Molecular Biology

M: Largest DNA bands (closest to the wells)
N: Smallest DNA bands (furthest from the wells)

Agarose gel electrophoresis is a widely used laboratory technique for separating and analyzing DNA, RNA, or proteins based on their size and charge. It utilizes an agarose gel matrix, an electrical field, and a buffer solution to separate molecules according to their molecular weight.

In agarose gel electrophoresis, DNA fragments are separated based on their size. Smaller fragments move faster through the gel matrix and migrate further from the wells (where the DNA is loaded) than larger fragments.

Thus, the correct option is C: M-Largest DNA bands & N-Smallest DNA bands

A nucleosome is the basic structural unit of chromatin, consisting of DNA wrapped around histone proteins.

The length of DNA in a nucleosome is ~146 base pairs (bp) tightly bound to the histone core, with an additional ~54 bp as linker DNA, totaling ~200 bp per nucleosome.

Let's match the scientists listed in Column I with their significant contributions listed in Column II:

• i. Griffith: Frederick Griffith conducted experiments in 1928 with *Streptococcus pneumoniae*, demonstrating that genetic material could be transferred from heat-killed virulent bacteria to live non-virulent bacteria, making them virulent. This phenomenon was termed the Transforming principle (r).
  So, i → r.
• ii. Jacob & Monad: François Jacob and Jacques Monod proposed the operon model in 1961 to explain the regulation of gene expression in prokaryotes, based largely on their studies of the lactose metabolism system in *E. coli*, known as the Lac operon (p).
  So, ii → p.
• iii. Meselson & Stahl: In 1958, Matthew Meselson and Franklin Stahl conducted an experiment using density gradient centrifugation and isotopes of nitrogen (¹⁴N and ¹⁵N) to demonstrate that DNA replicates semi-conservatively (s), where each new DNA molecule consists of one old strand and one newly synthesized strand.
  So, iii → s.
• iv. Hershey and Chase: Alfred Hershey and Martha Chase performed experiments in 1952 using bacteriophages labeled with radioactive isotopes (³²P for DNA and ³⁵S for protein). Their results showed that phage DNA, not protein, entered the bacterial host cell during infection, providing strong evidence that DNA is the genetic material (q).
  So, iv → q.

The correct matching is:

• i → r
• ii → p
• iii → s
• iv → q

Comparing this to the options provided:

• i-p, ii-q, iii-r, iv-s
• i-p, ii-s, iii-q, iv-r
• i-r, ii-p, iii-s, iv-q
• i-r, ii-q, iii-p, iv-s

The correct option is i-r, ii-p, iii-s, iv-q.

The process of transcription involves synthesizing an mRNA molecule using a DNA template strand. The RNA polymerase enzyme reads the template strand in the 3' to 5' direction and synthesizes the mRNA strand in the 5' to 3' direction.

The sequence of the mRNA will be complementary to the DNA template strand, with Uracil (U) replacing Thymine (T) in the RNA molecule.

The given DNA template strand is:

3'-ATGCTTCCGAAT-5'

We need to find the complementary mRNA sequence, remembering the base pairing rules (DNA → RNA):

• A (DNA) → U (RNA)
• T (DNA) → A (RNA)
• G (DNA) → C (RNA)
• C (DNA) → G (RNA)

Let's determine the complementary base for each nucleotide in the template strand, starting from the 3' end of the DNA to build the 5' end of the mRNA:

DNA Template: 3' - A T G C T T C C G A A T - 5'

Complementary mRNA: 5' - U A C G A A G G C U U A - 3'

So, the sequence of the corresponding transcribed mRNA is:

5' - UACGAAGGCUUA - 3'

Now, let's compare this with the given options:

• 5' - TAC GAA GGC CTT - 3' (Incorrect - contains T)
• 5' - UAC GAA GGC UUA - 3' (Correct)
• 3' - UAC GAA GGC UUA - 5' (Incorrect - wrong direction/polarity)
• 3' - TAC GAA GGC CTT - 5' (Incorrect - wrong bases and polarity)

The correct option is 5' - UAC GAA GGC UUA - 3'.

Transfer RNA (tRNA) molecules play a crucial role in protein synthesis by acting as adaptors between messenger RNA (mRNA) codons and amino acids.

