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Biochemistry
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Biochemistry 3380G
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Prof
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Dr. Chris Brandl – Course co-ordinator TRANSCRIPTION and GENE REGULATION REQUIRED: Readings Chapters 7 and 8 Overview: Topic 22. Humans have approximately 25,000 genes. (See Table 9-1, p317) Of these ~5,000-10,000 are expressed in a tissue at any one time. The ability to induce the proper genes at the proper times is critical in the growth and development of all organisms and for their ability to respond to their environment. The importance of gene expression becomes evident when we realize defects in gene regulation often result in disease. Some of the most notable examples come from cancers where many tumor suppressors (proteins required for the prevention of cancer) and oncogenes (a gene that makes a cell cancerous) are transcription factors. The importance of gene regulation also lends itself to potential avenues for cures. For example, inhibiting the gene expression of a pathogen (e.g. viruses) provides a clear and specific pathway for a cure. STEPS IN CONTROLLING GENE EXPRESSION The Central Dogma in its simplest form is (See figure 7-1): transcription translation DNA ----- > RNA ---- > Protein The Central Dogma defines transcription (copying of DNA into RNA) and translation (process of decoding an RNA to synthesize protein) as two key processes in gene expression. The RNA intermediary allows a step of amplification, since in contrast to DNA where unit copies exist, the RNAs for different messages can be produced at dramatically different levels (Figure 7-2). In addition RNA, unlike DNA, is unstable in the cellular environment. It can be rapidly degraded providing a mechanism to turn genes off. As we will see later, RNA also provides additional opportunities for regulation. Points of Gene Regulation in a EUCARYOTIC Cell (FIGURES 7-23, 7-40, 8-3) In the pathway from gene to protein there are many potential points of regulation. Several of these are common between procaryotes and eucaryotes; others are specific for eucaryotes. (* steps are NOT found for procaryotes). These include: 1. Rate of transcription. Transcription can be divided into stages: initiation, elongation and termination. Since it is the first, initiation is often a principal site of regulation. 2. Rate of RNA processing (the steps converting a newly transcribed RNA into a molecule that can be translated) For eucaryotes these RNA processing steps include: 5' 7-methylguanosine capping*, 3' polyadenylation, RNA splicing of introns*. RNA editing is another processing event. 3. Rate of transport of mRNA out of the nucleus*. (Why is this not an issue for procaryotes?) 4. Rate of mRNA degradation (see Fig. 8-26). 5. Rate of translational initiation (see Figure 8-25), elongation and termination insulin) phosphorylation (addition of a covalent phosphate group from ATP), glycosylationrotein (eg (addition of one or more sugar moieties), acetylation, (addition of an acetyl group). (Fig. 7-41) 7. Protein degradation (loss of the protein or turnover). The amount of protein present is determined by the balance of its rate of synthesis and its rate of loss. Topic 23: TRANSCRIPTION Key Definitions 1 1. The DNA sequence required for transcriptional initiation of a gene is called the promoter. The promoter includes the sequences that 1) recognize RNA polymerase and 2) recognize any gene specific regulatory factors. 2. The DNA sequence required for transcriptional termination is called the terminator. 3. Transcription is catalyzed by the enzyme RNA polymerase. 4. Gene specific regulatory proteins (or transcription factors) are the key molecules in the differential transcription of genes. Bacterial transcription is very similar to eucaryotic transcription in the basic process and the structure/function of the molecules involved. Because it is somewhat simpler, we will initially focus on bacterial transcription. RNA Polymerase (Fig. 7-7) The core bacterial RNA polymerase is a large enzyme which contains 4 subunits, 2 alpha subunits and single beta and beta subunits . The core enzyme will synthesize RNA from ends or nicks in DNA templates but it lacks promoter specificity (that is, it cannot recognize promoter sequences). The RNA polymerase holoenzyme will transcribe RNA specifically from promoters. In addition ’ to 2 alpha, beta and beta subunits it contains a sigma subunit. Sigma enhances recognition of promoter structures and decreases binding to nonpromoter DNA. Steps in the initiation of transcription (See Figure 7-9) 1. RNA polymerase recognizes and binds to the promoter DNA. This is called the closed complex. Promoter recognition is via the sigma factor. Sigma makes base specific contacts with the promoter. 