NCERT solutions for class 12 biology chapter 6 Molecular Basis of Inheritance: Do you know, what acts as a genetic material of living organisms? The answer is DNA, in solutions of NCERT class 12 biology chapter 6 molecular basis of inheritance, you will learn some important facts about DNA such that it acts as the genetic material in most of the organisms. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acids found in living systems. RNA though it also acts as genetic material in some viruses, mostly functions as a messenger. RNA has additional roles as well. It functions as an adapter, it is structural, and in some cases it acts as a catalytic molecule. CBSE NCERT solutions for class 12 biology chapter 6 molecular basis of inheritance will give you answers to the questions based on our molecular basis of inheritance. If you want solutions for any other class (6-12), then NCERT solutions are there for you.
NCERT solutions for class 12 biology chapter 6 molecular basis of inheritance contains most of the answers on DNA and RNA related questions. You will learn that DNA stores genetic information, RNA mostly helps in the transfer and expression of information. Though DNA and RNA both function as genetic material, DNA being chemically and structurally more stable is a better genetic material. However, RNA is the first to evolve and DNA was derived from RNA.
Here are the important topics of solutions for NCERT class 12 biology chapter 6 molecular basis of inheritance given below:
6.1 The DNA
6.1.1 Structure of Polynucleotide Chain
6.1.2 Packaging of DNA Helix
6.2 The Search for Genetic Material
6.2.1 The Genetic Material is DNA
6.2.2 Properties of Genetic Material (DNA versus RNA)
6.3 RNA World
6.4.1 The Experimental Proof
6.4.2 The Machinery and the Enzymes
6.5.1 Transcription Unit
6.5.2 Transcription Unit and the Gene
6.5.3 Types of RNA and the process of Transcription
6.6 Genetic Code
6.6.1 Mutations and Genetic Code
6.6.2 tRNA– the Adapter Molecule
6.8 Regulation of Gene Expression
6.8.1 The Lac operon
6.9 Human Genome Project
6.9.1 Salient Features of Human Genome
6.9.2 Applications and Future Challenges
6.10 DNA Fingerprinting
In CBSE NCERT solutions for class 12 biology chapter 6 molecular basis of inheritance, you will also get solutions to the questions based on the Human genome project as it was a megaproject that aimed to sequence every base in the human genome. This project has yielded much new information among us. Many new areas and avenues have opened up as a consequence of the project. In NCERT solutions for class 12 biology chapter 6 molecular basis of inheritance, you will get explanation of DNA Fingerprinting. It is a technique to find out variations in individuals of a population at the DNA level. It works on the principle of polymorphism in DNA sequences.
In solutions for NCERT class 12 biology chapter 6 molecular basis of inheritance, you will get solutions to the questions based on two important nucleic acid which are also called as genetic materials for living organisms and these are:
- Deoxyribonucleic acid (DNA)
- Ribonucleic acid (RNA)
DNA and RNA are the two types of nucleic acids found in living systems whereas DNA is double-stranded and RNA is single-stranded, DNA acts as the genetic material in most of the organisms and RNA acts as genetic material in some viruses, mostly functions as a messenger. In NCERT solutions for class 12 biology chapter 6 molecular basis of inheritance, you will get solutions to the questions on Central dogma which consists of three important things that are:
After going through solutions for NCERT class 12 biology chapter 6 molecular basis of inheritance you will be able to answer all the questions which are given at the end of this chapter.
NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance: Solved Exercise Questions
Nitrogenous bases- Adenine, thymine, uracil and cytosine
Nucleosides- Cytidine and guanosine
Write down the sequence of mRNA.
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 semi-conservative mode of DNAreplication? Explain.
The property of DNA double helix which led Watson and Crick to hypothesise semi-conservative mode of DNA replication is that the two strands of DNA are antiparallel and complementary to each other in terms of their base sequences. This arrangement suggests that DNA replication is semiconservative. During replication, the two strands unwind and each strand acts as a template for the synthesis of a new strand. At the end of replication, DNA of parental types and recombinant types are formed.
