Chapter 11, Problem 1MCQ

If a restriction enzyme cuts between G and A whenever it encounters the sequence GAATTC, how many fragments will be produced when the enzyme digests DNA with the following sequences?

TGAGAATTCAACTGAATTCAAATTCGAATTCTTAGC

a.	Two
b.	Three
c.	Four
d.	Five

Summary Introduction

Introduction:

Restriction enzyme finds the specific sequence of interest and cut the DNA sequence. It results in the fragments of specific DNA sequence.

Answer to Problem 1MCQ

Correct answer:

GAATTC occurs three times in the given sequence. Thus, the restriction enzyme will cut three times, which will produce four fragments. Hence, the correct answer is option c.

Explanation of Solution

Reason for correct answer:

Option c. is given as, “Four.”

Restriction enzyme will cut the sequence where it will find the GAATTC and this sequence is present three times in the given sequence; TGAG/AATTCAACTG/AATTCAAATTCG/AATTCTTAGC. Hence, it will produce four fragments of the given sequence.

Reason for incorrect answer:

Option a. is given as, “Two.”

GAATTC is present three times in the sequence of interest, which will generate four fragments, not two. Hence, option a. is incorrect.

Option b. is given as, “Three.”

When restriction enzyme will cut the sequence three times, then four fragments will be produced. Hence, option b. is incorrect.

Option d. is given as, “Five.”

GAATTC occurs three times in the given sequence; thus, it will generate four fragments, not five. Hence, option d. is incorrect.

Hence, the options a., b., and d. are incorrect.

Conclusion

GAATTC is present three times in the sequence of interest. The restriction enzyme will cut the sequence three times and will generate four fragments. Thus, the correct option is c.