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