The substrate DNA you used in lab 19 was at a concentration of 0.1ug/ul. What vo
ID: 83727 • Letter: T
Question
The substrate DNA you used in lab 19 was at a concentration of 0.1ug/ul. What volume of stock DNA (0.5 ug/ul) did I use to make 200 ul of DNA substrate (enough for the whole class)? We need to make enough substrate DNA for 12 students. Each student requires 15 ul of substrate DNA at a concentration of 0.4 ug/ul. How much stock DNA (at 0.6 ug/ul) do we need? We need 100 of a buffer that contains 0.5ug/ul of component X and 15 ug/ul of component Y. Our stock solutions are at a concentration of 2.5 mg/ml component X and 30 mg/ml for component Y. What volume of each component we need to make the buffer? What volume of distilled water do we need to add to bring the final volume to 100 ul? Two components of buffer are in stock. Component X comes as a 60 mg/ml solution and component Y comes as a 12 mg/ml solution. What volume from each stock solution do we need to make 150 ml of buffer with component X at 2 ug/ul and component Y at 8 ug/ul? How much distilled water do we need to bring solution to 150 ml? We have a 10X solution of buffer component A and a 100X solution of buffer component B. We wish to make a final solution containing 50 units of an enzyme with the two buffer components at IX each... The enzyme comes at a concentration of 5,000 units/ml. The final volume is to be 100 ul. What volume of each buffer and enzyme do we need to do this? How much distilled water must add to bring we the final volume to the desired 100 ul?Explanation / Answer
1 From the dilution calculator
C1V1 =C2V2 we can calculate the volume of 0.5µg/µl needed to make the 0.1µg/µl of 200µl of DNA
0.5 * V1 = 0.1 * 200
V1 = 0.1*200/0.5
40 µL
2- given that there are 12 students and each require 15 µl of DNA so the total DNA volume required is 180 µl
each students requires 0.4µg/µl of DNA
Stock DNA is 0.6 µg/µl
We know that
C1V1 =C2V2
0.6 * V1 = 0.4 * 180
V1 = 0.4* 180/ 0.6
120 µl of 0.6 µg/µl of Stock DNA.
3- Required concentration of X is 0.5µg/µl and its stock concentration is 2.5 mg/µl. Final volume of solution is 100µl
We know that 1 mg = 1000µg
So from C1V1 = C2V2
2500 * V1 = 0.5 * 100
V1 = 0.5 *100 / 2500
V1 = 0.02 µl
Required concentration of Y is 15µg/µl and its stock concentration is 30 mg/µl
C1V1 =C2V2
30000 * V1 = 15 * 100
V1 = 15 *100/30000
= 0.05 µl
Final volume of solution is 100 µl
Distill water required will be 100 – (0.02+0.05)
100 – 0.07
99.93 µl.
4- Required concentration of X is 2 µg/µl and its stock concentration is 60 mg/µl. Final volume of solution is 150µl
We know that 1 mg = 1000µg
So from C1V1 = C2V2
60000 * V1 = 2 * 150
V1 = 2 *150 / 60000
V1 = 0.005 µl
Required concentration of Y is 8 µg/µl and its stock concentration is 12 mg/µl
C1V1 =C2V2
12000 * V1 = 8 * 150
V1 = 8 *150/12000
= 0.1 µl
Final volume of solution is 100 µl
Distill water required will be 150 – (0.1+0.005)
150 – 0.105
149.895 µl.
5- Required concentration of X is 1X and its stock concentration is 10X. Final volume of solution is 100µl
So from C1V1 = C2V2
10 * V1 = 1 * 100
V1 = 100/10
V1 = 10 µl
Required concentration of Y is 1X and its stock concentration is 100 X
C1V1 =C2V2
100 * V1 = 1 * 100
V1 = 100/100
= 1 µl
Given that concentration of enzyme is 5000unit/ ml Or 5 unit / µl. required concentration of enzyme is 50 unit/ml or 0.05unit/ µl.
5 * V1 = 0.05 * 100
V1 = 1 µl.
So we need 1 µl. of enzyme in 100 µl reaction
Distil water is
100- (10+ 1 + 1)
88 µl.
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