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

1: Group the Following As Nitrogenous Bases and Nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine

 Ans:

• Nitrogenous bases are the molecules that mainly consist of nitrogen with other chemical components and they act as bases. It includes adenine, thymine, uracil, and cytosine.

• Nucleosides are made up of nitrogenous bases and sugar. It includes cytidine and guanosine.

2:  If a Double Stranded DNA has 20% of Cytosine, Calculate the Percent of Adenine in the DNA.

Ans: In the double helix model or in double-stranded  DNA the ratio between the adenine and thymine molecule is the same, whereas the ratio between the guanine and cytosine is the same. 

In 100% of DNA, if the percent of cytosine is 20% then the percent of guanine is also equal to 20%. By adding the percentage of cytosine and guanine total of 40% are present and the remaining 60% of DNA is formed by adenine and thymine.

Thus in DNA, there is 30% of adenine and 30% of thymine.

So, the percent of adenine in DNA is 30%.

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

Ans: The DNA double strands are complementary strands where one base pair forms pairing with other base pairs like adenine with thymine and cytosine with guanine. The given sequence of one strand is  5'- ATGCATGCATGCATGCATGCATGCATGC - 3'

In 3' to 5' direction the sequence will form like,

3'- TACGTACGTACGTACGTACGTACGTACG - 5'

Therefore,  in 5' to 3' direction the sequence of the complementary strands will  form  like,

5'- GCATGCATGCATGCATGCATGCATGCAT - 3'

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.

Ans: In the transcription unit the coding strand does not code for anything therefore the sequence remains the same, only the thymine get replaced by uracil.

If the given sequence is   5'- ATGCATGCATGCATGCATGCATGCATGC-3'

Then the  sequence of mRNA form will be 5' - AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'

5: Which Property of DNA Double Helix led Watson and Crick Hypothese Semi-Conservative Mode of DNA Replication? Explain.

Ans: The following  properties of DNA double helix led Watson and Crick hypothesis semi-conservative model of DNA replication,

1. The antiparallel and complementary nature of DNA double-strand with respect to the base sequence. Due to this, each strand acts as a template strand for the synthesis of new strands.

2. Semi conservative nature of DNA due to the presence of one parental strand and one newly synthesised strand. 

3. The sequence of the base of the template strand forms the sequence of the daughter strand due to the complementary nature of DNA.

(Image will be uploaded soon)

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.

Ans: There are mainly  two types  of nucleic acid polymerases are present, such as

• DNA-dependent DNA polymerases- It is the main polymerase enzyme that helps in replicating the parental strands(Template DNA) by which new strands of DNA form.

• DNA-dependent RNA polymerase- It is the main polymerase enzyme in the transcription process where the RNA gets formed by copying one strand of the DNA.

7: How Did Hershey and Chase Differentiate Between DNA and Protein in Their Experiment While Proving That DNA Is the Genetic Material?

Ans: Alfred Hershey and Martha Chase in 1952 work with the bacteriophage and E.Coli to find out the genetic material between DNA and protein by the following process.

1. To find out the genetic material between DNA and protein, they cultured some viruses in radioactive phosphorus and some in radioactive sulfur.

2. In radioactive phosphorus, the virus consists of radioactive DNA but not protein, whereas in radioactive sulfur the virus consists of radioactive protein but not radioactive DNA.

3. This radioactive virus and radioactive phages are then inserted into E.Coli which cause the spread of infection.

4. The viral coat gets remove by agitating the phages in the blender, due to this the virus particle get separated from bacteria in a centrifuge.

5. The virus-carrying DNA transmitted the infection to bacteria whereas the virus-carrying protein does not. 

Therefore it concluded that between DNA and protein DNA is the genetic material.

(Image will be uploaded soon)

 8:Differentiate Between the Following

1)  Repetitive DNA and satellite DNA

      Ans:

2)  mRNA and tRNA

       Ans:

3)  Template strand and coding strand

9: List Two Essential Roles of the Ribosome During Translation.

Ans: During translation, the two essential roles performed by the ribosomes are as follow:

• The synthesis of protein takes place by the ribosome because it acts as a cellular factory. It is divided into two subunits in its inactive state, one is a large subunit and one is a small subunit. In large subunits bind amino acids and when the mRNA enters in small subunits and by this the synthesis of protein begins.

