DNA molecule contains one parental and one newly synthesised strand, making the process semi-conservative. This model of replication was later experimentally proved by Meselson and Stahl (1958) in E. coli.

Q6. 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.
Answer: Depending on the type of template (DNA or RNA) and the kind of nucleic acid formed (DNA or RNA), nucleic acid polymerases are of two main types.

1. DNA-dependent DNA polymerase uses a DNA template to synthesise a new DNA strand during the process of replication.
2. DNA-dependent RNA polymerase uses a DNA template to synthesise RNA during transcription. In bacteria, a single RNA polymerase enzyme is responsible for forming all types of RNA such as mRNA, tRNA, and rRNA. In eukaryotes, there are three different RNA polymerases: RNA Polymerase I forms rRNA, RNA Polymerase II forms mRNA, and RNA Polymerase III forms tRNA and 5S rRNA.

Q7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer: Hershey and Chase proved that DNA is the genetic material by using radioactive isotopes to distinguish DNA from protein. They grew bacteriophages in two different media — one containing radioactive phosphorus (³²P), which labels DNA, and another containing radioactive sulfur (³⁵S), which labels protein. When these labelled viruses infected bacteria, they observed that radioactivity from ³²P entered the bacterial cells, showing that DNA had entered, while radioactivity from ³⁵S remained outside, showing that proteins did not enter. From this, they concluded that DNA, not protein, carries the genetic information.

Q8. Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand
Answer:

(a) Repetitive DNA and Satellite DNA: The difference between Repetitive DNA and Satellite DNA are-

1. Repetitive DNA consists of short DNA sequences repeated many times in the genome, whereas satellite DNA forms distinct bands separated from the main DNA during density gradient centrifugation.
2. Repetitive DNA includes microsatellites and minisatellites, while satellite DNA is classified based on base composition, sequence length, and repeat number.
3. Repetitive DNA generally does not code for proteins, whereas satellite DNA is highly polymorphic and forms the basis of DNA fingerprinting.

(b) mRNA and tRNA: The difference between mRNA and tRNA are-

1. mRNA acts as a template carrying genetic information for protein synthesis, whereas tRNA serves as an adapter that brings amino acids to the ribosome.
2. The structure of mRNA contains codons read continuously during translation, while tRNA has an anticodon loop complementary to the mRNA codon and an amino acid binding site.
3. mRNA determines the sequence of amino acids in a protein, whereas tRNA ensures the correct amino acid is added to the growing chain.

(c) Template Strand and Coding Strand: The difference between Template Strand and Coding Strand are-

1. The template strand has 3′ → 5′ polarity, whereas the coding strand has 5′ → 3′ polarity.
2. The template strand serves as the actual template for RNA synthesis, while the coding strand is not transcribed.
3. The RNA sequence is complementary to the template strand, whereas it is identical to the coding strand except that thymine (T) is replaced by uracil (U).

Q9. List two essential roles of ribosome during translation.
Answer: The two essential roles of ribosome during translation are-

1. Platform for Synthesis: Ribosome binds to the mRNA and provides a platform for joining amino acids. The large subunit has two sites for subsequent amino acids to bind closely for peptide bond formation.
2. Catalytic Role: The ribosome acts as a catalyst for peptide bond formation. Specifically, 23S rRNA in bacteria acts as a ribozyme (RNA enzyme) for this function.

Q10. 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?
Answer: The lac operon synthesizes the enzyme β-galactosidase which hydrolyzes lactose into galactose and glucose.

When lactose (the inducer) is present, the operon is active.
As the bacteria consume lactose, its concentration decreases.
Once lactose is depleted, the repressor protein becomes active again, binds to the operator region, and prevents RNA polymerase from transcribing the operon. Hence, the lac operon shuts down when lactose is used up.

Q11. Explain (in one or two lines) the function of the following:
(a) Promoter
(b) tRNA
(c) Exons
Answer:

• (a) Promoter: It is a DNA sequence that provides the binding site for RNA polymerase. Its activity determines the rate of transcriptional initiation.
• (b) tRNA: Functions as an adapter molecule that brings amino acids and reads the genetic code by forming complementary base pairs with mRNA codons through its anticodon loop.
• (c) Exons: Exons are coding (expressed) sequences in split genes of eukaryotes. They appear in the mature or processed RNA.

Q12. Why is the Human Genome Project called a mega project?
Answer: The Human Genome Project (HGP) was called a mega project due to its enormous scale and requirements.

1. Size and Cost: The human genome has approximately 3 billion base pairs, requiring an estimated cost of US $9 billion.
2. Data Volume: Storing the sequence of a single human cell would require about 3300 books of 1000 pages each.
3. Computational Need: The massive data led to the development of bioinformatics for data storage, retrieval, and analysis.

Q13. What is DNA fingerprinting? Mention its application.
Answer: DNA Fingerprinting is a quick method to compare DNA sequences of individuals by identifying variations in repetitive DNA regions.

Applications od DNA Fingerprinting are:

1. Forensic Identification: Helps match DNA from individuals in criminal investigations.
2. Paternity Testing: Used to determine biological parentage.
3. Genetic Diversity: Helps in studying population and genetic diversity.

Q14. Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Answer:

(a) Transcription: It is the process of copying genetic information from one strand of DNA into RNA. Only a segment of DNA is copied, and RNA polymerase binds to the promoter, elongates the strand, and stops at the terminator region.
(b) Polymorphism: Refers to variation at the genetic level. When more than one allele occurs in a population at a frequency >0.01, it is called DNA polymorphism — an inheritable mutation found commonly in a population.
(c) Translation: The process of polymerising amino acids to form a polypeptide. The order of amino acids is determined by the sequence of bases in mRNA. It occurs at ribosomes where peptide bonds form between amino acids.
(d) Bioinformatics: A branch of biology that developed with the Human Genome Project; it uses computational tools for storage, retrieval, and analysis of genetic data.