Question

The domestic cat genome contains 2.9×109 base pairs. The length of linker DNA in mammals is 50 base pairs. Approximately how many nucleosomes are required to organize the 10-nm-fiber structure of the genome?

Answer 1

Number of nucleosomes in
`2.9 * 10^9`bp is equal to
`1.47 * 10^7`

Step-by-step explanation:

For wounding one nucleosome, total length of DNA required is equal to
`146` bp

The length of linker DNA in mammals is equal to
`50` bp

Thus , the total length of DNA that confides between two nucleosome is equal to the sum of wounding length of DNA and the linker length

`= 146 + 50\\= 196`bp

Thus, in
`196`bp length of DNA, the total number of nucleosomes is equal to
`1`

Thus, number of nucleosomes in
`2.9 * 10^9`bp is equal to

`(2.9* 10^9)/(196) \\1.47 * 10^7`