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Molecular Basis of Inheritance - DNA Replication

Grade 12CBSEBiology

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

🔑Concepts

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DNA replication is semi-conservative, meaning each daughter DNA molecule retains one parental strand and one newly synthesized strand, as proven by Meselson and Stahl using 15N^{15}N and 14N^{14}N isotopes.

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The replication process begins at a specific site called the Origin of Replication (OriOri). Because DNA molecules are long, the whole molecule cannot be unzipped at once; instead, a small opening called the 'Replication Fork' is formed.

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The enzyme DNA-dependent DNA polymerase catalyzes the polymerization of deoxynucleotides only in the 5′→3′5' \rightarrow 3' direction. This creates a constraint on the replication fork.

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On the template strand with 3′→5′3' \rightarrow 5' polarity, replication is continuous, forming the 'Leading Strand'. On the template strand with 5′→3′5' \rightarrow 3' polarity, replication is discontinuous, forming small 'Okazaki fragments'.

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The enzyme DNA Ligase joins the discontinuously synthesized Okazaki fragments. DNA Helicase is responsible for unwinding the double helix by breaking HH-bonds between base pairs.

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Deoxyribonucleoside triphosphates (dNTPsdNTPs) serve a dual purpose: they act as substrates for polymerization and provide energy for the reaction through the hydrolysis of high-energy phosphate bonds (PPi→2PiPP_i \rightarrow 2P_i).

📐Formulae

Direction of synthesis: 5′→3′\text{Direction of synthesis: } 5' \rightarrow 3'

Base Pairing: A=T (2 H-bonds), G≡C (3 H-bonds)\text{Base Pairing: } A = T \text{ (2 } H\text{-bonds), } G \equiv C \text{ (3 } H\text{-bonds)}

Fraction of Hybrid DNA (after n generations): 22n\text{Fraction of Hybrid DNA (after } n \text{ generations): } \frac{2}{2^n}

Fraction of Light DNA (after n generations): 2n−22n\text{Fraction of Light DNA (after } n \text{ generations): } \frac{2^n - 2}{2^n}

💡Examples

Problem 1:

If E.coliE. coli was allowed to grow for 8080 minutes in a medium containing 14N^{14}N, starting from a parent cell with pure 15N^{15}N DNA, what would be the proportion of light, hybrid, and heavy DNA? (Assume E.coliE. coli divides every 2020 minutes).

Solution:

  1. Number of generations n=8020=4n = \frac{80}{20} = 4.
  2. Total DNA molecules = 24=162^4 = 16.
  3. Hybrid molecules (15N−14N^{15}N-^{14}N) = 22 (always constant after first replication).
  4. Light molecules (14N−14N^{14}N-^{14}N) = 16−2=1416 - 2 = 14.
  5. Heavy molecules (15N−15N^{15}N-^{15}N) = 00.
  6. Ratio (Light:Hybrid) = 14:214:2 or 7:17:1.

Explanation:

In semi-conservative replication, the two original 15N^{15}N strands are always preserved as one half of two 'hybrid' molecules. All subsequent synthesis uses the available 14N^{14}N in the medium.

Problem 2:

Calculate the energy source for the polymerization of DNA. Why are dNTPsdNTPs used instead of dNMPsdNMPs?

Solution:

The reaction is: dNTP+(DNA)n→Polymerase(DNA)n+1+PPi+energydNTP + (DNA)_n \xrightarrow{Polymerase} (DNA)_{n+1} + PP_i + \text{energy}. The hydrolysis of the pyrophosphate (PPiPP_i) provides the ΔG<0\Delta G < 0 required for the endergonic polymerization process.

Explanation:

dNTPsdNTPs (Deoxyribonucleoside triphosphates) contain high-energy terminal phosphate bonds similar to ATPATP. dNMPsdNMPs lack these high-energy bonds and cannot provide the thermodynamic drive for phosphodiester bond formation.