GATE 2024 Ecology and Evolution Question Paper is available for download. The exam was successfully conducted by IISc/IITs on February 4 in the afternoon session from 2:30 PM to 5:30 PM. As per the student’s initial reactions, the GATE 2024 Ecology and Evolution Question Paper was reported as Moderate. The Ecology section was reported as Moderate to Challenging, the Evolution section as Moderate, and the General Aptitude section as Easy to Moderate.
GATE 2024 Ecology and Evolution Question Paper with Answer Key PDF
Candidates can download the GATE 2024 Ecology and Evolution Question Paper with Answer Key PDFs using the link below.
GATE Ecology and Evolution 2024 Questions with Solutions
GENERAL APTITUDE
Question 1:
If ‘→’ denotes increasing order of intensity, then the meaning of the words [simmer → seethe → smolder] is analogous to [break → raze → ]. Which one of the given options is appropriate to fill the blank?
(A) obfuscate
(B) obliterate
(C) fracture
(D) fissure
Correct Answer: (B) obliterate
View Solution
- Step 1: Analyze the given analogy. The words simmer → seethe → 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.
This relationship indicates increasing intensity in terms of energy or destruction.
- Step 2: Analyze the second analogy. The words break → 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.
Question 2:
In a locality, the houses are numbered in the following way:The house-numbers on one side of a road are consecutive odd integers starting from 301, while the house-numbers on the other side of the road are consecutive even numbers starting from 302. The total number of houses is the same on both sides of the road. If the difference of the sum of the house-numbers between the two sides of the road is 27, then the number of houses on each side of the road is:
(A) 27
(B) 52
(C) 54
(D) 26
Correct Answer: (A) 27
View Solution
- 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, . . . ,(301 + 2(n − 1)).The sum of these house numbers, Sodd, is given by the formula for the sum of an arithmetic progression: Sodd= n * (First term + Last term)/2. The last term is: 301 + 2(n − 1) = 301 + 2n − 2 = 299 + 2n. Sodd= n * (301 + (299 + 2n))/2 = n * (600 + 2n)/2 = n * (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, . . . ,(302 + 2(n − 1)). The sum of these house numbers, Seven, is: Seven= n * (302 + (302 + 2(n − 1)))/2. Seven= n * (302 + (300 + 2n))/2 = n * (602 + 2n)/2 = n * (301 + n).
- Step 2: Difference between the sums. The difference between the sums of the two sides of the road is given as 27: Seven − Sodd= 27. Substitute the expressions for Seven and Sodd: n * (301 + n) − n * (300 + n) = 27. Simplify: n(301 + n − 300 − n) = 27. n * 1 = 27. Therefore, n = 27.
Question 3:
For positive integers p and q, with p/q ≠ 1, (pq)/q = pq-1. Then, which of the following is true?
(1) qp= pq
(2) qp= p2q
(3) √q = √p
(4) pq= qp
Correct Answer: (4) p
q= q
p
View Solution
- We are given the equation: (pq)/q = pq-1. Multiplying both sides by q to clear the denominator : pq = q*pq-1. Now divide both sides by pq-1(assuming p ≠ 0) : pq/pq-1 = q, which simplifies to pq-(q-1) = q, thus p = q1/q . This implies that pq = qp. Thus, the correct answer is option (4).
Question 4:
Which one of the given options is a possible value of x in the following sequence? 3, 7, 15, x, 63, 127, 255
(A) 35
(B) 40
(C) 45
(D) 31
Correct Answer: (D) 31
View Solution
- 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 → 7 = 2 × 3 − 1, 7 → 15 = 2 × 7 − 1, 15 → x = 2 × 15 − 1. The next term is: x = 2 × 15 − 1 = 30 − 1 = 31.
- Step 2: Verify the continuation of the pattern. Check the subsequent terms using x = 31: 31 → 63 = 2 × 31 − 1, 63 → 127 = 2 × 63 − 1, 127 → 255 = 2 × 127 − 1. The pattern holds, confirming that x = 31.
Question 5:
On a given day, how many times will the second-hand and the minute-hand of a clock cross each other during the clock time 12:05:00 hours to 12:55:00 hours?
(1) 51
(2) 49
(3) 50
(4) 55
Correct Answer: (1) 51
View Solution
- 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.
Question 6:
In the given text, the blanks are numbered (i)–(iv). Select the best match for all the blanks.
