Organisms and Populations — Class 12 Biology practice questions

Question 1

Read the following passage and answer the questions that follow:

A wildlife biologist is studying two animal populations in the Corbett National Park. She records the following data for a tiger population over two consecutive years:

• Year 1: Population size (N) = 120 tigers; Birth rate = 18 per year; Death rate = 12 per year; Immigration = 4 per year; Emigration = 6 per year.
• Year 2: The population shows a dramatic decline — rangers discover that a severe outbreak of canine distemper virus has killed many individuals. Surviving tigers are now restricted to isolated forest patches due to new highway construction cutting through the reserve. The biologist also notes that the age pyramid of the tiger population has shifted to show a very large post-reproductive class and a very small pre-reproductive class.

She compares this with a population of spotted deer (prey) in the same reserve, which is currently showing a J-shaped growth curve despite abundant resources.

Accepted Answer

Sub-part (i): [1 mark]
Calculate the population growth rate (dN/dt) for the tiger population in Year 1.

Formula: dN/dt = (Birth rate + Immigration) − (Death rate + Emigration)

dN/dt = (18 + 4) − (12 + 6)
= 22 − 18
= +4 tigers per year                                                    ✦ 1 mark

(The population is increasing by 4 individuals per year.)

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Sub-part (ii): [1 mark]
What does the age pyramid described for the Year 2 tiger population indicate about the future of this population? Name this type of age pyramid.

• The pyramid type described = Declining / Regressive pyramid              ✦ ½ mark
• Interpretation: A large post-reproductive class and a very small pre-reproductive class means very few individuals will be available for future reproduction. This indicates the population size will DECREASE further in coming years — the population has a poor growth prospect and faces a risk of local extinction.   ✦ ½ mark

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Sub-part (iii): [1 mark]
The highway construction has restricted tigers to isolated forest patches. Explain ONE population-level consequence of this habitat fragmentation on the genetic health of the tiger population.

• Small, isolated populations undergo genetic drift — random changes in allele frequencies.         ✦ ½ mark
• This leads to inbreeding (mating between closely related individuals) → inbreeding depression → reduced genetic diversity → lower adaptability to disease, environmental change and reduced reproductive fitness. This makes the population even more vulnerable to extinction.         ✦ ½ mark

(Acceptable alternative: Founder effect — if a small group gets isolated, they carry only a fraction of the original gene pool → reduced genetic variation in the isolated sub-population.)

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Sub-part (iv): [1 mark]
The spotted deer population is showing a J-shaped growth curve. However, a biologist argues that this cannot continue indefinitely in nature. Name the growth model that will EVENTUALLY apply to this deer population and write the equation that describes it. Identify the term in the equation that is absent in the J-shaped growth model.

• Growth model that will eventually apply = Logistic growth (S-shaped / sigmoid growth curve)     ✦ ¼ mark

• Equation:
  dN/dt = rN × [(K − N) / K]                                            ✦ ½ mark

Where:
  r = intrinsic rate of natural increase
  N = population size at time t
  K = carrying capacity (maximum population size the environment can sustain)

• The term ABSENT in J-shaped (exponential) growth model = (K − N)/K
  This is the environmental resistance term (also written as the logistic term). It represents the unutilised resources / unused carrying capacity. As N approaches K, this term approaches zero, slowing growth.    ✦ ¼ mark

Question 2

Read the following passage carefully and answer the questions that follow:

Dr. Priya, an ecologist, was studying two species of flour beetles — Tribolium castaneum (Species A) and Oryzaephilus surinamensis (Species B) — in a series of controlled laboratory microcosms containing flour as the only food resource. In Experiment 1, she cultured each species separately in identical containers of flour. Both populations grew rapidly and reached stable carrying capacities. In Experiment 2, she placed both species together in the same container with the same total amount of flour. After 30 weeks, Species A was thriving and had actually increased its population beyond its previous carrying capacity, while Species B had declined to local extinction. Interestingly, when Dr. Priya examined the flour under a microscope, she noticed that Species A had been consuming the eggs and pupae of Species B, in addition to feeding on flour.

(i) Identify the type of interspecific interaction occurring between Species A and Species B in Experiment 2. Justify your answer using + / − notation. (1 mark)

(ii) Based on the outcome of Experiment 2, which ecological principle does this experiment demonstrate? State the exact principle with its condition. (1 mark)

(iii) Compare the population growth of Species A in Experiment 1 (alone) versus Experiment 2 (with Species B). Explain why Species A exceeded its earlier carrying capacity in Experiment 2, using ecological reasoning. (2 marks)

Accepted Answer

MARKING SCHEME — Total: 4 marks

─────────────────────────────────────────
(i) Type of interaction and justification [1 mark]
─────────────────────────────────────────
• The interaction is PREDATION (+ / −).
  — Species A (+): benefits — obtains additional nutrition from eggs and pupae of Species B, and also faces reduced competition for flour.
  — Species B (−): harmed — its offspring are consumed, leading to population decline and eventual extinction.

