Question

. For each of the restriction enzymes listed below: (i) Approximately how many restriction fragments would result from digestion of the human genome (3 x 109 bases) with the enzyme? (ii) State whether the fragments produced by digestion with each enzyme would have sticky ends with a 5’ overhang, sticky ends with a 3’ overhang, or blunt ends. (The recognition sequence for each enzyme is given in parentheses, where N means any of the four nucleotides. ^ marks the site of cleavage.)

a. EcoRV (GAT^ATC)

b. HpaII (C^CGG)

c. DrdI (GACNNNN^NNGTC)

Answer 1

Given,

Size of Human Genome=3*109 =3,000,000,000 bp

(i) EcoRV (GATATC)

As this restriction site consists of 6 base pairs,

the probability of finding the site in a genome will be (P)= 1/46

Hence, total number of sites in a given genome=genome size* P

= 3*109 *1/46

= 732421.87

i.e. 732421 sites of EcoRV enzyme may be present in the given human genome

Since the human genome is linear,

For linear DNA No. of fragments= No. of restriction site + 1

Therefore, 732421 sites within the genome will produce 732422 fragments.

for example, two sites within the linear genome will produce three fragments and three sites within the genome will produce four fragments.

Also, this restriction enzyme will produce a blunt end because the cleavage site is exactly in the middle.

(b) Hpall (CCGG)

As this restriction site consists of 4 bp

the probability of finding this site (P)= 1/44

Hence, the total number of sites in a given genome=genome size* P

= 3*109 *1/44

= 11718750

i.e. 11718750 sites of EcoRV enzyme may be present in the given human genome

Since the human genome is linear

For linear DNA No. of fragments= No. of restriction site + 1

Therefore, 11718750 sites within the genome will produce 711718751 fragments.

for example, two sites within the linear genome will produce three fragments and three sites within the genome will produce four fragments.

Also, this restriction enzyme will produce a sticky end with 5 prime overhang

(c) Drdl (GACNNNNNNGTC)

Number of Non-ambiguous bases=6

Probability of N= 1

the probability of finding the site in a genome will be (P)= 1/46 *1

Hence, the total number of sites in a given genome=genome size* P

= 3*109 *1/46

= 732421.87

i.e. 732421 sites of EcoRV enzyme may be present in the given human genome

Since the human genome is Linear

For linear DNA No. of fragments= No. of restriction site + 1

therefore, 732421 sites within the genome will produce 732422 fragments.

Also, this restriction enzyme will produce a sticky end with 3' overhang.