101. Three stages of a cultured plant tissue are recorded.
| Stage | Net growth rate | Resource condition | Curve region |
|---|
| P | Low but positive | Resources abundant | Initial region |
| Q | Rapidly increasing | Resources still sufficient | Steep middle region |
| R | Near zero | Resources strongly limiting | Plateau |
The correct phase assignment is:
ⓐ. P stationary, Q lag, R exponential
ⓑ. P lag, Q exponential, R stationary
ⓒ. P exponential, Q stationary, R lag
ⓓ. P lag, Q stationary, R exponential
Correct Answer: P lag, Q exponential, R stationary
Explanation: Stage P lies in the initial region and has a low but positive growth rate despite abundant resources, matching the lag phase in which the system is beginning active growth. Stage Q occupies the steep middle region and shows rapid increase while resources remain adequate, identifying the exponential phase. Stage R forms a plateau with nearly zero net growth and strong resource limitation, which characterises the stationary phase. The phase assignment requires combining rate, resource state and graph position rather than using one clue alone. Rate, resource state and curve position together assign P to lag, Q to exponential and R to stationary. Together the stages trace the normal sigmoid progression from slow establishment through rapid geometric increase to a resource-limited stationary condition. The table joins biological cause with curve shape: abundant resources permit the steep middle rise, whereas depletion reduces net increase near the plateau.
102. Two identical tissue cultures have just begun to enter the stationary phase. Fresh nutrient medium is supplied to P, while Q receives no addition. P resumes rapid increase and reaches a later plateau; Q remains nearly unchanged. The result supports the conclusion that:
ⓐ. Nutrient replenishment converts the sigmoid pattern into arithmetic growth
ⓑ. Fresh nutrients prevent the culture from reaching another plateau
ⓒ. resource limitation caused the plateau, and replenishment restored growth
ⓓ. The plateau reflects a fixed internal limit unaffected by medium renewal
Correct Answer: resource limitation caused the plateau, and replenishment restored growth
Explanation: The cultures are initially equivalent, and the principal changed condition is the addition of fresh nutrient medium to P. Its renewed increase shows that at least part of the culture retained growth capacity when the original plateau began. The contrasting response of Q links the stationary condition with continuing resource limitation. Replenishment permits further active growth and delays the next plateau, although another limit may eventually develop. The experiment does not establish that all cells were dead at the first plateau; such a conclusion conflicts with the recovery of P. It also does not show a permanent conversion to arithmetic growth. The comparison demonstrates that the stationary phase of a sigmoid curve can arise when environmental resources restrict an otherwise growth-capable biological system. The later plateau in P also shows that replenishment postpones limitation rather than granting unlimited growth.
103. Which of the following statements about sigmoid growth are valid?
I. It commonly occurs in cultured cells, plant tissues, plant organs and whole plants under natural limitations.
II. Its slope changes as growth passes through slow, rapid and stationary regions.
III. It represents unrestricted geometric growth throughout the entire observation period.
IV. Its plateau necessarily proves that every cell in the system is dead.
ⓐ. I and III only
ⓑ. I and II only
ⓒ. II and IV only
ⓓ. III and IV only
Correct Answer: I and II only
Explanation: Sigmoid growth is common when a biological system initially has enough resources for acceleration but later encounters limitations. Cultured cells, developing tissues, organs and whole plants may all show this pattern. The curve begins with a relatively slow phase, becomes steep during rapid growth and then approaches a plateau as the net rate declines. This changing slope makes statement II valid. Unrestricted geometric growth would continue as an accelerating exponential curve instead of developing a stationary region, so statement III is not valid. A plateau also does not establish that every cell is dead. Growth may slow when nutrients, space or another requirement becomes limiting even while many cells remain viable. Together, statements I and II accurately describe both the occurrence and graphical behaviour of sigmoid growth. A sigmoid curve is defined by the ordered lag, rapid-growth and slowing phases, not by indefinite exponential increase.
104. A graph plots the size of a developing plant organ on the y-axis against time on the x-axis. The curve begins gradually, rises steeply and later approaches a plateau. Why is this record sigmoid rather than indefinitely exponential?
