Molecular Basis of Inheritance ncert solutions: Class 12th biology chapter 5 ncert solutions
Textbook | NCERT |
Class | Class 12 |
Subject | Biology |
Chapter | Chapter 5 |
Chapter Name | Molecular Basis of Inheritance class 12 ncert solutions |
Category | Ncert Solutions |
Medium | English |
Are you looking for Ncert Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance? Now you can download Ncert class 12 biology chapter 5 questions and answers pdf from here.
Question 1: Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Solution 1: Nitrogenous bases present in the list are Adenine (A), Thymine (T), Uracil (U), Cytosine (C).
Nucleosides present in the list are Cytidine (Cytosine + Ribose), Guanosine (Guanine + Ribose).
Question 2: If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Solution 2: In a double-stranded DNA molecule, the base pairing rules follow Chargaff’s rule, which states that:
- Cytosine (C) pairs with Guanine (G), so the percentage of cytosine is equal to the percentage of guanine.
- Adenine (A) pairs with Thymine (T), so the percentage of adenine is equal to the percentage of thymine.
Given:
- The DNA has 20% cytosine (C).
- According to Chargaff’s rule, this means the DNA also has 20% guanine (G).
Now, the total percentage of cytosine and guanine together is:
C + G = 20% + 20% = 40%
Since the total percentage of all bases must add up to 100%, the remaining 60% must consist of adenine (A) and thymine (T). Therefore, the percentage of adenine (A) is half of that:
\(\text{A} + \text{T} = 60\%\)
\(\text{A} = \frac{60\%}{2} = 30\%\)
The percentage of adenine (A) in the DNA is 30%.
Question 3: 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 3: To write the sequence of the complementary strand in the 5′ to 3′ direction, we first apply the base pairing rules:
- Adenine (A) pairs with Thymine (T).
- Thymine (T) pairs with Adenine (A).
- Guanine (G) pairs with Cytosine (C).
- Cytosine (C) pairs with Guanine (G).
Now, the original strand is:
5′ -ATGCATGCATGCATGCATGCATGCATGC-3′
The complementary strand in the 3′ to 5′ direction would be:
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
To express the complementary strand in the 5′ to 3′ direction, we simply reverse it:
5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′
Question 4: 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 4: The mRNA sequence is complementary to the template strand of DNA, but it is identical to the coding strand (except that uracil (U) replaces thymine (T)).
Given the sequence of the coding strand: 5’− ATGCATGCATGCATGCATGCATGCATGC-3’
The mRNA sequence will be the same as the coding strand, except that T (thymine) will be replaced with U (uracil).
mRNA sequence: 5’ − AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’
Question 5: Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Solution 5: Ans: The antiparallel, double-stranded nature of the DNA molecule led Watson and Crick to hypothesise semi-conservative mode of DNA replication. They suggested that the two strands of DNA molecule uncoil and separate, and each strand serves as a template for the synthesis of a new (complementary) strand alongside it.
The template and its complement, then form a new DNA double strand, identical to the original DNA molecule. The sequence of bases which should be present in the new strands can be easily predicted because these would be complementary to the bases present in the old strands. A will pair with T, T with A, C with G, and G with C.
Thus, two daughter DNA molecules identical to the parent molecule are formed and each daughter DNA molecule consists of one old (parent) strand and one new strand. Since only one parent strand is conserved in each daughter molecule, this mode of replication is said to be semiconservative. Meselson and Stahl and Joseph Taylor, later proved it by experiments.
Question 6: 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 6: Nucleic acid polymerases are enzymes that synthesize nucleic acids (DNA or RNA) from a nucleic acid template. Depending on the chemical nature of the template (DNA or RNA) and the type of nucleic acid synthesized, the following types of nucleic acid polymerases can be identified:
1. DNA-Dependent DNA Polymerase:
- Template: DNA
- Product: DNA
- Function: This enzyme synthesizes a new DNA strand using an existing DNA strand as a template. It is involved in DNA replication, ensuring that a copy of the DNA is made before cell division.
- Example: DNA polymerase in prokaryotes (e.g., DNA polymerase I, II, III in bacteria), and eukaryotic DNA polymerases (e.g., polymerase α, δ, ε).
2. DNA-Dependent RNA Polymerase:
- Template: DNA
- Product: RNA
- Function: This enzyme synthesizes RNA using a DNA template during the process of transcription. It reads the DNA strand and assembles RNA nucleotides in the correct sequence.
- Example: RNA polymerase in prokaryotes (single enzyme), RNA polymerase I, II, III in eukaryotes.
3. RNA-Dependent DNA Polymerase (Reverse Transcriptase):
- Template: RNA
- Product: DNA
- Function: This enzyme synthesizes DNA using an RNA template. It is crucial in reverse transcription, commonly seen in retroviruses (e.g., HIV), where the viral RNA is reverse-transcribed into DNA to integrate into the host genome.
- Example: Reverse transcriptase in retroviruses.