Text: From the ancient Athenian arena to the modern Olympic stadiums, athletics (i) the potential for a spectacle. The crowd (ii) with bated breath as the Olympian artist twists his body, stretching the javelin behind him. Twelve strides in, he begins to cross-step. Six cross-steps (iii) in an abrupt stop on his left foot. As his body (iv) like a door turning on a hinge, the javelin is launched skyward at a precise angle.
(A) (i) hold (ii) waits (iii) culminates (iv) pivot
(B) (i) holds (ii) wait (iii) culminates (iv) pivot
(C) (i) hold (ii) wait (iii) culminate (iv) pivots
(D) (i) holds (ii) waits (iii) culminate (iv) pivots
Correct Answer: (D) (i) holds (ii) waits (iii) culminate (iv) pivots
View Solution
- 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.
Question 7:
Three distinct sets of indistinguishable twins are to be seated at a circular table that has 8 identical chairs. Unique seating arrangements are defined by the relative positions of the people. How many unique seating arrangements are possible such that each person is sitting next to their twin?
(1) 12
(2) 14
(3) 10
(4) 28
Correct Answer: (1) 12
View Solution
- 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 * 1 = 12. 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: 12.
Question 8:
The chart given below compares the Installed Capacity (MW) of four power generation technologies, T1, T2, T3, and T4, and their Electricity Generation (MWh) in a time of 1000 hours (h). The Capacity Factor of a power generation technology is:
![Installed Capacity (MW) of four power generation technologies,]()
Capacity Factor = Electricity Generation (MWh) / (Installed Capacity (MW) × 1000 (h))
Which one of the given technologies has the highest Capacity Factor?
(A) T1
(B) T2
(C) T3
(D) T4
Correct Answer: (D) T4
View Solution
- Step 1: Calculation of Capacity Factor for each technology. Using the formula for Capacity Factor and the values extracted from the chart: CFT1 = 0.0005, CFT2 = 0.0005385, CFT3 = 0.0002727, CFT4 = 0.0009
Question 9:
In the 4 × 4 array shown below, each cell of the first three columns has either a cross (X) or a number, as per the given rule.
![4 × 4 array shown below, each cell of the first three columns]()
Rule: The number in a cell represents the count of crosses around its immediate neighboring cells (left, right, top, bottom, diagonals).
As per this rule, the maximum number of crosses possible in the empty column is:
(A) 0
(B) 1
(C) 2
(D) 3
Correct Answer: (C) 2
View Solution
- 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.
Question 10:
During a half-moon phase, the Earth-Moon-Sun form a right triangle. If the Moon-Earth-Sun angle at this half-moon phase is measured to be 89.85°, the ratio of the Earth-Sun and Earth-Moon distances is closest to:
(A) 328
(B) 382
(C) 238
(D) 283
Correct Answer: (B) 382
View Solution
- 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 ∠MES = 89.85°. - The Earth-Moon distance (dEM) as one leg of the triangle. - The Earth-Sun distance (dES) as the hypotenuse.
- Step 2: Trigonometric relation. Using the cosine rule in the right triangle: cos(∠MES) = dEM / dES . Rearranging for the ratio dES / dEM: dES / dEM= 1 / cos(89.85°).
- Step 3: Calculate cos(89.85°). Since 89.85° is very close to 90°, cos(89.85°) can be approximated using a calculator: cos(89.85°) ≈ 0.002617.
- Step 4: Compute the ratio. Substitute cos(89.85°) ≈ 0.002617: dES/dEM= 1 / 0.002617 ≈ 382.
Ecology and Evolution
Question 11:
The molecular clock model assumes that mutation rates are:
(A) equal for all genes.
(B) constant for a gene.
(C) variable across geographical regions.
(D) variable across geological time.
Correct Answer: (B) constant for a gene.
View Solution
- 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.
Question 12:
The intermediate disturbance hypothesis was proposed to explain patterns of:
(A) species redundancy.
(B) species diversity.
(C) species dispersal.
(D) species extinctions.
Correct Answer: (B) species diversity.
View Solution
- 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.
Question 13:
A few years ago, a very small population of zebrafish became isolated by a newly built dam. As a result, which statement is most likely to be true about this population of zebrafish now?
(A) Genetic variability is low.
(B) Fixation of genotypes due to drift is low.
(C) Inbreeding is low.
(D) Mutation rate is high.
Correct Answer: (A) Genetic variability is low.
View Solution
- 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.
Question 14:
A researcher measures the heights of 200 randomly selected individuals of a tree species in a forest. Which one of the following is NOT a measure of variability in the sample?
