D1.2 PROTEIN SYNTHESIS

Outline the process of transcription

Outline the process of transcription

1. Transcription is the synthesis of RNA using a DNA template

2. Transcription factors (family of proteins) bind to the promoter sequences & enhancer sequences/region of DNA

3. RNA polymerase:

   • binds to the promoter (site on DNA) upstream/before the gene

   • breaks hydrogen bonds between nitrogenous base pairs

   • separates and unwinds the two strands of DNA

   • reads the antisense strand/non-coding strand of DNA

   • Adds RNA nucleotides using complementary base pairing (A/U, G/C) of the antisense/non-coding strand

4. RNA nucleotides are added to the 3’ end of the sugar of the new strand, in which RNA polymerase works in the 5’ to 3’ direction → elongation

5. Covalent phosphodiester bond between the nucleotides (3’ and phosphate group) forms the sugar-phosphate backbone

6. As RNA polymerase reaches a terminator/ termination sequence of DNA:

   • RNA polymerase breaks away

   • mRNA strand is released

   • DNA winds back together

Recall the processes in protein synthesis

1. Transcription: synthesis of RNA (mRNA, tRNA, rRNA) using a DNA template strand

2. Translation:

Compare and contrast the transcriptional modification in eukaryotic and prokaryotic cells

EUKARYOTIC TRANSCRIPTION:

1. Occurs within the nucleus and mRNA undergoes post-transcriptional modifications(before leaving the nucleus and reaching the ribosome)

2. Introns/noncoding regions are removed from the mRNA

3. Exons/coding regions are spliced together

   • 5’ Guanosine cap is added to the 5’ end → aids in attachment of the small ribosomal subunit

   • 3’ poly A tail is added to the 3’ end → sequences of adenine nucleotides that prevent degradation

PROKARYOTIC TRANSCRIPTION

1. Occurs within the cytoplasm

2. No introns or exons

3. No post-transcriptional modifications/introns are not present in bacterial DNA and mRNA

4. mRNA moves directly to the ribosome

Discuss the role of hydrogen bonding and complementary base pairing in transcription

1. Complementary base pairing between nitrogenous bases:

   • Allos for genetic information to be replicated (in DNA replication) and expressed (transcription of DNA to form mRNA and translation with protein synthesis using mRNA strand

   • Adenine pairs with Thymine (DNA) or Uracil (RNA), and Cytosine pairs with Guanine

   • Hydrogen bonds form between complementary nitrogenous base pairs

2. RNA polymerase adds new RNA nucleotides to the mRNA strand based on CBP between template/antisense strand of DNA and mRNA during transcription

3. mRNA strand will have complementary base sequence to the template strand and identical sequences (with Uracil replacing Thymine) to the sense strand/coding strand of DNA

Outline the directionality of the strands in transcription

1. READ Antisense/non-coding/complementary strand: 3’ to 5’

2. DNA Sense/coding/complementary strand: 5’ to 3’

3. RNA polymerase adds to the 3’ end, but in a 5’ to 3’ direction (it’s anti-parallel)

Discuss the stability of DNA templates

1. The stability of the DNA template is due to:

   • Pyrimidine-purine bonding with nitrogenous bases by complementary base pairing → consistent width

   • Covalent bonding between nucleotides within the sugar-phosphate backbone

   • Hydrogen bonds between nitrogenous base pairs/two strands of DNA → maintains the integrity

2. Single strands of DNA are used as templates (non-coding/antisense) to copy the genetic code without changing the base sequence

