Biology Molecular Basis of Inheritance Chapter 6 – Exercise

Q.1 Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Ans. Nitrogenous bases: Adenine, Thymine, Uracil, Cytosine

Nucleosides: Cytidine, Guanosine

Q.2 If a double stranded DNA has 20% of cytosines, calculate the percent of adenine in the DNA.

Ans. In a DNA molecule, the number of cytosine molecules is equal to guanine molecules and the number of adenine molecules are equal to Thymine molecules. Thus,

If a double stranded DNA has 20% of cytosine, it has 20% guanine. Thus, cytosine plus guanine make 40% of the total bases. The remaining 60% includes both adenine and thymine which are in equal amounts. Therefore, the percent of adenine is 30%.

Q.3 If the sequence of one strand of DNA is written as follows:


Write down the sequence of complementary strand in 5’→3′ direction

Q.4 If the sequence of coding strand in a transcription unit is written as follows:


Write down the sequence of mRNA.

Ans. If the coding strand in a transcription unit is – 5′-ATGCATGCATGCATGCATGCATGCATGC-3′. Then the template strand will be 3′- TACGTACGTACGTACGTACGTACGTACG − 5’ and the sequence of bases in mRNA will be-


Q.5 Which property of DNA double helix led Watson and crick to hypothesize semiconservative mode of DNA replication, explain.
Ans. Watson and Crick suggested that the two strands of double helix DNA molecule uncoil and separate and each of the two strands serves as a template for the synthesis of a new (complementary) strand alongside it. The template (i.e., parental strand) and its complement (i.e., the daughter strand) then form a new DNA double strand, identical to the original DNA molecule. The sequence of bases which should be present in the new strands can be easily predicted because these would be complementary to the bases present in the old strands. A will pair with T, T with A, C with G and G with C. Thus two daughter DNA molecules are formed from the parent DNA molecule and these are identical to the parent molecule. Since each daughter DNA molecule consists of one old (parental) strand and one new strand, this mode of replication is said to be semiconservative (i.e., the new DNA molecule has conserved half of the parent molecule).
Q.6 Depending upon the chemical nature of template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.

Ans.  Types of nucleic acid polymerases are:

  1. DNA-dependent DNA polymerase uses a DNA template for synthesis of DNA
  2. DNA-dependent RNA polymerase uses a DNA template for synthesis of RNA
  3. RNA dependent DNA polymerase uses RNA template for synthesis of DNA (reverse transcription)
Q.7 How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Ans. Hershey and Chase used different radioactive isotopes to label DNA and protein. They used radioactive sulphur (35S) to identify protein and radioactive phosphorus (32P) to identify nucleic acids.
Q.8 Differentiate between the following:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

(c) Template strand and Coding strand

Ans. (a) Difference between Repetitive DNA and Satellite DNA

Repetitive DNA Satellite DNA
A DNA molecule consists of a large number of base pairs arranged in a definite sequence. Some of these sequences are coding, called exons. Others are non-coding called introns. There are some specific regions in DNA molecules where a small stretch of non-coding DNA sequence is repeated several hundred or thousand times. The stretch of DNA which is repeated several times is called repetitive DNA. These repetitive DNA stretches are found at many places throughout the length of DNA and are specific in each individual. The repetitive DNA, when separated from bulk genomic DNA by the technique of density gradient centrifugation, form minor peaks whereas the bulk DNA forms the major peak. The minor peaks are referred to as satellite DNA. Thus, satellite DNA represents highly repetitive nature of small stretches of DNA having distinctive base composition. The so called satellite bands, quite separate from the major band of bulk DNA, have proved a useful genetic marker in DNA finger printing.

