Molecular Basis of InheritanceClass 12 Biology NCERT Solutions
14 Solutions
Generated by KedovoAI
Solution 1 of 14
Q1EXERCISES
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Solution
Based on their chemical structure, the given molecules are grouped as follows:
Nitrogenous Bases: These are the fundamental purine or pyrimidine structures.
- Adenine
- Thymine
- Uracil
- Cytosine
Nucleosides: These are formed when a nitrogenous base is linked to a pentose sugar.
- Cytidine (Cytosine + sugar)
- Guanosine (Guanine + sugar)
Q2EXERCISES
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Solution
The calculation is based on Erwin Chargaff's rules for double-stranded DNA.
Given:
- Percentage of Cytosine (C) = 20%
To Find:
- Percentage of Adenine (A)
Principle (Chargaff's Rule):
- The amount of Adenine (A) is equal to the amount of Thymine (T). So, A = T.
- The amount of Guanine (G) is equal to the amount of Cytosine (C). So, G = C.
- The total percentage of all four bases is 100%. So, A + T + G + C = 100%.
Calculation:
-
According to the rule, the percentage of Guanine (G) is equal to the percentage of Cytosine (C). Therefore, G = 20%.
-
The total percentage of Guanine and Cytosine is:
-
The remaining percentage is for Adenine and Thymine:
-
Since the percentage of Adenine (A) is equal to the percentage of Thymine (T):
Final Answer: The per cent of adenine in the DNA is 30%.
Q3EXERCISES
If the sequence of one strand of DNA is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of complementary strand in 5'->3' direction.
Solution
To find the sequence of the complementary strand, we apply the base-pairing rule (A with T, and G with C) and consider the anti-parallel nature of the DNA strands.
Given Strand:
Step 1: Write the complementary sequence in the 3' to 5' direction.
We pair each base with its complement:
A pairs with T
T pairs with A
G pairs with C
C pairs with G
Original (5'→3'): A T G C A T G C A T G C A T G C A T G C A T G C A T G C
Complementary (3'→5'): T A C G T A C G T A C G T A C G T A C G T A C G T A C G
So, the complementary strand in the 3'→5' direction is:
Step 2: Reverse the sequence to get it in the 5' to 3' direction.
To write the sequence in the standard 5'→3' direction, we read the sequence from Step 1 from right to left.
Final Answer: The sequence of the complementary strand in the 5'→3' direction is:
Q4EXERCISES
If the sequence of the coding strand in a transcription unit is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of mRNA.
Solution
During transcription, the RNA polymerase synthesises an mRNA strand that is complementary to the template strand. The coding strand of the DNA has the same sequence as the mRNA, except that Thymine (T) in the DNA is replaced by Uracil (U) in the mRNA.
Given Coding Strand (5'→3'):
To Find:
- Sequence of mRNA (5'→3')
Principle:
The mRNA sequence is identical to the coding strand sequence, with the substitution of Uracil (U) for every Thymine (T).
Derivation:
We simply replace every 'T' in the coding strand sequence with a 'U'.
Coding Strand: 5'-ATGCATGCATGCATGCATGCATGCATGC-3'
mRNA Sequence: 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'
Final Answer: The sequence of the mRNA is:
Q5EXERCISES
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Solution
The property of the DNA double helix that led Watson and Crick to hypothesise the semi-conservative mode of DNA replication was complementary base pairing.
Explanation:
Watson and Crick observed that the two polynucleotide chains of a DNA molecule are not identical but are complementary to each other. Adenine (A) on one strand always pairs with Thymine (T) on the opposite strand, and Guanine (G) always pairs with Cytosine (C).
This specific pairing suggested a simple and elegant copying mechanism. They proposed that during replication, the two strands would separate, and each strand would then act as a template for the synthesis of a new, complementary strand. For example, wherever an 'A' existed on the template strand, a 'T' would be added to the new strand, and wherever a 'C' existed, a 'G' would be added, and so on.
