Class 12 Biology Chapter 13 Organisms and Populations

Key Concept: Regulation of Population Size, Survivorship

c) Increase of 25
[Solution Description]
To find the net change, consider all additions and losses:
- Additions: Births + Immigration = 40 + 15 = 55
- Losses: Deaths + Emigration = 20 + 10 = 30
Net change = Additions - Losses = 55 - 30 = 25
Thus, the population increases by 25 individuals.

Key Concept: Population Growth, Natality

c) 0.693 per hour
[Solution Description]
The problem describes exponential growth with no mortality (death rate = 0). Given:
Initial population ($N_0$) = 500 cells, Final population ($N_t$) = 4,000 cells, Time ($t$) = 3 hours.
Using the exponential growth formula:
$N_t = N_0 e^{rt}$
Taking natural logarithm on both sides:
$\ln(N_t) = \ln(N_0) + rt$
$r = \frac{\ln(N_t / N_0)}{t}$
Substituting values:
$r = \frac{\ln(4000 / 500)}{3} = \frac{\ln(8)}{3} \approx \frac{2.079}{3} \approx 0.693 \, \text{per hour}$
So, the birth rate per hour is approximately 0.693.

Key Concept: Energy allocation, Population dynamics

c) 50 kJ/day
[Solution Description]
The energy budget equation is:
$\text{Energy Intake} = \text{Maintenance} + \text{Growth} + \text{Reproduction} + \text{Heat Loss}$
Plugging in the given values:
$1000 \, \text{kJ/day} = 600 \, \text{kJ/day} + 200 \, \text{kJ/day} + \text{Reproduction} + 150 \, \text{kJ/day}$
Combine the known expenditures:
$600 + 200 + 150 = 950 \, \text{kJ/day}$
Subtract this from total energy intake to find reproduction energy:
$\text{Reproduction} = 1000 - 950 = 50 \, \text{kJ/day}$
Thus, the correct answer is 50 kJ/day.

Key Concept: Energy transfer efficiency, Population energy budget

b) 5 kJ/m$^2$/year
[Solution Description]
First, calculate the energy available at the second trophic level (herbivores). Since only about 10% of energy is transferred between trophic levels:
$\text{Energy at second trophic level} = 5000 \, \text{kJ/m}^2/\text{year} \times 0.10 = 500 \, \text{kJ/m}^2/\text{year}$
Next, calculate the energy available at the third trophic level (primary carnivores):
$\text{Energy at third trophic level} = 500 \, \text{kJ/m}^2/\text{year} \times 0.10 = 50 \, \text{kJ/m}^2/\text{year}$
Finally, calculate the energy available for the tertiary consumer (fourth trophic level):
$\text{Energy at fourth trophic level} = 50 \, \text{kJ/m}^2/\text{year} \times 0.10 = 5 \, \text{kJ/m}^2/\text{year}$
Thus, the correct answer is 5 kJ/m$^2$/year.

Key Concept: Migration, Emigration, Fecundity

b) 715
[Solution Description]
Step-by-step solution:
1. Initial population = 1200
2. Temporary migrants = 300 (of which 80% will return, so 20% won't)
Permanent loss from migration = 20% of 300 = 60
3. Permanent emigrants = 200
4. Total permanent reduction = 60 + 200 = 260
5. Remaining population after migrations = 1200 - 300 - 200 = 700
6. Births = 5% of remaining population = 0.05 × 700 = 35
7. Spring population before returns = 700 - 260 + 35 = 475
8. Returning migrants = 80% of 300 = 240
9. Final spring population = 475 + 240 = 715
Note: The question asks about "when spring begins" which implies before migrant return, but options suggest after return. Following full cycle, answer is 715.

Key Concept: Population Dispersal, Migration, Population Density

b) Increases by 100 individuals
[Solution Description]
To calculate the net change in population density, we use the formula:
$\text{Population change} = \text{Natality} + \text{Immigration} - \text{Mortality} - \text{Emigration}$
Given:
Initial density = 500 (not needed for change calculation)
Natality = 200
Immigration = 100
Mortality = 150
Emigration = 50
Calculation:
Net change = 200 (natality) + 100 (immigration) - 150 (mortality) - 50 (emigration)
Net change = 300 - 200 = +100
Therefore, the population increases by 100 individuals.

Key Concept: Sex Ratio, Population Density

b) Males have higher survival rates under resource scarcity
[Solution Description]
High population density often leads to increased competition for resources. Males may have higher survival rates or reproductive advantages under such conditions, leading to a skewed sex ratio. Another possibility is that females may produce more male offspring under stress conditions caused by high density to balance future reproduction dynamics.
The most plausible explanation is that high population density induces physiological changes favoring male offspring production or survival.

Key Concept: Sex Ratio, Factors Influencing Sex Ratio

c) 50%
[Solution Description]
At 28°C, 70% of offspring are female, so 30% are male. At 32°C, 30% are female, so 70% are male. Since incubation occurs equally at both temperatures, we can average the proportions.
Total proportion of females = $\frac{70% + 30%}{2} = \frac{100%}{2} = 50%$.
Therefore, the expected proportion of females is 50%.

Key Concept: Survivorship Curve, Age Distribution

c) Between 50% and 60%
[Solution Description]
For a stable population, the age pyramid is bell-shaped, meaning the pre-reproductive and reproductive groups are nearly equal. The survivorship data shows:
1. 60% survive to reproductive age (so pre-reproducitve must be higher than that).
2. 20% reach post-reproductive age (implying 80% mortality till then).
Let’s assume a total population of 100 for simplicity.
Since 60% reach reproductive stage, the pre-reproductive group must be larger than or equal to 60.
For a stable population, the pre-reproductive and reproductive groups are close in size. Thus, the pre-reproductive group is likely slightly more than 50% (as some die before reproduction). But exact numbers depend on mortality rates at each stage.
Given the context, the most logical inference is that the pre-reproductive group accounts for the majority (>50%) of the population but not excessively so (like in expanding populations).

Key Concept: Age Pyramids, Population Growth

b) 3,680
[Solution Description]
To solve this problem, we use the exponential growth formula: $N_t = N_0 e^{rt}$. Here, $N_0 = 10,000$, $r = -0.02$, and $t = 50$ years. Plugging in these values:
$N_{50} = 10,000 \times e^{-0.02 \times 50}$
Calculate the exponent first: $-0.02 \times 50 = -1$
Now compute $e^{-1}$ which is approximately 0.3679
Multiply to get the final population size: $10,000 \times 0.3679 = 3,679$
Therefore, the population size after 50 years is approximately 3,679.

Key Concept: Carrying Capacity, Environmental Resistance

b) 1170
[Solution Description]
The logistic growth equation is $\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$. However, to find the population size after time $t$, we use the integrated form:
$N(t) = \frac{K}{1 + \left(\frac{K - N_0}{N_0}\right)e^{-rt}}.$
Here, $K = 5000$, $N_0 = 1000$, $r = 0.2$, and $t = 1$ year. Plugging in the values:
$N(1) = \frac{5000}{1 + \left(\frac{5000 - 1000}{1000}\right)e^{-0.2 \times 1}} = \frac{5000}{1 + 4e^{-0.2}}.$
Calculating $e^{-0.2}$ (approximately 0.8187):
$N(1) = \frac{5000}{1 + 4 \times 0.8187} = \frac{5000}{1 + 3.2748} = \frac{5000}{4.2748} \approx 1169.88.$
Approximating to the nearest whole number, the population size is 1170.

Key Concept: Exponential Growth, Logistic Growth

c) 2460
[Solution Description]
The problem involves exponential growth, where the population grows without any limiting factors. The formula for exponential growth is $N_t = N_0 e^{rt}$. Here, $N_0 = 1000$, $r = 0.3$ per hour, and $t = 3$ hours. Plugging in the values:
$N_t = 1000 \times e^{0.3 \times 3} = 1000 \times e^{0.9}.$
Calculating $e^{0.9}$ (approximately 2.4596):
$N_t = 1000 \times 2.4596 = 2459.6.$
Since population size must be an integer, we approximate to 2460.

