REGULATION GENE EXPRESSIN EUKARYOTES

Question 1

learning objectives

Answer 1

• discuss major differences between prokaryotes and eukaryotes transcription
• provide general overview of postranscriptional regulation
• RNA capping and how affects gene expression
• explain how alternative splicing results in more than one protein isoform
• provide general overview to post translational modifications
• explain how protein phosphorus can effect protein interactions

Question 2

E VS P gene expression

Answer 2

EUKARYOTES - transcription and translation are uncoupled due to nucleus membrane

PROKARYOTES - Transcription and translation occur simultaneously

Question 3

eukaryote extra processes

Answer 3

1. mRNA undergoes 5’end capping, polyadenylation and splicing before reaching ribosome
2. mRNA are generally monocistronic

Question 4

3 RNA POLYMERASE

Answer 4

RNA Polymerase I: Synthesizes rRNA (ribosomal RNA), except for 5S rRNA.

RNA Polymerase II: Synthesizes mRNA (messenger RNA) and some snRNAs (small nuclear RNAs).

RNA Polymerase III: Synthesizes tRNA (transfer RNA), 5S rRNA, and other small RNAs.

Question 5

promoters

Answer 5

in eukaryotes, promoters are longer and more complex than in prokaryotes

contain promoter-proximal elements like enhancer/activators and silencer/repressors

Question 6

RNA polymerase

Answer 6

it is bound to DNA in both on and off states, so requires TFs to activate or repress

• activators bind to DNA sites called enhancers and start transcription
• repressors bind to sites called silencers, prevent transcription

Question 7

post transcriptional regulation steps

Answer 7

from RNA to mRNA, 3 steps:

1. 5’ end capping - to protect from degradation
2. 3’end polyadenylation - protect from degradation - adds a poly-A tail
3. • splicing - removes introns and joins exons to form mRNA

the 5’end capping and 3’ end polyA interact with each other to protect from nucleases

Question 8

3’ end polyA

Answer 8

as translation continues, poly A tail degrades and no longer interacts with 5’cap end.

overtime mRNA loses more As and protection until nucleases degrades whole molecule

Question 9

why are eukaryotic genes so long

Answer 9

eukaryotic genes have many introns with short stretches of exons - splicing brings these together

Question 10

splicing process

Answer 10

spliceosome - a ribonucleoprotein complex responsible for splicing , recognising the 5’ and 3’ sequences of INTRONS and removes them.

Question 11

alternative splicing

Answer 11

explains how single gene can generate more than one mRNA transcription - expanding the proteome - regulated process

spliceosome may cut out an exon to change protein structure

• mutations can lead to cancer

Question 12

example of alternative splicing - sex determination in drosophila

Answer 12

gene called doubles gene in pre mRNA
- alternative splicing will either splice the female exon or male exons out to determine sex

Question 13

post-translation modifications PTMs

Answer 13

PTM allow for quick response to environmental signals, thus saving the need for transcription and translation

reversible reactions, acting and shutting off the response does not require synthesising and degrading of proteins

Question 14

example PTM - protein phosphorylation

Answer 14

kineses transfer phosphate from ATP to serine, threonine or tyrosine

phosphates remove phosphate groups

phosphorylation changes the proteins charge, affecting its interaction, either in a positive or negative way

Question 15

phospphorlayion examples

Answer 15

DNA replication enzymes are activated by phosphor at beginning of S phases and then are degraded, this allows proteins to increase and then rapid decline of protein

Question 16

advantages of using a model organism - yeast

Answer 16

Easy to breed and maintain in labs.

Short generation times.

Large number of offspring for reliable sample sizes.

Well-studied genomes.

Efficient manipulation of genomes.

Less ethical concerns compared to human research.

Fewer genes, making studies simpler.

Question 17

yeast S.Cerevisiae model organism

Answer 17

Easy to breed.

Cell cycle duration: 1.5 hours.

Exists in both haploid and diploid states.

Undergoes meiosis and forms spores.

No ethical concerns.

Ideal for studying fundamental biological processes (cell cycle, DNA repair, etc.)

Question 18

yeast mating types

Answer 18

Haploid cells: Types “a” and “α” can mate to form diploid cells.

Diploid cells: Cannot mate.

Mating signal: Mating occurs when an α-cell secretes A-factor, triggering the opposite sex cell (a-cell) to respond and mate.

Question 19

pheromone signalling in yeast

Answer 19

If no α-cell is nearby, cells continue their cell cycle.

If an α-cell is present, mating is triggered through
pheromone signaling.

The pheromone binds to a receptor, transmitting a signal to the Ste5 scaffold protein.

Ste5 scaffold brings 3 kinases (Ste11, Ste7, Fus3) together, initiating a phosphorylation cascade.

Question 20

kinase cascade in yeast

Answer 20

The kinase cascade is activated when Ste11 phosphorylates Ste7, which then phosphorylates Fus3.

The cascade amplifies signals exponentially, ensuring a fast cellular response.

This process is common in yeast mating and DNA damage responses

Question 21

G1 CELL CYCEL ARREST

Answer 21

G1 arrest prevents progression into the cell cycle to ensure proper mating.

Mating between cells in G1 and G2 would result in a triploid cell with 3 chromosome copies, which cannot undergo meiosis or produce offspring.

Question 22

MAP kinase signalling pathways

Answer 22

MAPK (Mitogen-Activated Protein Kinase) pathways respond to:
- Mating pheromones.
- High osmolarity.
- Low nitrogen conditions.

MAPK cascade: A conserved mechanism found across eukaryotes for regulating various cellular responses.

Question 23

mating type switching in yeast

Answer 23

Mother cells can switch mating types, while daughter cells cannot.

This ensures mating pairs form and successful reproduction occurs.

Diploids: Have better fitness and can survive adverse conditions through sporulation.

Question 24

A-CELL (haploid) regulation

Answer 24

Mcm1 activates A-specific genes (positive regulation).

Alpha 1 is needed to activate αlpha-specific genes but is switched off.

Haploid-specific genes are usually expressed by default.

Question 25

ALPHA - CELL (haploid) regulation

Answer 25

Alpha 2 represses A-specific genes (negative regulation).

Alpha 1 activates αlpha-specific genes (positive regulation).

Haploid-specific genes are expressed without an activator.

Question 26

DIPLOID A/ALPHA cell regulation

Answer 26

Alpha 1 is not expressed.

Alpha 2 represses A-specific genes.

A1 and Alpha 2 prevent the expression of haploid-specific genes (negative regulation).

Question 27

mating type specific genes in yeast

Answer 27

Haploid cells express both haploid-specific and their own mating-type specific genes.

A-cells express A-specific genes.

Alpha-cells express αlpha-specific genes.

Diploid cells suppress both A- and α-specific genes.

Question 28

phosphorylation and amino acids

Answer 28

can only occur on specific amino acids that have an OH group as need this to be an acceptor

Question 29

why do eukaryotic genomes require multiple origins of replication?

Answer 29

• they are much larger than prokaryotic genomes
• they are organised into linear chromosomes
• they replicate more slowly than prokaryotic

Question 30

RNA POLYMERASE 2

Answer 30

this is responsible for transcribing all protein coding genes in eukaryotes