The statistical frequency of the occurrence of a particular restrictio...
Explanation:
Given that the restriction enzyme cleavage site is 6 bases long, we need to calculate the statistical frequency of its occurrence.
The number of possible combinations of a 6-base sequence is calculated as 4^6 since each base can be one of the four nucleotides (A, T, G, C). Therefore, there are 4^6 = 4096 possible combinations.
To calculate the statistical frequency, we need to divide the total number of bases in the genome by the number of possible combinations of the 6-base sequence.
Let's assume the total number of bases in the genome is 'N'.
Statistical frequency = N / 4096
To find the option that represents the correct statistical frequency, we need to compare the given options to the expression N / 4096.
Comparison:
a) once every 24 bases: This would mean N / 24 = 4096, which implies N = 98,304. This does not satisfy the condition.
b) once every 256 bases: This would mean N / 256 = 4096, which implies N = 1,048,576. This does not satisfy the condition.
c) once every 1024 bases: This would mean N / 1024 = 4096, which implies N = 4,194,304. This does not satisfy the condition.
d) once every 4096 bases: This would mean N / 4096 = 4096, which implies N = 16,777,216. This satisfies the condition.
Therefore, the correct answer is option 'D', as it represents the statistical frequency of the occurrence of a particular restriction enzyme cleavage site that is 6 bases long once every 4096 bases.