AHSEC Class 12 Biology Chapter: 5 Molecular Basis of Inheritance Notes 2025

Get AHSEC Class 12 Biology Chapter: 5 AHSEC Class 12 Biology Chapter: 5 Molecular Basis of Inheritance Notes 2025 Important Questions Answers 2025. In this Post we have Prepared for you Assam Board Biology Chapter: 5 Solution as Per AHSEC Latest Syllabus and exam patterns.

AHSEC Class 12 Biology Chapter: 5 Molecular Basis of Inheritance Notes 2025 Solution Can be a great resource for your exam preparation. Use this chapter’s Notes/Solution as a guideline or reference.

An Overview of AHSEC Class 12 Biology Notes for 2025

Name of BoardAHSEC
Class:Assam Board Class 12
Subject:AHSEC Class 12 Biology
Number of Chapter:05
Chapter NameMolecular Basis of Inheritance
Content Type:Text, Images and PDF Format
Academic Year:2024-25
Medium:English
Available Solution Link:AHSEC Class Biology Notes

AHSEC Class 12th Biology Chapter: 5 MOLECULAR BASIS OF INHERITANCE

A. FILL IN THE BLANKS (1 MARK)

1. In DNA is a long polymer of_______.

Ans: Deonyribonucleotides

2. Uracil is present in____.

Ans: RNA

3. Purine and pyrimidine are collectively known as______.

Ans: Nucleotides

4. The distance between two consecutive base pairs is________.

Ans : 3.4Å

5. The chromatin that is more densely packed and stains dark is called_______.

Ans: Heterochromatin

6. DNA replication is termed as________.

Ans: Semi conservative

7. The DNA- dependent DNA polymerases catalyse polymerization in_______direction.

Ans: 5/→ 3/

8. A gene is defined as the_____unit of inheritance.

Ans: Functional

9. _______is a segment of DNA coding for a polypeptide.

Ans: Gene

10. AUG codes for______.

Ans: Methionine

11. refers to the process of polymerization of amino acids to form a______

Ans: Translation, polypeptide

12. The______codes for the repressor in lac operon.

Ans: t-gene

13. and terminator segments flank the structural gene.

Ans. Promoter

14. An mRNA has some additional sequences that are not translated and are refrred as-

Ans. Untranslated regions (UTR)

15. During transcription RNA poluymerase binds to-

Ans. Promoter

B. TRUE OR FALSE (1 MARK)

1. Central Dogma was proposed Watson.

Ans: False

2. RNases are RNA digesting enzymes.

Ans: True

3. DNA duplication is semiconservative.

Ans: True

4. The process of formation of DNA from RNA is called transcription.

Ans: False

5. The discontinuously synthesized fragments are joined by the enzyme DNA polymerase.

Ans. False

C. VERY SHORT ANSWER QUESTIONS (1-MARK)

1. What is satellite DNA?

Ans. Part of DNA which contains repeated sequences is called satellite DNA.

2. What is central Dogma?

Ans. Central dogma states that genetic mformation litaws from DNA to mRNA (transcription) and then from miRNA to protein (Translation) always unidirectionally.

3. Define autoradiography?

Ans. Autoradiography is a technique used to determine the location of a radioactive isotope that has been supplied to living cells using photographic film.

4. What is an inducer in lac operon?

Ans. An inducer is a chemical substance which after coming in contact with the repressor changes it into a non-DNA binding site so as to free the operator gene. In lac operon inducer is lactose.

5. Name the bacteria used by Griffith for his transformation experiment.

Ans. Streptococcus pneumoniae.

6. Who proved that DNA in chromosomes also replicate semiconservatively?

Ans., Mathew Meselson and Franklin Stahl.

7. How many structural genes are present in lac operon?

Ans. In lac operon three structural genes, Z, Y and A are present.

8. What is translation?

Ans. Translation is the process in which the coded genetic message brought by mRNA from DNA is changed into a polypeptide chain.

9. What is an intron?

Ans. Introns are the regions of a gene which do not form part of mRNA and are removed during the processing of mRNA.