• During translation ribosomes also act as catalysts for peptide bond formation and this catalyst is a ribozyme.

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?

Ans:

• In the lac operon, lactose acts as an inducer that regulates switching on and off of the operon. Apart from this lac operon also consists of one regulatory gene and three structural genes which are z,y and a. 

• The regulatory gene I code for repressor and in structural gene the z code for beta-galactosidase, y for permease and a-gene for transacetylase. The three structural genes metabolised lactose.

• When lactose was added to the E.Coli medium, it gets transported into the cell by the action of permease and shows binding with the repressor, due to this the RNA polymerase gets bind with the promoter region, hence the synthesis of products of the structural gene get initiated which also led to the metabolization of glucose and galactose. 

• Due to the metabolism of lactose, the level of lactose gets decreased and the formation of a repressor starts. The binding of the repressor to the operator prevents the transcription of RNA polymerase. This is called negative regulation because of the stoppage of the transcription process.

(Image will be uploaded soon)

11: Explain(in one or two lines) the Functions of the Following:

(a)  Promoter

Ans: (1) It initiates the process of transcription.

         (2) It provides the binding site for RNA polymerase.

(b)  tRNA

Ans: (1) It reads the genetic code of messenger RNA.

         (2) During translation it carries a specific ribosome to mRNA to initiate the process.

(c)  Exons

Ans: (1) It is the coding sequence of DNA that transcribes proteins.

(2) In between the long sequence of axons introns are present which disappear in    the mature one.

12: Why is the Human Genome Project called a Megaproject?

Ans: Human Genome Project is called a megaproject because of the following reasons:

1. The human genome has 3×109 base-pair approximately and if it is required 3 US dollars per base pair then its estimated cost will go approximately to 9 billion US dollars.

2. If its sequence were stored in a book in the typed form then each page will consist of 1000 letters and each book will consist of approximately 1000 pages which led to the formation of 3300 books from a single human cell.

 For all this, high computational devices are required for data storage, retrieval and analysis. 

13: What is DNA Fingerprinting? Mention its Application.

Ans: The technique which is used to identify and analyse the variation in DNA in every individual is known as DNA fingerprinting.

The various applications of DNA fingerprinting are as follow:

1. In forensic science, it is used for identifying potential crime suspects.

2. It is used for finding out paternity and family relationships.

3. It is used for the identification and protection of commercial crop varieties and livestock.

4. It is used to find out the evolutionary relationship and linkage between the various organisms.

14: Briefly Describe the Following:

1. Transcription:

Ans: The process of formation of mRNA from the DNA template is known as transcription. In this process, a single strand of DNA gets copied into mRNA. It starts at the promoter region of the template DNA and stops or terminates at the terminator region of the template. Between these two regions, a transcription unit is present. The process of transcription is catalysed by DNA dependent RNA polymerase.

• Initiation, elongation and termination are the three main processes of transcription.

• At the promoter region of the template strand the DNA dependent RNA polymerase and initiation factor-like (σ) bind and the process of transcription gets initiated.

• Due to enzymes the DNA double helix unwinds and then one of the strands, called sense strand, starts mRNA synthesis and this strand is called the template strand. The process of transcription remains continuous until the terminator region.

• As the transcription enzymes reach the terminator region, the enzyme and the newly synthesised mRNA are released.  The termination takes place by the termination factor(ρ).

(Image will be uploaded soon)

2. Polymorphism

Ans: Polymorphism is a form of genetic variation in which distinct nucleotide sequences can exist at a particular site in a DNA molecule in a population. This heritable mutation is observed at a high frequency in a population. It arises due to mutation either in a somatic cell or in germ cells. The germ cell mutation can be transmitted from parents to their offspring. This results in the accumulation of various mutations in a population, leading to variation and polymorphism in the population. This plays a very important role in the process of evolution and speciation. Polymorphism in DNA sequences is the basis for gene mapping and DNA fingerprinting.

3. Translation

Ans: -Translation is the process of polymerizing amino acids to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain.

