Molecular Biology

Sickle cell anemia is caused by a point mutation in the -globin gene of hemoglobin, where the codon GAG (glutamate) is mutated to GTG (valine). This single base substitution changes the amino acid and leads to the disease.

Mutations are changes in the DNA sequence (genotype) that can lead to changes in the protein expression and traits (phenotype).
Thus, both can be affected.

The codons for the given amino acids are: Methionine (Start) – AUG, Phenylalanine – UUU/UUC, Arginine – CGC, Glycine – GGC, Phenylalanine – UUC. The sequence given in option 1 correctly matches these codons.

Probes are gene-specific short sequences, single-stranded DNA/RNA fragments, and they are used in colony hybridization to identify specific DNA sequences.

- DNA Ligase joins Okazaki fragments → C - i
- DNA Polymerase catalyzes DNA polymerization in one direction → A - iv
- Splicing removes introns to make mRNA → B - iii
- Ribozyme is an RNA enzyme → D - ii

Nucleosomes are the repeating units of chromatin that appear as "beads-on-string" under an electron microscope. Each nucleosome contains about 200 base pairs of DNA wrapped around histone proteins.

In DNA, nucleotides are linked by phosphodiester bonds to form the backbone of the strand, while complementary nitrogenous bases are joined via hydrogen bonds across the strands. Peptide bonds are found in proteins, not nucleic acids.

The Lac Operon model explains gene regulation in prokaryotes and was introduced by François Jacob and Jacques Monod in 1961.
They demonstrated how genes could be turned on/off based on environmental conditions, particularly the presence or absence of lactose in \textit{E. coli}.
Their work established foundational concepts in molecular biology about:
- Operons
- Regulatory genes
- Repressors
- Inducible gene systems

Let:
- Tall = dominant (T)
- Dwarf = recessive (t)
- Heterozygous tall = Tt
- Dwarf = tt
Cross: Using a Punnett square: Resulting progeny:
- 2 Tt (tall)
- 2 tt (dwarf)
So, 50% of the progeny are dwarf.

Magnesium ions (Mg²⁺) are essential for stabilizing and associating the small and large subunits of prokaryotic ribosomes. Other ions like Na⁺, K⁺, and Ca²⁺ are not directly involved in this association.

Step 1: Analyze Assertion (A)
In prokaryotes, the DNA is located in a region called the nucleoid. It is organized in large loops that are stabilized by proteins. Hence, Assertion (A) is true. Step 2: Analyze Reason (R)
DNA is negatively charged due to the phosphate backbone. In prokaryotes, this DNA is associated with proteins that are positively charged, allowing stabilization. So Reason (R) is also true. Step 3: Examine the explanation relationship
The reason explains why the DNA can be looped and held by proteins — the electrostatic interaction between negatively charged DNA and positively charged proteins.

Step 1: Analyze each statement
- I. Purines and cytosine are common in DNA and RNA — This is correct. Purines (adenine, guanine) and cytosine are found in both DNA and RNA.
- II. Thymine is present only in RNA — Incorrect. Thymine is found only in DNA, not RNA.
- III. Uracil is present in DNA at the place of Thymine — Incorrect. Uracil is found in RNA, not DNA.
- IV. Adenine and Guanine are Purines — This is correct. Step 2: Conclusion
Statements II and III are incorrect.

Understand each linkage and structure in the context of DNA.
A. 3'-5' phosphodiester linkage connects one nucleotide to another via phosphate groups. So, it matches with II: Nucleotide + Nucleotide.
B. N-Glycosidic Linkage connects the nitrogenous base to the sugar. Hence, it matches with I: Nitrogen base + Sugar.
C. 5' end of the polynucleotide refers to the end with a free phosphate group. So, it matches with IV: Free phosphate moiety.
D. Backbone of DNA is formed by repeated sugar-phosphate groups. So, it matches with III: Sugar + Phosphate. Step 2: Match accordingly. A → II
B → I
C → IV
D → III

Step 1: Analyze Statement I.
This is true. Chromatin packaging at higher levels involves non-histone chromosomal proteins along with histones to provide structural support and regulation. Step 2: Analyze Statement II.
This is true. In retroviruses (like HIV), reverse transcription occurs — genetic information flows from RNA to DNA. Step 3: Analyze Statement III.
This is false. The identification of DNA (not RNA) as the genetic material was demonstrated by Avery, MacLeod, and McCarty using Streptococcus pneumoniae. RNA is not the genetic material in this context. Step 4: Analyze Statement IV.
This is false. RNA polymerase III, not II, transcribes 5S rRNA. RNA polymerase II transcribes mRNA precursors.