4 Math Moment 1. You have a protein stock that has the concentration 46 HM. The

Question

4 Math Moment 1. You have a protein stock that has the concentration 46 HM. The molecular weight of the protein is 18 kDa. What is the concentration of the protein in mg/mL? 2) Let's say you want to load 10 g of protein into a well. For the proten in the question above, how many L of protein sample should be loaded into one well? To 3. We must not put the protein in the well by itself but rather first mix it with denaturing buffer. It is important to denat the protein with this buffer and by heating before running the sample on an SDS-PAGE because this way t not influence the rate at which the protein migrates and the gel results are due to size only. You are us buffer. This means that you must dilute it 4-fold to reach your final sample concentration (V2VI e 3D shape does ing a 4X sample 4).You plan to add 10 mple buffer to your sample. What volume of protein solution should you add the sample buffer to in order to have of sa the final concentration of the sample buffer be 1X? 4. You are using a 4x sample buffer. You will mix 10 L of sample buffer with 30 L of protein sample. Your protein sample has the concentration 0.828 mg/mL. You want there to be 10 g of protein in the sample after you mix protein with sample buffer, whiat dilution ofyour protein do you need to make in order to be able to use 30 L of protein and achieve 10 g of protein in your sample? 5. You do not know the concentration of your protein but you suspect it to be in the range 0.2 mg/mL to 2 mg/mL. You know you ideally want to have around 10 ug of protein in a well but this is an approximation. If you assume your protein

Explanation / Answer

(3)

Since the sample buffer needs to be diluted 4 times, this means that the final volume of the mixture should be four times the volume of the buffer added.

Volume of buffer added = 10 uL

So , volume of mixture should be: 4*10 = 40 uL

So, volume of protein solution that needs to be added = 40-10 = 30 uL

(4)

Protein conc = 0.828 mg/mL = 828 ug/mL

Desired protein mass = 10 ug

Desired protein conc = Desired protein mass/Desired protein volume = 10/30 ug/uL = 0.33 g/L = 0.33 mg/mL

So,

required dilution = Original conc/Desired conc = 0.828/0.333 = 2.5 times dilution approximately

Hope this helps !

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