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Ch. 15 - Recombinant DNA Technology and Its Applications
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoSanders 3rd EditionCh. 15 - Recombinant DNA Technology and Its ApplicationsProblema 2a
Capitolo 15, Problema 2a

The human genome is 3×10⁹ bp in length.
How many fragments would be predicted to result from the complete digestion of the human genome with the following enzymes: Sau3A (˘GATC), BamHI (G˘GATCC), EcoRI (G˘AATTC), and NotI (GC˘GGCCGC)?

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Step 1: Understand the problem. The question asks how many fragments would result from the complete digestion of the human genome (3×10⁹ base pairs) using specific restriction enzymes. Each enzyme recognizes a specific sequence of DNA and cuts at that sequence. To solve this, we need to calculate the frequency of each recognition sequence in the genome and then determine the number of fragments produced.
Step 2: Calculate the probability of finding each recognition sequence in the genome. For a random sequence of DNA, the probability of finding a specific base at a given position is 1/4 (since there are four bases: A, T, G, C). For a recognition sequence of length n, the probability of finding that sequence is (1/4)^n. For example, Sau3A recognizes a 4-base sequence (˘GATC), so its probability is (1/4)^4.
Step 3: Determine the expected number of recognition sites for each enzyme. Multiply the probability of finding the recognition sequence by the total number of base pairs in the genome (3×10⁹ bp). For example, for Sau3A, the expected number of recognition sites is (1/4)^4 × 3×10⁹.
Step 4: Calculate the number of fragments produced. Each recognition site corresponds to a cut in the DNA, and the number of fragments produced is equal to the number of recognition sites plus one. Perform this calculation for each enzyme: Sau3A (˘GATC), BamHI (G˘GATCC), EcoRI (G˘AATTC), and NotI (GC˘GGCCGC).
Step 5: Summarize the results. After calculating the expected number of fragments for each enzyme, compare the values to understand how the length of the recognition sequence affects the number of fragments produced. Longer recognition sequences result in fewer cuts and larger fragments, while shorter sequences produce more cuts and smaller fragments.

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Restriction Enzymes

Restriction enzymes are proteins that cut DNA at specific sequences, known as recognition sites. Each enzyme recognizes a unique sequence of nucleotides, allowing for targeted cleavage of DNA. Understanding how these enzymes work is crucial for predicting the number of fragments generated from a given DNA sequence, such as the human genome.
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DNA Fragmentation

DNA fragmentation refers to the process of breaking down long DNA molecules into smaller pieces. When a restriction enzyme cuts the DNA, it creates fragments whose number and size depend on the frequency of the enzyme's recognition sites within the DNA. Analyzing the human genome's sequence helps estimate how many fragments will result from digestion with specific enzymes.
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Genome Size and Recognition Sites

The size of the genome, measured in base pairs (bp), is essential for understanding how many times a restriction enzyme can cut the DNA. The frequency of recognition sites for each enzyme determines the expected number of fragments. For example, if an enzyme recognizes a sequence that occurs every 1,000 bp, it would theoretically produce many fragments from a 3 billion bp genome.
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Genomics Overview
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