UNIT 3 - MOLECULAR GENETICS-1 PDF

SBI4U1 – 2024 UNIT 3 – MOLECULAR GENETICS – LESSON LIST LESSON TITLE HOMEWORK COMPLETED History of DNA Textbook:...

SBI4U1 – 2024 UNIT 3 – MOLECULAR GENETICS – LESSON LIST LESSON TITLE HOMEWORK COMPLETED History of DNA Textbook: 1 Chapter 6.1 (pages 270-272) Page 279 (#2, 4, 7, 9) Chapter 6.2 (pages 273-279) Textbook: Page 290 (#2, 3, 4, 6, 7) DNA Replication 2 Chapter 6.4 (pages 282-290) Worksheet: DNA Replication From Gene to Protein Textbook: 3 Chapter 7.1 (pages 312-318) Page 318 (#3, 4, 5, 6, 8, 9, 10) Transcription Textbook: 4 Chapter 7.2 (pages 319-324) Page 324 (#1-4, 6, 7, 9) Textbook: Page 331 (#1-4, 6) Translation Worksheets: 5 Chapter 7.3 (pages 325-331) Transcription and Translation Review (1) Transcription and Translation Review (2) Worksheets: Regulating Gene Expression 6 Control Mechanisms Chapter 7.4 (pages 332-339) Prokaryotes vs. Eukaryotes Textbook: Page 345 (#1, 2, 5) Genetic Mutations 7 Chapter 7.5 (pages 340-345) Worksheet: DNA Mutations Biotechnology: Manipulating and Cloning DNA Textbook: Chapter 8.1 (pages 366-375) Page 375 (#3, 4) 8 Chapter 8.2 (pages 376-385) Page 385 (#6) Chapter 8.3 (pages 386-390) Page 390 (#8) DNA Sequencing Textbook: 9 Chapter 8.2 (pages 376-385) Page 385 (#1, 2) 1 SBI4U1 – 2024 UNIT 3 – MOLECULAR GENETICS – EVALUATION LIST ASSESSMENT DESCRIPTION DUE DATE SUBMITTED 1 2 3 4 5 6 7 8 9 10 11 12 13 14 -drawing task? -codon code – transcription, translation and create monster task? 2 SBI4U1 – 2024 1. HISTORY OF DNA DNA (deoxyribonucleic acid – Carrier of genetic information in all living organisms on Earth All organisms, from very simple to very complex, use DNA to pass information for their construction and operation from one generation to the next. Within DNA are the instructions necessary to build all necessary proteins DNA is passed from generation to generation in the form of chromosomes. The size and number of chromosomes in a cell are specific to each species o E.g., Humans = 46, Turkeys = 82, Fruit Flies = 8, Sea Stars = 36, Potatoes = 48 Proteins, which are made up of amino acids, are found throughout an organism. o Some proteins are common to all forms of life, while others are specific to an individual species. Proteins have chemical and physical roles o E.g., proteins are enzymes responsible for the catalyzation of all cellular reactions When a particular protein is needed, the portion of DNA (the gene) that codes for this protein is activated. o The nucleotide sequence is copied (transcribed) into a molecule of RNA (ribonucleic acid). o The RNA then moves to the cytosol, where its sequence is translated by the ribosomes into amino acid chains called polypeptides. o Elsewhere in the cell, polypeptides are further modified to form functional proteins o This is protein synthesis 3 SBI4U1 – 2024 HISTORIACAL DISCOVERIES: GENETIC MATERIAL There are many scientists that have contributed to the discovery of genetic material including: Miescher (1869), Griffith (1928), Avery, McLeod, and McCarty (1944), and Hershey & Chase (1952). 1. FRIEDRICH MIESCHER Discovered DNA Collected white blood cells from pus Lysed cells and isolated nuclei Found a substance he called nuclein, which would later be identified as DNA Present in every cell type he tested High in phosphorus At the time, people thought proteins contained our hereditary information as they were complex and carried out biological functions (20 amino acids VS. 4 nucleotides) He collected pus from bandages of his patients – collected an unknown substance high in phosphorus Mendel was doing his experiments around this time but his work was not realized until much later (1900) 2. FREDERICK GRIFFITH Studied the bacterium that causes pneumonia, Streptococcus Used two different strains of pneumonia bacterium o Smooth (S strain) – contained a capsule that surrounded each cell and caused the bacterial colonies to look smooth and glossy when grown on agar o Rough (R strain) – lacked smoo0th capsule and formed rough and irregular colonies when cultured 4 SBI4U1 – 2024 Experiment: Could not identify exactly how this phenomenon worked Called it transformation o He defined it as change in a cell’s function by the transfer of an unknown substance o Modern definition: change in genotype and phenotype due to assimilation of external DNA by a cell 3. AVERY, MCLEOD, AND MCCARTY Continued Griffith’s idea by using R- and S- strains of pneumococcus bacteria Purified DNA from the S-strain bacteria Added it to non-virulent R-strain bacteria Determine if protein, RNA, or DNA that carried genetic material 5 SBI4U1 – 2024 4. HERSHEY & CHASE What they knew: Virus composed of DNA and protein Virus infection reprograms a host cell to produce more virus Wondered which viral components is responsible for the reprogramming: DNA or protein Used a bacteriophage to infect bacteria HISTORICAL DISCOVERIES: DNA STRUCTURE There are many scientists that have contributed to the discovery of DNA structure including: Levene (1909), Chargaff (1947), Wilkins & Franklin (1952), and Watson & Crick (1953). 1. PHOEBUS LEVENE Determined that: DNA is made up of chains of nucleotides A nucleotide is a phosphate linked to sugar linked to one of four nitrogenous bases 2. ERWIN CHARGAFF Isolated DNA from different organisms and measured numbers of each base o Adenine, Thymine, Cytosine, and Guanine Disagreed with other scientists that DNA are made up of equal parts of each base Found that the quantities of adenine matched the amount of thymine, same was found between cytosine and guanine o Number of A’s = Number of T’s o Number of C’s = Number of G’s 6  SBI4U1 – 2024 3. WILKINS & FRANKLIN Already knew: Hereditary molecule is DNA. DNA molecules are polynucleotides. DNA has a double-helix structure with alternating sugar-phosphate molecules with bases in the middle. What they don’t know: How are the bases arranged in the middle? Used x-ray diffraction of crystalline DNA fiber Discovered DNA structure is double helix Backbone of alternating phosphate and sugars Nitrogenous bases are in the middle of the molecule Bases are at right angles to the backbone 4. WATSON & CRICK Gathered all the information that have been acquired until now Found that the DNA molecule could only be stable if the strands antiparallel Their model also showed that the bases were held together with hydrogen bonds o Determined complementary base-pairing rules o Adenine to Thymine o Cytosine to Guanine 7 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 6.2 on page 279 of your textbook: #2, 4, 7, 9 2. The nitrogenous base content of a sample of DNA was found to be 32 % adenine. Determine the amounts of the other three bases in this sample. 4. Write the complementary strand for the following sequence: GTGACTAACAGTGGCCAT 7. How did the reproductive behaviour of bacteriophages allow Hershey and Chase to conduct their experiment? 9. Watson and Crick did not actually conduct any experiments with DNA. Do you think they can be considered scientists? Explain your reasoning. 