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Nady [450]
3 years ago
11

The domestic cat genome contains 2.9×109 base pairs. The length of linker DNA in mammals is 50 base pairs. Approximately how man

y nucleosomes are required to organize the 10-nm-fiber structure of the genome?
Biology
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

Number of nucleosomes in 2.9 * 10^9bp 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\\= 196bp

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

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

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

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