DNA and Genomics Flashcards
What is the constant width between the sugar-phosphate backbone?
2nm
How many base pairs does 1 complete turn of the double helix have? What is the distance of a complete turn?
1 complete turn of the double helix has 10 base pairs and spans a distance of 3.4nm
Explain how Fig3.1 shows semi-conservative replication
- Parent DNA double helix** separates** into 2 single strands that serve as templates for synthesis of the complementary daughter strand
- New DNA molecule now contains one parental and one newly synthesized strand
(diagram showing percentage of DNA in each band at different generations)
100%(15N-15N) –> 100% (14N-15N) –> 50% (14N-14N) , 50% (14N-15N) –> 75% (14N-14N),25%(14N-15N)
Describe how these results provided evidence for semi-conservative replication [3m]
- In generation 0, only heavy 15N-15N DNA molecules were present which appeared as the lowest band in the caesium chloride solution
- During semi-conservative replication, original 15N-15N DNA strands were unzipped* and served as templates for the formation of new strands. Since only 14N DNA was present in the medium, resulting DNA molecules in generation 1 were hybrid* DNA molecules where each molecule was made up of one original 15N strand and one new 14N strand This hybrid DNA molecule thus appeared as an intermediate band *
- Each hybrid DNA molecule from generation 1 unzipped * and were used as templates for DNA replication. Hence 50% of DNA in generation 2 was made up of hybrid 14N-15N DNA which appeared as an intermediate band and 50% was made up of light DNA 14N-14N which appeared as an uppermost band.
What is conservative replication?
➔2 parental strands re-associate after acting as templates, thus restoring the original double helix. The other DNA molecule consists of 2 newly synthesized strands.
What is dispersive replication?
➔Parental DNA molecule is fragmented and dispersed. Daughter molecules are made up of a mixture of old and newly synthesized parts.
What is the main role of DNA?
What characteristics of DNA make it a suitable store of information?
main role: store information and pass it on from one generation to the next
Characteristics:
1. It can be replicated accurately➔daughter cells have identical copies of DNA as the parent cell
HOW?:
•Weak hydrogen bonding between the two strands allow them to separate and act as a template for new strand synthesis
(Adenine forms 2 hydrogen bonds with thymine and cytosine forms 3 hydrogen bonds with guanine through complementary base pairing)
2) It is a stable molecule ➔can be passed on to the next generation without loss of the coded information HOW?:
Collectively, numerous hydrogen bonds* hold the two strands of DNA together and adjacent nucleotides in each strand are joined by strong covalent phosphodiester bonds*
3) There is a backup of code
HOW?: DNA is double-stranded* and one strand to serve as a template*** for the repair(!! take away this idea) *of the other if a mutation occurs on either one
4) Coded information can be readily utilised/accessed
HOW?:
•Weak hydrogen bonding allows the template strand to separate from the non-template strand allowing transcription to take place
•Complementary base pairing allows the faithful transfer of info from DNA to RNA in transcription, which will be translated to protein subsequently
Describe the process of DNA replication
Main processes
- unzipping of parental strand
- addition of primer
- synthesis of daughter strand
- end of replication
note: the dna polymerase reads the DNA template in the 3’ to 5’ direction
- Before DNA replication, free deoxyribonucleoside triphosphates are manufactured in the cytoplasm and transported into the nucleoplasm via nuclear pores.
- DNA replication occurs at S phase of interphase**
UNZIPPING OF PARENTAL STRAND
- Replication begins at specific points of the DNA molecule each of which is known as an origin of replication***(ori ).
- Helicase binds to origin of replication. It disrupts hydrogen bonds between complementary base pairs, causing parental strands to unzip and separate.
- Single-strand binding proteins* keep the strands apart *so that they can serve as templates for the synthesis of new strands.
- Topoisomerase relieves “overwinding” strain** ahead of replication forks** by breaking, swivelling, and rejoining* DNA strands.
ADDITION OF PRIMER
- RNA primer** is added to each template (parental) strand by the enzyme primase**.
- RNA primer provides a free 3’ OH end** for DNA polymerase** to recognize and start DNA synthesis of the complementary daughter strand.
- DNA polymerase can only add deoxyribonucleotides(DNA nucleotides) to a pre-existing 3’OH end ** of a nucleotide.
SYNTHESIS OF DAUGHTER STRANDS
- DNA polymerase* uses the parental strand as a template and aligns the free activated deoxyribonucleoside triphosphates (dNTPs) **in a sequence complementary to that of the parental strand.
