৪৯তম বিসিএস ⎯ প্রাণ রসায়ন [৬০১]

c) Cholesterol
Explanation: Derived lipids are substances obtained by hydrolysis of simple or complex lipids, such as cholesterol, steroid hormones, and fat-soluble vitamins.

Explanation: Simple lipids include fats and oils made of fatty acids + alcohol (e.g., glycerol). Complex lipids (e.g., phospholipids, glycolipids) contain additional groups.

a) Simple lipid
Explanation: Waxes are esters of long-chain fatty acids and long-chain alcohols, falling under simple lipids.

Explanation: Glycolipids are lipids containing sugar residues (e.g., cerebrosides, gangliosides) and are important in cell recognition.

Answer: c) Phospholipid
Explanation: Sphingomyelin is a phospholipid containing sphingosine backbone, fatty acid, phosphate, and choline.

Answer: গ) Glycogen
Explanation: Glycogen is a carbohydrate (storage polysaccharide), not a lipid.

 Answer: c) Triacylglycerol
Explanation: Triacylglycerols (fats) store more energy per gram (9 kcal/g) than carbohydrates or proteins.

 Answer: b) Steroid
Explanation: Steroid hormones (e.g., cortisol, estrogen, testosterone) are derived from cholesterol.

Answer: c) They are poor conductors of heat
Explanation: Fat tissue prevents rapid heat loss, maintaining body temperature.

Answer: b) Transport of lipids in blood
Explanation: Lipoproteins (HDL, LDL, VLDL) transport hydrophobic lipids through the aqueous bloodstream.

Answer: b) Sphingomyelin
Explanation: Sphingomyelin is a major component of the myelin sheath around neurons, enabling fast conduction.

Answer: b) Iodine value test
Explanation: The iodine number measures the degree of unsaturation (double bonds) in fats and oils.

Answer: a) Oxidation of double bonds
Explanation: Unsaturated fatty acids undergo oxidation, producing foul-smelling aldehydes and ketones.

 Answer: b) Shorter fatty acid chains
Explanation: Saponification number is inversely related to fatty acid chain length.

 Answer: c) Degree of unsaturation
Explanation: Oils (liquid at room temp) have more unsaturated fatty acids than fats (solid at room temp).

Answer: b) Saturated fats
Explanation: Hydrogenation converts double bonds to single bonds, making oils solid (e.g., margarine).

Answer: b) Linoleic acid
Explanation: Linoleic, linolenic, and arachidonic acids are essential fatty acids required in diet.

Answer: b) Dermatitis and growth retardation
Explanation: EFAs are needed for skin health, growth, and membrane integrity.

 Answer: c) Fish oils and flax seeds
Explanation: Omega-3 fatty acids (EPA, DHA, ALA) are abundant in marine oils and plant seeds.

Answer: c) Oleic acid
Explanation: Oleic acid is a monounsaturated fatty acid that can be synthesized by the body.

 Answer: b) Structural framework
Explanation: Phospholipids form bilayers, providing fluidity and selective permeability.

Answer: b) Cell recognition and signaling
Explanation: Glycolipids on the outer membrane help in cell–cell interactions and immune recognition.

Answer: c) Both (a) and (b)
Explanation: Cholesterol stabilizes membrane fluidity by preventing extreme rigidity or excessive fluidity.

Answer: c) Cardiolipin
Explanation: Cardiolipin is essential for mitochondrial electron transport and ATP synthesis.

Answer: b) Niemann-Pick disease
Explanation: Niemann-Pick results from accumulation of sphingomyelin due to defective sphingomyelinase.

Answer: a) Glycerol + fatty acids + phosphate group + alcohol
Explanation: Typical phospholipids contain glycerol backbone, 2 fatty acids, phosphate, and an alcohol (e.g., choline).

 Answer: b) Glycerol
Explanation: Both triglycerides and most phospholipids are glycerol-based.

 Answer: c) 27
Explanation: Cholesterol is a C27 sterol with a four-ring steroid nucleus.

 Answer: b) Sphingosine + sugar + fatty acid
Explanation: Cerebrosides are glycolipids made of sphingosine backbone, one sugar residue, and one fatty acid.

Answer: C Explanation: Purines (Adenine and Guanine) are characterized by a two-ringed structure—a pyrimidine ring fused to an imidazole ring. Pyrimidines (Cytosine, Uracil, and Thymine) are single six-membered heterocyclic rings. Base pairing determines hydrogen bonds (A-T has 2, G-C has 3),

Answer: B
Explanation: A β-N-glycosidic bond is formed between the anomeric carbon (C1') of the sugar and the N9 of a purine or N1 of a pyrimidine. Phosphodiester bonds link nucleotides together, peptide bonds link amino acids, and hydrogen bonds hold the two DNA strands together.

Answer: B Explanation: A nucleoside consists of a nitrogenous base and a pentose sugar. A nucleotide is a nucleoside that has one or more phosphate groups attached to the sugar's 5' carbon.

