Molecular Genetics - Biology 30 - PDF

BIOLOGY 30 Molecular Genetics UNIT C Chapter 18 TOPIC 1: DNA STRUCTURE & REPLICATION ○ Griffith’s transformation experiment ○ The Hershey-Chase experiment ○ DNA structure ○ Semiconservative replication ○ Steps & enzymes involved in DNA replication ○ The Meselson-Stahl experiment Textbook pgs. 660-666 Remember from Bio 20 ○ Carbohydrates: plant sugars that are used to give us energy during cellular respiration ○ Lipids: fats, provide long term energy storage ○ Proteins: growth and repair and make up enzymes (ase) ○ made up of amino acids ○ Nucleic acids : genetic material ○ Code for DNA and RNA ○ Focusing on nucleic acids and how that information codes for making proteins Requirements of Genetic Material ○ It must be able to replicate ○ It must be able to control expression of traits ○ It must be able to change in a controlled way Genes and Genome Gene: functional sub-unit of DNA that directs the production of one or more polypeptides (protein molecules) Genome: sum of all the DNA that is carried in each cell of the organism DNA, genes and Genomes (2:12) ○ https://www.youtube.com/watch?v=ict Am2wSwtY&t=1s History: The “Transforming Principle” In 1928, an experiment was conducted by Frederick Griffith which demonstrated that bacteria can be re-programmed through a process called transformation… o Griffith’s experiment suggested that something inside the cell must be responsible for its programming, and that this molecule could be transferred from cell to cell. o The molecule that was responsible for the cell’s programming (aka the hereditary material) was still NOT known Transforming Principle given to the substance that could be transferred from nonliving cells to living cells, causing the living cell to show characteristics of the nonliving cell. Detail of Griffith’s Experiment In 1928, British bacteriologist Frederick Griffith conducted a series of experiments using Streptococcus pneumoniae bacteria and mice. Griffith wasn't trying to identify the genetic material, but rather, trying to develop a vaccine against pneumonia. In his experiments, Griffith used two related strains of bacteria, known as R and S. ✔ R strain When grown in a petri dish, the R bacteria formed colonies, or clumps of related bacteria, that had well-defined edges and a rough appearance. The R bacteria were nonvirulent (did not cause sickness when injected into a mouse) ✔ S strain Formed colonies that were rounded and smooth. The smooth appearance was due to a polysaccharide, or sugar-based, coat produced by the bacteria {this coat protected the S bacteria from the mouse immune system, making them virulent (capable of causing disease)}. Mice injected with live S bacteria developed pneumonia and died. As part of his experiments, Griffith tried injecting mice with heat-killed S bacteria (that is, S bacteria that had been heated to high temperatures, causing the cells to die). Unsurprisingly, the heat-killed S bacteria did not cause disease in mice. The experiments took an unexpected turn, when harmless R bacteria were combined with harmless heat-killed S bacteria and injected into a mouse. Not only did the mouse develop pneumonia and die, but when Griffith took a blood sample from the dead mouse, he found that it contained living S bacteria! Griffith concluded that the R-strain bacteria must have taken up what he called a "transforming principle" from the heat-killed S bacteria, which allowed them to "transform" into smooth-coated bacteria and become virulent. History: 1952 Hershey-Chase Experiment: DNA is the molecule that does the Transforming Principle” o In 1952, Alfred Hershey and Martha Chase conducted a series of experiments to prove that DNA, and not protein, was the hereditary material. o This involved radio-labelling the proteins and DNA associated with a particular virus and allowing it to infect a bacterium to see which factor would be “injected” to transform the bacterium. The Hershey-Chase Experiment:DNA as Hereditary Material https://www.youtube.com/watch?v=ZtSfFqqhEIY DNA: The Hereditary Material DNA is the nucleic acid molecule that governs the processes of heredity in all plant and animal cells. o Every cell in your body (with the exception of your gametes ½ DNA) contains the same DNA within its nucleus o The way that DNA is expressed, however, differs from cell to cell History: 1952 Discovering the Structure of DNA: https://www.youtube.com/watch?v=BIP0lYrdirI Rosalind Franklin Had a Ph.D. in science. Many universities and scientists at the time we’re trying to figure out the structure of Franklin took “pictures” of DNA. X-Ray Crystallography machine took Photo 51. 