Ncert Solutions For Class 12 Biology Chapter 6

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Ncert Solutions Class 12 Biology Chapter 6 Download Pdf

Class 12 Biology NCERT Solutions Chapter 6: Adda 247 provides NCERT Solutions for Class 12 Biology Chapter 6 which is for the students who want to go ahead in life and achieve great marks in their examinations. The NCERT Solutions for class 12 Biology Chapter 6 are provided by the teachers who are experts of their subjects. The solutions are set according to the rules formulated by the NCERT class 12 biology and in the language that can be understood by every student. By these reading the solutions students can build up a strong base easily. The NCERT class 12 Biology solutions covers the chapters 1 to 16 with the important questions and the answers in a detailed way.

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NCERT Solutions for Class 12 Biology Chapter 6 – Molecular basis of inheritance

The chapter provides information about Molecular basis of inheritance. DNA is a long polymer of deoxyribonucleotides. The length of DNA is usually defined as number of nucleotides (or a pair of nucleotide referred to as base pairs) present in it. DNA is an acidic substance in the nucleus. It was first identified by Friedrich Meischer in 1869. He named it as ‘Nuclein” In 1953 double helix structure of DNA was given by James Watson and Francis Crick, based on X-ray diffraction data produced Maurice Wilkins and Rosalind Franklin.

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Important Questions of Ncert Solutions Class 12 Biology chapter 6: Molecular Basis of Inheritance

Question 1. Group the following as nitrogenous bases and nucleosides:

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

Answer: DNA is a polymer of nucleotides that are made up of sugar, a nitrogenous base and a phosphate moiety. DNA has four bases; two purines and two pyrimidines. The two purines are namely adenine and guanine and two pyrimidines are cytosine, thymine. A nucleotide unit without phosphate group is called nucleoside. Nitrogenous bases = Adenine, thymine, uracil and cytosine. Cytidine and Guanosine are the nucleosides.

Question 2. If a double stranded DNA has 20 per cent of cytosine, calculate the percent of adenine in the DNA.

Answer: In a DNA molecule, the number of cytosine molecules is equal to guanine molecules and the number of adenine molecules is equal to thymine molecules. Thus, if a double-stranded DNA has 20% cytosine, it has 20% guanine. Thus, C + G makes 40% of the total bases. The remaining 60% includes both adenine and thymine which are in equal amounts. So, the percentage of adenine is 30%.

Question 3. If the sequence of one strand of DNA is written as follows:

5′ -ATGCATGCATGCATGCATGCATGCATGC-3′

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

Answer: DNA is a two-stranded molecule. Each strand is a polynucleotide composed of A(adenosine), T (thymidine), C (cytidine), and G (guanosine) residues polymerized by “dehydration” synthesis in linear chains with specific sequences. Each strand has polarity which runs from 5′ to 3′.

In DNA, A pirs with T and G pairs with C.

Fo through the original 5′ to 3′ sequence pairing each A with T ad each C wth G.

While RNA is a single strand molecule. This Strand also consists polynucleotide composed A, C, G and U (uridine).

For a complementary strand replace A with U (because RNA does not consist T), T with A,

C with G, and G with C.

Therefore, if the sequence of one strand of DNA is written as:

5′ -ATGCATGCATGCATGCATGCATGCATGC-3′

The sequence of the complementary strand in 3′ -> 5′ is as follows:

3′ – TACGTACGTACGTACGTACGTACGTACG – 5′

Subsequently, sequence of complementary strand in 5’→3′ direction is written as:

5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′

Question 4. 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:

If the coding strand in a transcription unit is 5ʹ-ATGCATGCATGCATGCATGCATGCATGC-3ʹ

Then, the template strand in 3 ‘to 5’ direction would be

3ʹ-TACGTACGTACGTACGTACGTACGTACG-5ʹ

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

5ʹ-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3ʹ

Question 5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

Answer: 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 semiconservative. 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.

ncert solutions for biology class 12 chapter 6 — Ncert solution for class 12 biology chapter 6

Question 6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.

Answer: There are two types of nucleic acid polymerases:

DNA-dependent DNA polymerase
DNA-dependent RNA polymerase

DNA-dependent DNA polymerase use a DNA template to synthesize a new DNA strand and DNA-dependent RNA polymerase use a DNA template to synthesize a new RNA strand.

Question 7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Answer: Hershey and Chase 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 subjected to the process of centrifugation.

