# if the length of E.coli DNA is 1.36mm, calculate the no.of base pairs in E.coli

The distance between two consecutive base pairs is 0.34 × 10 -9 m

The total length of double helix DNA in an organism can be calculated by multiplying the total number of base pairs with the distance between two consecutive base pairs.

Total length of double helix DNA = total number of bp × distance between two consecutive bp

Let, the total number of bp be x

• 1.36mm = x × 0.34 × 10 -9 m

• x = 1.36 ×10 -3 / 0.34 × 10 -9 m

• x = 4 × 10 6 bp

• The number of base pairs in the given strand of DNA of E.coli are 4 X 10 6 bp

• 104

In 3.4 nm,10 bp are present.

hence for 1.36 mm 4*106 base pairs would be present

• 16

How..????

• -2

4.6 * 106

• -4

the distance between each bp is .34 nm. Thus, total no. of b present = total length of DNA (nm) / 0.34 nm

For E.coli, tot. no. of bp = 1.36 x 109 / 0.34 = 3.91 x 109

• -1

actually @saloni hav used correct method only the units and calculations are to b corrected

i.e 1.35mm=1.35 x 10 -3 m and 0.34nm= 0.34 x 10 -9

therefore total no. of base pairs =1.35 x 10 -3 / 0.34 x 10-9

=4 x 10 6

• 25

i.e 1.35mm=1.35 x 10 -3 m and 0.34nm= 0.34 x 10 -9

therefore total no. of base pairs =1.35 x 10 -3 / 0.34 x 10-9

=4 x 10 6

• 7
What are you looking for?