Key Concept: Environmental stochasticity, Organism size/lifespan

c) Grasshoppers are more affected by environmental stochasticity due to their small size and short lifespan.
[Solution Description]
Small, short-lived organisms (grasshoppers) are more susceptible to environmental stochasticity because they reproduce quickly and have limited tolerance to fluctuations. Large, long-lived organisms (tortoises) buffer environmental changes due to slower life histories and stored resources. The correct answer must link body size/lifespan to stochasticity tolerance.

Key Concept: Environmental stochasticity, Density-dependent regulation

c) Rabbit population is influenced by environmental stochasticity; deer population is density-dependent.
[Solution Description]
The rabbit population experiences environmental stochasticity because its fluctuations are driven by unpredictable rainfall (external factor), independent of population density. The deer population is regulated by density-dependent factors like food availability (internal factor). Small, short-lived organisms (like some insects) are more affected by environmental stochasticity, but this question contrasts regulation mechanisms. The correct answer must highlight the external vs internal control distinction.

Key Concept: Supportive Capacity, Assimilative Capacity

b) High levels of pollution
[Solution Description]
Supportive capacity refers to the ability to provide resources without causing lasting damage, while assimilative capacity is the ability to absorb wastes without ill effects. If a region has low assimilative capacity, it means it cannot handle waste efficiently, leading to pollution or ecological harm despite having sufficient resources.

Key Concept: Logistic Growth Equation, Environmental Stochasticity

a) $\frac{rK}{4}$
[Solution Description]
To find the population growth rate when $N = \frac{K}{2}$, substitute $N$ into the logistic growth equation:
$\frac{dN}{dt} = r \left(\frac{K}{2}\right) \left(1 - \frac{\frac{K}{2}}{K}\right) = r \left(\frac{K}{2}\right) \left(1 - \frac{1}{2}\right) = \frac{rK}{4}.$
The growth rate is one-fourth of the product of the intrinsic growth rate and the carrying capacity.

Key Concept: Population Growth Curve, Carrying Capacity

c) S-shaped curve with carrying capacity
[Solution Description]
The described pattern represents an S-shaped (sigmoid) growth curve, where exponential growth slows down due to environmental resistance and stabilizes at the carrying capacity (K), the maximum sustainable population size.

Key Concept: Population Growth, Natality and Mortality

c) 20
[Solution Description]
First, calculate the birth rate:
$B = \frac{N_n}{t} = \frac{50}{500} = 0.1$ per deer per year.
Next, calculate the death rate:
$D = \frac{N_d}{t} = \frac{30}{500} = 0.06$ per deer per year.
Net increase = Birth rate - Death rate = 0.1 - 0.06 = 0.04.
Total increase = Net increase $\times$ Initial population = 0.04 $\times$ 500 = 20.

Key Concept: Logistic growth phases, density-dependent factors

c) Deceleration Phase
[Solution Description]
Logistic growth has four phases:
1. Lag Phase: Slow initial growth, minimal competition.
2. Exponential Phase: Rapid growth, but resources are still abundant.
3. Deceleration Phase: Growth slows as resources become limited, leading to increased competition.
4. Stable Phase: Population stabilizes at carrying capacity.
Density-dependent factors like competition for nutrients and space have the highest impact during the Deceleration Phase because resource scarcity becomes significant, slowing down population growth.

Key Concept: Logistic growth equation, carrying capacity

c) 187.5 individuals per year
[Solution Description]
The logistic growth equation is given by:
$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$.
Step 1: Substitute the given values into the equation.
$\frac{dN}{dt} = 0.5 \times 500 \left(1 - \frac{500}{2000}\right)$.
Step 2: Calculate $\frac{N}{K}$.
$\frac{500}{2000} = 0.25$.
Step 3: Compute $1 - \frac{N}{K}$.
$1 - 0.25 = 0.75$.
Step 4: Multiply to find $\frac{dN}{dt}$.
$\frac{dN}{dt} = 0.5 \times 500 \times 0.75 = 187.5$.
The instantaneous rate of population growth is 187.5 individuals per year.

Key Concept: Environmental resistance, Exponential growth limitations

c) Due to limitations caused by environmental resistance
[Solution Description]
Exponential growth occurs only under ideal conditions where resources are unlimited and environmental resistance factors (like predation, competition, disease) are absent. However, in nature:
1) Resources like food and space are finite.
2) Environmental resistance factors (predators, diseases, competition) limit population growth.
3) As population increases, these factors reduce the growth rate, preventing indefinite expansion.
4) Eventually, the population stabilizes when birth rates equal death rates, leading to logistic growth.

Key Concept: Exponential growth, Population growth rate

c) 7389 cells
[Solution Description]
The exponential growth formula is given by:
$N(t) = N_0 e^{rt}$
Here, $N_0 = 1000$, $r = 0.5 \, \text{hour}^{-1}$, and $t = 4$ hours.
Plugging in the values:
$N(4) = 1000 \times e^{0.5 \times 4}$
$N(4) = 1000 \times e^2$
Calculating $e^2$ (approximately 7.389):
$N(4) = 1000 \times 7.389 \approx 7389$
So, the population after 4 hours is approximately 7389 cells.

Key Concept: Carrying Capacity, Environmental Resistance

c) The environment cannot support more individuals
[Solution Description]
When a population reaches its carrying capacity ($K$), the resources such as food, space, and other necessities become limited. The environmental resistance factors like competition for resources, predation, and disease increase, causing the population growth rate to decline until it stabilizes. At this point, the birth rate equals the death rate, and $\frac{dN}{dt} = 0$. Thus, the population stops growing because the environment cannot support more individuals.

Key Concept: Exponential Growth, Logistic Growth

b) 490
[Solution Description]
To solve this, we use the logistic growth equation:
$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$
Given: $r = 0.5$, $N_0 = 100$, $K = 10,000$. For small time intervals, we can approximate the population growth using discrete steps. After 1 hour:
$N_1 = N_0 + rN_0 \left(1 - \frac{N_0}{K}\right) = 100 + 0.5 \times 100 \times \left(1 - \frac{100}{10,000}\right) = 100 + 50 \times 0.99 = 149.5$
After 2 hours:
$N_2 = 149.5 + 0.5 \times 149.5 \times \left(1 - \frac{149.5}{10,000}\right) = 149.5 + 74.75 \times 0.98505 = 222.12$
After 3 hours:
$N_3 = 222.12 + 0.5 \times 222.12 \times \left(1 - \frac{222.12}{10,000}\right) = 222.12 + 111.06 \times 0.9778 = 330.60$
After 4 hours:
$N_4 = 330.60 + 0.5 \times 330.60 \times \left(1 - \frac{330.60}{10,000}\right) = 330.60 + 165.30 \times 0.96694 = 490.46$
The population after 4 hours is approximately 490.

Key Concept: Density-independent growth, Environmental stochasticity

b) 8 weeks
[Solution Description]
In density-independent growth, the population follows the exponential growth model:
$N(t) = N_0 e^{rt}$
We need to find the time $t$ when $N(t) = N_{\text{pre-flood}}$. Let’s assume the pre-flood population was $N_{\text{pre-flood}}$. Since 70% were wiped out, the remaining population is 30% of $N_{\text{pre-flood}}$, so $N_0 = 0.3 N_{\text{pre-flood}} = 300$.
Therefore, $N_{\text{pre-flood}} = \frac{300}{0.3} = 1000$.
Plugging these into the equation:
$1000 = 300 e^{0.15t}$
Divide both sides by 300:
$\frac{1000}{300} = e^{0.15t}$
Simplify the fraction:
$3.\overline{3} = e^{0.15t}$
Take the natural logarithm of both sides:
$\ln(3.\overline{3}) = 0.15t$
Calculate $\ln(3.\overline{3})$:
$\ln(3.\overline{3}) \approx 1.204$
Solve for $t$:
$t = \frac{1.204}{0.15} \approx 8.03 \text{ weeks}$
It will take approximately 8 weeks for the population to recover.

