Class 12 Biology Portal

Answer: While both are survival strategies used by organisms to escape stressful environmental conditions over time, they differ significantly in their nature and the organisms that use them:

• Hibernation (Winter Sleep): It is a state of reduced metabolic activity and suspended animation occurring mostly in winter. Organisms escape winter cold by finding warm shelters.
  Example: Polar bears, frogs, and lizards.
• Diapause: It is a stage of suspended development encountered by various species under unfavorable environmental conditions, independent of seasons. It can occur at any stage of life history (like egg, larva, or pupa).
  Example: Many species of zooplankton in lakes and ponds, and certain insects.

Answer: No, the marine fish will not be able to survive in a freshwater aquarium.

Step-by-Step Explanation:

1. Hypertonic vs. Hypotonic Environment: A marine fish is adapted to live in seawater, which has a high salt concentration (hypertonic environment). Its body fluids are osmotically balanced with hypertonic water. Freshwater has a very low salt concentration (hypotonic environment).
2. Endosmosis: When placed in fresh water, water will continuously enter the fish's body through its gills and skin due to endosmosis (osmotic inflow of water).
3. Osmoregulatory Failure: To cope with the massive influx of water, the fish's kidneys would have to strain heavily, leading to severe dilution of its body fluids, swelling of cells, and ultimate tissue rupture or death.

Answer: Certain microbes, specifically known as Archaebacteria (e.g., Thermus aquaticus found in hot springs and deep-sea hydrothermal vents), can easily survive temperatures above $100^{\circ}\text{C}$ due to specialized structural adaptations:

• Thermal Resistant Enzymes: They possess highly stable, heat-resistant enzymes (like Taq polymerase) and proteins that do not denature or break down at extreme temperatures.
• Unique Cell Wall Structure: Their cell walls contain a complex branched-chain lipid structure rather than standard membrane lipids. This branching reduces membrane fluidity and prevents the cell membrane from melting down in extreme heat.

Answer: An individual organism is a single living unit, whereas a population is a group of individuals of the same species living in a specific geographical area. The attributes unique to a population are:

• Birth Rate (Natality): Calculated as per capita births, whereas an individual simply experiences birth.
• Death Rate (Mortality): Calculated as per capita deaths, whereas an individual simply experiences death.
• Sex Ratio: A population consists of a specific percentage of males and females (e.g., 55% males and 45% females), whereas an individual is strictly either male or female.
• Age Distribution / Pyramids: A population contains individuals belonging to different age groups (pre-reproductive, reproductive, post-reproductive) simultaneously.
• Population Density: The total number of individuals per unit area or volume at a given time.

Answer: To find the intrinsic rate of increase ($r$) for an exponentially growing population that doubles, we use the exponential growth equation:

$$N_t = N_0 e^{rt} \quad \text{}$$

Step-by-Step Calculation:

Given that the population doubles, $N_t = 2N_0$ at time $t = 3$ years. Substitute these values into the formula:

$$2N_0 = N_0 e^{r \times 3} \quad \text{}$$ $$2 = e^{3r} \quad \text{}$$

Taking the natural logarithm ($\ln$) on both sides:

$$\ln(2) = 3r \quad \text{}$$

We know that $\ln(2) \approx 0.693$:

$$0.693 = 3r \quad \text{}$$ $$r = \frac{0.693}{3} \quad \text{}$$ $$r = 0.231 \quad \text{}$$

Final Value: The intrinsic rate of increase ($r$) of the population is 0.231 per year (or approximately 23.1%).

Answer: Plants cannot run away from their predators (herbivores). Therefore, they have evolved various morphological and chemical defense mechanisms to protect themselves:

1. Morphological (Structural) Defenses:

• Thorns and Spines: Modified leaves or stems that cause physical injury to herbivores (e.g., Acacia, Cactus, and Bougainvillea).
• Hairy Leaves: Modified surface hairs that make it difficult for insects to feed or lay eggs (e.g., hairy leaves in cotton repel jassids).

2. Chemical Defenses:

• Cardiac Glycosides: Highly toxic substances produced by weeds like Calotropis. If a herbivore eats this plant, it affects their heart function and can be fatal.
• Secondary Metabolites: Plants produce chemicals like nicotine, caffeine, quinine, strychnine, and opium. These are defense mechanisms to make the plant distasteful or toxic to grazing animals.

Answer: The interaction between an orchid plant and a mango tree is classified as Commensalism.

Explanation:

• Commensalism is a type of interspecific interaction where one species benefits while the other species remains completely unaffected (neither harmed nor benefited).
• In this relationship, the orchid benefits by getting physical support, better access to sunlight, and air at a higher location on the tree canopy.
• The mango tree derives no benefit and suffers no harm ($0$) because the orchid is an epiphyte; it absorbs its own moisture and nutrients from the air and does not steal nutrients from the mango tree's vascular bundles.

