Pro and Eu Genome Flashcards

1
Q

Differentiate between Eukaryotic cells and prokaryotic cells

take note:
6 points of comparison

A

refer to summary notes

point of comparison:
1. Cell size - EU: Larger / Pro: Smaller

  1. Nucleus - EU: Nucleus with nuclear envelope present / Pro: No true nucleus
  2. Genetic material - EU: Linear DNA associated with histone proteins; found in membrane bound nucleus; no plasmids / PRO: Circular DNA associated with few histone-like proteins; found in a region of the cytoplasm known as the nucleoid region; plasmids present
  3. Ribosome for protein synthesis - EU: 80s ribosomes; may be bounded to ER or free in cytoplasm
    Pro- 70s ribosomes; NO ER present; ribosomes free in cytoplasm
  4. Organelles - EU: many membrane-bound organelles present / PRO: no membrane bound organelles
  5. Cell walls - EU: composed of cellulose in plants and chitin in fungi / PRO: Composed of peptidoglycan
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Compare the structure and organisation of the GENOME of prokaryotic and eukaryotic cells

A

Refer to summary

points of comparison:

  1. Size
  2. Appearance (multiple linear vs single, circular molecule)
  3. Molecule (both made up of double helix DNA)
  4. Association with proteins (yes - large amount of proteins -i.e. histones/scaffold proteins vs yes- relatively less histone-like proteins)
  5. Level of DNA packaging - EU: High level of DNA packaging - negatively charged DNA is wound around 8 positively charged histone proteins twice to form nucleosomes with linker DNA joining adjacent nucleosomes –> which coils around itself to form a 30nm fibre (interphase chromatin) –> which forms looped domains when associated with scaffold proteins forming 300nm fibre –> supercoiling present to form chromosome at metaphase vs PRO: low ; DNA double helix folded into looped domains by protein-DNA associations –> undergoes supercoiling with the help of DNA gyrase and topoisomerase)
  6. Location (nucleus vs nucleoid region)
  7. Extrachromosomal DNA - (yes - mitochondrial and chloroplast circular DNA vs yes- plasmids)
  8. number of genes ( 25,000 vs 4,500)
  9. Non-coding regions
    - presence of introns
    - presence of promoters
    - presence of repeated sequences (i.e. telomeres and centromeres)
    - presence of enhancers and silencers
  10. presence of operons ( very few vs many)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Structure and function of introns

A

Structure:
- non coding DNA sequences found within a gene, specifically between exons in a specific segment of DNA (also present in pre-mRNA)

-only in eukaryotes

Function:
-enables alternative RNA splicing to occur where different exons of a single pre-mRNA can be joined such that different mature mRNAs are produced

-one gene can now code for more than one polypeptide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Structure and function of promoters

A

Structure:

  • located just upstream of the transcription start site of a gene
  • has critical elements
    e. g.1.TATA box (located upstream of transcription start site) e.g. 2. CAAT and GC boxes(located upstream of TATA box but not always present)

Function:
1. recognition site for the binding of general transcription factors** (GTFs) which then recruit RNA polymerase**
2. to form transcription initiation complex which initiates transcription**
3. Has critical elements, TATA box** that determines the precise location of transcription start site

(deletion will result in transcription starting at a variety of locations, resulting in a truncated/ non-functional protein)

-has critical elements CAAT and GC** box–> t0 improve efficiency of promoter* by recruiting General Transcription Factors** and RNA polymerase** to the promoter

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Structure and function of enhancers and silencers

A

Structure:
- usually located far away from the promoter

Function:

enhancers:
- recognition and binding site for activators
1. when bound with specific transcription factors** known as activators, promotes assembly of transcription initiation complex at promoter* as bending of spacer DNA allows binding of activators with RNA polymerase and GTF
2. when bound with specific transcription factors, activator** may recruit histone acetyltransferase** and chromatin remodelling complexes** to decondense chromatin*
3. increase frequency of transcription*

Qn: Describe the role of a silencer control element in transcription. [2m]

  1. allow binding of specific transcription factors** called repressors** by preventing assembly of TIC*** at promoter
  2. when bound with specific transcription factors known as repressors, may recruit histone deacetylase chromatin remodelling complex ** to condense chromatin
  3. decreases the frequency of transcription

silencers:
- recognition and binding site for repressors (specific transcription factor)
- -decrease the frequency of transcription by preventing the assembly of the transcription initiation complex

