Week 4 - Transfer of genetic information in eukaryotic and prokaryotic cells.

Question 1

Describe gene expression in eukaryotes

Answer 1

Transcription % translation

is gene expression.
Read gene and translate into proteins

Question 2

Define gene expression

Expressed genes include

Answer 2

Definition:
Information encoded in a gene translated into protein or RNA.
2 steps:
transcription and translation

Expressed genes include both

• Genes transcribed into mRNA then translated into proteins
• Genes transcribed into RNA (e.g. rRNA and tRNA) but NOT translated into protein.

Question 3

Short about transcription and translation

https://www.youtube.com/watch?v=oefAI2x2CQM&t=65s

https://www.youtube.com/watch?v=oefAI2x2CQM&t=65s

Answer 3

Transcription: – process - making a copy of genetic information stored in DNA strand →complementary strand of mRNA, with help from RNA polymerases.

Translation: protein biosynthesis.
mRNA transports coded info → gets decoded, (prod.)→ produce specific sequences of amino acids in pp-chain.

Question 4

Schematic rep. of difference in gene expression
Eukary. vs prokary.

Name the main differences in the image

Answer 4

Main difference is determ,ined by difference in cell structure.

We can see that since prokary. dont have membranebounded organells → all processes happen in same place.
Meaning = Both
- copy of info and synthesis of mRNA
- proteinsynthesis
takes place in cytoplasm. No spacial division of these processes and quite parallel in time (still in a sequence but happens after each other)

Eukaryoptic cell: divided in time and place.
Synthesis of mRNA in nucleua

Question 5

Compare transcription in eukaryotes and prokaryotes
Combine with flashcard above?

Answer 5

Question 6

Transcription
Intitiation

Answer 6

the cluster of proteins assembles on the promoter sequence at the upstream (5’) end of a gene and forms the transcription initiation complex

HOW? LOOK AT THE IMAGE BELOW
6.42 lecture 4!

During initiation of transcription RNA polymerase II forms a transcription bubble and begins
polymerization of ribonucleotides (rNTPs) at the start site, which is located within the promoter 5’ UTR region

Cellular RNA polymerases unwinds and separates approx. 14 base pairs of DNA around the transcription bubble (where one of the strands serves as a template for RNA synthesis)
- this allows the RNA polymerase to access the template strand and begin synthesizing the RNA molecule

Transcription initiation is considered completed when the RNA polymerase has succesfully added and linked the first two ribonucleotides (rNTP) via a phosphodiester bond
(a type of chemical bond that connects the sugar of one nucleotide with the phosphate group of another forming the backbone of the RNA molecule)

Question 7

RNA polymerase in eukaryotes and in prokaryotes

Why do eukary. have 3 RNA polymerase

Answer 7

In eukaryotes exist 3 RNA polymerases:
◦ RNA polymerase II – initiate transcription in vast majority of protein coding genes. Task is to form transcriptional bubble.
◦ RNA polymerase I – initiate transcription in rRNA precursor coding gene
◦ RNA polymerase III – initiate transcription in other RNA coding genes

In prokaryotes
◦ exists 1 RNA polymerase with less complex structure

We have 3 polymerase because of regulation and process is different between protein-coding and rna-coding genes.

Question 8

10.40 lect. 4

Question 9

What is a transcription bubble?

Answer 9

8.20 lecture 4

Question 10

Transcription
Elongation

Answer 10

• During the stage of strand elongation, RNA polymerase moves along the template DNA one base at a time, opening the double-stranded DNA in front of its direction of movement and
  hybridizating the strands behind it
• One ribocleotide at a time is added to the 3’ end of the growing (nascent) RNA chain during strand elongation by the protein complex polymerase
• Approximately eight nucleotides at the 3’ end of the growing RNA strand remain base-paied
  to the template DNA strand in the transcription bubble
  ◦ Template strand: 3’-5’ (antisense strand)
  ◦ Coding strand: 5’-3’ (sense strand)
• Synthesis of the RNA transcript proceeds in 5’ - 3’ direction, wheres the strand of the gene which is transcribed and serves a template is read in 3’ - 5’ direction
• Transcription continues trough both – exonic and intronic regions of the gene in the presence of elongation factors (stimulate elongation)
• The elongation complex, comprising RNA polymerase, template DNA and the growing (nascent) RNA strand, is very stable
• RNA synthesis occurs ate the rate ~ 1000nt/min at 37°C, the elongation complex must
  remain intact for more as 24h

Question 11

Transcription
Termination
(3/3)

Answer 11

• During transcription termination the completed RNA molecule, or primary transcript is released from the RNA polymerase and the polymerase dissociates from template DNA
• AATAAA sequence is the signal for the bound RNA polymerase to terminate transcription

(Not AAUAAA)

Question 12

Sense and anti-sense strand?
Explain without looking at image which is the template strand.