A tRNA molecule has a characteristic structure, often represented in 2D as a cloverleaf and in 3D as an L-shape. Key features include:

• Anticodon Loop: Contains a sequence of three nucleotides (the anticodon) that is complementary to a specific mRNA codon. This loop recognizes and binds to the codon on the mRNA molecule during translation.
• Acceptor Stem: This is a region at one end of the molecule formed by base pairing between the 5' and 3' end strands. It terminates with a specific sequence at the 3' end.
• 5' end: Typically ends with a guanine residue.
• 3' end: This end extends slightly beyond the double-helical acceptor stem and always terminates with the nucleotide sequence CCA (Cytosine-Cytosine-Adenine).

The amino acid binding site is located at the 3' end of the tRNA molecule. Specifically, the amino acid corresponding to the tRNA's anticodon is covalently attached to the terminal Adenosine residue of the CCA sequence.

Therefore, the amino-acid binding site is located at the 3' end.

In molecular biology, vectors are used to transfer foreign DNA into host cells for various purposes, such as cloning or expressing genes of interest. To ensure that the desired DNA has been successfully inserted into the host cells, a selectable marker is often included in the vector. The most common selectable markers are antibiotic resistance genes.

These antibiotic resistance genes provide a way to select for cells that have taken up the vector and acquired the desired DNA. The vector typically contains a gene that confers resistance to an antibiotic, such as ampicillin or kanamycin. When the vector is introduced into the host cells, the cells are typically grown in a medium containing the corresponding antibiotic. Only the cells that have successfully incorporated the vector, along with the antibiotic resistance gene, will be able to survive and grow in the presence of the antibiotic.

Therefore, the presence of an antibiotic resistance gene in a vector allows for the selection and growth of transformed cells that have taken up the vector and acquired the foreign DNA.

• The Lac operon model was proposed by François Jacob and Jacques Monod in 1961.
• This model explains the regulation of lactose metabolism in bacteria, specifically in Escherichia coli (E. coli).
• The Lac operon consists of a promoter (where RNA polymerase binds to initiate transcription), an operator (a regulatory region where the repressor protein binds), and structural genes that code for enzymes involved in lactose metabolism (e.g., lacZ, lacY, and lacA).

Key points of the Lac operon model:

• Induction: The presence of lactose causes the repressor to detach from the operator, allowing the RNA polymerase to transcribe the structural genes and thus produce enzymes needed to metabolize lactose.
• Repression: In the absence of lactose, the repressor protein binds to the operator and blocks transcription, preventing unnecessary production of the enzymes.
• The discovery of the Lac operon model helped in understanding gene regulation and enzyme synthesis in prokaryotic cells.
• Jacob and Monod's work was significant in the field of molecular biology and earned them the Nobel Prize in Physiology or Medicine in 1965.

Thus, the Lac operon model was elucidated by François Jacob and Jacques Monod.

• During transcription, the DNA strand with a 3' → 5' polarity serves as the template for RNA synthesis.
• The RNA polymerase synthesizes the mRNA in the 5' → 3' direction by reading the DNA template strand in the 3' → 5' direction.
• The RNA polymerase catalyzes the addition of nucleotides to the growing mRNA strand, which occurs in the 5' → 3' direction, while it reads the template in the opposite direction (3' → 5').

The correct answer is (B) : Enzyme DNA dependent RNA polymerase always catalyse the polymerisation in 5'→3' direction.

• Monocistronic genes refer to genes that encode for a single protein product in eukaryotes, as opposed to operons in prokaryotes, which may encode multiple proteins.
• Eukaryotic genes are called split genes because they consist of exons and introns. The exons are the coding regions that are expressed, while introns are the non-coding regions that are transcribed but removed during RNA splicing.
• Exons are interrupted by introns (Option D). This means that the exons (coding regions) are separated by introns (non-coding regions) within the gene. After transcription, introns are spliced out, leaving the exons to form the final messenger RNA (mRNA) that codes for the protein.

The correct answer is (D) : Exons are interrupted by Introns.

Histone proteins are a class of proteins that play a crucial role in DNA packaging in eukaryotic cells. They are positively charged due to the presence of basic amino acids in their structure.

The basic amino acids that contribute to the positive charge of histones are arginine and lysine.

• Arginine contains a positively charged guanidino group in its side chain, which is important for interactions with the negatively charged phosphate backbone of DNA.
• Lysine has an amino group (-NH₂) that is positively charged under physiological conditions and helps in binding to the DNA.

These positive charges help histones interact with the negatively charged DNA, aiding in the formation of nucleosomes, which are the building blocks of chromatin.

Thus, histone proteins are rich in arginine and lysine residues, which makes them positively charged.

The correct answer is (C) : Arginine and Lysine