FPSC 2025 Zoology Exam Solutions Dashboard

FPSC 2025 Zoology Solutions

Complete Exam Solutions Dashboard

FPSC Competitive Examination 2025 – Zoology (BS-17)

Examination Information

Paper: Zoology
Time Allowed: 3 Hours
Maximum Marks: 100 (Part I: 20, Part II: 80)
Instructions: Attempt ONLY FOUR questions from PART-II

Question 1

What is the term for the process by which a caterpillar becomes a butterfly?

  1. Fragmentation
  2. Budding
  3. Metamorphosis
  4. None of these

Solution:

Correct Answer: C) Metamorphosis

Detailed Explanation: Metamorphosis is a biological process by which an animal physically develops after birth or hatching, involving a conspicuous and relatively abrupt change in the animal’s body structure through cell growth and differentiation. In insects like butterflies, this involves complete metamorphosis (holometabolism) with four distinct stages:

  • Egg: The first stage where embryonic development occurs
  • Larva (Caterpillar): The feeding and growth stage
  • Pupa (Chrysalis): The transformation stage where tissues are broken down and reorganized
  • Adult (Butterfly): The reproductive stage

This process is regulated by hormones including juvenile hormone and ecdysone.

Question 2

The monosaccharide which is readily absorbed through the intestine independent of Sodium ions is:

  1. Fructose
  2. Glucose
  3. Galactose
  4. None of these

Solution:

Correct Answer: A) Fructose

Detailed Explanation: Fructose is absorbed through facilitated diffusion via GLUT5 transporters in the intestinal epithelium, which does not require sodium ions or energy expenditure. In contrast:

  • Glucose and Galactose: Absorbed via sodium-dependent glucose transporters (SGLT1) that require sodium ions for active transport against concentration gradient
  • Mechanism: SGLT1 co-transports glucose/galactose with sodium ions using the sodium gradient maintained by Na+/K+ ATPase pump

This difference in absorption mechanisms is clinically significant in conditions like glucose-galactose malabsorption.

Question 3

The biodiversity at a geographic scale is called:

  1. Beta diversity
  2. Alpha diversity
  3. Gamma diversity
  4. None of these

Solution:

Correct Answer: C) Gamma diversity

Detailed Explanation: In ecology, biodiversity is measured at three spatial scales:

Type Definition Scale Example
Alpha Diversity Species diversity within a specific habitat or community Local scale Number of species in a single forest
Beta Diversity Species diversity between different habitats Regional scale Species turnover between forest and grassland
Gamma Diversity Total species diversity across a large geographic area Landscape/Continental scale Total species in entire mountain range

Gamma diversity = Alpha diversity × Beta diversity

Question 4

______ is the lag between the creation of the habitat fragment and the decrease in species richness to the equilibrium level.

  1. Fragment destruction
  2. Relaxation time
  3. Retention time
  4. None of these

Solution:

Correct Answer: B) Relaxation time

Detailed Explanation: Relaxation time is a key concept in island biogeography and conservation biology:

  • Definition: The time lag between habitat fragmentation and the eventual decline in species diversity to a new equilibrium level
  • Mechanism: When a habitat is fragmented, species don’t disappear immediately but undergo gradual extinction due to reduced habitat area, edge effects, and isolation
  • Significance: Explains why extinctions continue even after habitat protection measures are implemented
  • Example: In forest fragments, large predators and specialized species may take decades to disappear completely

This concept was developed by Robert MacArthur and E.O. Wilson in their Theory of Island Biogeography.

Question 5

Which co-enzyme is NOT involved in oxidative decarboxylation of pyruvic acid?

  1. CoA-SH
  2. Mg++
  3. Biotin
  4. None of these

Solution:

Correct Answer: C) Biotin

Detailed Explanation: The oxidative decarboxylation of pyruvate to acetyl-CoA is catalyzed by the pyruvate dehydrogenase complex (PDC), which involves:

  • Three Enzymes:
    • Pyruvate dehydrogenase (E1)
    • Dihydrolipoyl transacetylase (E2)
    • Dihydrolipoyl dehydrogenase (E3)
  • Five Coenzymes:
    • Thiamine pyrophosphate (TPP)
    • Lipoamide
    • Coenzyme A (CoA-SH)
    • Flavin adenine dinucleotide (FAD)
    • Nicotinamide adenine dinucleotide (NAD+)
  • Cofactor: Mg2+ (required by E1 component)

Biotin is involved in carboxylation reactions (adding CO₂), not oxidative decarboxylation (removing CO₂).

Question 6

The flexible rod along the dorsal midline in an embryo of chordates is replaced by:

  1. Nerve chord
  2. Sensory organs
  3. Vertebral column
  4. None of these

Solution:

Correct Answer: C) Vertebral column

Detailed Explanation: The notochord is a defining characteristic of chordates:

  • Embryonic Structure: Flexible rod of mesodermal cells along dorsal midline
  • Function: Provides structural support and signaling center for embryonic development
  • Fate in Vertebrates: In most vertebrates, the notochord is replaced by the vertebral column (spine) during development
  • Exceptions: In some primitive vertebrates (hagfish, lampreys) and invertebrate chordates (tunicates, lancelets), the notochord persists throughout life

Question 7

The pigment(s) that contain various forms of iron is/are:

  1. Myoglobin
  2. Hemoglobin
  3. Both (A) & (B)
  4. None of these

Solution:

Correct Answer: C) Both (A) & (B)

Detailed Explanation: Both hemoglobin and myoglobin are hemoproteins containing iron:

Pigment Structure Iron Content Function
Hemoglobin Tetramer (4 subunits: 2α + 2β) 4 heme groups, each with Fe2+ Oxygen transport in blood
Myoglobin Monomer (single polypeptide) 1 heme group with Fe2+ Oxygen storage in muscles

Both contain heme prosthetic groups where iron is in ferrous (Fe2+) state for oxygen binding.