What is the study of inheritance and variation called?

Genetics is the study of inheritance and variation, as discussed in the chapter. In layman’s language, one aspect of genetics is the scientific study of variations passed down from parent to progeny. A common example can help you understand the concept of inheritance and variation. We frequently notice that children have several characteristics that are similar to their parents because they ‘inherited’ these characteristics from their parents. Similarly, because of a ‘variation’ in their genes, the children do not look exactly like their parents.

What are the important questions of Chapter 5 Principles of Inheritance and Variation of Class 12 Biology?

Students can access the Important Questions Class 12 Biology Chapter 5 Principles of Inheritance and Variation from the Extramarks website. These questions cover all the main topics of the chapter. Students can get a good score if they thoroughly prepare these questions and practise NCERT textbook chapter end exercises. These questions were created by analysing past year’s questions as well as the updated recent syllabus. As a result, these questions will assist students in getting the highest possible score in the Biology board exam.

Home > CBSE > CBSE Important Questions Class 12 Biology Chapter 5 – Molecular Basis of Inheritance

CBSE Important Questions Class 12 Biology Chapter 5 – Molecular Basis of Inheritance

Class 12 Biology Chapter 5 Molecular Basis of Inheritance is the fifth chapter of the NCERT Class 12 Biology textbook. It explains the structure of DNA and RNA, how DNA is packed into nucleosomes and chromosomes, the Meselson and Stahl experiment that proved semi-conservative replication, transcription (DNA to RNA), the genetic code, translation (RNA to protein), post-transcriptional processing in eukaryotes, gene regulation and the Human Genome Project.

These CBSE important questions for Class 12 Biology Chapter 5 cover the nucleosome, the parts of a transcription unit (promoter, terminator, coding and template strands), splicing, capping and tailing, writing the mRNA from a given template or coding strand, translation with diagrams, the roles of the ribosome, the semi-conservative model, the Meselson and Stahl experiment and the features of the DNA double helix. The set has multiple-choice, short-answer and long-answer questions with answers for board exam revision.

Q1. Study the diagram given below.

Diagram of a nucleosome with parts labelled a, b, c and d

(a) What is the diagram representing?
(b) Name the parts a, b and c.
(c) In eukaryotes, the DNA molecules are organised within the nucleus. How is the DNA molecule organised in a bacterial cell in the absence of a nucleus?

Answer: (a) This is the diagram of a nucleosome, the basic unit of chromatin in eukaryotic cells.
(b) a – Histone octamer
b – DNA
c – H1 histone
(c) Bacteria do not have a defined nucleus, but the DNA is not scattered throughout the cell. The DNA is held with some proteins in a region termed the ‘nucleoid’. The DNA in the nucleoid is organised in large loops held by proteins.

Q2. In the following diagram, the two DNA strands represented are ready for transcription.

Two DNA strands of a transcription unit with parts marked 1 to 4

A. Label the parts marked 1 to 4 and state their functions in transcription.
B. Which one of the two strands of DNA has a nucleotide sequence similar to the mRNA that will be transcribed, and why?

Answer: A.
1 – Terminator: It is located towards the 3’-end of the coding strand. It usually defines the end of the process of transcription; here the RNA polymerase detaches from the DNA.
2 – Promoter: It is located towards the 5’-end of the coding strand. It provides the binding site for RNA polymerase and decides the start point of transcription.
3 – Coding strand: All the reference points used while defining a transcription unit are made with respect to the coding strand. It is the non-template strand.
4 – Template strand: The template strand is the strand of DNA that the RNA polymerase uses as a guide to build the complementary mRNA.
B. The coding strand has a nucleotide sequence similar to the mRNA. The mRNA is synthesised as the complement of the template strand, so it matches the coding strand, except that uracil (U) is present in place of thymine (T).

Q3. Briefly describe the post-transcriptional modification events occurring in eukaryotes.

Answer: 1. Splicing: Primary RNA transcripts (hnRNA) have both exons and introns copied in them. It becomes necessary to remove the introns so that the transcript has only expressible sequences in it. This is achieved by a process known as RNA splicing, in which snRNPs cut out the introns and leave the exons, which are then joined.
2. Capping: The hnRNA undergoes two more processing steps, capping and tailing, before taking part in protein synthesis. Capping involves the addition of one 7-methyl guanosine at the 5’ end of the hnRNA.
3. Tailing: Tailing involves the addition of a number of (several hundred) adenine nucleotides (poly-A tail) to the 3’ free end of the mRNA transcript.

mRNA showing 7-methyl guanosine cap at the 5' end and poly-A tail at the 3' end

Q4. (a) Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below:

Hypothetical template strand 3'-ATGCATGCATAC-5'

(b) Write the RNA strand transcribed from the above transcription unit along with its polarity.

