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Molecular Biology - DNA Replication, Transcription, and Translation

Grade 11IBBiology

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

🔑Concepts

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DNA replication is semi-conservative, meaning each new DNADNA molecule consists of one original strand and one newly synthesized strand, as proven by the Meselson-Stahl experiment using 15N^{15}N and 14N^{14}N isotopes.

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Helicase unwinds the double helix and separates the strands by breaking hydrogen bonds, while DNADNA Polymerase III links nucleotides together in a 5′→3′5' \rightarrow 3' direction to form a new strand.

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Transcription is the synthesis of mRNAmRNA copied from the DNADNA base sequences by RNARNA polymerase. The enzyme moves along the antisense strand in a 5′→3′5' \rightarrow 3' direction.

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Translation is the synthesis of polypeptides on ribosomes. The amino acid sequence is determined by mRNAmRNA according to the genetic code, where one codon (three bases) codes for one amino acid.

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The genetic code is universal and degenerate; almost all organisms use the same code, and multiple codons can code for the same amino acid (e.g., GGUGGU, GGCGGC, GGAGGA, and GGGGGG all code for Glycine).

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In PCRPCR (Polymerase Chain Reaction), TaqTaq DNADNA polymerase is used because it is heat-stable and does not denature at the high temperatures required to separate DNADNA strands.

📐Formulae

namino_acids=nnucleotides3−1 (excluding the stop codon)n_{amino\_acids} = \frac{n_{nucleotides}}{3} - 1 \text{ (excluding the stop codon)}

[A]+[G]=[T]+[C] (Chargaff’s Rule for double-stranded DNA)[A] + [G] = [T] + [C] \text{ (Chargaff's Rule for double-stranded DNA)}

%A=%T,%G=%C\%A = \%T, \%G = \%C

5′→3′ (Direction of synthesis for both DNA and RNA)5' \rightarrow 3' \text{ (Direction of synthesis for both DNA and RNA)}

💡Examples

Problem 1:

A specific mRNAmRNA molecule consists of 450450 nucleotides that are used for translation. Calculate the maximum number of amino acids in the resulting polypeptide chain.

Solution:

450÷3=150450 \div 3 = 150 amino acids (including the stop codon). If we exclude the stop codon, the polypeptide contains 149149 amino acids.

Explanation:

Each codon consists of 33 nucleotides and codes for 11 amino acid. Since the stop codon does not code for an amino acid, it is subtracted from the total count of codons.

Problem 2:

If a sample of double-stranded DNADNA contains 22%22\% Cytosine (CC), calculate the percentage of Adenine (AA) in the sample.

Solution:

If C=22%C = 22\%, then G=22%G = 22\%. Total C+G=44%C+G = 44\%. Therefore, A+T=100%−44%=56%A+T = 100\% - 44\% = 56\%. Since A=TA = T, A=56%2=28%A = \frac{56\%}{2} = 28\%.

Explanation:

According to Chargaff's rule, in double-stranded DNADNA, the amount of Guanine equals Cytosine, and Adenine equals Thymine.

DNA Replication, Transcription, and Translation Revision - Grade 11 Biology IB