This process would result in two new DNA molecules, each identical to the original parent molecule. Crucially, each of the new DNA molecules would consist of one original (parental) strand and one newly synthesised strand. This model, where each new DNA molecule conserves half of the original molecule, is known as the semi-conservative mode of replication.
Q6EXERCISES
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Solution
Nucleic acid polymerases are enzymes that synthesise nucleic acid polymers (DNA or RNA) using an existing template. They can be classified based on the type of template they use and the type of nucleic acid they synthesise. The main types are:
-
DNA-dependent DNA polymerase: This type of enzyme uses a DNA template to synthesise a new complementary DNA strand. It is the primary enzyme involved in DNA replication. (Template: DNA → Product: DNA)
-
DNA-dependent RNA polymerase: This enzyme uses a DNA template to synthesise a complementary RNA strand. It is the key enzyme in the process of transcription. (Template: DNA → Product: RNA)
-
RNA-dependent DNA polymerase: This enzyme, also known as reverse transcriptase, synthesises a complementary DNA strand using an RNA template. This process is called reverse transcription and is found in retroviruses like HIV. (Template: RNA → Product: DNA)
-
RNA-dependent RNA polymerase: This enzyme synthesises a new RNA strand using an RNA template. It is found in some viruses that have an RNA genome, which they need to replicate. (Template: RNA → Product: RNA)
Q7EXERCISES
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Solution
Alfred Hershey and Martha Chase differentiated between DNA and protein by using radioactive isotopes to specifically label each molecule. Their experiment exploited the unique chemical composition of DNA and proteins.
Basis of Differentiation:
- DNA contains phosphorus (P) in its sugar-phosphate backbone, but it does not contain sulfur (S).
- Proteins contain sulfur (S) in certain amino acids (methionine and cysteine), but they do not contain phosphorus (P).
Experimental Steps:
-
Labeling: They prepared two batches of bacteriophages (viruses that infect bacteria):
- Batch 1: Viruses were grown in a medium containing radioactive phosphorus (P). This resulted in phages with radioactively labeled DNA.
- Batch 2: Viruses were grown in a medium containing radioactive sulfur (S). This resulted in phages with radioactively labeled protein coats.
-
Infection: Both batches of radioactive phages were allowed to infect separate cultures of E. coli bacteria. The phages attached to the bacterial surface and injected their genetic material into the cells.
-
Blending: After infection, the cultures were agitated in a blender. This process sheared the viral coats off the surface of the bacteria.
-
Centrifugation: The cultures were spun in a centrifuge. The heavier bacterial cells formed a pellet at the bottom, while the lighter viral particles remained suspended in the supernatant (liquid).
Results and Conclusion:
- In the culture infected with P-labeled phages (radioactive DNA), the radioactivity was found inside the bacterial cells (in the pellet). This indicated that DNA had entered the bacteria.
- In the culture infected with S-labeled phages (radioactive protein), the radioactivity was found in the supernatant with the viral particles. This indicated that the protein coats did not enter the bacteria.
This unequivocal proof demonstrated that DNA, not protein, is the genetic material that is passed from the virus to the bacteria to direct the synthesis of new viruses.
