Molecular Basis Of Inheritance

Molecular Basis of Inheritance: A Summary

Molecular biology is the study of biological molecules, particularly DNA, RNA, and proteins, and how they interact to control cellular processes.

DNA: The Genetic Material

Structure: DNA is a double-stranded molecule composed of nucleotides, each containing a sugar, a phosphate group, and a nitrogenous base (adenine, thymine, guanine, or cytosine). The two strands are antiparallel and held together by hydrogen bonds between complementary base pairs (A-T and G-C).
Replication: DNA replication is the process by which a new DNA strand is synthesized using the existing strand as a template.

RNA: The Messenger Molecule

Types: There are three main types of RNA:
- Messenger RNA (mRNA)
- Transfer RNA (tRNA): Transports amino acids to the ribosomes for protein synthesis.
- Ribosomal RNA (rRNA): Forms part of the ribosome, the cellular machinery responsible for protein synthesis.
Transcription

Genetic Code

Codons: The genetic code is a triplet code, meaning that three consecutive nucleotides (called a codon) specify one amino acid in a protein.
Universality: The genetic code is nearly universal, meaning that it is the same in all organisms.

Protein Synthesis

Translation: The process by which mRNA is translated into a protein at the ribosomes. This requires the cooperation of tRNA molecules, which bring amino acids to the ribosome according to the genetic code.

Regulation of Gene Expression

Transcriptional Control: The regulation of gene expression often occurs at the level of transcription, where factors such as transcription factors can bind to DNA to promote or inhibit gene expression.
Post-Transcriptional Control: Gene expression can also be regulated after transcription, through processes such as alternative splicing, mRNA stability, and translation control.

Human Genome Project

Goals: The Human Genome Project aimed to sequence the entire human genome, which was completed in 2003.
Significance: The Human Genome Project has revolutionized our understanding of human genetics and has led to advances in medicine, biotechnology, and other fields.

Exercise

1. Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine

Ans :

Nitrogenous Bases – Adenine, Uracil and Cytosine, Thymine;

Nucleosides – Cytidine, guanosine

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

Ans :

The total percentage of cytosine and guanine in a double-stranded DNA molecule is always equal, as they form complementary base pairs (C-G).

Therefore, if the DNA has 20% cytosine, it also has 20% guanine.

The remaining 60% of the DNA is made up of adenine and thymine. Since adenine and thymine also form complementary base pairs (A-T), they are present in equal proportions.

So, to calculate the percentage of adenine:

Total percentage = 100%
Cytosine + Guanine = 20% + 20% = 40%
Adenine + Thymine = 100% – 40% = 60%
Adenine = Thymine = 60% / 2 = 30%

Therefore, the DNA has 30% 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.

Ans :

If the sequence of one strand of DNA is written as follows:

5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′

The sequence of the complementary strand in 5′ —> 3′ direction will be:

5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′

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.

Ans :

The mRNA sequence will be identical to the coding strand, except that thymine (T) will be replaced with uracil (U).

Therefore, the mRNA sequence will be:

5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

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

Ans :

The complementary base pairing between the two strands of DNA was the key property that led Watson and Crick to hypothesize the semi-conservative mode of DNA replication.

In the DNA double helix, the two strands are held together by hydrogen bonds between specific base pairs: adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). This complementary base pairing ensures that the two strands are exact copies of each other.

Watson and Crick proposed that during DNA replication, the two strands of the double helix would separate, and each strand would serve as a template for the synthesis of a new complementary strand. This process would result in two new DNA double helices, each containing one old strand and one new strand. This is the semi-conservative model of DNA 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

Ans :

DNA-dependent DNA polymerase:

Template: DNA
Product: DNA
Function: Replicates DNA during cell division.
Examples: DNA polymerase I, II, and III in prokaryotes; DNA polymerase α, β, δ, and ε in eukaryotes.

DNA-dependent RNA polymerase:

Template: DNA
Product: RNA
Function: Transcribes DNA into RNA for gene expression.
Examples: RNA polymerase I, II, and III in eukaryotes; RNA polymerase in prokaryotes.

RNA-dependent RNA polymerase:

Template: RNA
Product: RNA
Function: Synthesizes RNA molecules using an RNA template.
Examples: Found in some viruses, such as retroviruses like HIV.

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

Ans :

Hershey and Chase differentiated between DNA and protein in their experiment by labeling each component with a radioactive isotope.

Radioactive Labeling:
- DNA: Phosphorus-32 (32P) was used to label DNA, as phosphorus is a component of DNA but not protein.
- Protein: Sulfur-35 (35S) was used to label protein, as sulfur is a component of protein but not DNA.
Infection: Bacteriophages (viruses that infect bacteria) were labeled with either 32P or 35S and allowed to infect bacteria.
Blending and Centrifugation: After infection, the bacteria were blended to separate the viral particles from the bacterial cells. The mixture was then centrifuged to separate the heavier bacterial cells from the lighter viral particles.
Analysis:
- Radioactive DNA: The radioactive 32P was found inside the bacterial cells, indicating that DNA had been transferred from the virus to the bacteria.
- Radioactive Protein: The radioactive 35S was found outside the bacterial cells, indicating that the protein coat of the virus remained outside the cell.

