Chapter 5 : Molecular Basic of Inheritance

Class 12 biology Chapter 5 : Molecular Basic of Inheritance Exercise questions and Answers :

1 Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Answer : The given items as nitrogenous bases and nucleosides:

Nitrogenous Bases	Nucleosides
Adenine   Thymine   Uracil   Cytosine	Cytidine    Guanosine

Nitrogenous bases are the actual nitrogen-containing molecules that are the building blocks of nucleosides, which consist of a nitrogenous base attached to a sugar molecule. In this table, Adenine, Thymine, Uracil, and Cytosine are the nitrogenous bases, while Cytidine and Guanosine are nucleosides.

2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

Answer :  In a double-stranded DNA molecule, the amount of adenine (A) always equals the amount of thymine (T), and the amount of guanine (G) always equals the amount of cytosine (C). This is known as Chargaff's rules.

If the DNA has 20 percent cytosine (C), then it also has 20 percent guanine (G) because they always pair together. Together, they make up 40 percent of the DNA.

Since the total percentage of the four nucleotide bases in DNA must add up to 100 percent, we can calculate the percentage of adenine (A) as follows:

Total percentage of DNA = 100%

Total percentage of Cytosine (C) and Guanine (G) = 40%

Remaining percentage for Adenine (A) and Thymine (T) = 100% - 40% = 60%

Because adenine (A) always pairs with thymine (T), you can divide this remaining percentage by 2 :

Percentage of Adenine (A) = 60% / 2 = 30%

So, the DNA contains 30 percent adenine.

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.

Answer : The complementary strand of DNA, we need to match each base on the original strand with its complementary base:

Original Strand: 5' - ATGCATGCATGCATGCATGCATGCATGC - 3'

Complementary Strand (in the 5' → 3' direction): 3' -TACGTACGTACGTACGTACGTACGTACG-5'

So, the sequence of the complementary strand in the 5' → 3' direction is "TACGTACGTACGTACGTACGTACGTACG."

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.

Answer : The mRNA sequence transcribed from the coding strand of DNA, we need to replace thymine (T) with uracil (U) because RNA uses uracil instead of thymine. The mRNA sequence corresponding to the given coding strand: 5' - AUGCAUGCAUGCAUGCAUGCAUGCAUGC - 3'

So, the mRNA sequence is transcribed from the coding strand .

5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

Answer: Watson and Crick hypothesized the semi-conservative mode of DNA replication based on the property of DNA's structural symmetry and complementary base pairing. They observed that DNA consists of two anti-parallel strands, with each strand serving as a template for the synthesis of a new complementary strand. This allowed for the conservation of one original strand in each newly synthesized DNA molecule, ensuring the retention of genetic information. The complementary base pairing (A with T and C with G) further supported this model by facilitating accurate replication, with each daughter molecule containing one parental and one newly synthesized strand, hence "semi-conservative" replication.

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.

Answer : Nucleic acid polymerases are enzymes that catalyze the synthesis of nucleic acid chains. They can be categorized based on the chemical nature of the template (DNA or RNA) and the nature of the nucleic acids synthesized (DNA or RNA).

The main types of nucleic acid polymerases are :

(i)  DNA-dependent DNA polymerase: These enzymes synthesize DNA using a DNA template. Examples include DNA polymerase I, DNA polymerase II, DNA polymerase III, and DNA polymerase IV.

(ii)  RNA-dependent DNA polymerase (reverse transcriptase): This enzyme synthesizes a DNA strand from an RNA template. It is commonly found in retroviruses and is essential for reverse transcription. HIV reverse transcriptase is a well-known example.

(iii)  DNA-dependent RNA polymerase: These enzymes synthesize RNA using a DNA template. RNA polymerase I, RNA polymerase II, and RNA polymerase III are examples found in eukaryotes, each responsible for transcribing different types of RNA.

(iv)  RNA-dependent RNA polymerase: These enzymes synthesize RNA using an RNA template. They are primarily found in viruses that have RNA genomes and are essential for replicating the viral RNA.

These various polymerases play critical roles in DNA replication, repair, transcription, and viral replication, depending on the specific type of nucleic acid synthesis required.

7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Answer : Hershey and Chase conducted their experiment to differentiate between DNA and protein in order to prove that DNA is the genetic material. The experimental statement are :

(i) They used two different types of radioactive isotopes, radioactive phosphorus () and radioactive sulfur (), to label the DNA and protein in the bacteriophages, respectively. They knew that DNA contains phosphorus but not sulfur, while protein contains sulfur but not phosphorus.

(ii) They grew some bacteriophages in a medium containing radioactive phosphorus () and others in a medium containing radioactive sulfur ( ). This allowed them to specifically label either the viral DNA or the viral protein.

(iii) These labeled bacteriophages were then allowed to infect E. coli bacteria. When a bacteriophage infects a bacterial cell, it injects its genetic material into the cell.

(iv) After allowing the infection to proceed, Hershey and Chase separated the viral protein coats from the bacterial cells by agitating them in a blender. This step was crucial because it removed any external components of the virus that might have remained attached to the bacterial surface.

(v) They then separated the remaining bacterial cells from the virus particles by using a centrifuge.

(vi) Bacteria that were infected with viruses labeled with radioactive phosphorus (),) became radioactive because the viral DNA had entered the bacterial cells. In contrast, bacteria infected with viruses labeled with radioactive sulfur (),) did not become radioactive because the viral protein remained outside the bacterial cells.

