Table 1 Component Blank 1 2 3 4 5 6 7 8 Water 1.00 0.80 0.60 0.40 0.20 Glucose S
ID: 859096 • Letter: T
Question
Table 1
Component
Blank
1
2
3
4
5
6
7
8
Water
1.00
0.80
0.60
0.40
0.20
Glucose Std
0.20
0.40
0.60
0.80
1.00
Sucrose Std
1.00
Diluted extract
0.50
1.00
Buffer
1.00
1.00
1.00
1.00
1.00
1.00
0.50
Invertase
1.00
1.00
1.00
Data Table:
Blank
.2
.6
.8
1
Tube 6
Tube 7
Tube 8
Abs 540nm
0
0.124
0.303
0.408
0.521
0.372
0.091
0.324
y = 0.5071x + 0.0075
Use your standard curve to calculate the amount of glucose(mg) produced from the standard sucrose solution. Compare this with the amount of glucose expected from the amount of sucrose used?
Component
Blank
1
2
3
4
5
6
7
8
Water
1.00
0.80
0.60
0.40
0.20
Glucose Std
0.20
0.40
0.60
0.80
1.00
Sucrose Std
1.00
Diluted extract
0.50
1.00
Buffer
1.00
1.00
1.00
1.00
1.00
1.00
0.50
Invertase
1.00
1.00
1.00
Explanation / Answer
this one is the equation of straight line which you you might have got for standard solutions you used for calibration
y=mx +c
here Concentration=x and Absorbance A=y
you might have determied the absorbance of your unknown sample (y-intercept value) whose value could be pluged in this equation which gives you the value of concentration of unown sample
y = 0.5071x + 0.0075
here 0.5071=dy/dx=m is slope of straight line and 0.0075=c is the x-intercept when y=0 i.e the absorbance of blank solution
In the end you can compare this amount of glucose produced with the amount that might have been produced by the addition of sucrose
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