10. What is an exon?

Ans. Exons are the regions of a gene, which become part of mRNA and code for the different regions of proteins.

11. What are the three major types of RNAs present in bacteria.

Ans. mRNA (messenger RNA), tRNA (transfer RNA) and rRNA (ribosonal RNA)

12. Name the enzyme used to catalyse the polymerization of deoxymcleotides.

Ans. DNA polymerase.

13. Name the enzyme that joins the discontinuous fragments of DNA.

Ans. DNA ligase.

14. Who proposed the semiconsertative DNA replication.

Ans. Watson and Crick.

15. How many bases code for an amino acid.

Ans. Three

16. Give the site of protein synthesis.

Ans. Ribosome

17. What is the function of tRNA?

Ans. tRNA acts as the adapter molecule as it take part in transferring amino acide from cellular pool to ribosomes for polymerisation to form polypeptides.

18. Who proposed the operon model?

Ans. Francois Jacob & Jacque Monod.

19. What is a bacteriophage?

Ans. Bacteriophage is a virus that infects a bacteria.

20. What is a euchromatin?

Ans. Euchromatin is the part of a chromosome that is loosely packed and stain lighter than the other regions and are transcriptionally more active.

21. Define chromatin.

Ans. Chromatin are thread like stained bodies in the nucleus.

22. Give the definition of nucleosome.

Ans. Nucleosome is submicroscopic subunit of chromatin which is formed by wrapping of DNA over a core of histone proteins.

23. Which base triplet code for the amino acid phenylalanine?

Ans. UUU and UUC.

24. Name three non sense codon.

Ans. UAA, UAG and UGA.

25. What are the hosts for cloning DNA fragments?

Ans. The hosts for cloning DNA fragments are bacteria and yeast.

26. What are exons?

Ans. The ceding DNA regions in the primary mRNA of eukaryotes are called exons.

D. SHORT ANSWER QUESTIONS (2 MARKS)

1. Who coined the term genetic code? What does it mean?

Ans. George Gamow coined genetic code.

Genetic code means the relationship between the sequence of amino acids in a polypeptide and nucleotide sequence of DNA or mRNA.

2. Discuss the structure of nucleosome.

Ans. The structure that is formed when negatively charged DNA is wrapped around the positively charged histone octamer. A typical nucleosome contains 200bp of DNA helix. It constitutes the repeating unit of chromatin and are seen as ‘beads on string.’

3. Define genetic material.

Ans. Genetic material is that substance which not only controls the inheritance of traits from one generation to the next but is also able to express its effect through the formation and functioning of the traits.

4. What are the four features of genetic materials?

Ans. The four features are

(i) It should be able to generate its replica.

(ii) It should be chemically and structurally stable.

(iii) It should provide the scope for slow changes (mutation) that are required for evolution.

(iv) It should be able to express itself in the form of Mendelian characters.

5. Why DNA replication is stated to be discontinuous?

Ans. DNA replication is semidiscontinuous because one strand called leading strand grows continuously while the other strand called lagging strand is formed discontinuously in fragments (Okazaki fragments).

6. How is the translation of mRNA terminated?

Ans. mRNA sequence is elongated till the termination codon UAG, UAA or UGA reach the A site of ribosome. These codon do not code for any amino acid and thus translation stops.

7. What is transcription?

Ans. The process of copying genetic information from one strand of the DNA into RNA is called transcription. The principle of complementarity governs the process of transcription except the adenosine now forms base pair with uracil instead of thymine.

8. Name the transcriptional units in DNA.

Ans. A transcription unit in DNA has three regions

(i) A promotor

(ii) The structurel gene

(iii) A terminator

9. What are exons?

Ans. Exons are the regions of a gene, which become part of mRNA and code for the different regions of proteins.

10. Why hnRNA is required to undergo splicing.

Ans. hnRNA undergoes splicing in order to remove introns which are in non coding sequences and exons are joined to form functional mRNA.

11. What is a satellite DNA?

Ans. The repetitive DNA sequences who do not code for any proteins but form a large portion of human genome and show high degree of polymorphism is called satellite DNA.