2. Polymerase unwinds the DNA strands at the transcriptional start site. This complex of polymerase and unwound promoter DNA is called the open complex. 3. The first NTP is brought to the template. No primer is required. Base pair rules apply. 4. Using NTPs (A,G, C,U) as substrates, chain elongation begins and proceeds in a 5'-3' direction. Phosphodiester bonds are formed and pyrophosphate is released. 5. After the incorporation of 5-10 nucleotides, sigma falls off. 6. The transcription bubble moves downstream with the template DNA reannealing behind. 7. Chain elongation continues until a terminator is reached and the polymerase falls off. Structure of Bacterial Promoters (See Figure 7-10) There are recognition sites on bacterial DNA that signal the recruitment of RNA polymerase holoenzyme. Relative to the start site of transcription that occurs at +1, these are found centred at approximately -35 and -10. The consensus sequences for the -10 and -35 sequences are: -10 consensus: TATATT; -35 consensus: TTGACA (Note: When written like this, by convention it is the coding strand in the 5' to 3' direction.) The -10 and -35 sequences when properly spaced are sufficient to recruit the holoenzyme to the promoter. As outlined below other transcription factors play critical roles in regulation. Note that the -10 and -35 sequences as written above are consensus sequences. Not all promoters have exactly the same sequences. How is Transcription Differentially Regulated? There are two key reasons why some bacterial promoters are transcribed more than others. 1. Strength of the basic promoter elements. Not all -10 and -35 sequences are equally active; they bind RNA polymerase holoenzyme with different affinities. 2. Gene specific regulatory proteins bind specific DNA sequences are found in one or more promoters and serve to activate or repress transcription. Examples: Trp repressor as the name suggests is an example of a repressor. The Catabolite Activator Protein (CAP) regulates transcription of the lac operon and other genes involved in carbon metabolism. CAP is an example of an activator protein. 2 The Trp Operon (See Figure 8-6) The genes required for tryptophan biosynthesis in E. coli are transcribed as a single unit from a common promoter. From the single transcript 5 proteins are translated. This type of gene structure is called an operon and is common in bacteria but not found in eucaryotes. As shown in Figure 8-7 (Alberts et al), RNA polymerase is recruited to the Trp promoter by the – 10 and –35 sequences allowing transcription when tryptophan concentrations are low in the media. When tryptophan concentration is high, the Trp repressor binds tryptophan. A conformational change occurs in the repressor's structure such that it can bind the operator site within the promoter. Binding of the repressor sterically inhibits access of RNA polymerase to the promoter. The Trp Repressor (Protein Science Fig 8-18) Trp repressor contains 107 amino acid residues. It has a helix turn helix motif that is required for DNA binding. Amino acid side chains on helix 5 make base specific contacts with the major groove of its operator sequence. Trp repressor binds DNA as a dimer. Positive Control and the Lac Operon (Figure 8-9) Other gene specific regulatory proteins can enhance transcription. Increased transcription is generally accomplished by increasing the rate of recruitment of RNA polymerase or the activity of RNA polymerase. Both enhanced recruitment and/or enhanced activity of polymerase are usually achieved by protein-protein interactions between the activator protein and the enzyme. The Lac operon contains the genes required for the metabolism of lactose in E.coli. Its expression of lac is controlled by two signals. The first is negative regulation similar to what is seen for the Trp operon. In the absence of lactose the Lac repressor binds the promoter, inhibiting the action of RNA polymerase. The Lac repressor binds lactose when lactose is present in the growth media. Lactose binding to the Lac repressor, results in a conformational change in the protein so that it no longer binds the operator sequence in the promoter allowing transcription. The second mechanism for regulating the Lac operon involves control by glucose. All cells preferentially use glucose as their carbon source. In the presence of glucose, the genes required for the metabolism of other carbon sources are shut off. This is known as catabolite repression. In E.coli, catabolite repression occurs through the induction of genes in the absence of glucose. When the concentration of glucose in the media is low, the intracellular concentration of cAMP rises. At high concentrations of cellular cAMP the E. coli Catabolite Activator Protein (CAP) binds cAMP. A conformational change occurs in CAP upon binding cAMP that allows the protein dimer to bind DNA in a site-specific fashion. At the Lac operon, the CAP binding site is upstream (5') of the -35 sequence. Through direct protein-protein interactions with polymerase, CAP stimulates the recruitment of RNA polymerase to the promoter. 