Part 2. Practice Dilution Problems (1) How much Sodium Chloride (NaCI, MW- 58.44
ID: 280456 • Letter: P
Question
Part 2. Practice Dilution Problems (1) How much Sodium Chloride (NaCI, MW- 58.44 )is required ofa?. M solution? to make the desired volumes .01mis i56445- A) 10 ml A?) ??.0i mic Macl B) 100 ml C) 1 L D) 10 L (2) How much SDS (Sodium Dodecyl Sulfate) is required to make the désired volumes of a s58 10% w/v solution in ddH2O? A) 10 ml B) 100 ml 100mL ? = 109 sps C) 1 L D) 10 L (3) How much of a 4 M NaCI solution will you need to make the following solutions? A) 10 ml of a 1M NaCl solution B) 100 ml of a 100 mM NaCl solution au m4Odf (4) Ydu are following a protocol which involves the addition of two different DNA molecules. DNA sample # 1 is at a concentration of 50 ug/ml and DNA sample # 2 is at a concentration of 250 ng/ul. The protocol says to mix 500 ng of each molecule together and bring the volume up to a final of 50 ul with sterile ddH20. Determine how many ul of each DNA to dispense into a tube and the additional volume of ddH20 needed (5) You have just built a 10 KL (Kiloliter) swimming pool. As you fill your new swimming pool with water you decide to save money on chemical treatment supplies by adding the appropriate chemicals in solid form. Based on the desired concentration [mg/L] of the 3 chemicals listed below, determine how much mass (in Kg) of each chemical you need to add to the swimming pool while you fill it up. Also, determine the total cost to treat your pool if each chemical is priced in bulk at $10/ Kg at your local pool store supplier. A) Sodium Hypochlorate: 500 mg/L B) EDTA (EthylDiamine TrisAcetate): 100 mg /L C) Sodium Chloride: 25 mg / L D) Total Cost Calculation: (6) Set up the following 100 ul PCR reaction based on the information given. Stocks 100 nM primers 10 pM primers 20 mM dNTPs 2 mM dNTPs 200 mM MgC12 2 mM MgC12 10 U/ul Taq Pol. 1 U/ul Taq PCR reaction Dilution Factor Volume (100 ul reaction 1 ug/ul DNA10 ng/ul DNA DISCUSSION: 1. Did you encounter any difficulties or problems with making dilutions? 2. Did you learn a new method for making dilutions?Explanation / Answer
Question 4) Concentration of sample 1 is 50 ug/ml=(50*1000) ng/ml=50000 ng/ml
So, 50000 ng is present in 1000 ul.
500 ng is present in =(1000*500)/50000=10 ul
Concentration of sample 2 is 250 ng/ul, so 500 ng is present in 2 ul.
Final volume to be made=50 ul, so ddH2O required=50-(10+2)=38 ul
So the required volumes of sample 1, sample 2 and ddH2O are 10 ul, 2 ul and 38 ul, respectively.