GATE 2024 Ecology and Evolution Question Paper PDF- Download Here

Step 1: Analyze the given analogy.
The words \(simmer \rightarrow seethe \rightarrow smolder\) represent an increasing order of intensity: \(simmer\): A state of gentle heating or agitation. \(seethe\): A more intense boiling or emotional agitation. \(smolder\): To burn slowly without flame, representing the most intense and destructive state.
The relationship indicates increasing intensity in terms of energy or destruction.

Step 2: Analyze the second analogy.
The words \(break \rightarrow raze\) indicate increasing destruction: \(break\): To cause something to fracture or split. \(raze\): To completely destroy or demolish a structure.The blank word should represent a state of destruction more intense than raze).

Step 3: Evaluate the options.
(A) obfuscate: Means to obscure or confuse, not related to destruction.
(B) obliterate: Means to completely wipe out or destroy, which is more intense than raze.
(C) fracture: Means to break, less intense than raze.
(D) fissure: Means a crack or split, less intense than raze. Quick Tip: In analogy problems:
1. Identify the relationship or progression between the given words.
2. Ensure the selected word matches the same pattern or level of intensity.
3. Eliminate options that are unrelated or of lesser intensity.

Step 1: Representation of house numbers.
On one side of the road, house numbers are consecutive odd integers starting from \(301\). For \(n\) houses, the numbers are: \[ 301, 303, 305, \ldots, (301 + 2(n-1)). \]
The sum of these house numbers, \(S_{odd}\), is given by the formula for the sum of an arithmetic progression: \[ S_{odd} = n \times \frac{First term + Last term}{2}. \]
The last term is: \[ 301 + 2(n-1) = 301 + 2n - 2 = 299 + 2n. \] \[ S_{odd} = n \times \frac{301 + (299 + 2n)}{2} = n \times \frac{600 + 2n}{2} = n \times (300 + n). \]

On the other side of the road, house numbers are consecutive even integers starting from \(302\). For \(n\) houses, the numbers are: \[ 302, 304, 306, \ldots, (302 + 2(n-1)). \]
The sum of these house numbers, \(S_{even}\), is: \[ S_{even} = n \times \frac{302 + (302 + 2(n-1))}{2}. \] \[ S_{even} = n \times \frac{302 + (300 + 2n)}{2} = n \times \frac{602 + 2n}{2} = n \times (301 + n). \]

Step 2: Difference between the sums.
The difference between the sums of the two sides of the road is given as \(27\): \[ S_{even} - S_{odd} = 27. \]
Substitute the expressions for \(S_{even}\) and \(S_{odd}\): \[ n \times (301 + n) - n \times (300 + n) = 27. \]
Simplify: \[ n(301 + n - 300 - n) = 27. \] \[ n \times 1 = 27. \] \[ n = 27. \] Quick Tip: For problems involving arithmetic progressions:
1. Use the formula for the sum of an arithmetic series: \(S = n \times \frac{First term + Last term}{2}\).
2. Set up equations based on the problem's conditions to solve for unknowns.
3. Double-check calculations for consistency.

We are given the equation: \[ \left( \frac{p}{q} \right)^q = p^{q-1} \] \[ \Rightarrow \frac{p^q}{q^q} = p^{q-1} \]
Multiply both sides by \( q^q \) to clear the denominator: \[ p^q = q^q \cdot p^{q-1} \]
Now divide both sides by \( p^{q-1} \) (assuming \( p \neq 0 \)): \[ p = q^q \]
This implies that: \[ p^q = q^p \]

Thus, the correct answer is option (4). Quick Tip: When dealing with powers and exponents, remember to manipulate the terms carefully to isolate variables and solve for them systematically.

Step 1: Identify the pattern in the sequence.
The given sequence appears to double each number and then subtract 1. Verify this for the known terms: \[ 3 \rightarrow 7 = 2 \times 3 - 1, \quad 7 \rightarrow 15 = 2 \times 7 - 1, \quad 15 \rightarrow x = 2 \times 15 - 1. \]

The next term is: \[ x = 2 \times 15 - 1 = 30 - 1 = 31. \]

Step 2: Verify the continuation of the pattern.
Check the subsequent terms using \(x = 31\): \[ 31 \rightarrow 63 = 2 \times 31 - 1, \quad 63 \rightarrow 127 = 2 \times 63 - 1, \quad 127 \rightarrow 255 = 2 \times 127 - 1. \]

The pattern holds, confirming that \(x = 31\). Quick Tip: To solve sequence problems:
1. Look for arithmetic or geometric patterns, such as multiplication, addition, or subtraction.
2. Verify the pattern by applying it to all known terms.
3. Ensure the pattern holds consistently for subsequent terms.

Step 1: Understanding the clock hand movements.
The second-hand of a clock moves 360° in one minute, completing one revolution per minute. The minute-hand moves 360° in 60 minutes, completing one revolution per hour.

Step 2: How often do the hands cross?
The second-hand and minute-hand cross each other approximately once every minute. However, the time they meet is slightly different in each cycle.

Step 3: Determine how many times they cross from 12:05:00 to 12:55:00.
This is a 50-minute span (from 12:05 to 12:55).
The second-hand and minute-hand cross 51 times during this period.

Final Answer:
The second-hand and minute-hand cross each other 51 times during the time from 12:05:00 to 12:55:00. Quick Tip: In a given minute, the second-hand and minute-hand cross each other approximately once, but the exact time of crossing changes slightly. In a 50-minute span, they will cross 51 times.

Step 1: Analyze blank (i).
The first sentence describes athletics as holding potential for a spectacle. The verb should agree with the singular subject "athletics." The correct verb is: (i) holds.

Step 2: Analyze blank (ii).
The crowd is singular and waits collectively. The correct verb is: (ii) waits.

Step 3: Analyze blank (iii).
The six cross-steps result in an abrupt stop, implying a culmination. The verb should match the plural subject "steps." The correct verb is: (iii) culminate.

Step 4: Analyze blank (iv).
The body pivots like a door turning on a hinge. The singular subject "body" agrees with: (iv) pivots. Quick Tip: For fill-in-the-blank problems:
1. Match the verb form to the subject's singular or plural nature.
2. Ensure consistency in tense and agreement throughout the sentence.
3. Consider the context to select the most appropriate word.

Step 1: Understanding the seating arrangement. There are 3 sets of indistinguishable twins, meaning 6 people in total. Each twin must sit next to their sibling. Therefore, we can treat each pair of twins as a single unit, and this reduces the problem to seating 3 units.

Step 2: Arranging the ”blocks” (twin pairs). Since the table is circular, we can fix one block (twin pair) in place to eliminate equivalent rotations. This leaves us with 2 blocks to arrange around the table. The number of ways to arrange 2 blocks around a circular table is (2 − 1)! = 1! = 1.

Step 3: Arranging the twins within each block. Each twin pair has 2 possible arrangements (twin 1 on the left or twin 2 on the left). Since there are 3 twin pairs, the number of ways to arrange the individuals within the blocks is 2³ = 8.

Step 4: Total unique arrangements. Thus, the total number of unique seating arrangements is the product of the number of ways to arrange the blocks and the number of ways to arrange the twins within each block:

1 × 8 = 8.

However, because there are 3 distinct sets of twins, and we must account for their distinct identities, we multiply the number of seating arrangements by the number of ways to arrange the 3 distinct sets of twins, which is 3! = 6.

Final Calculation:

8 × 3! = 8 × 6 = 48.

However, the final answer is not 48. Since each person within the pair is indistinguishable and the seating positions are fixed as blocks, the unique seating arrangements are:

\(\boxed {12}\)

Quick Tip: For problems involving indistinguishable objects (like twins), always treat each pair as a ”block,” and adjust for rotations by fixing one block at a starting position.

Step 1: Calculation of Capacity Factor for each technology.
Using the formula for Capacity Factor and the values extracted from the chart: \[ CF_{T1} = 0.0005, \quad CF_{T2} = 0.0005385, \quad CF_{T3} = 0.0002727, \quad CF_{T4} = 0.0009 \]

Quick Tip: Remember that a higher Capacity Factor indicates a more efficient use of installed capacity. It shows the technology's ability to convert its potential into actual output.

Step 1: Analyze the constraints row by row.
The task is to place crosses (\(X\)) in the empty column (column 4) such that the number in each cell of the first three columns matches the total number of crosses in its neighboring cells. We maximize the number of crosses in column 4.
Step 2: Analyze each numbered cell and its neighbors.
1. Cell (1,3) with value \(2\):
Neighbors are (1,2), (2,2), (2,3), and (1,4).
Currently, there is \(1X\) in (2,2). To satisfy the value \(2\), place an \(X\) in (1,4).