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Structure and function of telomeres

Qn: outline the role of telomeres [4m]
1. structure - telomeres are non-coding** tandem repeat sequences** found at both ends** of linear chromosomes

A

Structure:
1. found at both ends/terminals of linear, eukaryotic chromosomes
2-non-coding tandem repeat DNA** sequences (a specific sequence of nucleotides occurring many times in a row)
3. in humans, each repeat has the sequence 5’ TTAGGG3’
4. have a single stranded region at their 3’ ends known as the 3’ overhang (due to a limitation of DNA polymerase,this region of DNA does not have a complementary strand)

Functions:
(MAIN ROLE)
1. Telomeres ensure that genes are not eroded and vital genetic information are not lost with every round of replication due to the end replication problem.

–> Due to the end replication problem, the ends of chromosomes shorten* with every round of replication*
Since telomeres are non-coding, shortening of chromosomal ends leads to shortening of the telomeres
The genes within the chromosome will thus not be eroded with each round of replication hence preventing the loss of vital genetic information or any deleterious effects

  1. Telomeres protect and stabilise the terminal ends of chromosomes by forming a loop using 3’ overhang
    - This prevents single-stranded terminal end of one chromosome from annealing to a complementary single-stranded terminal end of another chromosome, prevent fusing of 2 chromosomes.

The formation of the loop also prevents the cell’s DNA repair machinery from detecting the chromosome as damaged DNA(i.e. double stranded breaks)and trigger apoptosis.

  1. Telomeres allow their own extension as they have a 3’ overhang which provides an attachment point for the correct positioning of the enzyme telomerase
    - Although telomeres shorten with every round of DNA replication, telomerase activity in germ cells, embryonic stem cells and cancer cells can maintain telomere length.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Explain how the end-replication problem will lead to the erosion of the ends of the chromosomes? [3m]

A
  1. During the replication of DNA, the RNA primer** at the 5’ end** of the lagging strand is removed*;
  2. but the gap cannot be filled with complementary* deoxyribonucleotides;
  3. Due to the absence of an existing 3’-OH **group for DNA polymerase to add nucleotides
  4. Hence, over repeated cycles of DNA replication, there will be gradual shortening of the ends of the chromosomes
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Describe the function of the telomerase reverse transcriptase. [3m]

key: telomerase is an enzyme!! (what are the keywords)

A

take note of point 4!!!

  1. Telomerase has an active site* that is complementary in conformation and charge* to a specific* telomeric DNA sequence;
  2. Using telomerase RNA as a template, telomerase reverse transcriptase forms a complementary DNA sequence through complementary base pairing*; (whereby adenine base pair with uracil, thymine with adenine, cytosine with guanine, and guanine with cytosine)
  3. Catalyzes the formation of phosphodiester bonds* between deoxyribonuceotides;
  4. translocation in the 5’ to 3’ direction to produce a series of tandem repeats of GGTTAG, thus elongating the telomere/DNA;
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Structure and functions of centromeres

A

Structure:

  1. Constricted region on chromosome where kinetochore microtubules attach during nuclear division
  2. non-coding DNA made up of a series of tandem repeat sequences

Function:

  1. allow sister chromatids to adhere to each other
  2. allow kinetochore proteins to attach and which in turn allow spindle fibres to attach so that sister chromatids/homologous chromosomes can align along the metaphase plate and subsequently be separated to opposite poles. Ultimately, allows proper alignment and segregation of chromosomes.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Describe the eukaryotic gene regulation at the genomic level

4 types

Recall:
Heterochromatin - highly compacted DNA where DNA winds more tightly around histones
–> result in silencing of genes / inactive gene expression as it limits access of RNA pol and GTF to promoters of genes and thus prevent formation of transcription initiation complex

Euchromatin - less compacted DNA where DNA winds less tightly around histones
–>allows access of RNA pol and GTFs to promoters of genes hence allowing the formation of TIC

A
  1. Chromatin remodelling complex

Chromatin remodelling complex are protein complexes which
➔ alter structure of nucleosomes temporarily
1) (-) can cause DNA to be more tightly coiled around histones ➔inhibits transcription as it prevents assembly of transcription initiation complex
2) (+) can cause DNA to be less tightly coiled around histones ➔ promotes transcription as it promotes assembly of transcription initiation complex

