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Molecular Biology - DNA replication, transcription and translation

Grade 12IBBiology

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

🔑Concepts

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DNA replication is semi-conservative, meaning each new DNA 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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The enzyme Helicase unwinds the double helix and separates the strands by breaking hydrogen bonds between A=TA=T and G≡CG \equiv C base pairs.

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DNA Polymerase III adds nucleotides in a 5′→3′5' \rightarrow 3' direction, requiring a primer synthesized by RNA Primase to provide a free 3′3'-OH group.

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Replication is continuous on the leading strand and discontinuous on the lagging strand, where it forms Okazaki fragments that are eventually joined by DNA Ligase.

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Transcription is the process by which an RNA sequence is produced from a DNA template. RNA Polymerase binds to a promoter region and synthesizes mRNA in a 5′→3′5' \rightarrow 3' direction.

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The 'Antisense' strand (template strand) is complementary to the mRNA, while the 'Sense' strand (coding strand) has the same sequence as the mRNA (with UU replacing TT).

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Translation occurs at the ribosome, where mRNA codons are read in the 5′→3′5' \rightarrow 3' direction. tRNA molecules carry specific amino acids and possess anticodons that pair with mRNA codons via complementary base pairing.

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

📐Formulae

Chargaff’s Rule: [A][T]=[G][C]=1\text{Chargaff's Rule: } \frac{[A]}{[T]} = \frac{[G]}{[C]} = 1

Total Bases: [A]+[G]+[C]+[T]=100%\text{Total Bases: } [A] + [G] + [C] + [T] = 100\%

Purines (A+G)=Pyrimidines (C+T)\text{Purines } (A+G) = \text{Pyrimidines } (C+T)

Nucleotide Addition: (DNA)n+dNTP→DNA Pol(DNA)n+1+PPi\text{Nucleotide Addition: } (DNA)_n + dNTP \xrightarrow{\text{DNA Pol}} (DNA)_{n+1} + PP_i

💡Examples

Problem 1:

A double-stranded DNA molecule contains 18%18\% Adenine (AA). Calculate the percentage of Cytosine (CC) present in this molecule.

Solution:

C=32%C = 32\%

Explanation:

According to Chargaff's rule, [A]=[T][A] = [T]. Therefore, [T]=18%[T] = 18\%. The sum of [A]+[T]=18%+18%=36%[A] + [T] = 18\% + 18\% = 36\%. The remaining percentage for G+CG + C is 100%−36%=64%100\% - 36\% = 64\%. Since [G]=[C][G] = [C], the percentage of Cytosine is 64%2=32%\frac{64\%}{2} = 32\%.

Problem 2:

Given the DNA antisense strand sequence: 3′−TAC GGC ATT−5′3'-TAC \, GGC \, ATT-5', determine the resulting mRNA sequence and the corresponding tRNA anticodons.

Solution:

mRNA: 5′−AUG CCG UAA−3′5'-AUG \, CCG \, UAA-3'; tRNA anticodons: 3′−UAC−5′3'-UAC-5', 3′−GGC−5′3'-GGC-5', 3′−AUU−5′3'-AUU-5'

Explanation:

mRNA is synthesized complementary to the antisense strand in the 5′→3′5' \rightarrow 3' direction. TT pairs with AA, AA with UU (in RNA), CC with GG, and GG with CC. tRNA anticodons are complementary to the mRNA codons.

Problem 3:

Identify the direction of movement of the replication fork and the synthesis direction of the lagging strand.

Solution:

Replication fork moves towards the unwinding DNA; Lagging strand synthesis is 5′→3′5' \rightarrow 3' (moving away from the fork).

Explanation:

While the replication fork opens the DNA, DNA Polymerase III can only add nucleotides to the 3′3' end of the new strand. Thus, the lagging strand is synthesized in short bursts (5′→3′5' \rightarrow 3') in the direction opposite to the fork's movement.

DNA replication, transcription and translation Revision - Grade 12 Biology IB