3. A copy of the coding strand is contained within the RNA base sequence

4. In somatic cells that don’t divide (ex. neurons), base sequences must be conserved throughout the life of the cell

Outline gene expression

1. Gene expression is the process of producing a protein/polypeptide using information in a gene

2. A gene is a sequence of bases within DNA that contain the information to produce a polypeptide

3. Transcription and translation are the processes required to produce a protein/polypeptide

4. Transcription is the synthesis of mRNA using a DNA template

5. Translation is the synthesis of polypeptides using mRNA

6. All cells contain the same genes, but not all genes are expressed → differentiation

Recall the non-coding sequences of DNA that do not code for polypeptides

1. Introns: non-coding regions that are removed from RNA → can bind to DNA to regulate gene expression

2. Regulators of gene expression: promotor regions & operator regions

3. Telomeres: repetitive sequences of DNA → prevent degradation of chromosomes after replication

4. Genes for rRNA and tRNA in eukaryotes

Discuss the alternative splicing of exons

1. Alternative splicing of exons → can produce multiple variants of a protein in a single gene

2. Exons can be rearranged in a number of ways

Define translation

1. Translation is the synthesis of polypeptides from mRNA at ribosomes

2. The amino acid sequence of the polypeptide is determined by mRNA sequence

3. The mRNA sequence is determined by the sequences of bases in a gene

Outline translation

INITIATION

1. mRNA binds to the ribosome: the small ribosomal subunits attach to the mRNA at the 5’ end

2. The first codon in mRNA is the start codon, AUG

3. tRNA binds to the ribosome at the site where the anti-codon (tRNA) corresponds to the codon (mRNA)

4. tRNA brings an amino acid to the ribosome

   • Initiator tRNA molecule complementarybasepairs with AUG and brings the amino acid methionine at the P site

   • large ribosomal subunit attaches with energy provided by GTP (guanosine triphosphate)

   • a newtRNA for the nextcodon brings an amino acid to the A site

ELONGATION

1. Anti-codon (3 base pair sequence) on the tRNA complementary base pairs and binds with the codon

2. The amino acid from the P site is transferred to the amino acid in the A site by the enzyme peptidyl transferase

3. A peptidebond is formed between two amino acid molecules of consecutive tRNA molecules (P and A site)

TRANSLOCATION

1. Ribosome moves along the mRNA in a 5’ to 3’ direction and linkage of amino acids by peptide bonding to the growing polypeptide chain

2. tRNA in the P site shifts to the E site: exits and picks up new amino acid

3. tRNA in the A site moves to the P site: a new tRNA molecule enters the A site

4. the process continues along the mRNA molecule → elongates polypeptides chain

TERMINATION

1. When the ribosome reaches a stopcodon, a releasefactor is introduced

2. Polypeptide is hydrolysed and large & small ribosomal subunits & tRNA molecules breaksaway

Discuss the initiation of the ribosome and the initiation of tRNA

RIBOSOME

1. The smallribosomalsubunit attaches to the mRNA at the 5’ end

2. The initiator tRNA molecule complementary base pairs with AUG & bringsaminoacid methionine

3. The largeribosomal subunit attaches with the energy provided by GTP, guanosine triphosphate

4. A new tRNA brings an aminoacid to the A site

tRNA

1. There are specific tRNA-activating enzymes (amino acyl that attach amino acids to tRNA molecules

2. ATP is hydrolysed to AMP → allows for met to bind → AMP is released → tRNA attaches → barged/activated tRNA molecule

3. The amino acid is attached to the 3’ end of the tRNA with the base sequence CCA-OH

4. The enzymes tRNA, ATP and amino acids are free-floating molecules in the cytoplasm

Discuss the role of mRNA in translation

1. Contains codons that carry the genetic information to the ribosome

2. Codons in mRNA complementary base pair with the anti-codon in tRNA

3. Codons are 3 base pair sequences that correspond with one amino acid in a polypeptide

4. Binds with the ribosome so that genetic information can be translated

Discuss the structure and function of ribosomes in translation

STRUCTURE

1. Consists of rRNA and proteins, and organised into tertiarystructure/globularproteins

2. Eukaryotic ribosomes are 80s and prokaryotic ribosomes are 70s

3. Two units, small and large

   • large has three binding sites

   • E(exit), P (peptidyl) and A (aminoacyl)

FUNCTION:

1. Small ribosomal subunit:

• contains a bindingsite for mRNA

• initiation begins with the small ribosomal subunits attaching to the mRNA and the initiator met - tRNA molecule