12. Where and when replication does occur?

Ans. Replication occurs in the nucleus of cell during S-phase of the cell cycle.

13. How does DNA express its biological information.

Ans. The mechanism of gene expression involves biochemical genetics. It consists of synthesis of specific RNAs, polypeptides, structural proteins, proteinaceous biochemicals or enzymes which control the structure or functioning of specific traits. Certain genes form rRNAs, tRNAs and other small RNAs. Other gene transcribe mRNAs which contain coded information for synthesis of polypeptide.

14. Who proposed the Semiconservative DNA replication? 

Ans. Since there are 64 triplet codons and only 20 amino acids, the incorporation of same amino acids must be influenced by more than one codon. Out of 64 codons 3 are stop codons. Only two codons specify a single amino acids. They are AUG for methionine and UGG for tryptophan. Three amino acids have 6 codons each, 5 amino acids 4 codons, one amino acid with 3 cocoons and amino acids are specified for 2 codons each.

E. SHORT ANSWER QUESTION (3 MARKS)

1. Draw the structure of tRNA. 

Ans: 

(tRNA- the adaptger molecule) 

2. Describe the transformation experiment of Griffith. What was his conclusion?

Ans. In 1928 Frederick Griffith discovered the process of transformation in a bacterial strain, streptococcus pneumoniae which caused pneumonia in mammals. It has two strains disease causing S-strain and non pathogenic R-strain. He observed that

(i) When mouse is infected with S-strain, it died.

(ii) When infected with heat killed S-strain, no diseasee symptoms appeared.

(iii) When infected with R-strain. no disease appeared in mice & it survived.

(iv) When infected with heat killed S-strain and living R-strain disease appeared in the mice. He concluded that the R-strain bacteria had somehow been transformed by the heat killed S-strain bacteria. Some transforming principle transferred from the heat killed S-strain had enabled the R-strain to synthesise a smooth polysaccharide coat & become virulent. This is due to the transfer of the genetic material.

3. Write a note on biochemical characterization of transforming principle of Griffith.

Ans: Oswald Avery, Colin Macleod and Maclyn Mc Carty worked to determine the biochemical nature of transforming principle. They killed S type bacteria and separated DNA, proteins and carbohydrates. DNA component was subdivided into two; one with hydrolysing enzyme DNA- ase and the other without it. They added these different components over culture medium supporting Rtype bacteria. After an interval the bacteria were analysed. There was no change in the cultures having carbohydrates, proteins and DNA with DNA ase. The fourth culture having S – type DNA without DNAase was found to have some bacteria of S- type. They must have been formed from R – Type bacteria with the DNA of S type. Therefore DNA represents the genetic material.

4. Write the differences between DNA and RNA

Ans:

FeatureDNARNA
Full NameDeoxyribonucleic AcidRibonucleic Acid
SugarDeoxyriboseRibose
StructureDouble-stranded helixMostly single-stranded
BasesAdenine (A), Thymine (T), Cytosine (C), Guanine (G)Adenine (A), Uracil (U), Cytosine (C), Guanine (G)
FunctionStores genetic informationTransmits genetic information
LocationFound in the nucleusFound in the nucleus and cytoplasm
StabilityMore stableLess stable
Enzymes involvedDNA polymerase, DNA ligaseRNA polymerase, reverse transcriptase
TypesOne main type (double-stranded DNA)Several types (mRNA, tRNA, rRNA, etc.)
Role in protein synthesisProvides template for mRNA synthesisMessenger RNA (mRNA) carries the genetic code from DNA to ribosomes for protein synthesis

5. Briefly discuss the process of DNA replication

Ans: Involves a number of enzymes of which the main enzyme is DNA dependent DNA polymerase, that catalyses the polymerisation of deoxy nucleotides. The process involves the separation of intertwined strands of DNA from a particular point called origin of replication. Since the two strands cannot be separated in its entire length, replication occurs with small opening of the DNA helix; the Y shaped structure formed is called replication fork. The DNA polymerase catalyse polymerisation of the nucleotides only in 5’→3′ direction. Consequently on one of the template strands (3’5′) the synthesis of DNA is continuous while on the other template strand (5′-3′) the synthesis of DNA is discontinuous. The discontinuously synthesised strands are later joined together by the enzyme DNA ligase.