Answer 12

Question 13

Explain transcription

Answer 13

Question 14

Complete the sentence

Primary conscripts of protein coding genes are precursor genes (pre-mRNAs) that must go several modifications termed as …. to form a ….

Answer 14

Primary conscripts of protein coding genes are precursor genes (pre-mRNAs) that must go several modifications termed as (RNA processing) to form a (functional mRNA)

Question 15

Explain mechanism and biological role of RNA processing:
5’ capping
3’ modification
splicing
Alternatives splicing
explain the role of involved enzymes

Answer 15

At the 5’ end of a growing RNA chain appearing from the surface of RNA
polymerase II, it is immediately acted on by several enzymes that together synthesize the
5’ cap (7-methylguanylate). This process is called capping.
This cap protects the mRNA from enzymatic degradation, assists in export to cytoplasm and has a role in the initiation of the translation in the cytoplasm.

Processing at the 3’ end of pre-mRNA involves cleavage by an endonuclease to provide a
free 3’-OH group to which adenosines are added by an enzyme called
poly(A)-polymerase.

The final step in the processing of mRNA molecules is RNA splicing in which the internal cleavage of a transcript to remove the introns, the ligation of coding exons. Splicing is ensured by a protein complex called spliceosome.

Alternative splicing

SIMPLE

Process which inlude the exons with pre mRNA transcript differently joined or skipped → diversity when it comes to synthesis protein.
One gene can synthesize different amino acids → alternative splicing gives protein isoforms.

CANT CUT OUT first and last exon

Question 16

Splicesome looks for sequence

Answer 16

A great guy ugly …. A great guy
Nice guys finish last!!!!

AGGU … AGG

Question 17

What is she yappin abt?
26.00 och framåt med sina bilder??

Answer 17

26.00

Question 18

rRNA and ribosomes in translation

Answer 18

rRNA and ribosomes

• the ribosome is RNA-protein complex in the cell which directs elongation of a polypeptide
  at a rate of 3 – 5 amino acids added per second
• these complex structures physically moves along an mRNA molecule, catalyze the assembly
  of amino acids into polypeptide chains. They also bind tRNAs and various accessory proteins, necessary for protein synthesis
• in all cells, each ribosome consists of a large and a small subunit. The two subunits contain
  rRNAs of different lengths, as well as a different set of proteins

Question 19

45.11 lecture 4

what is said about this image

Answer 19

Difference in size between prokaryptic and eukaryptic ribosome: result of number difference.

Question 20

tRNA

Answer 20

• Transfer RNA is the key to decoding the codons in mRNA. Each type of amino acid has its own subset of tRNAs, which bind the amino acid and carry it to the growing end of a polypeptide chain.
• Each specific tRNA molecule contains a three-nucleotide sequence, an
  anticodon, that can base-pair with its complementary codon in the mRNA.

To participate in protein synthesis, a tRNA molecule must become chemically linked to a particular amino acid via a high energy bond, forming an aminoacyl-tRNA. The anticodon in the tRNA base-pairs with a codon in mRNA and the activated amino acid can be added
to the growing polypeptide chain. The process is ensured by the enzyme
aminoacyl-tRNA-synthetase.

Question 21

Why is tRNA and rRNA not transcribed into proteins?

Answer 21

Because tRNA transports aminoacid sequences from the cytoplasm to the ribosome

rRNA conncects aminoacid sequences with peptidebonds.

Question 22

How many type of RNA polymerase are there and what do they synthesize

Difference between RNA polymerase in eukaryotes and prokaryotes

Question 23

What does this picture depict?

Fill in all the boxes!

Answer 23

Question 24

Translation

Initiation 1/3

48.50

Answer 24

Initiation of translation
* Translation begins with the binding of the small ribosomal subunit to a specific sequence on
the mRNA chain

• The small subunit binds via complementary base pairing between one of its internal subunits
  and the ribosome binding site (~10 nt sequences on the mRNA, approx. 5 to 11 nt upstream the initiating codon, AUG)
• A specific tRNA molecule (tRNA-Met) recognizes and binds to the initiator codon (AUG) – P site. All other tRNAs bind to A site of ribosomes
• The large subunit binds, forming the initiation complex
• The process is mediated by initiation factors (proteins)
• After initiation Met is removed from polypeptide sequence

Question 25

52.00

Question 26

Translation

Elongation 2/3

Answer 26

Elongation
* correct positioning of initiation complex by step-wise addition of amino acids by the inframe translation of the mRNA

• the elongation factors are required
• The process involves entry of each succeeding aminoacyl-tRNA, formation of a peptidebond, and the movement of the ribosome one codon at a time along the mRNA

Question 27

1. Explain all the steps
2. What binds to A site and what binds to P site?

Answer 27

Question 28

Translation

Termination 3/3

Answer 28

Termination
* The STOP code is signal for termination of translation

• The process is assisted by two release factors
• Translation completes the flow of genetic information within the cell. The sequence of nucleotides in DNA has now been converted to the sequence of amino acids in a **polypetide chain **

Question 29

Compare translation in prokaryotes vs Eukaryotes

Think:
* Location
* First aminoacid?
* Initiation factor numbers?
* Release factor numbers?
* How stable is mRNA?

