If the length of E.coli DNA is 1.36 mm, can you calculate the number of base pairs in E.coli?
Answer
Verified
448.8k+ views
Hint: DNA is a polymer that is made up of the repeating unit nucleotides. These nucleotides contain nitrogenous bases that pair with each other and are known as base pairs (bps). The DNA of Escherichia coli is double-stranded circular DNA.
Complete answer:
The length of DNA is usually represented in terms of the number of nucleotides (or a pair of nucleotides referred to as base pairs) present in it. The structure of DNA is such that between two consecutive base pairs, a distance of 0.34 nm (0.34×${ 10 }^{ -9 }$ m) is present. Thus, one can find the total length of a DNA of an organism by multiplying the total number of base pairs with a distance between two consecutive base pairs. Since we are given the length of the DNA of E.coli as 1.36 mm, we can divide this value by the distance between two consecutive base pairs to get the total number of base pairs present.
Total number of base pairs= 1.36mm/0.34x${ 10 }^{ -6 }$mm
Total number of base pairs= 4x${ 10 }^{ 6 }$ base pairs
Note: -In Escherichia coli, a clearly defined nucleus is not present, and thus the DNA is present in the cytoplasm of the cell and helps together with positively charged proteins in a region known as the nucleoid.
-The number of base pairs in Bacteriophage lambda is 48502 base pairs, in Bacteriophage $\phi $ ×174 is 5386 base pairs, and in a haploid cell of a human is 3.3 x ${ 10 }^{ 9 }$ base pairs.
-The length of DNA in a diploid human cell is calculated as 6.6 × ${ 10 }^{ 9 }$ bp × 0.34 × ${ 10 }^{ -9 }$m/bp, and it comes out to be approximately 2.2 meters.
Complete answer:
The length of DNA is usually represented in terms of the number of nucleotides (or a pair of nucleotides referred to as base pairs) present in it. The structure of DNA is such that between two consecutive base pairs, a distance of 0.34 nm (0.34×${ 10 }^{ -9 }$ m) is present. Thus, one can find the total length of a DNA of an organism by multiplying the total number of base pairs with a distance between two consecutive base pairs. Since we are given the length of the DNA of E.coli as 1.36 mm, we can divide this value by the distance between two consecutive base pairs to get the total number of base pairs present.
Total number of base pairs= 1.36mm/0.34x${ 10 }^{ -6 }$mm
Total number of base pairs= 4x${ 10 }^{ 6 }$ base pairs
Note: -In Escherichia coli, a clearly defined nucleus is not present, and thus the DNA is present in the cytoplasm of the cell and helps together with positively charged proteins in a region known as the nucleoid.
-The number of base pairs in Bacteriophage lambda is 48502 base pairs, in Bacteriophage $\phi $ ×174 is 5386 base pairs, and in a haploid cell of a human is 3.3 x ${ 10 }^{ 9 }$ base pairs.
-The length of DNA in a diploid human cell is calculated as 6.6 × ${ 10 }^{ 9 }$ bp × 0.34 × ${ 10 }^{ -9 }$m/bp, and it comes out to be approximately 2.2 meters.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Class 12 Question and Answer - Your Ultimate Solutions Guide
Master Class 12 Economics: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
What are the major means of transport Explain each class 12 social science CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
What is a transformer Explain the principle construction class 12 physics CBSE
Explain sex determination in humans with the help of class 12 biology CBSE