It gave an idea that DNA acted as hereditary material which was transmitted from bacteriophage to bacteria. Bacteria which were infected with bacteriophages had radioactive proteins.

Question 8. Differentiate between the following:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

(c) Template strand and Coding strand

Answer: (a) Repetitive DNA and satellite DNA

Repetitive DNA		Satellite DNA
	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
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.
It is a linear molecule.	It has clover leaf shape.

(c) Template strand and coding strand

Template strand	Coding strand
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).
It runs from 3’ to 5’.	It runs from 5’to 3’.

Question 9. List two essential roles of ribosome during translation.

Answer: The important functions of ribosome during translation are as follows.

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.

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. Then, why does lac operon shut down some time after addition of lactose in the medium?

Answer: Inducer is a chemical (substrate, hormone or some other metabolite) which after coming in contact with the repressor, changes the latter into non-DNA binding state so as to free the operator gene. The inducer for lac-operon of Escherichia coli is lactose (actually allolactose, or metabolite of lactose). RNA polymerase is recognized by promoter gene. It passes over the freed operator gene and then catalyses transcription of mRNAs over the structural genes. The mRNA pass into the cytoplasm and form particular proteins or enzymes. Out of the three enzymes produced by lac-operon, permease is meant for bringing lactose inside the cell. galactosidase ( = lactase) breaks lactose into two components, glucose and galactose. The enzyme like lactase or galactosidase which is formed in response to the presence of its substrate is often called inducible enzyme. Inducible operon systems generally occur in catabolic pathways. The lac-operon will not, however, remain operative indefinitely despite presence of lactose in the external environment. It will stop its activity with the accumulation of glucose and galactose in the cell beyond the capacity of the bacterium for their metabolism.

Features of the NCERT Solutions for Class 12 Biology Chapter 6 – Molecular basis of inheritance

NCERT Solutions of class 12 have been answered based on the important information on the question.

The columns are used wherever necessary.
Solutions are solved point wise and accurately answered point to point.

Frequently Asked Questions on NCERT Solutions for Class 12 Biology Chapter 6

What are the advantages of referring NCERT Solutions for class 12 Biology chapter 6?

Students referring the NCERT Solutions of class 12 by Adda 247 find the solutions helpful during the exams. The solutions are prepared by the experts in an interactive manner keeping in mind the students. The students’ perspective is kept in mind while preparing the solutions. It helps in completing the syllabus on time and also provides notes for the revision prior to the exam.

What are the advantages of referring NCERT in competitive exams like JEE and AIPMT?

Most of the competitive exams like NEET, JEE etc. follow the basic NCERT books for designing their question papers. NCERT serves as the base for every book prepared for NEET and JEE. The competitive exams are based on the CBSE syllabus applied in XI and XII classes and NCERT books strictly follow CBSE syllabus. In addition to this, NCERT books play an important role in clearing out the theoretical concepts. Every topic given in NCERT books is explained in such a way to help students make their basics and fundamentals strong and clear.

How to read NCERT books more efficiently?

Given below are the important points which must be followed while reading the NCERT books in an efficient manner:

Go through each topic thoroughly by understanding the meaning and significance of each line mentioned in that particular topic.

Ask your teacher if any doubts.

Note down the important topics to revise at the time of examination.

Solve all the exercise questions given at the end of each chapter. These questions are important for understanding the concepts in a better way.

Is it mandatory to solve all the NCERT questions mentioned at the end of each chapter?

The questions and answers mentioned in NCERT textbooks at the end of each chapter are quite important not only for examination but also for understanding the concepts in a better way. These questions aim to test the students’ understanding and learning over the topics that they have learned in the chapter.

Solving the NCERT exercise problems will help to

clear all the concepts and formulae you learned in a chapter
get comfortable with different types of questions that might be asked in exams
get enough practice which is key to succeed in Mathematics exam
improve your accuracy and speed

What are the important concepts covered in the Chapter 6 of NCERT Solutions for Class 12 Biology?

The concepts involved in the chapter 6 of NCERT Solutions are –

6.1 –The DNA

6.2 – The Search for Genetic Material
6.3 – RNA World
6.4 – Replication
6.5 – Transcription

6.6 – Genetic Code

6.7 – Translation

6.8 – Regulation of Gene Expression

6 .9 – Human Genome Project

6.10 -DNA Fingerprinting

These concepts are created by the faculty at Adda 247. The solutions are available at Adda 247 in the PDF format which can be downloaded by the students.