Q8EXERCISES
Differentiate between the followings:
(a)
Repetitive DNA and Satellite DNA
(b)
mRNA and tRNA
(c)
Template strand and Coding strand
Solution
(a) Repetitive DNA and Satellite DNA
| Feature | Repetitive DNA | Satellite DNA |
|---|---|---|
| Definition | A general term for stretches of DNA sequences that are repeated many times in the genome. | A specific category of repetitive DNA that forms small, separate peaks from the bulk genomic DNA during density gradient centrifugation. |
| Scope | It is a broad category that includes both satellite DNA and other dispersed repeats. | It is a sub-type of repetitive DNA, classified based on base composition, length, and number of repeats (e.g., mini-satellites, micro-satellites). |
| Function | Most do not code for proteins but play roles in chromosome structure, dynamics, and evolution. | Forms a large portion of the human genome and heterochromatin; the high degree of polymorphism in satellite DNA is the basis for DNA fingerprinting. |
(b) mRNA and tRNA
| Feature | mRNA (messenger RNA) | tRNA (transfer RNA) |
|---|---|---|
| Function | Carries the genetic code from DNA in the nucleus to the ribosome in the cytoplasm, serving as the template for protein synthesis. | Acts as an adapter molecule that reads the codons on the mRNA and brings the corresponding specific amino acid to the ribosome. |
| Structure | A long, linear single-stranded molecule. Its sequence contains codons. | A small, compact molecule with a clover-leaf secondary structure and an L-shaped tertiary structure. It has an anticodon loop and an amino acid acceptor end. |
| Codon/Anticodon | Contains codons (triplets of bases) that specify amino acids. | Contains an anticodon loop with a base sequence complementary to an mRNA codon. |
(c) Template strand and Coding strand
| Feature | Template Strand | Coding Strand |
|---|---|---|
| Role in Transcription | Acts as the template for RNA polymerase to synthesise a complementary RNA strand. | Does not act as a template for transcription. |
| Polarity | Has a 3' → 5' polarity with respect to the direction of transcription. | Has a 5' → 3' polarity with respect to the direction of transcription. |
| Sequence Relationship to RNA | Its sequence is complementary to the transcribed RNA (with A pairing to U, and G pairing to C). | Its sequence is identical to the transcribed RNA, except that Thymine (T) in the DNA is replaced by Uracil (U) in the RNA. |
| Other Name | Also known as the anti-sense strand. | Also known as the sense strand. |
Q9EXERCISES
List two essential roles of ribosome during translation.
Solution
The ribosome is the cellular machinery responsible for protein synthesis (translation). Its two essential roles are:
-
Providing a Platform for Translation: The ribosome provides a structural framework and binding sites for both mRNA and the charged tRNA molecules. It holds the mRNA template in the correct position so that its codons can be read sequentially, and it has sites (A, P, and E sites) for tRNAs to bind and deliver amino acids.
-
Catalyzing Peptide Bond Formation: The ribosome acts as an enzyme (a ribozyme) to form the peptide bonds that link amino acids together into a polypeptide chain. Specifically, the 23S rRNA in bacteria (and 28S rRNA in eukaryotes) within the large ribosomal subunit catalyzes this reaction.
Q10EXERCISES
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Solution
The lac operon shuts down after some time because the inducer, lactose, is consumed by the enzymes produced by the operon itself.
Explanation:
-
Induction: When lactose is added to the medium, it enters the E. coli cells and acts as an inducer. It binds to the lac repressor protein, inactivating it. This allows RNA polymerase to transcribe the structural genes (z, y, a).
-
Metabolism: The transcription and translation of these genes lead to the synthesis of enzymes, including β-galactosidase (coded by the z gene). β-galactosidase hydrolyzes lactose into glucose and galactose, which the bacterium uses for energy.
-
Removal of Inducer: As β-galactosidase breaks down the lactose, the concentration of the inducer (lactose/allolactose) in the cell decreases.
-
Repression: Once all the lactose has been metabolized, there is no longer an inducer to bind to the repressor protein. The repressor protein returns to its active shape, binds to the operator region of the lac operon, and physically blocks RNA polymerase from transcribing the genes. This switches the operon off, and the synthesis of the enzymes stops.
This feedback mechanism ensures that the cell only produces the enzymes for lactose metabolism when lactose is actually present, thus conserving energy.
Q11EXERCISES
Explain (in one or two lines) the function of the followings:
(a)
Promoter
(b)
tRNA
(c)
Exons
Solution
(a) Promoter: A specific DNA sequence located at the 5'-end (upstream) of a structural gene that serves as the binding site for RNA polymerase, thereby initiating the process of transcription.