Answer: (a) A transcription unit contains (1) a promoter, (2) the structural gene and (3) a terminator. In a transcription unit, the promoter is located towards the 5’ end (upstream) of the coding strand (non-template strand), and the terminator is located towards the 3’ end (downstream) of the coding strand.
Given template strand: 3’-ATGCATGCATAC-5’
Complementary coding strand: 5’-TACGTACGTATG-3’

Transcription unit showing promoter, coding strand, template strand and terminator

(b) RNA is transcribed from the template strand. The RNA is complementary to the template, but uracil is present in RNA in place of thymine.
mRNA: 5’-UACGUACGUAUG-3’

Transcribed mRNA strand 5'-UACGUACGUAUG-3'

Q5. Describe the process of translation up to the formation of a three amino acid long chain with diagrams only.

Diagrams of translation: charging of tRNA, initiation, elongation to a tripeptide and termination

Answer: The diagrams show: 1. Activation of tRNA (aminoacylation) – methionine is attached to its tRNA (anticodon UAC) by aminoacyl tRNA synthetase. 2. Initiation – the small ribosomal subunit binds the mRNA at the start codon AUG, the Met-tRNA occupies the P-site, and the large subunit joins to form the complete ribosome. 3. Elongation – the second charged tRNA (serine) enters the A-site, a peptide bond forms between the two amino acids, and the ribosome moves one codon (translocation); the third charged tRNA (glutamic acid) then enters the A-site and a tripeptide is formed. 4. Termination – at the stop codon, release factors free the polypeptide and the ribosomal subunits separate.

Q6. Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics

Answer: (a) Transcription: The process of copying genetic information from one strand of the DNA into RNA is termed transcription. The segment of DNA which is transcribed is the transcription unit and consists of a promoter, the structural gene and a terminator. All three types of RNA, namely mRNA, tRNA and rRNA, are synthesised by transcription. DNA-dependent RNA polymerase catalyses the reaction using the template strand, which has 3’ to 5’ polarity. It binds to the promoter and initiates transcription. The process follows the base complementarity rule, except that adenine pairs with uracil instead of thymine (elongation). Once the polymerase reaches the terminator sequence, the RNA polymerase along with the nascent RNA falls off the template (termination). RNA polymerase by itself is capable of only elongation; it takes the help of an initiation factor and a termination factor. In bacteria, the mRNA does not need further processing, so translation can begin at the same time. In eukaryotes there are two extra complexities: different RNA polymerases make different kinds of RNA, and the primary transcript (with exons and introns) is non-functional and undergoes splicing, capping and tailing.
(b) Polymorphism: It is a form of genetic variation in which a DNA molecule has distinct nucleotide sequences at a particular site. It is a heritable mutation and can be present in a population at a high frequency. It can arise due to mutation in somatic or germ cells. Mutations in germ cells are transmitted from parents to offspring. This causes accumulation of various mutations in a population, leading to variation and polymorphism.
(c) Translation: The process of polymerisation of amino acids to form a polypeptide chain, using the information in the sequence of bases in the mRNA, is called translation. It is an energy-driven process. The amino acids are first activated in the presence of ATP and linked to their cognate tRNA (aminoacylation of tRNA). When two such charged tRNAs come close to each other, a peptide bond forms between the amino acids. The ribosome is the cellular machinery that carries this out; it consists of structural RNAs and about 80 different proteins, and has two subunits that stay separate until the smaller subunit meets an mRNA. The larger subunit has two sites for the charged tRNAs to bind, bringing the amino acids close together, and the ribosome also acts as a catalyst (ribozyme) for peptide bond formation. A translational unit in mRNA is the sequence flanked by the start codon (AUG) and the stop codon; mRNA also has untranslated regions (UTRs) that help in efficient translation. During elongation, charged tRNAs bind codon by codon through their anticodons, and the ribosome moves along the mRNA adding amino acids one by one until a stop codon is reached. A release factor then binds the stop codon and the complete polypeptide is released.
(d) Bioinformatics: The application of computer technology to manage biological information is called bioinformatics. It combines computer science, statistics, mathematics and engineering to process biological data. Computers gather, store, analyse and integrate the information, which is then used for developing methods in biological systems. A major activity is developing software tools to generate useful biological knowledge from existing data. Bioinformatics grew with the Human Genome Project, when a very large amount of data (the sequence of about three billion base pairs) became publicly available, and it has helped gene discovery programmes across the world. With next-generation sequencing technologies generating massive data for many animals and plants, bioinformatics has become all the more relevant.

Q7. List two essential roles of the ribosome during translation.

