16) Look at the results for tubes 5-9. Plot these on the graph paper provided (low tech) or if you choose to plot with Excel (high tech), then you may do so...Enzyme Activity (Absorbance) is on one axis and Time is on the other (you select which is the y-axis and which is the x-axis - bear in mind which is the dependent variable and which is the independent variable). 17) What conclusion can you draw about the way the graph looks? 18) Which tubes show peroxidase activity after 5 min?

Curren'S Math For Meds: Dosages & Sol
11th Edition
ISBN:9781305143531
Author:CURREN
Publisher:CURREN
Chapter10: Reconstitution Of Powdered Drugs
Section: Chapter Questions
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The data presented below was from an actual experiment performed by Bio 102 students.
Competitive inhibition was assessed on PEROXIDASE, an enzyme found in turnips. Peroxidase
works on H2O2 to detoxify the hydrogen peroxide by generating water and molecular oxygen as
the products. Guaiacol binds with oxygen to yield a brown coloration that can be detected with a
spectrophotometer. In other words, one of the products can be detected and measured. The
greater the amount of O2, the more intense the coloration, the greater the absorbance at 470nm.
Hydroxylamine has a shape similar to H2O2. Evaluate the tube conditions in the first table and
correlate that with the spec results in the second table. Answer the questions below the tables.
Table 11.5
Experimental Conditions to Test the Inhibition of Hydroxylamine on Peroxidase Activity
Hydrogen
Peroxide
(3%)
Distilled
Guaiacol
(25 mM)
Turnip
Extract
Hydroxylamine
(10%)
Tube
Water
5.9 ml.
0.1 ml
2
5.8 ml.
0.2 ml.
3
5.7 ml.
0.1 ml.
0.2 ml.
4.9 ml.
0.1 ml
1.0 ml.
4,7 ml.
0.1 ml.
0.2 ml.
1.0 ml.
6.
4.2 ral.
0.1 ml
0.2 ml.
1.0 ml.
0.5 ml.
3.7 ml.
0.1 ml.
0.2 ml.
15 ml.
05 ml.
3.2 al.
0,1 ml.
0.2 ml.
2.0 ml.
05 ml.
2.2 ml.
0.1 ml.
0.2 ml.
3.0 ml.
05 ml.
2 |Page
Table 11.6
Absorbance at 470 nm of Peroxide/Peroxidase Solutions
0.0
0.5
min
1.0
1.5
min
2.0
min
2.5
min
3.0
3.5
4.0
4.5
min
5.0
min
Tube
min
min
min
min
min
0.0j2 0.D100.0I0 0.010 0.010 0.010 0.01d o-0j0l o.01ol0.010 0.010
0,008 0.008 0-008 0-008 o.008 0.008 o.008 0008 0-008 o.008 0.008
0:01le 0.01i 0.01) 0.011 0.DII 0.DI1 0.011 0,DIl OI011 0-011 o.01
l0:0640.0590:058 0,058 0,059 0.059 0059|0:059 0.059 0.0590.059
1.999 |1.999 | 1.999 |1.999|1999|1L999 1,999|1999|1999 1.999|1999
0.132 0.203 0,288 0,340 0140001434 0.450014600.46201454 01488
l0.448/0.688D.950 1,170 1,3301.500 1,6001,70117801.852 1910
0.378 0.778 1,130 1.420 1.660 1.850 1.999 1.9991.999 1.999 1.999
0.825 1,5601.999 1.999 1.999 1991.99¶|19991.999 1.999 1.999
4
8
13)Write a null hypothesis and an alternate hypothesis Ter the impact of the inhibitor en
enzyme aetivity:
14) Summary: What are the fellowing eompunents in the experiment?
Substrate
Ar-
Enzyme
C.Inhibitor
. Indieator
15) Leek at the esults for tubes +4. Explain the results that were obtained for these tubes.
16) Look at the results for tubes 5-9. Plot these on the graph paper provided (low tech) or if
you choose to plot with Excel (high tech), then you may do so....Enzyme Activity
(Absorbance) is on one axis and Time is on the other (you select which is the y-axis and
which is the x-axis – bear in mind which is the dependent variable and which is the
independent variable).
17) What conclusion can you draw about the way the graph looks?
18) Which tubes show peroxidase activity after 5 min?
19) Could you still detect peroxidase activity in the presence of the inhibitor? Why?
20) Where do you think the inhibitor had its impact? Why is it able to do what it has done?
Hint: I tell you above the tables.
How do organisms exploit the use of inhibitors?