  1. (-) DNA methylation
    ➔addition of a methyl group by DNA methylases to selected cytosine nucleotides located in the CG sequences

inhibits transcription by

1) blocking the binding of transcription factors at the promoter and hence preventing the formation of the transcription initiation complex
2) recruiting DNA-binding proteins(e.g. transcriptional repressors, histone deacetylases and repressive chromatin remodelling complexes)

  1. Histone acetylation(+) /deacetylation (-)
  • acetylation works in concert with chromatin remodelling complex, allowing formation of the TIC
    (histone acetylation)
    ➔addition of acetyl group** to lysine residues on histones** , catalysed by histone acetyltransferase**
    ➔removes positive charge* on histones
    ➔decreases electrostatic interaction* between –vely charged DNA and histones –> loosen* chromatin
    ➔promoter* region is more accessible to RNA polymerase** and general transcription factors
    ➔promotes transcription as it promotes assembly of transcription initiation complex**

(histone deacetylation)
➔removal of acetyl groups from histones by histone deacetylase
➔restores +ve charge on histones
➔restores tighter interaction between DNA & histones ➔promoter region is less accessible to RNA polymerase and general transcription factors
➔inhibits transcription as it prevents assembly of transcription initiation complex

  1. Gene Amplification
    - refers to the replication of a specific gene multiple times to create more copies of that gene
    - gene amplification results in a cell which has normal number of all other genes except for the gene of interest which exist in high copy number
    - during transcription and translation, increased copies of mRNA and increased copies of the required protein will be obtained
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Describe eukaryotic regulation at the transcriptional level

A

(i) Promoter
(similar to structure and function)
Structure:
-located just upstream of the transcription start site of a gene
-has critical elements
e.g.1.TATA box (located upstream of transcription start site) e.g. 2. CAAT and GC boxes(located upstream of TATA box but not always present)

Function:

  • recognition & binding site for general transcription factors (GTFs) which then recruit RNA polymerase to form transcription initiation complex which initiates transcription
  • TATA box –> determines precise location of transcription start site (deletion will result in transcription starting at a variety of locations, resulting in a truncated/ non-functional protein)
  • CAAT and GC box–> improve efficiency of promoter by helping to recruit GTFs and RNA pol to the promoter

(ii) Enhancers and silencers - Regulatory proteins : Activators and repressors

1) Activators** - specific transcription factor
➔bind to enhancer sequence**
➔promote assembly of transcription initiation complex as BENDING OF SPACER DNA** allows binding of activators with RNA polymerase* and/or GTFs* at the promoter

➔transcription frequency increases => Bound activator may RECRUIT histone acetyltransferase and chromatin remodelling complex to decondense chromatin & increase accessibility of promoter to general transcription factors and RNA polymerase)

-to promote assembly of transcription initiation complex at the promoter
- increase the efficiency of transcription
2) Repressors
➔bind to silencers
(interfere with action of activator)
➔prevent assembly of transcription initiation complex as bending of spacer DNA allows repressor bind to general transcription factors hence preventing activators from binding to general transcription factors
(changing the chromatin structure)
➔transcription frequency decreases
–> Bound repressor may recruit histone deacetylase and repressive chromatin remodelling complex to decrease accessibility of promoter to general transcription factors and RNA polymerase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Describe prokaryotic regulation at the transcriptional level

A

Activators and repressors
1. Repressors (in both lac and trp operon)
➔binds to the operator, preventingRNA polymerase from binding to the promoter
➔transcription frequency decreases
➔Negative gene regulation

  1. Activators
    ➔is the activated Catabolite Activator Protein (CAP) which binds to the CAP binding site at the promoter of the lac operon and increases the affinity of RNA polymerase to the promoter
    ➔transcription frequency increases
    ➔Positive gene regulation(CAP is activated by binding to cAMP)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Describe eukaryotic regulation at the post-transcriptional level

A

1) Addition of 5’ Cap
- Addition of 7- methylguanosine nucleotide is added to the 5’ end of the pre-mRNA
- -occurs shortly after transcription begins i.e. it occurs co-transcriptionally