4. RNA-Dependent RNA Polymerase:
- Template: RNA
- Product: RNA
- Function: This enzyme synthesizes RNA using an RNA template, playing a key role in the replication of RNA viruses, such as RNA viruses (e.g., coronaviruses, influenza virus).
- Example: RNA polymerase found in RNA viruses.
Question 7: How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Solution 7: In 1952, Alfred Hershey and Martha Chase conducted an experiment using bacteriophages (viruses that infect bacteria) to prove that DNA is the genetic material. To differentiate between DNA and protein, they used radioactive isotopes to label these two molecules separately in the phage, allowing them to track which component entered the bacterial cells during infection.
Key Steps in the Hershey-Chase Experiment:
1. Labeling DNA and Protein:
DNA labeled with radioactive phosphorus (³²P): DNA contains phosphorus but proteins do not. Hershey and Chase incorporated radioactive ³²P into the DNA of the bacteriophages to label the DNA specifically.
Protein labeled with radioactive sulfur (³⁵S): Proteins contain sulfur (in amino acids like methionine and cysteine), but DNA does not. They incorporated radioactive ³⁵S into the protein coat of the phages to label the proteins.
2. Infection of Bacteria:
The labeled bacteriophages were allowed to infect Escherichia coli (E. coli) bacteria. During infection, the phage attaches to the bacterial surface and injects its genetic material into the host cell, leaving the phage’s protein coat outside.
3. Blender Experiment:
After allowing the bacteriophages to inject their genetic material into the bacteria, Hershey and Chase used a blender to separate the phage protein coats from the bacterial cells.
They then centrifuged the mixture. The heavier bacterial cells formed a pellet at the bottom, while the lighter viral coats (which remained outside the bacteria) stayed in the liquid supernatant.
4. Tracking Radioactivity:
- They measured the radioactivity in both the bacterial pellet and the supernatant.
- In the experiment where phages were labeled with ³²P (DNA), most of the radioactivity was found in the bacterial pellet, indicating that the DNA had entered the bacteria.
- In the experiment where phages were labeled with ³⁵S (protein), most of the radioactivity remained in the supernatant, showing that the protein did not enter the bacterial cells.
Component | Labeled Isotope | Location of Radioactivity | Conclusion |
---|---|---|---|
DNA | ³²P (Phosphorus) | Found in the bacterial pellet | DNA enters the bacterial cells |
Protein | ³⁵S (Sulfur) | Found in the supernatant (liquid) | Protein remains outside the bacterial cells |
Thus, Hershey and Chase demonstrated that DNA, not protein, is the material responsible for heredity.
Question 8: Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand
Solution 8: (a) Repetitive DNA vs. Satellite DNA
Repetitive DNA | Satellite DNA |
---|---|
Sequences of DNA that are repeated multiple times in the genome. | A specific type of repetitive DNA that forms distinct bands when separated by density gradient centrifugation. |
Can include satellite DNA, as well as interspersed repeats (e.g., SINEs, LINEs). | A subclass of repetitive DNA, further divided into microsatellites, minisatellites, and macrosatellites. |
Found throughout the genome, including coding and non-coding regions. | Often found in specific areas such as centromeres and telomeres. |
May have structural roles or contribute to genome evolution; some are non-functional. | Primarily structural, playing roles in chromosome stability and segregation. |
Broad category, not necessarily forming visible bands in experiments. | Detected as distinct bands in a density gradient due to their highly repetitive nature. |
(b) mRNA vs. tRNA
mRNA (Messenger RNA) | tRNA (Transfer RNA) |
---|---|
Carries the genetic information from DNA to the ribosome for protein synthesis. | Brings specific amino acids to the ribosome during protein synthesis. |
Single-stranded linear molecule. | Cloverleaf structure with loops, including an anticodon arm. |
Serves as the template for the sequence of amino acids in a protein. | Matches the anticodon with codons on mRNA to add the correct amino acid to the growing peptide chain. |
Highly variable depending on the gene being expressed. | Shorter, usually about 70-90 nucleotides long. |
Contains codons that determine the sequence of amino acids in a protein. | Contains an anticodon that pairs with the codon on mRNA. |
(c) Template Strand vs. Coding Strand
Template Strand | Coding Strand |
---|---|
The DNA strand that is used as a template by RNA polymerase to synthesize mRNA. | The DNA strand that has the same sequence as the mRNA (except for thymine being replaced by uracil). |
Acts as the direct template for mRNA synthesis; RNA polymerase reads this strand. | Not used directly in transcription but has the same nucleotide sequence as the produced mRNA (with T instead of U). |
Read in the 3′ to 5′ direction during transcription. | Runs in the 5′ to 3′ direction, like the resulting mRNA. |
Complementary to the mRNA sequence (A-U, T-A, C-G, G-C). | Identical to the mRNA sequence (except T is replaced by U in RNA). |
Also called the non-coding strand, antisense strand, or minus strand. | Also called the sense strand, non-template strand, or plus strand. |
Question 9: List two essential roles of ribosome during translation.
