NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance

Q1. Group the following as nitrogenous bases and nucleosides:

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

Answer:

Nitrogenous bases- Adenine, thymine, uracil and cytosine

Nucleosides- Cytidine and guanosine

Q2. If a double-stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

Answer:

According to Chargaff's rule, A=T and G=C. As there are 20% cytosines, the guanine percentage will also be 20%. The G+C count is 20+20= 40%. Hence, the remaining 60% will be shared equally between adenine and thymine. So, the percentage of adenine and thymine are 30% each.

Q3. If the sequence of one strand of DNA is written as follows:
5'-ATGCATGCATGCATGCATGCATGCATGC-3'

Write down the sequence of the complementary strands in 5′→3′ direction.

Answer:

About base pairs, the DNA strands are complementary to each other. So, if the sequence of DNA is

5'-ATGCATGCATGCATGCATGCATGCATGC-3'

The sequence of the complementary strands will be

3`- TACGTACGTACGTACGTACGTACGTACG - 5`.

In the 5' - 3' strand, it can be written as

5'- GCATGCATGCATGCATGCATGCATGCAT-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 sequence of mRNA is the same as the coding strand of DNA except that thymine is replaced by uracil. Thus, the sequence of mRNA will be

5' - AUGCAUGCAUGCAUGCAUGCAUGCAUGC - 3'.

Q5. Which property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication? Explain.

Answer:

The characteristic of the DNA double helix that prompted Watson and Crick to speculate a semi-conservative mode of DNA replication is that the two DNA strands are antiparallel and complementary to one another in the sense of their base sequences. It follows from this organization that DNA replication is semiconservative. When replication occurs, the two strands separate, and each strand serves as a template for the production of a new strand. By the completion of replication, parental types and recombinant types of DNA are produced.

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 upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), two types of nucleic acid polymerases are found

1. DNA-dependent DNA polymerase - It uses DNA as a template to synthesise new strands of DNA.

2. DNA-dependent RNA polymerase - It uses DNA as a template to synthesise RNA.

Q7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Answer:

Alfred Hershey and Martha Chase (1952) used a virus that infects bacteria, bacteriophages. They labeled DNA and protein with various radioactive isotopes. In their experiment, in one preparation, the protein component was rendered radioactive and in the other, nucleic acid (DNA) was rendered radioactive. These two phage preparations were permitted to infect the culture of E.coli.

As soon as after infection, prior to cell lysis, the E.coli cells were mildly shaken in a blender, loosening the attached phage particles and the culture was centrifuged. The denser infected bacterial cells were pelleted to the bottom and lighter viral particles were in the supernatant. When bacteriophage with radioactive DNA was used to infect E.coli, radioactivity was found in the pellet. If a bacteriophage with a radioactive protein coat was employed to infect E.coli, the radioactivity was present in most of the supernatant. They demonstrated in their experiment that protein does not enter the bacterial cell, whereas DNA gets transformed. Thus, they established that DNA is the genetic material.

Q8. Differentiate between the following:

(a) Repetitive DNA and Satellite DNA

Answer:

(b) mRNA and tRNA

Answer:

(c) Template strand and Coding strand

Answer:

Q9. List two essential roles of the ribosome during translation.

Answer:

The ribosome is a complex structure made up of ribonucleoproteins. It consists of two subunits i.e. a larger subunit and a smaller subunit. The essential roles of ribosomes during translation are as follows:

1. The ribosome acts as the site of protein synthesis. The larger subunit of the ribosome acts as an amino acid binding site, while the small subunit attaches to the mRNA.

2. The ribosome acts as a catalyst for forming peptide bonds.

Q10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down sometime after the addition of lactose in the medium?

Answer:

Lac operon is a piece of DNA that contains one regulatory gene (i ) and three structural genes (y,z and a). Of the three structural genes, the z gene specifies the enzyme beta-galactosidase that catalyzes the hydrolysis of the disaccharide lactose into galactose and glucose, its monomeric units.