[Award 1 mark for: correctly naming 'Predation' AND stating + for Species A and − for Species B, or equivalent correct justification. Accept 'Predation superimposed on Competition' for full credit if justified.]

─────────────────────────────────────────
(ii) Ecological principle demonstrated [1 mark]
─────────────────────────────────────────
• The experiment demonstrates GAUSE'S COMPETITIVE EXCLUSION PRINCIPLE.
• Exact statement: Two species competing for the same limiting resource CANNOT coexist indefinitely; the competitively superior species will eventually eliminate (exclude) the inferior one.
• Condition: Both species must be competing for the SAME limiting resource (here: flour as sole food).

[Award 1 mark for: naming the principle correctly AND stating the condition (same limiting resource / niche overlap). 'Competitive exclusion principle' alone without Gause's name is also acceptable.]

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(iii) Comparison and ecological explanation [2 marks]
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Comparison (½ mark):
• In Experiment 1 (alone): Species A grew logistically and stabilised at its carrying capacity (K) set by the flour resource alone.
• In Experiment 2 (with Species B): Species A exceeded its earlier carrying capacity — its final population size was HIGHER than K recorded in Experiment 1.

Ecological explanation (1½ marks):
• In Experiment 2, Species A gained an ADDITIONAL food source — the eggs and pupae of Species B — that was NOT available in Experiment 1.
  → This effectively expanded the total available resource base (energy input) for Species A beyond flour alone.
• Simultaneously, as Species B declined, interspecific competition for flour was progressively REDUCED, freeing up more flour for Species A.
• Both effects together — additional prey resource + reduced competition — allowed Species A's population to be supported at a density higher than what flour alone could sustain, i.e., its effective carrying capacity increased in the mixed culture.
• This illustrates that carrying capacity (K) is not a fixed absolute value but depends on the TOTAL available resources and the INTENSITY OF COMPETITION; when a species simultaneously acts as a predator on its competitor, its realised resource base expands.

[Award:
  ½ mark — correct comparison (Exp 1: reached K; Exp 2: exceeded earlier K)
  ½ mark — identifies additional prey/food source as reason
  ½ mark — identifies reduced interspecific competition as contributing factor
  ½ mark — links both factors to an effectively higher carrying capacity in Experiment 2]

─────────────────────────────────────────
VALUE POINTS SUMMARY (for quick marking)
─────────────────────────────────────────
(i)  Predation / (+/−) with justification .............. 1
(ii) Gause's Competitive Exclusion Principle + condition . 1
(iii) Comparison of K in Exp 1 vs Exp 2 ............... ½
      Additional food resource (eggs/pupae of B) ....... ½
      Reduced interspecific competition for flour ....... ½
      Effective/realised K increases explanation ........ ½
                                            TOTAL = 4 marks

Question 3

Read the following passage carefully and answer the questions that follow.

A team of ecologists studied two plant populations (Population A and Population B) of the same species growing in two different habitats — a forest clearing and a dense forest interior. They recorded the following data after 5 years:

| Parameter | Population A (Forest Clearing) | Population B (Dense Forest Interior) |
|---|---|---|
| Initial population size (N₀) | 200 | 200 |
| Final population size (N) | 800 | 250 |
| Birth rate (per individual per year) | 0.60 | 0.30 |
| Death rate (per individual per year) | 0.10 | 0.25 |
| Immigration | Negligible | Negligible |
| Age structure | Mostly pre-reproductive | Mostly post-reproductive |

The ecologists also noted that Population A showed a J-shaped growth curve, while Population B showed an S-shaped (sigmoidal) growth curve. In Population B, the carrying capacity (K) of the habitat was estimated to be 260 individuals.

(i) Calculate the per capita rate of increase (r) for BOTH Population A and Population B using the formula r = birth rate − death rate. State which population is growing faster and give ONE ecological reason for this difference based on the data provided. (2 marks)

(ii) Population B is showing logistic growth (S-shaped curve) with its population size (N = 250) close to the carrying capacity (K = 260). Explain what will happen to the growth rate of Population B as N approaches K, and state the term used to describe the maximum population size a habitat can support. Also, identify ONE feature from the table that suggests Population B's growth will slow down further in the near future. (2 marks)

Accepted Answer

MARKING SCHEME — CASE STUDY (4 marks)

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Part (i) — 2 marks
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Calculation of per capita rate of increase (r):

Formula: r = Birth rate − Death rate

• Population A:
  r = 0.60 − 0.10 = 0.50 per individual per year          [½ mark]

• Population B:
  r = 0.30 − 0.25 = 0.05 per individual per year          [½ mark]

• Population A is growing faster (r = 0.50 >> r = 0.05).  [½ mark]