Key Concept: Density-dependent growth, Logistic equation

c) 16 rabbits/month
[Solution Description]
To find the population growth rate $\frac{dN}{dt}$, we use the logistic equation:
$\frac{dN}{dt} = rN\left(1 - \frac{N}{K}\right)$
Substitute the given values: $r = 0.1$, $N = 200$, $K = 1000$.
$\frac{dN}{dt} = 0.1 \times 200 \left(1 - \frac{200}{1000}\right)$
Simplify the fraction inside the parentheses:
$1 - \frac{200}{1000} = 1 - 0.2 = 0.8$
Now multiply the terms:
$\frac{dN}{dt} = 0.1 \times 200 \times 0.8 = 16$
The population growth rate at this moment is approximately 16 rabbits per month.

Key Concept: Potential Natality, Realized Natality

b) 100 offspring/month
[Solution Description]
Potential natality is the maximum possible reproduction rate under optimal conditions, while realized natality is the actual birth rate under existing conditions.

Key Concept: Biotic Potential, Population Growth Rate

b) 0.21 per year
[Solution Description]
To find the population growth rate, use the formula:
$r = \frac{(No.\ of\ births + No.\ of\ immigrants) - (No.\ of\ deaths + No.\ of\ emigrants)}{No.\ of\ individuals\ in\ population/time}$

Key Concept: Population Density, Per Cent Cover

d) Percentage ground area covered
[Solution Description]
The banyan tree's large canopy makes per cent cover or biomass a more meaningful measure than counting individual plants. The herbs cover 10 m², while the banyan tree alone covers 30 m². Thus, the banyan tree dominates in terms of spatial coverage despite having fewer individuals.

Key Concept: Population Density, Indirect Methods

c) 1.5 tigers/km²
[Solution Description]
First, find the total number of unique tigers using pug mark data. If there are 120 pug marks and each tiger leaves 4 pug marks, then:
$\text{Number of Tigers} = \frac{120}{4} = 30$
Next, calculate population density by dividing the number of tigers by the area:
$D = \frac{30}{20} = 1.5 \, \text{tigers/km}^2$

Key Concept: Population Dispersal, Migration

b) A herd of wildebeests moves seasonally between the Serengeti and Masai Mara.
[Solution Description]
Migration involves the mass two-way movement of an entire population, where organisms return to their original area after moving away, often due to seasonal changes or unfavorable conditions. This is distinct from emigration (one-way outward movement) and immigration (one-way inward movement).

Key Concept: Population Density, Age Distribution

c) The population is rapidly growing with a total of 10,000 individuals.
[Solution Description]
The age pyramid with a broad base indicates a high percentage of young individuals, suggesting a rapidly growing population. Population density ($D$) is given by $D = N/a$, where $N$ is the total number of individuals and $a$ is the area. Here, $D = 50$ individuals/km$^2$ and $a = 200$ km$^2$. Thus, the total population size $N = D \times a = 50 \times 200 = 10,000$ individuals.

Key Concept: Population Dispersal, Carrying Capacity

c) Population will increase but stabilize near 2000.
[Solution Description]
Current population after migration:
$1500 + 300 - 100 = 1700$.
Since 1700 < 2000 (carrying capacity), the population will initially grow rapidly due to available resources and then slow down as it approaches 2000 (S-shaped curve).
The correct option must state that the population will increase but approach stability near 2000.

Key Concept: Population Growth, Natality and Mortality

a) 1.2%
[Solution Description]
First, calculate the birth rate (B) and death rate (D).
Birth rate formula: $B = \frac{N_n}{t} = \frac{50}{5} = 10$ per year.
Death rate formula: $D = \frac{N_d}{t} = \frac{20}{5} = 4$ per year.
Net growth = Birth rate - Death rate = $10 - 4 = 6$ per year.
Per capita growth rate = $\frac{6}{500} = 0.012$ or 1.2%.

Key Concept: Age Structure

a) Expanding (Broad Base)
[Solution Description]
An expanding population has a broad base, indicating a high proportion of young individuals. This suggests rapid growth due to high birth rates and low mortality in younger age groups. Such pyramids are typical in developing countries with increasing populations.

Key Concept: Population Growth

b) 165
[Solution Description]
Given: $N_0 = 100$, $r = 0.05$, $t = 10$ years.
Using the exponential growth formula:
$N_t = N_0 e^{rt}$
Substituting the values:
$N_{10} = 100 \times e^{0.05 \times 10} = 100 \times e^{0.5}$
Calculating $e^{0.5} \approx 1.6487$:
$N_{10} = 100 \times 1.6487 = 164.87$
The population size after 10 years is approximately 165.

Key Concept: First Law of Thermodynamics in Bioenergetics

c) 1200 J
[Solution Description]
According to the First Law of Thermodynamics:
$\Delta U = Q - W$
where $\Delta U$ is the change in internal energy, $Q$ is the heat added to the system (energy gained), and $W$ is the work done by the system (energy lost).
Given $Q = 2000\,J$ and $W = 800\,J$, substituting:
$\Delta U = 2000 - 800 = 1200\,J$
Thus, the change in internal energy is 1200 J.

Key Concept: Population Dispersal - Factors Influencing Dispersal

c) Permanent climate change
[Solution Description]
The factors influencing population dispersal include search for food, avoidance of predators, prevention of overcrowding, breeding behavior, and physical reasons. "Permanent climate change" is not listed as a direct factor in the syllabus.

Key Concept: Population Dispersal - Migration

c) Mass two-way movement with periodic returns to the original area
[Solution Description]
Migration refers to the two-way movement of an entire population where organisms return to their original area, often during seasonal or unfavorable conditions. This is a key adaptation for survival and affects population density and genetic diversity.

Key Concept: Factors Affecting Sex Ratio

c) Temperature and photoperiod
[Solution Description]
The sex ratio can be affected by environmental factors such as temperature, photoperiod, minerals, growth substances, and population density.

Key Concept: Sex Ratio Definition

c) The distinction between male and female individuals
[Solution Description]
The sex ratio refers to the distinction between male and female individuals in a population. It represents the proportion of males to females within that population.

Key Concept: Age Structure Analysis

c) Low percentage of young individuals and shrinking population
[Solution Description]
An urn-shaped age pyramid shows a low percentage of young individuals compared to reproductive and post-reproductive groups, indicating a declining population due to drastically reduced birth rates.

Key Concept: Age Pyramid Interpretation

a) Pyramid with Broad Base (Expanding)
[Solution Description]
A broad-based age pyramid signifies a high percentage of young individuals, indicating exponential population growth as each successive generation is more numerous than the preceding one. This is typical of expanding populations with high birth rates.

Key Concept: Exponential and Logistic Growth

d) 6400
[Solution Description]
For exponential growth, the population at time t is given by $N(t) = N_0 e^{rt}$, where r is the growth rate. Since the population doubles every 20 minutes, we can find r using the doubling time formula: $2N_0 = N_0 e^{rT_d}$, where $T_d = 20$ min. Simplifying, we get $2 = e^{rT_d}$. Taking natural logarithm on both sides, $\ln 2 = rT_d$, so $r = \frac{\ln 2}{20} \approx 0.03466$ per minute. Now, for t = 2 hours = 120 minutes, $N(t) = 100 \times e^{0.03466 \times 120} \approx 100 \times e^{4.1592} \approx 100 \times 64 = 6400$.

Key Concept: Population Growth Curve

c) The maximum sustainable population size
[Solution Description]
The carrying capacity (K) represents the maximum population size that an environment can sustain indefinitely given the available resources. In the logistic growth model, as the population N approaches K, the term $\left(1 - \frac{N}{K}\right)$ approaches zero, reducing the growth rate $\frac{dN}{dt}$ to zero.