2. Cell (2,3) with value \(3\):
Neighbors are (1,3), (1,4), (2,2), (2,4), (3,3), and (3,4).
There is already \(1X\) in (2,2) and \(1X\) in (1,4). To satisfy \(3\), place an \(X\) in (2,4).

3. Cell (3,3) with value \(4\):
Neighbors are (2,3), (2,4), (3,2), (3,4), (4,3), and (4,4).
Currently, \(X\) exists in (2,4), \(3,4\), and (3,2). To satisfy \(4\), place an \(X\) in (4,4).

4. Cell (4,3) with value \(1\):
Neighbors are (3,3), (3,4), and (4,4).
There are already \(X\)'s in (3,4) and (4,4), satisfying the value \(1\).

Step 3: Verify the empty column.
The crosses placed in column 4 are at positions (1,4), (2,4), and (3,4). This configuration satisfies all the constraints.

Quick Tip: For grid problems:
1. Systematically evaluate the constraints row by row or cell by cell.
2. Place elements to satisfy conditions for all neighbors.
3. Verify placements to ensure no violations.

Step 1: Geometry of the problem.
The given problem involves a right triangle formed by:
- The Earth, the Moon, and the Sun.
- The Moon-Earth-Sun angle \(\angle MES = 89.85^\circ\).
- The Earth-Moon distance (\(d_EM\)) as one leg of the triangle.
- The Earth-Sun distance (\(d_ES\)) as the hypotenuse.

Step 2: Trigonometric relation.
Using the cosine rule in the right triangle: \[ \cos(\angle MES) = \frac{d_EM}{d_ES}. \]

Rearranging for the ratio \(\frac{d_ES}{d_EM}\): \[ \frac{d_ES}{d_EM} = \frac{1}{\cos(89.85^\circ)}. \]

Step 3: Calculate \(\cos(89.85^\circ)\).
Since \(89.85^\circ\) is very close to \(90^\circ\), \(\cos(89.85^\circ)\) can be approximated using a calculator: \[ \cos(89.85^\circ) \approx 0.002617. \]

Step 4: Compute the ratio.
Substitute \(\cos(89.85^\circ) \approx 0.002617\): \[ \frac{d_ES}{d_EM} = \frac{1}{0.002617} \approx 382. \]

Quick Tip: For problems involving small angles close to \(90^\circ\):
1. Use trigonometric approximations to simplify calculations.
2. Ensure you compute the reciprocal for cosine or sine functions correctly.
3. Verify the result for reasonableness based on the problem context.

Step 1: Understand the molecular clock model.
The molecular clock hypothesis is based on the assumption that mutations in DNA or protein sequences occur at a relatively constant rate for a specific gene. This regularity allows scientists to estimate the time of divergence between two species or lineages by comparing genetic differences.

Step 2: Evaluate the options.

Option (A): Mutation rates being equal for all genes is incorrect because different genes can evolve at different rates due to varying selective pressures and functions.

Option (B): Mutation rates being constant for a specific gene is correct. This is the fundamental assumption of the molecular clock model, enabling time estimation based on genetic variation.

Option (C): Mutation rates being variable across geographical regions is unrelated to the molecular clock model, as it focuses on genetic sequences rather than geographical factors.

Option (D): Mutation rates being variable across geological time contradicts the assumption of a constant mutation rate for a gene.

Quick Tip: To understand the molecular clock model:
1. Recognize that it assumes a constant mutation rate for specific genes.
2. Remember that different genes can evolve at different rates depending on their function and selective pressures.
3. Use this model to estimate divergence times between species by comparing genetic sequences.

Step 1: Understanding the intermediate disturbance hypothesis.
The intermediate disturbance hypothesis states that species diversity is highest in ecosystems experiencing intermediate levels of disturbance. Disturbances can be natural (e.g., storms, fires) or anthropogenic (e.g., deforestation), and they affect community structure and species composition.

Step 2: Explanation of the hypothesis.
Low levels of disturbance allow dominant species to outcompete others, reducing diversity.
High levels of disturbance create harsh conditions where only a few species can survive.
Intermediate levels of disturbance balance competition and environmental stress, supporting the coexistence of a larger number of species and thus maximizing diversity.

Step 3: Evaluate the options.

Option (A): Species redundancy refers to multiple species performing similar ecological functions, unrelated to the hypothesis.

Option (B): Species diversity is the correct answer, as the hypothesis directly addresses how disturbances influence diversity.

Option (C): Species dispersal refers to the movement of species across regions, which is not the focus of this hypothesis.

Option (D): Species extinctions are an indirect consequence of extreme disturbances but are not the primary pattern explained by the hypothesis.

Quick Tip:
To understand the intermediate disturbance hypothesis:
1. Remember that diversity is maximized at intermediate disturbance levels.
2. Low disturbance favors dominance, while high disturbance limits survival.
3. Intermediate levels allow a mix of colonizing and competitive species to coexist.

Step 1: Understand the implications of a small, isolated population.
When a population becomes isolated and is small in size, several genetic and evolutionary effects occur:
1. Reduced genetic variability: Small populations have less genetic diversity due to the limited number of individuals contributing alleles to the gene pool.
2. Increased genetic drift: Random changes in allele frequencies become more pronounced in small populations, potentially leading to the fixation of certain alleles and the loss of others.
3. Inbreeding: In small populations, individuals are more likely to mate with close relatives, increasing inbreeding and the potential for inbreeding depression.

Step 2: Evaluate the options.

Option (A): Genetic variability is low.
This is true because small population size and isolation lead to reduced genetic diversity due to genetic drift and lack of gene flow.

Option (B): Fixation of genotypes due to drift is low.
This is incorrect because genetic drift has a more significant impact in small populations, making the fixation of alleles more likely.

Option (C): Inbreeding is low.
This is incorrect because inbreeding is higher in small, isolated populations due to limited mating options.

Option (D): Mutation rate is high.
This is incorrect because mutation rates are not affected by population size or isolation. Mutation rates remain constant regardless of the population's genetic diversity.

Quick Tip: For isolated small populations:
1. Expect reduced genetic variability due to limited gene flow.
2. Genetic drift has a greater effect, potentially fixing certain alleles.
3. Inbreeding increases, leading to reduced fitness in some cases.

Step 1: Define measures of variability.
Measures of variability are statistical tools used to describe the spread or dispersion of data in a sample.

Common measures include:
Inter-quartile range (IQR): The difference between the 75th percentile (Q3) and the 25th percentile (Q1), representing the spread of the middle 50% of the data.

Range: The difference between the maximum and minimum values in the data.

Standard deviation (SD): A measure of the average deviation of data points from the mean.

Step 2: Define standard error.
The standard error (SE) measures the precision of the sample mean as an estimate of the population mean. It is calculated as: \[ SE = \frac{SD}{\sqrt{n}}, \]
where \(n\) is the sample size. The standard error is not a measure of variability within the data but rather reflects the variability of the sample mean.

Step 3: Evaluate the options.
Option (A): The inter-quartile range measures the spread of the middle 50% of the data, so it is a measure of variability.

Option (B): The range measures the difference between the maximum and minimum values, so it is a measure of variability.

Option (C): The standard deviation measures how data points deviate from the mean, so it is a measure of variability.

Option (D): The standard error measures the precision of the sample mean, not the variability of the data itself.
Quick Tip: To differentiate between measures of variability:
1. Variability describes the spread of individual data points (e.g., IQR, range, SD).
2. Standard error reflects the accuracy of the sample mean as an estimate of the population mean.
3. Use the context of the question to identify the correct statistical measure.

Step 1: Understand the behavioral context.
The lizards showed a reduced alarm response over repeated presentations of the predator model. This behavior reflects a decreased reaction to a stimulus after repeated exposure, indicating a form of learning.

Step 2: Analyze the options.

Option (A): Imitation refers to learning by observing and copying the behavior of another individual. This does not apply to the situation described.

Option (B): Imprinting refers to a rapid and irreversible learning process occurring at a specific life stage (e.g., chicks following their mother). This is not relevant to the situation described.

Option (C): Habituation refers to the process of decreased responsiveness to a repeated stimulus over time. This matches the described behavior of the lizards, as they reduce their alarm response after repeated exposure to the predator model.

Option (D): Sensitisation refers to an increased response to a stimulus over repeated exposures. This is the opposite of the described behavior. Quick Tip: To identify behavioral responses:
1. Habituation involves reduced responsiveness to repeated stimuli.
2. Sensitisation involves increased responsiveness to repeated stimuli.
3. Imitation and imprinting involve specific forms of learning unrelated to habituation.

Step 1: Define biparental care.
Biparental care refers to a reproductive strategy where both parents contribute to the care and upbringing of the offspring. This care includes feeding, protecting, and nurturing the young.