1952 Photo 51 was evidence DNA’s structure was in the form of a double Helix. DNA. From this X-ray crystallographic image, the shape and dimensions of the DNA molecule could be deduced. Rosalind Franklin died due to cancer before she could share in the Nobel Prize for her contributions to solving the secondary structure of DNA. Watson and Crick ○ Built DNA models from data. ○ X pattern suggested the structure of DNA was a helix. One of Franklin’s colleagues stole her picture and showed it to two American scientists. ○ Watson and Crick used the data to build first structure of DNA. ○ Watson and Crick are recognized as scientists who discovered the structure of DNA. ○ Franklin produced the data that lead to their discovery. ○ Watson, Crick, and Wilkins were awarded the 1962 Nobel Prize for Physiology and Medicine. ○ Franklin never received award. ○ Franklin died in 1958 from cancer. ○ Nobel prices aren’t awarded after someone has died. DNA Overview https://www.youtube.com/watch?v=8m6hHRlKwxY Nucleotides ○ 4 different types of nucleotides ○ Adenine (A) ○ Guanine (G) ○ Thymine (T) ○ Cytosine (C) The Structure of DNA DNA molecules form a double helix, with a sugar-phosphate backbone held together by complementary nitrogenous base pairs. The “rules” for each DNA molecule include… 1. Hydrogen bonds form between nitrogenous base pairs 2. Adenine (A) always pairs with Thymine (T) forming two hydrogen bonds 3. Cytosine (C) always pairs with Guanine (G) fomring three hydrogen bonds 4. Each base is bonded to a sugar group. 5. Sugar groups are bonded to phosphates. 6. The grouping that exists between a phosphate molecule, a sugar molecule, and a single nitrogenous base is collectively referred to as a nucleotide NUCLEOTIDE ○ Sugar and Phosphate group are identical in each nucleotide. ○ Pyrimidines: Single ring nucleotide. ○ C, T ○ Purines: Double ring nucleotide. ○ A, G DNA Structure Hydrogen bond Base pair Ribbon model Partial chemical structure Computer model Structure A G C T T of DNA A GDNA C A T TPractice GGA C C T DNA T C GA A T C GT A A C C T GGA The Structure of DNA DNA strands are also said to be antiparallel, meaning that they run in the opposite direction as one another: The 5’ and 3’ ends run in opposite directions This property has important implications for DNA replication & protein synthesis 5’ vs. 3’? https://www.youtube.com/watch? v=IV53GZGr11g Chargaff Rule ○ Amount of A in a sample of DNA is equal to the amount of T. Amount of C is equal to amount of G. ○ A% + T% + C% + G% =100% Nucleotide Proportion % Composition A 34% C G T Question 1 You find some new Remember bacteria and discover A% = T% the percentage of G% = C% thymine is 12%, what is the percentage of A% + T% + G% + C% =100% adenine? 12% Question 2 You sequence the genome of a dog and find the Remember percentage of thymine is 10%, what is the A% = T% percentage of cytosine? G% = C% 100% 80%/2 = 40% A% + T% + G% + C% =100% -20% = A/T G = 40% 80% = C/G C = 40% 40% Question 3 A geneticist sequences the genome of a fish and finds the percentage of guanine is 34%, what is the percentage of adenine? G = 34% 100% 32%/2 = C = 34% -68% 16% G/C=68% 32% = A/T A = 16% T = 16% Re-cap: The DNA Molecule (watch for review) Introduction DNA Replication https://www.youtube.com/watch?v=Qqe4thU-os8 Steps in DNA Replication DNA Replication During S phase of interphase, a cell replicates its entire genome in order to produce a new cell with an identical set of DNA. Each strand of DNA serves as a template for the creation of its complementary strand DNA replication is thus said to be semi-conservative; that is, each new molecule of DNA contains one strand of the original DNA molecule (one of new and one of the old) The Meselson-Stahl Experiment The most beautiful experiment in biology Semi-Conservative: every new molecule of DNA contains ONE strand of the original complementary DNA molecule and one new parent strand https://www.youtube.com/watch?v=4gdWOWjioBE Nucleotides Parental Both parental Two identical molecul strands serve daughter e as templates molecules of DNA of DNA STEP 1: INITIATION 1. A group of enzymes known as helicases bind to a specific sequence of DNA referred to as the replication origin 2. The helicases then “cut” and unravel the double helix. The y-shaped area that forms where the DNA is split is known as the replication fork, while the unwound region is known as the replication bubble. 