DNA Replication, Transcription and Translation

DNA Replication, Transcription and Translation DNA Replication Explain how DNA replication is semi-conservative DNA replication is semi-conservative because when a new double-stranded DNA molecule is formed: ………………………………………………………………………………………………………………………………………………….......... • One strand is from the original template molecule (i.e. conserved) ………………………………………………………………………………………………………………………………………………….......... • One strand is newly synthesised (i.e. not conserved) ………………………………………………………………………………………………………………………………………………….......... Outline how the results of the Meselson-Stahl experiment supported semi-conservative DNA replication Meselson and Stahl treated DNA with a heavier nitrogen isotope (15N) and then replicated in the presence ………………………………………………………………………………………………………………………………………………….......... of a lighter nitrogen isotope (14N) - so template DNA and newly synthesised DNA could be differentiated. ………………………………………………………………………………………………………………………………………………….......... The results supported a semi-conservative model of DNA replication: ………………………………………………………………………………………………………………………………………………….......... • After one division, all molecules contained both 15N and 14N ………………………………………………………………………………………………………………………………………………….......... • After two divisions, some molecules contained both 15N and 14N, while other molecules only contained 14N ………………………………………………………………………………………………………………………………………………….......... Describe the role of the following enzymes in DNA replication Helicase unwinds and separates double-stranded DNA molecules Helicase: ………………………………………………………………………………………………………………………………………… (by breaking the hydrogen bonds between the complementary base pairs) ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... DNA polymerase synthesises a new strand (complementary to the template strand) DNA Polymerase: ……………………………………………………………………………………………………………………………. Nucleotides align opposite their partner, and DNA Pol III covalently joins them together ………………………………………………………………………………………………………………………………………………….......... DNA Pol III synthesises a new strand in a 5’ - 3’ direction ………………………………………………………………………………………………………………………………………………….......... Explain the significance of complementary base pairing in the conservation of base sequence Free nucleotides can only align opposite their complementary base partner (A=T, G=C) ………………………………………………………………………………………………………………………………………………….......... This means a newly synthesised strand will be identical to the complementary partner of a template strand ………………………………………………………………………………………………………………………………………………….......... Hence, base sequence is conserved ………………………………………………………………………………………………………………………………………………….......... State when and where DNA replication happens in a typical eukaryotic cell DNA replication occurs within the nucleus, during the S phase of interphase ………………………………………………………………………………………………………………………………………………….......... Describe the purpose and process of the polymerase chain reaction (PCR) PCR is used to rapidly amplify minute quantities of DNA ……………………………………………………………………………………………………… It involves a thermal cycler and three repeating steps: ……………………………………………………………………………………………………… • Denaturation - DNA is heated to separate strands ……………………………………………………………………………………………………… • Annealing – Primers are introduced to designate copying points ……………………………………………………………………………………………………… • Elongation - Taq polymerase* synthesises new strand ……………………………………………………………………………………………………… These three steps double the amount of DNA, so a typical reaction of ……………………………………………………………………………………………………… 30 cycles will produce over 1 billion copies of desired DNA sequence ……………………………………………………………………………………………………… * Taq polymerase is heat resistant and so doesn’t denature during PCR ……………………………………………………………………………………………………… Transcription Define transcription Transcription is the process by which an RNA sequence is produced from a DNA template (gene) ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Distinguish between sense and antisense strands The antisense strand is the DNA strand that IS transcribed (complementary to eventual RNA sequence) ………………………………………………………………………………………………………………………………………………….......... The sense strand is the strand that is NOT transcribed (identical to RNA sequence - except T in place of U) ………………………………………………………………………………………………………………………………………………….......... Outline the role of RNA polymerase in the process of transcription RNA polymerase unwinds and separates the double stranded DNA and then synthesises a new RNA strand ………………………………………………………………………………………………………………………………………………….......... based on the antisense template – the RNA strand is then released and DNA double helix reforms ………………………………………………………………………………………………………………………………………………….......... (when RNA polymerase separates the DNA strands, free nucleotides align opposite their complementary ………………………………………………………………………………………………………………………………………………….......... base partners and RNA polymerase covalently joins them together) ………………………………………………………………………………………………………………………………………………….......... Convert the following DNA sequence into an RNA transcript T A C A A A T T C G T A C T G C A C T C C G G A