NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance
NCERT solutions for class 12 biology chapter 6 Molecular Basis of Inheritance:
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Are you in search of NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance? Your search ends here as we provides detailed solutions to each chapter of NCERT Solutions for Class 12 Biology that are in accordance with the most recent CBSE Syllabus 2022–23. Do you know, what acts as a genetic material of living organisms?
The answer is DNA, in solutions of the molecular basis of inheritance ncert, 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.
Students can also access NCERT Solutions for Class 12 for Physics, Mathematics, Chemistry, and Biology. 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.
Also Read,
- NCERT Notes for class 12 biology chapter 6 Molecular Basis of Inheritance
- NCERT Exemplar for class 12 biology chapter 6 Molecular Basis of Inheritance
NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance: Solved Exercise Questions
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:
By Chargaff's rule, we know that A=T and G=C. Since there are 20% cytosines, so guanines percentage will also be 20%. The G+C count is 20+20= 40%. Therefore, the remaining 60% will be contributed equally by adenine and thymine. Thus, 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 complementary strand in direction.
Answer:
With regard to base pairs the DNA strand are complementary to each other. So, if the sequence of DNA is
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
The sequence of the complementary strand will be
3`- TACGTACGTACGTACGTACGTACGTACG - 5`.
In 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'.
Answer:
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.
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 a DNA as a template to synthesise new strand of DNA.
2. DNA dependent RNA polymerase- It uses DNA as a template to synthesise RNA.
Answer:
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.
Q8. Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
Answer:
Repetitive DNA | Satellite DNA |
Repetitive DNA refers to DNA sequences containing small segments that are repeated many times | DNA sequences containing highly repetitive DNA. |
Answer:
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
Answer:
Template strand | 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.
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. 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.
Answer:
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.
Q11. Explain (in one or two lines) the function of the followings:
Answer:
Function of promoter
Promoter gene refers to the site where the RNA polymerase enzyme binds and transcription of mRNA starts.
Answer:
Function of tRNA
tRNA plays a major role in the process of protein synthesis. It reads the genetic code present on mRNA.
Q12. Why is the Human Genome project called a mega project?
Answer:
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 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:
Answer:
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.
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 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.
14 . (c) Translation
Answer:
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.
Answer:
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.
Important topics of solutions for NCERT class 12 biology chapter 6 molecular basis of inheritance:
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 Replication
6.4.1 The Experimental Proof
6.4.2 The Machinery and the Enzymes
6.5 Transcription
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
6.7 Translation
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
NCERT Solutions for Class 12 Biology Chapter 6 – Molecular Basis of Inheritance
Chapter 6 Biology Class 12 NCERT solutions are well-structured in accordance with the CBSE syllabus for 2022-23. Solutions for the molecular basis of inheritance ncert contains most of the answers to 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.
In the molecular basis of inheritance ncert, 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 the molecular basis of inheritance ncert, you will get an 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 molecular basis of inheritance ncert, you will get solutions to the questions based on two important nucleic acid which is 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 molecular basis of inheritance ncert pdf, you will get solutions to the questions on Central dogma which consists of three important things that are:
- DNA
- mRNA
- Protein
After going through solutions for the molecular basis of inheritance pdf, you will be able to answer all the questions which are given at the end of this chapter.
Highlights of NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance
class 12 biology chapter 6 NCERT solutions highlights are given below:
- The answers are written in the best possible way using simple language that makes solutions of the molecular basis of inheritance class 12 NCERT easier to read and understand
- In the NCERT molecular basis of inheritance, Points are used to frame answers to help understand quickly.
- Solution content for chapter 6 biology class 12 is derived from the textbook by the subject expert.
- Molecular basis of inheritance class 12 NCERT solutions are as per the latest CBSE Syllabus and guidelines.
- Molecular basis of inheritance class 12 NCERT pdf download links are readily available and easily accessible for free.
- Important topics are listed in the molecular basis of inheritance class 12 pdf.
And if still, you have any doubt or problems in getting the correct answers then the molecular basis of inheritance ncert solutions will help you. But don't forget to try at least once, so that you can answer those questions by yourself then check your answers with the help of solutions for NCERT class 12 biology chapter 6 molecular basis of inheritance.