6. How did Hershey and Chase proved that DNA is the genetic material?

Ans. Hershey and Chase grew some bacteriophage virus on a medium that contained radioactive phosphorus (32P) and some in another medium with radioactive sulphur (35S). Viruses grown in the presence of 32p contained radioactive DNA, while viruses grown in presence of 35S contained radioactive protein. Both the radioactive virus types were then allowed to infect E-col: separately. Soon after infection the bacterial cells were gently agitated in blender to remove viral coats from the bacteria. The culture was also centrifuged to separate the viral particle, from the bactierial cell.

It was observed that bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Therefore DNA is the genetic material.

7. How is protein synthesis initiated in a cell?

Ans. The ribosome in its inactive state exists as two subunits- a large subunit and a small subunit when the small subunit encounters the mRNA translation begines. The mRNA binds to the small subunit of ribosome, following base pair rule between the bases of mRNA and those on rRNA which is catalysed by initiation factors. There are 2 sites on the larger subunit the P- site and the A-site. The small subunit (with the tRNA) attaches to the large subunit in such a way that the initiation codon.( AUG) comes on the P-site. The initiator tRNA (methionyl tRNA) binds to the P-site.

8. Draw a labelled schematic sketch of replication fork of DNA.

Ans:– 

9. Write 3 goals of Human Genome Project.

Ans. 1) To identify all the approximately 20,000-25,000 genes in human DNA.

2) To determine the sequences of the 3 billion chemical base pairs that make up human DNA.

3) To store this information in databases.

10. What is meant by semiconservation of DNA replication?

Ans. Watson and Crick suggested that the two strands of DNA would separate and each acts as a template for the synthesis of a new complementary strand. After complete replication each DNA molecule would have one parental and one newly synthesised strand. This is called semiconservation of DNA replication.

11. What is Frame shift insertion?

Ans. Frame shift insertion is a type of mutation in which insertion of one or two bases changes the reading frame from the point of insertion. Insertion of three or its multiple bases insert one or more multiple codon hence one or multiple amino acids and reading frame remains unaltered from that point onwards. This forms the genetic basis of proof that codon is a triplet and it is read in a contiguous manner.

12. What are the functions of DNA polymerases.

Ans. (i) DNA polymerase I is used in proof reading and repair.

(ii) DNA polymerase II is specialised repair enzyme.

(iii) DNA polymerase III is highly efficient with the ability to polymerise and actually takes part in replication.

13. What are the functions of mRNA and tRNA.

Ans. Functions of mRNA are-

(i) mRNA carries coded information for translation into polypeptide formation.

(iii) It has a cap region for attachment to ribosome.

(iv) mRNA strand contains a non-coding region, coding region and termination codon. Function of tRNA are

(i) tRNA is adapter molecule which is used for transferring amino acid to ribosomes for synthesis of polypeptides.

(ii) They hold peptidyl chains over the mRNAs.

(iii) The initiator tRNA initiates protein synthesis as well as brings the first amino acid.

14. Elucidated the lac operaon printing? Mention its application.

Ans. Lac operon is on inducible operon where lactose is the inducer. It is the substrata for the enzyme ẞ galactosidase. The components of lac operon and their functions are as follows-

(i) Structural genes: There are three structural genes (z, y, a) which transcribe a polycistronic mRNA. Gene ‘Z’ codes for ẞ- galactosidase, that catalyses the hydrolysis of lactose into galactose & glucose. Gene ‘Y’ codes for permease which increases the permeability of the cell to ẞ-Galactosidase gene ‘a’ codes for transacetylase.

(ii) Promoter: It is a sequence of bases near to structural gene. It is the site where RNA polymerase binds for transcription.

(iii) Operator: It is a sequence of bases of DNA near the promoter where a repressor always binds. It functions as a switch for the operon.

(iv) Repressor: It is a protein coded by i gene. It binds to the operator and prevents the RNA polymerase from transcribing.

