1 2 3 (1 mark) 4 (1 mark) 5 6 (1 mark) Dil’n of sample A 562 for each sample [GF
ID: 87359 • Letter: 1
Question
1
2
3 (1 mark)
4 (1 mark)
5
6 (1 mark)
Dil’n of sample
A562for each sample
[GFPS65T] in diluted sample (g/ml). Find this by solving the eq’n
Actual [GFPS65T] in the undiluted sample (mg/ml)
Vol. of the GFPS65T eluate (ml)
Total mg of GFPS65T in entire sample volume
1/1*
0.299
3
1/1
0.287
3
1/1
0.300
3
1/3
0.115
3
1/3
0.112
3
1/3
0.113
3
1/5
0.075
3
1/5
0.076
3
1/5
0.081
3
Amount of GFPS65T in the sample (mg) (averaged from samples) (1 mark)
=
Use the data in Columns 1 and 2, and the equation you defined (y=0.0011x) , to calculate the amount of GFPS65T recovered. Show working
1
2
3 (1 mark)
4 (1 mark)
5
6 (1 mark)
Dil’n of sample
A562for each sample
[GFPS65T] in diluted sample (g/ml). Find this by solving the eq’n
Actual [GFPS65T] in the undiluted sample (mg/ml)
Vol. of the GFPS65T eluate (ml)
Total mg of GFPS65T in entire sample volume
1/1*
0.299
3
1/1
0.287
3
1/1
0.300
3
1/3
0.115
3
1/3
0.112
3
1/3
0.113
3
1/5
0.075
3
1/5
0.076
3
1/5
0.081
3
Amount of GFPS65T in the sample (mg) (averaged from samples) (1 mark)
=
Explanation / Answer
for 1/1 = 0.299 = 0.0011x .,
x = 0.299/0.0011 = 271.82 (g/ml) since dilution is 1:1 there fore concentration is half of actual =543 .64 g/ml
since 1 mg = 1000 g = 543.64 * 10 -3 mg/ml 0.54 mg/ml
since total volume = 3ml and in 1ml = 0.54 there fore in 3ml (entire sample volume) = 1.62 mg/3ml
0.287 = 0.0011x = 260.90 (g/ml) since dilution is 1:1 there fore concentration is half of actual = 521.81 g/ml
= 0.52 mg/ml
= 1.56 mg/3ml
for 1/1 when Y = 0.300; = .001x
0.300 = 0.0011x = 272.72 (g/ml) since dilution is 1:1 there fore concentration is half of actual = 545.45 g/ml
= 0.54 mg/ml
= 1.62 mg/3ml
for 1/3; Y= 0.115; = 0.0011x
= .115= 0.011x = 104.54 (g/ml) since dilution is 1:3 there fore concentration is one third of actual = 316.62 g/ml
= 0.31 mg/ml
= 0.93 mg/3ml
for 1/3 when Y = 0.112
= 0.112= 0.0011x = 101.81 (g/ml) since dilution is 1:3 there fore concentration is one third of actual = 305.45 g/ml
= 0.30 mg/ml
for 1/3when Y= 0.113 = 0.90mg/3ml
= 102.72 (g/ml) since dilution is 1:3 there fore concentration is one third of actual = 308.18 g/ml
= 0.30 mg/ml
= 0.90mg/3ml
for 1/5 when y= 0.075 =
68.18 (g/ml) since dilution is 1:5 there fore concentration is one fifth of actual = 340.90 g/ml
= 0.34 mg/ml
= 1.02 mg/3ml
for 1/5 when Y = 0.076
69.09 (g/ml) since dilution is 1:5 there fore concentration is one fifth of actual = 345.45 g/ml
= 0.34 mg/ml
= 1.02mg/3ml
for 1/5 when Y = 0.081 =
73.63 (g/ml) since dilution is 1:5 there fore concentration is one fifth of actual = 368.18 g/ml; 0.36 mg/ml
= 1.08 mg/3ml
for 1/5 when Y= 0.081
73.63 (g/ml) since dilution is 1:5 there fore concentration is one fifth of actual = 368.18 g/ml; 0.36 mg/ml
= 1.08mg/ml
Dil’n of sample
A562for each sample
[GFPS65T] in diluted sample (g/ml). Find this by solving the eq’n
Actual [GFPS65T] in the undiluted sample (mg/ml)
Vol. of the GFPS65T eluate (ml)
Total mg of GFPS65T in entire sample volume
1/1
0.299
271.82 (g/ml)
0.54 mg/ml
3
1.62 mg
1/1
0.287
260.90 (g/ml)
0.52 mg/ml
3
1.56 mg
1/1
0.300
272.72 (g/ml)
0.54 mg/ml
3
1.62 mg
1/3
0.115
104.54 (g/ml
0.31mg/ml
3
0.93 mg
1/3
0.112
101.81 (g/ml)
0.30 mg/ml
3
0.90 mg
1/3
0.113
102.72 (g/ml)
0.30 mg/ml
3
0.90 mg
1/5
0.075
68.18 (g/ml)
0.34 mg/ml
3
1.02 mg
1/5
0.076
69.09 (g/ml)
0.34 mg/ml
3
1.02mg
1/5
0.081
73.63 (g/ml)
0.36 mg/ml
3
1.08 mg
Dil’n of sample
A562for each sample
[GFPS65T] in diluted sample (g/ml). Find this by solving the eq’n
Actual [GFPS65T] in the undiluted sample (mg/ml)
Vol. of the GFPS65T eluate (ml)
Total mg of GFPS65T in entire sample volume
1/1
0.299
271.82 (g/ml)
0.54 mg/ml
3
1.62 mg
1/1
0.287
260.90 (g/ml)
0.52 mg/ml
3
1.56 mg
1/1
0.300
272.72 (g/ml)
0.54 mg/ml
3
1.62 mg
1/3
0.115
104.54 (g/ml
0.31mg/ml
3
0.93 mg
1/3
0.112
101.81 (g/ml)
0.30 mg/ml
3
0.90 mg
1/3
0.113
102.72 (g/ml)
0.30 mg/ml
3
0.90 mg
1/5
0.075
68.18 (g/ml)
0.34 mg/ml
3
1.02 mg
1/5
0.076
69.09 (g/ml)
0.34 mg/ml
3
1.02mg
1/5
0.081
73.63 (g/ml)
0.36 mg/ml
3
1.08 mg
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