Gene y specifies enzyme permease, which makes the cell more permeable. The gene a coded for the enzyme transacetylase. Lactose referred to as the inducer is the substrate for the enzyme beta-galactosidase and controls the turning on and off of the operon. The regulatory gene, i codes for the repressor of the lac operon.

Lactose can bind to the repressor and inactivate it. When lactose is attached to the repressor, the RNA polymerase attaches to the Lac Operon's promoter. Due to this, the transcription of three structural genes occurs and they develop their respective enzymes. In addition, these enzymes break down the lactose and result in the production of glucose and galactose. When lactose metabolism is at its maximum, repressor protein is released to bind with the operator gene. Due to this, the process of transcription of Lac Operon is arrested. Thus, Lac Operon shut down at some point after lactose addition to the medium.

Q11. Explain (in one or two lines) the function of the following:

(a)Promoter

Answer:

Function of promoter

The promoter gene refers to the site where the RNA polymerase enzyme binds and transcription of mRNA starts.

(b) tRNA

Answer:

Function of tRNA

tRNA plays a major role in the process of protein synthesis. It reads the genetic code present on mRNA.

(c) Exon

Answer:

Exons are the coding sequences on DNA that transcribe into proteins.

Q12. Why is the Human Genome project called a mega project?

Answer:

The Human Genome project is called a mega project because its specific goal is to sequence every base pair in the human genome. HGP took approximately 13 years to accomplish in 2006. The main aim of HGP was to develop new technology and generate new information in genomic studies. Because of this project, many new areas have opened in genetics. Thus HGP is a mega project.

Q13. What is DNA fingerprinting? Mention its application.

Answer:

DNA fingerprinting is a very easy and quick way to compare the DNA sequence of any two individuals. It includes identifying differences in some specific regions in DNA sequences called repetitive DNA sequences. In these regions, a small stretch of DNA is repeated many times and they are specific for every individual. The technique of fingerprinting was initially developed by Alec Jeffrey.

Applications of DNA Fingerprinting

1. It is used in forensic science to identify individuals.

2. It can be used to establish paternity or maternity-related disputes.

3. DNA fingerprinting is used to establish evolutionary relationships between organisms.

Q14. Briefly describe the following:

(a) Transcription

Answer:

It is the process of replication of genetic information from a strand of DNA into mRNA. Transcription involves only one strand that is replicated into the RNA. During replication, the adenine position is occupied by uracil. DNA transcription contains a transcription unit. The transcription unit has a promoter, the structural gene and a terminator. The 3'-5' polarity strands serve as a template and are referred to as template strands whereas the other strand is referred to as coding strands. A schematic organization of a transcription unit is represented below.

The promoter is located at the 5’ end and it binds the enzyme RNA polymerase to start transcription. The sigma factor also helps in the initiation of the process of transcription. The terminator is located at 3’end of the coding strand and usually defines the end of transcription where the rho factor will bind to terminate transcription.

(b) Polymorphism

Answer:

Polymorphism- It refers to a special kind of genetic variation in which nucleotide sequence can exist at a particular site in a DNA molecule. This heritable mutation is caused by a mutation in either the somatic cell or the germ cell. It ultimately results in the accumulation of various mutations at one site. Polymorphism brings revolution in the process of finding a location on the chromosome for disease-associated sequences and tracing human history.

14. (c) Translation

Answer:

Translation- The process of amino acid polymerisation to form a polypeptide chain is called translation. The sequence and order of the amino acids within a polypeptide chain rely on the sequence of mRNA. Translation consists of three steps, i.e. initiation, elongation and termination. In the process of initiation, the ribosome attaches at the beginning codon, which is AUG. Ribosomes then continue to travel codon by codon along the mRNA to add onto the polypeptide chain. Ultimately, the release factors recognize the stop codon, which results in the end of translation and release of the polypeptide from the ribosome.

(d) Bioinformatic

Bioinformatics- It is a recent and very effective area in the field of biology. In this discipline, the knowledge from the DNA sequences is used for solving various doubts regarding organisms which can not be studied on them in real-time. Therefore, we can say that we utilize the biological information stored in the DNA of an organism.