Question

An enzyme that cleaves DNA recognizes an eight base-pair unique DNA sequence. The probable number of times this enzyme will cleave a 70 kilobase pair random DNA sequence is
A. 1
B. 2
C. 3
D. 4

Answer 1

One can say that the sequencing of DNA is essential to apply to the human genome. It is because that DNA sequencing can permit researchers to arrange qualities and genomes. It is said that the researchers can search for changes that cause illness whenever qualities are distinguished and investigated from grouping data. They also included these lines giving significant clinical data.

Complete answer:
As indicated by the question, the limitation grouping for the catalyst is an eight base pair long arrangement. If a self-assertive DNA grouping of 8 base pairs as XXXXXXXX, where "X" can be any of four nucleotides (A/T/G/C); there are four opportunities for one "X," and the complete arrangement is eight base sets.
For each of the eight bases; the likelihood of having the given succession is ($4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 65536$ base sets), which implies that the given restriction site happens for each 65,536 base sets. A restriction site is a succession of roughly 6–8 base sets of DNA that ties to a given limitation catalyst. These limitation proteins, of which there are many, have been confined from microbes. Their regular capacity is to inactivate attacking infections by severing the viral DNA.
For a DNA portion of 70 kb, the chance of having the given arrangement $= \dfrac{70000}{65536} =$ around 1.

Hence, the correct answer is option (A).

Note: Restriction proteins perceive short DNA arrangements and separate twofold abandoned DNA at explicit locales inside or adjoining these groupings. In the question, an eight base pair long arrangement is given. Consequently, the restriction site will happen at 65,536 base sets. Restriction destinations, or acknowledgment regions, are situated on a DNA particle-containing explicit arrangements of nucleotides, which are perceived by limitation proteins. The arbitrary DNA arrangement will be $\dfrac{70000}{65536}$ i.e., 1.