ⓐ. The organ maintains a constant absolute growth rate throughout
ⓑ. Relative growth rate increases as the plateau is approached
ⓒ. Metabolic activity declines abruptly when rapid growth begins
ⓓ. Internal and environmental limits restrict further growth
Correct Answer: Internal and environmental limits restrict further growth
Explanation: The axes show how organ size changes with time. During the early region, the slope is small, indicating slow initial growth. The steep middle region reflects rapid increase as cells divide or enlarge under favourable conditions. Continued unrestricted geometric growth would keep the curve accelerating upward. Instead, the slope decreases and the organ approaches a plateau. This pattern shows that further enlargement is being restricted by developmental limits, resource availability or other environmental conditions. An individual organ may also approach its mature size even while the whole plant retains growth capacity elsewhere. The plateau does not require complete loss of metabolism, and it cannot be produced by equal absolute additions, which would generate a straight line. Changing limitation across time creates the characteristic S-shaped record. The plateau reflects a declining net rate as organ-specific limits or environmental constraints become increasingly important.
105. A cultured plant tissue initially grows slowly, then increases rapidly and finally shows little net biomass gain. Cells taken from the plateau phase remain viable and resume rapid growth after transfer to fresh nutrient medium. Which conclusion is justified without overinterpreting the result?
ⓐ. Plateau-phase cells can retain the capacity to resume growth when conditions improve
ⓑ. Every cell in the plateau phase was dividing at its maximum rate
ⓒ. Transfer proves that nutrient shortage was the only possible cause of the plateau
ⓓ. The culture permanently changes from sigmoid to arithmetic growth after transfer
Correct Answer: Plateau-phase cells can retain the capacity to resume growth when conditions improve
Explanation: The transferred cells resume rapid growth, so at least a viable fraction retained the capacity to divide and enlarge. The plateau in the original culture therefore cannot be equated automatically with irreversible death of all cells. Fresh medium supplies an improved environment in which this retained capacity is expressed. The result does not establish that one specific nutrient was the sole limiting factor, because transfer also changes several features of the culture environment and may relieve more than one constraint. Nor does it show that every stationary-phase cell had been dividing rapidly before transfer. The strongest conclusion stays close to the evidence: cells from a culture with little net biomass gain can remain developmentally capable of renewed growth. A stationary phase describes the observed net growth pattern, not necessarily permanent loss of viability or a fixed future growth form.
106. Three growth records are described. Record P is a straight line with constant positive slope. Record Q rises exponentially while conditions remain unrestricted. Record R begins slowly, becomes steep and later approaches a plateau. The correct classification is:
ⓐ. P geometric, Q sigmoid, R arithmetic
ⓑ. P sigmoid, Q arithmetic, R geometric
ⓒ. P arithmetic, Q geometric, R sigmoid
ⓓ. P arithmetic, Q sigmoid, R geometric
Correct Answer: P arithmetic, Q geometric, R sigmoid
Explanation: A straight size-time line with constant positive slope indicates that equal absolute increments are being added during equal time intervals. That is the graphical signature of arithmetic growth, so P is arithmetic. An exponentially rising curve under unrestricted conditions reflects multiplicative increase in which the absolute addition becomes larger as the growing system enlarges; Q is geometric. Record R changes slope across time. Its slow beginning, rapid middle increase and final plateau form the S shape of sigmoid growth. The three categories are related but not interchangeable. A biological system may show geometric behaviour during the steep part of a sigmoid record, yet the complete curve includes later limitation and cannot be classified as unrestricted geometric growth throughout. The visual distinction rests on slope behaviour: constant, continuously steepening and eventually flattening trajectories correspond to arithmetic, geometric and sigmoid growth.
107. Arrange the events during a transition from early geometric growth to partly arithmetic growth in an embryo.
P. Most early daughter cells retain division capacity.
Q. Rapid early embryonic divisions begin.
R. Some cellular derivatives lose division capacity and start differentiating.
S. Continued growth includes a maintained dividing line and specialised products.
ⓐ. P → R → Q → S
ⓑ. Q → P → R → S
ⓒ. Q → S → P → R
ⓓ. R → Q → S → P
Correct Answer: Q → P → R → S
Explanation: Early embryonic growth begins with rapid cell divisions, so Q comes first. When most daughter cells retain the capacity to divide, the dividing population can increase geometrically; this establishes P as the next event. Development does not remain indefinitely in that state. Some descendants later leave the actively dividing population, lose division capacity and begin differentiation, represented by R. Growth then includes both continued production from retained dividing cells and accumulation of specialised derivatives, giving the partly arithmetic feature described in S. The order follows changes in daughter-cell fate rather than a simple decline in division rate. Early multiplication expands the embryonic cell population, while later restriction of division permits tissue organisation and functional specialisation. The transition occurs when daughter-cell fates cease to be uniformly mitotic and an increasing share enters differentiation.