• Ecological reason (any ONE acceptable):
  — Population A has a much higher birth rate (0.60) and a much lower death rate (0.10) compared to Population B, indicating more favourable resource availability in the forest clearing (less competition / more light / less crowding).
  OR
  — Population A shows J-shaped (exponential) growth, meaning resources are not yet limiting; Population B is near its carrying capacity (K = 260) and experiences greater intraspecific competition, which suppresses birth rate and increases death rate.
                                                            [½ mark]

(1×2 = 2 marks)

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Part (ii) — 2 marks
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• As N approaches K, the growth rate (dN/dt) of Population B will DECREASE and approach ZERO.
  Reason: The term (K − N)/K in the logistic growth equation approaches zero as N → K, meaning the population has used up almost all available resources and the habitat can support very few additional individuals.
                                                            [1 mark]

• The maximum population size a habitat can support is called the CARRYING CAPACITY (K).
                                                            [½ mark]

• Feature from the table that suggests further slowing of growth:
  The age structure of Population B is 'mostly post-reproductive' — a population dominated by post-reproductive individuals will have very few individuals capable of reproducing, so the birth rate will decline further, causing the growth rate to slow down even more.
  (Accept: high death rate of 0.25 compared to birth rate of 0.30, giving a very low r = 0.05, indicating the population is already barely growing.)
                                                            [½ mark]

(1×2 = 2 marks)

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TOTAL: 4 marks
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Question 4

Read the following case and answer the questions that follow:

A group of ecology students set up a controlled experiment in a laboratory. They introduced 20 Paramecium caudatum individuals into a culture flask containing a fixed amount of nutrient medium. They recorded the population size every 24 hours over a period of 20 days. Initially, the population grew rapidly. However, after day 12, the growth rate slowed and the population stabilised at around 200 individuals. The students plotted their data on a graph.

(a) Name the type of population growth curve the students obtained. Draw and label the curve, marking clearly: (i) the lag phase, (ii) the exponential (log) phase, (iii) the stationary phase, and (iv) the carrying capacity (K). [2 marks]

(b) What does the term 'carrying capacity (K)' mean in the context of this experiment? State its value as observed by the students. [1 mark]

(c) Write the equation that describes this type of population growth. Identify what the term (K − N)/K represents in the equation. [1 mark]

Accepted Answer

MARKING SCHEME — Total: 4 marks

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(a) Type of curve + Labelled diagram [2 marks]
─────────────────────────────────────────

The type of growth curve obtained is the LOGISTIC GROWTH CURVE (Verhulst–Pearl logistic growth) — also called the S-shaped curve or sigmoid curve. (1 mark)

Diagram (1 mark for correctly drawn and labelled figure):

Population
         Size (N)
              |
          K ──|─────────────────────────── (Carrying capacity = 200)
              |                 ___________
              |            ___/   (iii) Stationary phase
              |          _/
              |        _/ (ii) Exponential / Log phase
              |      _/
              |  ___/
              | / (i) Lag phase
              |/___________________________
                          Time (days)

Labels required on diagram:
• (i)  Lag phase       — initial slow growth (days 1–3 approx.)
• (ii) Exponential / Log phase — rapid increase in population size
• (iii) Stationary phase — population stabilises; growth rate ≈ 0
• K    — Carrying capacity marked as a horizontal dashed line at the top

(Award 1 mark for correct S-shaped curve drawn with all four labels clearly marked. An unlabelled or incorrectly shaped curve earns 0 marks for this component.)

─────────────────────────────────────────
(b) Meaning of carrying capacity + observed value [1 mark]
─────────────────────────────────────────

Carrying capacity (K) is the maximum number of individuals that a given habitat (or environment) can support / sustain at a given time, given the available food, space, and other resources.

In this experiment, K = 200 individuals. (1 mark)

(Award 1 mark for correct definition AND the value 200. Accept equivalent phrasing such as 'the maximum population size the culture flask can support'.)

─────────────────────────────────────────
(c) Equation + meaning of (K − N)/K [1 mark]
─────────────────────────────────────────

Equation for logistic growth:

dN/dt = rN × (K − N)/K

Where:
• dN/dt = rate of change of population size
• r     = intrinsic rate of natural increase
• N     = current population size
• K     = carrying capacity

The term (K − N)/K represents the unutilised opportunity for growth / the fraction of carrying capacity still available / the environmental resistance acting on the population.

(Award 1 mark for the correct equation AND correct identification of (K − N)/K. Accept 'remaining resources available for growth' or 'degree to which the environment can support further growth'.)

─────────────────────────────────────────
MARK SUMMARY:
(a) Name of curve = 1 mark; Labelled diagram = 1 mark → 2 marks
(b) Definition + value of K → 1 mark
(c) Equation + meaning of (K−N)/K → 1 mark
TOTAL = 4 marks (1×4=4)
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Question 5

Assertion (A): In a population showing logistic growth, the rate of population growth (dN/dt) is maximum when N = K/2.
Reason (R): At N = K/2, the value of the logistic growth equation dN/dt = rN[(K−N)/K] yields its highest possible output because the term (K−N)/K is at its maximum at this point.

