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to estimate the number of cleavage sites in a particular piece of dna with a known size, you can apply the formula, n/4n where n is the number of base pairs in the target dna and n is the number of bases in the recognition sequence of the restriction enzyme. if the recognition sequence for sau3ai is gatc and the haemophilus influenzae genome contains approximately 1,830,000 bp , how many cleavage sites would you expect?

User Dxpelou
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Final answer:

The Haemophilus influenzae genome contains approximately 1,830,000 base pairs, and the recognition sequence for sau3ai is GATC. Therefore, the number of cleavage sites would be 457,500.

Step-by-step explanation:

To estimate the number of cleavage sites in a particular piece of DNA, we can use the formula n/4n.

In this case, n represents the number of base pairs in the target DNA, and n represents the number of bases in the recognition sequence of the restriction enzyme. The recognition sequence for sau3ai is GATC.

The Haemophilus influenzae genome contains approximately 1,830,000 base pairs. Therefore, the number of cleavage sites would be:

1830000 / 4 = 457500

User Andrew Barr
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