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Molecular Basis of Inheritance - The Search for Genetic Material (DNA versus RNA)

Grade 12CBSEBiology

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

🔑Concepts

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Transforming Principle: Frederick Griffith (1928) observed that SS-strain (virulent) Streptococcus pneumoniae killed mice, while RR-strain (non-virulent) did not. Heat-killed SS-strain mixed with live RR-strain killed the mice, suggesting a 'transforming principle' moved from SS to RR.

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Biochemical Characterization: Avery, MacLeod, and McCarty (1933-44) proved that DNA is the transforming agent. They used proteases, RNases, and DNases; only DNase inhibited the transformation, proving DNA is the genetic material.

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Hershey-Chase Experiment (1952): Used Bacteriophage T2T_2 labeled with 32P^{32}P (to label DNA) and 35S^{35}S (to label protein). They proved that only 32P^{32}P entered the bacteria, confirming DNA as the genetic material.

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Stability of DNA vs RNA: DNA is chemically less reactive and structurally more stable than RNA because DNA lacks the reactive 2′−OH2'-OH group present in the ribose sugar of RNA. RNA has 2′−OH2'-OH at every nucleotide, making it labile and easily degradable.

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Thymine vs Uracil: DNA contains Thymine (55-methyl uracil), which confers additional stability compared to Uracil found in RNA.

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Mutation and Expression: RNA can mutate at a faster rate, making viruses with RNA genomes (e.g., HIV) evolve quickly. DNA is better for the storage of genetic information, while RNA is better for the transmission of genetic information.

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Chargaff's Rules: In a double-stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equal to one.

📐Formulae

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

[A]+[G]=[T]+[C][A] + [G] = [T] + [C]

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

Pitch of DNA helix=3.4 nm\text{Pitch of DNA helix} = 3.4 \text{ nm}

💡Examples

Problem 1:

In a sample of double-stranded DNA, the content of Adenine (AA) is found to be 20%20\%. Calculate the percentage of Cytosine (CC).

Solution:

According to Chargaff's rule, [A]=[T][A] = [T]. Since [A]=20%[A] = 20\%, then [T]=20%[T] = 20\%. The sum of [A]+[T]=20%+20%=40%[A] + [T] = 20\% + 20\% = 40\%. The remaining 100%−40%=60%100\% - 40\% = 60\% must represent [G]+[C][G] + [C]. Since [G]=[C][G] = [C], then [C]=60%2=30%[C] = \frac{60\%}{2} = 30\%.

Explanation:

Chargaff's rule states that the amount of purines equals the amount of pyrimidines in double-stranded DNA.

Problem 2:

Why did Hershey and Chase use 35S^{35}S and 32P^{32}P to label the bacteriophage?

Solution:

They used 35S^{35}S to label protein because sulfur is a constituent of certain amino acids (Cysteine and Methionine) but is absent in DNA. They used 32P^{32}P to label DNA because phosphorus is a major constituent of the DNA backbone (phosphate groups) but is absent in proteins.

Explanation:

This specific labeling allowed them to track whether the protein coat or the DNA core of the virus entered the bacterial cell during infection.