A group of biochemical researchers measured the enzymatic breakdown rate of cellulose (in … — ACT

A group of biochemical researchers measured the enzymatic breakdown rate of cellulose (in $\text{mg/L}\cdot\text{min}$ ) by three bacterial strains (Strain X, Strain Y, and Strain Z) across three different pH environments ( $5.0$ , $6.0$ , and $7.0$ ) at a constant temperature of $37^\circ\text{C}$ . The results are recorded in the table below:

Strain	Breakdown Rate at pH $5.0$	Breakdown Rate at pH $6.0$	Breakdown Rate at pH $7.0$
Strain X	$12$	$18$	$24$
Strain Y	$15$	$25$	$35$
Strain Z	$8$	$14$	$20$

True or False: The average enzymatic breakdown rate across all three pH environments for Strain Y is $5\text{ mg/L}\cdot\text{min}$ greater than the average breakdown rate for Strain X.

Answer: Answer

Answer

The statement is False because the average breakdown rate for Strain Y (

25\text{ mg/L}\cdot\text{min}

) exceeds that of Strain X (

18\text{ mg/L}\cdot\text{min}

) by

7\text{ mg/L}\cdot\text{min}

, not

5\text{ mg/L}\cdot\text{min}

The correct response is False because computing the mean breakdown rates yields

18\text{ mg/L}\cdot\text{min}

for Strain X and

25\text{ mg/L}\cdot\text{min}

for Strain Y. Subtracting

18

from

25

gives a difference of

7\text{ mg/L}\cdot\text{min}

, rendering the statement claiming a difference of

5\text{ mg/L}\cdot\text{min}

mathematically incorrect.

Step-by-Step Solution

Calculate the average breakdown rate for Strain X across all three pH levels.

Average for Strain X =

\frac{12 + 18 + 24}{3} = \frac{54}{3} = 18\text{ mg/L}\cdot\text{min}

To determine the overall mean performance of Strain X across the tested environments.

Calculate the average breakdown rate for Strain Y across all three pH levels.

Average for Strain Y =

\frac{15 + 25 + 35}{3} = \frac{75}{3} = 25\text{ mg/L}\cdot\text{min}

To determine the overall mean performance of Strain Y across the tested environments.

Subtract the average rate of Strain X from the average rate of Strain Y.

Difference =

25 - 18 = 7\text{ mg/L}\cdot\text{min}

To test the claim that the average for Strain Y is

5\text{ mg/L}\cdot\text{min}

greater than Strain X.

Key Concept

Calculating mean values from tabular experimental data and finding differences between summary statistics.

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