Question

The following sequence of 30 nucleotides corresponds to one of the two strands of a double stranded DNA:

5’ GATGTGATCAGACCGGGTGCACTCTAATCT 3’

a) This sequence has two perfect palindromes that consist of 6 base pairs each. What is the sequence of these two palindromes?

b) Show both strands of your FIRST palindrome (indicate the 5’-3’ polarity)

c) Show both strands of your SECOND palindrome (indicate the 5’-3’ polarity)

Assume that the two palindromes are recognized by “6-cutter” restriction enzymes, and that each enzyme creates blunt ends. Assuming the 30bp DNA sequence is digested with BOTH enzymes, and that the resulting dsDNA fragments are stable (i.e. the complementary strands remain annealed):

1. How many fragments would you expect upon digestion?
2. How long would these fragments be (in bp)?

Answer 1

Question 1

a) The two palindromes of 6 base pair are: 5’-TGATCA-3’ & 5’-GTGCAC-3’.

Palindromes are the words which read same from both the sides. The order of arrangement of words is same from both side that is the reverse of word will be the same as the read from front.

The restriction enzymes are the specific enzymes which search the DNA strands and cuts it at the specific part. Every restriction enzyme has a specific place where they cut.

When the restriction enzyme cuts the palindromes it either cuts it and leave the sticky ends or it leaves the blunt ends. In sticky ends some nucleotide are left unpaired and hence other DNA can directly attach to it with the complementarity. But when blunt ends are produced the ends have paired nucleotide base pairs.

b) Two strands of first palindrome:

5’ – TGATCA - 3’

3’ – ACTAGC - 5’

c) Two strands of second palindrome:

5’ – GTGCAC - 3’

3’ – CACGTG - 5'

QUESTION 2

a. Three sequences will be produced in the end of the 6-cutter restriction enzyme activity as the enzymes gives the blunt ends there will be no handing or sticky part left and the enzyme will cut the strand in the middle of the palindromic sequence that is after the 3^rd nucleotide. As the enzyme will cut the DNA sequence at two places so there will be three parts will be produced.

The restriction enzymes are the specific enzymes which search the DNA strands and cuts it at the specific part. Every restriction enzyme has a specific place where they cut.

When the restriction enzyme cuts the palindromes it either cuts it and leave the sticky ends or it leaves the blunt ends. In sticky ends some nucleotide are left unpaired and hence other DNA can directly attach to it with the complementarity. But when blunt ends are produced the ends have paired nucleotide base pairs.

The sequence of three fragments will be

5’ – GATGTGA – 3’

5’ – TCAGACCGGGTG – 3’

5’ – CACTCTAATCT - 3’

b. The three sequences produced will be of 7 base pairs, 12 base pairs and 11 base pairs.

7 BASE PAIRS SEQUENCE

5’ – GATGTGA – 3’

3’ – CTACACT – 5’

12 BASE PAIRS SEQUENCE

5’ – TCAGACCGGGTG – 3’

3’ – AGTCTGGCCCAC- 5’

11 BASE PAIRS SEQUENCE

5’ – CACTCTAATCT - 3’

3’ – GTGAGATTAGA – 5’