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Molecular Basis of Inheritance - DNA Packaging

Grade 12CBSEBiology

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

🔑Concepts

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DNA is a negatively charged polymer due to the presence of phosphate groups (PO43−PO_4^{3-}). In eukaryotes, it is packaged with positively charged basic proteins called Histones.

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Histones are rich in the basic amino acid residues LysineLysine and ArginineArginine, which carry positive charges in their side chains.

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The basic unit of DNA packaging in eukaryotes is the Nucleosome. It consists of a histone octamer (two molecules each of H2AH2A, H2BH2B, H3H3, and H4H4) wrapped by approximately 200200 base pairs (bpbp) of DNA.

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The 'beads-on-a-string' structure seen under an electron microscope represents nucleosomes in chromatin.

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Higher-level packaging of chromatin requires an additional set of proteins collectively referred to as Non-histone Chromosomal (NHCNHC) proteins.

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Chromatin is classified into two types: Euchromatin (loosely packed, stains light, and is transcriptionally active) and Heterochromatin (densely packed, stains dark, and is transcriptionally inactive).

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In prokaryotes like E.coliE. coli, DNA is not scattered but is held with some proteins in a region called the nucleoid.

📐Formulae

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

Number of Nucleosomes=Total number of base pairs200\text{Number of Nucleosomes} = \frac{\text{Total number of base pairs}}{200}

Distance between two consecutive base pairs=0.34 nm=0.34×10−9 m\text{Distance between two consecutive base pairs} = 0.34 \text{ nm} = 0.34 \times 10^{-9} \text{ m}

💡Examples

Problem 1:

If the length of E.coliE. coli DNA is 1.36 mm1.36 \text{ mm}, calculate the total number of base pairs present in the DNA.

Solution:

The number of base pairs (bpbp) is calculated as 1.36×10−3 m0.34×10−9 m/bp=4×106 bp\frac{1.36 \times 10^{-3} \text{ m}}{0.34 \times 10^{-9} \text{ m/bp}} = 4 \times 10^6 \text{ bp}.

Explanation:

We divide the total length of the DNA by the distance between two consecutive base pairs (0.34 nm0.34 \text{ nm}) to find the total count of base pairs.

Problem 2:

Calculate the number of nucleosomes present in a diploid human cell containing 6.6×109 bp6.6 \times 10^9 \text{ bp}.

Solution:

Number of nucleosomes =6.6×109200=3.3×107= \frac{6.6 \times 10^9}{200} = 3.3 \times 10^7.

Explanation:

One typical nucleosome contains approximately 200 bp200 \text{ bp} of DNA. Dividing the total genomic DNA by 200200 gives the total number of nucleosome units.