(v) Inducer: Lactose is the inducer that inactivates the

[F] LONG ANSWER QUESTIONS (5 Marks)

1. Give a brief account of DNA replication.

Ans. Involves a number of enzymes of which the main enzyme is DNA dependent DNA polymerase, that catalyses the polymerisation of deoxy nucleotides. The process involves the separation of intertwined strands of DNA from a particular point called origin of replication. Since the two strands cannot be separated in its entire length, replication occurs with small opening of the DNA helix; the Y shaped structure formed is called replication fork. The DNA polymerase catalyse polymerisation of the nucleotides only in 5’→3′ direction. Consequently on one of the template strands (3’5′) the synthesis of DNA is continuous while on the other template strand (5′-3′) the synthesis of DNA is discontinuous. The discontinuously synthesised strands are later joined together by the enzyme DNA ligase.

2. Describe the structure of DNA with a neat labelled diagram.

Ans. DNA is made up of two polynucleotide chains where the backbone is made up of sugar and phosphate groups and the nitrogenous bases project towards the centre. The two chains have antiparallel polarity, one chain has 5′-3′ and the other 3’5′ polarity. The bases are paired through hydrogen bonds. Adenine forms two hydrogen bonds with Thymine and Guanine is bonded with cytosine with three hydrogen bonds. The diameter of the DNA duplex is 20A°. The distance between the base pairs in a helix is 0.34nm and a complete turn contain approximately 10 base pairs. Thus the pitch of the helix is 3.4nm and the two strands are right handed coiled.

3. Discuss how the DNA double helix packed to form chromatin in eukaryotes.

Ans: In eukaryotes DNA is stabilised with the help of a set of positively charged basic proteins called histones. Histones arc rich in basic amino acids, lysine and arginine. Therefore they carry positive charges on their side chains. Histones which are positively charged attract the negatively charged phosphate containing backbones of DNA double strand. Histones are organised to form a unit of eight molecules called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. Nucleosomes constitute the repeating unit of chromatin, which are thread like coloured bodies in the nucleus. The nucleosomes in chromatin are seen as beads on string which is packaged to form chromatin fibres, that are further coiled and condensed at metaphase stage of cell division to form chromosomes.

4. How Meselson and Stahl proved that DNA replication is semiconservative?

Ans: Meselson and Stahl performed the following experiment in 1958.

(i) They grew E.coli in a medium containing 15NH, CL (15N is the heavy isotope of N₂) as the only N₂ source for many generations. As a result, 15N got incorporated into newly synthesised DNA. This heavy DNA molecule could be distinguished from the normal DNA by centritugation in a caesium chloride density gradient.

(ii) Then they transferred the cells into a medium with normal 14NH4Cl and took samples at various definite time intervals as the cells multiplied and extracted the DNA that remained as double stranded helices. The various samples were separated independently on CsCl gradients to measure the densities of DNA.

(iii) Thus the DNA that was extracted from the culture one gernation after the transfer from 15N to 14N medium had an intermediate density. DNA extracted from the culture after another generation was composed of equal amounts of this hybrid DNA and of light DNA so it was proved the DNA replication is semiconservative.

5. Discuss the proces of translation in detail.

Ans: Translation is the process of polymerisation of amino acids to form a polypeptide. The various steps involved are (i) Activation of amino acids: In this step a particular amino acide becomes activated and attached to the 3′ end of a specific tRNA molecule. The reaction is catalysed by the enzyme amino acyle tRNA synthetase.

(ii) Initiation of polypeptide synthesis: The ribosome in its enactive state exists as two subunits a large subunit and a small subunit when the small subunit encounters the mRNA translation begins. The mRNA binds to the small subunit of ribosome following base pair rule between the bases of mRNA and those on rRNA catalysed by initiation factors. There are two sites on the larger subunit, the P-site and the A-site. The small submit (with the tRNA) attaches to the large submit in such a way that the initiation codon (AUG) comes on the P site. The initiator tRNA (metheonyl tRNA) binds to the P site.

