Nucleic acid and gene expression (wk3) Flashcards
Basic understanding of the genetic sources of proteins
-Nucleotides-ribo and deoxyribo-nucleotides -> They are the building blocks of nucleic acids (DNA and RNA)
* Sugar -> 5 carbon pentose sugar. Ribose in RNA. Deoxyribose sugar (no OH group on carbon 2 in DNA)
* Bases -> Purines (2-ring): adenine and guanine. Pyrimidines (1-ring): thymine (uracil in RNA) and cytosine. Nitrogen 9 of the base bonds with Carbon 1 of sugar.
* Phosphoryl group -> Oxygen of phosphate binds to Carbon 5 of sugar.
The structure of nucleotides and biopolymers
(DNA and RNA)
-Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA) are biopolymers of nucleotides
-Nucleotides are joined by phosphodiester bonds between adjacent carbons of deoxyribose sugars (3’ and 5’)
-Only variable groups in this sequence are the bases – so describing primary structure of DNA requires reporting the base sequence (e.g. ACGTACC)
-DNA and RNA are slightly acidic due to the deprotonation of the hydroxyl group (circled) at physiological pH
-DNA has a distinct secondary structure (double helix), whereas RNA exists in a single-stranded structure
-The structure of DNA is key in controlling the flow of genetic information
The structure of nucleotides and biopolymers: DNA and RNA
-The DNA double helix
-The DNA Double Helix -> In 1953, Watson and Crick crystallised the structure of DNA for the 1st time. Experiments showed that a DNA molecule was composed of 2 polynucleotide chains wrapped around each other in a helix, with complementary pairing of base pairs (A:T & C:G) via hydrogen bonds (secondary structure). Histone proteins anchor and support DNA to allow dense packaging within cells as chromosomes (10 billion base pairs). Watson and Crick predicted how DNA could transfer information – unzip both strands and copy to make 2 identical copies of the DNA molecule.
The process of DNA replication (semi-conservative replication)
-DNA replication 1-4
-DNA replication -> Is an important part of the cell cycle that enables cells to divide and duplicate their genetic materials (mitosis).
1. Separation of strands -> Helicase breaks hydrogen bonds to unzip DNA molecule
2. Primer binding -> A small strip of RNA (primer) ‘kick-starts’ replication
3. Elongation -> DNA polymerase adds complementary base pairs in the 5’ to 3’ direction. This builds a new DNA molecule that wraps around the parental strand
4. Termination -> Exonuclease removes the original primers and the bases and then re-added to complete a new DNA double helix
The process of DNA replication (semi-conservative replication)
-DNA mutations
-DNA mutations -> 99.9% of the genome in humans is identical – 0.1% variation is due to the gradual appearance of DNA mutations. Bases can be inserted or deleted incorrectly resulting in a change in the protein formed – this can be ‘bad or ‘good’. An advantageous mutation forms the bases of Darwin’s theory of evolution – a more efficient property that increases the chance of survival. Training specific stimuli can increase the expression of specific genes and therefore makes particular proteins more abundant.
The process of DNA transcription
-Types of RNA (3)
-Ribonucleic acid (RNA) is a single-stranded polynucleotide chain involved in coding genes
-Types of RNA:
* Messenger (mRNA) -> Decodes DNA
* Ribosomal (r RNA) -> Decodes m RNA
* Transfer (t RNA) -> Facilitates protein formation
The process of DNA transcription
-DNA unwinding
-DNA unwinds and RNA polymerase replicated DNA template to form m RNA primary transcript (5’ to 3’):
* Promoters -> DNA region next to transcription site that docks polymerase
* Enhancers -> DNA regions that attract the transcription factor complex proteins
* Transcription Factor Complex -> Proteins that regulate the rate of m RNA formation (+/- stimulus)
The process of DNA transcription
-mRNA splicing
-m RNA splicing ->
* Primary m RNA transcript is not the final mature m RNA that then codes for protein translation
* Processing in the nucleus splices introns – exercises, non-coding regions of m RNA.
* Remaining exons form mature m RNA and exons
* How the 21,000 m RNA transcripts lead to the translation of more proteins -> Alternate m RNA splicing can account for different isoforms of proteins i.e. tropomyosin (aids contraction of actin-myosin in both skeletal and smooth muscle)
Exercise and gene expression
-mRNA copies and gene expression
-The amount and/or types of cell signals will modulate the number of m RNA copies made by:
1. Regulating promoter and enhancer activity
2. Altering transcription factor activity
3. Controlling access of RNA polymerase to the gene
-Gene expression is governed by signals sent to the cell after exercise
-The signals produced from different types of exercise will result in a different region of the gene being copied and different m RNA and proteins being produced:
* PGC – 1a4 – activated following strength training
* PGC – 1a1 – activated following endurance training
The process of protein translation
-Translation 1
-Translation 1 ->
* Translation is the formation of a polypeptide chain (protein) from a mature m RNA transcript
* m. RNA binds to the small subunit of ribosomal RNA (r RNA) where it’s decoded and a signal sent by the ribosomal proteins to the large subunit, where translation begins
-More complementary base paring -> A 3 base sequence on a m RNA transcript (codon) complements a 3 base sequence on a transfer RNA (t RNA) molecule (anti-codon), which binds to the large subunit of r RNA holding an amino acid. The interaction facilitates the formation of a new polypeptide chain via peptide linkages C-N
The process of protein translation
-Translation 2
-Translation 2 ->
* The ribosome moves along the m RNA transcript (5’ to 3’ direction) forming multiple codon and ant-codon pairings with ‘charged’ tRNA molecules
* A peptide bond (C-N) is formed between each new amino acid carried by the t RNA, resulting in a polypeptide chain being formed in the opposite direction to the m RNA movement
The process of protein translation
-Translation 3
-Translation 3 ->
* 4 base pairs alone could not directly encode 20 amino acids
* As shown on the last slide, 3 bases on a t RNA molecule (anti-codon) correspond to a specific amino acid
* The chart details the codon sequence on m RNA that attracts this anti-codon with a specific amino acid (64 combinations)
Identify features of amino acids that make up proteins
-Quality control of proteins
-Quality control of proteins ->
* Similar to the primary m RNA transcript, the polypeptide chain that is formed in the cytosol (primary structure) is modified
* The endoplasmic reticulum contains various proteins that fold the protein to form its tertiary structure
* The Golgi apparatus then govern post-translational modifications (e.g. glycosylation, phosphorylation)
* Proteins have specific peptide sequences that can then guide them to specific locations within the cell: 1. Cell release (e.g. insulin, adrenaline), 2. Traffic to a specific organelle (e.g. nucleus, mitochondria) or the cell membrane