8. Differentiate between the followings:

(a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA

(c) Template strand and Coding strand

Ans :

(a) Repetitive DNA and Satellite DNA

Feature	Repetitive DNA	Satellite DNA
Definition	DNA sequences that are repeated multiple times	Repetitive DNA with a high AT content
Classification	Tandem repeats or interspersed repeats	Tandem repeats
Location	Found throughout the genome	Often found in the centromere region
Density Gradient	May or may not form distinct bands	Forms dense bands in cesium chloride density gradient

(b) mRNA and tRNA

Feature	mRNA	tRNA
Function	Carries genetic information from DNA to ribosomes	Transports amino acids to ribosomes
Structure	Single-stranded	Cloverleaf structure
Codons	Contains codons that specify amino acids	Contains an anticodon that matches the mRNA codon
Location	Cytoplasm	Cytoplasm

(c) Template strand and Coding strand

Feature	Template Strand	Coding Strand
Role in Transcription	Serves as a template for RNA synthesis	Not used as a template
Orientation	Antiparallel to the mRNA transcript	Identical to the mRNA transcript (except for T/U substitution)
Relationship to mRNA	Complementary to mRNA	Identical to mRNA (except for T/U substitution)

9.List two essential roles of ribosome during translation

Ans :

Two essential roles of ribosomes during translation:

Decoding mRNA: Ribosomes read the mRNA sequence in codons (triplets of nucleotides). Each codon specifies a particular amino acid. The ribosome decodes the mRNA sequence and ensures that the correct amino acid is brought to the growing polypeptide chain by tRNA molecules.
Peptide Bond Formation: Ribosomes catalyze the formation of peptide bonds between the amino acids brought by tRNA molecules. This process links the amino acids together to form a polypeptide chain, which will eventually become a 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?

Ans :

The shutdown of the lac operon after the addition of lactose in the medium is due to a mechanism called catabolite repression.

Catabolite repression is a regulatory mechanism that prevents E. coli from utilizing less efficient carbon sources (like lactose) when a more preferred carbon source (like glucose) is available.

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

(a) Promoter

(b) tRNA

(c) Exons

Ans :

(a) Promoter: A DNA sequence that signals the start of transcription, where RNA polymerase binds to initiate gene expression.

(b) tRNA (Transfer RNA): Carries specific amino acids to the ribosomes during protein synthesis, matching the mRNA codons to ensure correct amino acid sequence.

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

Ans :

The Human Genome Project (HGP) was called a mega project due to its massive scale, complexity, and global collaboration. Here’s why:

Scale: The human genome is incredibly large and complex, containing approximately 3 billion base pairs. Sequencing and analyzing such a vast amount of genetic information required significant technological advancements and computational power.
Complexity: The HGP involved developing new technologies and techniques for sequencing DNA, storing and analyzing the data, and interpreting the results. This required a multidisciplinary approach involving scientists from various fields, including biology, genetics, computer science, and mathematics.
Global Collaboration: The HGP was a truly international effort, involving scientists from around the world. Countries contributed to the project, sharing data and resources to accelerate progress.

13.What is DNA fingerprinting? Mention its application.

Ans :

It involves analyzing specific regions of DNA that are highly variable between individuals, known as DNA polymorphisms.

Applications of DNA fingerprinting:

Forensic Science: DNA fingerprinting is widely used in criminal investigations to identify suspects or victims based on DNA evidence found at crime scenes.
Paternity Testing: DNA fingerprinting can be used to establish paternity or maternity by comparing the DNA profiles of individuals.
Immigration and Citizenship: It can help determine familial relationships for immigration and citizenship purposes.
Wildlife Conservation: DNA fingerprinting is used to track wildlife populations, identify endangered species, and combat illegal wildlife trade.
Medical Research: It can be used to study genetic diseases, track the spread of infectious diseases, and identify genetic risk factors for certain conditions.
Personal Identification: DNA fingerprinting can be used for personal identification, such as in disaster relief efforts or for identifying remains.

14. Briefly describe the following:

(a) Transcription

(b) Polymorphism

(c) Translation

(d) Bioinformatics

Ans :

(a) Transcription: The process of copying the DNA sequence into an RNA sequence, which serves as a messenger for protein synthesis.

(b) Polymorphism: A variation in the DNA sequence that occurs in a significant proportion of a population.

(c) Translation: The process of converting the mRNA sequence into a protein sequence at the ribosomes. This involves the decoding of mRNA codons and the assembly of amino acids into a polypeptide chain.

(d) Bioinformatics: The use of computer technology to analyze biological data, such as DNA sequences, protein structures, and gene expression patterns. Bioinformatics tools are essential for understanding complex biological systems and developing new medical treatments.