Key Concept: Population Growth Rate

c) 0.1
[Solution Description]
To find the population growth rate ($r$), substitute the given values into the formula:
$r = \frac{(150 + 0) - (100 + 0)}{500} = \frac{50}{500} = 0.1$

Key Concept: Environmental Stochasticity

c) Random fluctuations in population size caused by changing environmental conditions
[Solution Description]
Environmental stochasticity refers to unpredictable fluctuations in population size due to varying environmental conditions like weather. Small, short-lived organisms exhibit more pronounced fluctuations compared to larger, long-lived species. The question tests understanding of this concept.

Key Concept: Environmental Stochasticity

b) A small, short-lived insect species.
[Solution Description]
Environmental stochasticity refers to unpredictable fluctuations in population numbers due to changing environmental conditions such as weather. Small, short-lived organisms are more susceptible to these fluctuations because they have shorter lifespans and lower resistance to environmental variations compared to large, long-lived organisms that can tolerate a wider range of conditions.

Key Concept: Logistic Growth Equation

d) The population growth rate becomes zero.
[Solution Description]
The logistic growth equation is given by:
$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$
When $N = K$, we substitute $N$ with $K$ in the equation:
$\frac{dN}{dt} = rK \left(1 - \frac{K}{K}\right) = rK (1 - 1) = rK \times 0 = 0$
Therefore, the population growth rate becomes zero when $N = K$.

Key Concept: Natality and Mortality

b) 0.20
[Solution Description]
Realized natality is the actual birth rate under existing conditions. Using the formula for birth rate:
$B = \frac{N_n}{t}$
Here, $N_n = 100$ and the initial population is 500. Thus:
$B = \frac{100}{500} = 0.2 \text{ bacteria per bacterium per week}$
The realized natality is 0.2 per bacterium per week.

Key Concept: Population Growth

b) 0.25
[Solution Description]
The birth rate $B$ is calculated using the formula:
$B = \frac{N_n}{t}$
where $N_n$ is the number of new individuals added, and $t$ is the time unit (1 year). Here, $N_n = 250 - 200 = 50$ rabbits. Therefore:
$B = \frac{50}{200} = 0.25 \text{ rabbits per rabbit per year}$
The birth rate per rabbit per year is 0.25.

Key Concept: Phases of Logistic Growth

b) Exponential phase
[Solution Description]
The logistic growth curve has four phases: lag phase, exponential phase, deceleration phase, and stable phase. The highest relative growth rate occurs during the **exponential phase**, where resources are abundant, and the population grows rapidly. As the population approaches the carrying capacity ($K$), growth slows down.

Key Concept: Logistic Growth Model

b) $\frac{rK}{4}$
[Solution Description]
The rate of population growth $\frac{dN}{dt}$ is given by the logistic equation: $\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$. Given that $N = \frac{K}{2}$, substitute this value into the equation:
$\frac{dN}{dt} = r \left(\frac{K}{2}\right) \left(1 - \frac{\frac{K}{2}}{K}\right)$
Simplify the term inside the parentheses:
$1 - \frac{\frac{K}{2}}{K} = 1 - \frac{1}{2} = \frac{1}{2}$
So, the equation becomes:
$\frac{dN}{dt} = r \left(\frac{K}{2}\right) \left(\frac{1}{2}\right) = \frac{rK}{4}$

Key Concept: Limitations of Exponential Growth

d) Unlimited resources
[Solution Description]
For a population to exhibit exponential growth, the following conditions must be met:
- Unlimited resources (food and space).
- Absence of environmental resistance (no predators, diseases, or adverse climatic conditions).
- The population growth rate is density-independent.
Therefore, the correct condition among the options is "Unlimited resources."

Key Concept: Exponential Growth Equation

d) 272
[Solution Description]
The exponential growth equation is given by:
$N = N_0 e^{rt}$
where:
- $N$ = final population size
- $N_0$ = initial population size (100)
- $r$ = intrinsic rate of increase (0.5/year)
- $t$ = time (2 years)
Substituting the values:
$N = 100 \times e^{0.5 \times 2} = 100 \times e^1 \approx 100 \times 2.7183 \approx 271.83$
Rounding to the nearest whole number, the population size after 2 years is approximately 272.

Key Concept: Exponential Growth Model

d) 165
[Solution Description]
The exponential growth equation is given by:
$N(t) = N_0 e^{rt}$
Where:
$N_0 = 100, \quad r = 0.05, \quad t = 10$
Substituting the values:
$N(10) = 100 \times e^{0.05 \times 10} = 100 \times e^{0.5}$
Calculating $e^{0.5} \approx 1.6487$:
$N(10) = 100 \times 1.6487 \approx 164.87$
Thus, the population size after 10 time units is approximately 165 individuals.

Key Concept: Logistic Growth Model

c) 360
[Solution Description]
The logistic growth equation is given by:
$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$
Substituting the given values:
$\frac{dN}{dt} = 0.3 \times 2000 \left(1 - \frac{2000}{5000}\right)$
Simplifying inside the parentheses:
$1 - \frac{2000}{5000} = 1 - 0.4 = 0.6$
Multiplying further:
$\frac{dN}{dt} = 600 \times 0.6 = 360$
Therefore, the population growth rate is approximately 360 individuals per unit time.

Key Concept: Environmental Resistance

d) Flood
[Solution Description]
Density-independent factors affect populations regardless of their size, such as natural disasters or extreme weather conditions. These do not depend on the population density.

Key Concept: Density-dependent growth

b) $\frac{rK}{4}$
[Solution Description]
The logistic growth equation is $\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$. Given $N = \frac{K}{2}$, substitute into the equation:
$\frac{dN}{dt} = r \left(\frac{K}{2}\right) \left(1 - \frac{\frac{K}{2}}{K}\right) = r \left(\frac{K}{2}\right) \left(1 - \frac{1}{2}\right) = r \left(\frac{K}{2}\right) \left(\frac{1}{2}\right) = \frac{rK}{4}.$

Key Concept: Intrinsic Rate of Growth

b) 230
[Solution Description]
Given the intrinsic rate of growth ($r_m$) is 0.15 per individual per month, and the current population size ($N$) is 200. Under ideal conditions, the population growth can be approximated by the equation:
$\frac{dN}{dt} = r_m N$
For a small time interval (one month), the change in population ($\Delta N$) is approximately:
$\Delta N = r_m \times N \times \Delta t = 0.15 \times 200 \times 1 = 30$
Therefore, the new population size after one month will be $200 + 30 = 230$.

Key Concept: Biotic Potential

b) 0.04 per individual per year
[Solution Description]
To calculate the population growth rate ($r$), we use the formula:
$r = \frac{(No.\ of\ births + No.\ of\ immigrants) - (No.\ of\ deaths + No.\ of\ emigrants)}{No.\ of\ individuals\ in\ population/time}$
Here, number of births = 30, number of deaths = 10, immigrants = 0, emigrants = 0, and initial population = 500. Time period = 1 year. Substituting the values:
$r = \frac{(30 + 0) - (10 + 0)}{500/1} = \frac{20}{500} = 0.04$
The population growth rate is 0.04 per individual per year.

Key Concept: Alternative Measures of Population Density

c) When dealing with numerous small herbs and few large trees
[Solution Description]
Biomass is a more meaningful measure of population density when individual counts are misleading due to variations in size or ecological impact. For example, a single large mango tree may have a greater ecological role than numerous small herb plants despite having a low numerical density.

Key Concept: Population Density Calculation

c) 25 trees per km$^2$
[Solution Description]
The formula for population density is $D = \frac{N}{a}$, where $N$ is the total number of individuals and $a$ is the area. Here, $N = 5000$ trees and $a = 200$ km$^2$. So, $D = \frac{5000}{200} = 25$ trees per km$^2$.