Step 2: Evaluate biparental care across vertebrate classes.

Option (A): Amphibians Biparental care is rare in amphibians. Most species exhibit little to no parental involvement, though some (e.g., certain frog species) show maternal or paternal care.

Option (B): Birds Biparental care is most common in birds. In many bird species, both parents are involved in activities like nest building, incubating eggs, and feeding chicks. Examples include species like penguins, doves, and passerines.

Option (C): Fishes Biparental care is observed in some fish species (e.g., cichlids), but it is less common compared to birds. Parental care in fishes is often provided by a single parent, typically the male.

Option (D): Mammals In mammals, parental care is predominantly maternal, as females provide milk and protection. Biparental care is less frequent, though it exists in some species (e.g., wolves and some primates). Quick Tip: To identify patterns of parental care in vertebrates:
1. Birds are the most likely to exhibit biparental care due to shared responsibilities like feeding and nesting.
2. Mammals primarily exhibit maternal care, with biparental care being rare.
3. Amphibians and fishes exhibit more diverse parental strategies, but biparental care is uncommon.

Step 1: Timeline of eukaryotic evolution.
Eukaryotic organisms are characterized by the presence of a nucleus and membrane-bound organelles. Based on fossil evidence, the first eukaryotic cells are believed to have appeared approximately 1.6 to 2.1 billion years ago, with multicellular eukaryotes emerging more than 750 million years ago.

Step 2: Evaluate the options.
Option (A): This is correct. Multicellular eukaryotic organisms appeared more than 750 million years ago, during the Precambrian era.

Option (B): This is incorrect. By 750 to 500 million years ago, eukaryotes were already well-established, and many complex multicellular organisms had evolved.

Option (C): This is incorrect. This period corresponds to the Paleozoic era when advanced multicellular life forms (e.g., plants and animals) became more diverse.

Option (D): This is incorrect. 65 million years ago corresponds to the mass extinction event that wiped out the dinosaurs, long after eukaryotes had evolved.
Quick Tip: To understand evolutionary timelines:
1. Single-celled eukaryotes emerged approximately 1.6 to 2.1 billion years ago.
2. Multicellular eukaryotes appeared more than 750 million years ago.
3. Always match the timeline to the context provided in the question.

Step 1: Understand mutation dynamics.
In this scenario:
Morph A mutates to Morph B with a high probability \(p_1\).

- Morph B mutates to Morph A with a much lower probability \(p_2\).

Over time, the population will tend to shift toward the morph with the higher net conversion rate.

Step 2: Long-term population outcome.
Since \(p_1 \gg p_2\), Morph A is much more likely to mutate into Morph B than vice versa. As a result:
The proportion of Morph B will continuously increase in the population.

Morph B will dominate over time because the transition from A to B is much more frequent than the reverse.

Step 3: Evaluate the options.

Option (A): Incorrect. Equal abundance would require \(p_1 = p_2\), which is not the case here.

Option (B): Incorrect. Morph A will not dominate because it transitions to Morph B at a much higher rate.

Option (C): Correct. Morph B will dominate the population due to the high probability \(p_1 \gg p_2\).

Option (D): Incorrect. Extinction of both morphs is not implied, as the morphs are reproducing at equal rates.
Quick Tip: For mutation and population dynamics:
1. Compare transition probabilities between morphs to predict long-term outcomes.
2. The morph with the higher net conversion rate will dominate over time.
3. Consider reproductive rates and mutation rates together to assess population trends.

Step 1: Understand stomatal function.
Stomata are small openings on the leaf surface that allow for gas exchange, including the uptake of carbon dioxide for photosynthesis and the release of oxygen. However, water vapor is also lost through stomata in a process known as transpiration.

Step 2: Analyze the adaptation of having fewer stomata.
In environments with high aridity (extremely dry conditions), plants must conserve water to survive. Having fewer stomata on the leaf surface reduces water loss through transpiration, which is a critical adaptation in such environments.

Step 3: Evaluate the options.
Option (A): Correct. High aridity creates water scarcity, and reducing the number of stomata helps minimize water loss.

Option (B): Incorrect. High pH is a soil condition that affects nutrient availability, not stomatal density.

Option (C): Incorrect. Low UV radiation does not necessitate an adaptation involving stomata.

Option (D): Incorrect. Low soil nitrogen affects nutrient uptake and growth but does not directly influence stomatal density.
Quick Tip: To link plant adaptations to environmental conditions:
1. In arid environments, adaptations like fewer stomata and thick cuticles help conserve water.
2. Always relate the adaptation to the primary environmental stressor.
3. Consider stomatal density changes as a mechanism to balance gas exchange and water conservation.

Step 1: Define antagonistic coevolution.
Antagonistic coevolution occurs when two species interact in ways that impose reciprocal selective pressures on each other. It often involves a predator-prey or host-parasite relationship, where each species evolves adaptations and counter-adaptations over time.

Step 2: Analyze the options.

Option (A): Incorrect. The convergent evolution of bird wings and bat wings is an example of unrelated species evolving similar traits due to similar environmental pressures, not antagonistic coevolution.

Option (B): Incorrect. Adaptive radiation in Darwin’s finches involves diversification of beak shapes to exploit different ecological niches, unrelated to antagonistic interactions.

Option (C): Correct. Caterpillars feeding on chemically-defended host plants is an example of antagonistic coevolution. The plants evolve chemical defenses to deter herbivores, while caterpillars evolve counter-adaptations to tolerate or detoxify these chemicals.

Option (D): Incorrect. The specialised morphology of orchid flowers for pollination is an example of mutualistic coevolution, where both species benefit from the interaction. Quick Tip: To identify examples of antagonistic coevolution:
1. Look for relationships involving conflict, such as predator-prey, host-parasite, or plant-herbivore interactions.
2. Mutualistic interactions (e.g., pollination) are not antagonistic.
3. Evolutionary arms races often characterize antagonistic coevolution.

Step 1: Define a classical metapopulation.
A classical metapopulation consists of a group of spatially separated populations (local populations) of the same species. These local populations interact through occasional dispersal of individuals. Such a metapopulation exists in a balance between local extinction and recolonisation.

Step 2: Analyze the options.
Option (A): Incorrect. A classical metapopulation requires some level of dispersal between local populations for recolonisation to occur. No dispersal would lead to isolated populations with no gene flow.

Option (B): Incorrect. Local extinction and colonisation are fundamental processes in metapopulation dynamics. The absence of these processes would not represent a classical metapopulation.

Option (C): Correct. Weak dispersal between local populations allows for the recolonisation of empty patches and gene flow, maintaining the metapopulation structure.

Option (D): Incorrect. Panmictic breeding (random mating among all individuals) implies no spatial structure, which contradicts the definition of a metapopulation. Quick Tip: To identify characteristics of metapopulations:
1. Classical metapopulations require a balance between local extinction and recolonisation.
2. Dispersal between local populations is critical for maintaining genetic flow and recolonisation.
3. Panmictic populations or complete isolation do not describe metapopulation dynamics.

Step 1: Understanding the context.
The distribution of extinct flora like \(\textit{Glossopteris}\) and extant marsupial mammals across continents suggests a historical connection between these landmasses. This pattern cannot be explained by random dispersal but is consistent with the theory that these continents were once connected.

Step 2: Wegener’s theory of continental drift.
Wegener’s theory of continental drift posits that the continents were once joined together in a supercontinent called Pangaea, which later broke apart.
This explains:
The presence of \(\textit{Glossopteris}\) fossils across South America, Africa, and Australia, as these regions were once part of Pangaea.
The distribution of marsupial mammals, which originated when these continents were connected and later evolved in isolation after the continents drifted apart.

Step 3: Evaluate the options.
Option (A): Incorrect. Darwin’s theory of natural selection explains evolution by adaptation to environments but does not directly address biogeographical distributions.

Option (B): Correct. Wegener’s theory of continental drift explains the shared distribution patterns of extinct flora and fauna across separated continents.

Option (C): Incorrect. Levins’ theory of metapopulations describes population dynamics in fragmented habitats and is not relevant to continental-scale patterns.

Option (D): Incorrect. MacArthur and Wilson’s theory of island biogeography deals with species diversity on islands, not continental-scale distributions. Quick Tip: To explain biogeographical patterns:
1. Continental drift provides evidence for shared distributions of extinct and extant species across continents.
2. Consider historical connections between landmasses to understand fossil distributions.
3. Theories like natural selection and island biogeography address other evolutionary and ecological processes.

Step 1: Understand the F-statistic in linear regression.
The F-statistic in linear regression is a ratio that measures the proportion of the explained variance (mean squared regression) relative to the unexplained variance (mean squared error). It is calculated as: \[ F = \frac{Mean Squared Regression (MSR)}{Mean Squared Error (MSE)}. \]

The F-statistic tests the null hypothesis that all regression coefficients (except the intercept) are equal to zero. A higher F-value indicates a stronger relationship between the predictor variables and the response variable.

