1. Human mtDNA consists of approximately how many base pairs?

  1. 1650
  2. 16500
  3. 165000
  4. 1650000

Answer: (2) 16500

Explanation: In humans, mitochondrial DNA spans about 16,500 DNA building blocks (base pairs)
-> It contains 37 genes, all of which are essential for normal mitochondrial function.

2. The approximate number of branches in amylopectin is

  1. 10
  2. 20
  3. 40
  4. 80

Answer: (4) 80

-> Amylopectin: water-soluble polysaccharide and highly branched polymer of α-glucose units found in plants.
-> It is one of the two components of starch, and another component is amylose.

3. Which one is the largest particulate of the cytoplasm?

  1. Lysosomes
  2. Mitochondria
  3. Golgi apparatus
  4. Entoplasmic reticulum

Answer: (2) Mitochondria

Explanation: Mitochondria is the largest particulate component of the cytoplasm
-> About 15% to 20% of dry weight of the cell.

4. The general formula of polysaccharides is

  1. (C6H10O5)n
  2. (C6H12O5)n
  3. (C6H10O6)n
  4. (C6H20O6)n

Answer: (1) (C6H10O5)n

Explanation: No explanation is needed.

5. Two sugars that differ from one another only in configuration around a single carbon atom are termed

  1. Epimers
  2. Anomers
  3. Optical isomers
  4. Stereoisomers

Answer: (1) Epimers

-> Epimers are carbohydrates that vary in one position for the placement of the -OH group. Example: D-glucose and D-galactose.
-> Anomers: Cyclic stereoisomer, such as sugar, whose sole conformational difference involves the arrangement of atoms or groups in the aldehyde or ketone group.
-> Optical isomers: Compounds with the same number and kinds of atoms, and bonds but different spatial arrangements of the atoms. Two optical isomers are L and D isomers.

6. Enzymes are absent in

  1. Virus
  2. Algae
  3. Fungi
  4. Cyanobacteria

Answer: (1) Virus

Explanation: Viruses are non-living. Feeds on the host body and makes use of the enzyme machinery of the host for their replication.

7. Rapid Conversion of α- D- glucose to β-D- glucose in solution is known as

  1. Fluxionality
  2. Epimerization
  3. Mutarotation
  4. Racemization

Answer: (3) Mutarotation

Explanation: Mutarotation is defined as the change in the optical rotation because of the change in the equilibrium between two anomers when the corresponding stereocenters interconvert.

8. Allergic reactions are mediated by

  1. IgA
  2. IgG
  3. IgD
  4. IgE

Answer: (4) IgE

Explanation: Allergic reactions can be grouped into two classes.
-> Immunoglobulin E (IgE): most common and best understood is mediated by a class of antibodies called immunoglobulin E (IgE)
-> Non-IgE mediated: symptoms to appear more slowly and reaction does not necessarily involve antibodies but instead, cell reactions of the immune system.
-> Diagnose for IgE-mediated allergy: blood test to identify allergen-specific IgE or a skin prick test which results in a local inflammatory reaction after administration of the trigger allergen.
Reference: Allergy by Immunology British Society

9. ABABA……. represents the arrangement layer of

  1. BCC
  2. FCC
  3. CCP
  4. HCP

Answer: (4) HCP

Explanation: Hexagonal close-packed (hcp) has ABABABA (every other layer is the same) arrangement. In this arrangement, each sphere has twelve neighbors.

10. Synthesis of glucose from fat is known as

  1. Glycogenolysis
  2. Glycolysis
  3. Saponification
  4. Gluconeogenesis

Answer: (4) Gluconeogenesis

-> Glycolysis: cytoplasmic pathway that breaks down glucose into two three-carbon compounds (pyruvic acid) and generates energy.
-> Gluconeogenesis: synthesis of glucose from nonsugar precursors, such as lactate, pyruvate, and the carbon skeleton of glucogenic amino acids.
-> Glycogenolysis: biochemical pathway in which glycogen breaks down into glucose-1-phosphate and glucose.
-> Saponification: formation of a metallic salt of a fatty acid; such a salt is called soap. In general, a base (for example, sodium hydroxide) reacts with free fatty acid to form glycerol and soap.

11. Regions of chromosomes that are transcriptionally active are known as

  1. Euchromatin
  2. Chromatin
  3. Telomere
  4. Centromere

Answer: (1) Euchromatin

Explanation: During most of a cell’s life cycle, the chromosomes exist in an unraveled
a linear form that can be transcribed to code for proteins.

-> Region of chromosomes that are transcriptionally active is known as euchromatin.
● These chromosomes undergo normal chromosome condensation and decondensation
during the cell cycle.
● These regions account for most of the genome and lack repetitive DNA.

-> The transcriptionally inactive portions of chromosomes, such as centromeres, are
heterochromatin regions and remain condensed throughout the cell cycle.
● These chromosomes are generally not sequenced due to complex repeat patterns (repetitive DNA) that can be found at centromeres, much of the Y chromosome’s long arm, and the short arms of the acrocentric chromosomes.

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