Transcription Is a Process That Genetic Information

¶ … Transcription is a process that genetic information on the DNA copies into RNA and the DNA acts as the template for the new molecules of RNA. Transcription process begins with the DNA double helix unwinding as the hydrogen bonds holding the opposing bases breaks and the DNA strands are uncoupled. The process occurs within the cytoplasm of a prokaryote and in the nucleus of eukaryotic cells. Transcription process consists of three steps; initiation, elongation, termination, and are regulated by transcription factors that include protein products of the genes. The protein products regulate at postranscriptional levels every time.

Initiation of transcription begins with enzyme RNA polymerase that identifies and attaches to DNA at the promoter and transcription of the DNA template starts. An initiation complex forms by association of 50 proteins different from each other required by RNA polymerase II. RNA polymerase synthesizes polynucleotides of RNA from the template of DNA. Transcription occurs only on one of the DNA strands in the gene (Latchman, 2009). The polymerase enzyme bind to the promoter and the helix unwinds making the two strands separate. The eukaryotic gene expression has a sequence of elements located far from the transcription start site and the elements can be upstream, downstream or in a transcription unit. A transcription initiation site can be determined by the DNA protein interaction that several components of the initiation transcription complex. Polymerase I, II, and III enzymes required for the process of transcription to start making it one of the most conserved protein in eukaryotes like yeast and complex ones like humans. Amino acids bind to the TATA box stimulating the initiation of transcription.

The elongation process occurs when the enzyme moves along the template DNA strand adding nucleotides to the 3' end of the growing chain. RNA polymerase detaches the strands and attaches nucleotides following the pairing rules of bases. Cytosine (C) bonds with guanine (G) and adenine (A) with uracil (U). Transcription unit consist of a triplet of bases that code for specific amino acids. Enhancers increase the promoter's activity while lacking promoter activity themselves. Enhancers have various binding sites for interaction to mediate transcription process. During the transcription process, there is interaction of histones with DNA that depends on conformational changes triggered by interaction with activators and coactivators.

A newly formed transcription unit ships out of the ribosome from the nucleus if processed by a series of enzymes. The addition of a 5' cap to the 5' end takes place helping to protect the RNA strand from degradation by enzymes that bind the RNA strand to the ribosome. At the 3' end, a poly (A) tail adds itself protecting the RNA from hydrolytic enzyme degradation and helps to release the RNA into the cytoplasm of the cell. Small nuclear ribonucleoproteins (snRNPs) functions in the removal of introns that are non-coding regions of the gene. RNA transcription, splicing and polyadenylation occur as a continuous process coordinated by interaction of processing factors of the transcription process (Latchman, 2009).

Termination of transcription occurs at specific sequence of bases on the template DNA strand. RNA polymerase moving along the template DNA releases the messenger RNA polymer and gets detached from DNA as it reaches the terminator sequence. The mRNA moves from the nucleus to the cytoplasm where it exists as a single strand unlike DNA that is double stranded. Transcription lead to creation of three products namely; mRNA that carry genetic information for manufacturing of polypeptides, rRNA which perform a structural function for ribosomes and tRNA which deliver amino acids to the ribosome to be assembled into proteins. During termination, the RNA product of transcription released shows a complementary image of the sequence of bases in the DNA template strand.

Question 2. Why do you suppose E. coli has three different DNA polymerases? Why do eukaryotes have more DNA polymerases?

The first DNA polymerase to gain characterization came from the E. coli and was DNA Polymerase 1. In each cell, there exist around 400 molecules of that enzyme. The weight of the enzyme is 103 kDa and is that of one huge protein. There are other two DNA polymerases with the same particular characteristics. The reason why the E. coli has these three different DNA polymerase is for the effective working of the E. coli provided the fact that these DNA polymerse may fail to carry out its enzymatic activity thus requiring the action of the other (Singer, 2011).

Further explanation of this is that since the location of these polymerases is recognized, the "5-to-3," which is the third of the three DNA polymerases, it is possible to remove it by using the enzyme protease that consequently cuts DNA pol 1. Like any other DNA polymerase, the first DNA polymerase needs a primer used to start tearing. In the E. coli, the replication happens in a form that cannot take place without the presence of all the polymerases. The replication in this stage is like any other because it happens in an order from 3 to 1 that enables the proteins to participate actively in the body processes.