108. Use the lineage described below. In the early region, each dividing embryonic cell produces two daughters and both continue dividing. In a later region, one branch continues mitosis while several derivatives leave the lineage and specialise. This arrangement represents:
ⓐ. an early geometric pattern followed by a partly arithmetic pattern
ⓑ. an early arithmetic pattern followed by unrestricted geometric growth
ⓒ. stationary growth followed by reversible physical swelling
ⓓ. secondary growth followed by primary elongation
Correct Answer: an early geometric pattern followed by a partly arithmetic pattern
Explanation: In the early lineage, both daughters remain mitotically active. Repetition of that fate increases the number of dividing cells multiplicatively, which is the cellular basis of geometric growth. The later lineage behaves differently. A continuing branch retains division capacity, while other products leave the active line and specialise. Each later cycle can maintain the dividing source while adding differentiated derivatives, producing an arithmetic component. The description concerns a transition in developmental cell fate, not a shift between primary and secondary organ growth. It likewise does not describe a stationary state, since mitosis and production of specialised cells continue. The lineage illustrates how an embryo can expand rapidly at first and later organise its cells into distinct developmental pathways.
109. During early embryonic development, a change causes nearly all daughter cells to retain mitotic capacity for longer than usual. The most likely immediate consequence is:
ⓐ. an earlier stationary phase caused by complete loss of division
ⓑ. rapid differentiation with a smaller dividing population
ⓒ. conversion of all cell enlargement into reversible swelling
ⓓ. prolongation of geometric increase with delayed specialisation
Correct Answer: prolongation of geometric increase with delayed specialisation
Explanation: Geometric increase is supported when both daughters of a division remain capable of further mitosis. Extending this condition allows the dividing population to multiply for additional rounds, so rapid geometric expansion persists. Differentiation requires at least some derivatives to leave the active division programme and acquire specialised structures or functions. If nearly all daughters retain mitotic capacity, that transition is delayed. The condition does not imply that growth has stopped or that cells have entered a stationary phase. It likewise does not convert living growth into passive swelling. The prediction follows from the balance between proliferation and differentiation: maintaining division in most descendants expands cell number, while restricting departure from the dividing population postpones tissue specialisation. Geometric growth preserves a proportional relation, so the same relative rate acts on an increasingly large base.
110. Assertion: Embryonic growth may shift from an early geometric pattern to a partly arithmetic pattern as development proceeds.
Reason: This shift occurs when every daughter cell continues to retain division capacity.
ⓐ. Both are true, and the Reason correctly explains the Assertion
ⓑ. Both are true, but the Reason does not explain the Assertion
ⓒ. Only the Assertion is true; the Reason is a false statement
ⓓ. Only the Reason is true; the Assertion is a false statement
Correct Answer: Only the Assertion is true; the Reason is a false statement
Explanation: The assertion is true. During early embryonic development, many progeny cells may continue dividing, allowing a geometric increase in cell number. Later, developmental fates become unequal. Some cells retain division capacity, while others leave the active lineage and differentiate. This introduces an arithmetic component in which a continuing dividing source adds specialised products. That reason is false because retention of division capacity by every daughter would preserve geometric multiplication rather than create the transition. A partly arithmetic pattern requires restriction of division in at least some descendants. The shift is biologically important since an embryo must move beyond simple expansion of cell number and establish specialised tissues with different structures and functions.
111. An organ increases in area from \(6\,\text{cm}^2\) to \(18\,\text{cm}^2\) during the first \(4\,\text{hours}\) and then reaches \(24\,\text{cm}^2\) after another \(2\,\text{hours}\). Its absolute growth rate over the complete interval is:
ⓐ. \(3\,\text{cm}^2\,\text{h}^{-1}\)
ⓑ. \(4\,\text{cm}^2\,\text{h}^{-1}\)
ⓒ. \(6\,\text{cm}^2\,\text{h}^{-1}\)
ⓓ. \(9\,\text{cm}^2\,\text{h}^{-1}\)
Correct Answer: \(3\,\text{cm}^2\,\text{h}^{-1}\)
Explanation: Absolute growth rate measures the total increase in the chosen parameter divided by the elapsed time. The complete area increase is \(24-6=18\,\text{cm}^2\). The total duration includes both intervals, giving \(4+2=6\,\text{hours}\). Using \(\mathrm{AGR}=\frac{W_2-W_1}{t_2-t_1}\), the average absolute growth rate is \(\frac{18\,\text{cm}^2}{6\,\text{hours}}=3\,\text{cm}^2\,\text{h}^{-1}\). Using only the first or second interval would not represent the stated whole-period average. The result is an average across \(6\,\text{hours}\); the organ need not have expanded at exactly \(3\,\text{cm}^2\,\text{h}^{-1}\) during every individual hour. Using the starting and final areas across the full six hours automatically incorporates both subintervals. The calculation reports a whole-period average, so unequal rates within the two phases would not change this particular result. The unit preserves that leaf area, rather than length or mass, was measured.