8 SBI4U1 – 2024 2. DNA REPLICATION CONSERVATIVE VS. SEMICONSERVATIVE REPLICATION Semiconservative replication – a mechanism of DNA replication in which each of the two strands of parent DNA is incorporated into a new double-stranded DNA molecule Conservative replication – would involve copying the DNA molecule “as is”, leaving the two original (parent) strands together DNA REPLICATION – THE PROCESS STEP 1: STRAND SEPARATION Replication Origin Replication origins – specific sites on DNA where replication begins Replication begins at one fixed origin Replication proceeds bidirectionally until entire DNA is replicated There can be more than one replication origin 9 SBI4U1 – 2024 Helicase – an enzyme that bings to these origins and begins to unwind the two strands of DNA by breaking the hydrogen bonds between the complementary base pairs Replication Forks Y shaped structure At the origin sites, the two DNA strands separate forming a replication “bubble” with replication forks at each end The replication bubbles elongate as the DNA is replicated and eventually fuses o Replication Bubble – the separating of DNA in both directions during replication ▪ There are many replication bubbles along a DNA strand due to the length of the eukaryotic DNA Unwinding the DNA Helicase – enzyme that disrupts H bonds between two strands of DNA to separate the template DNA strands at the replication fork Single-strand binding proteins (SSBPs) – proteins that bind to unwound single-stranded regions of DNA to keep the template strands apart during replication Topoisomerase – enzymes that can break bonds in DNA and then reforms the bonds o Purpose is to release the twists in DNA that are generated during DNA replication Without multiple replication origins, which each produce a replication bubble, replicating the genome of a typical human would take an impractical amount of time 10 SBI4U1 – 2024 STEP 2: BUILDING THE COMPLEMENTARY STRANDS Priming the DNA Blue line: DNA to be copied Pink line: RNA nucleotides added (Primer) Light pink blob: enzyme that adds RNA (RNA polymerase) Priming DNA for replication Primer – a short segment of RNA needed to initiate DNA replication o All nuclei acids are formed in the 5’ to 3’ direction, even RNA Primase – an RNA polymerase (RNAP) which synthesizes the primer by adding ribonucleotides that are complementary to the DNA template Polymerase – enzymes that makes polymers Why is priming required? Due to different abilities of RNA polymerase (RNAP) and DNA polymerase (DNAP) RNA Polymerase: Can start a new chain without an existing end o Just needs a template DNA Polymerase: Can only add nucleotides o Can never start a new chain 11 SBI4U1 – 2024 DNA Polymerase III DNA polymerase III (DNAP III) catalyzes the elongation of DNA molecules by adding nucleotides to the 3’ end of a pre-existing nucleotide As each nucleotide is added, the last two phosphate groups are hydrolyzed DNA Polymerase I DNA polymerase I (DNAP I) replaces the RNA primer with DNA complementary to the template DNA Elongation DNA Polymerase III – elongates DNA strand DNA Polymerase I – replaces RNA with DNA 12 SBI4U1 – 2024 The Problem at the Fork Due to antiparallel nature of DNA One parental strand has its 3’ end at the fork while the other parental strand has its 5’ end at the fork DNA synthesis can only proceed in a 5’ → 3’ direction o A strand of DNA can only add nucleotides onto its 3’ end o DNAP must move along the parent strand from 3’ end to 5’ end The leading and lagging strands Leading strand – synthesized continuously Lagging strand – is synthesized in short, discontinuous segments of 1000-2000 nucleotides called Okazaki fragments Lagging Strand & Okazaki Fragments DNAP III synthesizes the DNA DNAP I replaces the RNA primer with DNA complementary to the parent strand template DNA ligase – joins broken pieces of DNA by catalyzing the formation of phosphodiester bonds 13 SBI4U1 – 2024 Figure 6 (a) As the replication fork opens, RNA primase attaches RNA primers to both the leading and lagging strands. These primers are oriented in opposite directions. (b) A DNA polymerase III enzyme attaches to the 39 end of each primer and begins assembling the new DNA strands in opposite (59 to 39) directions. (c) As the replication fork continues to open, the new leading strand continues to be assembled in one continuous process. On the lagging strand, a new RNA primer is attached and DNA polymerase III begins assembling a new DNA fragment. (d) The resulting DNA fragments are then joined by DNA ligase. (e) The entire process results in the formation of two new daughter DNA molecules: one consisting of a parent strand and the new leading strand, and the other formed by the other parent strand and the new lagging strand. STEP 3: REPAIR ERRORS DNA replication is not perfect – errors can and will occur! DNA polymerase II (DNAP II) – a slow working enzyme specialized in reading errors between replication events and will fix any errors in base-pairing The incorrect base-pairing may occur but DNAP II can detect the error as it is moving along the strand o It will stop and replace the incorrect base with the correct base 14 SBI4U1 – 2024 Table 1: Enzymes and Other Molecules Involved in DNA Replication Enzyme/Molecule Function Helicase Unwinds DNA helix Single-strand binding protein Stops the two separated parent strands from annealing (SSB) Cleave and then reattach one or two of the DNA strands to relieve tensions Topoisomerases created by the unwinding process RNA primase Places RNA primers on template strands RNA primers Acts as starting strands for DNA polymerase Several closely related enzymes that assemble nucleotides into new DNA strands; remove RNA primer nucleotides and replace them with DNA DNA polymerases nucleotides; proofread and repair replication errors and other damage to DNA molecules Forms the phosphodiester bond that joins the ends of DNA that make up DNA ligase the Okazaki fragments 15 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 6.4 on page 290 of your textbook: #2, 3, 4, 6, 7 2. Place the following enzymes in the order in which they are used in DNA replication: topoisomerase, DNA ligase, DNA polymerase III, RNA primase, helicase, DNA polymerase I. 3. What would happen if each of the following were not available for DNA replication? How would this affect the process? a) topoisomerase b) DNA ligase c) single-strand binding protein 4. Explain why the process of DNA replication is slower on the lagging strand than on the leading strand. 6. Why can both template strands not be replicated continuously? 7. Helicase separates and unwinds the DNA strands from each other. Why is an enzyme not required to bring the strands back together (re-anneal them)? 16 SBI4U1 – 2024 HOMEWORK: DNA REPLICATION Read the summary below and answer the questions that follow. DNA replication is the process in which a cell’s entire DNA is copied or replicated. The identification of the structure of DNA suggested that each strand of the double helix would serve as a template for synthesis of a new strand. DNA replication process occurs during the Synthesis phase (Interphase) of the eukaryotic cell cycle. As each DNA strand has the same genetic information, both strands of the double helix can serve as templates for the reproduction of a complementary new strand. The two resulting double helices, which each contain one "old" strand and one "new" strand of DNA, are identical to the initial double helix. DNA replication is said to be semi-conservative because of this process of replication, where the resulting double helix is composed of both an old strand and a new strand. The semi-conservative mechanism of replication was one of three models originally proposed for DNA replication: Semi-conservative replication would produce two copies that each contained one of the original strands and one new strand. Conservative replication would leave the two original template DNA strands together in a double helix, with the new DNA composed entirely of two new strands. DNA replication begins as an enzyme DNA helicase which breaks the hydrogen bonds holding the two strands together and forms a replication fork. The strands are held open by a single strand of binding proteins, preventing premature reannealing. Topoisomerase solves the problem caused by tension generated by winding/unwinding of DNA. This enzyme wraps around DNA and makes a cut permitting the helix to spin and relax. Once DNA is relaxed, topoisomerase reconnects broken strands. The resulting structure has two branching strands of DNA backbone with exposed bases. These exposed bases allow the DNA to be “read” by another enzyme, RNA primase, an RNA polymerase (RNAP) which synthesizes the primer by adding ribonucleotides that are complementary to the DNA template. Once a primer is added, DNA polymerase III can come in and start adding complementary nucleotides along the parent DNA. As DNA helicase continues to open the double helix, the replication fork grows. 