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

Answer:

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

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

$\frac{2.9* 10^9}{196} \\1.47 * 10^7$