The exploitation of biological processes for industrial and other purposes especially the genetic manipulation of microorganisms for the production of antibiotics , hormones, ETC

8 0

3 years ago

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What is cell specialization and differation?

kap26 [50]

Specialization is what each cells personal function is, and differentiation is how one cell is different from another in any way

6 0

3 years ago

Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

$C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}$

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

Best regards.

7 0

3 years ago