3 *PS Figure 8-19 Shows the Trp repressor dimer * PS Figure 8-18 The Trp repressor monomer. with the core of the protein held together via protein-protein interactions. *PS Figure 8-20 (a) Trp repressor in the absence of bound tryptophan. It cannot bind the operator DNA since the recognition helices are tilted inward. (b) Active Trp repressor (capable of binding its operator) when tryptophan is bound. Recognition helix 5 on each of the two monomers makes base specific contacts in consecutive major grooves of the DNA. Summary points for Regulated Transcription Regulation of transcription often occurs through the action of one or more site-specific DNA binding proteins. The mechanisms by which they regulate transcription are similar. 1. The DNA binding proteins recognize internal or environmental signals (e.g. tryptophan for the Trp repressor). * *Introduction to Protein Structureedition (1999) Carl Branden and John Tooze, Garland Publishing. 4 2. The regulatory proteins stimulate or repress the promoter binding or activity of RNA polymerase. 3. Activation is through protein-protein interactions. (e.g. CAP interacts with RNA polymerase). Differences in Transcriptional Regulation between Procaryotes and Eucaryotes Although generally very similar, there are differences in the process of transcription in procaryotes and eucaryotes. 1. Eucaryotes have three different RNA polymerase (bacteria have one). (Table 7-2) Pol I for ribosomal RNA rRNA Pol II for messenger RNA mRNA Pol III for tRNA and snRNA 2. Eucaryotes do not have operons. Genes are transcribed as single units (monocistronic). 3. Promoter Recognition is through a distinct set of proteins. The role of these transcription factors is similar to bacterial Sigma factor but more complex. One of the factors, the TATA-binding protein (TBP; Figure 8-11), is a component of the transcription factor TFIID. It binds a basal promoter element, the TATA-box, found ~20 basepairs upstream of the transcriptional start site +1. Refer to Figure 7-12. 4. Regulatory elements are often located many thousands of base pairs distant from +1. They may be brought into proximity of the promoter by DNA looping. (Figure 8-10) 5. Nucleosomes and higher order chromatin structure (Figures 5-22, 5-25, 5-27, 5-28, 8-11) have a profound effect (both positively and negatively) on determining the access of transcription factors to DNA. Much of eucaryotic transcriptional control involves the regulation of chromatin structure, which occurs in part through the post-translational modification of histones. 6. Combinational Control: groups of proteins work together to determine the expression of a single gene. Figure 8-12 summarizes many of these points: TBP and promoter recognition, DNA looping, co- ordinated action of multiple factors at eucaryotic promoters Other DNA binding motifs In bacteria we focused on two transcription factors with helix turn helix DNA binding motifs. Many other DNA binding motifs or folds have been characterized for eucaryotic transcriptional regulators. As shown in Figure 8-5, these include: Zn finger, Homeodomain, Leucine zipper, Helix loop helix. Like the helix turn helix these function as dimers (usually homodimers but sometimes heterodimers), and generally make contacts in the major groove. Regulation of the genes required for galactose metabolism in yeast. The genes required for galactose metabolism in yeast, GAL1, GAL2, GAL10 and GAL7 are controlled in both a positive and negative manner. They are not found in an operon as in the case of bacterial Trp or Lac operons, rather each gene is encoded by its own mRNA. 5 GAL10 promoter TATAAA +1 UAS GAL Positive regulation of the GAL genes. The genes required for galactose metabolism are turned on when yeast are grown in the presence of galactose. The key transcription factor is the protein Gal4. As a dimer, Gal4 specifically binds to a 17 bp sequence, making contacts with bases in the major groove. The GAL10 promoter has 4 such Gal4 recognition sites, which as a unit are referrGALt(Upstream Activating Sequence- galactose). Gal4 is constitutively expressed and ocGALin the presence or absence of galactose. However, it is only active in the presence of galactose. In the absence of galactose Gal4 is bound to a repressor protein called Gal80. Gal80 blocks the activation function of