task a) The restriction enzyme AluI cleaves at the sequence 5´- AGCT-3´, EcoRI c
ID: 272252 • Letter: T
Question
task
a) The restriction enzyme AluI cleaves at the sequence 5´- AGCT-3´, EcoRI cleaves at 5´-GAATTC-3´ and NotI cleaves at 5-GCGGCCGC-3´. What would be the average distance between cleavage sites for each enzyme on digestion of double-stranded DNA? Assume that the DNA contains equal proportions of A, G, C and T.
b) Taq1 recognizes the sequence 5´-TCGA-3, whereas MinI recognizes the sequence 5´-ATNAT-3´ where N represents any nucleotide. Which statement is true:
1) Both sequences will be present at the same frequency in random DNA
2) Taq1 sites will outnumber MinI sites in random DNA
3) MinI sites will outnumber Taq1 sites in random DNA
c) The recognition sequence for the restriction endonuclease HincII is GTYRAC
where R is any purine (A or G) and Y is any pyrimidine (C or T). In a truly random DNA sequence what will be the average distance between successive HincII cut sites?
d) The restriction endonuclease, XhoI, recognizes the sequence 5´-CTCGAG-3´ and cleaves between the C and the T on each strand.
The restriction endonuclease, SalI, recognizes the sequence 5´- GTCGAC-3´and cleaves between the G and the T on each strand.
Can the sticky ends produced after XhoI and SalI cleavage adhere to each other? Explain your reasoning.
Explanation / Answer
a) The restriction enzyme AluI cleaves at the sequence 5´- AGCT-3´, EcoRI cleaves at 5´-GAATTC-3´ and NotI cleaves at 5-GCGGCCGC-3´. What would be the average distance between cleavage sites for each enzyme on digestion of double-stranded DNA? Assume that the DNA contains equal proportions of A, G, C and T.
The answer is
DNA has composed with four base pairs, out of four only one base occupaies one position at a time. So, the probability of one base is 1/4.
So, the probability of AluI = AGCT = 1/4*1/4*1/4*1/4 = 1/256. which means the probability of AluI restriction site is 1 per 256 basepairs.
the probability of EcoRI = GAATTC = 1/4*1/4*1/4*1/4*1/4*1/4 = 1/4096. which means the probability of EcoRI restriction site is 1 per 4096 basepairs.
the probability of NotI = GCGGCCGC = 1/4*1/4*1/4*1/4*1/4*1/4*1/4*1/4 = 1/65536. which means the probability of EcoRI restriction site is 1 per 65536 basepairs.
b) Taq1 recognizes the sequence 5´-TCGA-3, whereas MinI recognizes the sequence 5´-ATNAT-3´ where N represents any nucleotide. Which statement is true:
1) Both sequences will be present at the same frequency in random DNA
2) Taq1 sites will outnumber MinI sites in random DNA
3) MinI sites will outnumber Taq1 sites in random DNA
The probability of Taq1 = TCGA = 1/4*1/4*1/4*1/4 = 1/256. which means the probability of TaqI recognistion site is 1 per 256 basepairs.
The probability of MinI = ATNAT = 1/4*1/4*1/4*1/4*1/4 = 1/1024. which means the probability of MinI recognistion site is 1 per 1024 basepairs.
So, the answer is 2) Taq1 sites will outnumber MinI sites in random DNA
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