D1.2 Protein Synthesis Flashcards

1
Q

[D1.2.4] Describe how transcription is a key stage for gene expression

A

Gene expression is the process which results in an observable effect on the organism due to the synthesis of polypeptides according to the base sequences of the gene. Transcription is the first stage in gene expression and the gene can either be switched on or off. All cells have the same genome, however, only some of the genes are expressed. The full range of RNA that can be made in a cell is called its transcriptome and different cells have different transcriptome.

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2
Q

[D1.2.8/ D1.2.9] Describe the degeneracy and universality of the genetic code

A

Information in DNA are stored in a coded form, where 3 base pairs make up 1 codon. The codons specify which amino acid is added to the polypeptide chain. This genetic code is degenerate because different codons can code for the same amino acid. The code is also universal because it is used by all living organisms and viruses.

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3
Q

[D1.2.1/ D1.2.2/ D1.2.12/ D1.2.13] Describe the location of where transcription occurs and the process

A

Transcription occurs in the nucleus, and it is the synthesis of RNA by using DNA as a template. Since RNA is single-stranded, transcription only occurs along one of the 2 strands of DNA. RNA with the information for the synthesis of proteins are called messenger RNA (mRNA).

To transcribe the gene repeatedly and make many copies of a desired base sequence, RNA polymerase:
- Binds to a site on the DNA which is the start of the gene that is being transcribed
- Unwinds the double helix structure of DNA and separates the strands into sense and antisense strands
- Moves along the antisense strand and place RNA nucleotides in 5’ to 3’ direction to form a strand of RNA
- Detaches the assembled RNA from the template strand and allow DNA’s double helix to form again

Transcription stops when the RNA polymerase reaches a terminating sequence that indicates the end of the gene. Since RNA has a complementary base to the template strand, its sequence is identical to the sense strand of the DNA apart from the fact that thymine is replaced by uracil.

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4
Q

[D1.2.1/ D1.2.2/ D1.2.12/ D1.2.13] Describe the promoter region, in reference to 2 types of sequences it can consist of

A

A promoter is a section of the DNA (between 100 to 1000 bases long) that initiates gene transcription. The base sequence of a promoter allows RNA polymerase and other transcription factors to bind. Within the promoter, there are repressor sequences that allow transcription factors to bind and prevent the binding of RNA polymerase. Thus, the transcription of the gene is prevented. On the other hand, promoter sequence can also have activator sequences that allow transcription factors to bind and encourage the binding of RNA polymerase, leading to the transcription of the gene.

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5
Q

[D1.2.3] Describe the stability of DNA and its importance for its use as a template for transcription

A

When RNA polymerase splits DNA into sense and antisense strands, there are no changes made to the base sequences of DNA. After transcription, hydrogen bonds between the complementary bases form again. Since the strands are detached for only a short period of time, there is little chance for chemical changes in the bases that can cause mutations. The base sequences in somatic cells (cells that doesn’t have genes passed on to offspring) must be conserved throughout the life of a cell.

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6
Q

[D1.2.14/ D1.2.15/ D1.2.16] Describe the role of post-transcriptional modification and why this process is absent in prokaryotes

A

In prokaryotes, RNA can be translated as soon as it has been transcribed. In eukaryotes, newly produced RNA is changed into mRNA before translation. This is post-transcriptional modification that occurs in the nucleus, before the mRNA leaves via nuclear pores, and into the cytoplasm for translation. This explains why there is a nuclear membrane in eukaryotes but not in prokaryotes - modification must be completed before mRNA encounters nuclease enzymes in the cytoplasm to protect the mRNA from becoming digested.

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7
Q

[D1.2.14/ D1.2.15/ D1.2.16] Describe the 3 post-transcriptional modification processes

A

To modify the RNA, extra nucleotides are added to each end:
- 5’ caps: a modified nucleotide with 3 phosphate groups and a methyl group added to 5’ end of the RNA. The base of the nucleotide is guanine
- Poly A tails: between 100 to 2000 adenine nucleotides are added to the 3’ end of the RNA. Translation stops before the ribosome reaches this

Eukaryotic genes that code for polypeptide contain 2 types of sequences:
- Exons: coding sequences that are expressed by translation into amino acid sequences
- Introns: non-coding sequences that are not expressed
○ The non-coding sequences can be for
§ Transcribing transfer RNA (tRNA) or ribosomal RNA (rRNA)
§ Regulating gene expression such as promoters, enhancers and silencers
§ Telomeres - structure at the end of chromosomes

Transcribed RNA coding for protein synthesis have alternating exons and introns. The introns are removed and digested, while axons are spliced together. This alternative splicing produces an uninterrupted base sequence for translation, and it can allow several variants of the polypeptide to be produced form the same primary transcript of a gene. The most frequent alternative splicing is exon skipping, where a sequences of bases in the mRNA is edited out, producing variants in the polypeptide. By producing variants of polypeptides with different structure and function, this increases the diversity of proteome, which is all the proteins that can be expressed in an organism.

