Theme 1 Module 4 Script PDF 2021

Theme 1 Module 4 Script 1 Slide 1: YOUR SUMMARY NOTES: Theme 1, Module 4: Nucleic Acids Slide 2: In this module, we will discuss how DNA was determined to be the molecule of heredit...

Theme 1 Module 4 Script 1 Slide 1: YOUR SUMMARY NOTES: Theme 1, Module 4: Nucleic Acids Slide 2: In this module, we will discuss how DNA was determined to be the molecule of heredity. We will also investigate its structure and location within the cell. Finally, we will look at the structure and function of the different RNAs found in the cell. Slide 3: Unit 1: the nucleic acids Slide 4: We have discussed three classes of macromolecules: lipids, carbohydrates, and proteins. These macromolecules are found throughout the cell and have specific functions related to their structure and chemical properties. The fourth class of macromolecule found in all cells includes the nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Let’s start with DNA. Where do we find DNA in the cell? In prokaryotic cells, the majority of DNA is contained in the nucleoid. There are also small circular DNA molecules called plasmids. These plasmids often carry only one or two genes. Examples include genes for antibiotic resistance. Plasmids can replicate independently of the core genome and can be transferred from one cell to another. This allows the rapid spread of genes that confer properties such as antibiotic resistance in a bacterial population. Slide 5: Chromosome is a word that describes the organization of a double-stranded DNA molecule in its association with proteins and RNAs. Usually, eukaryotic cells have large linear chromosomes and prokaryotic cells have smaller circular chromosomes, though exceptions have been observed. Also, organelles such as the mitochondria contain their own, smaller, chromosomes. The circular chromosome of E.coli seen here is supercoiled into about 100 loops and associated with proteins to form the nucleoid. This allows a bacterial chromosome that may be 1000 times longer than the diameter of a cell to fold up within the nucleoid.  Theme 1 Module 4 Script 2 Slide 6: Supercoiling of DNA in a circular molecule is the YOUR SUMMARY NOTES: coiling that occurs in addition to the coil of the helical DNA structure. This supercoiling preserves the double helix structure and compacts the DNA into a small space. We can envision this supercoiling property with an analogy. Take a rubber band and apply rotation at one end of this band. Soon, thereafter, you will see that there is supercoil that further tightly compacts the coiled band that you are spinning. This is analogous to the organization of the supercoiled DNA in prokaryotes. Slide 7: In recent years a large number of prokaryotic genomes have been completely sequenced, in over 400 bacterial strains as of 2013. Researchers then use this information to create maps that illustrate the position of all the genes on the prokaryotic chromosome. E. coli is a common bacterium with a circular chromosome, which carries the genes necessary for the normal functioning of the cell. It also has small circular plasmids. These extra- chromosomal DNA molecules carry a limited number of genes that may be required for survival in distinct environments such as in the presence of particular antibiotics. This genomic map illustrates the position of the genes on the prokaryotic chromosome. For example, there is a cluster of genes (operon) involved in the metabolism of the sugar, lactose, and another cluster of genes required for the synthesis of the amino acid, leucine. Slide 8: We take this knowledge for granted now but how did scientists discover that DNA was the molecule that contained genetic information? Slide 9: Checkpoint question #1 Slide 10: Unit 2: DNA as hereditary material Slide 11: Streptococcal bacteria are normal inhabitants of the human upper respiratory tract. Benign strains are harmless and not associated with any disease symptoms. Virulent strains, however, are associated with mild symptoms, such as the sore throat resulting from Strep throat or as the severe symptoms resulting from pneumonia.  