(A) Inter-quartile range
(B) Range
(C) Standard deviation
(D) Standard error
Correct Answer: (D) Standard error
View Solution
- 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 = SD / √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.
Question 15:
Individual lizards were repeatedly presented with a predator model. Over successive trials, they showed a reduction in the duration of their alarm response. Which one of the following is this an example of?
(A) Imitation.
(B) Imprinting.
(C) Habituation.
(D) Sensitisation.
Correct Answer: (C) Habituation.
View Solution
- 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.
Question 21:
A classical metapopulation at equilibrium is made up of local populations with:
(A) no dispersal between them.
(B) no local colonisation or extinction.
(C) weak dispersal between them.
(D) panmictically breeding individuals across populations.
Correct Answer: (C) weak dispersal between them.
View Solution
- 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.
Question 22:
Which one of the following theories is supported by the distribution patterns of extinct flora such as Glossopteris across South America, Africa, and Australia, and extant marsupial mammals across South America and Australia?
(A) Darwin’s theory of natural selection
(B) Wegener’s theory of continental drift
(C) Levins’ theory of metapopulations
(D) MacArthur and Wilson’s theory of island biogeography
Correct Answer: (B) Wegener’s theory of continental drift
View Solution
- Step 1: Understanding the context. The distribution of extinct flora like 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 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.
Question 23:
In linear regression, mean squared regression (effect variance) divided by mean squared error (error variance) is called the:
(A) p-value.
(B) F-statistic.
(C) t-statistic.
(D) R-squared value.
Correct Answer: (B) F-statistic.
View Solution
- 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 = 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.
Question 24:
The figure shows the time-series of atmospheric CO2 concentration on Earth (graph not-to-scale).
Which one of the factors given is the primary reason for the sudden increase in atmospheric CO2 concentration after 1950?
![time-series of atmospheric CO2 concentration]()
(A) Overfishing
(B) An increase in Arctic sea ice melting
(C) An increase in fossil fuel burning
(D) Volcanic eruptions
Correct Answer: (C) An increase in fossil fuel burning
View Solution
- Step 1: Understand the CO2 concentration trend. The graph shows a rapid increase in atmospheric CO2 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 CO2 levels. Option (B): Incorrect. Arctic sea ice melting is a consequence of global warming rather than a cause of increased CO2 levels. Option (C): Correct. Fossil fuel burning releases large amounts of CO2 into the atmosphere, which is the primary reason for the sudden increase in CO2 levels post-1950. Option (D): Incorrect. Volcanic eruptions release CO2, but their contribution to the observed increase post-1950 is negligible compared to human activities.
Question 25:
The population size at which net recruitment is the highest is also when the greatest amount can be harvested, while ensuring the long-term survival of the population. The amount harvested at this population size is known as:
(A) carrying capacity.
(B) maximum sustainable yield.
(C) maximum survival density.
(D) optimal recruitment.
Correct Answer: (B) maximum sustainable yield.
View Solution
- 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.
Question 26:
The variance in male mating success is Vm and that of females is Vf . Assuming that the sex ratio is 1:1, in which one of the following mating systems is Vm/Vf expected to be the greatest?
(A) Monogamy
(B) Random mating
(C) Polyandry
(D) Polygyny
Correct Answer: (D) Polygyny
View Solution
- Step 1: Understand variance in mating success. The variance in mating success (Vm and Vf ) describes how unequally individuals of a sex contribute to reproduction. The ratio Vm/Vf 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, Vm/Vf 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. Vm/Vf is not expected to be very large. Polyandry (Option C): In polyandry, females mate with multiple males. This increases Vf (female variance in mating success) relative to Vm, leading to a small Vm/Vf ratio. Polygyny (Option D): In polygyny, some males mate with many females while others mate with none. This creates a high Vm (male variance in mating success) compared to Vf , leading to the highest Vm/Vf ratio among the options.
Question 27:
Some air-breathing marine vertebrates such as whales, seals, and marine turtles possess adaptations for long, deep dives. Which one or more of the following is/are examples of such adaptations?
(A) Tolerance to hypoxia
(B) Slow heart rate
(C) High levels of haemoglobin
(D) Salt tolerance
Correct Answer: (A) Tolerance to hypoxia
(B) Slow heart rate, and(C) High levels of haemoglobin
View Solution
- 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.
Question 28:
Which one or more of the following statements about evolution is/are true?
(A) Evolution is change that is heritable across generations.
(B) Evolution occurs at the level of populations, not species.
(C) Evolution is a change in gene frequencies through time.