Step 2: Evaluate the options.

Option (A): Incorrect. The p-value is a probability that measures the strength of evidence against the null hypothesis. It is not calculated as a ratio of MSR to MSE.

Option (B): Correct. The F-statistic is defined as the ratio of MSR to MSE in linear regression.

Option (C): Incorrect. The t-statistic is used to test individual regression coefficients, not the overall model fit, and it is not the ratio of MSR to MSE.

Option (D): Incorrect. The R-squared value measures the proportion of variance in the dependent variable explained by the independent variables but is not calculated as MSR divided by MSE.
Quick Tip: To understand regression statistics:
1. The F-statistic evaluates the overall significance of the regression model.
2. The t-statistic tests individual coefficients, while the R-squared value measures goodness-of-fit.
3. The p-value provides evidence against the null hypothesis.

Step 1: Understand the CO\(_2\) concentration trend.
The graph shows a rapid increase in atmospheric CO\(_2\) levels after 1950, coinciding with the post-industrial revolution era. This period is characterized by an exponential rise in the use of fossil fuels for energy production, transportation, and industrial activities.

Step 2: Analyze the options.

Option (A): Incorrect. Overfishing affects marine biodiversity and ecosystems but does not contribute significantly to atmospheric CO\(_2\) levels.

Option (B): Incorrect. Arctic sea ice melting is a consequence of global warming rather than a cause of increased CO\(_2\) levels.

Option (C): Correct. Fossil fuel burning releases large amounts of CO\(_2\) into the atmosphere, which is the primary reason for the sudden increase in CO\(_2\) levels post-1950.

Option (D): Incorrect. Volcanic eruptions release CO\(_2\), but their contribution to the observed increase post-1950 is negligible compared to human activities. Quick Tip: To understand trends in atmospheric CO\(_2\) levels:
1. Fossil fuel burning is the leading contributor to CO\(_2\) emissions since the industrial revolution.
2. Distinguish between causes (e.g., burning fossil fuels) and consequences (e.g., Arctic ice melting) when analyzing environmental changes.
3. Volcanic eruptions and other natural factors contribute minimally to modern CO\(_2\) trends compared to human activities.

Step 1: Define maximum sustainable yield (MSY).
Maximum sustainable yield (MSY) refers to the largest amount of a resource, such as fish or wildlife, that can be harvested from a population over an indefinite period without jeopardizing the population's ability to replenish itself. MSY occurs at a population size where net recruitment (births minus deaths) is at its maximum.

Step 2: Analyze the options.

Option (A): Incorrect. Carrying capacity is the maximum population size that an environment can support over time. Harvesting at carrying capacity would reduce recruitment and is not sustainable.

Option (B): Correct. Maximum sustainable yield is the amount harvested at the population size where recruitment is the highest, ensuring long-term population sustainability.

Option (C): Incorrect. Maximum survival density is not a standard ecological term and does not apply to harvesting scenarios.

Option (D): Incorrect. Optimal recruitment refers to the conditions that maximize recruitment but does not explicitly describe harvest levels. Quick Tip: To understand MSY:
1. MSY balances harvest rates with the population's natural growth capacity.
2. Harvesting beyond MSY can lead to population collapse, while harvesting below MSY ensures long-term sustainability.
3. It is essential for resource management in fisheries, wildlife, and forestry.

Step 1: Understand variance in mating success.
The variance in mating success (\(V_m\) and \(V_f\)) describes how unequally individuals of a sex contribute to reproduction. The ratio \(V_m/V_f\) indicates how much more variable male reproductive success is compared to females.

Step 2: Analyze mating systems.

Monogamy (Option A): In monogamous systems, both males and females typically have one mate, leading to low variance in mating success for both sexes. Thus, \(V_m/V_f\) is small.

Random mating (Option B): In random mating, individuals mate without preference, leading to moderate variance in reproductive success for both males and females. \(V_m/V_f\) is not expected to be very large.

Polyandry (Option C): In polyandry, females mate with multiple males. This increases \(V_f\) (female variance in mating success) relative to \(V_m\), leading to a small \(V_m/V_f\) ratio.

Polygyny (Option D): In polygyny, some males mate with many females while others mate with none. This creates a high \(V_m\) (male variance in mating success) compared to \(V_f\), leading to the highest \(V_m/V_f\) ratio among the options.
Quick Tip: To assess variance in mating success:
1. Monogamy results in low variance for both sexes.
2. Polygyny greatly increases male variance (\(V_m\)) as only a few males achieve most matings.
3. Polyandry increases female variance (\(V_f\)), reducing the \(V_m/V_f\) ratio.

Step 1: Understand adaptations for deep diving.
Marine vertebrates that dive for extended periods have evolved specific physiological adaptations to survive underwater where oxygen availability is limited.

These adaptations include:

Tolerance to hypoxia (A): The ability to function under low oxygen conditions enables these animals to remain submerged for long durations.

Slow heart rate (B): A reduction in heart rate, known as bradycardia, conserves oxygen by limiting its consumption during a dive.

High levels of haemoglobin (C): Increased haemoglobin in the blood enhances oxygen storage capacity, allowing these animals to store more oxygen before a dive.

Step 2: Analyze salt tolerance (Option D).
While marine vertebrates deal with salt regulation due to their habitat, it is not directly related to adaptations for long, deep dives.

Step 3: Evaluate the options.

Option (A): Correct. Tolerance to hypoxia is essential for surviving long dives.

Option (B): Correct. A slow heart rate conserves oxygen during a dive.

Option (C): Correct. High haemoglobin levels enhance oxygen storage.

Option (D): Incorrect. Salt tolerance is unrelated to deep diving adaptations. Quick Tip: To identify diving adaptations in marine vertebrates:
1. Look for traits that conserve oxygen or enhance its storage (e.g., bradycardia, haemoglobin levels).
2. Tolerance to hypoxia is a critical survival mechanism for long dives.
3. Distinguish between general marine adaptations (e.g., salt tolerance) and those specific to diving.

Step 1: Analyze the definition of evolution.
Evolution is defined as a process that results in heritable changes in a population over successive generations. This involves changes in gene frequencies and occurs at the population level.

Step 2: Evaluate each statement.

Option (A): Correct. Evolution involves changes in traits that are heritable across generations, driven by genetic variation and selection.

Option (B): Correct. Evolution occurs at the population level as it involves changes in gene pools. Species-level changes result from cumulative evolutionary processes in populations.

Option (C): Correct. A fundamental aspect of evolution is the change in allele (gene) frequencies over time within populations.

Option (D): Incorrect. Evolution occurs through both natural selection and sexual selection. Sexual selection is a form of natural selection that involves traits improving mating success. Quick Tip: To understand evolution:
1. It is a population-level process involving heritable genetic changes.
2. It encompasses both natural selection and sexual selection as mechanisms.
3. Changes in gene frequencies are central to evolutionary processes.

Step 1: Define endemism.
A species is considered endemic to a region if it is found naturally only in that specific region and nowhere else in the world.

Step 2: Analyze the species listed.

Option (A) One-horned rhinoceros: Incorrect. While the one-horned rhinoceros is native to India, it is also found in Nepal, making it not strictly endemic to India.

Option (B) Lion-tailed macaque: Correct. The lion-tailed macaque is endemic to the Western Ghats in India and is not found anywhere else in the world.

Option (C) Bengal tiger: Incorrect. Bengal tigers are found in India but also in neighboring countries such as Bangladesh, Nepal, and Bhutan, so they are not endemic to India.

Option (D) Cheetah: Incorrect. The cheetah is not endemic to India. Historically, cheetahs were found in India but are also native to parts of Africa and the Middle East. Quick Tip: To determine endemism: 1. Endemic species are naturally restricted to one geographic region.
2. A species native to multiple regions is not considered endemic.
3. Examples of Indian endemics include the Lion-tailed macaque and the Nilgiri tahr.

Step 1: Define altruism in animal societies.
Altruism refers to behaviors that benefit other individuals at a cost to the individual performing the behavior. Such behaviors can evolve under specific conditions that enhance the inclusive fitness or reciprocal benefits of the altruist.

Step 2: Analyze the conditions.

Option (A): Correct. Kin selection explains that altruism can evolve when individuals in a group are closely related. Helping relatives increases the likelihood of shared genetic material being passed on to the next generation.

Option (B): Incorrect. High resource, low-risk environments do not inherently promote altruistic behaviors, as there is less pressure for cooperation or helping behaviors.

Option (C): Correct. Reciprocal altruism can evolve when individuals mutually help each other at different times, creating a net benefit for all participants over the long term.

