MCG - Molecular Basis of Gene Transcription Flashcards

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1
Q

If all the cells in your body contain the same set of genomic DNA in the nucleus (i.e. the same set of instructions), how can they be structurally and functionally different from one another?

A

Because, in eukaryotes, only some instructions are sent to the cytoplasm. These instructions differ between e.g. cell/tissue type.

Each cell receives a different message, and this reads a different part of the DNA, meaning that it will make different proteins, which will cause it to have a different structure and function.

The message they read is called mRNA, or messenger RNA. It is produces from DNA by transcription, and it translated by ribosomes to a protein.

In prokaryotes, the mRNA doesn’t need to have been completed for translation to occur, while in eukaryotes it does.

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2
Q

What are 3 differences between DNA and RNA?

A
  1. DNA is double stranded.
    RNA is single stranded (but can form secondary structures).
  2. In RNA, the sugar is ribose (not deoxyribose). Ribose has a hydroxyl group (OH) at the 2’ position.
  3. The four bases in DNA are ACGT. The four bases in RNA are ACGU. Uracil instead of thymine.
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3
Q

Describe RNA Polymerase.

A

This is what created the mRNA.

The DNA enters in the uptake channel, and they arrive at the active site where the DNA helix is unwound. The template strand is then used to create mRNA by complementary base pairing. The mRNA then exits through the RNA exit channel.

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4
Q

What are the functions of mRNA, rRNA and tRNA?

A

Messenger RNA (mRNA) is copied from DNA. It encodes proteins.

Ribosomal RNA (rRNA) is a structural and enzymatic component of the ribosome.

Transfer RNA (tRNA) delivers amino acids to the ribosome.

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5
Q

What are the three parts of transcription?

A
  1. Initiation
  2. Elongation
  3. Termination
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6
Q

Describe the process of transcription initiation.

A

The nucleotide after the start site of transcription is known as the +1 nucleotide, and the one before is the -1 nucleotide.

We have -10 and -35 boxes as eukaryotic DNA promoters.

There are only two hydrogen bonds in between A and T pairs, so these sequences of As and Ts will be relatively easy to pull apart.

These sequences serve to align RNA polymerase in the right place.

RNA polymerase has affinity for DNA, but this is not sequence specific.
Instead a ‘sigma factor’ binds to the -10 (Pribnow) and -35 boxes, and recruits RNA polymerase to the transcription start site.

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7
Q

Describe the process of transcription elongation.

A

RNA polymerase unwinds DNA as it moves.

RNA polymerase reads the DNA template, synthesizing mRNA by adding complementary nucleotides – no primer required.

After the first few nucleotides are complete, the sigma factor dissociates.

The average rate of synthesis is ~50 nt/sec.

The addition of nucleotides is in the 5’ → 3’ direction.

The mRNA has the same sequence as the non-template (coding) strand of DNA.

Like in DNA, the nucleotides in RNA are joined by covalent phosphodiester bonds.

When two parts are hydrolysed by RNA polymerase, it release inorganic pyrophosphate.

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8
Q

Describe transcription rho independent termination.

A

This is the most common method of transcription termination in E. coli.

The termination of transcription is controlled by GC-rich stop sequences in DNA.

The mRNA transcript forms a stem loop, reducing binding of mRNA to DNA template.

A string of weak A-U pairings help mRNA detachment from DNA template.

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9
Q

Describe the transcription minority termination method.

A

The RNA polymerase pauses at certain sequences, often GC-rich, probably because the strands are hard to separate (remember, there are 3 hydrogen bonds between each G and C pair).

Rho has helicase activity, and unwinds the DNA-RNA duplex. This separates nascent RNA from the DNA.

When RNA polymerase stalls, the rho ‘catches up’ with it and releases the RNA from the transcription machinery.

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10
Q

How many RNA polymerases are there in eukaryotes?

A

RNA Pol I: rRNA
RNA Pol II: mRNA, snRNA
RNA Pol III: 5S rRNA, tRNA

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11
Q

RNA Pol II has a C-terminal
domain (CTD) ‘tail’.

Why?

A

The C-terminal domain (CTD) of RNA pol II recruits processing factors to eukaryotic mRNAs

Only fully processed mRNA can be exported from the nucleus and is made available for translation.

Prior to processing, the primary transcript is called heterogeneous nuclear RNA (hnRNA) or pre-mRNA.

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12
Q

What processes make up co-/post-transcriptional processing?

A
  • capping
  • polyadenylation
  • splicing
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13
Q

Describe the capping process.

A

A 7-methylguanosine cap is formed at 5’ end by a triphosphate bond.

A guanosine base is added ‘by reaction’ with GTP.

This protects the transcript from degradation, and has roles in export from the nucleus and translation.

Splicing and capping of eukaryotic mRNAs often happens whilst transcription is ongoing
– it is co- transcriptional.

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14
Q

Describe the polyadenylation process.

A

In this reaction, multiple (50-250) adenosine (A) residues
are added to the 3’ end of mRNA.

This is not encoded by the DNA template.

It protects the transcript from degradation, and has roles in export from the nucleus and translation.

We also think the poly A tail attaches to the 5’ cap, making a circle of mRNA. This allows the ribosome to continue synthesising the protein without having to detach and reattach.

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15
Q

Describe the splicing process.

A

A complex of proteins (the spliceosome) removes the introns.

Splicing and capping of eukaryotic mRNAs often happens whilst transcription is ongoing
– it is co- transcriptional.

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16
Q

Are there any different ways for splicing to occur?

A

Some pre-mRNAs can be spliced in different ways, resulting in different protein products from the same gene.

This adds a lot of variation of what can be produced from our genes.