(D) Evolution occurs through natural selection, but not sexual selection.
Correct Answer: (A) Evolution is change that is heritable across generations.
(B) Evolution occurs at the level of populations, not species.
(C) Evolution is a change in gene frequencies through time.
View Solution
- 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.
Question 29:
Which one or more of the following mammal species is/are endemic to India?
(A) One-horned rhinoceros
(B) Lion-tailed macaque
(C) Bengal tiger
(D) Cheetah
Correct Answer: (B) Lion-tailed macaque
View Solution
- 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.
Question 30:
Under which one or more of the following conditions can altruism evolve in animal societies?
(A) Individuals in a group are closely related to each other.
(B) Individuals live in a high resource, low risk environment.
(C) Individuals in a group mutually help each other at different times.
(D) Mating opportunities are equally distributed among individuals.
Correct Answer: (A) Individuals in a group are closely related to each other. and (C) Individuals in a group mutually help each other at different times.
View Solution
- 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.
Question 31:
Two species of fruit bats (Species 1 and Species 2) eat fruits of varying sizes. The curves shown represent the ecological niche for these two species. If the curves for both species were to completely overlap, which one or more of the statements given would be correct?
![Two species of fruit bats]()
(A) There will be no resource competition between Species 1 and Species 2.
(B) One of the species may become extinct due to competitive exclusion.
(C) There will be little competition between Species 1 and Species 2.
(D) The two species will use identical resources.
Correct Answer: (B) One of the species may become extinct due to competitive exclusion. and (D) The two species will use identical resources.
View Solution
- 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.
Question 32:
During the process of succession in a community, species that are good colonizers are gradually replaced by species that are good competitors. Which one or more of the following statements is/are consistent with this pattern?
(A) Initially, there is great resource limitation.
(B) Keystone species must establish first to facilitate the later establishment of higher trophic level species.
(C) Trees are the climax stage of terrestrial communities and generally have low competitive ability, but high dispersal ability.
(D) For many taxa, there is a tradeoff between dispersal ability and local competitive ability.
Correct Answer: (D) For many taxa, there is a tradeoff between dispersal ability and local competitive ability.
View Solution
- 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.
Question 33:
An ornamental shrub species was brought from Japan in the early 1800s to India, where it was planted frequently in gardens and parks. The species persisted for many decades without spreading and then began to spread invasively fifty years ago. Which one or more of the following processes could have led to it becoming invasive?
(A) Evolutionary adaptation to the environment
(B) Open niches due to recent habitat degradation
(C) Climate change
(D) Recent introduction of a specialized herbivore of this shrub species
Correct Answer: (A) Evolutionary adaptation to the environment
(B) Open niches due to recent habitat degradation, and (C) Climate change
View Solution
- 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.
Question 34:
Male voles pair with either a single female (monogamous) or with two females (polygynous) during a given breeding season. The probability of a male being polygynous in a breeding season is 0.2. The reproductive success (number of offspring) of monogamous males is 2, and of polygynous males is 3. A male’s expected reproductive success in a breeding season is . (Round off to one decimal place)
Correct Answer: 2.2
View Solution
- 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 = (Pmonogamous × Successmonogamous) + (Ppolygynous × Successpolygynous) where: - Pmonogamous = 1−Ppolygynous = 1−0.2 = 0.8, - Successmonogamous = 2, - Successpolygynous = 3, and - Ppolygynous = 0.2.
- Step 2: Substitute the values and calculate. E = (0.8 × 2) + (0.2 × 3) E = 1.6 + 0.6 = 2.2
Question 35:
Consider a randomly breeding population of squirrels with two morphs – white-striped and brown-striped. In a population, 16% are white-striped individuals, while the rest are all brown-striped. The trait for stripes is governed by one gene where the allele for brown stripes is dominant. Assuming Hardy–Weinberg equilibrium, the frequency of the allele for white stripes would be . (Round off to two decimal places)
Correct Answer: 0.40
View Solution
- Step 1: Understand the problem. The trait for white stripes is recessive, and 16% of the population is white-striped (q2= 0.16). To find the frequency of the allele for white stripes (q), we take the square root of q2: q = √q2 = √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: p2 + 2pq + q2= 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.
Question 36:
Observations of algal species showed that their diversity was higher in pools where there were grazing snails compared to pools without snails. Which one of the following statements best explains this result?
(A) Snails feed preferentially on the more abundant algal species.
(B) Snails avoid feeding on algal species.
(C) Snails feed only on the less abundant algal species.