Also Check NCERT Books and NCERT Syllabus here:
- NCERT Books Class 12 Biology
- NCERT Syllabus Class 12 Biology
- NCERT Books Class 12
- NCERT Syllabus Class 12
NCERT Solutions for Class 12 Biology- Chapter wise
Chapter 1 | Reproduction in organisms |
Chapter 2 | Sexual Reproduction in Flowering Plants |
Chapter 3 | Human Reproduction |
Chapter 4 | Reproductive Health |
Chapter 5 | Principles of Inheritance and Variation |
Chapter 6 | Molecular basis of inheritance |
Chapter 7 | Evolution |
Chapter 8 | Human Health and Disease |
Chapter 9 | Strategies for Enhancement in Food Production |
Chapter 10 | Microbes in Human Welfare |
Chapter 11 | Biotechnology: Principles and Processes |
Chapter 12 | Biotechnology and its Applications |
Chapter 13 | Organisms and Populations |
Chapter 14 | Ecosystem |
Chapter 15 | Biodiversity and Conservation |
Chapter 16 | Environmental Issues |
NCERT Solutions for Class 12- Subject wise
NCERT solutions for Class 12 Maths |
NCERT solutions for Class 12 Chemistry |
NCERT solutions for Class 12 Biology |
NCERT solutions for Class 12 Physics |
NCERT Exemplar Class 12 Solutions
NCERT Exemplar Class 12 Chemistry Solutions |
NCERT Exemplar Class 12 Mathematics Solutions |
NCERT Exemplar Class 12 Biology Solutions |
NCERT Exemplar Class 12 Physics Solutions |
Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance
Question: Can I score full marks by practising the NCERT Solutions for Class 12 Biology Chapter 6?
Answer:
Yes, by practising the NCERT Solutions for Class 12 Biology Chapter 6, you can earn all of the points. The following is a list of the reasons why the solutions PDF are significant.
1. The answers help students confidently prepare for the Class 12 board exams.
2. Students can easily understand the concepts thanks to the step-by-step explanations provided for each question.
3. Each concept's diagrams encourage visual learning, which is crucial for exam success.
4. In keeping with the standards of the CBSE Board exams, the solutions produced are concise and clear.
Question: How to download solutions for molecular basis of inheritance ncert?
Answer:
To download solutions for the molecular basis of inheritance ncert, students can use the online webpage to pdf converter tools. To Score Well in the examination, follow the NCERT syllabus and exercise given in the NCERT Book. To practice more problems, students can refer to the NCERT exemplar
Question: What are the applications of DNA fingerprinting given in molecular basis of inheritance ncert?
Answer:
The applications of DNA fingerprinting given in molecular basis of inheritance ncert are:
- It helps in forensic science for the identification of crime suspects.
- It helps in identifying paternity and family relationships.
- It is also used in the identification and protection of the commercial varieties of crops and livestock.
- It is also used to finding the evolutionary history of any organism.
Question: What are the two essential roles of ribosome during translation?
Answer:
- mRNA bind on the surface of the ribosome on the groove of the smaller subunit of the ribosome
- It helps in joining amino acids by forming peptide bonds on it's 'P' site.
Question: What are the important topics of solutions for molecular basis of inheritance ncert?
Answer:
The important topics of solutions for the molecular basis of inheritance ncert are:
- DNA
- Structure of Polynucleotide Chain
- Packaging of DNA Helix
- Genetic Material is DNA
- Properties of Genetic Material (DNA versus RNA)
- RNA World
- Replication
- The Experimental Proof
- The Machinery and the Enzymes
- Transcription
- Transcription Unit
- Genetic Code
- Translation
- Regulation of Gene Expression
- Human Genome Project
- DNA Fingerprinting
Question: What are the important components of central dogma given in molecular basis of inheritance ncert?
Answer:
Important components of central dogma are:
- DNA
- mRNA
- Protein
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Questions related to CBSE Class 12th
i couldnt clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th
hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)
Hello dear,
If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.
As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.
Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.
Believe in Yourself! You can make anything happen
All the very best.
if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?
Hello Student,
I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.
You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.
All the best.
I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me
Hi,
You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.
I will have my boards in Feb 2023 (Class 12) and I still havent completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?
If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.