17 SBI4U1 – 2024 5' → 3 The two new strands of DNA are “built” in opposite directions, through either a leading strand or a lagging strand. The leading strand is the DNA strand that DNA polymerase III constructs in the 5' → 3' direction. This strand of DNA is made in a continuous manner, moving as the replication fork grows. The lagging strand is the DNA strand at the opposite side of the replication fork from the leading strand. It goes in the opposite direction, from 3' to 5'. DNA polymerase cannot build a strand in the 3' → 5' direction. Thus, this "lagging” strand is synthesized in short segments known as Okazaki fragments. On the lagging strand, RNA primase builds another RNA primer. DNA polymerase III is then able to use the new RNA primer to make DNA in the 5' → 3' direction. The RNA fragments are then degraded by an enzyme DNA polymerase I, and new DNA nucleotides are added to fill the gaps where the RNA was present. Another enzyme, DNA ligase, is then able to attach (ligate) the DNA nucleotides together, completing the synthesis of the lagging strand. Many replication forks develop along a chromosome. This process continues until the replication forks meet, and the all of the DNA in a chromosome has been copied. Each new strand that has formed is complementary to the strand used as the template. Each resulting DNA molecule is identical to the original DNA molecule. Errors DNA polymerase III has the ability to detect when an incorrect nucleotide is “accidentally” bonded on the parent strand. The enzyme will automatically halt the addition of more nucleotide and fix this error before continuing. Another enzyme, DNA polymerase II, has the ability to detect any errors on the new strand. The enzyme follows a sort-of “autocorrect” feature when fixing any errors. 18 SBI4U1 – 2024 In the table below, explain how the following mutation could affect DNA replication. Helicase is unable to detect DNA strands. RNA polymerase form primers that can’t form hydrogen bonds. Single-stranded binding proteins unable to detect DNA strands. Topoisomerase unable to bond with DNA. DNA polymerase III can travel bidirectional. DNA polymerase III unable to detect the difference between adenine or thymine. DNA polymerase I is unable to detect ribonucleotides. DNA polymerase II unable to bond with DNA. DNA ligase unable to form phosphodiester bonds. 19 SBI4U1 – 2024 3. FROM GENE TO PROTEIN ONE GENE-ONE ENZYME HYPOTHESIS ARCHIBALD GARROD Studied patients with an inherited condition called Alkaptonuria Urine turned black when exposed to air o Due to a build up of alkapton Found that this was due to an inability to breakdown alkapton In normal conditions, breakdown of alkapton is catalyzed by an enzyme Concluded that this defect must’ve caused by a faulty gene being passed down by the parents GEORGE BEADLE & EDWARD TATUM Experimented bread mould known as Neurospora crassa o Simple organisms that could grow in a minimal media that contained a few biological substances (salts, sugar, and vitamin biotin) o The mould could use these simple substances to synthesize the nutrients on their own Used X-ray to create mutations in genes Found that the mutated molds were unable to grow unless additional nutrients were added THE ONE GENE-ONE ENZYME HYPOTHESIS Beadle & Tatum hypothesized that there was a direct relationship between DNA and enzymes “one gene-one enzyme hypothesis” o One gene-one enzyme hypothesis – each gene is unique and codes for the synthesis of a single enzyme Based on the idea that chemical reactions take place in a series of steps (metabolic pathway) The hypothesis was inaccurate as not all proteins are enzymes, and proteins can be made of multiple subunits of polypeptide 20 SBI4U1 – 2024 Many proteins are not enzymes, and many proteins consist of more than one subunit. o Beadle and Tatum’s hypothesis was restated as the one gene–one polypeptide hypothesis as this subunit is a polypeptide CONNECTION BETWEEN DNA, RNA, AND PROTEIN Central dogma – the term given by Francis Crick in 1956 that describes the flow of information from DNA to RNA to protein. This process has two major steps: Transcription and Translation. THE FLOW OF INFORMATION: 1) DNA → mRNA Known as transcription – mechanism by which information coded in the nucleic acids of DNA is copied into nucleic acids of RNA; something rewritten in the same language Occurs in the nucleus 2) mRNA → protein Known as translation – mechanism by which the information coded in the nucleic acids of RNA is copied into the amino acids of proteins Occurs on the ribosomes (RER) in the cytoplasm 21 SBI4U1 – 2024 RNA: RIBONUCLEIC ACID Made up of building blocks called ribonucleotides – similar to DNA nucleotides (sugar, phosphate group, nitrogenous base) DNA RNA Deoxyribose sugar Ribose sugar Adenine, Thymine, Guanine, Cytosine Adenine, Uracil, Guanine, Cytosine Double-stranded Singled stranded Located in the nucleus Can be found anywhere in the cell 22 SBI4U1 – 2024 THE THREE MAJOR TYPES OF RNA Messenger RNA (mRNA) – the end product of the transcription of a gene; mRNA is translated by ribosomes into a protein Transfer RNA (tRNA) – a carrier molecule that binds to a specific amino acid and adds the amino acid to the growing polypeptide chain Ribosomal RNA (rRNA) – an RNA molecule within the ribosome that bonds the correct amino acid to the polypeptide chain Types of RNA Characteristics and key functions varies in length, depending on the gene that has been copied acts as the Messenger RNA intermediary between DNA and the ribosomes (mRNA) is translated into protein by ribosomes is the RNA version of the gene encoded by DNA functions as the delivery system of amino acids to ribosomes as they synthesize Transfer RNA proteins (tRNA) is very short, only 70 to 90 base pairs long Ribosomal RNA binds with proteins to form the ribosomes (rRNA) varies in length TRANSCRIPTION AND TRANSLATION – AN OVERVIEW TRANSCRIPTION RNA Polymerase – creates an RNA molecule with a base sequence that is complementary to one strand of the DNA sequence given o It is an enzyme that reads a DNA strand and creates a complementary strand of RNA Transcription follows the same basic rules of complementary base pairing and nucleic acid chemistry o I.e., since the DNA strand is read in the 3’ to 5’ direction, the mRNA will be formed in the complementary 5’ to 3’ direction Template Strand – One strand of the DNA and is read by RNA Polymerase o Is a DNA strand that is transcribed into a precursor mRNA molecule Precursor mRNA (or pre-mRNA) – the initial RNA transcription product o Cannot be used to price a protein; modified to become an mRNA strand which can now exit the nucleus and enter the cytosol, where ribosomes are found 23 SBI4U1 – 2024 DNA Template Strand DNA polymerase base pairing: o Adenine – Thymine o Cytosine – Guanine RNA polymerase base pairing: o Adenine – Uracil o Cytosine – Guanine If the DNA template strand has the sequence: 3’-CTAAGT-5’, what will the sequence of the mRNA strand transcribed? TRANSLATION The mRNA molecule associates with a ribosome As the ribosome moves along the mRNA, the amino acids coded for by the mRNA are delivered by tRNA to the ribosome The amino acids are joined together, one by one, to form the polypeptide encoded by the gene THE GENETIC CODE Genetic Code – the specific coding relationship between bases and the amino acids they specify; the genetic code can be expressed in terms of either DNA or RNA bases DNA