Accepted Answer

Correct option: (C) A is true but R is false.

• Assertion (A) is TRUE: In logistic (Verhulst-Pearl) growth, dN/dt = rN[(K−N)/K]. This is a product of two terms — rN (increases as N rises) and (K−N)/K (decreases as N rises). The product is maximised when N = K/2, giving maximum growth rate.

• Reason (R) is FALSE: The reasoning given in R is incorrect. At N = K/2, the term (K−N)/K = (K − K/2)/K = 0.5, which is NOT at its maximum — it is at its maximum value of 1 when N = 0 (i.e., at the very start). The reason the growth rate is maximum at N = K/2 is NOT because (K−N)/K is maximum there, but because the product rN × [(K−N)/K] is maximised at N = K/2 — a result of the two opposing factors (increasing N and decreasing (K−N)/K) balancing to give the highest combined value.

Hence, A is correct but R gives a wrong explanation — option (C).

Question 6

A wildlife ecologist is studying a population of tigers in a national park. She notices that the park has a large proportion of old tigers (post-reproductive age group) compared to young (pre-reproductive) and reproductive individuals. Based on this age structure, which of the following predictions about the future population trend is MOST appropriate?

Accepted Answer

Correct answer: (B)

Reasoning (CBSE marking scheme style):

• A population's age structure (age pyramid) is divided into three groups:
  – Pre-reproductive (young)
  – Reproductive
  – Post-reproductive (old)

• When the post-reproductive age group is disproportionately LARGE compared to the pre-reproductive and reproductive groups, the age pyramid takes a URN-SHAPED (declining) form.

• Since only the reproductive age group contributes to birth rate (natality), a small reproductive fraction means: births < deaths → population declines over time.

• Therefore, option (B) is correct: the population will decline because insufficient reproductive individuals are available to replace the dying post-reproductive members.

Why other options are wrong:
• (A) Incorrect — old individuals do not participate in reproduction; exponential growth requires a large pre-reproductive base, not a post-reproductive one.
• (C) Incorrect — stability requires roughly equal representation or a larger young cohort; a dominant post-reproductive group predicts decline, not stability.
• (D) Incorrect — reduced intraspecific competition alone does not cause population increase; reproduction rate is the key driver of future population size.

Question 7

Assertion (A): In a population showing logistic growth, the rate of population growth is maximum when the population size (N) is exactly equal to K/2 (where K is the carrying capacity).
Reason (R): In logistic growth, the factor (K − N)/K acts as an environmental resistance that decreases as N approaches K.

Accepted Answer

Correct option: (A) Both A and R are true, and R is the correct explanation of A.

Explanation:
• Assertion is TRUE: In logistic (S-shaped / sigmoid) growth described by the Verhulst–Pearl equation,
  dN/dt = rN × (K − N)/K
  The growth rate (dN/dt) is a product of rN and the term (K − N)/K. This is maximised when N = K/2, because at that point the two opposing factors — the biotic potential (rN, which increases with N) and the environmental resistance ((K − N)/K, which decreases as N increases) — are exactly balanced to give the maximum product. Above or below K/2, the growth rate is lower.

• Reason is TRUE: The term (K − N)/K represents environmental resistance (limiting factor). When N is small, (K − N)/K ≈ 1 → little resistance. As N → K, (K − N)/K → 0 → maximum resistance → growth rate → 0.

• R correctly explains A: The growth rate peaks at N = K/2 precisely because the environmental resistance factor (K − N)/K, which reduces the realised growth rate below the maximum possible (rN), is at the specific value (0.5) that maximises the overall product rN(K − N)/K.

[1 mark for option A]

Question 8

A wildlife biologist is studying two populations of the same deer species — one in a dense forest with abundant food and no predators, and another in an open grassland with limited food and several predators. After five years, the forest population shows a J-shaped growth curve, while the grassland population shows an S-shaped growth curve. Which of the following correctly explains the difference in growth patterns observed?

Accepted Answer

Correct Answer: (B)

Explanation (for examiner reference / 1 mark awarded for choosing B):

• In the forest population: food is abundant and predators are absent → resources are NOT limiting → population grows at its biotic potential → exponential / J-shaped growth described by dN/dt = rN.

• In the grassland population: food is limited and predators are present → these act as environmental resistance → population growth slows as it approaches the carrying capacity (K) → logistic / S-shaped (sigmoid) growth described by dN/dt = rN[(K−N)/K].

• When N approaches K, the term (K−N)/K → 0, so growth rate declines → the curve levels off at K — this is the pattern seen in the grassland population.