112. Four leaves are measured over different intervals.
| Leaf | Initial area | Final area | Time |
|---|
| P | \(10\,\text{cm}^2\) | \(22\,\text{cm}^2\) | \(4\,\text{days}\) |
| Q | \(30\,\text{cm}^2\) | \(42\,\text{cm}^2\) | \(6\,\text{days}\) |
| R | \(8\,\text{cm}^2\) | \(20\,\text{cm}^2\) | \(3\,\text{days}\) |
| S | \(20\,\text{cm}^2\) | \(35\,\text{cm}^2\) | \(5\,\text{days}\) |
The highest absolute growth rate is shown by:
ⓐ. leaf P
ⓑ. leaf Q
ⓒ. leaf S
ⓓ. leaf R
Correct Answer: leaf R
Explanation: Absolute growth rate requires the area gained and the duration. Leaf P gains \(12\,\text{cm}^2\) in \(4\,\text{days}\), giving \(3\,\text{cm}^2\,\text{day}^{-1}\). Leaf Q gains \(12\,\text{cm}^2\) in \(6\,\text{days}\), giving \(2\,\text{cm}^2\,\text{day}^{-1}\). Leaf R gains \(12\,\text{cm}^2\) in \(3\,\text{days}\), giving \(4\,\text{cm}^2\,\text{day}^{-1}\). Leaf S gains \(15\,\text{cm}^2\) in \(5\,\text{days}\), giving \(3\,\text{cm}^2\,\text{day}^{-1}\). Leaf R has the greatest area increase per unit time even though its total increase is not uniquely largest. All four total gains must first be converted to gains per day before they can be ranked. R combines the common \(12\,\text{cm}^2\) gain with the shortest interval, producing the largest quotient. The comparison shows why total gain and elapsed time must be interpreted together.
113. Organs X and Y are observed for the same time interval. X gains \(12\,\text{g}\) of dry mass, while Y gains \(8\,\text{g}\). No conclusion about their relative growth rates is requested. Which relation is certain?
ⓐ. \(\mathrm{AGR_X}=\mathrm{AGR_Y}\)
ⓑ. \(\mathrm{AGR_X}\gt \mathrm{AGR_Y}\)
ⓒ. \(\mathrm{AGR_X}\lt \mathrm{AGR_Y}\)
ⓓ. \(\mathrm{AGR_X}\) depends only on the initial masses
Correct Answer: \(\mathrm{AGR_X}\gt \mathrm{AGR_Y}\)
Explanation: Absolute growth rate is the total increment divided by the time interval. The two organs are observed for the same duration, so the denominator is common. X gains \(12\,\text{g}\), while Y gains \(8\,\text{g}\). Dividing both gains by the same positive time preserves their order, giving \(\mathrm{AGR_X}\gt \mathrm{AGR_Y}\). Initial mass is not required for this absolute comparison. It would become essential when calculating relative growth rate, where the increment is normalised to the starting size. The supplied information supports only an absolute-rate comparison. Because the elapsed-time denominator is identical and positive, the larger dry-mass increment necessarily gives the larger absolute rate. Numerical AGR values cannot be assigned without the duration, but their ordering is already fixed. Without initial masses, one cannot decide which organ grew more efficiently relative to its original size, but the larger dry-mass gain over equal time fixes the absolute-rate relation.