Answer 29

Question 30

runt 59.00

Question 31

Protein function 59.43

Answer 31

1. Primary structure:
   the amino acids linked by peptide bonds into linear chain, based on mRNA sequence
2. * Secondary structure
   various spatial arrangements resulting from the folding of localized parts of a
   polypeptide chain. Structure is stabilized by hydrogen bounds. The alpha-helix, the betasheet structure or random coils and turns exist as secondary protein structure
   stabilized by:
   ▪ hydrophobic interactions between the nonpolar side chains
   ▪ hydrogen bonds between polar side chains
   ▪ peptide bonds

→ these stabilizing forces hold together elements of secondary structure

3. Tertiary structure
   ◦ the overall conformation of a polypetide chain: the three-dimensional arrangement of all
   its amino acid residues

Question 32

Name of the 2 red circles in the top left corner?
- How to recognise them?

Explain the picture in general. Also look at the other 2 red circles.

Answer 32

C-terminal and n-terminal
C= carboxylgroup
N=Aminogroup

Question 33

Processing of proteins

Answer 33

The proper folding of proteins within cells is mediated by protein-chaperones.

Chaperones bind to the amino (N) terminus of the growing polypeptide chain, stabilizing it in an unfolded configuration until synthesis of the polypeptide is completed.

The completed protein is then released from the ribosome and is able to fold into its correct three-dimensional conformation.

Question 34

Isomerases

Answer 34

Two types of enzymes – isomerases

• catalyse protein folding by braking and re-forming covalent bonds
• from disulphide between cysteine residues
• catalyse the isomerization of peptide bonds between proline residues
  ◦ important in stabilizing the folded structures of many proteins

Question 35

Proteolysis

Answer 35

An important step in the maturation of many proteins is cleavage of the polypeptidechain – proteolysis

Proteolitic modifications play role in the translocation of many proteins across membranes by cleavage of amino-terminal sequences

Activation of enzymes or hormones by cleavage of larger precursors (e.g. insulin)

Question 36

What is talked about in this slide?
1h 8min

Question 37

Glycosylation

Answer 37

This is a process that modifies proteins through the addition of sugar. It is initiated in the endoplasmic reticulum before translation is complete. Some proteins in eukaryotic cells are modified through the attachment of lipids to the polypeptide chain. Improperly folded
or modified proteins are degraded in degradation pathways.

Question 38

What is talked about in this slide?
1h 10min 40 sek

Question 39

Student is able describe in the level of molecular mechanism; how improper protein folding can lead to human pathology
(to name diseases is not requested).

Answer 39

SUMMARY: Improper protein folding can lead to the accumulation of misfolded proteins and
aggregates overwhelming the cellular quality control mechanism, and disrupting normal cellular
functions. These processes can contribute to the development of a wide range of human pathologies,
particularly those involving neurodegeneration, metabolic disorders and systematic diseases.

• Errors in the folding process can result in misfolded proteins than often expose hydrophobic regions, which are normally buried in the interior of properly folded proteins. These exposed areas can lead to abnormal aggregation
• Misfolded proteins tend to aggregate, forming insoluble fibrils or plaques known as amyloids. These aggregates can accumulate in cells and extracellular spaces, interfering with
  cellular functions. E.g., they can disrupt cellular trafficking, impair organelle function and trigger inflamatory responses
• The accumulation of misfolded proteins can overwhel the cell’s quality control system, such
  as the proteasiome and lysosome, which are responsible for degrading and clearing misfolded proteins. This overload can lead to proteotoxic stess, imparing cell function and
  viability
• Misfolded proteins within the ER trigger a cellular response known as the unfolded protein response (UPR). While initially protective, chronic ER stress and UPR activation can lead to cell dysfunction and death, contribiuting to the pathology of various diseases
• Aggregates of misfolded proteins can physically interfere with the normal function of cellular components. They can disrupt the integrity of cellular membranes, inhibit the function of enzymes and receptors, and interfere with signal transduction pathways
• In some cases, the accumulation of misfolded proteins and the resulting cellular stress can activate apoptotic pathways, leading to programmed cell death. While apoptosis is a mechanism to prevent the propagation of damaged cells, excessive cell death in critical
  tissues can contribute to organ dysfunction and disease
• In the nervous system the accumulation of misfolded protein aggregates is particularly
  detrimental. Neurons are highly dependent on proper protein function for their survival and activity. Aggregates can impair synaptic function and lead to degeneration of neuronal cells,
  a hallmark of many neurodegenerative diseases