(D) Snails feed equally on all the algal species irrespective of algal abundance.
Correct Answer: (A) Snails feed preferentially on the more abundant algal species.
View Solution
- 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.
Question 37:
Which two of the following processes can result in a decline in heterozygosity in populations?
I) Inbreeding;
II) Genetic drift;
III) Mutation;
IV) Random mating
(A) I and II
(B) II and III
(C) I and III
(D) II and IV
Correct Answer: (A) I and II
View Solution
- 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.
Question 38:
Given below is a table with ecological observations and processes.
Ecological observations Processes
P) Bright spotted pigmentation in guppy males in low predation habitats I) Kin selection
Q) Vampire bats share blood meals II) Sexual selection
R) Cooperative breeding in African weaver birds III) Reciprocal altruism
Select the option that best matches each ecological observation with its corresponding process.
(A) P-III, Q-I, R-II
(B) P-II, Q-III, R-I
(C) P-II, Q-III, R-II
(D) P-II, Q-I, R-III
Correct Answer: (B) P-II, Q-III, R-I
View Solution
- 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.
Question 39:
An ecologist must determine whether (i) the means of two independent samples differ, and (ii) there is an association between two continuous variables.
Assuming that all samples are normally distributed, which one of the following options represents the most appropriate statistical tests for (i) and (ii), respectively?
(A) Spearman’s correlation; (ii) Shapiro-Wilk test
(B) Wilcoxon’s matched pairs signed rank test; (ii) chi-squared test
(C) t-test; (ii) Pearson’s correlation
(D) Kendall’s test of concordance; (ii) Kolmogorov-Smirnov test
Correct Answer: (C) t-test; (ii) Pearson’s correlation
View Solution
- 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.
Question 40:
Males of the swordtail fish Xiphophorus helleri possess long tails, while those of X. maculatus do not. Females of X. helleri prefer males with longer tails. Interestingly, experimental studies show that females of X. maculatus prefer X. maculatus males with attached artificial long tails over those without. If the long-tailed Xiphophorus species evolved from ancestors that lacked a long tail, which one of the following processes best explains the evolution of the observed preference among X. maculatus females?
(A) Kin selection
(B)Sensory bias
(C) Group selection
(D) Runaway selection
Correct Answer: (B) Sensory bias
View Solution
- 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 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 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 X. maculatus males.
Question 41:
Which one of the options given best matches vector to disease?
Vector Disease
I. Fleas P. Kyasanur Forest Disease
II. Ticks Q. Dengue
III. Mosquitoes R. Plague
(A) I-R; II-P; III-Q
(B) I-P; II-R; III-Q
(C) I-R; II-Q; III-P
(D) I-R; II-P; III-R
Correct Answer: (A) I-R; II-P; III-Q
View Solution
- 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 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 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).
Question 42:
Optimal foraging theory predicts whether a foraging animal will be risk-prone, risk-averse or risk-insensitive depending on a utility function that describes the value of each additional food item to the animal. Risk-prone foraging is expected when the utility increases disproportionately with each additional food item encountered. Which one of the graphs shown depicts a scenario where risk-prone foraging would be expected?
![Optimal foraging theory predicts]()
(A) P
(B) Q
(C) R
(D) S
Correct Answer: (C) R
View Solution
- 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.
Question 43:
There are two species, X and Y, with abundances x and y, respectively. Species X has growth rate α, and species Y has growth rate β. Assume that the sum of the species abundances is constant over time, i.e., x + y = 1. Let x and y follow the rate equations:
dx/dt = αx − φx,
dy/dt = βy − φy,
where φ is the average species fitness. Which one of the following options correctly represents the expression for φ?
(A) (αx2+βy2)/(α+β)
(B) αx + βy
(C) (αx+βy)/(x2+y2)
(D) 1/(αx+βy)
Correct Answer: (B) αx + βy
View Solution
- Step 1: Understand the definition of average species fitness (φ). The average fitness, φ, 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 φ. The average fitness is given by the sum of the contributions of both species: φ = αx + βy, where: - αx represents the contribution of species X (growth rate α multiplied by its proportion x), and - βy represents the contribution of species Y (growth rate β 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, φ = αx + βy. Option (C): Incorrect. The denominator x2 +y2is unnecessary and does not fit the definition of average fitness. Option (D): Incorrect. The reciprocal of αx + βy is not relevant to the calculation of φ.
Question 44:
The graphs shown represent the relationship between population size (N) and population growth rate dN/dt . Which one of the following growth curves represents a density-dependent population that experiences a strong Allee effect?
