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Brut [27]
3 years ago
15

Which cellular environments is conducive to the formation of disulfide bonds within or between proteins?

Biology
1 answer:
Solnce55 [7]3 years ago
3 0

Answer:

The rough ER

Explanation:

Generally the formation of native  disulphide bond is essential for the folding and maturation of  proteins in the secretory pathways.These bonds are formed in the rough E.R, and also in the intramembrane of the mitochondria.

Basically an oxdative parthways is required  for the disulfide bond to form, while a reductive pathway is needed for the formation of isomers for non-native disulphide bonds. These oxidative pathways takes place in the rE.R

Generally the disulphide bonds is usually formed by oxidation of  two cysteines bonds. it does not occur in the cytoplasm because it is always in reduced state due to the presence of disulphide reductase. The same explains why it does not occur in the Nuclei

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How much 3M NaOH (MW=40) can be made from 10.0mL of a 10M solution?
vaieri [72.5K]

Explanation:

The given data is as follows.

              M_{1} = 3 M,           V_{1} = ?

              M_{2} = 10 M,         V_{2} = 10.0 ml

Therefore, calculate the volume of given solution as follows.

              M_{1} \times V_{1} = M_{2} \times V_{2}

Now, putting the given values into the above formula as follows.

            M_{1} \times V_{1} = M_{2} \times V_{2}

            3 M \times V_{1} = 10 M \times 10.0 ml    

                   V_{1} = 33.3 ml

or,                             = 0.033 L        (as 1 ml = 0.001 L)

Since, molarity of NaOH is given as 3 M and its volume is calculated as 33.3 ml.

           Molarity = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1}{\text{volume in L}}

               3 M = \frac{mass}{40 g/mol \times 0.033 L}

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