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8
Q

[D1.2.5/ D1.2.6/ D1.2.7] Define translation and describe the 3 main components of translation

A

Translation is the process of synthesising polypeptides by linking different amino acids together, according to the base sequences in the mRNA formed by transcription. The 3 main components of translation include:

1. mRNA
mRNA has a site for ribosomes to bind to and sequence of codons that specifies the amino acids for the polypeptide. mRNA has the start codon AUG which initiates the start of translation, and end codons of UAG, UAA and UGA that indicate the end of translation. A table of codons show which amino acid would be added to the polypeptide chain

2. tRNA
tRNA translates the base sequence on mRNA into the corresponding amino acid for the polypeptide. tRNA molecules have an anticodon at one end and an attachment point for the amino acid on the other end. This anticodon must be complementary to the codon on the mRNA for the tRNA to bind to the ribosome and deliver its amino acid.

3. Ribosomes
Ribosomes are complex structures consisting of a small and large subunit. The small subunit has a binding site for mRNA. The large subunit has binding sites for tRNA (E, P, A sites) and also a catalytic site that makes the peptide bonds between the amino acids.
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9
Q

[D1.2.10/ D1.2.12/ D1.2.17] Describe the process of translation

A

Translation is a cycle of steps, where each cycle results in the addition of 1 amino acid to the polypeptide chain. The steps for translation:
1. An activating enzyme attaches the amino acid methionine to tRNA with the anticodon UAC. This tRNA is referred as the initiator protein.
2. Initiator tRNA binds to the small subunit of the ribosome. This produces a ternary complex that is composed of the initiator tRNA, the small subunit of ribosome and an amino acid.
3. Ternary complex binds to the 5’ end of the mRNA and slides along until the UAC is linked to a AUG codon by hydrogen bonds.
4. Large subunit of the ribosome binds to the small subunit. The initiator tRNA is at the P site of the ribosome.
5. tRNA with the anticodon complementary to the next codon binds to the A site. The tRNA binds to the mRNA by hydrogen bonds.
6. A peptide bond is formed between the amino acids carried by the initiator tRNA on the P site and the tRNA on the A site. The tRNA on the A site holds the formed dipeptide.
7. Ribosome moves in 5’ to 3’ direction along the mRNA, causing the initiator tRNA to move to the E site and exit the ribosome. The tRNA with the dipeptide is moved to the P site, and a new tRNA that has the complementary anticodon for the next codon enters the A site.
8. The cycle starts again by the amino acid being linked to the tRNA on the A site.

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10
Q

[D1.2.11] Explain how gene mutations occur and lead to diseases such as sickle cell anemia

A

Gene mutation is a change to a base sequence of a gene. A type of point mutation is a base substitution mutation, where a single base is changed to a different one, resulting in a completely different amino acid to be added. Diseases like sickle cell anemia can occur due to a base substitution mutation. The gene that codes for the beat-globin polypeptide in haemoglobin is called Hb. Most humans have the allele HbA, however, a mutation results in a HbS allele. The mutation results in red blood cells to be in the shape of a sickle, which blocks blood capillaries and reduces blood flow. This mutation can be inherited if it occurs in a gamete cell.

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11
Q

[D1.2.18] List the polypeptide modifications that can occur and reason for this modification

A

Modification of polypeptides after translation can turn them into functional proteins. These changes include:
- Removal of amino acid methionine from 5’ end of polypeptide
- Changes to the side chains of amino acids in the polypeptide
○ Ex. Phosphorylation or addition of carbohydrate molecules
- Folding of polypeptide and making intramolecular interactions to stabilise the tertiary structure
- Combing 2 or more polypeptides into quaternary structure proteins
- Combining non-polypeptide components into quaternary structure to form conjugated proteins

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12
Q

[D1.2.18] Describe how pre-proinsulin is modified to insulin

A

Pre-proinsulin can be modified to insulin by:
1. Gene encoding for insulin on chromosome II is transcribed into mRNA
2. mRNA is translated to synthesis a polypeptide called pre-proinsulin
3. Signal peptide and C peptide is cut out, to form insulin

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13
Q

[D1.2.19] Explain why proteins have a short lifespan and the function kf proteasomes in recycling amino acids

A

Most of the proteins produced by translation have a relatively short lifespan because:
- Structure of proteins are easily changed, resulting in the protein to lose tis function
- Protein may be denatured

Proteins that no longer have a function are broken down by proteasomes. Proteins that needs to be broken down are tagged with a chain of small proteins called ubiquitin, which signals proteasomes to digest this protein. This protein is broken down into its aminos acids through hydrolysis, which can be reused to synthesise new proteins.

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