(b) Difference between mRNA and tRNA

  1. mRNA is the longest and linear in shape
  2. mRNa carry information from DNA for protein synthesis
  3. They are numerous in number
  1. tRNa is the shortest and clover leaf-like, folded into L-shape
  2. They carry amino acids to mRNA codons for protein synthesis
  3. They are about 60 in number

(c) Difference between Template strand and Coding strand

They are two strands in a DNA double helix. One strand, which functions as template for RNA synthesis is known as template strand or sense strand. The complementary strand of template strand is called coding strand or antisense strand.

Q.9 List two essential roles of ribosome during translation.

Ans. These are:

  1. Ribosome provides the site for protein synthesis. The mRNA comes in contact with small sub-unit of ribosome and then protein synthesizing complex is formed. The large sub-unit provides sites for amino acid binding.
  2. Ribosome also acts as a catalyst for the formation of peptide bond.
Q.10 In the medium where E.coli was growing, lactose was added, which induced the lac-operon. But why does lac-operon shut down after some time after addition of lactose in the medium?
Ans. Lactose acts as an ‘inducer’ by switching on and off the operon. If lactose is present in the bacterial culture, it inactivates the repressor and induces transcription for the synthesis of enzyme beta-galactosidase. This enzyme acts on lactose so that the lactose is utilized. Absence of lactose (or inducer) causes synthesis of repressor mRNA that binds to the operator region and stops transcription. In this way the lac-operon is shut down.
Q.11 Explain (in one or two lines) the function of following : (a) Promoter (b) tRNa (c) Exons

Ans. (a) Promoter: Promoter gene lies adjacent to the operator gene and marks the site where the RNA polymerase enzyme binds.

(b) tRNA: tRNA molecules read the code and link it to the amino acids during protein synthesis.

(c) Exons: In split-gene arrangement of eukaryotic mRNA, the sequence of base that appear in processed RNA are called exons. They are interrupted by introns in unprocessed RNA

Q.12 Why is the Human Genome Project called a mega project?
Ans. Human Genome Project (HGP) was called a mega project because it had involved a lot of money, most advanced technologies, numerous computers, many scientists and a long span of time. The magnitude of the project can be imagined by visualizing that it was aimed to find out the complete DNA sequence of human genome which is said to have approximately 3 × 109 bp. The cost of the project can be imagined that if the cost of sequencing a bp is 3 dollars, sequencing of 3 × 109 bp would be a billion dollars. Further, if the data is to be stored in books, with each book having 1000 pages and each page having 1000 letters, some 3300 books will be required. Thus, high speed computational devices were used for storing and analysing the data.
Q.13 What is DNA fingerprinting? Mention its application.

Ans. DNA fingerprinting is a technique used to identify and analyze the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.


(1) It is used in forensic science to identify potential crime suspects.

(2) It is used to establish paternity and family relationships.

(3) It is used to identify and protect the commercial varieties of crops and livestock.

(4) It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

Q.14 Briefly describe the following:

(a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics

(a) Transcription: Transcription is the process in which RNA is synthesized from DNA.
Transcription is a 3 step process-
1) Initiation- RNA polymerase binds to DNA sequence called the promoter and providing the single-stranded template for transcription process.
2) Elongation- RNA polymerase reads the template strand and adds bases leading to elongation of the chain.
3) Termination- In this step terminator sequences mark the end of transcription.

(b) Polymorphism: If an inheritable mutation is observed in a population at high frequency, it is called DNA polymorphism.

(c) Translation:Translation is the process in which protein is synthesized from RNA.
Translation is a 3 step process-
1) Initiation- In initiation, the ribosome assembles around the mRNA that has to be read and the first tRNA carrying the amino acid methionine
2) Elongation- In this stage the amino acid chain is elongated
3) Termination- In this stage, the synthesized polypeptide chain is released

(d) Bioinformatics: It is the branch of science concerned with the management and analysis of biological information stored in databases. This branch is a recently developed science using information to understand biological phenomenon. It has developed after establishment of genetic engineering techniques, introductive of automated protein and DNA sequencing technologies and use of computers to store enormous data which could be easily accessed remotely.

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