Control of the Eukaryotic Genome

Define gene expression

1. Gene expression is the process in which genetic information encoded in a gene is used to direct the synthesis of a protein.

2. Gene expression involves the entire process of protein synthesis.

3. When a gene is expressed, the functional protein is produced in the cell and the phenotype is observed in the individual.

Define differential gene expression

1. Differential gene expression is the switching on and off of different genes in different cells within the same organism.

2. Spatial gene expression means that different sets of genes are expressed in different cells.

3. Temporal gene expression means that different sets of genes are expressed at different times.

4. Differential gene expression is needed for specialisation of cells.

Define DNA methylation

1. DNA methylation is the attachment of methyl groups to DNA bases at CpG sites, catalysed by enzyme DNA methyltransferase.

2. These regions are where a cytosine nucleotide is followed by a guanine nucleotide.

3. They are usually found near the promoter of a gene.

4. The common methylated base is cytosine which is converted to 5-methylcytosine.

Describe the significance of DNA methylation

1. DNA methylation is essential for the long-term inactivation of genes. This contributes to cellular differentiation and development in embryos.

2. Genes usually remain methylated through successive cell divisions as methylation patterns are passed on to subsequent generations of cells.

Define histone acetylation

1. Histone acetylation is the attachment of acetyl groups to certain amino acids of histone proteins, catalysed by enzyme histone acetyltransferase.

Define histone deacetylation

1. Histone deacetylation is the removal of acetyl groups from certain amino acids of histone proteins, catalysed by enzyme histone deacetylase.

Describe the significance of histone acetylation

1. When histones are acetylated, the change shape, become less positively charged and bind to DNA less tightly.

2. Transcription factors and RNA polymerase have easier access to the promoter of genes in the acetylated region.

3. TIC is formed.

4. Transcirption occurs.

Explain why genes are not expressed

1. DNA methylation and histone deacetylation causes DNA to bind more tightly to histones and nucleosomes to pack tightly together, forming heterochromatin.

2. Transcription factors and RNA polymerase cannot access the promoter of genes.

3. Transcription initiation complex is not formed.

4. Transcription does not take place.

5. Genes are not expressed.

Explain why genes are expressed

1. DNA demethylation and histone acetylation causes DNA to bind less tightly to histones and nucleosomes pack loosely together, forming euchromatin.

2. Transcription factors and RNA polymerase have access to the promoter of genes.

3. Transciption initiation complex is formed.

4. Transcription takes place.

5. Genes are expressed.

State the general features of transcription factors

1. Transcription factors have a protein-binding domain.

2. Transcription factors have a DNA-binding domain.

3. Shape of protein-binding domain of transcription factor is complementary to shape of other protein of transcription initiation complex.

4. Shape of DNA-binding domain of transcription factor is complementary to shape of specific DNA sequence.

State the structure of general transcription factors

1. General transcription factors assemble at all promoters used by RNA polymerase.

2. They are required for transcription to occur but do not increase or decrease the rate of transcription beyond basal level.

State the functions of general transcription factors

1. Position RNA polymerase correctly at the promoter and aid in the formation of the transcription initiation complex.

2. Helps to unwind and separate the 2 DNA strands for transcription to begin.

3. Helps to release RNA polymerase after transcription has begun.

State how transcription initiation complex is formed and gene is expressed

1. General transcription factors recognise and bind to the promoter of genes.

2. The first transcription factor has a TATA-binding protein that recognise and bind to the TATA box.

3. It distorts the DNA, causing the helix to partially unwind.

4. General transcription factor recruit RNA polymerase to bind to the promoter.

5. Transcription initiation complex is formed.

6. Transcription takes place.

Describe how activators and enhancers increase rate of transcription

1. Activators recognise and bind to enhancers to increase the rate of transcription in the following ways.

2. Recruit histone acetyltransferase. When histones are acetylated, they change shape and become less positively-charged and bind DNA less tightly. General transcription factors and RNA polymerase have easier access to the promoter of genes in the acetylated region. Transcription initiation complex is formed faster. Rate of transcription increases.

2. Recruit DNA bending protein. DNA bends to bring activator bound to enhancer closer to the promoter of genes. Activator interacts with proteins of transcription initiation complex. Transcription initiation complex is formed faster. Rate of transcription increases.

Describe how repressors and silencers decrease rate of transcription

1. Repressors recognise and bind to silencers to decrease rate of transcription in the following ways.

2. Block activator from binding to enhancer. Shape of DNA-binding domain of repressor is complementary to shape of enhancer. Repressor competes with activator to bind to the same or overlapping site where activator binds to DNA. Transcription initiation complex is formed slower. Rate of transcription decreases.

