NIOS Class 12 Biology Molecular Inheritance and Gene Expression

NIOS Biology Chapter 23 Solutions offer clear answers to terminal exercises for effective revision. The chapter covers important biological concepts essential for exam success. With structured and simple solutions, students can revise quickly and gain confidence in their preparation.

These answers act as a reliable guide for learners aiming to perform well in exams. This chapter plays a key role in strengthening understanding and ensuring success in NIOS Class 12 Biology studies.

NIOS Biology Chapter 23 Solutions

1. How did Hershey and Chase prove that DNA is the hereditary material?

• Hershey and Chase in 1952 used the T2 bacteriophage, a virus that infects bacteria, for their experiments.
• They labelled the protein coat of the virus with the radioactive isotope of sulphur 35S.
• When the virus was introduced into the bacteria, no radioactivity was found inside the bacteria as the viral coat was left outside.
• When they labelled viral DNA with 52P32 or radioactive phosphorus, radioactivity was found inside the bacteria.
• It became clear that new generations of the virus were reproduced inside bacteria because of viral DNA.

2. Explain (i) Transduction and (2) Lysogeny

Transduction

Transduction refers to the transfer of DNA from one bacterial cell into another bacterium

through the agency of a virus (bacteriophage). The viral DNA integrates and becomes part of bacterial DNA, which is now a new recombinant DNA. Sometimes the viral genome may become independent and carry host bacterial genes to another new host bacterium and recombine into its genome. This process of gene transfer is called transduction.

Lysogeny

Lysogeny is the process by which a virus that has undergone transduction enters a bacterium and integrates along with the bacterial genome to increase the number of viral particles. 

3. Describe the Watson and Crick model of DNA.

According to the Watson and Crick model

• The DNA molecule is a double helix consisting of two strands of DNA
• The arrangement of the two strands is antiparallel, which means that the sequence of nucleotides goes up in 5′ to 3′ direction in one strand and the other strand comes down in 3′ to 5′ direction. (3′ and 5′ refer to the carbon atom to which the phosphate group is attached).
• The backbone of the helix is made of sugar and phosphate. Nitrogenous bases are linked to the sugar.
• The bases of the two strands are linked by hydrogen bonds.
• Base pairing is very specific as per Chargaff’s rule. Adenine, a purine base, always pairs with thymine, a pyrimidine base. The purine base Guanine pairs with the pyrimidine Cytosine. These pairs of bases are called complementary bases.

There are two hydrogen bonds between A and T and three hydrogen bonds between G and C. A and T are complementary bases, and so are G and C.

In the DNA helix, a complete helical turn occurs after 3.4 nm (or 34Å). This complete turn encloses 10 base pairs. Each base pair lies 0.34 nm (3.4 Å) apart. The diameter of the double helical DNA molecule is 2.0 nm. 

4. Explain how replication takes place.

Replication may thus be defined as a mechanism for the transmission of genetic information from generation to generation.DNA replication occurs in the S-phase of the cell cycle. 

Replication occurs through the following steps:

• 1. Unwinding of the DNA double helix: The two strands of the replicating DNA molecule separate by the action of the enzyme Helicase. The topoisomerase enzyme keeps it open. The open part is the replication fork.

• 2. Synthesis of the primer: A Primer is a short RNA molecule of about 5 to 10 bases. It is formed in the presence of the enzyme primase. The primer provides a 3’-OH group for attachment of the new DNA strand.
• 3. Synthesis of new DNA strand: The opened strands of DNA form the template. New strands complementary to the template get synthesized. At the replication fork, a new DNA strand begins to synthesise, attaching itself to the primer, in the presence of the enzyme DNA polymerase. It begins synthesis from its 5’ end, and a DNA strand complementary to one of the unwound parental DNA strands gets synthesized. The new strand of DNA continues to be synthesized uninterrupted and is termed the leading strand.

5. Write a note on the Central Dogma.

Genes are in the nucleus and proteins are synthesised in the cytoplasm of the cell. The transfer of information from genes to the site of protein synthesis constitutes the Central Dogma. The central dogma operates in the following sequence. Information flows from DNA (a particular gene) to the particular protein through RNA.

DNA Transcription → RNA Translation → Protein

For protein synthesis, first, the information coded in DNA is copied as a complementary messenger RNA molecule. This is termed as Transcription. Messenger RNA carrying information moves out of the nucleus into the cytoplasm, attaches to the ribosomes to translate the information in the form of a protein. This is termed Translation.

In retroviruses, the genetic material is RNA. Therefore, during protein synthesis, it is first transcribed into a DNA molecule in the presence of the enzyme Reverse Transcriptase, and then the path of the central dogma is followed as shown below.