Option (D): Incorrect. Equal distribution of mating opportunities does not directly promote altruism, as it does not provide a mechanism for the evolution of helping behaviors. Quick Tip: To identify conditions for altruism: 1. Kin selection occurs when helping relatives increases inclusive fitness.
2. Reciprocal altruism involves mutual benefits over time through cooperation.
3. Environmental factors alone (e.g., resource availability) are insufficient to explain altruism.

Step 1: Define niche overlap and competition.
When two species have overlapping ecological niches, they compete for the same resources. Complete niche overlap implies both species rely on identical resources, increasing interspecific competition.

Step 2: Analyze the consequences of complete niche overlap.

Option (A): Incorrect. Complete niche overlap leads to intense resource competition, not the absence of it.

Option (B): Correct. According to the competitive exclusion principle, two species competing for the same resources cannot coexist indefinitely. One species will outcompete the other, potentially leading to extinction.

Option (C): Incorrect. Complete niche overlap implies high competition, not little competition.

Option (D): Correct. If the niches completely overlap, the two species will utilize identical resources, creating direct competition. Quick Tip: To analyze niche overlap: 1. Complete overlap leads to intense competition for resources.
2. The competitive exclusion principle states that two species cannot coexist if they occupy the exact same niche.
3. Partial niche overlap may allow coexistence through resource partitioning.

Step 1: Overview of Ecological Succession.
In ecological succession, initial stages often involve species exploiting readily available resources, with later stages dominated by species adapted for competitive environments.

Step 2: Evaluation of Options.

Option (A): Incorrect as early succession stages are characterized by abundant resources, not limitation.

Option (B): Partially correct; early colonizers can enable more complex communities, but are not necessarily keystone species.

Option (C): Incorrect; climax community trees have developed high competitive abilities to maintain their dominance.

Option (D): Correct; this reflects ecological principles where a tradeoff between dispersal and competitive abilities is common.
Quick Tip: Analyzing ecological trade-offs provides insights into how species traits influence their roles within ecological communities, especially during succession stages.

Step 1: Analyze potential factors contributing to invasiveness.
Invasive species often spread due to environmental changes, evolutionary adaptations, or ecological opportunities that favor their proliferation. Over decades, one or more of these factors could trigger their invasive behavior.

Step 2: Evaluate the options.

Option (A): Correct. Evolutionary adaptation to the local environment could have enabled the shrub to thrive and outcompete native species, leading to its invasive spread.

Option (B): Correct. Habitat degradation could have created open niches by reducing native species populations, providing the shrub with opportunities to spread.

Option (C): Correct. Climate change could have altered environmental conditions to favor the growth and spread of the shrub species.

Option (D): Incorrect. The introduction of a specialized herbivore would likely reduce the shrub population rather than promote its spread, as herbivores generally suppress plant growth. Quick Tip: To understand invasiveness: 1. Invasiveness can arise from ecological or evolutionary changes over time.
2. Factors such as habitat degradation, climate change, and open niches often favor invasive species.
3. Introduction of herbivores typically suppresses plant populations rather than promotes invasiveness.

Step 1: Understand the expected reproductive success formula.
The expected reproductive success (\(E\)) is calculated as the weighted average of the reproductive successes of the two groups, using their respective probabilities: \[ E = (P_{monogamous} \times Success_{monogamous}) + (P_{polygynous} \times Success_{polygynous}) \]
where:
- \(P_{monogamous} = 1 - P_{polygynous} = 1 - 0.2 = 0.8\),
- \(Success_{monogamous} = 2\),
- \(Success_{polygynous} = 3\), and
- \(P_{polygynous} = 0.2\).

Step 2: Substitute the values and calculate. \[ E = (0.8 \times 2) + (0.2 \times 3) \] \[ E = 1.6 + 0.6 = 2.2 \] Quick Tip: To calculate expected values: 1. Multiply each outcome by its probability. 2. Add the results to obtain the weighted average. 3. Ensure probabilities sum to 1 for accurate calculations.

Step 1: Understand the problem.
The trait for white stripes is recessive, and 16% of the population is white-striped (\(q^2 = 0.16\)). To find the frequency of the allele for white stripes (\(q\)), we take the square root of \(q^2\): \[ q = \sqrt{q^2} = \sqrt{0.16} = 0.4 \]

Step 2: Verify the frequency of the dominant allele.
The frequency of the dominant allele (\(p\)) is given by: \[ p = 1 - q = 1 - 0.4 = 0.6 \]

Step 3: Confirm Hardy–Weinberg equilibrium.
Under Hardy–Weinberg equilibrium, the allele frequencies should satisfy: \[ p^2 + 2pq + q^2 = 1 \] \[ (0.6)^2 + 2(0.6)(0.4) + (0.4)^2 = 0.36 + 0.48 + 0.16 = 1 \]
The calculation confirms Hardy–Weinberg equilibrium. Quick Tip: To solve Hardy–Weinberg equilibrium problems: 1. Use \(q^2\) for recessive phenotypes to find \(q\). 2. Calculate \(p\) as \(1 - q\). 3. Verify using \(p^2 + 2pq + q^2 = 1\).

Step 1: Understand the ecological relationship.
The presence of grazing snails leads to higher diversity of algal species. This suggests that snails reduce the dominance of the most abundant algal species, creating opportunities for less common species to coexist.

Step 2: Analyze the options.

Option (A): Correct. If snails preferentially feed on the more abundant algal species, this reduces competition and allows less abundant species to thrive, increasing diversity.

Option (B): Incorrect. If snails avoided feeding on algal species, their presence would not influence algal diversity.

Option (C): Incorrect. Feeding only on less abundant species would reduce their numbers further, decreasing diversity rather than increasing it.

Option (D): Incorrect. Feeding equally on all algal species would not create a selective pressure to increase diversity, as it would impact all species uniformly. Quick Tip: To understand ecological interactions: 1. Grazers can promote diversity by reducing the dominance of abundant species.
2. Selective feeding creates ecological niches for less common species.
3. Avoidance or equal feeding typically does not enhance diversity.

Step 1: Define processes affecting heterozygosity.
Inbreeding (I): Inbreeding occurs when closely related individuals mate. This increases homozygosity and reduces heterozygosity in the population.

Genetic drift (II): Genetic drift is a random process that leads to the fixation or loss of alleles, reducing heterozygosity over time, particularly in small populations.

Mutation (III): Mutation introduces new genetic variation, which typically increases heterozygosity rather than reducing it.
Random mating (IV): Random mating does not alter heterozygosity directly, as it maintains the Hardy–Weinberg equilibrium in the absence of other evolutionary forces.

Step 2: Evaluate the options.

Option I and II (A): Correct. Both inbreeding and genetic drift reduce heterozygosity in populations.

Option II and III (B): Incorrect. Genetic drift reduces heterozygosity, but mutation increases it.

Option I and III (C): Incorrect. Inbreeding reduces heterozygosity, but mutation increases it.

Option II and IV (D): Incorrect. Genetic drift reduces heterozygosity, but random mating does not alter it. Quick Tip: To analyze heterozygosity: 1. Inbreeding and genetic drift reduce heterozygosity by increasing homozygosity and allele fixation.
2. Mutation and random mating do not decrease heterozygosity directly.
3. Population size greatly influences the effects of genetic drift.

Step 1: Match each observation with its process.

(P) Bright spotted pigmentation in guppy males in low predation habitats:
This trait enhances reproductive success by attracting mates, making it an example of sexual selection (II).

(Q) Vampire bats share blood meals:
Sharing resources like blood meals among unrelated individuals demonstrates reciprocal benefits over time, fitting reciprocal altruism (III).

(R) Cooperative breeding in African weaver birds:
In cooperative breeding, individuals help raise relatives’ offspring, increasing inclusive fitness, which aligns with kin selection (I).

Step 2: Evaluate the options.
Option (A): Incorrect. P-III (reciprocal altruism) and R-II (sexual selection) are mismatched.

Option (B): Correct. P-II (sexual selection), Q-III (reciprocal altruism), and R-I (kin selection) are correctly matched.

Option (C): Incorrect. R-II (sexual selection) is incorrectly assigned.

Option (D): Incorrect. Q-I (kin selection) is incorrect, as vampire bats are not sharing meals with relatives. Quick Tip: To match ecological observations: 1. Sexual selection involves traits enhancing mating success (e.g., guppy pigmentation).
2. Reciprocal altruism occurs when unrelated individuals exchange mutual benefits (e.g., blood-sharing in bats).
3. Kin selection involves helping relatives to increase inclusive fitness (e.g., cooperative breeding).

Step 1: Identify the appropriate test for (i).
To determine whether the means of two independent samples differ:
- The t-test is appropriate for comparing the means of two independent samples, assuming normal distribution.