3. Block activator from binding to transcription factors. This prevents formation of transcription initiation complex. Transcription does not take place.

4. Interact with proteins of transcription initiation complex. This makes it more difficult for transcription initiation complex to be formed. Rate of transcription decreases.

5. Recruit enzyme histone deacetylase. When histones are deacetylated, they change shape and become more positively-charged and bind DNA more tightly. General transcription factors and RNA polymerase cannot access the promoter of genes. Transcription initiation complex is not formed. Transcription does not take place.

State the effects of rate of RNA processing during post-transcriptional modification

1. Rate of 5’ capping, 3’ polyadenylation and RNA splicing determines how much mature mRNA is formed and hence how much functional protein is produced within a given time.

Describe the effects of alternative RNA splicing during post-transcriptional modification

1. Alternative RNA splicing allows one gene to code for more than one polypeptide.

2. Different exons are removed together with introns to form different mature mRNA molecules from the same pre-mRNA.

3. Different cell types or organisms at different stages of development have different small nuclear ribonucleoproteins.

4. Different splice sites are recognised.

5. Different exons on the pre-mRNA are removed.

6. Different exons join together to form different coding sequences.

7. Different amino acid sequences formed leads to changes in folding of polypeptide chain into its specific 3D conformation.

Describe the effects of 5’ cap and 3’ poly(A) tail during post-transcriptional modification

1. Both 5’ cap and 3’ poly(A) tail facilitate the export of mature mRNA from nucleus to cytoplasm via nuclear pores for translation to occur.

2. Both 5’ cap and 3’ poly(A) tail protects the mature mRNA from degradation by enzyme ribonuclease in the cytoplasm.

3. Both 5’ cap and 3’ poly(A) tail facilitate the binding of eIFs and small ribosomal subunit to 5’ cap of mature mRNA for translation to occur.

4. mRNA with a longer 3’ poly(A) tail is more stable and has a longer half life. The duration of translation is longer and more functional protein is produced before the mature mRNA is degraded by enzyme exonuclease.

Describe what happens during RNA degradation

1. Enzyme exonuclease shortens the 3’ poly(A) tail until a shortened critical length is reached.

2. Enzymes remove the 5’ cap.

3. Once the 5’ cap is removed, exonuclease rapidly digest the mature mRNA from the 5’ end.

Describe the functions of eukaryotic initiation factors

1. eiFs are proteins needed to initiate translation by promoting proper binding of small ribosomal subunit to 5’ cap of mature mRNA, movement of small ribosomal subunit to the start codon, binding of aminoacyl-tRNA complex and joining of large ribosomal subunit.

Describe how translation is regulated

1. Availability of eIFs. eIFs are proteins needed needed to initiate translation by promoting proper binding of small ribosomal subunit to 5’ cap of mature mRNA, movement of small ribosomal subunit to the start codon, binding of aminoacyl-tRNA complex and joining of large ribosomal subunit.

2. Binding of poly(A) binding protein to 3’ poly(A) tail to promote translation.

3. Binding of translational repressors to 5‘ cap or 5’ UTR. This prevents the small ribosomal subunit from binding to 5’ cap o mature mRNA. Translation does not occur.

4. Binding of translational repressors to 3’ poly(A) tail. This prevents poly(A) binding protein from binding to eIFs at 5’ cap. This prevents small ribosomal subunit from binding to 5’ cap of mature mRNA.

Describe the function of ubiquitin

1. Ubiquitin is a small protein which identifies proteins for degradation by enzyme proteasome.

Describe how protein activity can be regulated

1. Chemical modifications via the addition of chemical groups to the polypeptide chain at the GA.

• Phosphorylation is the addition of phosphate groups to activate or deactivate proteins, catalysed by enzyme protein kinase.

• Glycosylation is the addition of one or more sugar monomers to form glycoproteins.

• Ubiquitination is the addition of ubiquitin, a small protein which identifies proteins for degradation by enzyme proteasome.

2. Modifications of amino acid sequence in the polypeptide chain. Inhibitory portions of the polypeptide chain are removed by enzyme protease. This occurs in proteins which are synthesised as a larger inactive protein precursor (proinsulin to insulin).

3. Protein degradation. The longer a protein stays in the cell before degradation, the longer it can carry out its functions and the greater the gene expression.