3. The single unwound strands of DNA serve as templates for replication Replication fork STEP 2: ELONGATION 4. A primer sequence synthesized by the primase enzyme signals DNA polymerase III where to begin. 5. DNA polymerase III begins to add nucleotides within the replication bubble, creating two new strands of DNA that are complementary to each existing strand. 6. Once replication is complete, the primers are removed by DNA polymerase I. 7. Elongation always takes place in the 5’ to 3’ direction, meaning that one strand will always be “leading” (replicated continuously) and the other will always be “lagging” (replicated in short segments known as Okazaki fragments) Okazaki fragments of the lagging strand are spliced together by an enzyme known as DNA ligase DNA polymerase molecule 3′ 5′ Daughter strand Parental DNA synthesized 5′ continuously 3′ Daughter 3′ strand 5′ synthesized in pieces 5′ 3′ DNA ligase Overall direction of replication STEP 3: TERMINATION 8. As soon as the newly formed strands are complete, they rewind automatically into their chemically stable helix structure 9. Replication proceeds until the two new DNA molecules separate; the dismantling of the replication complex is known as termination Key Enzymes ENZYME FUNCTION helicase Cleaves and unwinds DNA primase Synthesizes primer to begin elongation process DNA polymerase III III adds new nucleotides from 5’ end to 3’ end of existing strand DNA Polymerase I I removes primer sequences from DNA once replication is complete DNA ligase Splices together Okazaki fragments of lagging strand Quick Review 1. If a parental DNA strand read: 5’- ACGTAC- 3’ what would a newly synthesized complementary strand read? Answer: 3’- TGCATG- 5’ 2. Summarize the role of helicase, and DNA Polymerase III. Answer: i. helicase unzips the DNA ii. DNAP III adds nucleotides to the new strand of DNA 3. What direction does DNA P III move along the parental strand? Answer: It moves 3’ to 5’ 4. Why do Okazaki fragments occur on the lagging strand. Answer: fragments occur because DNAP III moves away from the replication fork, so it has to replicate in chunks Re-cap: DNA Replication for your review TOPIC 2: TRANSCRIPTION & TRANSLATION ○ Properties of DNA vs. RNA ○ Transcription of DNA into mRNA ○ Translation of mRNA into an amino acid sequence ○ Role of tRNA in translation ○ Role of rRNA in translation ○ Codons vs. anticodons ○ Introns vs. exons Textbook pgs. 667-675 Gene ○ Recall that DNA provides the “instructions” for each of our cells. Expression ○ Each segment of DNA (gene) is responsible for encoding a specific protein, which carries out a specific function. ○ Not every gene is expressed in each cell. This is what allows cells to specialize and function differently. Transcription ∙ 1st stage of gene expression ∙ Messenger RNA (mRNA) is made that is complementary to a segment of DNA Translation : ∙ 2nd stage of gene expression ∙ mRNA nucleotide sequence directs the synthesis (making) a polypeptide (chain of amino acids)with the help of transfer RNA (tRNA) Central Dogma In order for a segment of DNA to be expressed by a cell, 1. It must be transcribed into mRNA 2. translated into a sequence of amino acids (a protein). This is referred to as the “central dogma” of gene expression. DNA Transcription RNA Nucleus Cytoplasm Translation Protein Central Dogma of Biology PROTEINS Proteins are polypeptides made up of amino acids Amino acids are linked by peptide bonds Gene expression: the process by which DNA directs the synthesis of proteins Includes two stages: transcription and translation Occur in all organisms DNA strand Transcriptio n RNA Codon Translation Polypeptide Amino acid Differences Between DNA and RNA DNA RNA Molecule type double strand helix Single strand Site Mostly in nucleus, some in Throughout the cell mitochondria Sugar Deoxyribose Ribose Metabolic rate Stable Unstable (active) Base Thymine Uracil G-C G-C A-T A-U Types one DNA Four (mRNA, tRNA, rRNA, nRNA) Structure rigid (double helix) Less ridged (single) Length Long short Types of RNA for transcription