alphabet consists of four letters: “A”, “T”, “G”, and “C” RNA alphabet consists of four letters: “A”, “U”, “G”, and “C” While there are only four RNA bases, there are 20 amino acids 24 SBI4U1 – 2024 How is nucleotide information in an mRNA molecule translated into the amino acid sequence of a polypeptide? The four bases in an mRNA must be used in combinations of at least three to provide the capacity to code for 20 amino acids. Codon – the name of the 3-letter code o Are in the 5’ to 3’ order in the mRNA THE CODON CHART Of the 64 codons, 61 specify for amino acids o Also known as the sense codons o E.g., AUG = methionine AUG is usually the first codon translated into any mRNA in both prokaryotes and eukaryotes o AUG is called the start codon or the initiator codon Start Codon (initiator codon) – the codon that signals the start of a polypeptide chain and initiates translation Stop Codon – a codon that signals the end of a polypeptide chain and causes the ribosome to terminate translation o UAA, UAG, and UGA are stop codons Only methionine and tryptophan are indicated by a single codon o The rest are represented by at least two codons E.g., UGU and UGC both code for cysteine 25 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 7.1 on page 318 of your textbook: #3, 4, 5, 6, 8, 9, 10 3. Using a Venn diagram, compare the structure of DNA and RNA. 4. What are the three major classes of RNA, and what is the function of each? 5. Compare and contrast transcription and translation in terms of their purpose and location. 6. The sequence of a fragment of one strand of DNA is AATTGCATATACGGGAAATACGACCGG. Transcribe this sequence into mRNA. 26 SBI4U1 – 2024 8. The following mRNA strand is being used to assemble a polypeptide strand by a ribosome: 59-AUGCUUGCUCAUCGGGGUUUUAA-39 (a) Write out the amino acids that will be assembled, in their correct order. (b) Provide an alternative mRNA sequence with four or more changes that would translate to the same amino acid sequence. 9. Write out all possible RNA base codons that could code for the following amino acid sequences. (a) Arg-Trp (b) Leu-Pro (c) Met-Phe-Trp 10. Differentiate between a stop codon and a start codon. 27 SBI4U1 – 2024 4. TRANSCRIPTION During transcription, the DNA code is chemically rewritten as an RNA code. This occurs within the nucleus. Transcription is divided into three sequential processes: (1) Initiation, (2) Elongation, and (3) Termination INITIATION In both prokaryotes and eukaryotes, the process of transcription begins when the enzyme RNA polymerase binds to the DNA and unwinds it near the beginning of a gene Binding occurs at a promotor o Promoter – transcription start point where RNA polymerase (RNAP) will first bind to ▪ Located upstream of a gene sequence Composed of an area rich in Thymine and Adenine (TATA box) recognized by RNA Polymerase o Reason for this is because adenine and thymine only have 2 hydrogen bonds compared to cytosine and guanine (3 H- bonds) o Less energy required to break A-T bonds RNA Polymerase (RNAP) will recognize the promoter and begin unwinding DNA ELONGATION Once the RNA polymerase binds to the promoter and opens the DNA double helix, it starts to build the single-stranded RNA molecule by unwinding the region of DNA exposing about 10-20 base pairs RNA polymerase, unlike DNA polymerase, can begin making the complimentary copy without needing a primer to be already in place RNA is made in the 5’→3’ direction, using the 3’→5’ DNA strand as a template strand 28 SBI4U1 – 2024 The opposite strand of DNA—the strand that is not being copied—is known as the coding strand since it contains the same base-pair sequence as the new RNA strand (except thymine is replaced with uracil) o Coding Strand – the DNA strand that is not being copied but contains the same sequence as the new RNA molecule Direction of transcription: 5’ → 3’ (downstream) RNAP unwinds DNA as its moving along; DNA begins winding back at the end TERMINATION Multiple different mechanisms can be used to stop transcription In prokaryotes: Example 1: A protein may come in to stop the process Example 2: the mRNA may bind with itself to form hairpin loops which causes RNAP to stop In eukaryotes: RNAP may transcribe a terminator sequence containing a string of adenine, which will result in transcribing a string of complementary uracil bases Creates an area of weak interactions Nuclear proteins can bind to these regions and stop transcription Termination Sequence – a sequence of bases at the end of a gene that signals RNAP to stop transcribing 29 SBI4U1 – 2024 30 SBI4U1 – 2024 MULTIPLE TRANSCRIPTION MACHINERY Multiple RNAP can transcribe simultaneously on the same gene As long as the promoter is free of RNAP, then the next RNAP can bind and begin transcription No need to wait for an RNAP to complete transcribing the whole gene before the next RNAP begins POST-TRANSCRIPTIONAL MODIFICATIONS In prokaryotes, mRNA can be used directly o Can even begin translation before mRNA is full transcribed In eukaryotes, mRNA needs to be modified before leaving the nucleus o From precursor (pre-mRNA) to mature mRNA mRNA MODIFICATIONS: CAPPING Poly (A) tail A series of 50-250 adenine added to the 3’ end Added by an enzyme called poly-A polymerase Functions are to protect mRNA from degradation and facilitate leaving the nucleus 5’ cap A sequence of 7 guanines added to the 5’ end Functions are to protect mRNA from degradation and signals for ribosomes to attach 31 SBI4U1 – 2024 mRNA MODIFICATIONS: SPLICING Removal of introns (non-coding sequence of DNA or RNA); mature mRNA will contain only of exons (a sequence of DNA or RNA that codes for part of a gene) Occurs in a spliceosome (an enzyme-protein complex that removes introns from the mRNA) A region that includes part of the mRNA, small ribonucleoproteins (snRNPs), and other proteins o Small ribonucleoproteins (snRNPs) - a protein that binds to the introns and signals them for removal Step 1: snRNPs bind to splice sites Step 2: excises introns Step 3: rejoins exons TRANSCRIPTION – PROKARYOTES VS. EUKARYOTES Variable Prokaryotes Eukaryotes Location Transcription occurs throughout the cell. Transcription takes place in the nucleus. Different RNA polymerases are used to A single type of RNA polymerase transcribes transcribe genes that encode protein (RNA Enzymes all types of genes. polymerase II) and genes that do not encode protein (RNA polymerase I, III). Bases are added quickly (15 to 20 nucleotides Bases are added slowly (5 to 8 nucleotides per Elongation per second). second). The promoters are immediately upstream of The promoters are less complex than those in Promoters protein-coding genes, and they are more eukaryotes. complex than those in prokaryotes. A protein binds to the mRNA and cleaves it, or Nuclear proteins bind to the polyuracil site and Termination the mRNA binds with itself. terminate transcription. Introns and exons There are no introns. There are both introns and exons. Transcription results in pre-mRNA, which must Transcription results in mRNA ready to be be modified to protect the final mRNA from Product translated into protein by ribosomes. degradation in the cytosol and to remove introns. 32 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 7.2 on page 324 of your textbook: # 1-4, 6, 7, 9 1. List and describe the three stages of transcription. 2. If the DNA template strand has the sequence 3’-CAAATTGGCTTATTACCGGATG-5’, what would be the sequence of an RNA molecule transcribed from it? 3. Explain the role of each of the following in transcription. (a) promoter (b) RNA polymerase (c) spliceosomes 33 SBI4U1 – 2024 4. Differentiate between introns and exons. 