Answer: The mechanism is known as Brood Parasitism.

Step-by-Step Explanation:

1. Egg Mimicry: Over evolutionary periods, the eggs of the parasitic bird (Cuckoo/Koel) have evolved to closely mimic the host's eggs (Crow) in terms of color, size, and pattern.
2. Deception: Due to this striking structural similarity, the host crow fails to detect the foreign eggs as alien.
3. Rearing: The crow incubates and rears the cuckoo's eggs alongside its own, providing food and protection to the parasitic chicks.

Answer: Out of the Exponential and Logistic growth models, the Logistic Growth Model (Verhulst-Pearl Logistic Growth) is significantly more realistic.

Reasons:

• Limited Resources: In nature, no habitat has infinite resources (food and space) to support endless growth. Resources are always finite.
• Competition: As population density increases, individuals face intense competition for the limited food supply and space.
• Carrying Capacity ($K$): A habitat can support only a maximum possible number of individuals beyond which no further growth is physically sustainable. This limit is called Carrying Capacity.
• Growth Curve: It accurately reflects natural populations by showing an initial lag phase, followed by acceleration, deceleration, and finally an asymptote (leveling off) when it hits $K$, yielding a realistic Sigmoid (S-shaped) curve.

• Commensalism ($+, 0$): An interspecific interaction where one species benefits while the other is neither harmed nor benefited. E.g., Barnacles growing on the back of a whale.
• Parasitism ($+, -$): An interaction where one smaller species (parasite) depends nutritionally on a larger species (host) for food and shelter, causing physiological harm to the host. E.g., Cuscuta growing on hedge plants.
• Mutualism ($+, +$): An obligate, mutually beneficial interaction between two different species where both organisms thrive together. E.g., Lichens (symbiosis between algae and fungi).
• Amensalism ($-, 0$): An interaction where one species is inhibited or completely destroyed, while the other species remains completely unaffected. E.g., The mold Penicillium secretes penicillin which kills surrounding bacteria without affecting the mold itself.

Answer: The three vital characteristics of a population are:

1. Population Density ($N$): It represents the total number of individuals of a species per unit area or volume. It indicates the status of the habitat and resource availability. It can be measured using absolute counts or relative abundance (e.g., pugmarks and fecal pellets for tiger population counting).
2. Natality (Birth Rate): It refers to the per capita rate of birth in a population over a given time frame. It acts as a positive factor that increases population density ($+N$).
3. Mortality (Death Rate): It refers to the per capita rate of death in a population over a given timeframe. It acts as a negative factor that decreases population density ($-N$).

A) 0.8
B) 0.2
C) 0.5
D) 8

Answer: B) 0.2

Explanation: Increase in individuals $= 48 - 40 = 8$. Initial population $= 40$. Birth rate $= \frac{8}{40} = 0.2$ births per capita per year.

A) J-shaped curve
B) Hyperbolic curve
C) Sigmoid curve
D) Linear curve

Answer: C) Sigmoid curve

Explanation: Limited resources introduce a carrying capacity factor, which levels out the exponential growth into an S-shaped or Sigmoid curve.

Q3. Assertion (A): The Logistic growth model is considered more realistic than the Exponential growth model.
Reason (R): Resources like food and space are always finite in any natural habitat.

Answer: A) Both A and R are true, and R is the correct explanation of A.

Explanation: Resource limitation creates natural checks, leading directly to the logistic model's accuracy.

Answer: Gause's Competitive Exclusion Principle states that two closely related species competing for the exact same limited resources cannot co-exist indefinitely. The competitively inferior species will eventually be eliminated.

Answer:

• Expanding Pyramid: Has a broad base because the proportion of the pre-reproductive population is much higher than the reproductive and post-reproductive populations.
• Declining Pyramid: Has a narrow base (urn-shaped) because the pre-reproductive population is smaller than the reproductive population.

Answer: When resource availability is limited, population growth follows a logistic pattern. The differential equation is:

$$\frac{dN}{dt} = rN \left( \frac{K - N}{K} \right) \quad \text{}$$

Components Breakdown:

• $N$: Population density at time $t$.
• $t$: Time period.
• $r$: Intrinsic rate of natural increase (biotic potential).
• $K$: Carrying capacity (the maximum population size an ecosystem can support with its available resources).
• $\frac{K-N}{K}$: Environmental resistance factor that slows down growth as $N$ approaches $K$.

Growth Phases: 1. Lag Phase, 2. Acceleration Phase, 3. Deceleration Phase, 4. Asymptote ($\frac{dN}{dt} = 0$ when $N = K$).