Why other options are wrong:
• (A) — Labels are reversed; unlimited resources → exponential (J), not logistic.
• (C) — The shape of the curve is determined by whether K is a limiting factor, not merely by the value of K; both populations have the same r by definition (same species), but environmental resistance differs.
• (D) — In logistic growth, birth rate does NOT become zero at K; birth rate equals death rate, so net growth = 0, but both rates are still positive.

Question 9

Which of the following population attributes can be measured for an individual organism as well as for a population?

Accepted Answer

Correct Answer: (A) Birth rate (natality)

The NCERT Class 12 Biology textbook (Chapter 13 – Organisms and Populations) explicitly states that birth rate (natality) can be expressed for an individual organism as well as for a population. For example, if in a pond there were 20 lotus plants and 8 new plants were added through reproduction, the birth rate = 8/20 = 0.4 offspring per lotus per year — this is a per-individual expression. Age structure and sex ratio are attributes that describe a population as a whole and cannot be meaningfully assigned to a single individual. Therefore, birth rate (natality) is the population attribute that can be measured at both the individual and population level.

(1 mark for selecting option A)

Question 10

A wildlife biologist studying a small island notices that as the island's area increases, the number of bird species also increases. She plots log(species richness) against log(area) and obtains a straight line. Which of the following correctly represents the equation of this relationship?

Accepted Answer

Correct answer: (A) log S = log C + Z log A

**Value point (1 mark):**
The Species-Area relationship (given by Alexander von Humboldt) is expressed as:

S = CA^Z

where:
- S = Species richness
- A = Area
- C = Y-intercept (a constant relating to species richness of the smallest unit area)
- Z = slope of the line / regression coefficient (lies between 0.1 and 0.2 for small areas; 0.6–1.2 for large areas/continents)

On a logarithmic scale, this becomes:

**log S = log C + Z log A**

This gives a straight line when log S is plotted against log A, confirming option (A). (1×1=1)

Question 11

A marine ecologist studying a coral reef notices that two closely related damselfish species (Species A and Species B) initially coexist in the same zone, feeding on the same algae. Over 18 months of observation, Species B gradually disappears from the reef entirely, while Species A thrives. The ecologist concludes this outcome was inevitable given their ecological relationship. Which of the following best explains the theoretical basis for the ecologist's conclusion?

Accepted Answer

Correct Answer: (B)

Gause's Competitive Exclusion Principle states: Two species competing for the same limiting resource cannot coexist indefinitely in the same niche — the competitively superior species will eventually eliminate (or competitively exclude) the inferior one.

Why (B) is correct:
• Both damselfish species occupy the same niche (same zone, same food resource = same limiting resource).
• The interaction is competition (−/− interaction: both species are negatively affected by each other's presence).
• The inevitable outcome of complete niche overlap is local extinction of the inferior competitor — exactly as observed (Species B disappears).
• This is the classic prediction of Gause's principle, demonstrated experimentally with Paramecium aurelia and P. caudatum.

Why other options are incorrect:
• (A) Allee effect: refers to reduced fitness at LOW population density (cooperative behaviours break down) — not the same as competitive exclusion; no mention of cooperative behaviour or density-dependent reproductive failure.
• (C) Amensalism (+/0 or 0/−): one species is harmed, the other is unaffected — this is NOT competition, and no evidence of toxic secretion is given in the scenario.
• (D) Founder effect: a type of genetic drift occurring when a small founding population colonises a new area — irrelevant here as both species are already established residents of the same reef.

Question 12

The table below shows the population data of a species of deer in a forest reserve over two consecutive years:

| Parameter | Year 1 | Year 2 |
|-----------|--------|--------|
| Population size (N) | 500 | 550 |
| Births (B) | 100 | — |
| Deaths (D) | 50 | — |
| Immigration (I) | 20 | — |
| Emigration (E) | 20 | — |

Using the data from Year 1, write the equation used to calculate population size at the end of Year 1 and verify the value given in Year 2.

Accepted Answer

Equation for population size (N t+1):

N(t+1) = N(t) + (B − D) + (I − E)          … (1 mark for correct equation with all four parameters)

Substituting Year 1 values:

N(t+1) = 500 + (100 − 50) + (20 − 20)
        = 500 + 50 + 0
        = 550                              … (1 mark for correct substitution and verified answer)

The calculated value (550) matches the given population size in Year 2, confirming the equation.

[Note: All four components — Natality (B), Mortality (D), Immigration (I), and Emigration (E) — together determine the population size at any given time. Here, immigration and emigration cancel out, so only net births (50) contribute to the increase.]

Question 13

A marine biologist observes that a small, isolated coral reef has a fish population of 500 individuals. Over the next year, 80 fish are born, 50 fish die, 30 fish immigrate into the reef, and 20 fish emigrate out. Calculate the population size at the end of the year. Name the attribute of population being calculated.