and translation As we go through transcription and translation there will be three key RNA molecules: 1. Messenger RNA (mRNA) Messenger RNA is synthesized during transcription using a DNA template mRNA carries information from the DNA (at the nucleus) TO the ribosomes (in the cytoplasm) 2. Ribosomal RNA (rRNA) rRNA helps form ribosomes Helps link amino acids together 3. Transfer RNA (tRNA) Transfer RNA molecules are important in the process of translation Each tRNA can carry a specific amino acid Can attach to mRNA via their anticodon A complementary codon to mRNA Allow information to be translated into a peptide sequence 1) Transcription During transcription, the information in a segment of double-stranded DNA is copied into a single strand of mRNA. Transcription takes place within the nucleus of a cell. Only one strand of the DNA (the coding or “sense” strand) is transcribed into mRNA. Transcription is initiated when RNA polymerase binds to the promoter region of the coding strand and begins to add complementary base pairs in the 5’ to 3’ direction. Thymine is replaced by uracil (U’s instead of T’s). Once RNA polymerase reaches a termination sequence, the mRNA strand is complete and can leave the nucleus to be translated into a protein. The Genetic Code DNA contains the sequence of nucleotides that codes for proteins ✔ The sequence is read in groups of three called the triplet code During transcription, only one DNA strand is being transcribed ✔ Known as the template strand (also known as the noncoding strand, minus strand, or antisense strand) mRNA molecules formed are antiparallel and complementary to the DNA nucleotides ✔ Base pairing: A=U and C=G ✔ The mRNA nucleotide triplets are called CODONS ▪ Codons code for AMINO ACIDS DNA Template Strand codon 3’ 5’ 5’ 3’ Before translation takes place, the mRNA sequence is modified to remove sequences of mRNA called introns. Introns are sequences of RNA that do not code for proteins. The remaining RNA sequences are referred to as exons. This modification process occurs only in eukaryotes (membrane bound nucleus). Strand to be transcribed DNA Transcription RNA Start Stop codon codon Translation Polypeptide Met Lys Phe ○64 different codon combinations ▪ 61 code for amino acids ▪ 3 are stop codons ▪ Universal to all life Redundancy: more than one codon code for each amino acid Transcription Practice DNA: G C A T T A G G C A T G G C C A RNA: C G U A A U C C G U A C C G G U Practice Questions 1. Use Table 18.3 Page 637 to find the amino acid that corresponds to each of the following codons: a) CCA b) AUG c) GCA d) AGG e) GUG 2. What three RNA codons serve as “STOP” signals? 1. 1. C 8. B First base base Second Third base Quick Check 1. If the DNA template strand reads: 3’-ACGAGA- 5’ what will the mRNA transcript read? Answer: 5’- UGCUCU-3’ 2. Using the mRNA transcript from #1, what amino acids will be produced during translation? (refer to a codon chart) Answer: cysteine (cys) and serine (ser) 2. 2. D ○ RNA before and after start and stop codons. ○ Protects the most important RNA from being damaged. 4. 4. B 2) Translation Once DNA is transcribed into mRNA, it then moves into the cytoplasm of the cell to be translated into a protein by the ribisomes. Ribosome reads mRNA strand. tRNA molecules bring amino acids to ribosome. tRNA molecules match with codons on mRNA. Amino acids bond together to form protein. Ribosomes are typically free-floating within the cytoplasm; these ribosomes produce proteins for use within the cell. Ribosomes consist of two subunits: a large subunit and a small subunit. The mRNA sequence is “sandwiched” between these two subunits when it is translated. Each codon to be translated arrives at the “A” site of the ribosome with the help of tRNA. It then moves to the “P” site, where the correct amino acid is attached to the growing polypeptide chain. It then exits at the “E” site. Bound ribosomes produce proteins for export. tRNA-binding sites Large subunit mRNA binding site Small subunit Next amino acid to be added to polypeptide Growing polypeptide tRNA mRNA Codons Codon ∙ A codon is a three nucleotide sequence of DNA or RNA that corresponds to a speciﬁc amino acid. ∙ Start codon is always methionine in eukaryotes (ﬁrst codon) ∙ Stop Codon : three stop codons that release the polypeptide from the ribosome ○Anticodon ∙ a sequence of three nucleotides