6. Compare and contrast DNA replication and transcription. How are they similar? How are they different? Present your answer in table form. 7. How is it possible for an organism to produce more proteins than it has genes for? 34 SBI4U1 – 2024 9. Suppose that you are provided with a sample of eukaryotic DNA. You divide the sample into three separate reaction mixtures and perform an experiment. Once transcription is complete, you analyze the base composition of mRNA from each mixture. You obtain the results in Table 2. Based on these results, answer the questions below. (a) Which strand of DNA served as the template for the synthesis of mRNA strand A? Which strand served as the template for the synthesis of mRNA strand B? Explain your reasoning. (b) Explain why the percentage of adenine is higher in the mRNA strands than in the DNA strands. 35 SBI4U1 – 2024 5. TRANSLATION Transcription uses the message coded in a DNA strand as a template for the synthesis of a complementary single-stranded mRNA molecule The next step in the process of creating a protein is translating the information coded in mRNA to a protein In translation, the encoded message is read, codon by codon, by a ribosome and, with the presence of transfer RNA molecules, the ribosome assembles one amino acid at a time into a polypeptide chain RECALL Gene Expression – the transfer of genetic information from DNA to RNA to protein READING THE GENETIC CODE 1. Find the sentence 2. Identify the three similarities/patterns that you see 3. How does each similarity relate to translating a genetic code Bisnetefthemanwastoooldforthewarenduidhdia Vuswktheboyhadoneredcarandoneoldhatendfgjhe beifuakbjkthecarwasredandtoooldforhernewjobendakebfiu 36 SBI4U1 – 2024 THE GENETIC CODE The final mRNA is transported to ribosomes where translation occurs Ribosomes read the mRNA in multiples of 3 nucleotides called codons For example: 5’-AUGCCCUCAUAG-3’ Each codon signals the cell to retrieve and add an amino acid to a growing chain of amino acids using transfer RNA (tRNA) READING THE GENETIC CODE Sentences Protein Synthesis (Translation) Each word is 3 letters long Each codon is 3 bases long Starts with “the” Starts with “AUG” Ends with “end” Ends with a stop codon CODONS Codon – a group of three base pairs that codes for an amino acid o Note that multiple codons can code for the same amino acid Start codon – the first codon of the mRNA; usually AUG which codes for methionine Stop codon – signals for the end of the polypeptide chain For example: 5’-AUGCCCUCAUAG-3’ 37 SBI4U1 – 2024 TRANSFER RNA (tRNA) Single stranded RNA of about 80 nucleotides, composed of: o An anticodon loop that is complementary to mRNA (binds to the mRNA codon) ▪ Anticodon – the complementary sequence of base pairs on a tRNA that corresponds to a codon on an mRNA o A 3’-region where an amino acid attaches Example: a tRNA that is linked to serine (Ser) pairs with the codon 5’-AGU-3’ in mRNA. The anticodon of the tRNA that pairs with this codon is 3’-UCA-5’ Process of adding an amino acid to the tRNA is known as aminoacylation o This forms a complex known as aminoacyl-tRNA RIBOSOMES Composed of ribosomal RNA (rRNA) and ribosomal proteins Comprised of 2 subunits: o Large (turquoise): catalyzes the peptide bonds between amino acids o Small (green): supports the mRNA 38 SBI4U1 – 2024 TRANSLATION: THE PROCESS There are three major stages of translation: (1) Initiation, (2) Elongation, and (3) Termination INITIATION In eukaryotes, ribosome small subunit recognizes and binds to the mRNA at the 5’ cap Ribosome will then move along until it reaches the translation start site (start codon AUG) Initiator tRNA carrying methionine will attach at the AUG codon o This will ensure that the following series of codons will be properly encoded (reading frame) 5’-CGAUGAGCUUGCGAUCGA-3’ RIBOSOMAL BINDING SITES 1 site where mRNA will bind to 3 sites where tRNA will bind to o A site: aminoacyl-tRNA site ▪ Holds the aminoacyl-tRNA carrying the next amino acid to be added o P site: peptidyl-tRNA site ▪ Holds the tRNA molecule carrying the growing polypeptide chain o E site: exit site ▪ Where tRNA molecule leaves the ribosome 39 SBI4U1 – 2024 ELONGATION Once initiator tRNA binds to the ribosome, incoming aminoacyl-tRNA binds to the A site H bonds form between the mRNA codon and tRNA anticodon This step requires energy in the form of GTP Ribosome catalyzes the formation of a peptide bond o Between the amino acid in the P-site to the amino acid in the A- site o Catalyzed by an enzyme called peptidyl-transferase Result: Polypeptide chain is longer by one amino acid Polypeptide chain is transferred to tRNA at the A site Ribosome moves along the mRNA: o tRNA from P site moves to E site and will subsequently leave the ribosome o tRNA from A site moves to P site to be ready for the next polymerization o A site is now empty and next aminoacyl-tRNA can bind to it Ribosome moves along mRNA in only one direction (5’ ⟶ 3’) Moves by 1 codon (3 nucleotides) at a time 40 SBI4U1 – 2024 TERMINATION At the stop codon, a protein release factor binds to the A site instead of an aminoacyl-tRNA Protein release factor will not hold an amino acid o Will add a water molecule instead of amino acid to polypeptide o The polypeptide chain will be released At the same time, the entire translation complex will disassemble PROTEIN SYNTHESIS IN EUKARYOTES AND PROKARYOTES POLYSOME Polysome – a complex that is formed when multiple ribosomes attach to the same mRNA in order to facilitate rapid translation In both prokaryotes and eukaryotes, a single strand of mRNA can be used to make multiple copies of a polypeptide simultaneously When this happens, a complex called a polysome is formed o When 1 molecule of mRNA has multiple ribosomes translating at the same time A Comparison of Translation in Prokaryotes and Eukaryotes 41 SBI4U1 – 2024 Variable Prokaryotes Eukaryotes mRNA can only be translated after exiting the nucleus to interact with ribosomes in the mRNA is translated by ribosomes in the cytosol cytosol Location as it is being transcribed from DNA some translation occurs in mitochondria and chloroplasts mRNA bases pair directly with a ribosomal binding site, just upstream of the start codon complex of Met–tRNA, with small ribosomal Initiation subunits, binds to an mRNA 59 cap and scans until it encounters the start codon mRNA 59 cap is involved Elongation 15 to 20 elongation cycles per second 1 to 3 elongation cycles per second Termination stop codon appears and a release factor binds so that the polypeptide is released mRNA strand can be translated by multiple mRNA strand can be translated by multiple Polysomes ribosomes simultaneously, even as it is being ribosomes simultaneously, but only in the transcribed from DNA cytosol POLYPEPTIDE TO PROTEIN The polypeptide that has been assembled by the ribosome is still not functional o The protein exists in an inactive state Since the shape of a protein defines its function, the polypeptide chain must be folded into the correct conformation Multiple processing reactions, carried out by specific enzymes, remove amino acids from the ends or interior of the chain and may add additional molecules, such as sugars, to the chain o These reactions activate the polypeptide, which then folds into its functional shape. 42 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 7.3 on page 331 of your textbook: # 1-4, 6 1. What are the key steps in the initiation of translation in eukaryotes and prokaryotes? 2. What is the role of tRNA in translation? 3. Why is there not a specific tRNA molecule for each possible codon? 4. List the possible anticodons for phenylalanine, alanine, and tyrosine. 6. Explain what occurs at the A, P, and E sites during the translation of mRNA into a polypeptide. 43 SBI4U1 – 2024 HOMEWORK: TRANSCRIPTION AND TRANSLATION REIVEW (1) 1. In what ways do the chemical structures of DNA and RNA differ? 2. What is a codon and what does it represent? 3. What is an anticodon? 