Accepted Answer

**Population size at end of year:**

Formula: N(t+1) = N(t) + (B − D) + (I − E)

Where:
- N(t) = initial population size = 500
- B (Births/Natality) = 80
- D (Deaths/Mortality) = 50
- I (Immigration) = 30
- E (Emigration) = 20

N(t+1) = 500 + (80 − 50) + (30 − 20)
N(t+1) = 500 + 30 + 10
**N(t+1) = 540**  ✦ (1 mark)

**Name of attribute:** Population density (population size) — it reflects the number of individuals of a species present in a given habitat at a given time.  ✦ (1 mark)

Question 14

A marine biologist observes that barnacles attached to the skin of a humpback whale are thriving and reproducing successfully, while the whale shows no measurable change in its swimming speed, feeding behaviour, or health. However, when the same barnacle species is found attached to a slow-moving sea turtle, the turtle's swimming efficiency decreases significantly due to increased drag. Identify the type of interaction between the barnacle and the whale, and explain why the same organism can show a different type of interaction with the sea turtle.

Accepted Answer

The interaction between the barnacle and the whale is commensalism (+/0): the barnacle benefits (gains a substrate for attachment and access to food-rich open waters as the whale moves), while the whale is neither benefited nor harmed. (1 mark)

The same barnacle shows a parasitism/predation-like harmful interaction (–) with the sea turtle because the turtle, being slow-moving and small, suffers a measurable fitness cost (reduced swimming efficiency due to drag), whereas the large, powerful whale can compensate for the barnacles' negligible mass without any physiological effect. This illustrates that the classification of an ecological interaction depends on the net effect on each partner — the same species pair can shift along the interaction spectrum depending on the host's size, physiology, and ability to tolerate the associated cost. (1 mark)

Question 15

The table below shows the population sizes of two species recorded over time in the same habitat:

| Time (weeks) | Species P | Species Q |
|---|---|---|
| 0 | 500 | 500 |
| 4 | 420 | 580 |
| 8 | 310 | 650 |
| 12 | 180 | 700 |

(i) Identify the type of population interaction shown between Species P and Species Q.
(ii) Give ONE reason to justify your answer.

Accepted Answer

(i) The type of population interaction shown between Species P and Species Q is **Predation** (or Parasitism) — a +/− interaction, where Species Q benefits (+) and Species P is harmed (−).

(ii) Justification: Species P shows a continuous decline in population size (500 → 420 → 310 → 180) while Species Q shows a continuous increase (500 → 580 → 650 → 700) over the same period in the same habitat. This indicates that Species Q benefits at the expense of Species P, which is characteristic of a +/− interaction such as predation or parasitism — NOT competition, since in competition (−/−) both species would be negatively affected, but here Species Q is clearly gaining.

Question 16

A wildlife biologist is studying two adjacent forest patches — Patch A (large, undisturbed) and Patch B (small, isolated due to a highway cutting through the forest). She observes that the tiger population in Patch B shows unusually high frequencies of certain genetic disorders over successive generations, even though food availability is similar in both patches. Meanwhile, a plant ecologist notes that a flowering plant species in Patch B rarely sets fruit despite having abundant pollinators.

(a) Name and explain the phenomenon responsible for the increasing genetic disorders in the tiger population of Patch B. (1 mark)
(b) Suggest one specific population interaction that could explain why the plant in Patch B fails to set fruit despite having pollinators, and justify your answer. (1 mark)
(c) The biologist calculates that Patch B tiger population has r = −0.03 per year. What does this negative value of 'r' indicate, and predict the fate of this population if conditions remain unchanged? (1 mark)

Accepted Answer

CBSE MODEL ANSWER (3 Marks)

(a) [1 mark]
Phenomenon: Genetic Drift (Bottleneck Effect)

Explanation: Patch B has a small, isolated tiger population due to the highway acting as a barrier, preventing gene flow from Patch A. In such small populations, allele frequencies change randomly by chance rather than by natural selection — this is called genetic drift. Over successive generations, rare deleterious (disease-causing) recessive alleles can increase in frequency purely by chance, and inbreeding within the small gene pool leads to their expression as homozygous genotypes, resulting in increasing genetic disorders.

[Award 1 mark for: naming Genetic Drift AND explaining it in the context of small/isolated population size leading to random changes in allele frequency. Accept 'inbreeding depression' as an additional supporting point if genetic drift is correctly named.]

(b) [1 mark]
Population Interaction: Competition (Interspecific Competition, −/− interaction)

Justification: An invasive or weedy plant species (favoured by the disturbed habitat created by the highway) may have colonised Patch B and competes with the native flowering plant for the attention of shared pollinators (exploitation competition for pollinator services). Although pollinators are abundant in number, they preferentially visit the more attractive or rewarding competitor species, resulting in insufficient or ineffective visits to the native plant. Consequently, the native plant fails to receive adequate compatible pollen and does not set fruit, despite pollinator presence.

[Award 1 mark for: correctly naming ONE specific population interaction (e.g., interspecific competition) AND providing a clear, consistent biological justification linking that interaction to pollination failure. Do NOT accept answers that mix two different interactions or contradict the premise that pollinators are present.]

