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Continuity and change - Protein synthesis

Grade 11IBBiology

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

🔑Concepts

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Protein synthesis is the process by which cells build proteins, following the 'Central Dogma' of molecular biology: DNA→transcriptionRNA→translationPolypeptideDNA \xrightarrow{\text{transcription}} RNA \xrightarrow{\text{translation}} \text{Polypeptide}.

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Transcription occurs in the nucleus. RNA polymeraseRNA \text{ polymerase} binds to the promoter region, unwinds the DNA, and adds RNARNA nucleotides in a 5′→3′5' \rightarrow 3' direction using the antisense strand as a template.

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Complementary base pairing rules for RNARNA: Adenine (AA) pairs with Uracil (UU), and Guanine (GG) pairs with Cytosine (CC). Note that Thymine (TT) is absent in RNARNA and is replaced by UU.

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Translation occurs at the ribosomes in the cytoplasm. mRNAmRNA binds to the ribosome, and tRNAtRNA molecules carry specific amino acids to the ribosome based on the interaction between the mRNAmRNA codon and the tRNAtRNA anticodon.

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The Genetic Code is triplet-based (1 codon=3 nucleotides1 \text{ codon} = 3 \text{ nucleotides}), universal (shared by almost all living organisms), and degenerate (multiple codons can code for the same amino acid, e.g., GGUGGU, GGCGGC, GGAGGA, and GGGGGG all code for Glycine).

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The start codon is usually AUGAUG (coding for Methionine), and translation ends when the ribosome reaches a stop codon (UAAUAA, UAGUAG, or UGAUGA).

📐Formulae

Number of nucleotides=3×Number of amino acids+3 (stop codon)\text{Number of nucleotides} = 3 \times \text{Number of amino acids} + 3 \text{ (stop codon)}

Number of codons=Number of nucleotides3\text{Number of codons} = \frac{\text{Number of nucleotides}}{3}

Possible codon combinations=4n=43=64\text{Possible codon combinations} = 4^n = 4^3 = 64

💡Examples

Problem 1:

A specific mRNAmRNA sequence is given as: 5′−AUG CCA UGG UAA−3′5'-AUG\,CCA\,UGG\,UAA-3'. Determine the sequence of the DNA antisense (template) strand used to produce this mRNAmRNA.

Solution:

The DNA antisense strand is complementary to the mRNAmRNA: 3′−TAC GGT ACC ATT−5′3'-TAC\,GGT\,ACC\,ATT-5'.

Explanation:

To find the antisense strand, we apply complementary base pairing: A→TA \rightarrow T, U→AU \rightarrow A, C→GC \rightarrow G, and G→CG \rightarrow C. The orientation is antiparallel, so the 5′5' end of mRNAmRNA corresponds to the 3′3' end of the DNA template.

Problem 2:

An insulin protein molecule contains 5151 amino acids. Calculate the minimum number of mRNAmRNA nucleotides required to code for this protein, including the stop codon.

Solution:

(51×3)+3=153+3=156 nucleotides(51 \times 3) + 3 = 153 + 3 = 156 \text{ nucleotides}

Explanation:

Each amino acid is coded by a triplet of nucleotides (a codon). Therefore, 5151 amino acids require 51×3=15351 \times 3 = 153 nucleotides. We must also add 33 nucleotides for the 'stop' codon, which does not code for an amino acid but is required to terminate translation.

Problem 3:

If a DNADNA coding (sense) strand has the sequence 5′−ATG CGT ACG−3′5'-ATG\,CGT\,ACG-3', what will be the resulting mRNAmRNA sequence?

Solution:

5′−AUG CGU ACG−3′5'-AUG\,CGU\,ACG-3'

Explanation:

The mRNAmRNA sequence is identical to the DNADNA sense (coding) strand, with the exception that all Thymine (TT) bases are replaced by Uracil (UU). The polarity (5′5' to 3′3') remains the same.