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Midterm

Biology 112 - Midterm 2 Learning Goals

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Department
Biology
Course
BIOL 112
Professor
All Professors
Semester
Fall

Description
Learning Goals – Week 5 1. Understand the central dogma and describe the step of information flow. The central dogma of biology states that information flows from the DNA (information storage) to the mRNA (information carries) through a process called transcription. The information is then translated through to proteins in a process called translated. In essence, DNA is transcripted to create mRNA, which is translated to form proteins (active cell machinery). 2. Understand the differences in chromosome, gene expression and arrangements between eukaryotic and prokaryotic cells. In prokaryotic cells, gene expression is a process that happened as the proteins are being made. Transcription and translation are happening at the same time. Transcription and translation happen in a different place and time in eukaryotic cells. Genetic material is supercoiled form in prokaryotic cells and is in the form or chromosomes condensed to chromatin in eukaryotic cells. 3. Describe the general features of gene organization on eukaryotic and prokaryotic chromosomes. In eukaryotic chromosomes, genes are wrapped around histones to form a structure called a nucleosome. No organization in prokaryotes, genes are supercoiled in the interior of the cell. 4. Describe the role of RNA molecules as a link between genes and proteins. RNA molecules are somewhat an intermediate molecule between genes and proteins. They can perform tasks specific to both the molecules, but to a less degree. RNA can hold information, but not to the ability of genes. This allows it to transfer information from the nucleus to the ribosome quickly. It then degrades. It can also act as enzymes, but it can’t catalyze as many reactions has proteins. This is because proteins consist of 21 different amino acids. RNA molecules are only made up of 4 varying nucleotides. 5. Identify the 5’ and 3’ polarity of DNA and RNA polymers and explain why such polarity is important. The 5’ end of a strand of DNA/RNA is the end where the 5’ carbon has nothing attached to it. Similarly, the 3’ end is the end where the 3’ carbon is not attached to anything. This is important because nucleotides always attached to the 3’ end. A DNA/RNA strand always synthesized in the 5’ 3’ direction. 6. Distinguish between the structures of monomers in DNA and RNA. Deoxyribonucleic acid is different from ribonucleic acid in a sense that I does not have a hydroxyl group attached to it 2’ carbon spot. Ribnucleotides have this –OH group at it’s 2’ carbon spot, but deoxyribonuclotides have no –OH group, but instead a O group. Another structural difference is that RNA has the pyrimidines U and C, DNA has the pyrimidines T and C. RNA carries U instead of T. 7. Relate DNA stability (denaturation and renaturation) to base stacking and hydrogen bonding between bases. The stability of DNA is achieved through two processes with the molecule. Complementary base paring and hydrophobic interactions between the base pairs. When a DNA molecule is denatured (heat), these bonds break. If the source of denaturation is removed, then the molecule will be renatured. This is because the when the base pairs match up, they will be more stable when undergoing an exothermic reaction. 8. Describe the relationship between DNA structure and the Watson-Crick base pair rules (Chargaff’s rules) Watson and Crick stated that from previous experiments, DNA has a sugar- phosphate backbone, the number of A = the number of T, the number of G = the number of C, the number of purines = the number of pyrimidines and that DNA has regular and repeating structure that is helical. These previous finding lead them to believe that Adenine pairs with Thymine and Guanine pairs with Cytosine. 9. Rationalize why polypeptides and RNA can act as enzymes, but DNA cannot. DNA is a simple structure that is too organized, too stable (double helix configuration), and doesn’t have any exposed chemical groups. Also, DNA only has a H group at its 2’ Carbon. This is unreactive when compared to –OH group on RNA. This contributes it DNA not being very reactive; therefore it cannot act as an enzyme. RNA can act as enzymes because it has structural and chemical diversity. It has exposed base pairs, which are made up one of four nucleotides. This is not nearly as much as the 21 A.A. found in polypeptides, but it is enough to allow for RNA to catalyze come reaction. 