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The E. coli DNA polymerase enzyme gives different domains with different activities on cleaving with protease subtilisin. Which of the following statements is correct with respect to the fragments generated?

(a) The smaller fragment is having C terminal and the larger fragment is having N terminal

(b) The smaller fragment is named as Klenow fragment and the intact molecule is called as Kornberg fragment

(c) The smaller fragment is having 5’-3’ exonuclease activity whereas the larger fragment is having 5’-3’ polymerase and 3’-5’ exonuclease activity

(d) Both the fragments are having 5’-3’ polymerase and 3’-5’ exonuclease activity

I had been asked this question in quiz.

The query is from Polymerases topic in portion Basics of Genetic Engineering of Genetic Engineering

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Correct option is (c) The smaller fragment is having 5’-3’ exonuclease activity whereas the larger fragment is having 5’-3’ polymerase and 3’-5’ exonuclease activity

The best explanation: The two fragments are generated of size 35kDa and 76kDa, with the N terminal in the smaller fragment. The larger fragment, which is of 76kDa is known as Klenow fragment and the intact molecule is called Kornberg fragment. The smaller fragment is having 5’-3’ exonuclease activity whereas the larger fragment is having 5’-3’ polymerase and 3’-5’ exonuclease activity.

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