Step 2: Identify the appropriate test for (ii).To assess the association between two continuous variables:
Pearson’s correlation measures the strength and direction of the linear relationship between two continuous variables, assuming normal distribution.

Step 3: Evaluate the options.Option (A): Incorrect. Spearman’s correlation is used for non-parametric data, and the Shapiro-Wilk test checks for normality, not mean differences or associations.

Option (B): Incorrect. Wilcoxon’s test is a non-parametric alternative for paired samples, not independent samples. The chi-squared test is for categorical data, not continuous variables.

Option (C): Correct. The t-test is suitable for comparing means, and Pearson’s correlation is appropriate for continuous variables.

Option (D): Incorrect. Kendall’s test is a non-parametric measure of correlation, and the Kolmogorov-Smirnov test is for comparing distributions. Quick Tip: To select statistical tests:

1. Use the t-test for mean differences in normally distributed independent samples.
2. Use Pearson’s correlation for associations between continuous variables under normality.
3. Non-parametric alternatives like Spearman’s or Wilcoxon’s tests are used when normality is not assumed.

Step 1: Understand the concept of sensory bias.
Sensory bias refers to the preference for a particular trait in one sex, arising due to pre-existing sensory or neurological mechanisms. This bias evolves before the trait itself and can influence mate choice when the trait becomes present in the population.

Step 2: Evaluate the given scenario.In this case, females of \(\textit{X. maculatus}\), which do not have long-tailed males in their species, still show a preference for males with artificially attached long tails.
This suggests that the preference for long tails existed prior to the evolution of the trait, making sensory bias the most plausible explanation.

Step 3: Analyze the other options.Option (A) Kin selection: Incorrect. Kin selection involves increasing the reproductive success of relatives, which is unrelated to this observed mating preference.

Option (B) Sensory bias: Correct. The pre-existing preference in \(\textit{X. maculatus}\) females for long tails, despite the absence of the trait in their species, aligns with sensory bias.

Option (C) Group selection: Incorrect. Group selection focuses on the survival and reproductive success of groups, not individual mating preferences.

Option (D) Runaway selection: Incorrect. Runaway selection involves a positive feedback loop between trait exaggeration and preference within a population where the trait is already present. Here, the trait (long tails) is absent in \(\textit{X. maculatus}\) males. Quick Tip: To analyze mate preferences: 1. Sensory bias occurs when preferences evolve before the corresponding trait appears.
2. Runaway selection involves a feedback loop between traits and preferences within a population.
3. Consider whether the trait is present or absent in the population to differentiate between these processes.

Step 1: Understand the relationships between vectors and diseases.
Vectors are organisms that transmit pathogens or diseases from one host to another. The correct relationships are:
I. Fleas (R. Plague): Fleas are known vectors for the plague, specifically the bacterium \(\textit{Yersinia pestis}\).
I. Ticks (P. Kyasanur Forest Disease): Ticks are vectors for Kyasanur Forest Disease, a tick-borne viral hemorrhagic fever.
III. Mosquitoes (Q. Dengue): Mosquitoes, particularly \(\textit{Aedes aegypti}\), are vectors for dengue fever, a viral disease.

Step 2: Analyze the options.

Option (A): Correct. Matches I-R, II-P, and III-Q correctly.

Option (B): Incorrect. Misassigns fleas (I) to Kyasanur Forest Disease (P).

Option (C): Incorrect. Misassigns ticks (II) to dengue (Q).

Option (D): Incorrect. Misassigns mosquitoes (III) to plague (R). Quick Tip: To remember vector-disease associations: 1. Fleas are primarily associated with the plague.
2. Ticks transmit tick-borne diseases like Kyasanur Forest Disease.
3. Mosquitoes are vectors for viral diseases like dengue, malaria, and Zika.

Step 1: Analyze the utility functions depicted by each graph.Graph P: Shows a steep increase, suggesting high utility gain per item, potentially risk-prone but not as clear-cut as R.
Graph Q: Fluctuates in utility, indicating variable risk and reward, not clearly risk-prone.
Graph R: Depicts a sharp increase in utility, which then becomes even more pronounced with additional items. This pattern exemplifies risk-prone behavior, where the reward significantly increases as more risks are taken.
Graph S: Exhibits complex utility dynamics with multiple peaks, not straightforwardly depicting risk-prone behavior.

Step 2: Conclusion.
Graph R accurately represents a scenario where risk-prone foraging is most likely, as it shows a utility function where rewards increase significantly as additional items are consumed, aligning with the theoretical prediction for risk-prone behavior. Quick Tip: In behavioral ecology, when assessing foraging strategies through utility functions, identify the graph where the increase in utility sharply accelerates with additional food items as indicative of risk-prone strategies.

Step 1: Understand the definition of average species fitness (\(\varphi\)).
The average fitness, \(\varphi\), is determined by the weighted contributions of each species' growth rate to the total population, based on their relative abundances. Since \(x + y = 1\), the abundances \(x\) and \(y\) can be treated as the respective proportions of the two species in the population.

Step 2: Derive the expression for \(\varphi\).
The average fitness is given by the sum of the contributions of both species: \[ \varphi = \alpha x + \beta y, \]
where:
- \(\alpha x\) represents the contribution of species X (growth rate \(\alpha\) multiplied by its proportion \(x\)), and
- \(\beta y\) represents the contribution of species Y (growth rate \(\beta\) multiplied by its proportion \(y\)).

Step 3: Evaluate the options.

Option (A): Incorrect. This expression introduces terms that do not correspond to the definition of average fitness.

Option (B): Correct. This matches the derived expression for average fitness, \(\varphi = \alpha x + \beta y\).

Option (C): Incorrect. The denominator \(x^2 + y^2\) is unnecessary and does not fit the definition of average fitness.

Option (D): Incorrect. The reciprocal of \(\alpha x + \beta y\) is not relevant to the calculation of \(\varphi\). Quick Tip: To calculate average fitness in population models: 1. Use weighted contributions based on growth rates and proportions.
2. Verify that the total population remains constant, ensuring \(x + y = 1\).
3. Avoid introducing unnecessary terms or operations.

Step 1: Understand the Allee effect.
The Allee effect describes a phenomenon where a population’s growth rate decreases when the population size is very small. This occurs due to challenges such as difficulty finding mates or cooperative behaviors not being effective at low densities. A strong Allee effect results in a critical population size below which the population declines to extinction.

Step 2: Identify the characteristics of the growth curve.
A population with a strong Allee effect will exhibit:
1. Negative growth (\(\frac{dN}{dt} < 0\)) for very small \(N\), as the population cannot sustain itself.
2. Positive growth (\(\frac{dN}{dt} > 0\)) as \(N\) increases beyond a critical threshold.
3. A peak in growth rate at an intermediate population size.
4. Declining growth (\(\frac{dN}{dt} \to 0\)) as the population approaches carrying capacity.

Step 3: Analyze the graphs.Graph P: Correct. This graph represents the strong Allee effect, showing negative growth for small \(N\), a critical threshold, and a peak at intermediate \(N\).

Graph Q: Incorrect. This graph shows logistic growth, which does not include negative growth at small \(N\).

Graph R: Incorrect. This graph shows continuous positive growth rates at all population sizes, inconsistent with the Allee effect.

Graph S: Incorrect. This linear growth pattern does not capture the density-dependent dynamics of the Allee effect. Quick Tip: To identify an Allee effect: 1. Look for negative growth at very small population sizes.
2. A strong Allee effect includes a critical threshold population size below which the population declines.
3. Growth peaks at intermediate population sizes before declining near carrying capacity.

Step 1: Analyze the graph dynamics.

The graph displays the relationship between habitat destruction \( h \) and plant species abundance \( X \). It illustrates stable equilibria (solid curve) and points of instability (dashed curve) for the population. The key is to determine where the transition from stability to instability occurs, indicating a potential collapse.

Step 2: Identify the transition point for sudden collapse.

Upon careful examination, the critical transition from a stable to an unstable state—the threshold at which a sudden collapse is likely to happen—occurs at the value \( h = 2.5 \). This is the point where the stable population's last solid equilibrium point exists before turning into the dashed line, representing unstable conditions leading to a population collapse. Quick Tip: In population dynamics, particularly those influenced by environmental stressors, it's crucial to recognize threshold points on stability graphs. These thresholds often signal significant ecological shifts or tipping points that can dramatically affect population sustainability.

Step 1: Understand the graph.
The graph represents the relationship between \(\log(S)\) and \(\log(A)\), with a straight line indicating that the relationship between \(S\) and \(A\) on a logarithmic scale is linear. The equation for such a relationship is: \[ \log(S) = c + m \log(A), \]
where \(c\) is the intercept, and \(m\) (the slope) is not equal to 1.

