NCERT Exemplar Class 12 Biology Chapter 6 Molecular Basis Of Inheritance

Last Updated: August 30, 2024Categories: NCERT Solutions

NCERT Exemplar for Class 12 Biology Chapter 6

Exemplar Class 12 Biology Chapter 6 helps students understand complex topics like the Molecular Basis of Inheritance. SimplyAcad offers this exemplar to cover all areas of Chapter 6 in the biology textbook. Subject experts have designed a variety of questions, including MCQs, very short, short, and long answer types. These questions help students become familiar with different exam patterns, giving them an advantage in their studies.

Students can easily access the NCERT Exemplar for Class 12 Biology below. It serves as a valuable resource for performing well in the upcoming 12th board exams. Additionally, SimplyAcad provides detailed exemplars for other Class 12 Science chapters as well.

MCQs and Solutions for Exemplar Class 12 Biology Chapter 6

Question 1:

In a DNA strand, the nucleotides are linked together by:

  • glycosidic bonds
  • phosphodiester bonds
  • peptide bonds
  • hydrogen bonds

Solution:

Option (b) is the answer.

Question 2:

A nucleoside differs from a nucleotide. It lacks the:

  • base
  • sugar
  • phosphate group
  • hydroxyl group

Solution:

Option (c) is the answer.

Question 3:

Both deoxyribose and ribose belong to a class of sugars called:

  • trioses
  • hexoses
  • pentoses
  • polysaccharides

Solution:

Option (c) is the answer.

Question 4:

The fact that a purine base always pairs through hydrogen bonds with a pyrimidine base in the DNA double helix leads to:

  • the antiparallel nature
  • the semiconservative nature
  • uniform width throughout DNA
  • uniform length in all DNA

Solution:

Option (c) is the answer.

Question 5:

The net electric charge on DNA and histones is:

  • both positive
  • both negative
  • negative and positive, respectively
  • zero

Solution:

Option (c) is the answer.

Question 6:

The promoter site and the terminator site for transcription are located at:

  • 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit
  • 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit
  • the 5′ (upstream) end
  • the 3′ (downstream) end

Solution:

Option (b) is the answer.

Question 7:

Which of the following statements is the most appropriate for sickle cell anaemia?

  • It cannot be treated with iron supplements
  • It is a molecular disease
  • It confers resistance to acquiring malaria
  • All of the above

Solution:

Option (d) is the answer.

Question 8:

Which of the following is true with respect to AUG?

  • It codes for methionine only
  • It is an initiation codon
  • It codes for methionine in both prokaryotes and eukaryotes
  • All of the above

Solution:

Option (d) is the answer.

Question 9:

The first genetic material could be:

  • protein
  • carbohydrates
  • DNA
  • RNA

Solution:

Option (d) is the answer.

Question 10:

With regard to mature mRNA in eukaryotes:

  • exons and introns do not appear in the mature RNA
  • exons appear but introns do not appear in the mature RNA
  • introns appear but exons do not appear in the mature RNA
  • both exons and introns appear in the mature RNA

Solution:

Option (b) is the answer.

Question 11:

The human chromosome with the highest and least number of genes in them are respectively:

  • Chromosome 21 and Y
  • Chromosome 1 and X
  • Chromosome 1 and Y
  • Chromosome X and Y

Solution:

Option (c) is the answer.

Question 12:

Who amongst the following scientists had no contribution to the development of the double-helix model for the structure of DNA?

  • Rosalind Franklin
  • Maurice Wilkins
  • Erwin Chargaff
  • Meselson and Stahl

Solution:

Option (d) is the answer.

Question 13:

DNA is a polymer of nucleotides which are linked to each other by 3’-5’ phosphodiester bond. To prevent the polymerisation of nucleotides, which of the following modifications would you choose?

  • Replace purine with pyrimidines
  • Remove/Replace 3′ OH group in deoxyribose
  • Remove/Replace 2′ OH group with some other group in deoxyribose
  • Both ‘b’ and ‘c’

Solution:

Option (b) is the answer.

Question 14:

Discontinuous synthesis of DNA occurs in one strand because:

  • DNA molecule being synthesised is very long
  • DNA dependent DNA polymerase catalyses polymerisation only in one direction (5’ → 3’)
  • it is a more efficient process
  • DNA ligase joins the short stretches of DNA

Solution:

Option (b) is the answer.