RNA (Reverse transcription)   →  DNA → mRNA → Protein

(genetic material of retrovirus)

6. State the properties of the genetic code.

The genetic code has the following characteristics:

• 1. The genetic code is a triplet code. This means that a sequence of 3 bases called a codon has the information for a particular amino acid. The sequence of codons determines the sequence of amino acids in a protein.
• 2. Genetic code is unambiguous, that is, a particular codon can code for only one amino acid.
• 3. Genetic code is commaless and non-overlapping. This means that it is read continuously from beginning to end.
• 4. The genetic code is degenerate. There are 20 amino acids that form the various proteins of living beings. But if 3 out of 4 nucleotides (each containing one of the four bases) form a codon, there can be 43 = 64 codons. Hence, more than one codon codes for a particular amino acid; that is, the code is degenerate. In fact, as you can see from Table 23.1 first two bases of the codons for the same amino acid are common, and the third one changes or wobbles. This is called the Wobble hypothesis.
• 5. The genetic code is read on the transcribed mRNA during protein synthesis.
• 6. AUG codon, codes for Methionine, and is the initiation codon as it is the first one to be transcribed from a cistron.
• 7. UAA, UAG, and UGA are stop codons, and anticodons of one of these three codons are present at the end of every cistron to terminate protein synthesis.
• 8. The genetic code is universal and common for almost all organisms on Earth.

7. Explain transcription in Eukaryotes and processing of hnRNA.

In Eukaryotes, a large molecule of RNA called hnRNA is synthesised in the nucleus when its sense strand is exposed. Catalysed by enzyme RNA polymerase, hnRNA is processed to form mRNA, which gets a cap at the 5’ end and a poly A tail, before leaving the nucleus.

Processing of hnRNA

hnRNA is large because eukaryotic genes contain coding sequences called exons and non-coding sequences called introns (I) in between exons. Both introns and exons (E) are transcribed into mRNA. During the processing of mRNA, introns are cut off and exons join to form mRNA.

• A nucleoside (recall from section 23.3) called methyl guanosine comes and attaches at the 5’ end of mRNA. This is called capping.
• A small piece of RNA having only nucleotides containing the base Adenine is attached at the 3’ end. This is called the poly A tail.
• The mRNA with cap and tail moves out of the pores in the nuclear membrane.

The process of formation of functional mRNA from hnRNA is termed RNA processing.

8. What do you mean by the regulation of genes?

Gene regulation refers to the control of gene expression. Gene expression in eukaryotes can be regulated at the level of transcription or processing of hnRNA into mRNA, or at translation or

post translation.

In Prokaryotes, the Lac-operon is an excellent example of control of gene expression in prokaryotes (bacteria). It is an inducible system and is switched on in the presence of the substrate lactose. Enzymes for metabolising lactose are galactosidase, permease, and transacetylase, and genes that code for them get switched on. In the absence of lactose, they remain switched off.

9. Explain how the lac operon gets switched on in the presence of lactose in E.coli.

In the presence of lactose

Regulator protein is attracted to lactose, the site opens; RNA polymerase finds promoter; genes z, y, a are switched on, transcription begins, and the three enzymes are synthesized inside the cell.

The above is an example of an inducible system. Repressible systems are also found in prokaryotes.

10. Name three levels at which regulation takes place in a eukaryotic cell.

• Transcription or Processing of hnRNA into mRNA
• Translation
• Post Translation

11. Write notes on:

(i) Types of mutations

(ii) Okazaki fragments

(iii) Chain termination during translation.

(i) Types of mutations

A heritable change in the structure, content, and organization of the genetic material that can be passed down to the next generation is termed a mutation. A mutation may occur in one gene when it is termed a point mutation or may affect a number of genes on a part of a chromosome when it is termed a chromosomal mutation.

• Gene mutation: A change that affects only one gene is called a gene mutation or point mutation. It can be a transition, transversion, frameshift, missense, nonsense, or silent mutation.
• Chromosomal mutation means a mutation in the number of genes. It can be a change in the number or structure of chromosomes.
  • The change in the number of genes can be aneuploidy or polyploidy.
  • The change in the structure of a chromosome can be a deletion, inversion, duplication, or translocation.

(ii) Okazaki fragments

DNA synthesis always takes place along the 5’ to 3’ direction. Therefore, the other new DNA strand gets synthesised in the direction opposite to the leading strand. This new strand, called the Lagging strand, builds up in small pieces in the presence of the enzyme DNA polymerase. Thus, the synthesis of the lagging strand is discontinuous. The new pieces of DNA are termed Okazaki fragments. In the presence of the enzyme ligase and the energy source ATP, the Okazaki pieces get joined together to form a DNA strand. 

(iii) Chain termination during translation.

When the stop codon on mRNA is reached, the polypeptide is synthesised. It leaves the ribosome, and the ribosome dissociates into its two subunits.

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