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If the length of E. coli 1.36 mm, can you calculate the number of base pairs in E.coli?
Most Upvoted Answer
If the length of E. coli 1.36 mm, can you calculate the number of base...
Its quite easy,


"First of all note this FACT-The
length between two
base pair is 0.34nm."


Total length of double helix DNA = Total no:of base
pairs × Distace between two base pairs.

Here, total no:of base pair is unknown, let us assume

it to be ¥.


1.36mm = ¥ × (0.34×10`9)

ie. (¥ = 4 ×10`6)


"i
hope
u
get
this"
Community Answer
If the length of E. coli 1.36 mm, can you calculate the number of base...
Calculation of the Number of Base Pairs in E. coli

To calculate the number of base pairs in E. coli, we need to consider the size of the E. coli genome and the average length of a base pair.

The Size of the E. coli Genome
- The E. coli genome consists of a single circular DNA molecule called the bacterial chromosome.
- The size of the E. coli genome is approximately 4.6 million base pairs (4.6 Mbp).

The Average Length of a Base Pair
- In DNA, there are four nucleotides: adenine (A), cytosine (C), guanine (G), and thymine (T).
- A base pair consists of two nucleotides, where A always pairs with T and C always pairs with G.
- The average length of a base pair in DNA is approximately 0.34 nanometers (nm).

Calculation
To calculate the number of base pairs in E. coli, we can use the following formula:

Number of base pairs = Size of the genome / Average length of a base pair

- Size of the genome = 4.6 million base pairs (4,600,000 bp)
- Average length of a base pair = 0.34 nm

Now, let's calculate the number of base pairs in E. coli:

Number of base pairs = 4,600,000 bp / 0.34 nm

To simplify the calculation, we convert the size of the genome from base pairs to nanometers:

Number of base pairs = 4,600,000 bp * 0.34 nm/bp

Number of base pairs = 1,564,000,000 nm

Therefore, the number of base pairs in E. coli is approximately 1,564,000,000.

Conclusion
The number of base pairs in E. coli is approximately 1,564,000,000. This calculation is based on the size of the E. coli genome, which is approximately 4.6 million base pairs, and the average length of a base pair, which is approximately 0.34 nanometers.
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If the length of E. coli 1.36 mm, can you calculate the number of base pairs in E.coli?
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