- Adenine base pairs with Thymine and vice versa. Guanine base pairs with Cytosine and vice versa.
- DNA polymerase catalyzes the formation of phosphodiester bonds** between adjacent daughter DNA nucleotides** of the newly synthesized strand**.
- As DNA polymerase moves along the template, it proofreads** the previous region for proper base pairing. Any incorrect deoxyribonucleotide is removed and replaced by the correct one***.
- The leading strand** is synthesized continuously** in the 5’ to 3’ direction**.
- The lagging strand is synthesized discontinuously. Its synthesis is similar to the leading strand, except that the lagging strand is synthesised in fragments known as Okazaki fragments**. Each fragment is initiated by an RNA primer before the addition of DNA nucleotides.
- A different DNA polymerase** then removes the RNA primer and replaces it with deoxyribonucleotides.
- DNA ligase* seals the nicks* by forming phosphodiester bonds* between adjacent nucleotides of each of the DNA fragments on the new strand.
END OF REPLICATION
18.Complementary parental and daughter strands rewind into a double helix.
19.Each resultant helix consists of one parental strand and one daughter strand*.
Hence this is called semi-conservative DNA replication**.
Describe the process of transcription
- initiation
- elongation
- termination
- post-transcriptional modification (only in eukaryotes)
INITIATION
- GTF first assemble along the promoter
- GTF recruit RNA polymerase **and position it correctly* on the promoter
- The complex* of RNA pol and transcription factors* is called the transcription initiation complex**
- RNA pol unzips* and separates* the DNA double helix at promoter by breaking H bonds** between complementary base pairs**
- Only one strand is used as the template* to synthesise complementary mRNA strand***.
ELONGATION
- Free ribonucleotides will bind by complementary base pairing** to deoxyribonucleotides on DNA template strand.
* 2.Cytosine,guanine,adenine,and uracil containing ribonucleotides will complementary base pair with guanine,cytosine,thymine and adenine containing deoxribonucleotides on the DNA template respectively - Cytosine forms 3 hydrogen* bonds with guanine, adenine form 2 hydrogen* bonds with thymine and uracil.
- RNA polymerase catalyses the formation of covalent phosphodiester bonds** between adjacent ribonucleotides, forming the sugar-phosphate backbone** of the growing mRNA transcript.
- mRNA strand is synthesized in the 5’ to 3’ direction ** as new ribonucleotides are added to the 3’OH end** of the growing mRNA strand
- As the transcription complex continues to move down the DNA double helix, unzipping the 2 strands,and synthesizing mRNA,the region of DNA that has just been transcribed*, reanneals.
TERMINATION
1.After RNA polymerase transcribes through the termination sequence, the mRNA chain is released and the RNA polymerase will dissociate, terminating transcription.
POST-TRANSCRIPTIONAL MODIFICATION
1.Addition of 7-methyl guanosine cap** to 5’ end of pre-mRNA**
➔protects mRNA from degradation** by ribonucleases, facilitates the export of mature mRNA from the nucleus to the cytoplasm & helps the cell recognize mRNA(amongst all other RNAs in cell) so that it can undergo subsequent steps such as splicing
2.RNA splicing involves spliceosomes which excise introns & join exons.
3.Synthesis of poly A tail* (polyadenylation) involves the cleaving of the pre-mRNA by an endonuclease* 10-35 nucleotides downstream of the polyadenylation signal and the addition of many adenine nucleotides* by the enzyme poly A polymerase downstream of the polyadenylation signal, AAUAAA.➔protects mRNA from degradation by ribonucleases, facilitates the export of mature mRNA from the nucleus to the cytoplasm and interacts with initiation factors(together with 5’cap) to form the translation initiation complex.
What is the role of mRNA?
1) Messenger RNA (mRNA) serves as a ‘messenger’ that,in eukaryotes,takes the information** out of the nucleus** via the nuclear pore** to the cytoplasm where translation takes place.
2) mRNA acts as a template for translation*
3) As each codon within the coding region of the mRNA represents an amino acid in a polypeptide, the sequence of codons will determine the polypeptide sequence**
Describe the process of translation
- amino acid activation
2. initiation
3. elongation
a) codon recognition
b) peptide bond formation
c) translocation
4. Termination
AMINO ACID ACTIVATION (not part of translation)
1)A specific amino acid is covalently attached to the 3’ CCA stem of a specific tRNA with a specific anticodon, catalysed by a specific aminoacyl-tRNA synthetase*.
2)There are 20 different aminoacyl tRNA synthetases. Each enzyme has an active site which is complementary to conformation and charge of (a) specific amino acid to be attached to the tRNA and (b) anticodon on RNA.