(b) tRNA (transfer RNA): An adapter molecule that reads the genetic code on mRNA via its anticodon and carries the corresponding specific amino acid to the ribosome for incorporation into a polypeptide chain.
(c) Exons: The coding or expressed sequences within a eukaryotic gene that are joined together after the removal of introns (splicing) to form the mature, functional mRNA molecule.
Q12EXERCISES
Why is the Human Genome project called a mega project?
Solution
The Human Genome Project (HGP) is called a mega project due to its immense scale, complexity, cost, and ambitious goals. Key reasons include:
-
Massive Genome Size: The project aimed to sequence the entire human genome, which consists of approximately 3.3 billion () base pairs.
-
High Financial Cost: The estimated cost was about US $3 per base pair, leading to a total project cost of approximately 9 billion US dollars.
-
Enormous Data Handling: The vast amount of sequence data generated required the development of high-speed computational devices and a new field of biology, Bioinformatics, for data storage, retrieval, and analysis.
-
Long Duration and International Collaboration: The project was a 13-year endeavor (1990-2003) that required the coordinated effort of scientists and research institutions from many countries, including the U.S., U.K., Japan, France, Germany, and China.
Q13EXERCISES
What is DNA fingerprinting? Mention its application.
Solution
DNA Fingerprinting:
DNA fingerprinting (also known as DNA profiling) is a laboratory technique used to identify and compare individuals based on their unique DNA sequences. It involves analyzing specific, highly variable regions of DNA called repetitive DNA (such as Variable Number of Tandem Repeats or VNTRs). Since these sequences show a high degree of polymorphism (variation) among individuals, they create a distinctive pattern, or 'fingerprint', for each person (except identical twins).
Applications of DNA Fingerprinting:
- Forensic Science: It is a powerful tool for criminal investigations, used to match DNA samples from a crime scene (e.g., blood, hair, saliva) with suspects.
- Paternity and Maternity Testing: It is used to resolve disputes of parentage by comparing the child's DNA pattern with that of the potential parents.
- Personal Identification: It can be used to identify victims of disasters or establish family relationships.
- Evolutionary and Population Genetics: It helps in studying genetic diversity, migration patterns, and evolutionary relationships among populations and species.
- Medical Diagnostics: It can be used to identify genes associated with inherited diseases.
Q14EXERCISES
Briefly describe the following:
(a)
Transcription
(b)
Polymorphism
(c)
Translation
(d)
Bioinformatics
Solution
(a) Transcription:
Transcription is the process of synthesising a complementary RNA molecule from a DNA template. It is the first step in gene expression. The enzyme DNA-dependent RNA polymerase binds to a promoter region on the DNA, unwinds the helix, and uses one of the DNA strands (the template strand) to build an RNA chain in the 5'→3' direction, following the rules of base complementarity (A pairs with U, G pairs with C).
(b) Polymorphism:
In the context of genetics, polymorphism refers to the occurrence of variations in the DNA sequence at a particular locus within a population. For a variation to be considered a polymorphism, the variant allele must occur at a frequency of more than 1% (or 0.01). These variations arise from mutations and are the basis for genetic diversity, evolution, and techniques like DNA fingerprinting.
(c) Translation:
Translation is the process by which the genetic information encoded in a messenger RNA (mRNA) molecule is used to synthesise a specific sequence of amino acids, forming a polypeptide or protein. This process occurs on ribosomes and involves tRNA molecules, which act as adapters to match mRNA codons with the correct amino acids. The ribosome moves along the mRNA, catalyzing the formation of peptide bonds between successive amino acids.
(d) Bioinformatics:
Bioinformatics is an interdisciplinary field that develops methods and software tools for understanding biological data. It combines biology, computer science, information engineering, mathematics, and statistics to analyze and interpret large and complex biological datasets, such as the enormous amount of sequence data generated by the Human Genome Project. Its applications include storing, retrieving, and analyzing genomic and proteomic information.