Chapter 6 Molecular Basis of Inheritance Solutions
Question - 1 : - Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine
Answer - 1 : -
Nitrogenous bases present in the list are adenine, thymine, uracil, and cytosine.
Nucleosides present in the list are cytidine and guanosine.
Question - 2 : - If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer - 2 : -
According to Chargaff’s rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
% A = % T and % G = % C
If dsDNA has 20% of cytosine, then according to the law, it would have 20% of guanine.
Thus, percentage of G + C content = 40%
The remaining 60% represents both A + T molecule. Since adenine and guanine are always present in equal numbers, the percentage of adenine molecule is 30%.
Question - 3 : - If the sequence of one strand of DNA is written as follows:
Answer - 3 : - 5'-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of complementary strand in 5'→3' direction
Answer
The DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is
5'- ATGCATGCATGCATGCATGCATGCATGC − 3’
Then, the sequence of complementary strand in direction will be
3'- TACGTACGTACGTACGTACGTACGTACG − 5’
Therefore, the sequence of nucleotides on DNA polypeptide in direction is
5'- GCATGCATGCATGCATGCATGCATGCAT− 3’
If the sequence of the coding strand in a transcription unit is written as follows:
Answer - 4 : -
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of mRNA.
Answer
If the coding strand in a transcription unit is
5’− ATGCATGCATGCATGCATGCATGCATGC-3’
Then, the template strand in 3’ to 5’ direction would be
3’ − TACGTACGTACGTACGTACGTACGTACG-5’
It is known that the sequence of mRNA is same as the coding strand of DNA.
However, in RNA, thymine is replaced by uracil.
Hence, the sequence of mRNA will be
5’ − AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’
Question - 5 : - Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer - 5 : -
Watson and Crick observed that the two strands of DNA are anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesized daughter strand.
Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication.
Question - 6 : - Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids
Answer - 6 : - synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer
There are two different types of nucleic acid polymerases.
(1) DNA-dependent DNA polymerases
(2) DNA-dependent RNA polymerases
The DNA-dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesizing RNA.
Question - 7 : - How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer - 7 : -
Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage.
They grew some bacteriophages on a medium containing radioactive phosphorus (32P) to identify DNA and some on a medium containing radioactive sulphur (35S) to identify protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated from the bacterial cell by blending and then subjected to the process of centrifugation.
Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.
Question - 8 : - Differentiate between the followings:
Answer - 8 : -
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand
Answer
(a) RepetitiveDNA and satellite DNA
| Repetitive DNA | Satellite DNA |
| 1. | Repetitive DNA are DNA sequences that contain small segments, which are repeated many times. | Satellite DNA are DNA sequences that contain highly repetitive DNA. |
| | |
(b) mRNA and tRNA
| mRNA | tRNA |
| 1. | mRNA or messenger RNA acts as a template for the process of transcription. | tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide. |
| 2. | It is a linear molecule. | It has clover leaf shape. |
(c) Template strand and coding strand
| Template strand | Coding strand |
| 1. | Template strand of DNA acts as a template for the synthesis of mRNA during transcription. | Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA). |
| 2. | It runs from 3’ to 5’. | It runs from 5’to 3’. |
Question - 9 : - List two essential roles of ribosome during translation.
Answer - 9 : -
The important functions of ribosome during translation are as follows.
(a) Ribosome acts as the site where protein synthesis takes place from individual amino acids. It is made up of two subunits.
The smaller subunit comes in contact with mRNA and forms a protein synthesizing complex whereas the larger subunit acts as an amino acid binding site.
(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s r-RNA in bacteria acts as a ribozyme.
Question - 10 : - In the medium where E. coli was growing, lactose was added, which induced the lac operon.
Answer - 10 : - Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer
Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolize lactose into glucose and galactose.
In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolized into glucose and galactose.
After sometime, when the level of inducer decreases as it is completely metabolized by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.