A eukaryotic cell contains organelles bound within a membrane. These cells have a very intricate cytoskeleton. These cells have a very complicated metabolism system, a reason why they require a more DNA than the normal prokaryotic cells. This complicated metabolism system ensures that the body transports more nutrients their way. Every protein contained in the cell comes from a gene. The proteins include the cell membrane, the cell organelles among other things. This means that eukaryotic cells require more proteins to support all this. It is thus another reason why the cell uses more DNA as opposed to the prokaryotic cells (Singer, 2011).

Question 3. Detail the process of translation. Include as many factors as you can comprehend.

Translation entails the transformation of information in the messenger RNA (mRNA) into sequences of amino acids making it an important pathway in expression of genes. The proteins produced take part in cell metabolism and coordination of overall cellular events. The translation process occurs at the ribosome and mRNA encounters transfer RNA (tRNA) molecule with the appropriate amino acids. The codon in the mRNA exposes itself towards the tRNA, and translation takes place when the ribosome moves along the mRNA. Translation process occurs in three steps; chain initiation, chain elongation and chain termination.

Protein synthesis takes place at the initiation stage where there are tRNAs having anticodons that the start codon AUG recognize and bind to the mRNA on the ribosomes small subunit. A functional ribosome forms by the addition of a large ribosomal subunit (Alvis, 2010).

The start codon AUG located in the P-site in eukaryotes, and in bacteria Shine Dalgarno sequence in prokaryotic cells. Bacteria have formylmethionine (fMet) as its first amino acid and three initiation factors; IF1, IF2, and IF3 which take part in the placement of fMet tRNA at the AUG initiation codon.

Elongation involves elongation factors that bind charged tRNAs that and can be recognized to the A-site of the ribosome, only aminoacyl-tRNAs gets attached. The bacterial cell contains several elongation factors (EF-Tu) which protect the charged tRNAs from hydrolysis that take place in the cytoplasm. The second tRNA inserted into the ribosome recognizes the UCC codon positioned in the mRNA.

The elongation factor EF-Tu recognizes the aminoacyl tRNAs ester moiety but cannot differentiate between different amino acid species. According to the information on the genetic code, the amino acid serine attaches to the tRNA. tRNA retains its position by hydrogen bonds located between the anticodon and codon bases, and peptide bond holds enzyme attached methionine and serine in position. The reaction requires energy supplied by guanosine triphosphate (GTP) and adenosine triphosphate (ATP). The release of the first tRNA occurs dropping the methionine molecule on the amino acid chain, and the tRNA moves one codon right exposing the codon (GCC) and tRNA attaches to amino acid alanine. The ribosome gets excited by the serine attached tRNA making chain elongation process continue as the ribosome moves to expose the next codon. During the elongation process, shifting of tRNA from the A-site to the P-site occurs with the help of elongation factor (EF-G) and hydrolysis of guanosine triphosphate. The elongation cycle repeats itself until a termination codon is reached (Alvis, 2010).

Chain termination involves the addition of tRNAs and transfer of elongating polypeptides to the incoming charged tRNA until the ribosome reaches the stop codon (UGA). Termination occurs since there is no tRNA to recognize the stop codons. The termination codons lines up at the A-site and release factors RF1 and RFR2 bind to the codon catalysing the hydrolysis of the peptidyl-moiety in the P-site to terminate protein synthesis. Releasing factors bind to the regions where the tRNAs normally attach. Release of polypeptides sets in and the large and small ribosomal subunits disassembled to mark the termination of the translation process. Polypeptides twist into their primary and secondary structures during protein synthesis made sure by chaperons and cytoplasmic proteins that cause perfect folding process. Dissociation of RF1 and RF2 from the ribosome gets triggered by RF3 and guanosine triphosphate hydrolysis. The termination complex consists of terminal RNA and mRNA that associate with polypeptide chains that had hydrolyzed

Question 4 There are 64 codons total. Why do you suppose some amino acids have only one or two codons while others have more? What is the wobble?