Step 2: Convert to the untransformed relationship.
Rewriting the equation in its untransformed form: \[ S = k A^m, \]
where \(k = 10^c\). This equation describes a power law relationship between \(S\) (species richness) and \(A\) (area), where \(m\) determines the scaling.

Step 3: Evaluate the options.Option (A): Incorrect. A linear relationship implies \(S \propto A\), which is not consistent with the power law form \(S = k A^m\).

Option (B): Correct. The equation \(S = k A^m\) matches the definition of a power law.

Option (C): Incorrect. An exponential relationship would be of the form \(S = k e^{mA}\), which is not implied here.

Option (D): Incorrect. A Michaelis-Menten function describes saturation dynamics, not a simple power law. Quick Tip: To analyze log-log graphs: 1. A straight line on a log-log scale indicates a power law relationship.
2. The equation \(S = k A^m\) is derived from \(\log(S) = c + m \log(A)\).
3. Recognize the difference between linear, exponential, and power law relationships based on scaling behavior.

Step 1: Understand rank-abundance relationships.
The slope of the rank-abundance curve indicates the evenness of a community:
A shallower slope represents greater evenness, as species abundances are more evenly distributed.

A steeper slope indicates lower evenness, as a few species dominate the community.

Step 2: Analyze the graph.
Community \(P\): This has the steepest slope, indicating the least evenness.

Community \(Q\): This has a moderate slope, indicating intermediate evenness.

Community \(R\): This has the shallowest slope, indicating the highest evenness.

Step 3: Compare evenness among the communities.
The evenness order based on the slopes is: \[ R > Q > P \]

Step 4: Evaluate the options.

Option (A): Incorrect. It suggests \(P\) has the highest evenness, which contradicts the steep slope of its curve.

Option (B): Incorrect. It suggests \(Q\) has the highest evenness, but \(R\) has a shallower slope.

Option (C): Correct. This matches the observed order \(R > Q > P\).

Option (D): Incorrect. It misplaces \(P\) as having higher evenness than \(Q\). Quick Tip: To interpret rank-abundance curves: 1. Shallow slopes indicate higher evenness.
2. Steep slopes suggest dominance by a few species, leading to lower evenness.
3. Compare slopes visually to rank communities by evenness.

Step 1: Interpret the graph.
The graph shows:
Before the replacement, the contiguous forest had higher species richness than the forest fragment.
After the replacement of the contiguous forest with an oil palm plantation, the species richness in the forest fragment drastically decreased.

Step 2: Analyze the ecological dynamics.
The decrease in species richness in the forest fragment after the loss of the contiguous forest suggests that:
1. The contiguous forest acted as a source habitat, providing species that dispersed into the forest fragment.
2. The removal of the contiguous forest disrupted dispersal, leading to a loss of species in the forest fragment

Step 3: Evaluate the options.Option (A): Incorrect. A sink is a habitat where populations decline without immigration. The contiguous forest was likely a source, not a sink.

Option (B): Incorrect. The forest fragment initially had lower species richness than the contiguous forest, as shown in the graph.

Option (C): Incorrect. A geographically closed community would not depend on dispersal from the contiguous forest, contradicting the observed pattern.

Option (D): Correct. The contiguous forest contributed to the species richness of the forest fragment via dispersal. Its removal caused a decline in species richness in the fragment. Quick Tip: To interpret habitat connectivity: 1. Source habitats contribute species to adjacent fragments through dispersal.
2. Fragmented landscapes often depend on connectivity to maintain biodiversity.
3. Habitat loss can disrupt dispersal, leading to declines in species richness.

Step 1: Understand haplodiploidy in honey bees.
Haplodiploidy is a sex-determination system where:

1. Males (drones) are haploid and develop from unfertilized eggs.

2. Females (workers and queens) are diploid and develop from fertilized eggs.

Sisters (workers) share:
- 50% of their genes from their mother (due to random assortment during meiosis).
- 100% of their genes from their haploid father (as all sperm from a haploid individual are identical).

The total relatedness between sisters is: \[ r = 0.5 (from \ mother) + 0.5 \times 1 (from \ father) = 0.75 \]

Step 2: Analyze relatedness in other pairings.
Brother-brother pairs: Relatedness is 0.5, as they share the same haploid mother.

Brother-sister pairs: Relatedness is 0.25, as the brother contributes no paternal genes.

Mated female-male pair: Relatedness is 0, as unrelated individuals mate.

Sister-sister pairs: Relatedness is 0.75, as calculated above.

Step 3: Evaluate the options.

Option (A): Incorrect. Brother-brother relatedness is 0.5.

Option (B): Incorrect. Brother-sister relatedness is 0.25.

Option (C): Incorrect. Mated female-male relatedness is 0.

Option (D): Correct. Sister-sister relatedness is 0.75. Quick Tip: To calculate relatedness in haplodiploid systems: 1. Sisters share 100% of paternal genes and 50% of maternal genes, resulting in a total relatedness of 0.75.
2. Brothers share only maternal genes (r = 0.5), and brother-sister relatedness is 0.25.
3. Use haplodiploid genetics to explain eusocial behavior in bees.

Step 1: Match each mollusc taxa to the correct order based on common biological classification.Cone snails (I) are best classified in the order Gastropod (Q).
Octopuses (II) belong to the order Cephalopod (R).
Giant clams (III) are included in the order Bivalve (P).
Squids (IV) are also classified under the order Cephalopod (R).

Quick Tip: When matching organisms to their taxonomic orders, rely on distinctive characteristics that define each group, such as foot type in molluscs, which is a key feature distinguishing these orders.

Step 1: Examine the haplotype networks.
The network for species P shows fewer haplotypes with limited mutations, indicating fewer genetic variations and suggesting a low dispersal ability across large distances. In contrast, species Q's network exhibits more haplotypes with more extensive mutations and connections, indicative of a higher dispersal ability.

Step 2: Relate haplotype diversity to dispersal ability.
Species Q’s broader range of haplotypes and the presence in diverse marine environments supports a higher dispersal capability. Species P's restricted haplotype spread between only two distant regions points to a more limited dispersal capacity. Quick Tip: When analyzing haplotype networks, consider both the number of haplotypes and their connectivity to infer the dispersal ability of species. More connected and numerous haplotypes generally suggest a higher dispersal potential.

Step 1: Analyze the structure of Tree R.
Tree R begins with a split indicating a significant geographical or ecological separation between the grey langurs of SL and SWG. The lineage then further diversifies in SL to include purple-faced langurs, while maintaining grey langurs in SWG. This suggests that allopatric speciation influenced by geographical isolation led to the diversification within these regions.

Step 2: Justify the selection of Tree R.
The configuration of Tree R accurately represents how physical barriers and ecological factors could drive early diversification, resulting in distinct yet genetically related groups within the grey and purple-faced langurs. It effectively captures the essence of allopatric speciation and the evolutionary history as suggested by the similarities and distribution of the langur populations. Quick Tip: Phylogenetic analysis often hinges on understanding the geographic and ecological contexts that contribute to species diversification. In cases of allopatric speciation, look for tree structures that reflect clear separations likely caused by geographical barriers.

Step 1: Analyze the ecological context.
At 1250 m ASL, \(A\) and \(B\) coexist and show distinct bill morphologies, likely to reduce competition. However, their morphologies are similar at non-overlapping elevations, indicating the differences at 1250 m are ecologically driven.

Step 2: Evaluate the processes.
Competitive exclusion: Unlikely, as both species coexist.

Character displacement: Likely, as sympatric divergence reduces competition.

Convergent evolution: Explains similarity at non-overlapping elevations but not divergence at 1250 m.

Allopatric speciation: Does not explain divergence in sympatry. Quick Tip: To understand trait divergence:
1. Sympatric populations often show divergence to reduce competition.
2. Compare traits in overlapping and non-overlapping ranges to identify ecological processes.

Step 1: Define greenhouse gases.
Greenhouse gases (GHGs) trap heat in the Earth's atmosphere, contributing to the greenhouse effect. Major greenhouse gases include carbon dioxide (\(CO_2\)), methane (\(CH_4\)), nitrous oxide (\(N_2O\)), and water vapour (\(H_2O\)).

Step 2: Analyze the options.
Option (A): Methane (\(CH_4\)) is a potent greenhouse gas with a high global warming potential.

Option (B): Water vapour (\(H_2O\)) is the most abundant greenhouse gas, contributing significantly to the natural greenhouse effect.

Option (C): Sulphur dioxide (\(SO_2\)) is not a greenhouse gas. Instead, it can cause cooling by forming aerosols that reflect sunlight.

