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Genetics and Evolution - Molecular Basis of Inheritance

Grade 12ICSEBiology

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

🔑Concepts

DNA as Genetic Material: The structure of DNADNA is a double helix proposed by Watson and Crick. It consists of two polynucleotide chains where the backbone is constituted by sugar-phosphate, and the bases project inside. The two chains have anti-parallel polarity (535' \rightarrow 3' and 353' \rightarrow 5').

Chargaff's Rule: In a double-stranded DNADNA, the ratio between Adenine (AA) and Thymine (TT), and Guanine (GG) and Cytosine (CC) are constant and equal to one. Thus, A+G=T+CA+G = T+C.

Nucleosome Structure: In eukaryotes, DNADNA is wrapped around a positively charged histone octamer to form a nucleosome. A typical nucleosome contains 200 bp200 \text{ bp} of DNADNA helix.

Central Dogma: Proposed by Francis Crick, it states that genetic information flows from DNARNAProteinDNA \rightarrow RNA \rightarrow \text{Protein}. In some viruses, reverse flow occurs via Reverse Transcription (RNADNARNA \rightarrow DNA).

Semiconservative Replication: DNA replication is semiconservative, meaning each daughter DNADNA molecule retains one parental strand and one newly synthesized strand, as proven by Meselson and Stahl using 15N^{15}N.

Transcription Unit: A transcription unit in DNADNA is defined by three regions: a Promoter, the Structural gene, and a Terminator. The strand with 353' \rightarrow 5' polarity acts as the template strand.

Genetic Code: The code is a triplet (6464 codons). AUGAUG has dual functions: it codes for Methionine (metmet) and acts as an initiator codon. UAA,UAG,UGAUAA, UAG, UGA are stop (nonsense) codons.

Lac Operon: A polycistronic structural gene is regulated by a common promoter and regulatory genes. In the Lac Operon, the ii gene codes for the repressor, and structural genes (z,y,az, y, a) code for β\beta-galactosidase, permease, and transacetylase respectively.

📐Formulae

Length of DNA=Total number of base pairs×0.34×109 m/bp\text{Length of DNA} = \text{Total number of base pairs} \times 0.34 \times 10^{-9} \text{ m/bp}

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

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

A=T and GC (Hydrogen bonding counts: 2 for A-T, 3 for G-C)A = T \text{ and } G \equiv C \text{ (Hydrogen bonding counts: 2 for A-T, 3 for G-C)}

💡Examples

Problem 1:

If a double stranded DNADNA has 20%20\% of Cytosine (CC), calculate the percentage of Adenine (AA) in the DNADNA.

Solution:

According to Chargaff's rule, G=CG = C. If C=20%C = 20\%, then G=20%G = 20\%. Therefore, G+C=40%G + C = 40\%. The remaining 60%60\% must be A+TA + T. Since A=TA = T, the percentage of A=60%2=30%A = \frac{60\%}{2} = 30\%.

Explanation:

Chargaff's rule implies that the total purines equal total pyrimidines in double-stranded DNA.

Problem 2:

Calculate the length of DNADNA of a mammalian cell if the total number of base pairs is 6.6×109 bp6.6 \times 10^9 \text{ bp}.

Solution:

Length=(6.6×109 bp)×(0.34×109 m/bp)=2.244 meters\text{Length} = (6.6 \times 10^9 \text{ bp}) \times (0.34 \times 10^{-9} \text{ m/bp}) = 2.244 \text{ meters}.

Explanation:

The distance between two consecutive base pairs is approximately 0.34 nm0.34 \text{ nm} (0.34×109 m0.34 \times 10^{-9} \text{ m}).

Molecular Basis of Inheritance - Revision Notes & Key Diagrams | ICSE Class 12 Biology