Q7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Alfred Hershey and Martha Chase (1952) worked with a virus that infects bacteria called bacteriophages. They used different radioactive isotopes to label DNA and protein. In their experiment, in one preparation, the protein part was made radioactive and in the other, nucleic acid (DNA) was made radioactive. These two phage preparations were allowed to infect the culture of E.coli. Soon after infection, before the lysis of cells, the E.coli cells were gently agitated in a blender, to loosen the adhering phage particles and the culture was centrifuged. The heavier infected bacterial cells pelleted to the bottom and the lighter viral particles were present in the supernatant. It was found that when bacteriophage containing radioactive DNA was used to infect E.coli, the pellet contained radioactivity. If bacteriophage containing radioactive protein coat was used to infect E.coli, the supernatant contained most of the radioactivity. Their experiment showed that protein does not enter the bacterial cell while DNA gets transformed. Hence, they proved that DNA is the genetic material.
(b) mRNA and tRNA
|mRNA or messenger RNA
||tRNA or transfer RNA
|It acts as the template for the translation of proteins.
||tRNA acts as the adapter molecule that carries specific amino acid to mRNA for the synthesis of a polypeptide.
(c) Template strand and Coding strand
|It is the template for the synthesis of mRNA during transcription
||Strand of DNA having the same sequence as mRNA (thymine in DNA is replaced by uracil in RNA
|Its polarity is 3' to 5'
||Its polarity is 5' to 3'
Q9. List two essential roles of ribosome during translation.
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. Ribosome acts as the site of protein synthesis. The larger subunit of ribosome act as an amino acid binding site while small subunit attaches to the mRNA.
2. 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 addition of lactose in the medium?
Lac operon refers to a segment of DNA which consists of one regulatory gene (i ) and three structural genes (y,z and a). Among the three structural genes, the z gene code for enzyme beta-galactosidase, that is responsible for the hydrolysis of the disaccharide lactose into its monomeric units, galactose and glucose. Gene y code for enzyme permease, which increases the permeability of the cell. The gene a encode for enzyme transacetylase. Lactose which is called the inducer is the substrate for enzyme beta-galactosidase and it regulates switching on and off of the operon. The regulatory gene, i code for the repressor of the lac operon. Lactose can bind to the repressor an inactivate it. When lactose is bound to the repressor, the RNA polymerase binds to the promoter of the lac operon. As a result of this, the transcription of three structural genes takes place and they form their respective enzymes. Further, these enzymes metabolize the lactose and lead to the formation of glucose and galactose. When the lactose metabolism is at its highest, the repressor protein is set free to bind with the operator gene. As a result of this, the transcription of lac operon stops. Therefore, lac operon shut down sometime after addition of lactose in the medium.
Function of tRNA
tRNA plays a major role in the process of protein synthesis. It reads the genetic code present on mRNA.
Exons are the coding sequences on DNA that transcribe into proteins.
Q12. Why is the Human Genome project called a mega project?
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.
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 sequence called as 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 in order 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:
Transcription- It refers to the process of copying genetic information from one strand of DNA into mRNA. In transcription, only one strand is copied into the RNA. While copying, the place of adenine is taken up by uracil. The transcription of DNA includes a transcription unit. The transcription unit is consists of a promoter, the structural gene and a terminator. The strands that have polarity 3’-5' act as a template and called template strand while the other strand is called coding strand. A schematic structure of a transcription unit is given below.
The promoter is located at the 5’ end and it binds the enzyme RNA polymerase to start transcription. Sigma factor also helps in initiation of the process of transcription.The terminator is located at 3’end of coding strand and usually defines the end of transcription where rho factor will bind to terminate transcription.
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 due to a mutation in either somatic cell or in 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.
Translation- The process of polymerisation of amino acids into a polypeptide chain is referred to as translation. The order and sequence of amino acids in a polypeptide chain is dependent upon the base sequence of mRNA. Process of translation involves three steps i.e. initiation, elongation and termination. During the initiation, the ribosome binds to the mRNA at the start codon which is AUG. Ribosomes further move from codon to codon along the mRNA for elongation of the polypeptide chain. In the end, the release factors bind to the stop codon, leading to the termination of translation and release of the polypeptide from the ribosome.
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 doubt 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.
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