Answer: A Explanation: Chargaff's rules state that the amount of adenine equals thymine (A=T) and the amount of guanine equals cytosine (G=C). Therefore, the total number of purines (A+G) must equal the total number of pyrimidines (C+T). This makes the ratio (A + G) / (C + T) equal to 1.

Answer: C Explanation: 5-methylcytosine is a common epigenetic modification of DNA in eukaryotes. This methylation is crucial for gene regulation, often leading to transcriptional silencing. While modified bases are abundant in tRNA, 5-methylcytosine is a hallmark modification of DNA.

Answer: C Explanation: While hydrogen bonds between complementary base pairs (A-T and G-C) are the primary force holding the strands together, the overall stability of the double helix is also significantly influenced by hydrophobic and base-stacking interactions between adjacent bases.

Answer: B Explanation: The structural framework, or backbone, of a nucleic acid strand is formed by alternating sugar and phosphate residues. These are linked by phosphodiester bonds between the 3' carbon of one sugar and the 5' carbon of the next.

Answer: D Explanation: The B-DNA is the classic Watson-Crick model and is the predominant form in the cell under physiological conditions. It is a right-handed helix with distinct major and minor grooves, and its base pairs are nearly perpendicular to the helix axis. Z-DNA is left-handed, and A-DNA has tilted base pairs.

Answer: B
Explanation: A pseudoknot is a common and important tertiary structure in RNA. It is formed when a single-stranded loop region base-pairs with a complementary sequence outside of that loop, creating a complex, knotted-like structure. Pseudoknots are functionally important in ribozymes and telomerase RNA.

Answer: B
Explanation: The formation of Z-DNA is particularly favored by alternating purine-pyrimidine sequences, especially alternating G and C residues. High salt concentration and negative supercoiling can also promote its formation.

Answer: A
Explanation: A G-quadruplex (or G4) is a secondary structure formed in nucleic acid sequences that are rich in guanine. It consists of a square arrangement of four guanine bases (a G-tetrad) stabilized by Hoogsteen hydrogen bonds, with these tetrads stacked on top of each other. They are often found in telomeres and promoter regions.

Answer: B
Explanation: The edges of the base pairs are exposed in the major and minor grooves. The major groove is wider and exposes more of the base pairs, making it the primary site for the binding of sequence-specific proteins like transcription factors, which can "read" the DNA sequence without unwinding the helix.

Answer: C
Explanation: Ribosomal RNA (rRNA) is the most abundant type, making up over 80% of the total cellular RNA. This is because it is a stable, structural component of ribosomes, and a cell needs a vast number of ribosomes to carry out protein synthesis.

Answer: D
Explanation: The amino acid is covalently attached to the free 3'-hydroxyl group of the adenosine residue at the end of the 3' acceptor stem. The sequence at this end is typically CCA. The anticodon loop pairs with the codon on the mRNA.

Answer: D
Explanation: The pentose sugar in RNA is ribose, which has a hydroxyl (-OH) group on the 2' carbon. The sugar in DNA is deoxyribose, which has only a hydrogen (-H) at that position. This 2'-OH group makes RNA more susceptible to hydrolysis and thus less stable than DNA. While C is generally true, it's a structural, not chemical, difference.

Answer: B
Explanation: The spliceosome, which removes introns, is a large complex of small nuclear RNAs (snRNAs) and proteins. The catalytic activity—the breaking and forming of phosphodiester bonds during splicing—is carried out by the snRNA components, not the proteins. This is a prime example of a ribozyme. The peptidyl transferase activity in the ribosome (rRNA) is another major example.

Answer: A
Explanation: The anticodon on the tRNA pairs with the codon on the mRNA in an antiparallel fashion. The base pairing rules are A with U, and G with C. Therefore, the anticodon 3'-UAC-5' will bind to the mRNA codon 5'-AUG-3'.

Answer: B
Explanation: snRNAs, in complex with proteins to form small nuclear ribonucleoproteins (snRNPs or "snurps"), are the core components of the spliceosome. The spliceosome is the machinery responsible for removing introns from a pre-mRNA transcript.

Answer: B
Explanation: Eukaryotic pre-mRNAs undergo processing, which includes the addition of a 7-methylguanosine cap to the 5' end. This cap is added via an unusual 5'-5' triphosphate linkage and is critical for mRNA stability, export from the nucleus, and initiation of translation. The poly(A) tail is at the 3' end.

Answer: D
Explanation: miRNAs are small non-coding RNA molecules that play a crucial role in post-transcriptional gene regulation. They typically bind to complementary sequences in the 3' untranslated region (UTR) of target mRNAs, leading to either translational repression or degradation of the mRNA.

Answer: B
Explanation: According to Chargaff's rules, G = C and A = T. If G = 20%, then C must also be 20%. Together, G + C = 40%. The remaining percentage must be A + T, which is 100% - 40% = 60%. Since A = T, they must each be half of 60%, so T = 30%.