- INITIATION
(specifically for eukaryotes) :
a) Small ribosomal subunit, eukaryotic initiation factors(eIFs) and initiator tRNA(carrying methionine) form a complex.
b)The complex recognizes & binds to the 5’ 7 methyl guanosine cap of the mRNA& moves in the 5’to 3’ direction* along the mRNA in search of the start codon, AUG.
3)Binding of large ribosomal subunit will complete ribosome forming translational initiation complex.
4)This positions initiator tRNA at peptidyl-tRNA binding site (P site) leaving aminoacyl-tRNA binding site* (A site) vacant for incoming aminoacyl-tRNA molecules. - ELONGATION
A) Codon recognition
1)A second aminoacyl-tRNA witha specific anticodon* and corresponding amino acid, complementary base pairs* with a specific mRNA codon at the A site by forming hydrogen bonds.
B) Peptide bond formation
1) A peptide bond* is formed between adjacent amino acids, catalysed by peptidyl transferase on the large ribosomal subunit.
2) Methionine dissociates** from initiator tRNA and as a result remains bound to second amino acid at A site.
C) Translocation
1) The ribosome translocates* in 5’ to 3direction, shifting first tRNA to exit site(E site)allowing it to be released into cytosol.
2) tRNA with growing polypeptide chain* is now at P site*
3) A site will hold a new incoming aminoacyl-tRNA* with an anticodon complementary to next codon on mRNA**.
4) The process is repeated until a stop codon is reached.
- TERMINATION
1) Termination begins when the stop codon (UAA/ UAG/ UGA) reaches the A site.
2) Release factors will enter A site causing hydrolysis of bond between polypeptide chain and tRNA in P site.
3) Polypeptide is released from ribosome and will fold into its secondary and tertiary structures.
4) The ribosome disassembles into its subunits.
What is the structure and function of tRNA?
structure:
- it folds back upon itself and held in shape by hydrogen bond** between complementary base pairs at certain regions to form a 3D L-shaped structure
- -> it has 3 loops
- -> on one of the loops, 3 specific unpaired triplet bases* form an anticodon that binds to a specific mRNA codon via complementary base pairing**
- -> on the 3’ end with CCA stem, is the attachment site for specific amino acid that corresponds to the anticodon
Function:
They bring in SPECIFIC* amino acids* in a sequence corresponding to the sequence of codon in mRNA to the growing polypeptide.
It can facilitate translation due to:
1)its ability to bind to a specific single amino acid **during amino acid activation
2)the ability of the anticodon to base-pair with the mRNA codon** during translation
Structure and function of rRNA
1) rRNA associates with a set of proteins to form ribosomes
2) rRNA is the main constituent of the interface between the large and small subunits of the ribosome Thus the small ribosomal subunit can bind to the mRNA as complementary base pairing can occur between the rRNA in the mRNA binding site of the small ribosomal subunit and themRNA
3) rRNA is the main constituent of the P site (peptidyl-tRNA binding site) and A site (amino-acyl tRNA binding site) on the large ribosomal subunit Hence rRNA enables the binding of aminoacyl-tRNAs to the P site and A site by complementary base pairing
4) An rRNA molecule (peptidyl transferase)on the large ribosomal subunit also catalyses the formation of the peptide bond between the amino group of the new amino acid in the A site and the carboxyl end of the growing polypeptide in the P site
Define gene mutation
DEFINITION: A gene mutation arises as a result of a change in the sequence of nucleotide bases in the DNA of a gene
What are the 4 types of gene mutation? Explain each briefly. (- description - result of mutation - effect on protein)
- Substitution
: replacement of one nucleotide by another
: resulting in 1 codon changed
: minor or major effect on protein depending on whether new amino acid synthesized/ change in properties/ position in the protein (critical region?) - Inversion
: A segment of nucleotides separates from the allele and rejoins at the original position but it is inverted
: resulting in 1 or more codon changed
: (effect on protein): minor/major depending on whether a frameshift occurs/ properties of new aa and their positions in the protein - Insertion
: one or several nucleotides are inserted into a sequence
: shifts reading frame from point of mutation
: (effect on p) major ; unless the number of nucleotides inserted or deleted are a multiple of 3 then there will be a change only in the primary sequence but a frame shift will not result - Deletion
: one or several nucleotides are removed from a sequence
: shifts reading framework from point of mutation
: (effect on p):major ; unless the number of nucleotides inserted or deleted are a multiple of 3 then there will be a change only in the primary sequence but a frame shift will not result