Question 15:

Which of the following steps in transcription is catalysed by RNA polymerase?

  • Initiation
  • Elongation
  • Termination
  • All of the above

Solution:

Option (d) is the answer.

Question 16:

Control of gene expression in prokaryotes takes place at the level of:

  • DNA-replication
  • Transcription
  • Translation
  • None of the above

Solution:

Option (b) is the answer.

Question 17:

Which of the following statements is correct about the role of regulatory proteins in transcription in prokaryotes?

  • They only increase expression
  • They only decrease expression
  • They interact with RNA polymerase but do not affect the expression
  • They can act both as activators and as repressors

Solution:

Option (d) is the answer.

Question 18:

Which was the last human chromosome to be completely sequenced?

  • Chromosome 1
  • Chromosome 11
  • Chromosome 21
  • Chromosome X

Solution:

Option (a) is the answer.

Question 19:

Which of the following are the functions of RNA?

  • It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
  • It carries amino acids to ribosomes.
  • It is a constituent component of ribosomes.
  • All of the above.

Solution:

Option (d) is the answer.

Question 20:

While analysing the DNA of an organism, a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule, it can be concluded that:

  • it is a double-stranded circular DNA
  • It is single-stranded DNA
  • It is a double-stranded linear DNA
  • No conclusion can be drawn

Solution:

Option (b) is the answer.

Question 21:

In some viruses, DNA is synthesised by using RNA as a template. Such a DNA is called:

  • A-DNA
  • B-DNA
  • cDNA
  • rDNA

Solution:

Option (c) is the answer.

Question 22:

If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of N15/N15: N15/N14: N14/N14 containing DNA in the fourth generation would be:

  • 1:1:0
  • 1:4:0
  • 0:1:3
  • 0:1:7

Solution:

Option (d) is the answer.

Question 23:

If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is: 5′ – A T G A A T G – 3′, the sequence of bases in its RNA transcript would be:

  • 5′ – A U G A A U G – 3′
  • 5′ – U A C U U A C – 3′
  • 5′ – C A U U C A U – 3′
  • 5′ – G U A A G U A – 3′

Solution:

Option (a) is the answer.

Question 24:

The RNA polymerase holoenzyme transcribes:

  • the promoter, structural gene and the terminator region
  • the promoter and the terminator region
  • the structural gene and the terminator region
  • the structural gene only

Solution:

Option (d) is the answer.

Question 25:

If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be:

  • 5′ – UAC – 3′
  • 5′ – CAU – 3′
  • 5′ – AUG – 3′
  • 5′ – GUA – 3′

Solution:

Option (b) is the answer.

Question 26:

The amino acid attaches to the tRNA at its:

  • 5′ – end
  • 3′ – end
  • Anticodon site
  • DHU loop

Solution:

Option (b) is the answer.

Question 27:

To initiate translation, the mRNA first binds to:

  • The smaller ribosomal sub-unit
  • The larger ribosomal sub-unit
  • The whole ribosome
  • No such specificity exists.

Solution:

Option (a) is the answer.

Question 28:

In E.coli, the lac operon gets switched on when:

  • lactose is present and it binds to the repressor
  • repressor binds to the operator
  • RNA polymerase binds to the operator
  • lactose is present and it binds to RNA polymerase

Solution:

Option (a) is the answer.

 

Very Short Answer Questions Exemplar Class 12 Biology Chapter 6

Question 1:

What is the function of histones in DNA packaging?

Solution:

DNA wraps itself around a protein called histone. Histones pack into structural units called nucleosomes.

Question 2:

Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?

Solution:

Heterochromatin is a highly packed form of DNA in the chromosomes and stains dark. Euchromatin is a loosely packed form of DNA in the chromosomes and stains lightly.

Euchromatin has transcriptionally active regions of DNA. Heterochromatin has transcriptionally inactive regions of DNA.

Question 3:

The enzyme DNA polymerase in E. coli is a DNA-dependent polymerase and also has the ability to proof-read the DNA strand being synthesized. Explain. Discuss the dual polymerase.

