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Molecular Basis of Inheritance - Structure of DNA and RNA

Grade 12CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A nucleotide has three components: a nitrogenous base, a pentose sugar (ribose in RNA, and deoxyribose in DNA), and a phosphate group. The base is linked to the 1′1' carbon of the pentose sugar via an NN-glycosidic linkage.

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Nitrogenous bases are of two types: Purines (Adenine and Guanine) and Pyrimidines (Cytosine, Thymine, and Uracil). In DNA, Thymine (55-methyl uracil) is present, whereas in RNA, Uracil (UU) is found in place of Thymine (TT).

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A phosphate group is linked to the 5′5' carbon of a nucleoside through a phosphoester linkage. Two nucleotides are linked through a 3′−5′3'-5' phosphodiester linkage to form a dinucleotide.

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The double-helix model of DNA (Watson and Crick) features two polynucleotide chains with anti-parallel polarity: one in 5′→3′5' \rightarrow 3' direction and the other in 3′→5′3' \rightarrow 5' direction.

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Base pairing is specific: Adenine pairs with Thymine via two hydrogen bonds (A=TA=T), and Guanine pairs with Cytosine via three hydrogen bonds (G≡CG \equiv C).

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In prokaryotes (e.g., E.coliE. coli), DNA is held with some proteins in a region called 'nucleoid'. In eukaryotes, DNA packaging involves histones (positively charged, basic proteins) forming a nucleosome, which contains roughly 200200 bp of DNA helix.

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The Central Dogma of molecular biology states that genetic information flows from DNA→RNA→ProteinDNA \rightarrow RNA \rightarrow Protein.

📐Formulae

[A]+[G]=[T]+[C][A] + [G] = [T] + [C] (Chargaff's Rule: Purines = Pyrimidines)

[A][T]=[G][C]=1\frac{[A]}{[T]} = \frac{[G]}{[C]} = 1

Length of DNA=Total number of bp×0.34×10−9 m/bp\text{Length of DNA} = \text{Total number of bp} \times 0.34 \times 10^{-9} \text{ m/bp}

Pitch of Helix=3.4 nm (per 10 base pairs)\text{Pitch of Helix} = 3.4 \text{ nm (per } 10 \text{ base pairs)}

💡Examples

Problem 1:

If a double-stranded DNA has 20%20\% of Cytosine, calculate the percentage of Adenine in the DNA.

Solution:

30%30\%

Explanation:

According to Chargaff's rule, the amount of Guanine equals Cytosine, so [G]=20%[G] = 20\%. The total [G]+[C]=20%+20%=40%[G] + [C] = 20\% + 20\% = 40\%. The remaining 60%60\% must be divided equally between Adenine and Thymine because [A]=[T][A] = [T]. Therefore, [A]=60%2=30%[A] = \frac{60\%}{2} = 30\%.

Problem 2:

Calculate the length of the DNA of a bacteriophage ϕ×174\phi \times 174 if it contains 53865386 nucleotides (Note: it is single-stranded).

Solution:

1831.24 nm1831.24 \text{ nm} or 1.83 \mum1.83 \text{ \mu m}

Explanation:

Since it is single-stranded, we simply multiply the number of nucleotides by the distance between consecutive bases (0.34 nm0.34 \text{ nm}). Length =5386×0.34 nm=1831.24 nm= 5386 \times 0.34 \text{ nm} = 1831.24 \text{ nm}.

Problem 3:

In a DNA molecule, if the distance between two consecutive base pairs is 0.34 nm0.34 \text{ nm} and the total number of base pairs in a mammalian cell is 6.6×109 bp6.6 \times 10^9 \text{ bp}, what is the approximate length of the DNA?

Solution:

2.2 metres2.2 \text{ metres}

Explanation:

Length =(6.6×109 bp)×(0.34×10−9 m/bp)= (6.6 \times 10^9 \text{ bp}) \times (0.34 \times 10^{-9} \text{ m/bp}). The 10910^9 and 10−910^{-9} cancel out, leaving 6.6×0.34≈2.244 m6.6 \times 0.34 \approx 2.244 \text{ m}.