Key Concept: Age Pyramids

c) Urn-shaped figure
[Solution Description]
An urn-shaped age pyramid indicates a declining population because it shows a low percentage of young individuals (pre-reproductive group), moderate reproductive group, and relatively larger post-reproductive group. This suggests that the birth rate has reduced significantly, leading to a shrinking population.

Key Concept: Population Density

c) 50 trees per square kilometer
[Solution Description]
To calculate population density, use the formula: $D = \frac{N}{a}$. Here, $N = 250$ and $a = 5$. So, $D = \frac{250}{5} = 50$ trees per square kilometer.

Key Concept: Population Growth

d) 800 bacteria
[Solution Description]
Exponential growth follows the formula:
$N_t = N_0 \times 2^{(t/T)}$
where $N_t$ is the final population, $N_0$ is the initial population (50), $T$ is the doubling time (1 hour), and $t$ is the total time (4 hours). Substituting the values:
$N_t = 50 \times 2^{4} = 50 \times 16 = 800$

Key Concept: Population Density

d) 20 trees per km$^2$
[Solution Description]
To find the population density, we use the formula:
$D = \frac{N}{a}$
where $D$ is the population density, $N$ is the number of individuals (500 trees), and $a$ is the area (25 km$^2$). Substituting the given values:
$D = \frac{500}{25} = 20 \text{ trees per km}^2$

Key Concept: Age Pyramids

c) Expanding (Broad Base)
[Solution Description]
An expanding age pyramid has a broad base, indicating a high proportion of young individuals and a lower proportion of older individuals. This structure suggests rapid population growth due to high birth rates or a large number of offspring produced per individual.

Key Concept: J-shaped Growth Curve

c) J-shaped
[Solution Description]
Under ideal conditions, a population can grow exponentially, doubling in size during a constant time period. This results in a J-shaped growth curve because there are no limiting factors to slow the growth. This type of growth is rare in nature but can occur in controlled environments like bacterial populations.

Key Concept: Bioenergetics Definition

b) Bioenergetics
[Solution Description]
According to the syllabus, energy transformations that occur within a population are referred to as bioenergetics. This concept is fundamental to understanding how energy influences population dynamics, growth, and ecological balance.

Key Concept: Energy Flow in Population

b) To evaluate observed fluctuations in density and determine the population's role in its community
[Solution Description]
The primary purpose of studying energy flow in a population is to understand how energy is utilized and transferred within the population, which helps in evaluating observed fluctuations in density and determining the role of the population within its community. This provides insights into population stability and interactions within ecosystems.

Key Concept: Population Dispersal - Migration

c) Seasonal two-way movement
[Solution Description]
Migration involves mass two-way movement of an entire population, where organisms return to their original area after moving away. It is often seasonal and periodic, as mentioned in the syllabus.

Key Concept: Population Dispersal - Emigration

d) Outward movement for permanent residence
[Solution Description]
Emigration refers to the outward movement of an organism from one place or country to another for permanent residence. It is a one-way movement, as per the syllabus.

Key Concept: Calculation of Sex Ratio

c) 125
[Solution Description]
To calculate the sex ratio, use the formula:
$\text{Sex Ratio} = \left( \frac{\text{Number of Males}}{\text{Number of Females}} \right) \times 100$
Substituting the given values:
$\text{Sex Ratio} = \left( \frac{150}{120} \right) \times 100 = 125$

Key Concept: Definition of Sex Ratio

c) The proportion of males to females, expressed as males per 100 females
[Solution Description]
The sex ratio is defined as the proportion of males to females in a population and is expressed as the number of males per 100 females.

Key Concept: Population Attributes

b) Equal pre-reproductive and reproductive groups
[Solution Description]
A stable age pyramid has a bell-shaped structure, where the pre-reproductive and reproductive age groups are nearly equal in size, indicating slow and stable population growth. The post-reproductive group remains the smallest.

Key Concept: Age Pyramid Types

a) Expanding pyramid
[Solution Description]
An expanding age pyramid has a broad base, indicating a high percentage of young individuals and exponential growth due to high birth rates. This is commonly seen in rapidly growing populations like yeast or house flies.

Key Concept: Logistic Growth

c) The maximum sustainable population size
[Solution Description]
The carrying capacity (K) in the logistic growth equation signifies the maximum population size that an environment can sustain indefinitely given the available resources. As (N) approaches (K), the growth rate slows down, leading to an S-shaped curve.

Key Concept: Exponential Growth

c) J-shaped (exponential) growth curve
[Solution Description] The J-shaped (exponential) growth curve occurs when a population grows rapidly under ideal conditions with no environmental resistance, doubling in size over constant time intervals. Factors like unlimited resources and absence of predators allow such growth. The equation for exponential growth is $\frac{dN}{dt} = rN$.

Key Concept: Organism Response to Environmental Stochasticity

b) Small, short-lived organisms
[Solution Description]
Small, short-lived organisms exhibit more pronounced fluctuations compared to large, long-lived ones because they are less tolerant to environmental variation.

Key Concept: Environmental Stochasticity

c) Unpredictable fluctuations in population size due to environmental conditions
[Solution Description]
Environmental stochasticity refers to unpredictable fluctuations in population size due to varying environmental conditions like weather. It is not influenced by population size but by external factors.

Key Concept: Logistic Growth Equation

d) $K$
[Solution Description]
The logistic growth equation is $\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$, where $K$ is the carrying capacity representing the maximum sustainable population size.

Key Concept: Carrying Capacity Definition

b) The maximum number of individuals the environment can support indefinitely
[Solution Description]
The carrying capacity is defined as the maximum number of individuals that an environment can support sustainably without causing lasting damage. It is a fixed limit denoted by $K$.

Key Concept: Mortality and Death Rate

b) 0.2 individuals per fruitfly per month
[Solution Description]
The death rate is calculated using the formula:
$D = \frac{N_d}{t}$
Here, $N_d$ (number of deaths) = $8$, and $t = 1$ month.
So, $D = \frac{8}{40} = 0.2$ individuals per fruitfly per month.

Key Concept: Natality and Birth Rate

b) 0.3 offspring per rabbit per year
[Solution Description]
The birth rate is calculated using the formula:
$B = \frac{N_n}{t}$
Here, $N_n$ (number of new individuals added) = $65 - 50 = 15$, and $t = 1$ year.
So, $B = \frac{15}{50} = 0.3$ offspring per rabbit per year.

Key Concept: Carrying Capacity

c) Maximum sustainable population size
[Solution Description]
Carrying capacity ($K$) is the maximum sustainable population size that an environment can support indefinitely, determined by factors such as food availability, space, and predation.

Key Concept: Logistic Growth Phases

b) Exponential (Log) Phase
[Solution Description]
The exponential (log) phase in logistic growth occurs when the population size increases rapidly because resources are plentiful, and competition is low. This is represented by the steepest part of the S-shaped curve.

Key Concept: Intrinsic Rate of Increase

c) The maximum possible growth rate under ideal conditions
[Solution Description]
The intrinsic rate of increase ($r_m$) is the maximum possible growth rate of a population when all individuals are producing the maximum number of offspring and all deaths are due to old age. It represents the theoretical maximum growth rate under ideal conditions with no environmental resistance.

Key Concept: Exponential Growth Formula

c) $\frac{dN}{dt} = rN$
[Solution Description]
The exponential growth of a population under ideal conditions is described by the differential equation where the rate of change in population size with respect to time is proportional to the current population size. This is given by $\frac{dN}{dt} = rN$, where $\frac{dN}{dt}$ is the change in population size, $r$ is the intrinsic rate of increase, and $N$ is the number of individuals.

Key Concept: Logistic Growth

b) Population growth slows down and reaches stability
[Solution Description]
The logistic growth equation accounts for limited resources by including a term $\left(1 - \frac{N}{K}\right)$. Here, $N$ is the current population size, $K$ is the carrying capacity, and $r$ is the intrinsic growth rate. As $N$ approaches $K$, the term $\left(1 - \frac{N}{K}\right)$ becomes very small (close to 0), slowing down the population growth rate. Eventually, when $N = K$, $\frac{dN}{dt} = 0$, meaning the population stops growing.