(role)
- 1)helps the cell to recognize mRNA(amongst other RNAs) so that subsequent steps such as splicing and polyadenylation can occur
- 2)acts as a signal to export mRNA out of nucleus(after certain proteins bind to it)
- 3)stabilises and protects the growing pre-mRNA chain from degradation by ribonucleases

2) RNA splicing
1)When the introns (non-coding regions within a gene) are excised and exons (coding regions within a gene) are joined together by spliceosomes (a complex of snRNA and proteins:snRNPs) which recognize the sequences at intron-exon boundaries (points of excision)
➔so that functional proteins can be produced

2)Alternative splicing➔where different exons of a single pre-mRNA can be joined together such that different mature mRNAs and hence different proteins can be produced

3) Addition of poly A tail
- The 3’end of a pre-mRNA is cleaved by an endonuclease downstream of the polyadenylation signal(AAUAAA).
Then, a poly-A polymerase will then add a long sequence of adenine nucleotides to 3’ end of the pre-mRNA, forming a poly(A) tail
-occurs immediately after transcription

(role)
1)acts as a signal to export mature mRNA out of nucleus
2)stabilises and protects the mature mRNA from degradation by ribonuclease
➔more proteins can be made
-3)interacts with eukaryotic initiation factors and with the 5’ cap for initiation of translation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Describe eukaryotic regulation at the translational level

A
  1. mRNA half-life/ stability
    mRNA half-life is determined by the length of its poly-A tail. The longer the poly-A tail, the longer the time mRNA can be used as a template to make proteins.
  • The poly-A tail is removed by ribonucleases in the 3’ to 5’ direction until a critical length is reached which will trigger removal of the 5’cap and degradation of the mRNA from the 5’end too.
    2. Formation of translation initiation complex

(i) Binding of translation repressors
1) During translation initiation, small ribosomal subunit binds to 5’cap of mRNA
- -> This can be prevented by binding of translational repressor to (1) 5’cap, (2) 5’UTR (untranslated region), (3) 3’ UTR which interferes with the interaction between the 3’poly-A tail, the5’cap,initiation factors, and the small ribosomal subunit which are needed for translation initiation

(ii) Anti-sense RNAs
Anti-sense RNAs which are complementary to part of the mRNA to be degraded will bind to the mRNA.
This double stranded RNA will
(a) block translation of the mRNA and
(b) will be targeted for degradation by ribonucleases

(iii) Initiation factors
During translation initiation, eukaryotic translation initiation factors facilitate the binding of the small ribosomal subunit to the 5’cap

The availability of activated eukaryotic translation initiation factors is determined by whether or not they are phosphorylated*.
Some initiation factors are activated by phosphorylation while others are inactivated by phosphorylation. Without activated translations initiation factors, translation cannot begin.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Describe the eukaryotic regulation at the post-translational level

A

1) Regulation of protein activity
Covalent modification/cleavage (eg. attachment of prosthetic groups, glycosylation, disulfide bond formation) of polypeptides make them functional proteins
- these modifications occur when polypeptides pass through rough endoplasmic reticulum (RER) and the Golgi Apparatus (GA)

2) Formation of functional proteins
Phosphorylation /dephosphorylation of translation initiation factors can activate/deactivate the protein and hence up / down regulate its activity.

3) Protein degradation
Protein degradation by proteasome determines how long a protein remains in a cell.
1. Proteins tagged with ubiquitin
2. Target proteins tagged with ubiquitin enter proteasome*
3. Enzymes of proteasome hydrolyse peptide bonds of protein to produce small peptides
4. Which can be further hydrolysed
5. Ubiquitin molecule released* and reused

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

qn 11(ii): Explain the significance of telomerase in germ cells. [2m]

A
  1. Function of telomerase: Telomerase allows telomere length to be maintained* from generation to generation, preventing the loss of vital genetic information* through the shortening at chromosomal ends/ prevent telomeres from reaching critical length
  2. Germ cells need to undergo continuous cell division*( mitosis and meiosis) to allow many replication cycles to occur
17
Q

Describe how spliceosome produces a mature mRNA [2m]

A
  1. snRNA in snRNP recognise and bind to specific splice sites at the intron-exon boundaries* via complementary base pairing
  2. These snRNPs associate together to form spliceosomes which are involved in the excision of introns* and joining of exons* via cleaving of phosphodiester bonds between adjacent nucleotides*