Questions related to nucleic acids

1. Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Ans: Nitrogenous bases are : Adenine, Thymine, Uracil and Cytosine

And nucleosides are Cytidine and Guanosine.

2. If a double stranded DNA has 20 per cent of cytosine, calculate the per

cent of adenine in the DNA.

According to Erwin Chargaff’s preposition the ratio of adenine to thymine remains constant and equals unity. So in the above case the percentage of Cytosine+Guanine = 20+20= 40%. Hence remaining 60% is made up of Adenine and Thymine. So individual percentages of Adenine and Thymine is 30%. Hence %age of adenine = 30%.

3. If the sequence of one strand of DNA is written as follows:

5

Write down the sequence of complementary strand in 5

Ans: The given sequence is 5

3’-TACGTACGTACGTACGTACGTACGTACG-5’

The bottom one is the complementary strand in 3’---- 5’ direction, When we give this strand in the 5’ – 3’ direction it appears as -

5’-GCATGCATGCATGCATGCATGCATGCAT-3’

4. Why does the distance between two polynucleotide chains remain almost constant throughout the DNA.

Ans Within DNA a Purine (a bulky nitrogenous base) gets always paired with Pyrimidine (simpler nitrogenous base). This maintains the uniformity in distance throughout the chain.

5.If the length of E. coli DNA is 1.36 mm, can you calculate the number of base pairs in E.coli?

The distance between two base pairs in a nucleotide sequence is .34nm = .34*10-9m.

I m = 1000mm

.34*10-9m = .34*10-6 mm

Length of E coli DNA = 1.36 mm

Hence the number of base pairs = 1.36 mm/34*10-6 mm

The answer for this is 4*106 basepairs.

1. Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Ans: Nitrogenous bases are : Adenine, Thymine, Uracil and Cytosine

And nucleosides are Cytidine and Guanosine.

2. If a double stranded DNA has 20 per cent of cytosine, calculate the per

cent of adenine in the DNA.

According to Erwin Chargaff’s preposition the ratio of adenine to thymine remains constant and equals unity. So in the above case the percentage of Cytosine+Guanine = 20+20= 40%. Hence remaining 60% is made up of Adenine and Thymine. So individual percentages of Adenine and Thymine is 30%. Hence %age of adenine = 30%.

3. If the sequence of one strand of DNA is written as follows:

5

*'*-ATGCATGCATGCATGCATGCATGCATGC-3*'*Write down the sequence of complementary strand in 5

*'*→3*'*direction.Ans: The given sequence is 5

*'*-ATGCATGCATGCATGCATGCATGCATGC-3*'*3’-TACGTACGTACGTACGTACGTACGTACG-5’

The bottom one is the complementary strand in 3’---- 5’ direction, When we give this strand in the 5’ – 3’ direction it appears as -

5’-GCATGCATGCATGCATGCATGCATGCAT-3’

4. Why does the distance between two polynucleotide chains remain almost constant throughout the DNA.

Ans Within DNA a Purine (a bulky nitrogenous base) gets always paired with Pyrimidine (simpler nitrogenous base). This maintains the uniformity in distance throughout the chain.

5.If the length of E. coli DNA is 1.36 mm, can you calculate the number of base pairs in E.coli?

The distance between two base pairs in a nucleotide sequence is .34nm = .34*10-9m.

I m = 1000mm

.34*10-9m = .34*10-6 mm

Length of E coli DNA = 1.36 mm

Hence the number of base pairs = 1.36 mm/34*10-6 mm

The answer for this is 4*106 basepairs.