Solution:

There are three types of DNA polymerase present in the bacteria which add nucleotides in the 5’-3’ direction. DNA III polymerases can proofread the newly formed strand and can sense the wrong base insertions.

Question 4:

What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesized DNA?

Solution:

DNA III polymerases can proofread the newly formed strand and can sense the wrong base insertions. An additional complication will be created at the replicating fork. The left-out strand of the DNA which was 5’-3’ has to be synthesized in the opposite direction as short stretches in a discontinuous manner.

Question 5:

Given below is the sequence of the coding strand of DNA in a transcription unit 3 ‘A A T G C A G C T A T T A G G – 5’ write the sequence of:

a) its complementary strand
b) the mRNA

Solution:

a) Complementary strand: 5’-T T A C G T C G A T A A T C C-3’
b) The mRNA: 5’ A A U G C A G C U A U U G G-3’

Question 6:

What is DNA polymorphism? Why is it important to study?

Solution:

The difference in the nucleotide sequence between the individuals is called DNA polymorphism. It is important for genetic variation and very useful for criminal cases to find the culprit.

Question 7:

Based on your understanding of the genetic code, explain the formation of any abnormal haemoglobin molecule. What are the known consequences of such a change?

Solution:

This abnormal haemoglobin molecule formation occurs because one codon GAG gets replaced by GUG, which means in the codon GAG adenosine gets replaced by uracil. This change leads to the incorporation of valine in the beta haemoglobin chain instead of Glutamic acid at the 6th position.

Question 8:

Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/position of the eye(s) etc. Comment.

Solution:

It happens due to uncoordinated regulation of gene expression in the gene sets associated with organ development.

Question 9:

In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxyribonucleoside triphosphates, but only deoxyribonucleotides are added during DNA replication. Suggest a mechanism.

Solution:

The DNA polymerase is highly specific to recognize only deoxyribonucleoside triphosphates, which ensures that only deoxyribonucleotides are added during DNA replication.

Question 10:

Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.

Solution:

  • Helicases: Separate the two DNA strands at the replication fork using energy.
  • DNA Clamp: Promotes the replication of the DNA by binding the DNA polymerase enzyme to the template strand and preventing it from disassociating.
  • Single-Stranded Binding Proteins: Prevent the single strand of DNA from getting digested by the nucleases and from forming secondary structures.

Question 11:

Name any three viruses which have RNA as the genetic material.

Solution:

  • Tobacco Mosaic Virus
  • Human Immunodeficiency Virus (HIV)
  • Influenza Virus

Short Answer Type Questions Exemplar Class 12 Biology Chapter 6

Question 1:

Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material.

Solution:

In 1928, Frederick Griffith experimented with two strains of bacteria Streptococcus Pneumoniae, the S-strain (smooth strain) and the R-strain (rough strain). When injected into mice, only the S-strain caused the disease and was virulent due to its polysaccharide coating, which protected it from the mouse’s immune system. The R-strain was non-virulent. The transformation of the non-virulent R-strain into the virulent S-strain indicated a genetic change. This experiment showed that DNA was the genetic material responsible for the transformation.

Question 2:

Who revealed the biochemical nature of the transforming principle? How was it done?

Solution:

Oswald Avery, Colin McLeod, and Maclyn McCarty revealed the biochemical nature of the transforming principle in 1933-34. They purified the biochemicals (DNA, RNA, and Protein) from the heat-killed S-strain cells and found that only DNA was responsible for the transformation of the non-virulent R-strain to the virulent S-strain.

Question 3:

Discuss the significance of the heavy isotope of nitrogen in the Meselson and Stahl experiment.

Solution:

The heavy isotope of nitrogen (N15) in the Meselson and Stahl experiment showed that DNA is of intermediate density between N14 and N15. This indicated that DNA replication is semi-conservative, with each daughter DNA molecule containing one original and one newly synthesized strand.

Question 4:

Define a cistron. Giving examples, differentiate between monocistronic and polycistronic transcription units.

Solution:

A cistron is a segment of DNA that contains the genetic code for a single polypeptide chain.

  • Monocistronic transcription unit: Contains a single cistron and codes for a single protein. It is found mainly in eukaryotes.
  • Polycistronic transcription unit: Contains multiple cistrons and codes for several proteins. It is found mainly in prokaryotes.