(iii) Elongation of polypeptide chain: A second tRNA charged with an appropriate amino acid binds to the A-site of the ribosome. A peptide bond is formed between the carbonyl group of metheonine and the amino group of the second amino acid. The ribosome moves from codon to cedon along the mRNA in the 5′-3′ S direction and amino acid are added that joined together to form polypeptide.

(iv) Termination: When one of the termination codon reaches the A site, it does not code for any amino acid. As a result polypeptide synthesis stops and the polypeptide is released by the release factor.

6. Discuss the various applications of DNA fingerprinting. 

Ans: (i) DNA fingerprinting is used as a tool in forensic tests to identify criminals.

(ii) To identify the true biological father a mother in case of disputes. 

(iii) It is used to determine population and genetic diversities to study evolution.

(iv) It can be used to study the breeding pattern of animals facing the danger of extinction.

(v) It can identify racial groups, their origin, historical migrations and invasions.

7. Give five salient features of Human genome project.

Ans: (i) The human genome contains 3164.7 million nucleotide bases.

(ii) The average gene consists of 3000 bases with largest known human gene being dystrophin at 2.4 million bases.

(iii) The total number of genes is estimated at 30,000-much lower than previous estimate of 80,000-1, 40,000 genes. Almost all the nucleotide bases are exactly the same in all people.

(iv) The functions are unknown for over 50% of the discovered genes.

(v) Less than 2% of the genome codes for eroteens.

8. Write the salient goals of Human genome Project.

Ans: (i) To identify all the approximately 20,000-25,000 genes in human DNA.

(ii) to determine the sequences of the 3 billion chemical bases pairs that make up human DNA.

(iii) Store this information in databases.

(iv) Improve tools for data analysis.

(v) Transfer related technologies to other sectors, such as industries.

(vi) To solve the ethical, legal and social issues that may arise from the project. 

9. Discuss the methodologies involved in HGP.

Ans: The methods involve two major approaches

(i) Expressed sequence Tags (ESTs): This method focusses on identifying all the genes that are expressed as RNA.

(ii) Sequence annotation: It is an approach of simply sequencing the whole set of genome that contains all the coding and non coding sequences and later assigning different regions in the sequence with functions.

For sequencing first the total DNA from cell is isolated and broken down in relatively small sizes as fragments. These DNA fragments are cloned in suitable host using suitable vectors. When hacteria is used as vector they are called bacterial artificial chromosomes (BAC) and when yeast is used as vector, they are called yeast artificial chromsome (YAC). Frederick Sanger developed a principle according to which the fragments of DNA are sequenced by automated DNA sequences. On the basis of overlapping regions on DNA fragments, these sequences are arranged accordingly. For alignment of these sequences, specialised computer based programmes were developed. Finally the genetic and physical maps of the genome were constructed by collecting information about certain repetitive DNA sequences and DNA polymorphism, based on endonuclease recognition sites.

10. Discuss in detail the lac operon developed by Jacob and Monod.

Ans: Lac operon is an inducible operon. Where lactose is the inducer, it is the substrate for the enzyme ẞ- gatactosidase The components of lac operon and their functions are as follows-

(i) Structural genes: There are three structural genes (z, y, a) which transcribe a polycistronic mRNA. Gene ‘Z’ codes b- galactosidase, that catalyses the hydrolysis of lactose into galactose and glucose Gene ‘y codes for permease, which increases the permeability, of the cell to ẞ- galactosides (lactose). Gene ‘a’ codes for transacetylase, the catalyses the transacetylation of lactose into its active form

(ii) Promoter: It is a sequence of bases near to the structural genes, it is the site where RNA polymerase binds for transcription.

(iii) Operator: It is a sequence of bases of DNA near the promoter, where a repressor always binds. It functions as switch for the operon. a

(iv) Repressor: It is a protein coded by i gene. It binds to the operator and prevents the RNA polymerase from transcribing (v) Inducer: Lactose is the inducer that inactivates the repressor and prevents it from binding to the operator. This allows an access from the RNA polymerase to the promoter and transcription. Thus the substrate lactose regulates the lac operon. 