The process of translation involves three steps :

Initiation

Elongation

Termination

During the initiation of the translation, tRNA gets charged when the amino acid binds to it using ATP. The start (initiation) codon (AUG) present on mRNA is recognized only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in a large subunit for the attachment of subsequent amino acids. The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it

initiates the process of translation. During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave the space for binding of another charged tRNA. The amino acid brought by tRNA gets linked with the previous amino acid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codons (VAA, UAG, and UGA), the process of translation gets terminated. The polypeptide chain is released and the ribosomes get detached from mRNA.

(Image will be uploaded soon)

4. Bioinformatics

Ans: Bioinformatics is the application of computational and statistical techniques to the field of molecular biology. It solves the practical problems arising from the management and analysis of biological data. The field of bioinformatics developed after the completion of the human genome project (HGP). This is because an enormous amount of data has been generated during the process of HGP that has to be managed and stored for easy access and interpretation for future use by various scientists. Hence, bioinformatics involves the creation of biological databases that store the vast information of biology.

It develops certain tools for easy and efficient access to information and its utilization. Bioinformatics has developed new algorithms and statistical methods to find out the relationship between the data, predict protein structure and their functions, and cluster the protein sequences into their related families.

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

These NCERT Solutions are centered around the molecules that constitute particles and biological components like proteins, and DNA and RNA. The Molecular Basis of Inheritance Class 12 NCERT Solutions PDFs are of utmost importance for your exam preparation. Our subject matter experts have prepared these NCERT Solutions as per the CBSE guidelines. By going through these NCERT Biology PDF for Molecular Basis of Inheritance, you will be able to revise all the topics covered in the chapter thoroughly. The PDF is downloadable for free.  

NCERT Solutions of Ch 6 Molecular Basis of Inheritance

Class 12th Biology Chapter 6 NCERT Solutions are solely made for helping students take that extra step towards achieving a higher score. Students should learn and practise all the topics covered in this properly to secure a good marks in the exam. Students should also solve the exercises in the back of the chapter and verify their answers by comparing them with the NCERT Solutions for Chapter 6. In this way, they will be able to improve their knowledge-base. 

The PDF consists of various topics which come under this chapter, such as DNA, genetic structure, Dogma & RNA, DNA packaging, DNA replication, RNA (as a separate topic), etc. The chapter even delves further into the nuances of the Human Genome Project and genetic code. 

The NCERT Solutions for Class 12 Biology's Molecular Basis of Inheritance make reliable revision notes for the exam preparation. The notes themselves are brief and are aimed at providing students with reference points after the student has studied the chapter.

NCERT Biology Class 12 Weightage

Students will have to attempt two sections, one section which carries a total weightage of 70 marks, the other which consists of 30 marks in the Biology paper. Following the latest marking scheme, this chapter’s weightage might vary between 9 and 15 marks, depending upon the types of questions asked in the exam. Also, keep in mind the possibility of this chapter’s importance from the practical examination’s point-of-view, as you may get questions from this chapter for experiments that come under Part A.

CBSE Class 12 Biology Chapter 6 NCERT Solutions

CBSE Class 12 Biology Chapter 6 NCERT Solutions cover the following topics.

• DNA

• Structure 

• Central Dogma

• DNA Replication

• RNA

• Genetic Code

• Human Genome Project

• Goals of HGP

• Methodologies of HGP

• Salient features of HGP

• Modular Basis of Inheritance

• DNA role in the modular basis of inheritance

• Regulation of gene expression

• Prokaryotic Gene regulation

• Discovery of DNA

• Properties of Genetic Material

• Salient features of the genetic code

• Genetic code and mutations

Benefits of NCERT Solutions

Going through the NCERT Solutions will give students multiple added benefits. Most of these benefits will only come to fruition when the student appears in the examination well-read. Also, they should be aware of the types of questions that are generally asked in the examination. 

• The NCERT solutions are made to ensure students clearly understand the use of appropriate terminology. 

• Students should go through these NCERT Solutions for a thorough revision of all the topics covered in this chapter.

• These NCERT Solutions provide precise answers to all the questions that are given at the end of the chapter.

• Since the NCERT Solutions are prepared in reference to the latest CBSE guidelines, students can follow and incorporate the format of these solutions while writing their answers. 

• By referring to these NCERT Solutions students will get a clear idea of all the important topics covered in the solutions. It will boost the confidence of students for their exam.