Extra

Question 1

How is the 5’ cap synthesised?

Answer 1

• Hydrolysis of the triphosphate on the end of the 5’ nucleotide of the mRNA to a diphosphate
• This diphosphate reacts with the α-phosphate of a molecule of GTP, to form a 5’-5’ phosphate linkage
• The cap is further modified at the N7 position in the purine ring to form a 7-methylguanylate cap

Question 2

Which enzyme is involved in poly A tail synthesis?

Answer 2

PAP

Question 3

Where do mutations in genetic disorders occur?

Answer 3

• gene promoters
• poly-A sequence
• RNA capping
• splice donor/acceptor sequences
• bad proteins after translations

Question 4

Thalassaemia

Answer 4

• Imbalance in the relative amount of α- and β- globin chains in haemoglobin
• It can cause severe anaemia, extramedullary haematopoiesis, hepatomegaly, hepatospenomegaly
• Craniofacial features represent extramedullary haematopoiesis
• Iron overload – elevated GI absorption of iron due to chronic anaemia, results in
  hepatic fibrosis, darkening of skin, cardiomyopathy, endocrinophaties i.e. diabetes etc.
• In β-thalassaemia, there is a relative deficiency of β-globin chains. Several types feature splice site mutations in the β-globin gene

Question 5

Initiation

Answer 5

• The ribosomal subunits dissociate into the 40S and the 60S
• The pre-initiation complex is assembled. This is a complex containing Met-tRNA + initiation factor (eI2s) + 40S subunit + GTP
• Only Met-tRNA can bind to the 40S subunit alone. The 40S subunit is involved in tRNA and mRNA recognition
• The complex binds mRNA at AUG, which is complementary to UAC, the anticodon on Met-tRNA
• This initiator Met-tRNA binding sets the frame for the translation
• eIF-4E and G bind to the cap and are recognised by the pre-initiation complex
• The 60S subunit binds, and GTP is hydrolysed to GDP
  The eIF-2 and GDP dissociate to be reused. The translation complex and initiation is complete

Question 6

Elongation

Answer 6

• A new tRNA carries the second amino acid to the A (aminoacyl) site
• The peptide bond is formed between the two amino acids by peptidyl transferase on the 60S
  subunit
• The second tRNA is translocated to the P (peptidyl) site and the first tRNA is dissociated
• Elongation factors are proteins that promote the movement of the ribosome along mRNA using GTP
• There is a new cycle until a stop codon is encountered

Question 7

Termination

Answer 7

• The recognition of a stop codon by release factors, which are proteins, bind to the A site containing the stop codon
• The release of the peptide chain via peptidyl transferase that catalyses the transfer of the completed protein chain to water and releases it from the ribosome
• The dissociation of release factors and ribosomes

Question 8

Signal sequence

Answer 8

First 20-24 amino acids. They are a sequence of hydrophobic amino acids (e.g. leucine, isoleucine, phenylalanine, tryptamine, tyrosine and alanine) to direct these proteins to the endoplasmic reticulum

Question 9

Secretory and transmembrane proteins

Answer 9

• The signal sequence is recognised by a protein-RNA complex - a signal-recognition particle (SRP).
• This halts translation
• The SRP binds to a receptor on the rER surface, and translation can resume
• The protein is synthesised as above, and the newly synthesised protein is pulled in via a pore into the lumen of the endoplasmic reticulum
• Transmembrane proteins have a hydrophobic sequence that holds them in the membrane
• The signal sequence is cleaved by a signal peptidase and the folding occurs

Question 10

What are the modifications made to insulin?

Answer 10

• Preproinsulin is translated which has an N-terminus signal sequence
• Enters the ER, the signal sequence is cleaved and three disulphide bonds form (Proinsulin)
• The C chain is cleaved producing insulin

Question 11

Which things should be considered when designing a primer?

Answer 11

Length - specificity (20 atleast)
Base composition - no tandem repeats = hairpins
No complementary at 3’ end as primers join to each other

Question 12

What determines the amount of energy needed to remove a probe?

Answer 12

• Length
• Base composition
• Chemical environment - lots of positive charges can stabilise DNA negative charge

Question 13

What does TFIID carry?

Answer 13

TATA binding protein TBP Accessory factors

Question 14

Which end of the intron is the splice acceptor sequence?

Answer 14

3’ end of the intron

Question 15

Which end of the intron is the splice donor sequence?

Answer 15

5’ end

Question 16

Where is the bond formed between A and G in the lariet structure?

Answer 16

5’ phosphate on the G and the 2’ OH on the A