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NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Edited By Irshad Anwar | Updated on Sep 14, 2023 08:39 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance: Your search ends here, as we provide detailed questions and answers for the Molecular Basis of Inheritance NCERT that are in accordance with the most recent CBSE Syllabus 2022–23. Do you know what acts as the genetic material of living organisms? The answer is DNA. In order to find solutions to such questions and study for the board exams, students can use these NCERT Solutions for Class 12. You will learn some important facts in Molecular Basis of Inheritance NCERT Solutions talk about DNA, which acts as the genetic material in most organisms. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acids found in living systems. Continue reading to learn more about NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance.

Molecular Basis of Inheritance NCERT Solutions: Class 12 Students will learn about RNA, which, 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. NCERT Solutions for Class 12 Biology Chapter 6 - Molecular Basis of Inheritance provide Questions and answers based on our Chapter 6 Biology Class 12. If you want solutions for any other class (6–12), then NCERT solutions are there for you.

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According to the CBSE Syllabus 2023–24, the Molecular Basis of Inheritance NCERT chapter has been renumbered as Chapter 5.

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NCERT Solutions for Class 12 Biology Chapter 6 - Molecular Basis of Inheritance (Solved Exercise)

The Class 12 Biology Chapter 6 NCERT Solutions are given in the sequence below as per the NCERt Textbook data:

Molecular Basis of Inheritance Class 12 Questions and Answers:

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.

Molecular Basis of Inheritance Class 12 Solutions:

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

Write down the sequence of complementary strand in 5'\rightarrow 3' 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'.

Molecular Basis of Inheritance Class 12 NCERT Solutions:

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

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.

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 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.

NCERT Solutions for Class 12 Biology Chapter 6:

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) 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.

Class 12 Biology Chapter 6 question answer:

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.


(b) mRNA and tRNA

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'


Class 12 Molecular Basis of Inheritance NCERT Solutions:

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.

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?

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.

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Molecular Basis of Inheritance Class 12 Solutions:

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

(a) Promoter

Answer:

Function of promoter

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) Exo

Answer:

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

Molecular Basis of Inheritance Class 12 NCERT Solutions

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.

Class 12 Biology Chapter 6 NCERT Solutions

Q14 . Briefly describe the following:

(a) Transcription

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.

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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.

(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 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.

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(d) Bioinformatic

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.

NCERT Solutions for Class 12 Biology Chapter 6 molecular basis of Inheritance

Important Topics in NCERT Solutions for Class 12 Biology Molecular Basis of Inheritance:

Section
Topic Name
6Molecular Basis of Inheritance
6.1The DNA
6.2The Search for Genetic Material
6.3RNA World
6.4Replication
6.5Transcription
6.6Genetic Code
6.7Translation
6.8Regulation of Gene Expression
6.9Human Genome Project
6.10DNA Fingerprinting
6.11Summary
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Important Subtopics in NCERT Solutions for Class 12 Biology Molecular Basis of Inheritance

  • 6.1.1 - Structure of Polynucleotide Chain
  • 6.1.2 - Packaging of DNA Helix
  • 6.2.1 - The Genetic Material is DNA
  • 6.2.2 - Properties of Genetic Material (DNA versus RNA)
  • 6.4.1 - The Experimental Proof of Replication
  • 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
  • 6.6.1 - Mutations and Genetic Code
  • 6.6.2 - tRNA– the Adapter Molecule
  • 6.8.1 - The Lac operon
  • 6.9.1 - Salient Features of Human Genome
  • 6.9.2 - Applications and Future Challenges
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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 2023-24. 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.

Molecular basis of inheritance NCERT also provide 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 Class 12 solutions, 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.

Class 12 Biology Chapter 6 NCERT solutions have questions and answers 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)

In molecular basis of inheritance Class 12 NCERT Solutions, you will get questions on Central dogma which consists of three important things that are:

  1. DNA
  2. mRNA
  3. Protein

After going through Biology Chapter 6 Class 12, 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

The key features of Class 12 Biology Chapter 6 NCERT Solutions are listed below:

  • The questions and answers are written in the best possible way using simple language that makes solutions of the Ch 6 Biology Class 12 easier to read and understand
  • In the Class 12 biology ch 6 question answer, 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.
  • Class 12 molecular basis of inheritance NCERT solutions 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 given solutions for NCERT Class 12 biology chapter 6 molecular basis of inheritance.

Also Check NCERT Books and NCERT Syllabus here:

NCERT Solutions for Class 12 Biology- Chapter wise

NCERT Solutions for Class 12- Subject wise

NCERT Exemplar Class 12 Solutions


We hope that you will ace in your board examinations and competitives exams with the help of NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance.

Frequently Asked Question (FAQs)

1. What are the important components of central dogma given in molecular basis of inheritance ncert?

Important components of central dogma are:

  • DNA
  • mRNA
  • Protein
2. What are the important topics of solutions for molecular basis of inheritance ncert?

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 
3. What are the two essential roles of ribosome during translation?
  • 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.
4. What are the applications of DNA fingerprinting given in molecular basis of inheritance ncert?

 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.
5. How to download solutions for molecular basis of inheritance ncert?

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  

6. Can I score full marks by practising the NCERT Solutions for Class 12 Biology Chapter 6?

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.

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  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

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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

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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