A C A A C T A U G U U U A A G C A U G A C G U G A G G C C U U G U U G A ………………………………………………………………………………………………………………………………………………….......... Translation Define translation Translation is the process of protein synthesis, whereby genetic information encoded by mRNA is ………………………………………………………………………………………………………………………………………………….......... translated into an amino acid sequence (i.e. polypeptide) at the ribosome ………………………………………………………………………………………………………………………………………………….......... Describe the function of the genetic code (including degeneracy) The genetic code is the set of rules by which information encoded by mRNA is translated into polypeptides ………………………………………………………………………………………………………………………………………………….......... • It identifies the specific amino acid encoded by each triplet of mRNA bases (codon) ………………………………………………………………………………………………………………………………………………….......... • There are 64 possible codon combinations (4 x 4 x 4), but only 20 possible amino acids ………………………………………………………………………………………………………………………………………………….......... • This means some codons code for the same amino acid (degeneracy) ………………………………………………………………………………………………………………………………………………….......... Identifying the key components of the process of translation Messenger RNA (mRNA) - Contains the genetic instructions M ……………………………………………………………………………………………… Ribosome – Site of translation R ……………………………………………………………………………………………… Codon - Triplet of bases denoting a specific amino acid C ……………………………………………………………………………………………… Anticodon - Complementary sequence on tRNA molecules A ……………………………………………………………………………………………… Transfer RNA (tRNA) - Transfers amino acids to ribosome T ……………………………………………………………………………………………… Amino acid - Monomeric component of a polypeptide chain A ……………………………………………………………………………………………… Peptide bond - The covalent bond formed between amino acids P ……………………………………………………………………………………………… Polypeptide - The end product of translation P ……………………………………………………………………………………………… Hint: Mr Cat App Summarise the process of translation • Ribosome binds to mRNA and moves along it in a 5’ - 3’ direction, reading the sequence in codons ………………………………………………………………………………………………………………………………………………….......... • Each codon encodes a specific amino acid, which is brought to the ribosome by tRNA molecules ………………………………………………………………………………………………………………………………………………….......... • Each tRNA is specific for a particular codon due to the presence of a complementary anticodon ………………………………………………………………………………………………………………………………………………….......... • The tRNA molecules bring the amino acids to the ribosome in an order determined by the codon sequence ………………………………………………………………………………………………………………………………………………….......... • The ribosome moves along the mRNA, joining the amino acids together via peptide bonds ………………………………………………………………………………………………………………………………………………….......... • Translation of a polypeptide begins at a START codon (AUG) and is terminated at a STOP codon ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Use the genetic code to convert the following DNA sequence into a polypeptide sequence TAC AAA TTC GTA CTG CAC TCC GGA ACA ACT UUU AAG CAU GAC GUG AGG CCU UGU UGA RNA: AUG …………………………………………………………………. Met-Phe-Lys-His-Asp-Val-Arg-Pro-Cys-STOP Protein: …………………………………………………………….. Outline the effect of the following mutations: His to Gln (missense mutation) GTA → GTT: ……………………………………………………… Arg to Arg (silent mutation) TCC → TCT: ……………………………………………………… Describe the different types of point mutations Silent mutations do not change the polypeptide sequence (possible due to degeneracy of the genetic code) ………………………………………………………………………………………………………………………………………………….......... Missense mutations involve a change in a single amino acid within the polypeptide sequence ………………………………………………………………………………………………………………………………………………….......... Nonsense mutations create a STOP codon (thus prematurely terminating the polypeptide chain) ………………………………………………………………………………………………………………………………………………….......... Outline the significance of a frameshift mutation Frameshift mutations (insertions, deletions) change the reading frame (meaning all codons are changed) ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Use the following image to explain how the universality of the genetic code allows for gene transfer Insulin gene Human Cell Recombinant Plasmid Bacteria Transgenic bacteria Grow in culture Extract insulin Plasmid The genetic code is universal, meaning (almost) all organisms follow the same set of genetic instructions ………………………………………………………………………………………………………………………………………………….......... This means that a DNA sequence from one organism can be successfully translated by another organism ………………………………………………………………………………………………………………………………………………….......... The gene for insulin is extracted from human cells and inserted into bacterial cells (via recombinant plasmid) ………………………………………………………………………………………………………………………………………………….......... The bacteria can now produce human insulin (bacteria divide quickly, allowing for effective mass production) ………………………………………………………………………………………………………………………………………………….......... This can be used to produce insulin treatments for type I diabetics …………………………………………………………………………………………………………………………………………………..........