forming a unit of genetic code in a transfer RNA molecule, corresponding to a complementary codon in messenger RNA. Table of mRNA codons and corresponding amino acids DNA 5’ TTCAGACCGATCTTA 3’ 3’ AAGTCTGGCTAGAAT 5’ TRANSCRIPTION Transcribe coding strand of DNA into mRNA Takes place in nucleus mRNA 5’ UUCAGACCGAUCUUA 3’ Separate mRNA molecule into codons Codons 5’ UUC AGA CCG AUC UUA 3’ TRANSLATION Translate codons into amino acids using table Takes place in cytoplasm Amino Acids 5’ phe/arg/pro/iso/leu 3’ Practice #1 Transcribe the following strand of DNA into mRNA: Translate your mRNA strand above into a sequence of amino acids: Practice #2 How many nucleotides are required to code for the following sequence of amino acids? Leu-Tyr-Arg-Trp-Ser Is it possible to determine the mRNA sequence that is responsible for producing the sequence of amino acids above? Explain… * Illustrates the redundancy of the genetic code! Practice #3 Identify a single-letter mutation that could occur in the following DNA sequence to produce a stop codon at some point in the amino acid: AACGAATATCGG Re-Cap: Types of Nucleic Acids NUCLEIC ACID STRUCTURE FUNCTION DNA Double helix Stores genetic info mRNA Linear single strand Carries genetic info from DNA in nucleus to ribosomes in cytoplasm for protein synthesis tRNA lobed Carries amino acids to the correct codon site during protein synthesis rRNA Linear single strand Combines to form a ribosome, the main structure required for protein synthesis Genetic Code has three important characteristics: 1. Genetic code is REDUNDANT (more than one codon can code for the same amino acid), only 3 codons DO NOT code for any amino acids as they are “STOP” signals to end the protein synthesis 2. Genetic code is CONTINUOUS (reads a series of 3 letter codons without spaces) A shift of one or two nucleotides could alter the codon and result in incorrect amino acid sequence 3. Genetic code is UNIVERSAL (almost all living organisms build proteins with the genetic code listed in the previous slide) Important for gene technology because a gene could be taken from one kind of organism and inserted into another A 1. 3412 1. TOPIC 3: MUTATIONS ○ Causes of mutations: Point mutation vs. frameshift mutation ○ Effects of mutations: Silent vs. missense vs. nonsense mutation ○ Mitochondrial DNA ○ Recombinant DNA ○ Gel electrophoresis ○ PCR Textbook pgs. 677-693 Mutations Mistakes in the coding of genetic information often take place during DNA replication, transcription and/or translation, resulting in the production of a different sequence than what was intended. These mistakes are referred to as mutations. Certain types of mutations are have more severe consequences than others. Mutations that occur in somatic cells (body cells) are not passed on to offspring (e.g. cancer) Mutations that occur in the germ line (reproductive cells)are passed on to offspring (e.g. trisomy 21) ○ Mutations can be – Spontaneous: due to errors in DNA replication or recombination – Induced by mutagens – High-energy radiation – Chemicals Copyright © 2009 Pearson Education, Inc. Causes of Mutations 1) POINT MUTATION CAG UUA CCC Chemical change that affects just one or a few nucleotides glu/leu/pro This involves the substitution of one nucleotide for another (e.g. a “G” instead of an “A”) CAG UUG CCC Often have very minor effects on the cell, as the amino acid that is produced does not change due to glu/leu/pro redundancy of the genetic code; thus, there is no change to the overall protein Mutation changes “A” to “G” in second codon, but does not result in a diﬀerent amino acid Causes of Mutations CAG UUA CCC 2) FRAMESHIFT MUTATION An insertion or deletion of a glu/leu/pro nucleotide that causes the entire reading frame of the gene to be altered CAGG UUA CCC Typically has more severe consequences, as it changes the protein entirely CAG GUU ACC C (not just a single amino acid) glu/val/thr Sickle-cell disease: changes the shape of red blood cells Normal hemoglobin DNA Mutant hemoglobin DNA mRNA mRNA Normal hemoglobin Sickle-cell hemoglobin Glu Val Effect of Mutations Mutations that have no effect on the cell’s overall functioning are called silent mutations Mutations that lead to a slightly altered but still functional polypeptide are called missense mutations Mutations that render the gene unable to code for a functional polypeptide are called