Genetic redundancy is a condition in which an individual suffers from a condition where a certain biochemical function cannot work as it is impaired by one or two genes. This condition makes the amino acids have only one or more codons. There are 4 bases and 21 amino acids. Four to the power of one gives one, 4 to the power of 2 gives 16, 4 to the power of three gives 64. The situation cannot work with the two bases because it would create a shortage of amino acids and this is unhealthy for the body. Thus, the body uses three bases to cater for the 21 amino acids.

In case of a point mutation inside the genome, the process may not actually mature to the protein getting mature. A perfect example is in the case ACC that has the responsibility of coding Threonine. If a person was to get ACA, the result would remain the same and the reason for this is that it would still code Threonine. This is where explanation of the wobble hypothesis comes in.

The Wobble Hypothesis states that the pairing of basis relaxes at the third position. The reason for this is so that a base can have the ability to pair up with another base other than the original ones. In some RNA, anticodons bear inosine at position three. Inosine has the ability to pair with three specific codons and thus a human being or any other organism does not require having over 60 RNA molecules, even 30 are enough (Singer, 2011)

Question 5. Detail the coliform test. What are the acceptable values and what do levels indicate?

There are different ways in which to detail the coliform test. Fecal coliforms comprise part of a larger group of coliforms that come from the origin of felicons in general. Human feces however have a number of aerogenes and other substances that are separate from the original coliform group and are termed as non-fecal segments. To differentiate between coliform and other products of the fecal origin, temperatures must be set at 39 degrees and higher. This is necessary for incubation to take place effectively. Fecal coliform can ferment carbohydrates at a temperature of 44 degrees in a time span of 24 hrs. It is important to leave the mixture unattended for that number of hours. The product that one attains after that is the total chloroform that does not ferment in that temperature (Campbell, 2009).

In order to enumerate fecal chloroform, one can go about this in two ways. The MPN (Most Probable Number) is one of the ways and it involves serial dilution in which there is observation of one bacterium instilled in a fermentation tube with conditions that support its survival. For this analysis to go perfectly, one requirement is five tubes of dimensions that do not exceed three decimals. Before acting on the tubes, the fermentation tube should portray gas production or lack of production. This is necessary in the determination of the initial number of bacteria that was in the tube initially.

The main aim of having the dilution is for the analysis to observe tubes that would produce gas eventually and the one that would not produce any gas. It is important to calculate the water to use when carrying out the dilution process. This might seem as a rather unnecessary step as it determines the width of the tubes to use during the distillation. Wastewater treatment water and can effectively work with dilutions of between 1 and 0.01ml. This is because the water is within 200/100ml of chlorinated effluents discharge and these are acceptable levels.

Question 6. Explain how to clone Eukaryotic genes. Include as many details as possible.

In the contemporary world of science, researchers carry out their work in order to advance knowledge and apply personal development. One of the most important aspects to acknowledge is the presence of Eukaryotic cells that are some of the latest microorganisms such as fungi and yeast. Modern scientists have even come up with ways through which they can clone these cells (Singer, 2011).

The most common method that cloning of these cells takes place is through structural gene sequences. Through this process, a complementary DNA and RNA receive transcribing through in vitro. After this process, they both become double-stranded. The procedure takes place inside the Escherichia coli plasmids by the tailing technique. Coli transformants that come from the DNA contain globin sequences which attainment is through hybridization of RNA to DNA caused by clones developed and then lysed in situ. This process takes place on nitrocellulose. In order to estimate the amount of globin sequence inserted coming from the DNA, fingerprint analysis of the RNA sequence combined to become a hybrid of purified chimeras takes place (Singer, 2011).

An alternative procedure of the same experiment states that when carrying out the cloning, the scientist does not recommend the use of DNA. This is because introns cannot receive splicing from prokaryotes. Instead of using DNA, c-DNA is the most appropriate one to use. The c-DNA comes from m-RNA through reverse transcription. As the m-RNA does not have its introns present, the c-DNA also lacks, this procedure however makes the process longer. Both procedures are correct but when choosing which the best to incorporate into an experiment, it is important to understand the type of product undergoing cloning.

Question 7. Explain how Polymerase Chain Reaction (PCR) works. Why is it such a powerful tool in diagnostics and other molecular biology applications?