Gal4, keeping Gal4 in an inactive state. When galactose is present Gal4 switches from an inactive to active state. This requires a third protein, Gal3. Gal3 binds galactose; the Gal3- galactose complex then interacts with the Gal80 repressor protein, effectively freeing the activation function of Gal4 from Gal80. Gal4 then is able to stimulate transcription and does so through direct protein-protein interactions with components of the transcriptional machinery and by altering chromatin structure. Gal4 inhibited by Gal80 80 80 80 80 4 4 4 4 TATAAA 6 Gal4 is active in the presence of 4 4 4 4 80 80 80 80 3 3 3 3 TATAAA Negative regulation of the GAL genes. All cells prefer to use glucose as their carbon/energy source. In fact they limit the expression of the genes required for the metabolism of other carbon sources in the presence of glucose. The process by which cells do this is called catabolite repression. An example of catabolite repression is the inhibition of the GAL genes in media containing glucose. This occurs whether or not galactose is present in the environment. Glucose repression of the GAL genes requires another DNA binding transcription factor called Mig1. In the presence of glucose Mig1 binds the GAL10 promoter. Promoter bound Mig1 recruits factors that modify the chromatin in the vicinity of GAL10 in such a way as to inhibit transcription. In the absence of glucose, Mig1 is phosphorylated. When phosphorylated, Mig1 can not enter the nucleus, and thus can not bind the promoter. Mig1 bound in the presence of glucose M i g 1 +1 Key Points to understand from GAL regulation. 1. The GAL genes are subject to positive and negative regulation. 2. DNA binding proteins are important for both positive and negative regulation. 3. Protein-protein interactions are important for regulation. 4. Regulation occurs through altering the transcriptional machinery and changes in chromatin structure. 5. The system is sensitive to environmental signals (glucose and galactose). These signals result in changes in gene expression. Topic 23 Review Questions 23-1. Bacteria contain more than one sigma factor. Why might these be important in gene expression? 23-2. Yeast induce the transcription of genes required for metabolism of galactose when galactose is present in their growing media. Outline two strategies by which this might occur. In one case assume positive regulation. In the second case assume negative regulation. 23-3. Contrast and compare the mechanisms of transcription and replication. 23-4. Protein-protein interactions are very important in gene regulation. Indicate protein-protein interactions that are important in gene expression. 23-5. Much of the early work on the regulation of the lac operon involved the identification and characterization of mutations that resulted in altered expression of the operon. For mutations with 7 the following characteristics, what would happen to expression of the lac operon in media containing lactose, glucose, or glucose and lactose? a) a mutation within the lac repressor such that it no longer binds to DNA b) a mutation within the lac repressor such that it no longer binds to lactose c) a mutation within the lac repressor such that it always binds the operator d) a mutation within the lac operator such that it no longer binds the lac repressor e) a mutation within the catabolite activator protein such that it no longer can bind DNA f) a mutation within the catabolite activator protein such that it is always bound to DNA Topic 24: RNA PROCESSING IN EUCARYOTES Readings 240 - 246 The primary RNA transcript must be processed to become a translatable mRNA. 1. Capping of the 5' end of the message with 7 methyl guanosine (Figure 7-16). Capping is required for RNA export from the nucleus, aids in stabilizing the mRNA from being degraded and acts as a translational signal. 2. Polyadelylation: the addition of a long A-tract to the 3' end of the RNA. The RNA is first cleaved ~ 30 bases following an AAUAAA sequence which is 3’ to the coding region. A string of A residues is then added. 5'_________________________AAUAAA (30 bases)_____________3' ^ cut 5'________________________________________AAAAAAAA(300) 3' Polyadenylation provides additional stability to the mRNA by reducing effects of 3' exonucleases, has a role in the nuclear export of the mRNA and in its translation. 