Key Concept: Exponential Growth

c) Intrinsic rate of growth
[Solution Description]
The given equation $\frac{dN}{dt} = rN$ describes exponential population growth, where $N$ is the number of individuals and $\frac{dN}{dt}$ is the rate of change of the population size with respect to time. The term $r$ represents the intrinsic rate of growth, which is the per capita growth rate when conditions are optimal (unlimited resources, no competition). It determines how fast the population grows.

Key Concept: Density-independent factors

d) Flood
[Solution Description]
Density-independent factors affect populations regardless of their size. Examples include natural disasters like floods, storms, or human-induced changes such as pollution. These factors do not depend on the population density and can impact small or large populations equally.

Key Concept: Density-dependent growth

c) Carrying capacity
[Solution Description]
The term used to describe the maximum number of individuals that an environment can sustainably support is called the carrying capacity (K). It is a key concept in density-dependent growth, where population size stabilizes when it reaches this limit due to factors like resource scarcity and competition.

Key Concept: Intrinsic Rate of Growth

b) $\frac{dN}{dt} = rN$
[Solution Description]
The maximum growth rate of a population, when all individuals are producing the maximum number of offspring and all deaths are from old age, is called the intrinsic rate of growth ($r_m$). This is explicitly mentioned as an important technical term in the syllabus.

Key Concept: Biotic Potential

d) Potential natality
[Solution Description]
The maximum possible rate of reproduction for a given species under optimal conditions is termed as potential natality, which is defined in the syllabus.

Key Concept: Indirect Measures of Population Density

b) When counting all individuals is difficult due to their sizes being vastly different
[Solution Description]
When counting individuals is impractical due to size differences, such as one large tree versus many small plants, percent cover or biomass becomes a better measure. For example, a single large mango tree with a wide canopy may have a low population density in numbers but high ecological impact.

Key Concept: Population Density Formula

d) $D = \frac{N}{a}$
[Solution Description]
The population density $D$ is calculated by dividing the total number of individuals $N$ by the area $a$ they occupy. The formula is:
$D = \frac{N}{a}$
Therefore, the correct answer is the option that represents this formula.

Key Concept: Age Distribution

c) Pyramid with broad base
[Solution Description]
A pyramid with a broad base indicates a high percentage of young individuals, which corresponds to a rapidly growing population with a high birth rate.

Key Concept: Population Density

d) $D = \frac{N}{a}$
[Solution Description]
Population density is calculated by dividing the total number of individuals (N) by the area (a) they occupy. The formula is given as $D = \frac{N}{a}$.

Key Concept: Natality

c) Number of new individuals added to a population per unit time
[Solution Description]
Natality refers to the birth rate or the number of new individuals added to a population per unit time. It is given by the formula:
$B = \frac{N_n}{t}$
where $B$ is the birth rate, $N_n$ is the number of new individuals, and $t$ is the time unit.

Key Concept: Population Density

c) $D = \frac{N}{a}$
[Solution Description]
Population density is calculated by dividing the total number of individuals in a population by the area they occupy. The formula is:
$D = \frac{N}{a}$
where $D$ is the population density, $N$ is the total number of individuals, and $a$ is the area.

Key Concept: Amensalism vs. Competition, Direct Interaction

c) Only one species is harmed in amensalism, unlike competition where both suffer.
[Solution Description]
Amensalism involves one-sided harm (Bacterium Y dies; X is unaffected), whereas competitive exclusion harms both species as they vie for the same resources. Here, X’s toxin directly inhibits Y without resource competition. The correct option must highlight the unilateral harm and absence of mutual negative effects.

Key Concept: Amensalism, Ecological Significance

c) Limits the distribution of certain plant species, altering community composition.
[Solution Description]
In amensalism, the fungal species (unaffected) releases chemicals that harm nearby saplings (affected). This reduces plant diversity in the vicinity by restricting the growth of competing species. It regulates community composition by favoring species resistant to these chemicals. The correct answer describes the impact on spatial distribution and species abundance.

Key Concept: Intraspecific Competition, Population Dynamics, Growth Models

c) Growth rate r decreases; population stabilizes near current level
[Solution Description]
For logistic growth near carrying capacity:
1) At 90% of K, the term $(K-N_0)/N_0$ becomes very small (0.11 when $N_0 = 0.9K$).
2) Intraspecific competition increases environmental resistance, directly affecting:
- Death rate increases (reducing effective r)
- Birth rate may decrease
3) The population growth rate slows as $e^{-rt}$ term dominates near K.
4) Population will stabilize slightly below original K as competition limits resources.
Main affected parameter is the intrinsic growth rate r.

Key Concept: Competition, Niche Overlap, Competitive Exclusion Principle

b) The species with higher feeding efficiency will dominate while the other declines
[Solution Description]
The competitive exclusion principle states that no two species can occupy the same niche indefinitely without competition eliminating one. Since the original scenario shows niche differentiation (different foraging heights), they coexist. If forced into the same niche:
1) Calculate population density impact ($D = N/a$) - both species' densities would decrease in the overlapping zone.
2) The species with lower competitive ability would either:
- Migrate (if possible)
- Face reduced fitness
- Potentially go extinct locally
3) Logistic growth equations show carrying capacity K would decrease for both as $\frac{K}{1 + (\frac{K-N_0}{N_0})e^{-rt}}$ is impacted by resource limitation.
The most likely outcome is dominance of the more efficient competitor and decline of the other.

Key Concept: Ectoparasites, Endoparasites, Adaptation

c) Degeneration of locomotory and sensory organs.
[Solution Description]
Endoparasites live inside the host's body and exhibit specific adaptations such as reduced sensory organs, loss of digestive systems, and anaerobic respiration due to the internal environment. Ectoparasites, living on the host's surface, retain features like aerobic respiration and sensory organs.
Step 1: Identify key differences between ecto- and endoparasites.
Step 2: Focus on structural adaptations unique to endoparasites (e.g., degeneration of organs).
Step 3: Match these traits to the given options.
Step 4: Eliminate options that describe ectoparasite features or general parasitic traits.

Key Concept: Parasitism, Haustoria, Host-Specificity

c) Haustoria are involved in attaching the parasite to the host and absorbing nutrients.
[Solution Description]
Haustoria are specialized structures in parasitic plants that penetrate the host's tissues to establish a connection with the host's vascular system for nutrient absorption. They are crucial for the survival of obligate parasites like $Cuscuta$, which lack chlorophyll and depend entirely on the host for nutrition.
Step 1: Understand what haustoria are—root-like structures that invade the host’s vascular system.
Step 2: Recognize their role in absorbing water, minerals, and organic nutrients directly from the host.
Step 3: Confirm that they are found in both partial (e.g., $Viscum$) and total parasites (e.g., $Cuscuta$).
Step 4: Eliminate incorrect options based on the function and presence of haustoria.

Key Concept: Predator-Prey Dynamics, Ecological Importance

b) Prey diversity will decrease due to competitive exclusion
[Solution Description]
Keystone predators maintain species diversity by controlling dominant prey species, preventing them from outcompeting others. Removal of such a predator leads to competitive exclusion, where dominant prey species proliferate and reduce overall diversity. This aligns with the concept seen in the Pisaster-starfish example, where removal resulted in reduced biodiversity.

Key Concept: Prey Defenses, Cryptic Coloration

b) The frogs will be more susceptible to predation due to lack of camouflage
[Solution Description]
Bright red coloration in a green environment makes the frogs highly visible to predators, increasing their chances of being preyed upon. Cryptic coloration (camouflage) helps prey blend into their surroundings, but bright colors disrupt this defense mechanism. Therefore, these frogs are likely to experience higher predation pressure compared to camouflaged individuals.