11. Draw a labelled diagrm of the double helical structure of a DNA straid.

Ans:

12. How did Griffith explain the transformation of strain bacteria to S-strain bacteria.

Ans: Griffith discovered the process of transformation in bacterial strain streptococcus pneumoniae which causes pneumonis in mammals. It has two strains. One type called S-type cause disease (virulent). It has capsule around its body. The other type called R type is without capsule and is non-virulent. He observed that

(I) When mouse is Injected with S-strain it died.

(ii) When injected with R-strain nor disease appeared and the mice survived.

(iii) When injected with heat killed S-stain no discase symptoms appeared.

(iv) When injected with heat killed S-strain and living R-strain disease appeared in the mice.

It was observed that a virulent bacteria turned to be virulent and killed the mice while the virulent bacteria that were injected along with it were totally killed. Even the bacteria isolated from dead mice showed that the non-capsulated avirulent type became virulent and capsulated type. It indicated that the non virulent bacteria acquired the genetic character of the virulent

Griffith concluded that the presence of the heat killed S-bacteria must have caused a transformation of the living R-bacteria, so as to restore them the capacity for capsule formation they had earlier lost by gene mutation.

13. Explain the southern blot hybridisation of DNA fingerprinting.

Ans: In this technique, a DNA molecule is cut into discrete fragments by a restriction enzyme. It is electrophoresed through an agarose gel which separates the various fragments according to size. The DNA is then denatured into single strands. A large piece of nitrocellulose paper is laid over the agarose gel and a few field of filter paper seated in buffer are placed under the gel Several layers of absorbent material such as filter papers are placed above nitrocellulose paper. This dry absorbent paper pulls the buffer up through the gel from the lower layer. While passing through the gel the buffer carries with it single stranded DNA which binds to the nitrocellulose membrane.

The nitrocellulose filter containing the DNA is first dried and then exposed to a solution by 32P labelled mRNA called molecular probe from the gene to be isolated. The radioactive mRNA hybridizes only with the single stranded DNA in restriction fragments that contain complementary sequences. The nitrocellulose filter is then removed and placed in contact with photographic film that when developed will reveal fragments from the original gel containing complementary sequences to the mRNA used in the assay. The procedure allows specific identification of restriction fragments containing DNA sequences to specific RNA molecules.

14. Elucidated the lac operaon printing? Mention its application.

Ans. Lac operon is on inducible operon where lactose is the inducer. It is the substrata for the enzyme ẞ galactosidase. The components of lac operon and their functions are as follows-

(i) Structural genes: There are three structural genes (z, y, a) which transcribe a polycistronic mRNA. Gene ‘Z’ codes for ẞ- galactosidase, that catalyses the hydrolysis of lactose into galactose & glucose. Gene ‘Y’ codes for permease which increases the permeability of the cell to ẞ-Galactosidase gene ‘a’ codes for transacetylase.

(ii) Promoter: It is a sequence of bases near to structural gene. It is the site where RNA polymerase binds for transcription.

(iii) Operator: It is a sequence of bases of DNA near the promoter where a repressor always binds. It functions as a switch for the operon.

(iv) Repressor: It is a protein coded by i gene. It binds to the operator and prevents the RNA polymerase from transcribing.

(v) Inducer: Lactose is the inducer that inactivates the repressor and prevents it from binding to the operator. This allows an access from the RNA polymerase to the promoter and transcription takes place. Thus the substrate lactose regulates the lac operon.

15. What is DNA finger printing? Mention its application.

Ans. DNA fingerprinting is the technique used for determining nucleotide sequences of some specific regions in DNA called as repetitive DNA, which are unique to each individual.

Applications :

1. DNA fingerprinting is a sure method of identification of criminals involved in various types of crimes.

2. The method can provide reliable information to do paternity testing.

3. It provides information as to human lineage and relationship with apes.

-0000-

Last Words on AHSEC Class 12 Biology Chapter: 5 Molecular Basis of Inheritance

The Chapter 5 of AHSEC Class 12 Biology deals with Molecular Basis of Inheritance for a brief understanding. You can Download this HS 2nd Year Biology Notes in PDF 2025.

Leave a comment