Question 5:

Give any six features of the human genome.

Solution:

  1. The genome has around 3164.7 million nucleotide bases.
  2. 99.9% of the nucleotide bases are the same in all humans.
  3. The largest gene is Dystrophin, having 2.4 million bases.
  4. Chromosome 1 has the most genes (2968), and Y has the least genes (231).
  5. Less than 2% of the genome has the coding sequence for proteins.
  6. Only 50% of the total discovered genes have known functions.

Question 6:

During DNA replication, why is it that the entire molecule does not open in one go? Explain the replication fork. What are the two functions that the monomers (dNTPs) play?

Solution:

The entire DNA molecule does not open in one go because it creates tension in the molecule, leading to the formation of supercoils. Replication occurs within a small opening of the DNA helix called the replication fork. Monomers provide energy for the polymerization reaction and supply deoxyribonucleotides for DNA replication.

Question 7:

Retroviruses do not follow central dogma. Comment.

Solution:

Retroviruses do not follow central dogma because their genetic material is RNA, not DNA. RNA is converted to DNA through reverse transcription using the enzyme reverse transcriptase, bypassing the traditional flow of genetic information from DNA to RNA to protein.

Question 8:

In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive bases increases from 0.34 nm to 0.44 nm. Calculate the length of the DNA double helix (which has 2×10⁹ bp) in the presence of a saturating amount of this compound.

Solution:

Distance between two consecutive base pairs: 0.44 nm or 0.44 x 10⁻⁹ m.

Length of the DNA: 2 x 10⁹ x 0.44 x 10⁻⁹ m

Length of DNA = 0.88 m

Question 9:

What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?

Solution:

If histones were mutated to be rich in acidic amino acids, they would become negatively charged and would not bind to the negatively charged DNA. This would prevent DNA from packing properly into chromatin, disrupting chromatin structure.

Question 10:

Recall the experiments done by Frederick Griffith, Avery, MacLeod, and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA, was the genetic material, would the heat-killed strain of Pneumococcus have transformed the R-strain into a virulent strain? Explain.

Solution:

If RNA was the genetic material, the heat-killed strain of Pneumococcus would not have transformed the R-strain into the virulent strain because RNA is not thermostable. The RNA would have been degraded by heat, preventing the transformation.

Question 11:

You are repeating the Hershey-Chase experiment and are provided with two isotopes: ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?

Solution:

³²P is radioactive and ¹⁵N is non-radioactive. If ¹⁵N were hypothetically radioactive, it would be found both inside the cell (incorporated into DNA) and in the supernatant (incorporated into amino acids). However, since ¹⁵N is non-radioactive, its presence wouldn’t be detected.

Question 12:

There is only one possible sequence of amino acids when deduced from given nucleotides. But multiple nucleotide sequences can be deduced from a single amino acid sequence. Explain this phenomenon.

Solution:

Some amino acids are coded by more than one codon, a phenomenon known as the degeneracy of the codon. Due to this degeneracy, multiple nucleotide sequences can be derived from a single amino acid sequence.

Question 13:

A single base mutation in a gene may not ‘always’ result in the loss or gain of function. Do you think the statement is correct? Defend your answer.

Solution:

Yes, the statement is correct. If the mutation occurs at the third base of a codon, it often does not lead to a phenotypic change. These mutations are known as silent mutations.

Question 14:

A low level of expression of the lac operon occurs at all times. Can you explain the logic behind this phenomenon?

Solution:

A low level of expression of the lac operon is necessary for the formation of permease within the bacteria. Permease allows lactose to enter the bacterium, which can then induce the operon fully.

Question 15:

How has the sequencing of the human genome opened new windows for the treatment of various genetic disorders? Discuss.

Solution:

Sequencing the human genome has provided a better understanding of genetic disorders, allowing for improved diagnosis, treatment, and prevention. It has also led to the development of personalized medicine and therapies targeted to specific genetic profiles.

Question 16:

The total number of genes in humans is far less (< 25,000) than the previous estimate (up to 1,40,000 genes). Comment.