Theme 1 Module 4 Script 3 Slide 12: YOUR SUMMARY NOTES: Scientific research begins with an observation which leads to one or often more questions. These questions are then used to formulate a hypothesis that can be tested experimentally. For example, Fred Neufeld made a number of observations in the 1920’s. There are two strains of the bacteria, Streptococcus pneumoniae. One strain, when injected into the mouse caused death. The other strain when injected into the mouse had no effect. The R strain is a harmless or benign member of the microbiome, in this case, the mouse microbiome. The S strain is a virulent form of the same bacteria. The question is, can information that determines the nature of the strain, whether benign or virulent, be transferred from one bacterial cell to another? To answer this question, an experiment was performed. In 1928, Frederick Griffith reported the results of his experiment in what is now called The Griffith's Experiment. Slide 13: Griffith heated the virulent bacteria to high temperatures in order to kill them. This is the same principle that is used when pasteurizing food through heating to eliminate or reduce pathogenic bacteria. Note that all of the macromolecules are still there (DNA, proteins, carbohydrates, lipids), the cell is simply dead. In Griffith's experiment, when the dead S-strain was injected into a mouse, the mouse was fine. Griffith hypothesized that the information that makes a cell virulent was still present, but that the cell carrying that information had been killed. Could that information still be used? Slide 14: Griffith set up an experiment with three controls. In the controls, we know what will happen based upon prior experience and our understanding of the biology of this system. Injecting the virulent strain should kill the mouse, while injecting the benign strain or the heat-killed virulent strain should not affect the mouse. Griffith was now ready to answer the biological question-could the living cells use the information from the dead cells? To test this, he killed the virulent strain and incubated the remains with living cells of the benign strain.  Theme 1 Module 4 Script 4 Slide 15: In the experiment designed to test his hypothesis Griffith incubated the remains of the killed cells from the virulent strain with living cells from the benign YOUR SUMMARY NOTES: strain and injected the combination into a mouse. What Griffith found was that the mouse died, indicating that the benign bacterial cells were now virulent! Griffith concluded that the cells of the benign strain had acquired information, or the ability to be virulent, from the dead cells. But what was it from the dead S-strain cells that was somehow being transferred to the non-virulent R-strain cells? In other words, what macromolecule carries the hereditary information? Slide 16: This process is called transformation. Transformation is a change in cell behaviour resulting from the incorporation of hereditary material from outside of the cell. Slide 17: It is important in science to look at previous experiments and observations and to build upon them. In 1944, Oswald Avery and his colleagues looked at the work of Griffith and decided to ask another question. Which macromolecule in the bacterial cell was holding all that important information that could cause transformation. Which molecule was carrying the hereditary information? There are five macromolecules in the cell that could be considered: Lipids, carbohydrates, protein, RNA, DNA. Lipids and carbohydrates were discounted. That left proteins, RNA, and DNA. Avery and his colleagues selectively eliminated each type of macromolecule from the cell extracts and then tested the remaining molecules for their ability to induce transformation. RNase is an enzyme that specifically degrades RNA molecules; DNase degrades DNA molecules; protease degrades proteins. Once again, the researchers have included the important control. In the first flask, the virulent heat-killed bacterial extracts are incubated with the nonvirulent bacteria with the expectation that transformation should occur.  Theme 1 Module 4 Script 5 Slide 18: As expected, in the control, the untreated extract YOUR SUMMARY NOTES: can transform nonvirulent cells into virulent cells. What these researchers discovered was that in the absence of protein or RNA, but in the presence of DNA, there was still transformation. In contrast, extracts treated with the enzyme that destroys DNA were unable to transform the benign bacterial cells. In the absence of DNA, there was no transformation. They concluded from these experiments that the macromolecule that determines the characteristics of the cell was carried in the DNA molecules of the cell. Slide 19: Checkpoint question #2 Slide 20: Unit 3: Let us review the structure of DNA. Slide 21: Data had accumulated to support the hypothesis that DNA was the hereditary molecule of the cell, but the structure of the molecule was unknown. A fundamental leap forward was made with the x- ray diffraction experiments carried out by biophysicist and x-ray crystallographer, Rosalind Franklin. Franklin used a technique called x-ray diffraction to aim x-rays at DNA to create images based on the diffraction of the x-rays by the atoms in the DNA molecule. This famous photo 51 was key to the discovery of the theoretical structure of DNA. It was from this single image that the helical nature of DNA was identified, and from which calculations were made to determine the dimensions of the molecule. Slide 22: Based on the X-ray crystallography images of the DNA molecule produced by other scientists, primarily Rosalind Franklin, this one-page paper published in the journal Nature in 1953 became one of the most significant contributions to cell biology. It was in this paper that James Watson and Francis Crick, working together in Cambridge, England, described the double helical structure of the DNA molecule based upon the X-Ray crystallography images of the molecule and a great deal of creativity. Let’s now look at the molecular makeup of the DNA molecule.  