Transcribed Image Text:The data presented below was from an actual experiment performed by Bio 102 students. Competitive inhibition was assessed on PEROXIDASE, an enzyme found in turnips. Peroxidase works on H2O2 to detoxify the hydrogen peroxide by generating water and molecular oxygen as the products. Guaiacol binds with oxygen to yield a brown coloration that can be detected with a spectrophotometer. In other words, one of the products can be detected and measured. The greater the amount of O2, the more intense the coloration, the greater the absorbance at 470nm. Hydroxylamine has a shape similar to H2O2. Evaluate the tube conditions in the first table and correlate that with the spec results in the second table. Answer the questions below the tables. Table 11.5 Experimental Conditions to Test the Inhibition of Hydroxylamine on Peroxidase Activity Hydrogen Peroxide (3%) Distilled Guaiacol (25 mM) Turnip Extract Hydroxylamine (10%) Tube Water 5.9 ml. 0.1 ml 2 5.8 ml. 0.2 ml. 3 5.7 ml. 0.1 ml. 0.2 ml. 4.9 ml. 0.1 ml 1.0 ml. 4,7 ml. 0.1 ml. 0.2 ml. 1.0 ml. 6. 4.2 ral. 0.1 ml 0.2 ml. 1.0 ml. 0.5 ml. 3.7 ml. 0.1 ml. 0.2 ml. 15 ml. 05 ml. 3.2 al. 0,1 ml. 0.2 ml. 2.0 ml. 05 ml. 2.2 ml. 0.1 ml. 0.2 ml. 3.0 ml. 05 ml. 2 |Page Table 11.6 Absorbance at 470 nm of Peroxide/Peroxidase Solutions 0.0 0.5 min 1.0 1.5 min 2.0 min 2.5 min 3.0 3.5 4.0 4.5 min 5.0 min Tube min min min min min 0.0j2 0.D100.0I0 0.010 0.010 0.010 0.01d o-0j0l o.01ol0.010 0.010 0,008 0.008 0-008 0-008 o.008 0.008 o.008 0008 0-008 o.008 0.008 0:01le 0.01i 0.01) 0.011 0.DII 0.DI1 0.011 0,DIl OI011 0-011 o.01 l0:0640.0590:058 0,058 0,059 0.059 0059|0:059 0.059 0.0590.059 1.999 |1.999 | 1.999 |1.999|1999|1L999 1,999|1999|1999 1.999|1999 0.132 0.203 0,288 0,340 0140001434 0.450014600.46201454 01488 l0.448/0.688D.950 1,170 1,3301.500 1,6001,70117801.852 1910 0.378 0.778 1,130 1.420 1.660 1.850 1.999 1.9991.999 1.999 1.999 0.825 1,5601.999 1.999 1.999 1991.99¶|19991.999 1.999 1.999 4 8 13)Write a null hypothesis and an alternate hypothesis Ter the impact of the inhibitor en enzyme aetivity: 14) Summary: What are the fellowing eompunents in the experiment? Substrate Ar- Enzyme C.Inhibitor . Indieator 15) Leek at the esults for tubes +4. Explain the results that were obtained for these tubes. 16) Look at the results for tubes 5-9. Plot these on the graph paper provided (low tech) or if you choose to plot with Excel (high tech), then you may do so....Enzyme Activity (Absorbance) is on one axis and Time is on the other (you select which is the y-axis and which is the x-axis – bear in mind which is the dependent variable and which is the independent variable). 17) What conclusion can you draw about the way the graph looks? 18) Which tubes show peroxidase activity after 5 min? 19) Could you still detect peroxidase activity in the presence of the inhibitor? Why? 20) Where do you think the inhibitor had its impact? Why is it able to do what it has done? Hint: I tell you above the tables. How do organisms exploit the use of inhibitors?
Nyll Hy pothesis :There is no change in enzyme activity of peroxidase with change in innibitor concentration.
Atternate Hypotnesis: Enzyme activity dcreases witn increase in
inhibitor concentration, witn time.
Summary: Components of the reaction
a) SubStrate : Hydrogen peroxide, HyDz
6) Enzy me : Peroxidase
c) Inhibitor: Hydroxylamine
dl Indicator: Guaiacol
Look at the veSult for tubes F4.Explain the vesults that were obtained for +hese tubes
In tubes 1,2,3 and 4, the en yme activity 1s constant witn time. The initial vate of the experiment is constant because there iS the pressence
Of a hvge
amount of Substrate. Initially, the fow enzyme molecules work an lange quantities of Substrates
and the initial and final Substrate concentration vemains the Same.
The enzyme activity is the same for 5 min.
Transcribed Image Text:Nyll Hy pothesis :There is no change in enzyme activity of peroxidase with change in innibitor concentration. Atternate Hypotnesis: Enzyme activity dcreases witn increase in inhibitor concentration, witn time. Summary: Components of the reaction a) SubStrate : Hydrogen peroxide, HyDz 6) Enzy me : Peroxidase c) Inhibitor: Hydroxylamine dl Indicator: Guaiacol Look at the veSult for tubes F4.Explain the vesults that were obtained for +hese tubes In tubes 1,2,3 and 4, the en yme activity 1s constant witn time. The initial vate of the experiment is constant because there iS the pressence Of a hvge amount of Substrate. Initially, the fow enzyme molecules work an lange quantities of Substrates and the initial and final Substrate concentration vemains the Same. The enzyme activity is the same for 5 min.
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