4. Compare and contrast the final products of DNA replication and transcription. 5. You have learned that there is a stop codon that signals the end of an amino acid chain. Why is it important that a signal to stop translation be part of protein synthesis? 6. Why does a cell need to carry out transcription before translation? 7. Explain how a gene directs the synthesis of a protein, Include in your explanation the words amino acid, anti-codon, codon, cytoplasm, DNA, mRNA, nucleotide, nucleus, protein, ribosome, RNA polymerase, tRNA, transcription, translation, 5’ cap, and poly-A tail. 44 SBI4U1 – 2024 8. In the cell how could a single changed base in mRNA affect the synthesis of proteins? 9. Describe the function of each of the following in protein synthesis: rRNA, mRNA and tRNA. 10. Considering that we are all made up of the same 4 nucleotides in our DNA, and the same 4 nucleotides in our RNA, and the same 20 amino acids in our proteins, why are we so different from each other? 11. Why does it make sense to use the word translation to describe protein synthesis? 12. Why would it not make sense to use the word translation to describe mRNA synthesis? 45 SBI4U1 – 2024 HOMEWORK: TRANSCRIPTION AND TRANSLATION REVIEW (2) 1. Complete the table: Types of RNA RNA Structure Function mRNA rRNA tRNA 2. For each of the following sequences, fill in either the DNA, the mRNA sequence, the tRNA anticodon, or amino acid sequences that have been left blank. DNA mRNA AUG ACU AGC UGG GGG UAU UAC UUU UAG tRNA AA DNA TAC CGC TCC GCC GTC GAC AAT ACC ACT mRNA tRNA AA 46 SBI4U1 – 2024 DNA mRNA tRNA UAC CAC CCC CGU AUG GCU GGG AAU AUC AA 3. Fill in the blanks. 47 SBI4U1 – 2024 48 SBI4U1 – 2024 6. REGULATING GENE EXPRESSION Not all proteins are always required by all cells. It would be inefficient for a cell to always transcribe and translate all its genes. Instead, both prokaryotic and eukaryotic cells regulate gene expression in response to their own life cycles and environmental conditions. For example: Human insulin is only required when the glucose level in the blood is high. o The E. coli enzyme that facilitates the breakdown of lactose is only transcribed and translated when the E. coli bacteria are exposed to lactose PROKARYOTE GENE CONTROL MECHANISMS Operons – A cluster of genes on the bacterial chromosome under the control of one promoter and one operator Promoter – Region on DNA where RNAP binds to start transcription Operator – Region on DNA where regulatory protein binds and controls access of RNAP to the genes o Can activate or repress transcription (on/off button) THE lac OPERON lac Operon – a cluster of genes that contains the DNA sequences to regulate the metabolism of lactose in prokaryotes Consists of three genes that code for the proteins involved in the metabolism of lactose (1) Promoter – the site where DNA transcription begins (2) Operator – the sequence of bases that control transcription (the region in the operon that regulatory factors bind to) (3) Coding regions for the various enzymes that actually metabolize the lactose 49 SBI4U1 – 2024 Repressor protein – a protein that binds to the operator to repress gene transcription o Located upstream from the operon o For the lac operon, this protein is called lacI protein or lac repressor which is found upstream that codes for the LACI protein WHEN LACTOSE IS ABSENT The system is turned off lacI-made repressor proteins binds to operator RNA polymerase can’t make mRNA WHEN LACTOSE IS PRESENT System is turned on LACI protein is inactive RNA polymerase can make mRNA Lactose as an inducer – a signal molecule that triggers the expression of an operon’s genes o Lactose will signal the cell to produce lactose-metabolizing enzyme 50 SBI4U1 – 2024 Trp OPERON Tryptophan is an important amino acid used to build proteins Similar structure to lac operon (promoter, operator, and genes to code for tryptophan synthesis) Has a repressor gene called trpR, a sequence found upstream that codes for a trp repressor WHEN TRYPTOPHAN IS ABSENT Repressor protein is not active and will not bind to the operator RNAP can bind to the promoter and continue with transcription 51 SBI4U1 – 2024 WHEN TRYPTOPHAN IS PRESENT Tryptophan will bind and activate the repressor protein Repressor protein will bind to operator and stop transcription Tryptophan is an example of a co-repressor (a signal molecule that binds to a regulatory protein to reduce the expression of an operon’s genes) o Signals the halt of its own production o Serves to repress rather than induce the expression of a set of genes 52 SBI4U1 – 2024 TRANSCRIPTIONAL REGULATION NEGATIVE REGULATION lac operon Repressor made in active form; inactivated by inducer Turns gene on trp operon Repressor made in inactive form; activated by corepressor Turns gene off PRE-TRANSCRIPTIONAL (EUKARYOTES) Histones: Histones wrap DNA into coils so that RNAP is unable to bind and properly transcribe it Only once general transcription factors build up at the promoter site, RNAP can initiate transcription Methylation: The addition of –CH3 (methyl) to the cytosine bases They silence genes by blocking transcription machinery from binding to DNA They will also recruit proteins to bind onto methylated DNA, which will further block RNAP Agouti Mice Mice fed a high-methyl diet o DNA found to be methylated o Agouti gene was turned off o Turned out brown, thin, and healthy 53 SBI4U1 – 2024 Mice fed a regular mouse diet o DNA found to be unmethylated (or low amounts) o Agouti gene was expressed o Mice turned out yellow, obese, and prone to cancer or diabetes (High nurtured mice) GR Gene in rats when active, produces a protein that helps the body relax after stress POST-TRANSCRIPTIONAL (EUKARYOTES) Influences gene expression by several mechanisms, including changes in pre-mRNA processing and the rate at which mRNAs are degraded Alternative Splicing Produces different mature mRNA by removing different combinations of introns TRANSLATIONAL REGULATION Occurs during protein synthesis by a ribosome Change in the length of the poly(A) tail of the mRNA molecules Specific enzymes can add or delete repeating sequences of adenine at the ends of the mRNA molecules This change in the length of the poly(A) tail may increase or decrease the time that is required to translate the mRNA into a protein 54 SBI4U1 – 2024 POST-TRANSLATIONAL REGULATION After the mRNA is translated and proteins are synthesized, the cell can still regulate expression by limiting the availability of functional proteins. Three methods are used: (1) Processing, (2) Chemical modification, and (3) Degradation FOUR LEVELS OF CONTROL OF GENE EXPRESSION IN EUKARYOTIC CELLS Type of Control Description Specific Examples Access to promoters is provided by loosening a DNA molecule from histones. regulates which genes are Activator and repressor proteins bind to the promoter transcribed (DNA to mRNA) or Transcriptional and enhance or decrease controls the rate at which the rate of transcription. transcription occurs Methyl groups are added to cytosine bases in the promoter. RNA polymerase cannot bind and transcribe. Alternative splicing occurs. Different combinations of controls the availability of mRNA introns are removed, and the remaining exons are molecules to ribosomes; pre- spliced together. mRNA molecules undergo Post-transcriptional Masking proteins bind to mRNA and inhibit further changes in the nucleus, processing. resulting in final mRNA before translation occurs The rate of degradation of mRNA is dependent on the need of the cell for the gene product. controls how often and how Variation of the length of the poly(A) tail is related to the Translational rapidly mRNA transcripts will be rate of translation. translated into proteins Processing occurs controls when proteins become The polypeptide is chemically modified to render it an fully functional, how long they active protein. Post-translational are functional, and when they The presence of hormones may lengthen or shorten the are degraded length of time that a protein is functional. Ubiquitin-tagged proteins are degraded. 