A team of plant physiologists investigated the transpiration rate of four plant species (S… — ACT

A team of plant physiologists investigated the transpiration rate of four plant species (Species W, X, Y, and Z) under two distinct light conditions—Low Intensity ( $150\text{ }\mu\text{mol/m}^2\text{/s}$ ) and High Intensity ( $800\text{ }\mu\text{mol/m}^2\text{/s}$ )—at a controlled temperature of $25^\circ\text{C}$ . The measured rates are shown in the table below:

Plant Species	Transpiration Rate at Low Intensity ( $\text{mg H}_2\text{O/cm}^2\text{/hr}$ )	Transpiration Rate at High Intensity ( $\text{mg H}_2\text{O/cm}^2\text{/hr}$ )
Species W	$1.2$	$4.8$
Species X	$2.5$	$6.5$
Species Y	$0.8$	$3.2$
Species Z	$1.5$	$5.5$

Based on the table, what was the average increase in transpiration rate (in $\text{mg H}_2\text{O/cm}^2\text{/hr}$ ) across all four plant species when light condition was increased from Low Intensity to High Intensity?

A
$1.5\text{ mg H}_2\text{O/cm}^2\text{/hr}$
$3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}$ Answer
C
$4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}$
D
$5.0\text{ mg H}_2\text{O/cm}^2\text{/hr}$

Answer

The average increase in transpiration rate across the four species is

3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

To find the average increase, first subtract the Low Intensity transpiration rate from the High Intensity transpiration rate for each of the four species: Species W (

4.8 - 1.2 = 3.6

), Species X (

6.5 - 2.5 = 4.0

), Species Y (

3.2 - 0.8 = 2.4

), and Species Z (

5.5 - 1.5 = 4.0

). Adding these four increases yields a total of

14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

. Dividing this sum by

4

species gives an average increase of

3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

Step-by-Step Solution

Calculate the difference between High Intensity and Low Intensity transpiration rates for each species.

Species W:

4.8 - 1.2 = 3.6\text{ mg H}_2\text{O/cm}^2\text{/hr}

; Species X:

6.5 - 2.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

; Species Y:

3.2 - 0.8 = 2.4\text{ mg H}_2\text{O/cm}^2\text{/hr}

; Species Z:

5.5 - 1.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

Determines the specific rate increase for each trial.

Sum the calculated rate increases across all four species.

3.6 + 4.0 + 2.4 + 4.0 = 14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

Aggregates the individual increases to calculate the total change.

Divide the total sum of increases by the number of species (

4

) to determine the mean increase.

14.0 / 4 = 3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

Yields the average (mean) change in transpiration rate per species.

Key Concept

Calculating mean changes and rate differences from tabular data

Estimated Time:1m 0s

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