Transcription & Translation

Question 1

Outline the ‘central dogma of molecular biology’

Answer 1

Idea proposed by Crick
Process by which the instructions in DNA are converted into a functional product - since DNA and proteins are structurally and chemically very different therefore DNA cannot be made directly to protein and a messenger molecule (RNA strand - diffusible) is needed
(Replication) DNA ->(transcription) RNA -> (translation) protein

Question 2

Definition of ‘gene’

Answer 2

A defined region/sequence of DNA that produces a diffusible product (RNA) that has some function

Question 3

Anatomy of prokaryotic genes - 3 regions

Answer 3

1. Coding sequence - stretch of DNA over gene that gives rise to protein
2. Promoter region - AT rich region where transcription starts (upstream/before coding sequence). It is the DNA segment recognised by RNA polymerase to start transcription
3. UTR (untranslated region) - region of the gene that will be transcribed to be part of mRNA but will not be translated so will not be part of the protein

Question 4

5’UTR vs. 3’UTR

Answer 4

5’ UTR = short (3 - few 100 nucleotides)

3’UTR = long (1000s of nucleotides)

Question 5

Functions of UTR

Answer 5

Regulatory functions (2 levels)

• Transcription level: part of DNA so involved in when to move from DNA to RNA (transcription)
• Translation level: part of mRNA so involved in when to move from mRNA to protein (translation)

Question 6

UTR in modifying mRNA in eukaryotic genes + function

Answer 6

2 types of modification - 1 on each side of mRNA
- 5’UTR side: 5’G cap (chemically modified G nucleotide)
- 3’UTR side: Poly-A-tail (long stretch of A nucleotides 20-250)
Function - both 5’G cap and poly-A-tail prevents mRNA degradation from enzymes

Question 7

what is the 3’UTR part of

Answer 7

Exon 4 - the last exon

Question 8

what is 5’UTR part of

Answer 8

Exon 1

Question 9

Functions of 5’G cap (4)

Answer 9

• Prevents mRNA degradation
• Regulates translation by providing a ribosome recognition and binding site
• regulation of nuclear export
• promote intron excision

Question 10

Functions Poly-A-tail

Answer 10

• Prevents mRNA degradation
• regulates translation
• regulates nuclear export

Question 11

Pre-mature mRNA vs. mature mRNA

Answer 11

Pre-mature mRNA has both intron and exon regions

Mature mRNA has intron regions cut out

Question 12

Important points for both prokaryotic and eukaryotic genes regarding non-coding regions (4)

Answer 12

• contain non-coding DNA regions upstream and downstream of the coding sequence (eukaryotes also have introns)
• the non-coding DNA regions can be transcribed but are not translated
• These non-coding DNA elements are involved in regulating gene expression
• changes/mutations in these non-coding gene sequences may ‘disrupt’ normal gene expression

Question 13

location of translation - prokaryotes vs. eukaryotes

Answer 13

prokaryotes - in cytoplasm

eukaryotes - in nucleus

Question 14

What are the 3 stages of transcription (same for translation)

Answer 14

Initiation, elongation, termination

Question 15

Outline initiation - the first step of transcription

Answer 15

1. Transcription factors bind to promotor region (TATA box) to signal RNA polymerase II to bind to this particular region
2. RNA pol II binds, forming a transcriptional initiation complex together with the transcription factors
3. RNA pol II unwind the DNA, the 2 DNA strands separate and RNA pol II starts transcription (RNA synthesis) without the need of a primer

Question 16

Outline elongation - the second step on transcription

Answer 16

RNA pol II uses the template strand which runs in the 3’ - 5’ direction, as a template and inserts complementary RNA nucleotides in the 5’ - 3’ direction (A-U, C-G)
1 DNA strand can be transcribed by many RNA polymerases at the same time

Question 17

Outline termination - the last step in transcription and state difference between prokaryotes and eukaryotes

Answer 17

RNA pol II encounters a particular sequence that code for termination and will disengage from the DNA template
In prokaryotes - the RNA strand will be produced and i ready for translation
In eukaryotes - the pre-mature mRNA strand (with introns) is produced and will undergo splicing to form mature mRNA

Question 18

Name the 2 ways that eukaryotic cell can produce multiple different proteins from one gene

Answer 18

Alternative splicing and post-translational modifications

Question 19

Alternative splicing

Answer 19

Exclusion of particular exons during pre-mRNA processing

Question 20

Post-translational modifications (2)

Answer 20

• Polypeptide processing = exclusion of particular regions of the translated (but still immature) protein chain
• Chemical modifications = attachment of biochemical functional groups to the translated (but still immature) protein which will affect the function and activity of the protein

Question 21

Another term for coding strand (5’ - 3’)

Answer 21

sense strand

Question 22

Another term for non-coding strand (3’ - 5’)

Answer 22

Anti-sense strand

Question 23

RNA polymerase (3 features)

Answer 23

• catalyst in transcription by forming phosphodiester bonds between ribonucleotides
• Has primase enzyme within itself (internal 3’ OH group) that it can use to start RNA synthesis
• also has helicase activity

Question 24

What is transcript

Answer 24

Newly synthesised RNA

Question 25

Key features of the genetic code (7)

Answer 25

- There are 20 amino acids but only 4 nitrogenic bases - Every 3 bases code for an amino acid -> triplet code (triplet codon hypothesis) - This provides 64 possibilities (4^3) - 61 out of 64 codons specify an amino acid - Most amino acids have more than 1 codon (redundancy) - 3 STOP codons (UAA, UAG, UGA) - 1 START codon (AUG)