![population size (N) and population growth rate dN]()
(A) P
(B) Q
(C) R
(D) S
Correct Answer: (A) P
View Solution
- 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 ( dN/dt < 0) for very small N, as the population cannot sustain itself. 2. Positive growth ( dN/dt > 0) as N increases beyond a critical threshold. 3. A peak in growth rate at an intermediate population size. 4. Declining growth ( dN/dt → 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.
Question 45:
The abundance (X) of a plant species with respect to the anthropogenic stressor habitat destruction (h) is shown. The solid and the dashed curves represent stable and unstable population equilibrium abundances, respectively.
In the absence of any stochasticity, and with increasing values of h, what is the value of h at which a sudden population collapse would occur?
![population size (N) and population growth rate dN]()
(A) 2.5
(B) 2
(C) 4
(D) 3
Correct Answer: (A) 2.5
View Solution
- 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.
Question 46:
Consider the graph shown, where S is species richness and A is area. S and A are log-transformed and the slope is not equal to 1.
The relationship between untransformed S and A follows a/an:
![species richness and A is area.]()
(A) linear relationship.
(B) power law.
(C) exponential relationship.
(D) Michaelis-Menten function.
Correct Answer: (B) power law.
View Solution
- 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 = kAm, where k = 10c. 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 ∝ A, which is not consistent with the power law form S = kAm. Option (B): Correct. The equation S = kAm matches the definition of a power law. Option (C): Incorrect. An exponential relationship would be of the form S = kemA, which is not implied here. Option (D): Incorrect. A Michaelis-Menten function describes saturation dynamics, not a simple power law.
Question 47:
The graph shows the rank-abundance relationships for species in three communities, P, Q, and R.
Which one of the following statements is true with respect to the evenness of the three communities?
![rank-abundance relationships for species]()
(A) P > Q > R
(B) Q > P > R
(C) R > Q > P
(D) R > P > Q
Correct Answer: (C) R > Q > P
View Solution
- 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.
Question 48:
The graph shows bird species richness in a large contiguous forest patch and a small adjacent forest fragment, before and soon after the large contiguous forest patch was replaced by an oil palm plantation.
Which one of the following options best explains the pattern shown?
![bird species richness in a large contiguous forest patch]()
(A) The contiguous forest is a sink and the forest fragment is a source for bird species.
(B) The forest fragment has higher species richness than the contiguous forest.
(C) The bird community in the forest fragment is geographically closed.
(D) The contiguous forest was contributing to forest fragment species richness via dispersal.
Correct Answer: (D) The contiguous forest was contributing to forest fragment species richness via dispersal.
View Solution
- 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.
Question 49:
Honey bees are haplodiploid, which means that the relatedness is, on average, expected to be 0.75 between:
(A) brother-brother pairs with the same parents.
(B) brother-sister pairs with the same parents.
(C) mated female-male pair.
(D) sister-sister pairs with the same parents.
Correct Answer: (D) sister-sister pairs with the same parents.
View Solution
- 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 × 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.
Question 50:
Match the mollusc taxa to their respective orders as shown in the table.
Mollusc Taxa Order
I. Cone snails P. Bivalve
II. Octopuses Q. Gastropod
III. Giant clams R. Cephalopod
IV. Squids
(A) I-P; II-Q; III-R; IV-Q
(B) I-Q; II-R; III-P; IV-R
(C) I-P; II-R; III-P; IV-Q
(D) I-P; II-R; III-Q; IV-R
Correct Answer: (B) I-Q; II-R; III-P; IV-R
View Solution
- 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).
Question 51:
A terrestrial species P is found in both India and West Africa and nowhere else, while a marine species Q is found in the Arabian Sea and the Bay of Bengal. The two species have similar generation times. An ecologist builds haplotype networks based on DNA sequences from these species, where each circle represents one haplotype and each dash (–) represents a mutation. Which one of the following inferences is best supported by the haplotype networks shown?
![A terrestrial species P is found in both India and West Africa]()
(A) P has high dispersal ability; Q has low dispersal ability.
(B) Q has high dispersal ability; P has low dispersal ability.
(C) P and Q have equal dispersal abilities.
(D) The genetic structure is not influenced by dispersal ability.
Correct Answer: (B) Q has high dispersal ability; P has low dispersal ability.
View Solution
- 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.