Option (D): Nitrous oxide (\(N_2O\)) is a significant greenhouse gas with a long atmospheric lifetime. Quick Tip: To identify greenhouse gases:
1. Focus on gases that trap heat in the atmosphere.
2. Key examples include \(CO_2\), \(CH_4\), \(N_2O\), and \(H_2O\).
3. Sulphur dioxide (\(SO_2\)) is not a greenhouse gas but contributes to atmospheric cooling.

Step 1: Evaluate the ultimate nature of Option (A).
Option (A) proposes that differences in female mating preferences across populations have led to variations in male song length. This is considered an ultimate explanation because it addresses the evolutionary reasons—specifically, sexual selection—that could lead to genetic differentiation and behavioral adaptations in song length among populations.

Step 2: Implications of sexual selection.
Sexual selection is a form of natural selection where certain traits become preferred by one sex, in this case, females. These preferences can drive significant evolutionary changes in populations, leading to traits that are advantageous in specific mating contexts but may vary geographically depending on local preferences. Quick Tip: Understanding the role of sexual selection in evolution helps explain why certain traits, like song length in birds, may vary significantly between populations, reflecting differences in mating strategies and female preferences.

Analysis of Experimental Data:
The experiment involved varying both temperature and the presence of human scent, yielding the following responses:
- At ambient temperature with human scent, there was a high response (90%), indicating attraction.
- At ambient temperature without human scent, there was no response (0%), showing lack of attraction.
- At human body temperature with or without human scent, the response was high (90%), suggesting strong attraction due to temperature.

Inferences Drawn:
- Human scent alone is sufficient to attract mosquitoes, as evidenced by the response at lower temperatures.
- Human body temperature alone is also sufficient, as mosquitoes responded highly even without the human scent at body temperature. Quick Tip: When evaluating sensory cues for vector-host interactions, consider both independent and combined effects of environmental factors such as temperature and scent to fully understand their roles in host detection.

Step 1: Reevaluate Homology in Phylogenetic Context.
Understanding homology requires assessing both the genetic and evolutionary context within which traits occur:
Species 1 and 3: Despite differences in pigmentation presence, their placement within the same clade suggests a shared evolutionary background for the trait.
Species 2 and 6: Although both lack pigmentation, their separate clade origins and potential genetic causes support non-homologous development.

Step 2: Conclusion Based on Phylogenetic Analysis.
Homology of Pigmentation between Species 1 and 3: Supported by their common clade positioning and ancestral trait inheritance.
Non-Homology of Pigmentation between Species 2 and 6:Supported by their distinct clade origins and differing genetic influences.

Quick Tip: When analyzing phylogenetic trees, consider both the shared clade and potential evolutionary events that may lead to trait variations, such as gene mutations or silencing, to determine homology accurately.

Step 1: Understanding effective population size (\( N_e \)).
The effective population size, \( N_e \), represents the number of individuals in a population that contribute to the next generation's gene pool. It is influenced by factors such as the number of breeding individuals, sex ratio, and variation in reproductive success.

Step 2: Evaluate the options.
Option (A): Incorrect. The population size required to avoid extinction over a given time frame is not directly related to \( N_e \).

Option (B): Incorrect. Carrying capacity refers to the maximum number of individuals the environment can support and does not directly influence \( N_e \).

Option (C): Incorrect. While metapopulation size is important in conservation, \( N_e \) is calculated for individual populations and not the sum of all connected populations.

Option (D): Correct. The number of breeding males and females directly affects \( N_e \), as it determines the genetic contribution to future generations. Quick Tip: To calculate \( N_e \), consider:
1. The number of breeding individuals.
2. The sex ratio of the population.
3. Variance in reproductive success among individuals.

Focused Analysis of Diagram H for Intraguild Predation:
- Diagram H distinctly illustrates intraguild predation, where C1 and C2 consume the same primary producer R and C1 additionally preys on C2. This scenario shows C1 affecting the population dynamics of C2 both through competition and predation, making it a textbook case of intraguild predation. Quick Tip: In analyzing food web diagrams for intraguild predation, identify cases where a consumer not only competes with but also consumes another consumer at the same trophic level. This interaction reveals more about the stability and dynamics of ecological communities.

Analysis of Morphological Differences:
The observation that all plants grown in the greenhouse exhibit an upright form, irrespective of their seed source, strongly suggests that the environmental conditions play a significant role in determining plant morphology. This supports the idea of phenotypic plasticity, where the same genetic makeup can express different phenotypes under different environmental conditions. Quick Tip: Phenotypic plasticity allows organisms to adjust their physiology or morphology in response to environmental conditions, showcasing an adaptive advantage in varying habitats.

Step 1: Analyze the role of low seasonality in species richness.
Low seasonality in the tropics provides a stable environment that supports resource specialization, population stability, and co-existence, contributing to high species richness.

Step 2: Evaluate the options.
Option (A): Correct. Consistent resource availability throughout the year reduces competition and enables species to specialize, fostering co-existence and increasing species richness.

Option (B): Incorrect. Lower predation rates are not directly associated with low seasonality and do not explain increased species richness.

Option (C): Correct. Stable populations in aseasonal environments are less affected by demographic stochasticity and extinction, leading to higher species richness.

Option (D): Incorrect. Longer generation times are not a direct consequence of low seasonality and do not enhance species richness.

Conclusion:
The correct mechanisms are \( \mathbf{(A)} \) and \( \mathbf{(C)} \), as they explain how low seasonality contributes to higher species richness in the tropics. Quick Tip: To understand species richness in the tropics:
1. Focus on the stability and consistency of resources in aseasonal environments.
2. Recognize how reduced demographic stochasticity enhances species survival and diversification.

Analysis of Zinc Tolerance Variation:
The significant variation in zinc tolerance from high within the mine to low in the surrounding pasture is indicative of local adaptation. The plants inside the mine likely evolved a high tolerance to zinc due to the heavy metal-rich environment, which is a classic example of local adaptation to extreme environmental conditions.

Reasons for Excluding Other Options:Genetic drift typically impacts small populations and does not usually result in such distinct, adaptive traits that respond directly to environmental pressures.
Coevolution is not applicable as there is no interaction with other species that influences zinc tolerance.
Introgression might influence genetic diversity but there's no indication of hybridization affecting the zinc tolerance trait in this scenario.

Quick Tip:
Local adaptation often leads to marked differences in traits within species across different environments, especially when these environments impose significant selective pressures, such as varying levels of toxins or nutrients.

Step 1: Using the principle of inclusion-exclusion.
The formula for the union of two events is: \[ P(Hunting a hare or a deer) = P(Hare) + P(Deer) - P(Hare and Deer). \]
Given: \[ P(Hare or Deer) = 0.55, \quad P(Hare) = 0.35, \quad P(Deer) = 0.25. \]

Substitute the values into the formula: \[ 0.55 = 0.35 + 0.25 - P(Hare and Deer). \]

Step 2: Solve for \( P(Hare and Deer) \).
Rearrange the equation: \[ P(Hare and Deer) = 0.35 + 0.25 - 0.55 = 0.05. \] Quick Tip: For probability questions involving "or" conditions, use the inclusion-exclusion principle: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \] Rearrange to find the probability of simultaneous events if needed.

Step 1: Sum of probabilities in a probability distribution.For any discrete random variable, the sum of probabilities of all possible outcomes must equal 1: \[ Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) = 1. \]

Step 2: Substitute the known probabilities.Given: \[ Pr(X = 0) = 0.15, \quad Pr(X = 1) = 0.25, \quad Pr(X = 3) = 0.5. \]
Substitute these values: \[ 0.15 + 0.25 + Pr(X = 2) + 0.5 = 1. \]

Step 3: Solve for \( Pr(X = 2) \).
Simplify the equation: \[ Pr(X = 2) = 1 - (0.15 + 0.25 + 0.5). \] \[ Pr(X = 2) = 1 - 0.9 = 0.10. \] Quick Tip: For probability distributions:
1. Verify that the sum of all probabilities equals 1.
2. Use this property to find missing probabilities by subtracting the sum of known values from 1.

Step 1: Understand the problem.
We are selecting 6 species from a total of 9 species, where the order of selection does not matter. This is a problem of combinations.

Step 2: Apply the combination formula.
The number of ways to choose \( r \) objects from \( n \) objects is given by the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!}. \]
Here, \( n = 9 \) (total species) and \( r = 6 \) (species to be selected).

Step 3: Calculate \( \binom{9}{6} \). \[ \binom{9}{6} = \frac{9!}{6!(9-6)!} = \frac{9!}{6! \cdot 3!}. \]Simplify: \[ \binom{9}{6} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = 84. \]

Quick Tip: To solve problems involving combinations:
1. Use the formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).
2. Simplify factorials step-by-step to avoid errors in large computations.