A group of microbiologists measured the enzymatic reaction rate (R, in \text{nmol/min}\cdo… — ACT

A group of microbiologists measured the enzymatic reaction rate ( $R$ , in $\text{nmol/min}\cdot\text{mg}$ ) of a newly discovered halophilic bacterial strain across various salinity levels ( $S$ , in parts per thousand, ppt) at three fixed temperatures ( $20^\circ\text{C}$ , $30^\circ\text{C}$ , and $40^\circ\text{C}$ ). The results are recorded in the table below.

Salinity ( $S$ , ppt)	Rate at $20^\circ\text{C}$	Rate at $30^\circ\text{C}$	Rate at $40^\circ\text{C}$
10	14.0	22.0	10.0
20	20.0	32.0	16.0
30	26.0	42.0	22.0
50	38.0	62.0	34.0

Based on linear interpolation and extrapolation of the data trends, arrange the four estimated reaction rates described below in order from lowest to highest.

1The estimated reaction rate at $30^\circ\text{C}$ and $5\text{ ppt}$ salinity
2The estimated reaction rate at $40^\circ\text{C}$ and $25\text{ ppt}$ salinity
3The estimated reaction rate at $20^\circ\text{C}$ and $60\text{ ppt}$ salinity
4The estimated reaction rate at $30^\circ\text{C}$ and $40\text{ ppt}$ salinity

Answer

The correct order from lowest to highest estimated reaction rate is: the estimated rate at

30^\circ\text{C}

and

5\text{ ppt}

(

17.0\text{ nmol/min}\cdot\text{mg}

), followed by the estimated rate at

40^\circ\text{C}

and

25\text{ ppt}

(

19.0\text{ nmol/min}\cdot\text{mg}

), followed by the estimated rate at

20^\circ\text{C}

and

60\text{ ppt}

(

44.0\text{ nmol/min}\cdot\text{mg}

), and finally the estimated rate at

30^\circ\text{C}

and

40\text{ ppt}

(

52.0\text{ nmol/min}\cdot\text{mg}

To place the estimated rates in correct order from lowest to highest, calculate each value using the constant linear rates of change shown in the table. At

30^\circ\text{C}

and

5\text{ ppt}

, extrapolating below

10\text{ ppt}

yields

17.0\text{ nmol/min}\cdot\text{mg}

. At

40^\circ\text{C}

and

25\text{ ppt}

, interpolating halfway between

20

and

30\text{ ppt}

yields

19.0\text{ nmol/min}\cdot\text{mg}

. At

20^\circ\text{C}

and

60\text{ ppt}

, extrapolating above

50\text{ ppt}

yields

44.0\text{ nmol/min}\cdot\text{mg}

. At

30^\circ\text{C}

and

40\text{ ppt}

, interpolating halfway between

30

and

50\text{ ppt}

yields

52.0\text{ nmol/min}\cdot\text{mg}

. Comparing these numerical values (

17.0 < 19.0 < 44.0 < 52.0

) produces the correct sequence.

Step-by-Step Solution

Determine the linear relationship (slope) for each temperature column.

For

20^\circ\text{C}

, the rate increases by

0.6\text{ nmol/min}\cdot\text{mg}

per ppt salinity. For

30^\circ\text{C}

, the rate increases by

1.0\text{ nmol/min}\cdot\text{mg}

per ppt salinity. For

40^\circ\text{C}

, the rate increases by

0.6\text{ nmol/min}\cdot\text{mg}

per ppt salinity.

Establishing the rate of change allows calculation of values between existing data points (interpolation) and beyond existing data points (extrapolation).

Calculate the numerical value for each item.

Item 1 (

30^\circ\text{C}, 40\text{ ppt}

42.0 + 10(1.0) = 52.0\text{ nmol/min}\cdot\text{mg}

.
Item 2 (

20^\circ\text{C}, 60\text{ ppt}

38.0 + 10(0.6) = 44.0\text{ nmol/min}\cdot\text{mg}

.
Item 3 (

40^\circ\text{C}, 25\text{ ppt}

16.0 + 5(0.6) = 19.0\text{ nmol/min}\cdot\text{mg}

.
Item 4 (

30^\circ\text{C}, 5\text{ ppt}

22.0 - 5(1.0) = 17.0\text{ nmol/min}\cdot\text{mg}

Obtaining exact numerical estimates enables precise relative comparison.

Sort the items by calculated value from lowest to highest.

17.0

(Item 4) <

19.0

(Item 3) <

44.0

(Item 2) <

52.0

(Item 1).

Direct numerical comparison determines the correct sequence.

Key Concept

Interpolation and Extrapolation of Data Trends

Estimated Time:1m 30s

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