2. Large ribosomal subunit:

• Contains the enzyme peptidyltransferase that joins amino acids together

• Attaches with the met-tRNA in the P site (which holds tRNA with the elongated polypeptide)

• A site is where the next tRNA molecule binds to the nextcodon and brings another amino acid

• Peptidebond forms between the two amino acids in the P and A sites

• Amino acid/polypeptide at the P site is transferred to the A site

3. Site of protein synthesis

• Free ribosomes synthesise proteins for use in the cell

• Bound ribosomes synthesise proteins for export out of the cell

Discuss the role of tRNA in translation

1. Consists of a single stranded tRNA folded in a cloverleaf shape

2. Contains the anticodon that complementary base pairs with the codon on mRNA

3. Anticodon is a 3 base pair sequence on tRNA → anticodon binds with the codon

4. Attached to the specific amino acid molecules — 3’ end consists of the base sequence, CCA, and is the location where the amino acid binds

5. Brings amino acid to the ribosome

Outline the features of the genetic code

1. The genetic code is used to determine the amino acids that correspond to the triplet codon — the order of nucleotides in mRNA determines the order of amino acids

2. Nucleotides of the mRNA are read in groups of three called codons

   • AUG - start codon in mRNA

   • UAA, UAG, UGA - stop codons in mRNA

   • tRNA with the anticodon carries the appropriate amino acid to the growing chain and complementary base pairs and binds to the codon in mRNA

3. Genetic code is degernerate/reduntant

   • 20 different amino acids, 64 codons

   • More than one codon for each amino acid

4. Genetic code is universal

   • same 20 amino acids for all living organisms

   • evidence for evolution/ LUCA is the last common ancestry

Define mutations

DEFINITION

1. Mutations are a change in the base sequence (nucleotide) of DNA

2. Base substitution mutations change a single nucleotide u

3. Ex. Mutation in the HBB gene can change the protein structure of haemoglobin

Outline mutations that are not harmful

1. A change in the base sequence/codon might not change the amino acid

2. There are multiple codons that code for each amino acid

3. The genetic code is degenerate

4. Mutagens (radiation and mutagenic chemicals) increase DNA mutations

5. DNA proofreading and repair mechanisms by DNA polymerase (I and III) may fail

Discuss the causes and consequences of sickle cell anaemia

1. DNA nucleotides, GAG are changed by base substitution into GTG

2. mutation is a base substitution mutation in the 6th codon

3. The codon GAG is mutated into GTG in the sense strand of DNA

4. The Hb^A allele of the gene was mutated forming a new allele/version of the gene called Hb^s

5. In the mRNA, its’s converted to GUG

6. When tRNA binds, the glutamic acid is converted to valine — changes amino acid sequence

CONSEQUENCES

1. Changes shape of beta-globin subunit (polypeptide) of haemoglobin → red boood cells become sickle shaped

2. Vital role in body: when sickle cell releases oxygen → becomes less soluble and crystallises out → impairs blood flow in capillaries

3. Causes anaemia (poor blood flow to peripheral tissues and tiredness) → red blood cells lives for only 8 days (80 days normally)

Outline post translational modifications/modifications of polypeptides in their functional state

Modified by the Golgi apparatus

1.  Methylation:  Addition of methyl groups to histone proteins

2. N-Acetylation:  Addition of acetyl group to the N terminus of the histone protein

3. Glycosylation:  Addition of a sugar

4. Lipidation:  Addition of lipid functional groups 

5.  Ubiquitation:  Addition of a ubiquitin protein

6. Phosphorylation:  Addition of a phosphate group

Discuss the recycling of amino acids

1. Proteins are constantly being synthesised by translation and broken down and recycled by proteasomes

2. Proteasomes break down proteins that are damanged, misfiled and no longer needed

3. Proteolysis is the hydrolysis of proteins into amino acids by by breaking peptide bonds

4. Proteins are marked for degradation by the attachment of ubiquitin (regulatory protein)