(c) [1 mark]
Indication: A negative value of 'r' (intrinsic rate of natural increase = −0.03 per year) indicates that the death rate exceeds the birth rate in the tiger population of Patch B, i.e., the population is declining.

Prediction: If conditions remain unchanged, the population will continue to decrease in size each year. Being already small and isolated, it will progressively lose individuals, genetic diversity will further erode through genetic drift and inbreeding, and the population will eventually face local extinction (extirpation) from Patch B.

[Award 1 mark for: stating that negative 'r' means deaths exceed births / population is declining AND predicting eventual local extinction if conditions remain unchanged.]

Question 17

A wildlife biologist is monitoring a population of spotted deer in a forest reserve. She records the following data over one year:

• Number of births = 120
• Number of deaths = 75
• Number of deer that moved INTO the reserve from adjacent forests = 30
• Number of deer that moved OUT of the reserve = 45
• Population size at the start of the year (N₀) = 600

(i) Calculate the net change in population size over the year. Show your working.
(ii) Calculate the population growth rate (r) for this population. (Use: r = (Births − Deaths) / N₀)
(iii) Based on the net change calculated, state whether this population is growing, declining, or stable, and identify ONE likely ecological reason for the observed emigration from the reserve.

Accepted Answer

MARKING SCHEME (3 marks)

─────────────────────────────────────
(i) Net change in population size [1 mark]
─────────────────────────────────────
Formula:
Net change (ΔN) = (Births + Immigration) − (Deaths + Emigration)

Substituting values:
ΔN = (120 + 30) − (75 + 45)
ΔN = 150 − 120
ΔN = +30

∴ The population INCREASED by 30 individuals over the year.
New population size = 600 + 30 = 630

[Award 1 mark for correct substitution AND correct answer of +30]

─────────────────────────────────────
(ii) Population growth rate (r) [1 mark]
─────────────────────────────────────
Formula given:
r = (Births − Deaths) / N₀

Substituting values:
r = (120 − 75) / 600
r = 45 / 600
r = 0.075 (per individual per year)

[Award 1 mark for correct calculation = 0.075; accept 7.5 × 10⁻² ]

Note: This is the intrinsic rate of natural increase considering only birth and death rates; it does not include migration.

─────────────────────────────────────
(iii) Population status + reason for emigration [1 mark]
─────────────────────────────────────
Status: The population is GROWING (net change = +30, which is positive). [½ mark]

Ecological reason for emigration (any ONE acceptable answer): [½ mark]
• Intraspecific competition — as population density increases, individuals compete for limited food, water, or shelter, causing some to emigrate to less crowded areas.
  OR
• Carrying capacity (K) being approached — when resources become limiting, emigration acts as a density-dependent regulatory mechanism.
  OR
• Territorial behaviour — dominant individuals drive sub-dominant deer out of the reserve.

[Award ½ mark for any ONE ecologically valid and clearly stated reason]

─────────────────────────────────────
SUMMARY OF VALUE POINTS
─────────────────────────────────────
• Correct formula and net change = +30 → 1 mark
• r = 0.075 (correct working shown) → 1 mark
• Population identified as growing + one valid ecological reason for emigration → 1 mark
Total = 3 marks

Question 18

A marine biologist studying a tropical coral reef ecosystem recorded the following observations over two years:

• The reef supports a population of Crown-of-Thorns starfish (Acanthaster planci), which feeds exclusively on coral polyps.
• In Year 1, the starfish population was small and coral cover was 65%.
• In Year 2, after a warm-water event reduced the population of the starfish's natural predator (the Triton's trumpet snail, Charonia tritonis), the starfish population exploded — coral cover dropped to 18%.
• Interestingly, small populations of cleaner shrimp continued to thrive on the reef, grooming fish clients and removing parasites from them.

(i) Identify the type of population interaction between the Crown-of-Thorns starfish and the coral polyps. Justify your answer using the +/− notation. (1 mark)

(ii) The Triton's trumpet snail is described as a 'keystone species' on this reef. Based on the observations, explain what is meant by this term and what ecological consequence followed the snail's decline. (1 mark)

(iii) Identify the type of population interaction between the cleaner shrimp and the fish. Using ONE feature from the passage, distinguish this interaction from the starfish–coral interaction in terms of effect on both partners involved. (1 mark)

Accepted Answer

CBSE MARKING SCHEME — SA (3 marks)

(i) Predation (+/−)
• The starfish (+) benefits by obtaining nutrition from the coral polyps.
• The coral polyps (−) are harmed / killed as they are consumed.
• Notation: Starfish = +, Coral polyps = − → Predation (one organism feeds on another living organism).
[Award 1 mark for correct identification 'Predation' WITH correct +/− assignment and brief justification; ½ mark if interaction named correctly but no justification given.]

