Immunoglobulin proteins consist of two heavy chains (50 kD each) and 2 light cha

Question

Immunoglobulin proteins consist of two heavy chains (50 kD each) and 2 light chains (25 kD each) joined by distribute bonds in the ER. After running an immunoglobulin sample on reducing SDS-PAGE, with the additional of 2-ME or DTT to the sample buffer, there are: A. 2 banes = 50 kD and 25 kD B. 1 band- 75 kD C. 1 band- 150 kD D. 2 bands- 50 kD and 75 kD After running an sample on non-reducing SOS PAGE there are A. 2 bands- 50 kD and 25 kD B. 1 bands 75 kD C. 1 band- 150 kD D. 2 bands- 5C kD and 75 kD What does one mole of (Avogadro's number of immunoglobulin molecules) weigh in grams? A. 150 g B. 75 g C. 150,000 g D. 1500 g E. 75,000 g What is the molar concentration of a 5 mg/ml immunoglobulin solution? Include the correct unit and circle your answer What two amino acid side chains are joined via disulfide bonds? A. profanes, because they have a cyclized side chain between the alpha-carbon and the ammo group B. methionine's, because their side chain is -CH2-CH2-SH-CH3. C. cysteine's, because their side chain is -CH2-SH. D. tyrosine's, after a phosphate group has been added to their -OH group. On the following page is a and Doolittle plot of a protein for questions 22-25 The y-axis shows a hydrophobicity score the x-axis shows the amino acid positions from 1 to 1400, with divisions every 200 acids.

Explanation / Answer

17 - ME or Mercaptoethanol and DTT or Dithiothreitol is used to break the disulfide linkage between the peptide. So we will get two bands one at 50 kDa and second at 25 kDa

so the option A is correct.

18 -

In immunoglobin light chain and heavy chain are attached to each other by disulfide linkage and also both the heavy chain are attached to each other by the disulfide linkage. So in non-reducing SDS-PAGE we will get only single band at kDa

SO the option C is correct

19-

the weight of One immunoglobulin is 150 kDa and in one mole there are 6.023*1023 molecules of immunoglobulins which will total weight 150000 g

so the option C is correct.

20 - 1 mole of immunoglobulin has 150000 g of immunoglobulin which means that 1 mole of immunoglobin is present in 1 liter of solution. Also, we can say that 150 g/ml will make 1 mole of immunoglobulin. 1 g is equal to 1000mg

so the 5mg/ml will have 5*10-3 mole of immunoglobulin.

21 - To have the disulfide linkage the amino acid should have terminal -SH group and we know that the only cysteine has the terminal -SH group.

SO the answer is option C

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