Theme 1 Module 4 Script 6 Slide 23: YOUR SUMMARY NOTES: The basic subunits of the DNA molecule are the nucleotides. A single nucleotide can be described by its three constituent components: a phosphate group, a 5-carbon deoxyribose sugar, and a nitrogenous base. Slide 24: While the deoxyribose sugar and the phosphate group are constant in each base, the four nucleotides that constitute DNA, are chemically distinguished by the specific nitrogenous base found in the molecule. Pyrimidines (cytosine and thymine) are aromatic heterocyclic molecules that have a single ring while purines (guanine and adenine) consist of two rings: a pyrimidine ring fused to an imidazole ring. For this reason, purines are larger than pyrimidines. Slide 25: When assembling a DNA molecule, a condensation reaction releases a water molecule and forms a covalent bond called a phosphodiester bond between two nucleotides. The nucleotides are linked along the phosphoribose backbone: the phosphate group on the 5’ carbon of the deoxyribose forms a bond with the hydroxyl group on the 3’ carbon on the next deoxyribose. The result is that the backbone has a 5’ to 3’ polarity created by these linkages and the nitrogenous bases stick out from the backbone. Slide 26: The resulting DNA strand forms as a sugar- phosphate backbone with the nitrogenous bases extending out from it. But we know that DNA is a double stranded molecule, how was this determined? Slide 27: Several interesting observations were made about the nucleotide complement within the DNA of cells. One came from the biochemical work of Erwin Chargaff in the mid-1900’s. This led to Chargaff's rules which state that DNA from any cell of all organisms should have a 1:1 ratio of pyrimidine and purine bases (this is the base Pair Rule) and, more specifically, that the amount of guanine is equal to cytosine and the amount of adenine is equal to thymine. This observation makes a powerful prediction about the structure of DNA. It suggests that there may be molecular pairing between purines and pyrimidines in the formation of DNA according to the rules: A pairs with T and C pairs with G.  Theme 1 Module 4 Script 7 Slide 28: YOUR SUMMARY NOTES: If Chargaff's rules are true, then a consistent spacing between two phosphoribose backbones should be maintained. Further structural images from Rosalind Franklin's X-ray diffraction experiments suggested that the DNA molecule existed in a helical structure. This led to the hypothesis that purines pair with pyrimidines in the DNA helix and that, based on the structure of the nitrogenous bases, the pairing allows for the creation of 2 hydrogen bonds between thymine and adenine and 3 hydrogen bonds between guanine and cytosine. Slide 29: All of this information came together to give us the familiar double helix structure of DNA determined by Watson and Crick. The two strands run anti-parallel to each other, with the bases facing inwards to give a diameter of 2.0 nm. The helix has a major groove and a minor groove, with one turn of the helix occurring every 10 nucleotides. Slide 30: Checkpoint question #3 Slide 31: Now let’s look at the structure of RNA Slide 32: All RNAs are synthesized in the same way, as a transcription of a sequence of DNA using an enzyme complex called RNA polymerase. RNAs can be classified into distinct categories: messenger RNA, ribosomal RNA and transfer RNA, and different enzymes (RNA polymerases) are used for the synthesis of the different classes of RNAs. If we want to take the information encoded in the DNA to make proteins, we need all three classes of RNA molecules Slide 33: There are two important biochemical differences between RNA and DNA. First, the 5-carbon sugar that makes up the backbone is ribose instead of deoxyribose. Ribose has a hydroxyl (- OH) group on the carbon at position 2’, while deoxyribose has a hydrogen (-H) at position 2’. We refer to the DNA nucleotides as deoxyribonucleotides and RNA nucleotides as ribonucleotides.  