Solution:

The initial overestimation was due to large portions of repetitive sequences in the human genome. When only non-repetitive coding regions are considered, the gene count is much lower.

Question 17:

Now, the sequencing of total genomes is getting less expensive day by day. Soon it may be affordable for a common man to get his genome sequenced. What is your opinion on the advantages and disadvantages of this development?

Solution:

Advantages:

  • Better diagnosis, treatment, and prevention of genetic disorders.
  • Enhanced understanding of biological systems.

Disadvantages:

  • Potential for genetic discrimination and privacy concerns.
  • Ethical issues related to gene patenting and the treatment of untreatable genetic disorders.

Question 18:

Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?

Solution:

DNA fingerprinting using VNTRs would not be appropriate for a bacteriophage because it lacks repetitive sequences like VNTRs. A bacteriophage has a small genome with coding sequences, so a pattern of bands would not form.

Question 19:

During in vitro synthesis of DNA, a researcher used 2’, 3’-dideoxy cytidine triphosphate as a raw nucleotide in place of 2’-deoxycytidine. What would be the consequence?

Solution:

Polymerization would not take place because the 3’OH group on the sugar is not present, preventing the formation of ester bonds between nucleotides.

Question 20:

What background information did Watson and Crick have to develop a model of DNA? What was their contribution?

Solution:

Background Information:

  • Chargaff’s law showed that Adenosine pairs with Thymine (A=T) and Guanine pairs with Cytosine (G=C).
  • Wilkins and Franklin’s X-ray diffraction data showed DNA has a diameter of 20 Å, a regular helix structure with 34 Å distance, and 10 base pairs per turn.

Contribution:

Watson and Crick proposed the double helix structure of DNA with complementary base pairing by hydrogen bonds.

Question 21:

What are the functions of (i) methylated guanosine cap, (ii) poly-A “tail” in a mature RNA?

Solution:

(i) Methylated guanosine cap: Helps bind mRNA to the smaller ribosomal subunit during the initiation of translation.

(ii) poly-A “tail:” Prolongs the life of mRNA.

Question 22:

Do you think that the alternative splicing of exons may enable a structural gene to code for several isoproteins from the same gene? If yes, how? If not, why?

Solution:

Yes, alternative splicing of exons can lead to the encoding of various proteins from a single gene by generating different combinations of exons. This allows for the production of multiple protein isoforms from a single gene.

Question 23:

Comment on the utility of variability in the number of tandem repeats during DNA fingerprinting.

Solution:

Variability in the number of tandem repeats (VNTRs) is unique for each individual, making them useful in DNA fingerprinting for identifying individuals in forensic cases.

Long Answer Type Questions Exemplar Class 12 Biology Chapter 6

Question 1

Give an account of the Hershey and Chase experiment. What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur, do you think the result would have been the same?

Solution:

Alfred Hershey and Martha Chase conducted an experiment in 1952 using bacteriophages to determine the genetic material. They grew bacteriophages in two separate mediums, one containing radioactive phosphorus (³²P) and the other containing radioactive sulfur (³⁵S). The radioactive phosphorus labeled the DNA, while the radioactive sulfur labeled the protein coat. After allowing the phages to infect E. coli bacteria, they found that only the radioactive DNA entered the bacterial cells, proving that DNA is the genetic material, not proteins.

Conclusion:

  • DNA is the genetic material that transfers from one cell to another, not protein.

If both DNA and proteins contained phosphorus and sulfur:

  • The experiment would not have yielded conclusive results as both DNA and proteins would have been labeled, making it impossible to distinguish between them.

Question 2

During evolution, why was DNA chosen over RNA as the genetic material? Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the light of biochemical differences between DNA and RNA.

Solution:

Desired Criteria for Genetic Material:

  1. Chemically and structurally stable.
  2. Able to replicate for inheritance.
  3. Capable of mutation and evolution.
  4. Expresses itself in Mendelian traits.
  5. Transferable from parent to progeny.

Biochemical Differences:

  • DNA: Less reactive and more stable chemically and structurally. Contains thymine instead of uracil, providing additional stability.
  • RNA: More reactive and less stable, making it prone to rapid mutation.
  • DNA: Double-stranded, providing a mechanism for error correction during replication.
  • RNA: Single-stranded, more prone to damage and errors.

