Radioactivity past paper assignment
Question 1:
In the fridge, from a previous experiment, there is a 50 mM [U-14C] glucose solution. A 20 uL aliquot was counted and gave 200 000 dpm. What is the specific activity of this preparation?
20 uL = 200 000 dpm (divide both sides by 20)
1 uL = 10 000 dpm
Glucose = 50 mM = 50 mmol / uL = 50 umol/ uL = 50 nmol / uL
Therefore in one uL = 50 nmol/ uL = 10 000 dpm (divide both side by 50) = 200 dpm
The specific activity of this preparation is 200 dpm/ nmol
Question 2. How much of this 50 mM [ U-14 C] glucose would you need to add to each flask to get the desired number of counts in each flask? …show more content…
Is it possible to predict what it might be?
Yes it is possible.
As there is 5 mM = 5 mmol/ ml = 5 umol/ml
Final volume = 2 ml
Therefore = 5 umol/ ml (times both side by 2)
= 10 umol/ 2ml
From the previous it was found that 5.23 umol are used in 25 minutes.
= 10 -5.23
= 4.7 umol/ 2ml (divide both sides by 2)
= 2.35 mM
After 25 mins the concentration is 2.35 mM of glucose.
Question 10
Consider the 15 min sample from flask 1. There was 18 045 dpm in the chloroform extract. If this sample has not been extracted in chloroform, but had been counted without lipid/aqueous separation, how many dpm would you have counted?
5 mM = 5 mmol/ ml
100 uL = 0.1 ml
Therefore 5 mmol/ 0.1 ml (times both side by 0.1)
= 0.5 mmol/ ml
= 0.5 umol/ ml
Using the specific activity 200 000 dpm/ umol
= 0.5 umol times it by 200 000)
= 100 000 dpm
Therefore 100, 000 dpm would have been counted if chlorofom had not been extracted.
Question 11
The 15 min sample from flask 1 contained 18 045 dpm. This could also be expressed as how many:
A. cpm
= 18 045 cpm as cpm and dmp can be used
Question 1:
In the fridge, from a previous experiment, there is a 50 mM [U-14C] glucose solution. A 20 uL aliquot was counted and gave 200 000 dpm. What is the specific activity of this preparation?
20 uL = 200 000 dpm (divide both sides by 20)
1 uL = 10 000 dpm
Glucose = 50 mM = 50 mmol / uL = 50 umol/ uL = 50 nmol / uL
Therefore in one uL = 50 nmol/ uL = 10 000 dpm (divide both side by 50) = 200 dpm
The specific activity of this preparation is 200 dpm/ nmol
Question 2. How much of this 50 mM [ U-14 C] glucose would you need to add to each flask to get the desired number of counts in each flask? …show more content…
Is it possible to predict what it might be?
Yes it is possible.
As there is 5 mM = 5 mmol/ ml = 5 umol/ml
Final volume = 2 ml
Therefore = 5 umol/ ml (times both side by 2)
= 10 umol/ 2ml
From the previous it was found that 5.23 umol are used in 25 minutes.
= 10 -5.23
= 4.7 umol/ 2ml (divide both sides by 2)
= 2.35 mM
After 25 mins the concentration is 2.35 mM of glucose.
Question 10
Consider the 15 min sample from flask 1. There was 18 045 dpm in the chloroform extract. If this sample has not been extracted in chloroform, but had been counted without lipid/aqueous separation, how many dpm would you have counted?
5 mM = 5 mmol/ ml
100 uL = 0.1 ml
Therefore 5 mmol/ 0.1 ml (times both side by 0.1)
= 0.5 mmol/ ml
= 0.5 umol/ ml
Using the specific activity 200 000 dpm/ umol
= 0.5 umol times it by 200 000)
= 100 000 dpm
Therefore 100, 000 dpm would have been counted if chlorofom had not been extracted.
Question 11
The 15 min sample from flask 1 contained 18 045 dpm. This could also be expressed as how many:
A. cpm
= 18 045 cpm as cpm and dmp can be used