Question 26

What are the 3 STOP codons

Answer 26

UAA, UAG, UGA

Question 27

What is the START codon and what else does it specify

Answer 27

AUG, this codon also specifies for methionine (Met) which is the first amino acid that is translated

Question 28

Adaptor hypothesis

Answer 28

The amino acid is carried to the template (mRNA) by an adaptor molecule (tRNA), and that the adaptor is the part which actually fits onto the RNA

Question 29

codon

Answer 29

triplet of bases on the mRNA that encodes for one amino acid

Question 30

anticodon

Answer 30

triplet of bases on the tRNA complementary tot he codon region of the mRNA

Question 31

Key features of tRNA (the 'adaptor' molecule) - (7)

Answer 31

- a single-strand of RNA - at least 1 tRNA for each amino acid - 70-80 nucleotides in length - 3' end = amino acid attachment site (binding region) - Anticodon = complementary to mRNA anticodon (interaction region) - 3D structure = upside down L - can contain I (Inosine)

Question 32

Importance of the upside down L structure of tRNA

Answer 32

Allows the tRNA to fit into the pockets of the ribosome which is the machinery that will carry out translation

Question 33

Total number of tRNA and explain how it is possible given that there are 64 codons

Answer 33

32 This is possible due to the 5' end containing I (Inosine) which is a base that can bind to A, U or C This allows for one anticodon to bind to more than 1 condon

Question 34

Outline the process of 'charging' tRNA (4)

Answer 34

The process by which an amino acid binds to a tRNA molecule 1. Enzyme aminoacyl-tRNA synthase recognises a specific amino acid and its corresponding tRNA molecule and has a specific site for the amino acid and anther specific site for the corresponding tRNA 2. Amino acid and corresponding tRNA binds to specific sites on aminoacyl-tRNA synthase 3. The enzyme will catalyse a chemical reaction and form a bond between the amino acid and the tRNA using ATP 4. Aminoacyl tRNA released (charged tRNA)

Question 35

Role of ribosomes in translation

Answer 35

site of protein synthesis using mRNA as a set of instructions

Question 36

Ribosome - prokaryote vs. eukaryote

Answer 36

same anatomy

Question 37

Anatomy of ribosome (4)

Answer 37

- contain both rRNA (ribosomal RNA) and proteins - contains small and large subunits which are separate unless performing translation - small ribosomal subunit = where the mRNA binds - large subunit (4 regions)

Question 38

State the 4 regions of the large ribosomal subunit and the role of each

Answer 38

``` A site (Aminoacyl - tRNA binding site) - where the charged tRNA carrying an amino acid will enter P site (Peptidyl-tRNA binding site) - where the tRNA carrying the growing polypeptide will enter E site (exit site) - where an empty tRNA (tRNA which has transferred its amino acid onto the growing chain of amino acids) will exit Exit tunnel (located above T site) - channel through which the growing polypeptide chain will exit ```

Question 39

Outline initiation - the first step of translation (4)

Answer 39

1. The small ribosomal subunits finds the initiation AUG codon on the mRNA 2. The AUG codon is positioned in the P site of the small ribosomal subunit 3. tRNA 'charged' with the amino acid Methionine (Met) binds to the P site 4. The large ribosomal unit attaches and completes the initiation complex - energy is required

Question 40

Outline elongation - second step of translation

Answer 40

The 'charged' tRNA, with an anticodon complementary to the A site codon, lands in the A site. Then 2 things happen at the same time 1. Peptide transfer - The ribosome will break the bond that binds the amino acid to the tRNA in the P site, transfer the amino acid to the newly arrived amino acid attached to the tRNA in the A site and form a peptide bond between them. Now we have a tRNA with the growing amino acid chain in the A site and an empty/uncharged tRNA in the P site 2. Translocation - When the tRNAs are bound to the mRNA in the P and A sites, the ribosome moves 3 nucleotides (1 codon) down the mRNA. The tRNA with the growing amino acid chain (which was in the A site) will now be in the P site so that the chain of amino acids can exit through the tunnel located above the P site. The 'uncharged' tRNA (which was in the P site) will now be in the E site and detaches from its anticodon and is expelled (Energy required) A new 'charged' tRNA with an anticodon complementary to the next A site codon enter the ribosome at the A site and the elongation process repeats itself

Question 41

Outline the process of termination - last step of transcription

Answer 41

1. When the ribosome reaches a STOP codon, a protein called release factor enters the A site 2. The release factor uses water to break the bond between the P site tRNA and the lastly added amino acid. This causes the polypeptide chain to detach from its tRNA and the newly made polypeptide (protein) is released 3. The small and large ribosomal subunits dissociate from the mRNA and each other (energy is required)

Question 42

When is energy required in each step of translation

Answer 42

Initiation - energy is needed to form the translation initiation complex Elongation - energy is needed to detach and expel the 'uncharged' tRNA from its anticodon and out through the E site Termination - energy is required to dissociate the small and large subunit from mRNA and from each other

Question 43

What is meant by polyribosomes

Answer 43

a single mRNA may be translated by several ribosomes at the same time

Question 44

Comparison of prokaryotic and eukaryotic translation + transcription

Answer 44

Prokaryotic - Transcription and translation both occur in cytoplasm - Transcription and translation are coupled Eukaryotic - Transcription in nucleus - Translation in cytoplasm - Transcription and translation are NOT coupled