Solution 9: Two essential roles of the ribosome during translation are as given below:
Since the large subunit of the ribosome has two different sites to attach to tRNA, it facilitates amino acids to come closer for peptide bond formation. Also, the ribosome behaves as a catalyst for the formation of the peptide bond. Example – 23s r-RNA acts as a ribozyme in bacteria
Ribosomes are sites where the synthesis of proteins occurs from individual amino acids. It consists of two subunits – the larger subunit serves as an amino acid binding site, whereas the smaller subunit attaches to the mRNA forming a protein-synthesising complex
Question 10: 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 10: Lac operon is switched on, on adding lactose in medium, as lactose acts as inducer and makes repressor inactive by binding with it. When the lac operon system is switched on, β-galactosidase is formed, which converts lactose into glucose and galactose. As soon as all the lactose is consumed, repressor again becomes active and causes the system to switch off (shut down).
Question 11: Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons
Solution 11: (a) Promoter: A promoter is a DNA sequence located upstream of a gene that provides a binding site for RNA polymerase, initiating the process of transcription.
(b) tRNA (Transfer RNA): tRNA carries specific amino acids to the ribosome during translation, matching its anticodon to the mRNA codon to ensure the correct amino acid is added to the growing protein chain.
(c) Exons: Exons are the coding regions of a gene that remain in the mature mRNA after splicing and are translated into proteins.
Question 12: Why is the Human Genome project called a mega project?
Solution 12: The Human Genome Project (HGP) is called a mega project due to its large scale, complexity, and far-reaching impact. Several reasons contribute to this designation:
1. Massive Scale: The project aimed to map and sequence the entire human genome, which consists of approximately 3 billion base pairs of DNA, spread across 20,000-25,000 genes. This made it one of the largest biological projects ever undertaken.
2. Collaborative Global Effort: The HGP involved the collaboration of numerous research institutions and scientists from various countries across the world. It was a highly coordinated, multinational effort, with countries like the U.S., U.K., Japan, France, and others participating.
3. High Costs and Resources: It required substantial financial investment, with the estimated cost being around $3 billion. The project also required advanced technology and resources, including specialized sequencing machines, computational tools, and vast data storage capacities.
4. Technological Advancements: The project led to the development of new technologies in DNA sequencing, bioinformatics, and computational biology, which were crucial for managing and analyzing vast amounts of genetic data.
5. Long Duration: The project spanned over a decade, officially starting in 1990 and completing in 2003, with additional time spent on post-project analysis and applications.
6. Impact on Biology and Medicine: The HGP had profound implications for fields like genomics, medicine, and biotechnology, opening new avenues for understanding genetic diseases, personalized medicine, and drug development. It continues to influence research and medical practices today.
Given these factors—its enormous scope, international collaboration, financial costs, and the long-lasting impact on science and medicine—the Human Genome Project is aptly called a mega project.
Question 13: What is DNA fingerprinting? Mention its application.
Solution 13: DNA fingerprinting, also known as DNA profiling or genetic fingerprinting, is a technique used to identify individuals based on their unique DNA sequence. It relies on analyzing specific regions of DNA that vary greatly between individuals, particularly short tandem repeats (STRs) or variable number tandem repeats (VNTRs), which are highly polymorphic.
Application
- Forensic Science: Used in criminal investigations to identify suspects or exonerate individuals based on DNA evidence from crime scenes.
- Paternity Testing: Helps in determining biological relationships, such as paternity or maternity.
- Identification of Remains: Useful in identifying victims of accidents, natural disasters, or wars when conventional identification is not possible.
- Genetic Disease Screening: Assists in identifying inherited genetic conditions or predispositions by comparing specific DNA markers.
- Wildlife Conservation: Used to track genetic diversity in endangered species and in cases of illegal wildlife trafficking by identifying species or individual animals.
- Personal Identification: In rare cases, DNA fingerprinting is used for personal identification, especially in legal or immigration disputes.
Question 14: Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Solution 14:
(a) Transcription: The process of copying genetic information from one strand of the DNA into RNA is known as transcription. RNA is assembled simply based on complementarity of the DNA strand, only uracil is substituted in place of thymine. Only a small segment of DNA that codes for a polypeptide is copied.
(b) Polymorphism: The variation in DNA arising through mutation at non-coding sequences is known as Polymorphism. Such variations are unique to specific sites of DNA and can occur due to deletion, insertion or substitution of bases. It can be observed by making fragments of DNA sample and separating them through electrophoresis.The polymorphism in a DNA sequence is the basis of genetic mapping of the human genome as well as DNA fingerprinting.
(c) Translation: It refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. It occurs in cytoplasm in both prokaryotes and eukaryotes.
(d) Bioinformatics: It is the application of computer science and information technology which deals with handling, storing of huge information of genomics, processing information, analyzing data and creating new knowledge.