Lecture 32 Flashcards

1
Q

Sterile mutants of yeast)

A
  • 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
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2
Q

Ligand binding and GTPases:

A
  • 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
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3
Q

Protein processing:

A
  • 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
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4
Q

Translational control:

A
  • Before you make the protein, so controls the timing of protein production
  • Can occur at multiple levels
  • Destabilising
  • Changing when/if translation starts
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5
Q

mTor Kinase:

A
  • 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
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6
Q

Iron regulatory protein:

A
  • 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
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7
Q

Post transcriptional control:

A
  • mRNA stability
  • mRNA translatability
  • This occurs after transcription
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8
Q

mRNA stability in RBC:

A
  • 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
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9
Q

mRNA stability in aspergillus nidulans:

A
  • 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
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10
Q

RNA localisation in saccaromyces cerivisiae:

A
  • 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
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11
Q

Post-transcriptional control:

A
  • RNA processing including alternate splicing
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12
Q

Alternate splicing - negative control:

A
  • 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
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13
Q

Alternate splicing - positive control:

A
  • 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
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14
Q

Dosage compensation in sex determination of drosophila:

A
  • 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!
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15
Q

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

A
  • 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
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16
Q

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

A

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
17
Q

DNA modification:

A
  • 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
18
Q

Hw does methylation operate?

A
  • 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
19
Q

DNA demethylase:

A
  • Does the opposite of de novo methylase, it takes methylase groups off
20
Q

How do DNA methylases act?

A
  • 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