Conclusion: Based on their findings, Hershey and Chase concluded that it was the DNA of the bacteriophage, not the protein, that entered the bacterial cells and played a role in the genetic replication of the virus. This experiment provided strong evidence that DNA is the genetic material responsible for carrying genetic information from the virus to the bacteria.

8. Differentiate between the followings:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

(c) Template strand and Coding strand

Answer : (a)The differences between repetitive DNA and satellite DNA are :

Characteristic	Repetitive DNA	Satellite DNA
Definition	Repeated sequences throughout the genome.	Short, repeated sequences in specific chromosome regions.
Function	Various functions, including structural and regulatory roles.	Often no clear function, but linked to chromosome structure.
Location	Dispersed throughout the genome.	Clustered in specific chromosome regions.

 (b)  The differences between mRNA and tRNA are :

Characteristic	mRNA (Messenger RNA)	tRNA (Transfer RNA)
Function	Carries genetic code for protein synthesis	Transfers amino acids to the ribosome
Structure	Single-stranded	Cloverleaf-shaped
Specific Role	Acts as a blueprint for protein synthesis	Delivers specific amino acids during translation

(c) The differences between the template strand and coding strand in DNA are :

Characteristic	Template Strand	Coding Strand
Function	Serves as a template for RNA synthesis during  transcription.	Complementary to the template strand during transcription.
Sequence Relation	Complementary to the coding strand.	Has the same sequence as the RNA transcript (except for T replaced by U in RNA).
Also Known As	Non-coding strand, antisense strand.	Sense strand

9. List two essential roles of ribosome during translation.

Answer :  Two essential roles of the ribosome during translation are:

(i) The ribosome reads the mRNA codons and ensures that the appropriate amino acids are added to the growing polypeptide chain, following the genetic code.

(ii) The ribosome facilitates the formation of peptide bonds between adjacent amino acids in the growing polypeptide, creating a linear chain that will fold into a functional protein.

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?

Answer :  The lac operon shuts down after the initial induction by lactose due to regulatory mechanisms. When lactose is introduced into the medium, it is transported into the bacterial cells through permease. Lactose then acts as an inducer, interacting with the repressor protein synthesized from the i gene. This interaction inactivates the repressor, allowing RNA polymerase to access the promoter and initiate transcription of the operon. However, as lactose is metabolized into glucose and galactose by beta-galactosidase, the inducer molecule allolactose becomes depleted. With diminishing allolactose levels, the lac repressor can once again bind to the operator region, inhibiting operon transcription, leading to the shutdown of the lac operon.

11. Explain (in one or two lines) the function of the followings:

(a) Promoter

(b) tRNA

(c) Exons

Answer : (a) Promoter: Initiates and regulates gene transcription by binding to RNA polymerase and determining the start site of transcription.

(b) tRNA: Carries specific amino acids to the ribosome during protein synthesis, matching them with mRNA codons.

(c) Exons: Encode the protein-coding information in eukaryotic genes and are retained in the mature mRNA after splicing to form the protein-coding sequence.

12. Why is the Human Genome project called a mega project?

Answer : The Human Genome Project (HGP) is termed a mega project due to its staggering scale and requirements. With an estimated base pairs in the human genome, and an initial cost of $3 per base pair, the project's estimated cost was about $9 billion. Storing the data in books would require 3300 books per single human cell. The immense data volume also mandated high-speed computational devices for storage, retrieval, and analysis. HGP played a pivotal role in the development of the field of Bioinformatics.

13. What is DNA fingerprinting? Mention its application.

Answer : DNA fingerprinting, also known as DNA profiling or genetic fingerprinting, is a technique used to identify and differentiate individuals based on their unique DNA patterns. It relies on the fact that the DNA of every individual (except identical twins) is distinct and can be used as a genetic "fingerprint" for identification purposes.

The application of DNA fingerprinting are :

Criminal Investigations: Solving crimes by matching DNA evidence to suspects.

Paternity Testing: Determining biological parentage in legal and personal matters.

Missing Persons: Identifying unidentified individuals and locating missing persons.

Immigration: Confirming family relationships for visa and immigration purposes.

Wildlife Conservation: Protecting endangered species by tracking and monitoring populations.

14. Briefly describe the following:

(a) Transcription

(b) Polymorphism

(c) Translation

(d) Bioinformatics

Answer :  (a) Transcription: Transcription is the process in molecular biology where a DNA sequence is used as a template to create a complementary RNA molecule. It occurs in the cell nucleus and is the first step in gene expression. The RNA molecule produced, known as messenger RNA (mRNA), carries genetic information from the DNA to the ribosome for translation.

(b) Polymorphism: Polymorphism refers to the coexistence of multiple forms or alleles of a particular gene within a population. These variations in DNA sequences can lead to differences in traits or characteristics among individuals. Polymorphisms are essential in genetics and can have implications for disease susceptibility, drug response, and evolutionary adaptation.

(c) Translation: Translation is the biological process where the information encoded in mRNA is used to synthesize a specific protein. It takes place in the ribosomes of the cell and involves the conversion of the mRNA's nucleotide sequence into a sequence of amino acids, which form the protein's primary structure. This process is vital for the synthesis of all cellular proteins.

(d) Bioinformatics: Bioinformatics is an interdisciplinary field that combines biology, computer science, and information technology to analyze and interpret biological data, particularly large datasets generated from genomics, proteomics, and other biological experiments. Bioinformatics tools and techniques are used to store, manage, and analyze biological information, leading to discoveries in areas such as genetics, drug development, and evolutionary biology.