# Slope of the line: Wheat germ extract

The pure anzyme was given at a 1mg/mL concentration.

The question is: How much (ug) of acid phosphatase enzyme was extracted from the wheat germ?

I believe the 1.022 is in (mg/mL). Dont I need to multiply this by a volume to get to a mass?
Or is this the mass?

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## Answers and Replies

tiny-tim
Homework Helper
Hi Thrax!

(hav a mu: µ )
The pure anzyme was given at a 1mg/mL concentration.

The question is: How much (ug) of acid phosphatase enzyme was extracted from the wheat germ?

I believe the 1.022 is in (mg/mL). Dont I need to multiply this by a volume to get to a mass?
Or is this the mass?

Your X (=1.022) is the ratio of two things that are the same, so it's a dimensionless number.

But since "The pure anzyme was given at a 1mg/mL concentration", that means the concentration of the acid enzyme must be 1.022 times that.

(uhh? how can something extracted have a higher concentration? obviously i'm misunderstanding something here )

Concentration is in mg/ml (mass per volume), so yes you need to multiply a concentration by a volume to get to a mass.

Maybe if i show you what i did it might help.

Extraction and Quantification of Acid Phosphatase Enzyme from Wheat
Germ
1. Obtain a 15 mL conical centrifugation tube and add 0.5 grams of raw
wheat germ to the tube.

2. Add 5 mL of 0.05M Sodium Acetate buffer, pH 4.5. Cap and shake the
tube to disperse the wheat germ into the buffer

3. Using the sonicator, disrupt the suspended wheat germ by sonicating for
30 seconds.

4. Shake the tube again and with a plastic transfer pipet, remove about 1 mL
of the suspension into a new microcentrifuge tube.

5. Centrifuge on high for 30 seconds and place tube in a tube rack. You will
be using the supernatant (which should contain acid phosphatase) in the
next experiment.

6. We need two rows of 6 wells on your plate. Mark one as pure enzyme and
the other as extract

7. Into both rows add 50 uL of 1M NaOH to each well.

8. Obtain two clean microcentrifuge tubes, label one pure enzyme and the
other as extract.

9. Into each tube, add 600 uL of 0.05M Sodium Acetate buffer, and 100 uL
of the substrate solution.

10. Notify your timer to pay attention!

11. Quickly but accurately add 60 uL of pure enzyme to the pure enzyme tube
and 60 uL of the extracted enzyme to the extracted enzyme tube.

12. Immediately mix well by shaking, start the timer counting up,
immediately remove 100 uL of each tube and put it in the first wells of
each row, combining it with the NaOH already in the well. This is a
time=0 time point

13. Repeat taking out 100uL samples from each tube and putting it in the
appropriate wells at time=2minutes, =4 minutes, =6 minutes, =8 minutes,
and finally, time=10 minutes.

14. On the plate reader, record absorbance values for each well at 405 nm.

15. plot and X-Y plot using the absorbance (on the Y) vs. time (on the X) and
do linear regression through the points.

16. Analyze the linear regression data to obtain the slope value (m) for each
line. This values tells how fast the reaction was proceeding. The more
enzyme, the faster the rate. We can use these values to quantify how
much enzyme is present in the reaction tubes.

17. We added 50 uL of pure active enzyme to the tube to produce the rate
observed in the pure enzyme timed reaction. The concentration was 1
mg/ml. How much active enzyme was in this tube?

18. By setting up a ratio between the slope/enzyme ug to the slope of the
extract reaction, we can determine how much active enzyme was in the
extract.

Last edited:
Hi tiny-tim

Thanks for the mu: µ,

Let me see if I understand this correctly. if the acid enzyme is 1.022 times the pure enzyme concentration of 1mg/mL, that would give 1.022 x 1mg/mL = 1.022 mg/mL correct? I believe the steeper the slope for the extract indicates that I extracted more than 1mg/mL worth of enzyme.

Slope of pure enzyme is equivalent to 1mg/mL = slope of extracted enzyme is equivalent to 1.022 mg/mL

I think this tells me that when i did the sonication on wheat grem, I just extracted more than the 1mg/mL of pure enzyme concentration that was given. Which when all is said and done, just proves that there was more wheat grem enzyme present (the slope was higher) than was present in the pure enzyme concentration. So, the 1.022 mg/mL concentration of extract holds more enzyme than the 1mg/mL of pure enzyme. Still fuzzy about the 1.022 units...could i just say that i extracted 1.022 mg and convert to µg since the 1.022 is unit less?

tiny-tim
Homework Helper
Hi Thrax!

I've honestly no idea what this experiment is about.

The 10.22 is unitless, but once you multiply it by the 1 mg/ml, you get a concentration. To get mass, you definitely need to find a volume somewhere to multiply it by.

Perhaps that's the volume of the sample?

Well, your not alone! ( :
I was going to post the steps before but i thought it would be to much. But, I'm post them so you can have a look see...Line 16 through 18 get to the heart of what i need to find. Trying to find what volume to multiple by is a bit tricky too. Mixed and diluted the enzymes with buffers and what not.

Extraction and Quantification of Acid Phosphatase Enzyme from Wheat
Germ
1. Obtain a 15 mL conical centrifugation tube and add 0.5 grams of raw
wheat germ to the tube.

2. Add 5 mL of 0.05M Sodium Acetate buffer, pH 4.5. Cap and shake the
tube to disperse the wheat germ into the buffer

3. Using the sonicator, disrupt the suspended wheat germ by sonicating for
30 seconds.

4. Shake the tube again and with a plastic transfer pipet, remove about 1 mL
of the suspension into a new microcentrifuge tube.

5. Centrifuge on high for 30 seconds and place tube in a tube rack. You will
be using the supernatant (which should contain acid phosphatase) in the
next experiment.

6. We need two rows of 6 wells on your plate. Mark one as pure enzyme and
the other as extract

7. Into both rows add 50 uL of 1M NaOH to each well.

8. Obtain two clean microcentrifuge tubes, label one pure enzyme and the
other as extract.

9. Into each tube, add 600 uL of 0.05M Sodium Acetate buffer, and 100 uL
of the substrate solution.

10. Notify your timer to pay attention!

11. Quickly but accurately add 60 uL of pure enzyme to the pure enzyme tube
and 60 uL of the extracted enzyme to the extracted enzyme tube.

12. Immediately mix well by shaking, start the timer counting up,
immediately remove 100 uL of each tube and put it in the first wells of
each row, combining it with the NaOH already in the well. This is a
time=0 time point

13. Repeat taking out 100uL samples from each tube and putting it in the
appropriate wells at time=2minutes, =4 minutes, =6 minutes, =8 minutes,
and finally, time=10 minutes.

14. On the plate reader, record absorbance values for each well at 405 nm.

15. plot and X-Y plot using the absorbance (on the Y) vs. time (on the X) and
do linear regression through the points.

16. Analyze the linear regression data to obtain the slope value (m) for each
line. This values tells how fast the reaction was proceeding. The more
enzyme, the faster the rate. We can use these values to quantify how
much enzyme is present in the reaction tubes.

17. We added 50 uL of pure active enzyme to the tube to produce the rate
observed in the pure enzyme timed reaction. The concentration was 1
mg/ml. How much active enzyme was in this tube?

18. By setting up a ratio between the slope/enzyme ug to the slope of the
extract reaction, we can determine how much active enzyme was in the
extract.