114. Two leaves are followed for \(4\,\text{days}\). Leaf P increases from \(10\,\text{cm}^2\) to \(18\,\text{cm}^2\), while leaf Q increases from \(40\,\text{cm}^2\) to \(52\,\text{cm}^2\). The strongest conclusion based only on absolute growth rate is:
ⓐ. P grows at \(8\,\text{cm}^2\,\text{day}^{-1}\)
ⓑ. P and Q have equal absolute growth rates
ⓒ. Q has the higher absolute growth rate
ⓓ. P has the higher absolute growth rate
Correct Answer: Q has the higher absolute growth rate
Explanation: Leaf P gains \(18-10=8\,\text{cm}^2\) during \(4\,\text{days}\), so its absolute growth rate is \(\frac{8}{4}=2\,\text{cm}^2\,\text{day}^{-1}\). Leaf Q gains \(52-40=12\,\text{cm}^2\) during the same interval, giving \(\frac{12}{4}=3\,\text{cm}^2\,\text{day}^{-1}\). Q adds more leaf area per day. This conclusion concerns absolute increase only. P begins much smaller, so a later relative-growth calculation could rank the leaves differently after normalising each increment to initial area. The common four-day denominator allows the \(8\,\text{cm}^2\) and \(12\,\text{cm}^2\) gains to be compared directly. This establishes only the absolute-rate ranking; initial areas would still be needed for a relative-efficiency comparison. The experiment supplies enough information to compare total area added per unit time, but the biological meaning must not be extended beyond the rate actually calculated.
115. A tissue increases in dry mass from \(20\,\text{g}\) to \(30\,\text{g}\) during \(5\,\text{days}\). Using initial mass as the common basis, its relative growth rate is:
ⓐ. \(0.02\,\text{day}^{-1}\)
ⓑ. \(0.05\,\text{day}^{-1}\)
ⓒ. \(0.20\,\text{day}^{-1}\)
ⓓ. \(0.10\,\text{day}^{-1}\)
Correct Answer: \(0.10\,\text{day}^{-1}\)
Explanation: The tissue gains \(30-20=10\,\text{g}\) during \(5\,\text{days}\). Relative growth rate expresses this increase per unit initial mass and per unit time. Using \(\mathrm{RGR}=\frac{W_2-W_1}{W_1(t_2-t_1)}\), substitution gives \(\mathrm{RGR}=\frac{10}{20\times5}=\frac{10}{100}=0.10\,\text{day}^{-1}\). The mass unit cancels when the increment is divided by the initial mass, leaving inverse time as the reporting unit. The corresponding absolute growth rate would be \(\frac{10\,\text{g}}{5\,\text{days}}=2\,\text{g}\,\text{day}^{-1}\), which represents absolute rather than relative growth. The denominator \(20\times5\) retains both the starting mass and the five-day interval required by the stated convention. The relative value indicates that the tissue adds dry mass at an average rate equal to \(10\%\) of its initial mass per day under the stated convention.
116. Assess the following statements concerning relative growth rate.
I. It expresses growth per unit time in relation to initial size.
II. It provides a common basis for comparing differently sized organs.
III. Equal absolute gains during the same interval always produce equal relative rates.
IV. It can serve as an index of growth efficiency.
ⓐ. I, II and IV only
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Relative growth rate normalises the increase to the initial size of the growing system. This makes statement I valid and permits comparison of organs that begin at different sizes, supporting statement II. The resulting value can be interpreted as an efficiency index since it asks how much increase occurred relative to the amount present initially. Statement III is not valid. Two organs can gain the same absolute amount over the same time yet have different relative rates when their initial sizes differ. The same increment forms a larger fraction of a small starting organ than of a large one. Taken together, statements I, II and IV capture the meaning and comparative use of relative growth rate, while statement III confuses equal absolute growth with equal size-normalised performance.
117. A graph plots area against time for leaves P and Q. The two records are straight and parallel, but P starts at \(5\,\text{cm}^2\) and Q starts at \(50\,\text{cm}^2\). Over the measured interval, both gain \(5\,\text{cm}^2\). Which interpretation is valid?
ⓐ. Q has a greater absolute and relative growth rate
ⓑ. P and Q have equal relative growth rates due to parallel slopes
ⓒ. equal absolute rates, with the higher relative rate in P
ⓓ. P has a greater absolute rate, but Q has the greater relative rate
Correct Answer: equal absolute rates, with the higher relative rate in P
Explanation: Parallel lines on the same area-time graph have equal slopes, so the leaves add area at the same absolute rate. Their common gain of \(5\,\text{cm}^2\) over the same interval confirms that equality. Relative growth considers the gain as a fraction of initial area. For P, the fraction is \(\frac{5}{5}=1\), equivalent to a \(100\%\) increase. For Q, it is \(\frac{5}{50}=0.1\), equivalent to a \(10\%\) increase. P has the greater relative rate even though the slopes are equal. The graph separates two comparisons: slope records absolute area addition per time, while the intercept provides the initial size needed for a proportional interpretation. Parallel lines establish equal absolute slopes, while the smaller initial area of P makes the same increment larger relative to its starting size.