Conclusion:

  • These biochemical differences make DNA more suitable as the genetic material over RNA.

Question 3

Give an account of post-transcriptional modifications of a eukaryotic mRNA.

Solution:

Post-transcriptional modifications are chemical alterations made to the primary RNA transcript, converting it into functional RNA.

Steps:

  1. Splicing: In eukaryotes, the primary transcript (hnRNA) contains both introns (non-coding regions) and exons (coding regions). Splicing removes introns and joins exons in a definite order to form mature mRNA.
  2. Capping: Addition of a methylated guanosine triphosphate nucleotide at the 5′ end of hnRNA, forming the cap structure.
  3. Tailing: Addition of adenylate residues at the 3′ end of the RNA, known as the poly-A tail.

After these modifications, hnRNA is processed into mRNA and transported out of the nucleus for translation.

Question 4

Discuss the process of translation in detail.

Solution:

Translation: The process of decoding mRNA to form an amino acid chain (protein) with the help of ribosomes in the cytosol of the cell.

Steps Involved:

  1. Initiation:
    • The smaller 40S ribosomal subunit with methionyl-tRNA scans the mRNA to find the start codon (AUG), specific to methionine.
    • The larger 60S ribosomal subunit binds to the mRNA, forming the complete ribosome.
    • The ribosome has two binding sites for tRNA:
      • P Site: Holds the growing peptide chain.
      • A Site: Holds the incoming aminoacyl-tRNA.
  2. Elongation:
    • Met-tRNA binds to the P site of the ribosome, while another aminoacyl-tRNA with an anticodon complementary to the next codon occupies the A site.
    • Peptidyl transferase forms a peptide bond between the amino acids, elongating the polypeptide chain.
    • The ribosome moves along the mRNA, and the uncharged tRNA leaves the P site, allowing the next aminoacyl-tRNA to enter the A site.
  3. Termination:
    • When the stop codon enters the A site, no tRNA binds to it.
    • The peptide bond between the polypeptide and tRNA is hydrolyzed, releasing the polypeptide into the cytoplasm.
    • The ribosomal subunits dissociate.

Question 5

Define an operon. Giving an example, explain an Inducible operon.

Solution:

Operon: A cluster of genes with related functions, regulated together, and transcribed as a single mRNA molecule.

Example: Lac Operon (Inducible Operon):

  • The lac operon is found in E. coli and is involved in the metabolism of lactose. It is an inducible operon, meaning it is usually off but can be turned on in the presence of an inducer (lactose).

Components of Lac Operon:

  • Structural Genes (z, y, a): Code for enzymes involved in lactose metabolism.
  • Promoter (P): Binding site for RNA polymerase.
  • Operator (O): Binding site for the repressor protein.
  • Regulatory Gene (i): Codes for the repressor protein.

Mechanism:

  • In the absence of lactose: The repressor protein binds to the operator, blocking RNA polymerase from transcribing the structural genes, keeping the operon off.
  • In the presence of lactose: Lactose acts as an inducer by binding to the repressor protein, causing it to change shape and release from the operator. RNA polymerase can then transcribe the structural genes, leading to the production of enzymes that metabolize lactose.

Question 6

‘There is a paternity dispute for a child’. Which technique can solve the problem? Discuss the principle involved.

Solution:

Technique: DNA fingerprinting

Principle:

  • DNA fingerprinting is based on the unique pattern of tandem repeats (Variable Number of Tandem Repeats – VNTRs) in an individual’s DNA. VNTRs are short sequences of DNA that are repeated multiple times in a specific pattern. The number and arrangement of these repeats are unique to each individual, making DNA fingerprinting a reliable method for identifying individuals and resolving paternity disputes.

Steps:

  1. DNA Extraction: DNA is extracted from the cells of the child, mother, and alleged father.
  2. DNA Fragmentation: The DNA is cut into fragments using restriction enzymes.
  3. Separation: The DNA fragments are separated using gel electrophoresis, creating a pattern of bands.
  4. Comparison: The band patterns of the child’s DNA are compared with those of the mother and alleged father. The child’s DNA will have bands that match both parents, allowing for the determination of paternity.