10.Contrast the different types of RNA molecules: rRNA, tRNA, mRNA. All three types are coded for by genes, and are made through the process of transcription. rRNA is ribosomal RNA and is found in ribosomes. Its function is to translate the mRNA “language” into amino acid “language”. It also interacts with the incoming tRNA. tRNA is responsible to transferring free nucleotides to the rRNA to make the protein during translation. It has a 3’ terminal site for amino acid attachment by the aminoacyl tRNA synthetase. mRNA is a short lived messenger RNA that is synthesized in a process called transcription. It is responsible for bringing the hereditary material to the rRNA so it can be translation to form proteins. 11.Identify the regions of DNA that can be used by proteins to recognize specific DNA sequences. The regions of DNA that can be used by proteins to recognize specific DNA sequences are bounded by one promoter and one terminator. Learning Goals – Week 6 1. Describe the step of information flow in transcription and translation. This process follows the central Dogma of Biology which states that information flows from DNA to mRNA to the Proteins. In transcription, hereditary material in the DNA is transcripted to form short-lived mRNA molecules. These information capsulated in these molecules are then translated into proteins in a process called translation. 2. Describe what is meant by a “consensus sequence” and be able to determine a consensus sequence from a simple data set. A consensus sequence is the one sequence that is most likely to occur; it is the “average” sequence. It is the sequence contacting the most frequent bases at each position. 3. Describe the role of RNA polymerase and promoters in transcription. Promoters are sections of DNA where transcription begins. The RNA polymerase/sigma holoenzyme binds on this specific sequence on the DNA to start transcription. In Prokaryotes, the sigma binds onto the -10 box and he -35 box. Once transcription has started, the sigma leaves to start another process. Learning Goals – Week 7 1. Describe the role of RNA polymerase and promoters in transcription. The role of RNA polymerase is to synthesis the production of mRNA in transcription. It essentially synthesizes the RNA from nucleotides using a DNA template. Promoters are short nucleotides sequences in DNA that binds RNA polymerase, enabling transcription to begin. In prokaryotic DNA, a single promoter often is associated with several continuous genes, in eukaryotic DNA, each gene generally has it’s own promoter. 2. Define what is meant by promoter, how if functions and how it is recognized by cellular enzymes. Promoters are particular stretches of DNA that have a recognized base sequence that the sigma protein can bind to. Promoters have two boxes, the -35 box and the - 10 box (both are upstream from the +1 site. The sigma recognizes these boxes and guides the RNA polymerase towards these boxes. Once the sigma attaches with the RNA polymerase to the boxes of the promoter sequence, transcription can begin. Afterwards the sigma releases itself. 3. Describe the process of transcription of a gene and operon. Transcription begins when the sigma guides RNA polymerase to DNA, RNA polymerase will then induce the DNA to uncoil. RNA polymerase will move along the DNA 3’  5’ direction to synthesize RNA is the 5’  3’ direction. Free nucleotides will flow in and RNA polymerase will bind these together to from one single mRNA molecule. 4. Describe the structure of the RNA polymerase including the “sigma factor”. The structure of the RNA polymerase is a large, globular and has several prominent channels running through it. RNA pol binds with a sigma factor to form a holoenzyme. This is because RNA pol alone cannot bind to the DNA promoter sites, it needs the help of sigma. 5. Describe why the genetic code is termed “redundant” and “non-overlapping”. The genetic code is termed redundant because more than one code can code more the same amino acid. It is termed non-overlapping because only one reading frame codes for the correct protein. Other reading frames will only code for gibberish at the protein level. 6. Describe the roles of the initiation codon, termination codons and the ribosomal binding sites on a Bacterial mRNA. The initiation codon (methionine) is the start codon, it is located on the mRNA is signals for a tRNA carrying methionine to hydrogen bond to the codon. When translation first starts out, the initiation codons are attached to the “P” slot in the middle of the ribosome. Termination codons signify for the detachment of the polypeptide from the ribosome. The codon induces the linkage between the amino acid the “p” slot and the polypeptide and frees the polypeptide. The ribosomal binding sites in bacteria mRNA are about 6 base pairs that bind upstream from the start codon. Their purpose is to attach the mRNA in the right place so that a
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