118. A small root increases from \(2\,\text{cm}\) to \(4\,\text{cm}\), while a larger root increases from \(20\,\text{cm}\) to \(24\,\text{cm}\) during the same interval. Which root shows greater growth efficiency relative to its initial length?
ⓐ. the larger root, because its absolute gain is \(4\,\text{cm}\)
ⓑ. the smaller root, because its gain is larger relative to initial length
ⓒ. both roots, because their relative increases are equal
ⓓ. neither root, because relative growth applies only to equal initial sizes
Correct Answer: the smaller root, because its gain is larger relative to initial length
Explanation: The small root gains \(4-2=2\,\text{cm}\). Relative to its initial length, this is \(\frac{2}{2}=1\), or a \(100\%\) increase. The larger root gains \(24-20=4\,\text{cm}\), which is the greater absolute increment. Its relative increase is only \(\frac{4}{20}=0.2\), or \(20\%\). The small root has the higher relative growth efficiency even though it adds less total length. This is precisely why relative growth rate is useful: it prevents a large starting organ from appearing more efficient merely because it can add a larger absolute amount. The common observation time cancels in the ranking, leaving each increment to be judged against its own initial length. Although the larger root adds twice as many centimetres, the smaller root doubles whereas the larger root rises by only one-fifth. The time interval is common, so comparing the size-normalised fractions is sufficient to establish the ranking.
119. Organs P and Q gain the same amount \(\Delta W\) during the same time \(\Delta t\), but \(W_{1P}\lt W_{1Q}\). Using \(\mathrm{RGR}=\frac{\Delta W}{W_1\Delta t}\), which relation follows?
ⓐ. \(\mathrm{RGR_P}\gt \mathrm{RGR_Q}\)
ⓑ. \(\mathrm{RGR_P}=\mathrm{RGR_Q}\)
ⓒ. \(\mathrm{RGR_P}\lt \mathrm{RGR_Q}\)
ⓓ. \(\frac{\mathrm{RGR_P}}{\mathrm{RGR_Q}}=\frac{W_{1P}}{W_{1Q}}\)
Correct Answer: \(\mathrm{RGR_P}\gt \mathrm{RGR_Q}\)
Explanation: The two organs have the same numerator \(\Delta W\) and the same time factor \(\Delta t\). Their relative growth rates differ only through the initial-size term in the denominator. Since \(W_{1P}\lt W_{1Q}\), the denominator for P is smaller. Dividing the same positive increment by a smaller positive denominator produces the larger value, so \(\mathrm{RGR_P}\gt \mathrm{RGR_Q}\). Their absolute growth rates are equal since \(\frac{\Delta W}{\Delta t}\) is common. The symbolic comparison captures the biological advantage of normalisation: an identical gain represents a greater proportional achievement for the initially smaller organ. With equal gain and time, relative growth varies inversely with initial size. The smaller denominator for P makes its proportional rate larger even though the two absolute rates are identical. No additional numerical values are needed once the relative ordering of the initial sizes is known.
120. Leaf A increases from \(5\,\text{cm}^2\) to \(10\,\text{cm}^2\), while leaf B increases from \(50\,\text{cm}^2\) to \(55\,\text{cm}^2\) during the same interval. Which comparison is accurate?
ⓐ. A has ten times the absolute growth of B, but both have equal relative growth
ⓑ. B has greater absolute and relative growth since its final area is larger
ⓒ. Both leaves show \(10\%\) relative increase and equal absolute growth
ⓓ. Both gain \(5\,\text{cm}^2\), but A shows \(100\%\) and B shows \(10\%\) relative increase
Correct Answer: Both gain \(5\,\text{cm}^2\), but A shows \(100\%\) and B shows \(10\%\) relative increase
Explanation: Leaf A gains \(10-5=5\,\text{cm}^2\), and leaf B gains \(55-50=5\,\text{cm}^2\). Since the time interval is the same, their absolute growth rates are equal. Relative increase uses initial area as the basis. For A, \(\frac{5}{5}\times100=100\%\). For B, \(\frac{5}{50}\times100=10\%\). The relative increase of A is \(10\) times that of B. Final area alone cannot measure efficiency, since B began much larger. Absolute growth compares the two \(5\,\text{cm}^2\) additions, whereas relative growth divides each addition by a different starting area. The equal increments therefore produce equal absolute rates but a tenfold contrast in proportional increase. Keeping these two comparisons separate prevents final size from being used as a substitute for growth efficiency. The comparison demonstrates that equal total additions can carry very different biological meanings when the starting sizes are unequal.