55 SBI4U1 – 2024 HOMEWORK: CONTROL MECHANISMS 56 SBI4U1 – 2024 57 SBI4U1 – 2024 HOMEWORK: PROKARYOTES VS. EUKARYOTES 58 SBI4U1 – 2024 7. GENETIC MUTATIONS Genetic mutations are changes in the DNA sequence, caused by various mechanisms. For example, synthetic chemicals, radiation, incorrect replication, and random mutations can change the structure and function of the genome SMALL-SCALE MUTATIONS Point Mutation – Occurs when one nucleotide (or a base pair) is altered Types of point mutations include: (1) Substitution – A base pair is replaced by a different base pair (2) Insertion – Addition of an extra base (3) Deletion – A base pair is removed (4) Inversion – Two adjacent base pair trade places The differences in the DNA of individuals within a population that are caused by point mutations are referred to as single nucleotide polymorphisms or SNPs The effects of small-scale mutations can range from being positive, through having no effect, to being severe Small-scale mutations can be categorized into four groups: (1) Missense mutations – mutation that changes a single amino acid in the coding sequence (2) Nonsense mutations – mutation that results in a premature stop codon (3) Silent mutations – mutation that does not alter the resulting sequence of amino acids (4) Frameshift mutations – a shift in the reading frame resulting in multiple missense and/or nonsense effects 59 SBI4U1 – 2024 MISSENSE MUTATION Change of a single base pair or group of base pairs results in the code for a different amino acid The protein that is synthesized will have a different sequence and structure, and it may be non-functional or function differently Can be beneficial if it creates a new, desirable effect NONSENSE MUTATION Change of a single base pair or group of base pairs results in a premature stop code in the gene The polypeptide is cut short and, most likely, will be unable to function SILENT MUTATION Change in one or more base pairs does not affect the functioning of the gene The mutated DNA sequence codes for the same amino acid as the non-mutated sequence, and the resulting protein is not altered FRAMESHIFT MUTATION Occurs when one or more nucleotides are inserted into or deleted from a DNA sequence, causing the reading frame of codons to shift in one direction or the other This results in multiple missense and/or nonsense effects The frameshift mutation “shifts” the reading frame by one or more steps, and every amino acid coded for after this mutation is affected Any deletion or insertion of base pairs in multiples of three does not cause frameshifts because the reading frame is unaltered E.g., Tay-Sachs disease is a result of the insertion of four base pairs. 60 SBI4U1 – 2024 EXAMPLE – SICK CELL ANEMIA Substitution missense: A → T Changes amnio acid glutamine to valine EXAMPLES – Classify these mutations Wild Type: Two men sat and had hot tea Two men Sat and had hot tea Frameshift insertion Two men sat and had hot sea Frameshift deletion Two me. Silent Two men sat and had hot tte a Missense Two mes ata ndh adh ott ea Nonsense LARGE-SCALE MUTATIONS Involves multiple nucleotides, entire genes, or whole regions of chromosomes Chromosomal Translocation – movement of entire genes of DNA from one chromosome to another o Translocation between two non-homologous chromosomes usually occurs when portions of each chromosome break off and exchange places. o If a DNA coding sequence is translocated adjacent to another coding sequence, this can result in an entirely new gene and a completely novel polypeptide chain. o Certain sequences of DNA, called transposable elements, move freely about the genome. o If transposable elements are inserted near an existing gene sequence, they can enhance, disrupt, or otherwise modify the expression of the gene. Inversion occurs when a portion of a DNA molecule, often containing one or many genes, reverses its direction in the genome. o This does not directly result in the loss of genetic material. However, if the break occurs in the middle of a coding sequence, the gene may be compromised. 61 SBI4U1 – 2024 Trinucleotide Repeat – A triplet of nucleotides o Example: CAG CAG CAG CAG. o Trinucleotide repeats are normal in the genome but sometimes a mutation can occur, in which these repeats become unstable and expand uncontrollably. ▪ Known as trinucleotide repeat expansion and increases in the number of repeats from one generation to the next. E.g., Huntington’s disease arises from a trinucleotide repeat. CAUSES OF GENETIC MUTATIONS SPONTANEOUS MUTATION Takes place naturally as a result of normal molecular interactions Examples: o DNA replication o Transposons (jumping genes) INDUCED MUTATION Caused by environmental agents Mutagens: a substance that increases the rate of mutation Examples: o Nitrous acid (HNO2), can turn Cytosine into Uracil o UV light is absorbed by thymine in DNA causing it to bind with adjacent nucleotides 62 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 7.5 on page 345 of your textbook: # 1, 2, 5. 1. Define the following terms: (a) mutation (b) frameshift mutation (c) point mutation (d) nonsense mutation (e) missense mutation 2. The normal form of a gene contains the nucleotide sequence 3’-ATACCCGCCTTTTCGTACTTCCTAG-5’ What type of mutation is each of the following, and what effect will this mutation have on the structure of the protein encoded in the gene? (a) Three extra adenine nitrogenous base pairs are inserted in positions 10, 11, and 12. (b) The third nucleotide is changed from an adenine to a thymine nucleotide. (c) The thymine nucleotide in position 10 is removed. 5. During the translation of a molecule of mRNA, the process is stopped prematurely. What type of mutation has occurred? What type of codon has been read by the ribosome? 63 SBI4U1 – 2024 HOMEWORK: DNA MUTATIONS 64 SBI4U1 – 2024 65 SBI4U1 – 2024 66 SBI4U1 – 2024 8. BIOTECHNOLOGY: MANIPULATING AND CLONING DNA ARTIFICAL SELECTION & SELECTIVE BREEDING Artificial selection – Identification by humans of desirable traits in plants and animals, and the steps taken to enhance and perpetuate those traits in future generations Selective breeding – The practice of mating individuals with desired traits as a means of increasing the frequency of those traits in a population. Artificial selection Used in agriculture to produce animals and crops with desirable traits Artificial selection appeals to humans since it is faster than natural selection and allows humans to mold organisms to their needs. Examples Meats sold today are the result of the selective breeding of chickens, cattle, sheep, and pigs. Many fruits and vegetables have been improved or even created through artificial selection, such as broccoli, cauliflower, and cabbage, which were all derived from the wild mustard plant through selective breeding. Breeds of dogs are mixed to produce mixed breeds such as a goldendoodle, labrador retriever, etc. GENETICALLY MODIFIED ORGANISMS (GMOs) Organisms whose genes have been directly manipulated by scientists, often by inserting or deleting one or more genes Inserted genes are typically from another species 67 SBI4U1 – 2024 BIOTECHNOLOGY The intentional manipulation of genetic material of an organism But why would we want to do this? To study cellular processes of an organism o Glowing gene from a jellyfish added to a tobacco plant To give one organism the trait(s) of another o Anti-freeze from fish blood into strawberries to survive through early frosts COMMON USES OF BIOTECHNOLOGY 1. Make “stuff” o Proteins, enzymes, medication, etc. o Food can