Theme 1 Module 4 Script 8 Slide 34: YOUR SUMMARY NOTES: The second difference between RNA and DNA is the nitrogenous bases used in RNA versus DNA. While RNA ribonucleotides, like DNA deoxyribonucleotides, contain cytosine, guanine, and adenine, the pyrimidine, uracil, replaces thymine in its ability to pair with adenine in RNA. The difference is that uracil has a hydrogen where thymine has a methyl group. Slide 35: When transcribing the information from a DNA molecule, new ribonucleotides are matched to the DNA template by hydrogen bonding according to the rules: U to A, C to G. In addition, the 3' hydroxyl (or 3'end) of the existing polymer attacks the high-energy phosphate bond of the new ribonucleotide. As a result, phosphodiester bonds, much like those seen in a DNA molecule backbone, are also formed in the RNA backbone. Slide 36: Here we see the synthesis of different RNA molecules using DNA as a template and the transcriptional enzyme machinery to direct the process of RNA polymerization. Let’s look further into the 3 types of RNA molecules that can be produced during transcription. Slide 37: Messenger RNA is the RNA copy of genes coding for proteins, they are single-stranded molecules that are then translated into proteins by the ribosomes. Shown here is the RNA polymerase that synthesized the single stranded RNA copy (green) of one strand, the template strand (red), of the double-stranded DNA molecule. Slide 38: A transfer RNA or tRNA is a functional RNA. This RNA is never translated. The sequence of the RNA is such that it folds into a 3-dimensional structure held together by base pairing between the ribonucleotides. Its function is to attach to amino acids and bring them to the ribosomes during translation.  Theme 1 Module 4 Script 9 Slide 39: Ribosomal RNA (rRNA) is another functional YOUR SUMMARY NOTES: RNA. It is transcribed from the ribosomal genes of which there are many copies in the cell, and it is a very large proportion of the total RNA in a cell. There are 3 different ribosomal RNA molecules that together form integral components of the ribosome. Slide 40: Checkpoint question #4 Slide 41: Unit 5: DNA in eukaryotes Slide 42: The eukaryotic cell is compartmentalized and contains many different organelles. Where do we find DNA? The nucleus is the home of the chromosomal DNA. The mitochondria and chloroplasts are membrane-bound organelles that, as we learned earlier, carry their own genetic material. Slide 43: The DNA molecule in the eukaryotic cell serves the same function as in the prokaryotic cell. The processes of transcription into RNAs in both prokaryotic and eukaryotic cells are essentially the same. The DNA codes for genes that are transcribed into the functional RNAs: tRNA and rRNA, as well as mRNAs that are templates for protein synthesis. The DNA molecules in the eukaryotic nucleus are linear molecules that are organized around proteins (called histones) to form chromatin. There are several levels of organization or compaction that allow the DNA molecule to be folded into the small space of the nucleus. A DNA molecule in just one human chromosome is about 6000 times longer than the width of an average human cell. This makes compaction important. A double-stranded helix is 2nm in width, but once folded around the histones, it forms a fibre that is 10 nm wide. Further coiling creates a 30 nm chromatin fibre, the structure commonly found in a transcriptionally active cell. In preparation for cell division, additional coiling at three distinct levels creates the metaphase chromosome seen at the top. This compacted chromosome can more easily be moved during the process of cell division.  Theme 1 Module 4 Script 10 Slide 44: Despite the many similarities between the DNA YOUR SUMMARY NOTES: contained within the prokaryotic nucleoid and the eukaryotic nucleus, it is important to note that the structures evolved independently and code for and make use of different types of proteins. One notable difference that can be made when comparing prokaryotes and eukaryotes is that the size of the eukaryotic chromosome is much greater than the size of the bacterial nucleoid. In addition, the eukaryotic genome is much greater in size relative to the prokaryotic genome. There are almost 5 million base pairs in the E. coli genome and about 4 290 protein coding genes. This compares to 6 billion base pairs in the human genome and an estimated 20 000 protein-coding genes. DESPITE these differences, we see the same nucleotides, same double helix, and similar genes coding for proteins such as those of the RNA polymerase and ribosomes. Slide 45: In this module we have identified that hereditary information is contained in DNA molecules. This DNA is folded into a double helical structure with conserved base-pairing rules. and is transcribed into RNA. We have also learned that RNA molecules have diverse functions in the cell including roles in protein synthesis. Finally, we have identified that the basic units of prokaryotic and eukaryotic genomes are the same despite differences in size and organization.