Answer: 1. The ribosome brings the mRNA and the charged tRNAs together. The small subunit binds the mRNA, and the large subunit has two sites (A and P) where the charged tRNAs bind, so that the amino acids come close to each other in the correct order for peptide bond formation.
2. The ribosome acts as a catalyst (ribozyme) for the formation of the peptide bond between adjacent amino acids of the growing polypeptide chain (peptidyl transferase activity of the rRNA of the large subunit).

Q8. Suppose evolution on earth had occurred in such a way that there were 96 amino acids instead of 20, DNA had 12 different types of bases and DNA synthesis occurred in the same manner as today. Then, what would be the minimum number of bases in a codon?
(a) 12
(b) 8
(c) 3
(d) 2

Answer: (d) 2
With 12 bases, a codon of n bases can form 12ⁿ combinations. For a unique code for each of the 96 amino acids, 12ⁿ must be at least 96. For n = 1, 12 combinations are not enough; for n = 2, 12 × 12 = 144 combinations are enough. So the minimum number of bases in a codon would be 2.

Q9. If the sequence of the coding strand in a transcription unit is written as 5’-ATGCATGCATGCATGCATGCATGCATGC-3’, write down the sequence of the mRNA.

Answer: mRNA sequence: 5’-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’
Explanation: The process of converting genetic information from one strand of the DNA into RNA is termed transcription. The template strand is used for mRNA synthesis. The coding strand does not code for the mRNA but has a sequence similar to the mRNA, except that uracil replaces thymine. So the mRNA is written by replacing every T in the coding strand with U.

Q10. Which property of the DNA double helix led Watson and Crick to hypothesise the semi-conservative mode of DNA replication? Explain.

Answer: The specific base-pairing property of nucleotides in the DNA double helix (adenine pairs with thymine and cytosine pairs with guanine) led Watson and Crick to hypothesise the semi-conservative mode of DNA replication. Because the two strands are complementary, they proposed that the two strands of a DNA molecule separate during replication. After separation, each strand acts as a template for the synthesis of a new strand, following the base-pairing rule. Thus, each new DNA double helix has one parental strand and one newly synthesised strand. This scheme is referred to as semi-conservative DNA replication.

Q11. (a) What did Meselson and Stahl observe when
(i) they cultured E. coli in a medium containing ¹⁵NH₄Cl for a few generations and centrifuged the content?
(ii) they transferred one such bacterium to the normal medium of ¹⁴NH₄Cl and cultured it for 2 generations?
(b) What did Meselson and Stahl conclude from this experiment? Explain with the help of diagrams.
(c) Which is the first genetic material? Give reasons in support of your answer.

Answer: (a) (i) After growing E. coli in ¹⁵NH₄Cl for several generations, the heavy isotope ¹⁵N was incorporated into the DNA. On centrifugation (CsCl density gradient), a single band of heavy DNA (¹⁵N/¹⁵N) was obtained.
(ii) After one generation (20 minutes) in the ¹⁴NH₄Cl medium, the DNA showed a single band of intermediate (hybrid) density. After two generations (40 minutes), the DNA showed two types of densities in equal amounts: one intermediate-density hybrid DNA (¹⁵N/¹⁴N) and one light DNA (¹⁴N/¹⁴N).
(b) This agrees perfectly with the semi-conservative mode of replication. Meselson and Stahl concluded that DNA replication is semi-conservative: each daughter double helix has one parental strand and one newly made strand.

Meselson and Stahl experiment showing heavy, hybrid and light DNA bands after the first and second replication cycles

(c) RNA is the first genetic material. RNA can both store information and act as a catalyst (ribozyme), so it could carry out essential life processes such as metabolism, translation and splicing. Its role in splicing and its catalytic ability make it the ideal candidate for the first genetic material. However, RNA is reactive and unstable, so DNA, which is more stable, later evolved from RNA as the genetic material.

Q12. List the salient features of the double helix structure of DNA.

Answer: Watson and Crick proposed the double helix structure of DNA. Its features are:
(i) In a DNA molecule, two polynucleotide chains are coiled to form a helix. The sugar-phosphate backbone lies on the outside, while the bases project inwards towards each other.
(ii) The two chains have anti-parallel polarity: one runs 5’ to 3’ and the other 3’ to 5’.
(iii) Complementary bases pair with each other through hydrogen bonds. A purine always pairs with a pyrimidine: adenine pairs with thymine through two hydrogen bonds, and guanine pairs with cytosine through three hydrogen bonds.
(iv) The helix is right-handed. Its pitch is 3.4 nm and there are about 10 base pairs in each turn.
(v) The plane of one base pair stacks over the other in the double helix. This stacking, along with hydrogen bonding, provides stability to the helix.

CBSE Important Questions Class 12 Biology Chapter 5 – Molecular Basis of Inheritance

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