nonsense mutations Original THE BOY CUT HIS LIP AND ATE THE HOT DOG Missense Mutation THE BOY CUT HIS FIP AND ATE THE HOT DOG Nonsense Mutation THE BOY CUT HIS Silent Mutation THE BOY CUT HIS LIP AND ATE THE HOT DOG Insertion Example THE BOY CUT HIS SLI PAN DAT ETH EHO TDO G Deletion Example THE BOY CUT HIS LIP ANA TET HEH OTD OG Mutations & Natural Selection DNA Technology and Genomics ○ Genetic engineering involves manipulating genes for practical purposes – Gene cloning leads to the production of multiple identical copies of a gene-carrying piece of DNA – Recombinant DNA is formed by joining DNA sequences from two diﬀerent sources – One source contains the gene that will be cloned – Another source is a gene carrier, called a vector – Plasmids (small, circular DNA molecules independent of the bacterial chromosome) are often used as vectors Copyright © 2009 Pearson Education, Inc. Steps in Cloning a gene ○ Steps in cloning a gene 1. Plasmid DNA is isolated 2. DNA containing the gene of interest is isolated 3. Plasmid DNA is treated with restriction enzyme (endonuclease) that cuts in one place, opening the circle 4. DNA with the target gene is treated with the same restriction enzyme and many fragments are produced 5. Plasmid and target DNA are mixed and associate with each other Copyright © 2009 Pearson Education, Inc. 6. Recombinant DNA molecules are produced when DNA ligase joins plasmid and target segments together 7. The recombinant DNA is taken up by a bacterial cell 8. The bacterial cell reproduces to form a clone of cells Copyright © 2009 Pearson Education, Inc. E. coli bacterium Plasmid Cell with DNA containing gene Bacterial 1 Isolate of interest chromosome plasmid 2 Isolate DNA 3 Cut plasmid DNA with enzyme Gene of interest 4 Cut cell’s DNA with same enzyme Gene of interest 5 Combine targeted fragment and plasmid DNA Examples of gene use 6 Add DNA ligase, which closes the circle with Genes may be inserted covalent bonds into other organisms Recombinant DNA Gene plasmid of interest 9 Genes or proteins 7 Put plasmid are isolated from the into bacterium cloned bacterium by transformation Recombinant bacterium Harvested proteins Examples of 8 Allow bacterium may be protein use to reproduce used directly Clone of cells E. coli bacterium Plasmid Cell with DNA containing gene Bacterial 1 Isolate of interest chromosome plasmid 2 Isolate DNA DNA Gene of interest E. coli bacterium Plasmid Cell with DNA containing gene Bacterial 1 Isolate of interest chromosome plasmid 2 Isolate DNA 3 Cut plasmid DNA with enzyme Gene of interest 4 Cut cell’s DNA with same enzyme Gene of interest E. coli bacterium Plasmid Cell with DNA containing gene Bacterial 1 Isolate of interest chromosome plasmid 2 Isolate DNA 3 Cut plasmid DNA with enzyme Gene of interest 4 Cut cell’s DNA with same enzyme Gene of interest 5 Combine targeted fragment and plasmid DNA E. coli bacterium Plasmid Cell with DNA containing gene Bacterial 1 Isolate of interest chromosome plasmid 2 Isolate DNA 3 Cut plasmid DNA with enzyme Gene of interest 4 Cut cell’s DNA with same enzyme Gene of interest 5 Combine targeted fragment and plasmid DNA 6 Add DNA ligase, which closes the circle with covalent bonds Recombinant DNA Gene plasmid of interest Recombinant DNA Gene plasmid of interest 7 Put plasmid into bacterium by transformation Recombinant bacterium Recombinant DNA Gene plasmid of interest 7 Put plasmid into bacterium by transformation Recombinant bacterium 8 Allow bacterium to reproduce Clone of cells Examples of gene use Genes may be inserted Recombinant into other organisms DNA Gene plasmid of interest 9 Genes or proteins 7 Put plasmid are isolated from the into bacterium cloned bacterium by transformation Recombinant bacterium Harvested proteins 8 Allow bacterium may be to reproduce used directly Clone of cells Examples of protein use Enzymes are used to “cut and paste” DNA ○ Restriction enzymes cut DNA at specific sequences – Different restriction enzymes binds to DNA at different nucleotide sequences – Many restriction enzymes make staggered cuts that produce restriction fragments with single-stranded ends called “sticky ends” – Fragments with complementary sticky ends can associate (bind) with each other, forming recombinant DNA ○ DNA ligase joins DNA fragments together Copyright © 2009 Pearson Education, Inc. Restriction enzyme recognition sequence 1 DNA