Key Concept: Parasitism, Amensalism

b) One involves nutrient derivation while the other involves chemical inhibition
[Solution Description]
Mistletoe growing on oak trees is an example of parasitism, where the parasite (mistletoe) benefits at the expense of the host (oak tree) without immediate lethal effects. On the other hand, Penicillium secreting antibiotics that inhibit bacterial growth represents amensalism, where one organism (bacteria) is harmed while the other (Penicillium) remains unaffected.
Step 1: Identify mistletoe-oak interaction - Parasitism (long-term harm to host).
Step 2: Identify Penicillium-bacteria interaction - Amensalism (one harmed, other unaffected).
Step 3: Differentiate between the two - Parasitism involves nourishment derivation, while amensalism involves inhibition without direct benefit.
Thus, the correct option highlights the contrast between nourishment-derived harm and non-beneficial inhibition.

Key Concept: Predation, Prey Defenses

d) Chemical defense leads to predator avoidance behavior
[Solution Description]
The butterfly feeding on poisonous weeds and becoming toxic to birds is an example of chemical defense in prey species. Monarch butterflies store toxins from milkweed plants, which makes them harmful or distasteful to predators like birds. This is a coevolutionary adaptation where the prey evolves defenses (toxicity) and the predator learns to avoid them.
Step 1: Identify the interaction - Predation (butterfly as prey, bird as predator).
Step 2: Understand the defense mechanism - Chemical defense via toxin accumulation.
Step 3: Evolutionary outcome - Birds avoid eating the toxic butterflies, leading to higher survival rates for defended prey.
Hence, the correct option is the one describing chemical defense and predator avoidance.

Key Concept: Mutualism, Plant-Pollinator Interactions

b) The orchid ensures pollination through pseudocopulation.
[Solution Description]
Orchids such as Ophrys employ 'sexual deceit' to attract male bees, which inadvertently pollinate the flowers while attempting to mate with the flower's petal that resembles a female bee. The correct answer is the option that highlights pollination as the primary benefit to the orchid.

Key Concept: Mutualism, Symbiotic Nitrogen Fixation

b) The plant receives fixed nitrogen for growth.
[Solution Description]
The symbiotic relationship between Rhizobium and legume plants involves the fixation of atmospheric nitrogen into ammonia, which the plant can utilize for growth. The plant provides carbohydrates and a protected environment for the bacteria. The correct answer is the option that states the plant receives fixed nitrogen from the bacteria.

Key Concept: Proto-cooperation, Commensalism

a) Proto-cooperation benefits both species, while commensalism benefits one without affecting the other.
[Solution Description]
Proto-cooperation involves mutual benefits to both species, though it is non-obligatory, while commensalism benefits only one species without affecting the other. The key difference lies in the mutual benefit aspect in proto-cooperation versus one-sided benefit in commensalism.

Key Concept: Proto-cooperation, Mutualism

c) Only 3 and 4
[Solution Description]
Statement 1 is incorrect because proto-cooperation is non-obligatory, unlike mutualism which can be obligatory. Statement 2 is incorrect because both species benefit in proto-cooperation, though it is not essential for survival. Statement 3 is correct as neither interaction involves harm. Statement 4 is correct as the hermit crab and sea anemone relationship is a classic example of proto-cooperation. Therefore, the correct combination is statements 3 and 4.

Key Concept: Commensalism, Epizoans, Ecological Relationships

d) Clown fish residing in sea anemone tentacles
[Solution Description]
The described scenario matches the oyster crab ($Pinnotheres\ ostreum$) example, which is classified as epizoic commensalism. Among the options, only the clown fish and sea anemone represent a similar protective commensalism where one species benefits (clown fish gets protection) while the other remains unaffected (anemone). Epiphytes and lianas involve plant-plant relationships, while cattle egret shows food-based commensalism.

Key Concept: Proto-cooperation, Commensalism

c) Sea anemones attaching to hermit crab shells for mobility and food, while the crab gains protection.
[Solution Description]
Proto-cooperation involves a beneficial but non-obligatory interaction between two species. An example is the hermit crab and sea anemone: the anemone gains mobility and food access, while the crab gets protection. In contrast, commensalism involves one organism benefiting without affecting the other, like epiphytes growing on trees. Proto-cooperation is favorable but not mandatory for survival, unlike mutualism.

Key Concept: Commensalism, Mutualism

b) The bacteria fix nitrogen for the plant, and the plant provides carbohydrates to the bacteria.
[Solution Description]
In mutualism, both organisms benefit. Here, $Rhizobium$ bacteria fix atmospheric nitrogen ($N_2$) into ammonia ($NH_3$), which the plant uses for growth, while the bacteria receive shelter and nutrients from the plant. In commensalism, only one organism benefits, and the other is neither helped nor harmed. For example, epiphytes grow on trees for support but do not harm or benefit the tree. The primary difference lies in the reciprocal benefit in mutualism versus unilateral benefit in commensalism.

Key Concept: Commensalism, Keystone Species

b) Orchid population will decline as their host trees are damaged by overgrazing herbivores.
[Solution Description]
Keystone predators regulate prey populations, maintaining ecological balance. Their removal can lead to overpopulation of herbivores, which may increase grazing pressure on trees. If trees are damaged or killed, the orchids relying on them for support (commensalism) would lose their habitat, negatively impacting their population.

Key Concept: Mutualism, Predation

c) Legume plants will suffer from nitrogen deficiency, leading to stunted growth.
[Solution Description]
The mutualistic relationship between $Rhizobium$ and legume plants is essential for nitrogen fixation. If a predator reduces the $Rhizobium$ population, the legumes will experience nitrogen deficiency, stunting their growth and reducing their productivity because they cannot independently fix atmospheric nitrogen.

Key Concept: Mutualism, Biotic Potential

c) 0.15
[Solution Description]
The net growth rate of species A is calculated by reducing its biotic potential by environmental resistance and then increasing it due to mutualism. However, the exact mutualistic benefit is not quantified here, so we focus on the reduction due to environmental resistance.
Growth rate of A after environmental resistance = $0.3 \times (1 - 0.5) = 0.15$.

Key Concept: Competition, Population Density

b) 500
[Solution Description]
The logistic growth equation is given by $\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$. The maximum growth rate occurs when $N$ is very small, so $\frac{dN}{dt} \approx rN$. Half of the maximum growth rate occurs when $\left(1 - \frac{N}{K}\right) = 0.5$. Solving for $N$, we get:
$1 - \frac{N}{K} = 0.5 \implies \frac{N}{K} = 0.5 \implies N = 0.5 \times K = 500$

Key Concept: Interspecific Competition, Population Density

c) The population density of species B will decrease
[Solution Description]
Interspecific competition occurs when two species compete for the same limited resource (insect prey). An increase in the population density of species A means greater competition for resources, which will likely lead to a decrease in the population density of species B due to reduced availability of food.

Key Concept: Mutualism, Predation

c) Mutualism
[Solution Description]
The interaction described is mutualism because both species benefit: the ants receive food (sugary secretions) from the aphids, and the aphids gain protection from predators. This is a symbiotic relationship where both organisms are interdependent.

Key Concept: Amensalism - Example

c) Penicillium inhibiting bacterial growth.
[Solution Description]
An example of amensalism involves one organism producing a substance that inhibits the growth or survival of another organism without any effect on itself.
The fungus Penicillium produces penicillin, which kills certain bacteria but does not affect Penicillium. Black walnut trees release juglone, preventing other plants from growing nearby without harming the tree itself.
The correct answer must involve such a one-sided negative interaction.

Key Concept: Amensalism - Definition

c) One species is harmed while the other remains unaffected.
[Solution Description]
Amensalism is an ecological interaction where one species is adversely affected while the other remains unaffected. This is different from competition, where both species are harmed, and parasitism, where one benefits at the expense of the other.
For example, Penicillium produces penicillin, which inhibits bacterial growth without affecting Penicillium itself. Similarly, black walnut trees release juglone to prevent the growth of nearby plants without any impact on themselves.
Therefore, the correct answer is the option that states one species is harmed while the other remains unaffected.