Question 7

Give an account of the methods used in sequencing the human genome.

Solution:

DNA Sequencing: The process of determining the precise order of nucleotides within a DNA molecule.

Methods Used:

  1. Maxam-Gilbert Method (Chemical Cleavage Method):
    • Developed in 1977 by Allan Maxam and Walter Gilbert.
    • DNA is chemically cleaved at specific positions using chemicals like dimethyl sulfate (for purines) and hydrazine (for pyrimidines).
    • The cleaved fragments are separated by size using gel electrophoresis to determine the sequence.
  2. Sanger Method (Chain Termination Method):
    • Developed by Frederick Sanger in 1977.
    • As a result, DNA is synthesized in the presence of dideoxynucleotides (ddNTPs), which terminate the chain when incorporated.
    • The resulting fragments are separated by size using gel electrophoresis, and the sequence is determined by reading the order of termination.
  3. High-throughput Sequencing (Next-Generation Sequencing):
    • Modern sequencing techniques that allow for the rapid sequencing of entire genomes.
    • Certainly it  involves fragmenting the DNA, amplifying the fragments, and sequencing them simultaneously.
    • Examples include Illumina sequencing, Pyrosequencing, and Ion Torrent sequencing.

Application: The human genome was sequenced using a combination of these methods, leading to the identification of approximately 3.2 billion base pairs and around 20,000-25,000 genes.

Question 8

List the various markers that are used in DNA fingerprinting.

Solution:

  • Restriction Fragment Length Polymorphisms (RFLPs): Variations in the length of DNA fragments produced by restriction enzyme digestion.
  • Random Amplified Polymorphic DNAs (RAPDs): DNA polymorphisms generated by amplifying random DNA segments using PCR.
  • Amplified Fragment Length Polymorphisms (AFLPs): DNA polymorphisms detected by amplifying restriction fragments.
  • Simple Sequence Repeats (SSRs) / Microsatellites: Repeated sequences of 1-6 base pairs used as markers.
  • Single Nucleotide Polymorphisms (SNPs): Variations in a single nucleotide at specific positions in the genome.

These markers are used to create unique DNA profiles for individuals. These are useful in forensic analysis, paternity testing, and genetic studies.

9. Replication was allowed to take place in the presence of radioactive deoxynucleotides precursors in E. coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result?

Solution:

Option (d) is the correct result.

NCERT Exemplar For class 12 biology molecular basis of inheritance

Students must practise these additional questions for their own benefits, the ncert exemplar are curated by the best subject-matter experts to boost your knowledge on the presented topic. Students can easily access the ncert exemplar for class 12 science by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks. 

 

Here are some other NCERT exemplar for class 12 biology: 

NCERT exemplar for class 12 biology Chapter 1 NCERT exemplar for class 12 biology Chapter 8
NCERT exemplar for class 12 biology Chapter 2 NCERT exemplar for class 12 biology Chapter 10
NCERT exemplar for class 12 biology Chapter 3 NCERT exemplar for class 12 biology Chapter 12
NCERT exemplar for class 12 biology Chapter 5 NCERT exemplar for class 12 biology Chapter 13
NCERT exemplar for class 12 biology Chapter 7 NCERT exemplar for class 12 biology Chapter 14

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

  • Why is the NCERT Exemplar important for Class 12 Biology students?
    • Ans. The NCERT Exemplar is important because it helps students understand difficult concepts, like the Molecular Basis of Inheritance. Certainly this practice boosts their performance in Class 12 board exams.
  • How can I access the NCERT Exemplar for Class 12 Biology Chapter 6?
    • Ans. Students can easily access the NCERT Exemplar by visiting our website, SimplyAcad. We offer all the resources needed, including detailed coverage of Chapter 6 and other chapters of Class 12 Biology.
  • What types of questions does the NCERT Exemplar for Chapter 6 include?
    • Ans. The NCERT Exemplar offers various types of questions, like MCQs, very short, short, and long answer questions. Subject experts have carefully crafted these questions to cover all parts of the chapter, making it easier for students to understand tough topics and prepare well for their exams.

editor’s pick

latest video

news via inbox

Nulla turp dis cursus. Integer liberos  euismod pretium faucibua

you might also like