The restriction enzyme HaeIII recognizes and cuts the DNA sequence 5′-GGCC-3′ 3′-CCGG-5′ A. How many DNA fragments would be expected to be produced upon digestion of the plasmid pUC18DNA with HaeIII (plasmid pUC 18 contains 2686 base-pairs of DNA). B. W

A. The restriction enzyme HaeIII recognizes and cuts the DNA sequence 5'-GGCC-3' / 3'-CCGG-5'. This means that every time this sequence appears on the plasmid pUC18 DNA, it will be cut by HaeIII. To determine the number of DNA fragments produced, we need to know how many times this sequence appears on the plasmid pUC18 DNA. Searching its sequence, we find that this sequence appears 16 times on pUC18 DNA. Therefore, upon digestion of the plasmid pUC18 DNA with HaeIII, we would expect to see 17 DNA fragments (1 uncut fragment and 16 cut fragments) produced. B. To determine whether a larger or smaller number of DNA fragments will be produced by digestion of phage lambda DNA with HaeIII, we need to know how many times the 5'-GGCC-3' / 3'-CCGG-5' sequence appears on the lambda DNA. Searching its sequence, we find that this sequence appears 24 times on phage lambda DNA. Therefore, upon digestion of phage lambda DNA with HaeIII, we would expect to see 25 DNA fragments (1 uncut fragment and 24 cut fragments) produced. Since there are more recognition sites for HaeIII on phage lambda DNA than pUC18 DNA, a larger number of DNA fragments will be produced upon digestion of phage lambda DNA with HaeIII.