(ii) A keystone species is one whose presence regulates community structure disproportionately relative to its own abundance — its removal causes dramatic ecological change.
• When the Triton's trumpet snail (keystone predator of the starfish) declined, the starfish population exploded unchecked → coral cover fell drastically from 65% to 18%, indicating collapse of the reef community structure.
[Award 1 mark for: correct definition/explanation of keystone species (½) + the ecological consequence stated from passage — reef/coral decline (½). Accept equivalent wording.]

(iii) The cleaner shrimp–fish interaction is Mutualism (+/+).
• Both partners benefit: the shrimp (+) obtains food (parasites/dead tissue); the fish (+) gets rid of ectoparasites / is groomed.

Distinction from starfish–coral interaction:

| Feature | Starfish–Coral (Predation +/−) | Cleaner Shrimp–Fish (Mutualism +/+) |
|---|---|---|
| Effect on Partner 1 | Starfish benefits (+) | Shrimp benefits (+) |
| Effect on Partner 2 | Coral polyps harmed/killed (−) | Fish also benefits (+) |

• In predation one partner is harmed/killed; in mutualism BOTH partners benefit — no harm to either.
[Award 1 mark for: correct identification 'Mutualism' (½) + valid distinction showing both-benefit vs one-harmed using passage context (½). The table is not compulsory; any clear comparative statement is acceptable.]

Question 19

In the context of interspecific interactions, differentiate between commensalism and mutualism. Give one example of each from animals, and state the effect of the interaction on each species involved using the +/0/− notation.

Accepted Answer

**Commensalism vs Mutualism (3 marks)**

**Commensalism (+/0):**
- An interaction in which **one species benefits** and the **other species is neither benefited nor harmed**.
- Notation: Species A = **+**, Species B = **0**
- **Example:** Barnacles (*Balanus*) attached to the back of a whale.
  - Barnacles **benefit** (+): gain a surface for attachment and are transported to nutrient-rich waters for filter-feeding.
  - Whale is **unaffected** (0): gains nothing and suffers no harm from the presence of barnacles.

**Mutualism (+/+):**
- An interaction in which **both species benefit** from the association.
- Notation: Species A = **+**, Species B = **+**
- **Example:** Fig tree (*Ficus*) and fig wasp.
  - Fig wasp **benefits** (+): fig tree provides shelter inside the fig and food (some seeds/ovules) for the wasp's larvae.
  - Fig tree **benefits** (+): the wasp is the **sole pollinator** of the fig — neither species can survive without the other (obligate mutualism).

**Summary Table:**

| Feature | Commensalism | Mutualism |
|---|---|---|
| Effect on Species 1 | + (benefits) | + (benefits) |
| Effect on Species 2 | 0 (unaffected) | + (benefits) |
| Dependency | Non-obligate (usually) | May be obligate |
| Animal example | Barnacle on whale | Fig and fig wasp |

*(Award 1 mark for correct definition + notation of commensalism with example; 1 mark for correct definition + notation of mutualism with example; 1 mark for correctly stating the effect on EACH species in both interactions.)*

Question 20

Describe any three types of population interactions observed in nature. Give one example for each type, clearly mentioning the effect (beneficial, harmful, or neutral) on each species involved.

Accepted Answer

Any THREE of the following population interactions are acceptable (1 mark each = 3 marks):

**1. Mutualism (+/+)**
- Both species benefit from the interaction.
- Example: Lichens — algal partner (phycobiont) provides food by photosynthesis; fungal partner (mycobiont) provides shelter, moisture, and minerals. Both organisms benefit.

**2. Commensalism (+/0)**
- One species benefits; the other is neither harmed nor benefited.
- Example: Barnacles growing on the back of a whale — barnacles gain transport and access to food-rich waters (benefit); whale is unaffected (neutral).

**3. Predation (+/−)**
- One species (predator) benefits by feeding on another species (prey), which is harmed.
- Example: Tiger (predator) feeding on deer (prey) — tiger benefits (+), deer is harmed (−).

**4. Parasitism (+/−)**
- Parasite lives on/in the host, benefiting at the host's expense.
- Example: Cuscuta (dodder plant) growing on hedge plants — Cuscuta (parasite) derives nutrition (+); host plant is harmed (−).

**5. Competition (−/−)**
- Both species are harmed because they compete for the same limiting resource.
- Example: Flamingos and fishes in a lake competing for the same zooplankton food resource — both populations are negatively affected.

**6. Amensalism (−/0)**
- One species is harmed; the other is unaffected.
- Example: Penicillium mould secretes penicillin that inhibits/kills bacteria growing nearby — bacteria are harmed (−); Penicillium is unaffected (0).

**Marking scheme:**
- Correct name of interaction with correct +/− notation: ½ mark
- Correct example with effect on both species stated: ½ mark
- × 3 interactions = 3 marks total

Organisms and Populations: Class 12 Biology Practice Questions

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