Question

Genomic DNA is digested with Alu I restriction enzyme that is a 4-base pair cutter. Assuming a random distribution of bases in the genome, what is the frequency with which it will cut the DNA?

• A. \(\dfrac{1}{4}\)
• B. \(\dfrac{1}{2^4}\)
• C. \(\dfrac{1}{256}\)
• D. \(\dfrac{1}{1296}\)

Hint:

For restriction enzymes: \[ \text{Frequency of occurrence of recognition site} = \frac{1}{4^n} \] where \(n\) is the number of base pairs in the recognition sequence. Thus: \[ 4\text{-bp cutter} \rightarrow \frac{1}{256} \] \[ 6\text{-bp cutter} \rightarrow \frac{1}{4096} \] \[ 8\text{-bp cutter} \rightarrow \frac{1}{65536} \] These are commonly asked values in biotechnology and molecular genetics.

Answer 1

The Correct Option is C

Concept: Restriction enzymes recognize specific DNA sequences known as recognition sites. If the bases in DNA are randomly distributed, each nucleotide position has an equal probability of containing any one of the four bases: \[ A,\;T,\;G,\;C \] Thus, the probability of finding a particular base at a given position is: \[ \frac{1}{4} \] For a restriction enzyme that recognizes a sequence of \(n\) specific base pairs, the probability of finding that exact sequence at any location is: \[ \left(\frac{1}{4}\right)^n \] Therefore, a restriction enzyme recognizing a \(4\)-base pair sequence will cut, on average, once every \[ 4^4 \] base pairs.

Step 1: Determine the recognition sequence length. The enzyme Alu I is a \(4\)-base pair cutter. Therefore, \[ n=4 \]

Step 2: Calculate the probability of occurrence of the recognition sequence. Since each position can be occupied by one of four bases, \[ P(\text{specific sequence}) = \left(\frac{1}{4}\right)^4 \] \[ = \frac{1}{256} \]

Step 3: Interpret the result. This means that, on average, the recognition sequence appears once in every \[ 256 \] base pairs. Hence, the frequency with which the enzyme cuts DNA is \[ \boxed{\frac{1}{256}} \]

Alternative Formula: For a restriction enzyme recognizing a sequence of length \(n\), \[ \text{Frequency of cutting} = \frac{1}{4^n} \] For \(n=4\), \[ \frac{1}{4^4} = \frac{1}{256} \] which confirms the answer.