Lecture 32

Question 1

Sterile mutants of yeast)

Answer 1

• Lack of cell surface receptors
• Which make up a heterochromatic complex
• They interact with the pheromones, and if there are mutations within this complex, it will not be able to recognise other pheromones, so the phosphorylation cascade will not be activated

Question 2

Ligand binding and GTPases:

Answer 2

• Exchange of GDP and GTP is a good way to turn proteins (especially GTP proteins) off and on)
• Intrinsic GTP activity is low without an effector protein, ( they are slow to hydrolyse the GT and add a phosphate)
• GDP is active, GAP activates it, so it increases its GTPase activity, hydrolyses it’s domain and is now inactive.
• Liberate GDP from it’s inactive form using guanine nucleotide dissociation inhibitors (GDI)
• Regulation of gene expression at the protein level by exchanging GTP and GDP

Question 3

Protein processing:

Answer 3

• Specifically cleaving proteins to give it different activity
• Specific proteases result in different polypeptide segments which can do different things compared to what it would do if it was cleaved by a different set of proteases

Question 4

Translational control:

Answer 4

• Before you make the protein, so controls the timing of protein production
• Can occur at multiple levels
• Destabilising
• Changing when/if translation starts

Question 5

mTor Kinase:

Answer 5

• A major regulatory kinase of transitional state
• mTor (mouse Tor kinase) controls when a particular mRNA is translated
• If a particular growth factor is present TorK will phosphorylate 4E-BP
• 4E-BP in the phosphorylated state cannot bind the initiation complex required to recruit other elongation factors and finally the ribosome, translation will occur!
• When 4E-BP is not phosphorylated it can bind, and the elongation factor 4E is kicked out, so it stops the formation of the ribosomal machinery, so no translation occurs

Question 6

Iron regulatory protein:

Answer 6

• Iron is required for iron kelators in the body, but too much is toxic, so it must be regulated
• an element can form a secondary structure called the iron regulatory element (a loop in mRNA)
• IRP will bind this secondary structure, preventing the ribosomal machinery assembling, so not translation occurs
• When iron levels increase, it binds to IRP so that it cannot bind the loop so it no longer inhibits the translational machinery, so transcripts are translated

Question 7

Post transcriptional control:

Answer 7

• mRNA stability
• mRNA translatability
• This occurs after transcription

Question 8

mRNA stability in RBC:

Answer 8

• RBC don’t have nuclei, so the RNA for RBC proteins comes from a progenitor
• RNA encoding hemoglobin must be stable so that they can be translated many times, without needing to replace it
• Stabilisation occurs because of the cell type the mRNA is deposited in

Question 9

mRNA stability in aspergillus nidulans:

Answer 9

• Ammonium can be sourced from the environment, but can also be expressed from the genome
• A transcriptional activator encoded by the areA gene is involved in positively acting to transcriptionally regulate genes involve din assimilating nitrogen sources other than ammonium.
• IT binds to the promoters of genes in the absence of ammonium, by creating an mRNA which is translated
• This mRNA has differentially stability based on the presence (10 mins half life) or absence (40 mins) of ammonium
• This difference in timing result due to the binding of de-adenylases on the 5’ polyA tail, which gets shorter and shorter and hence more unstable

Question 10

RNA localisation in saccaromyces cerivisiae:

Answer 10

• Mother cells can switch cell type
• If the mother creates two daughter cells, the bigger one is able to switch mating types
• This switching process occurs to HO, homothallic - a cell that can mate with itself, HO is dominant over Ho (must mate with the opposite mating type)
• If Ho gets turned off it can’t switch mating types
• Swi5: found in both mother and daughter cell types
• Ash1: only found the in daughter, not the mother. Ash1 is only translated in the daughter nucleus

Question 11

Post-transcriptional control:

Answer 11

• RNA processing including alternate splicing

Question 12

Alternate splicing - negative control:

Answer 12

• A primary transcript may splice an intron out in normal conditions
• In a different condition a protein may bind the intron/exon boundary, blocking the machinery from correctly splicing. A protein with a different function may result, or a nonsense protein may occur

Question 13

Alternate splicing - positive control:

Answer 13

• The primary transcript will usually leave the intron in and so there is no function or it has a particular action
• An activator protein that binds the RNA helps splicing occur properly, so that a different/functional protein is created by inducing splicing

Question 14

Dosage compensation in sex determination of drosophila:

Answer 14

• In mammals, everyone has one lot of X chromosome genes, by inactivating one X chromosome in females
• In flies, the X chromosome in males is upregulated (males are X, females are XX)
• Sex is determined by the ratio of X chromosomes to autosomes.
• Splicing at the sex lethal locus occurs due to having 2 X’s. This leads to female development.
• Only one X chromosome so turns off splicing in sex lethal, turning off the transformer gene. This leads to male development
• This is all post-transcriptional!

Question 15

The ratio of X to autosomes determines what happens at the sex lethal locus (males):

Answer 15

• Sxl gene, Tra gene, Dsx gene, male or female.
  FOR A MALE:
• In a male splicing occurs at the Sex lethal locus, producing a nonfunctional protein product
• The transformer gene produces a non-functional productdue to splicing
• The double sex gene is spliced from the 5’ of intron 1 to the 3’ of intron 2
• This is translated to produce a male-specific amino acid
• This repressed female differentiation genes and male development genes

Question 16

The ratio of X to autosomes determines what happens at the sex lethal locus (females):

Answer 16

FOR A FEMALE:

• The ratio of X chromosomes determines the level of sex lethal proteins
• A high level of this protein will block the splice site in the sex lethal protein, so it is a self-regulating system
• The sex lethal product binds to the 1st splice site of the transformer gene, so a functional transformer protein is produced
• Differential splicing occurs in double sex due to the binding of the transformer protein
• The double sex protein represses male differentiation genes and leads to female development

Question 17

DNA modification:

Answer 17

• Usually only occurs at CpG doublets
• In humans there is a lot of methylation. In drosophila there is almost no methylation. Methylation maybe isn’t the primary way of controlling gene expression. The effects are gene and organism dependent

Question 18

Hw does methylation operate?

Answer 18

• There are two classes of DNA methylases
• De novo methylase (unmehtylated DNA can be methylated) and maintenance methylase (maintain methylation after DNA replication)
• During replication the 2 strands separate, so one strand is methylated (it is a hemimethylated ds strand), but the matching G is not. Re-establishing methylation requires maintenance methylase to completely methylate the double strand
• ## Not maintaining methylation results in loss of methylation after replication

Question 19

DNA demethylase:

Answer 19

• Does the opposite of de novo methylase, it takes methylase groups off

Question 20

How do DNA methylases act?

Answer 20

• Co-valent changes on the DNA don’t in themselves do anything.
• They recruit proteins that can change the expression patterns of the strand they bind
• DNA methylation can result in silencing of gene expression
• Effect recruitment of histone modifying enzymes