be altered to have new traits o Cloning 2. Genetic screening o Crime scenes, paternity screenings, etc 3. Gene therapy GENETIC ENGINEERING Substituting or adding new genetic material into genomes o To create or alter new genes Insulin production o Insert human insulin producing gene into bacteria o Human-insulin-bacteria would begin producing human insulin Recombinant DNA o When a DNA strand contains genetic material from at least two different sources 68 SBI4U1 – 2024 RECOMBINANT DNA Able to “add” foreign genes into another organism’s DNA RECOMBINANT PLASMID RESTRICTION ENZYMES A method to isolate a DNA fragment that contains the gene of interest Restriction enzymes (or restriction endonucleases) o Biological “scissors” o Breaks the phosphodiester bonds within a nucleotide chain o Has a specific sequence and location that it can recognize and cut ▪ Palindrome – Same sequence on complementary strand in opposite orientation 69 SBI4U1 – 2024 Enzyme will cut the DNA molecule at the same site on both strands and create restriction fragments Restriction fragments are pieces of DNA created by restriction enzymes Resulting fragments can have either: o Sticky Ends – A single-stranded end of the restriction fragment o Blunt Ends – A straight end without any single- stranded regions RESTRICTION ENZYME: EcoRI DIGESTION RESTRICTION ENZYME: SmaI DIGESTION 70 SBI4U1 – 2024 FORMING RECOMBINANT DNA Bacterial genome contains one chromosomal DNA and many plasmids Plasmids are small, circular, and are self-replicating pieces of DNA independent from the bacterial chromosome Contains a small number of genes Usually the gene of interest is inserted into a bacterial plasmid called a cloning vector DNA LIGASE The enzyme used to seal gene of interest with plasmid Hydrogen bonds between complementary base pairs can be formed easily o DNA ligase helps to form the phosphodiester bonds between the backbones of the double strands Works best with sticky ends of DNA 71 SBI4U1 – 2024 TRANSFORMATION The successful introduction of DNA from another source is called transformation o “transform” recombinant DNA into bacterial cell Some bacterial cell may not naturally accept a plasmid that contains foreign DNA o One method is to place bacteria in a bath of calcium chloride o The calcium ions will stabilize the negative phosphate ions of the phospholipid bilayer o Heating and quickly cooling with cause a disruption in the membrane and allow the plasmid DNA to enter the bacteria o Cells are then incubated at an optimal growing temperature SELECTION To identify colonies of bacteria containing the recombinant DNA Bacterial Products Image Cause No transformation or transform No vector without vector Transform occurs but unsuccessful Empty vector ligation Transformation and ligation are recombinant successful NAMING RESTRICTION ENZYMES Restriction enzymes are named according to the organism from which it was identified EcoRI E – Escherichia co – coli R – strain RY13 I – 1st enzyme in this strain 72 SBI4U1 – 2024 POLYMERASE CHAIN REACTION Uses a machine called a thermocycler A method for producing more copies of a specific piece of DNA Help to make proteins, study, etc. DNA SEQUENCING Uses a machine called gel electrophoresis to visualize the DNA Machine separates DNA fragments according to their sizes 73 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 8.1 on page 375 of your textbook: # 3, 4 3. Why are restriction enzymes that produce DNA fragments with sticky ends more valuable to geneticists than restriction enzymes that produce DNA fragments with blunt ends? 4. What are the benefits of a bacterial cell that contains a plasmid? Complete the following questions from Chapter 8.2 on page 385 of your textbook: # 6 6. How and when is gel electrophoresis used in general society? Complete the following questions from Chapter 8.3 on page 390 of your textbook: # 8 8. Currently there is a big push from environmentalists and the public to include information about any genetically modified organisms used in our food. Use a t-chart to identify the pros and cons of including GMO information on food labels. 74 SBI4U1 – 2024 9. DNA SEQUENCING POLYMERASE CHAIN REACTION (PCR) Purpose: to increase the amount of DNA Create millions of copies of a specific DNA sequence synthetically “DNA replication in a tube” PCR Materials: DNA template Nucleotides DNA polymerase (Taq polymerase) DNA primers At which cycle do you get the correctly sized target sequence? 75 SBI4U1 – 2024 Taq POLYMERASE DNA polymerase isolated from bacteria (Thermus aquaticus) that live in hot springs Heat stable enzyme that can withstand extreme temperatures DNA REPLICATION VS. PCR Steps Natural PCR Unwinding Helicase Heat RNA primer DNA primer Priming Primase Annealing temperature DNA polymerase Taq polymerase Elongation Nucleotides Nucleotides End of chromosome End of DNA template Termination Meet another replication form Change in temperature GEL ELECTROPHORESIS Size separation of nucleic acids by moving them through a gel medium using an electric current DNA is negatively charged and will move towards the positive electrode Use a molecular marker with DNA fragments of known sizes to compare with the unknowns Ethidium bromide is a special dye used to stain the gel and to keep the DNA from floating out Separation by size Gel medium is semi-solid and provides resistance DNA movement The more concentrated the gel, the higher the resolving power of the gel Short DNA Moves through gel easily Travels further Long DNA Moves through gel slow er because its size gets caught in the web of polymers that make up the gel Does not move as far 76 SBI4U1 – 2024 SEQUENCING DNA Human Genome Project: an international collaborative research effort (1998-2003) to sequence and map all genes in human beings Frederick Sanger Developed a technique using dideoxynucleotides (ddNTP) called the Sanger Sequencing DIDEOXYNUCLEOTIDES (ddNTPs) Di = two, Deoxy = removed oxygen Have oxygens missing at both the 2’ and 3’ position The 3’OH is needed to react with the phosphate group on the 5’ end of the next nucleotide to form a phosphodiester bond What will happen if the 3’ carbon is missing an oxygen? SANGER SEQUENCING SETUP Method is based on DNA replication by DNA polymerase Materials: dATP ddATP dGTP ddGTP DNAP dCTP ddCTP Primer dTTP ddTTP Template DNA dNTPs (deoxyNTPs) ddNTPs (dideoxyNTPs) 77 SBI4U1 – 2024 What do you think will happen if you allow DNA replication to occur in the presence of both deoxyATP and dideoxyATP? dATP dGTP dCTP dTTP ddATP The newly synthesized strands of DNA have different lengths depending on where a ddNTP was incorporated into the strand Process is repeated for 3 more times but with different ddNTPs Once all four processes are completed, it vessel will strands of different length which will be separated using gel electrophoresis 78 SBI4U1 – 2024 SEQUENCE INTERPRETATION MODERN SEQUENCING Theory is exactly the same as Sanger sequencing but the method is simplified 4 different coloured fluorescent dyes are used to label the 4 different ddNTPs in a single test tube All 4 fluorescent ddNTPs can be run on one lane on a gel electrophoresis Fragments are still separated by size but shows up as coloured bands Colours have different wavelengths which can be read by a computer Computer translates the colours into the order of nucleotides 79 SBI4U1 – 2024 HOMEWORK: TEXTBOOK Complete the following questions from Chapter 8.2 on page 385 of your textbook: # 1, 2. 1. Explain the role of each of the following in the PCR. (a) the DNA primer (b) the choice of Taq polymerase (c) the cycling through three different temperatures 2. Explain, using at least two or three details, how PCR technology has revolutionized genetic testing. What are the benefits for society? What are some possible disadvantages for society? 80