Key Concept: Gause's Principle

b) One species will outcompete the other, leading to its extinction
[Solution Description]
Gause’s principle states that no two species can occupy the same niche indefinitely. If they do, competition will lead to one of three outcomes: extinction of one species, migration of one species, or niche differentiation to reduce competition.

Key Concept: Competition Types

c) Interference competition
[Solution Description]
The scenario describes interference competition because Species A actively prevents Species B from accessing the resource (insects near tree trunks) through aggressive behavior.

Key Concept: Parasitic Plants

d) $Orobanche$, a total root parasite.
[Solution Description]
Obligate root parasites derive all their nutrients from the host plant and lack chlorophyll. Examples include $Orobanche$, $Balanophora$, and $Rafflesia$. Among the options provided, the correct answer is the one referring to these types of plants.

Key Concept: Ectoparasites vs Endoparasites

c) Ectoparasites live externally on the host, whereas endoparasites live inside the host.
[Solution Description]
Ectoparasites live on the surface of the host's body, while endoparasites live inside the host's body. Additionally, ectoparasites can be temporary or permanent, whereas endoparasites are generally permanent. Ectoparasites respire aerobically, while endoparasites often respire anaerobically. The correct answer is the option that highlights this distinction.

Key Concept: Prey Defenses

b) A stick insect resembling a twig
[Solution Description]
Cryptic coloration, also known as camouflage, helps prey blend into their surroundings to avoid detection by predators. Among the given options, the stick insect resembling a twig is a classic example of cryptic coloration because its appearance mimics a part of the environment (twigs), making it difficult for predators to spot.

Key Concept: Parasitism vs. Predation

c) The host usually survives in parasitism but not in predation
[Solution Description]
In parasitism, the parasite derives nourishment from the host without killing it, whereas in predation, the predator kills and consumes the prey. Therefore, the key difference lies in whether the host/prey survives the interaction.

Key Concept: Predation

d) A wolf eating another wolf
[Solution Description]
Cannibalism is a special form of predation where the predator and prey are from the same species. Among the options, one animal eating another of its own kind fits this definition.

Key Concept: Mutualism - Root Nodules

c) The bacteria fix atmospheric nitrogen into a usable form for the plant.
[Solution Description]
In the symbiotic relationship between nitrogen-fixing bacteria (e.g., Rhizobium) and leguminous plants, the bacteria colonize the roots and form nodules. They convert atmospheric nitrogen into ammonia, which the plant can absorb and utilize for growth. In return, the bacteria receive carbohydrates and a protected environment from the plant.

Key Concept: Mutualism - Lichens

d) The alga supplies food to the fungus, while the fungus provides water, minerals, and protection to the alga.
[Solution Description]
In lichens, the algal component synthesizes organic food through photosynthesis, which is utilized by both the alga and the fungus. The fungal component provides water, minerals, and protection to the alga. This is an example of mutualism where both partners benefit from the interaction.

Key Concept: Proto-cooperation

c) To protect themselves from predators using the anemone's stinging tentacles.
[Solution Description]
Clown fish live among the stinging tentacles of sea anemones primarily for protection from predators. While the anemone does not derive any direct benefit from this association, the fish gains safety as predators avoid the anemone's stinging tentacles. This demonstrates a non-obligatory but beneficial interaction.
The correct answer highlights the protective advantage for the clown fish.

Key Concept: Proto-cooperation

d) The relationship is non-obligatory and mutually beneficial for both species.
[Solution Description]
Proto-cooperation involves a mutually beneficial relationship between two species, but it is not obligatory for their survival. Both organisms benefit from the interaction but can survive independently. An example includes the hermit crab and sea anemone, where the crab gains protection while the anemone gets mobility.
The correct statement should describe that the relationship is non-obligatory and beneficial to both species.

Key Concept: Epiphytes and Lianas

c) Commensalism, where the orchid benefits from support but does not harm the mango tree.
[Solution Description]
Epiphytes like orchids grow on other plants solely for support while deriving no nutritional benefit from them. The host plant (mango tree) is neither harmed nor benefited in this association. Hence, it is an example of commensalism.

Key Concept: Commensalism

c) Commensalism, where the cattle egret benefits without affecting the cattle.
[Solution Description]
The cattle egret benefits by foraging insects that are flushed out from the vegetation due to the movement of grazing cattle. The cattle remain unaffected by this interaction as they neither gain nor lose anything. This is a classic example of commensalism, where one species benefits while the other is neither benefited nor harmed.

Key Concept: Commensalism

c) Oyster crab residing in the mantle cavity of an oyster
[Solution Description]
Commensalism is an interaction where one organism benefits, and the other is neither harmed nor benefited. An example of commensalism is the relationship between epiphytes and their host plants. Epiphytes like orchids grow on trees for support but do not derive nutrients or water from them, leaving the host tree unaffected.

Key Concept: Mutualism

d) Rhizobium bacteria in root nodules of leguminous plants
[Solution Description]
Mutualism is a type of positive interaction where both species involved benefit. Lichens are a classic example of mutualism, consisting of a symbiotic association between algae and fungi. The algae provide food through photosynthesis, while the fungi offer protection and absorb nutrients.

Key Concept: Commensalism

b) Orchids growing on mango tree branches
[Solution Description]
Commensalism is an interaction where one organism benefits while the other remains unaffected. Epiphytes like orchids growing on mango trees obtain physical support and access to sunlight without harming or benefiting the tree. This fits the definition perfectly.

Key Concept: Mutualism

c) Fungi forming mycorrhizal associations with plant roots
[Solution Description]
Mutualism is a type of interspecific relationship where both species derive benefits. Lichens represent a classic example, consisting of a fungus and an alga (or cyanobacterium) living together. The fungus provides structure and protection, while the alga performs photosynthesis to supply nutrients.

Key Concept: Intraspecific Competition

d) They compete for the same limiting resources like food or mates.
[Solution Description]
Intraspecific competition occurs when individuals of the same species compete for limited resources such as food, space, or mates. This competition results in reduced fitness for the competing individuals because not all can access the necessary resources equally, leading to lower survival or reproductive success.

Key Concept: Keystone Species

c) A species whose removal significantly alters the community structure.
[Solution Description]
A keystone species is one whose presence and role have a disproportionately large effect on the structure and function of the community. Its removal can lead to significant changes in the ecosystem, affecting many other species. For example, sea otters are keystone species in kelp forests because they control sea urchin populations, preventing overgrazing of kelp.

Key Concept: Intraspecific Relationship: Competition

d) The stronger individual gains full access to the resource
[Solution Description]
Intraspecific competition occurs between members of the same species when they compete for a limited resource such as food, shelter, or mates. This competition can lead to reduced fitness for one or both individuals involved, as they expend energy competing rather than on growth or reproduction.

Key Concept: Interspecific Relationship: Mutualism

c) A lichen composed of a fungus and an alga
[Solution Description]
Mutualism is a type of interspecific relationship where both species benefit from the interaction. Lichens are a classic example of mutualism, consisting of a fungus and an alga or cyanobacterium living together. The fungus provides structure and protection, while the alga or cyanobacterium provides nutrients through photosynthesis.

Key Concept: Example of Amensalism

c) Microbes producing antibiotics that prevent bacterial growth.
[Solution Description]
The syllabus states: "Examples include microbes producing antibiotics that inhibit the growth of other microorganisms." This matches directly with one of the options.

Key Concept: Definition of Amensalism

c) One species is harmed while the other is unaffected.
[Solution Description]
Amensalism is an interaction where one species is harmed, and the other remains unaffected. This aligns with the syllabus definition: "Amensalism is an interaction between two living individuals of different species in which an organism does not allow other organisms to grow or live near it."