Question 52:
Grey langurs found in the southern Western Ghats (SWG) and grey langurs in Sri Lanka (SL) look very similar. Nilgiri langurs (found in SWG) and purple faced langurs (found in SL) also look similar. If allopatry played a role in the early diversification of this group (at point X in the tree), which one of the phylogenetic trees is most likely to be correct?
![Grey langurs found in the southern Western Ghats]()
(A) P
(B) Q
(C) R
(D) S
Correct Answer: (C) R
View Solution
- 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.
Question 53:
Two bird species, A and B, are found on a single mountainside. A is a low-elevation species, found between 500 m and 1500 m Above Sea Level (ASL), while B is a high-elevation species, found between 1000 m and 2000 m ASL. At 1250 m ASL, species A and B have very different bill morphologies, but the bill morphology of species A at 500 m is very similar to the bill morphology of species B at 2000 m ASL. Which one or more of the following explain(s) the difference in bill morphology at 1250 m ASL?
(A) Competitive exclusion
(B) Character displacement
(C) Convergent evolution
(D) Allopatric speciation
Correct Answer: (B) Character displacement
View Solution
- 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.
Question 54:
Which one or more of the following is/are greenhouse gas(es)?
(A) Methane
(B) Water vapour
(C) Sulphur dioxide
(D) Nitrous oxide
Correct Answer: (A) Methane, (B)Water vapour, and (D) Nitrous oxide.
View Solution
- 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 (CO2), methane (CH4), nitrous oxide (N2O), and water vapour (H2O).
- Step 2: Analyze the options. Option (A): Methane (CH4) is a potent greenhouse gas with a high global warming potential. Option (B): Water vapour (H2O) is the most abundant greenhouse gas, contributing significantly to the natural greenhouse effect. Option (C): Sulphur dioxide (SO2) is not a greenhouse gas. Instead, it can cause cooling by forming aerosols that reflect sunlight. Option (D): Nitrous oxide (N2O) is a significant greenhouse gas with a long atmospheric lifetime.
Question 55:
Males of the Indian robin in two populations sing songs of different lengths. Which one or more of the options given is/are an ultimate (not proximate) explanation(s) of the difference in song length between the two populations?
(A) Females prefer to mate with males that sing longer songs in one population but not in the other.
(B) The two populations have different forms of the gene that determines song duration.
(C) The two populations differ in hormone levels that activate the start and end of singing behaviour.
(D) Differences between populations in food availability during development affect neural circuitry that is involved in song production.
Correct Answer: (A) Females prefer to mate with males that sing longer songs in one population but not in the other.
View Solution
- 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.
Question 56:
Female Anopheles mosquitoes bite humans to get blood. Researchers performed an experiment to study whether female mosquitoes use temperature or scent, or both, when locating human hosts. They presented female mosquitoes with membranes kept at different temperatures. Some membranes had human scent applied to them. The response to each treatment was measured as the percentage of females that landed on the membrane (50 females for each treatment). The table shows the treatments and the corresponding responses.
(A) Human scent cues are necessary to locate human hosts.
(B) Human scent cues are sufficient to locate human hosts.
(C) Human body temperature cues are necessary to locate human hosts.
(D) Human body temperature cues are sufficient to locate human hosts.
Correct Answer: (B) Human scent cues are sufficient to locate human hosts. and (D) Human body temperature cues are sufficient to locate human hosts.
View Solution
- 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). - At ambient temperature without human scent, there was no response (0). - At human body temperature with or without human scent, the response was high (90).
- 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.
Question 57:
A phylogenetic tree for the evolution of two pigmentation traits in species of fish is shown for clades X, Y, and Z. Genes A and/or B, if mutated, can cause dark pigmentation in the body.
Which one or more of the following statements is/are correct?
![phylogenetic tree for the evolution of two pigmentation traits]()
(A) The character state ”pigmentation” is homologous between species 1 and 3.
(B) The character state ”pigmentation” is homologous between species 1 and 4.
(C) The character state ”pigmentation” is not homologous for species 6 and 7.
(D) The character state ”pigmentation” is not homologous between species 2 and 6.
Correct Answer: (A) The character state ”pigmentation” is homologous between species 1 and 3.,
(D) The character state ”pigmentation” is not homologous between species 2 and 6.
View Solution
- 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.
Question 58:
In conservation biology, which one or more of the following is/are used to calculate the effective population size, Ne?
(A) the population size required to avoid local extinction in the next 1000 years.
(B) the carrying capacity of the environment.
(C) the sum of the sizes of all connected populations in a metapopulation.
(D) the number of breeding males and females.
Correct Answer: (D) the number of breeding males and females.