Restriction enzyme cuts the DNA into fragments 2 Sticky end Restriction enzyme recognition sequence 1 DNA Restriction enzyme cuts the DNA into fragments 2 Sticky end Addition of a DNA fragment from 3 another source Restriction enzyme recognition sequence 1 DNA Restriction enzyme cuts the DNA into fragments 2 Sticky end Addition of a DNA fragment from 3 another source Two (or more) fragments stick together by base-pairing 4 Restriction enzyme recognition sequence 1 DNA Restriction enzyme cuts the DNA into fragments 2 Sticky end Addition of a DNA fragment from 3 another source Two (or more) fragments stick together by base-pairing 4 DNA ligase pastes the strands Recombinant 5 DNA molecule Recombinant DNA https://www.youtube.com/watch?v=C6DS2 0GktCI Recombinant DNA A molecule of DNA that includes genetic material from different sources is called recombinant DNA. It is produced through (by restriction the process of genetic engineering. endonucleases) Genetically engineering is used globally to produce organisms with desirable traits that could not be produced via selective breeding (GMO’s). (restriction fragment) E.g. mass-production of human insulin by bacteria. Restriction endonucleases “cut” the DNA sequence at a specific recognition site to allow for the insertion of a desired gene, which is “glued” into the genome using DNA ligase. Methylase protects DNA from cleavage by restriction enzymes. CRISPR Explained https://www.youtube.com/watch?v=UKbrwPL3wXE How CRISPR allows you to edit DNA (5:28) https://www.youtube.com/watch?v=6tw_JVz_IEc 7. 7 D. DNA Profiling DNA profiling is the analysis of DNA fragments to determine whether they come from a particular individual – Compares genetic markers from noncoding regions that show variation between individuals – Involves amplification (making many copies) of markers for analysis – Sizes of amplified fragments are compared Crime scene Suspect 1 Suspect 2 1 DNA isolated 2 DNA of selected markers amplified 3 Amplified DNA compared ○ Polymerase chain reaction (PCR) is a method of amplifying a specific segment of a DNA molecule Copyright © 2009 Pearson Education, Inc. Gel Electrophoresis Through a process known as gel electrophoresis, we can analyze and compare the DNA of any organism: Restriction enzymes cut DNA at specific points, forming a series of fragments DNA is negatively charged, so smaller fragments will migrate further towards the positive electrode The banding pattern of the gel allows us to compare samples of DNA Gel electrophoresis separates DNA molecules based on size – DNA sample is placed at one end of a porous gel – Current is applied and DNA molecules move from the negative electrode toward the positive electrode – Shorter DNA fragments move through the gel pores more quickly and travel farther through the gel – DNA fragments appear as bands, visualized through staining or detecting radioactivity or fluorescence – Each band is a collection of DNA molecules of the same length Mixture of DNA fragments of different sizes Longer (slower) Power molecules source Gel Shorter (faster) molecules Completed gel E.g. Samples #1 and #4 display the same banding pattern, indicating that the organisms they were taken from are closely related Applications of Gel Electrophoresis Who committed the crime? Who is the father? Recombinant cells and organisms can mass-produce gene products ○ Cells and organisms containing cloned genes are used to manufacture large quantities of gene products CONNECTION: Genetically modified organisms are transforming agriculture ○ Genetically modified (GM) organisms contain one or more genes introduced by artificial means ○ Transgenic organisms contain at least one gene from another species ○ GM plants – Resistance to herbicides – Resistance to pests – Improved nutritional profile ○ GM animals – Improved qualities – Production of proteins or therapeutics Copyright © 2009 Pearson Education, Inc. Should we be worried? Mitochondrial DNA DNA is not just found in the nucleus of the cell; it is also stored in the mitochondria. However, unlike nuclear DNA, mtDNA is only inherited from the mother (contained within the ova). By analyzing the mtDNA of a species, we can actually trace back lineages through time to identify a common ancestor 5. 5. D 6. 6. B D 9. A 10.

Molecular Genetics - Biology 30 - PDF

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