View Solution
- Step 1: Understanding effective population size (Ne). The effective population size, Ne, 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 Ne. Option (B): Incorrect. Carrying capacity refers to the maximum number of individuals the environment can support and does not directly influence Ne. Option (C): Incorrect. While metapopulation size is important in conservation, Ne 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 Ne, as it determines the genetic contribution to future generations.
Question 59:
In the foodweb diagrams shown, R represents the primary producer, C1 and C2 represent intermediate consumers, and P represents the top predator. Which one or more of these diagrams show(s) intraguild predation?
![foodweb diagrams]()
(A) E
(B) F
(C) G
(D) H
Correct Answer: (D) H
View Solution
- 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.
Question 60:
You are a plant ecologist studying a plant in the genus Veronica. You notice that, at open rocky sites, Veronica grows as a creeper spreading low to the ground, whereas in grasslands, the stem stands upright. You collect seeds from multiple populations in each habitat type and grow them under uniform conditions in a greenhouse. You find that all the plants grown in the greenhouse have stems that stand upright. Which one or more of the following explanations best support(s) your observations?
(A) The different morphologies in the natural habitat types are due to phenotypic plasticity.
(B) Inbreeding depression has led to the creeping form in the rocky sites.
(C) High gene flow between populations has restricted local adaptation in the two environments.
(D) The morphological differences between populations demonstrate that growth form is a polygenic trait.
Correct Answer: (A) The different morphologies in the natural habitat types are due to phenotypic plasticity.
View Solution
- 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.
Question 61:
One hypothesis for why the tropics have far greater species richness than higher latitudes is that the tropics are relatively aseasonal. Low seasonality can encourage high species richness through which one or more of the following mechanisms?
(A) Numerous resources are consistently available throughout the year, allowing different species to specialize on different resources, thereby minimizing competition and allowing co-existence.
(B) Low seasonality is associated with lower rates of predation, allowing large populations to thrive.
(C) Low seasonality is associated with more stable populations that are less vulnerable to demographic stochasticity and extinction.
(D) Low seasonality is associated with longer generation times, which enhances species richness.
Correct Answer: (A) Numerous resources are consistently available throughout the year, allowing different species to specialize on different resources, thereby minimizing competition and allowing co-existence. and
(C) Low seasonality is associated with more stable populations that are less vulnerable to demographic stochasticity and extinction.
View Solution
- 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 (A) and (C), as they explain how low seasonality contributes to higher species richness in the tropics.
Question 62:
The figure illustrates the soil zinc tolerance of the grass species Anthoxanthum along a transect from inside a mine to the middle of a pasture outside the mine.
Which one or more of the following processes explain(s) the observed pattern of zinc tolerance in this grass species?
![soil zinc tolerance of the grass species]()
(A) Genetic drift
(B) Local adaptation
(C) Coevolution
(D) Introgression
Correct Answer: (B) Localadaptation
View Solution
- 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.
Question 63:
In a forest, there are tigers, hare, and deer. On a given day, the probability of a tiger hunting a hare is 0.35, a deer is 0.25, and either a hare or a deer is 0.55. The probability of a tiger hunting both a hare and a deer on a given day is , (Round off to two decimal places).
Correct Answer: 0.05
View Solution
- 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, P(Hare) = 0.35, 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.
Question 64:
Consider a discrete random variable X that takes values from the set S = {0, 1, 2, 3}, being the number of individuals of a species within a habitat. Consider the probability distribution of X with P r(X = 0) = 0.15, P r(X = 1) = 0.25, and P r(X = 3) = 0.5, where P r denotes probability. The value of P r(X = 2) is . (Round off to two decimal places).
Correct Answer: 0.10
View Solution
- 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: P r(X = 0) + P r(X = 1) + P r(X = 2) + P r(X = 3) = 1.
- Step 2: Substitute the known probabilities. Given: P r(X = 0) = 0.15, P r(X = 1) = 0.25, P r(X = 3) = 0.5. Substitute these values: 0.15 + 0.25 + P r(X = 2) + 0.5 = 1.
- Step 3: Solve for P r(X = 2). Simplify the equation: P r(X = 2) = 1 − (0.15 + 0.25 + 0.5). P r(X = 2) = 1 − 0.9 = 0.10.
Question 65:
There are nine species of Impatiens (balsams) found in laterite plateaus of the northern Western Ghats, each with a distinct colour. If a plateau has exactly 6 species, then the number of possible colour combinations in the plateau is . (Answer in integer).