3. Splicing In eucaryotes protein coding sequences of genes can be interrupted by one or more noncoding sequences called introns (Figure 7-17). The coding sequences are called exons (expressed). Figure 7-14 shows the intron-exon structure of 2 genes. The origin of introns is unclear. It is not known if have procaryotes lost introns or if eucaryotes gained introns. Their utility, though, is clear. Introns provide the opportunity for differential splicing. That is, the mRNA can be put together in different ways generating functionally related but distinct gene products. Differential splicing is seen particularly in cases where tissue specific forms of a protein are created (Figure 7-21). How are introns removed from the primary RNA? Specific sequences within and surrounding the intron target the intron for removal. (See Figure 7- 19) These are the 5' junction, branch point (or acceptor site) and the 3' junction. Loss of these sites results in defects in splicing. In fact some of the thalasemias (genetic disorders resulting in anemia) are the result of the loss of splice junctions in the globin genes. The Spliceosome -- snRNPs (U1, 2, 4, 5, 6) Introns are removed as the result of the catalytic activity found in small nuclear ribonucleoprotein particles (snRNPs=snurps). snRNPs are complexes of RNA (small nuclear RNA) and protein. The RNA component has a recognition function, acting through base pairing with sequences on the precursor mRNA. snRNPs are also critical in arranging the ends into position. The assembly of snRNPs that catalyze splicing is called a splicesome. Splicing occurs by 2 transesterification reactions. (Fig. 7-20) Cleavage at the exon 1-intron boundary results from the attack of the 5' splice junction by adenine in the branch point. This generates a lariat structure as the result of the unique 2'-phosphodiester bond. In the second 8 reaction the 3'-OH of exon 1 reacts with the 3' splice junction cutting out the intron and joining the exons. The lariat is displaced and soon degraded. Self-Splicing Introns In pre mRNA in protozoa, mitochondria and chloroplasts, snRNPs are not required. The RNA is capable of catalyzing its own conversion to mRNA. This is one example of RNAs having catalytic potential. Another example is ribozymes, which catalyze the cleavage of other RNA molecules in a sequence specific fashion. RNA Export (Fig 7-22) In eucaryotic cells RNA is synthesized in the nucleus and translated in the cytoplasm. Mature mRNAs are transported from the nucleus to cytoplasm through nuclear pores, a highly organized set of proteins in the nuclear membrane (Fig. 15-8). Transport through nuclear pores is a regulated process that requires the recognition of proteins bound to the poly A-tail, the 5’ cap and often internally. Topic 24 Review Questions 24-1. Outline a strategy by which differential splicing of a pre-mRNA could occur. 24-2 How are RNA molecules important in determining the sites for splicing? Topic 25: TRANSLATION Readings 246 – 258 The decoding of the mRNA to produce a protein occurs on the ribosome. This process of translation is complex, requiring many RNA and protein factors. Genetic Code (See. Figure 7-24) The mRNA "spells out" the amino acid code in 3 letter "words" called codons. Each protein has a specific reading frame that is determined by where the decoding process begins (Fig. 7-25). The code is: Universal (found in all organisms) Nonoverlapping Commaless, no gaps (gaps were removed by splicing) 61 codons for 20 amino acids; therefore redundant Redundancy occurs at the 3rd position of the codon (wobble) 3 stop codons : 1 start codon Note also the following features of the code. Why might they have evolved? 1. More common amino acids (found often in proteins) have more codons. 2. Related amino acids have similar codons. For example: Gln Glu Asp CAA/ G GAA/ G GAC/U tRNA (Figure 7-28) Translation requires tRNA (transfer RNA) molecules, which act as the vehicle that bring amino acids to the growing peptide chain. They function in a codon specific fashion relying on base pairing rules. tRNAs are ~80 nucleotides in length and have a cloverleaf secondary structure. The anticodon of the tRNA hybridizes with the codon. The correct amino acid is covalently linked to the 3' end of the tRNA. Wobble results from the fact that accurate base pairing for some tRNAs only requires matching at the first 2 positions. Amino Acid Activation— Aminoacyl tRNA synthetases (Figure 7-29) Using ATP as the energy source, the carboxyl group of a specific amino acid is coupled to the 3' end of a specific tRNA in a high-energy bond. Amino acid activation by aminoacyl t-RNA synthatases is important for: 9 1. Providing an energy source for later peptide bond formation. 2. Providing specificity by matching the correct amino acid to the specific tRNA. Ribosomes (Figures 7-31, 7-32) Protein synthesis occurs on large multimeric protein RNA complexes called ribosomes. Ribosomes have two subunits, large and small. The small and large subunits are composed of both RNA (rRNA) and protein. In eucaryotes the small subunit contains 33 proteins and 1 RNA, whereas the large subunit contains 49 proteins and 3 RNAs. The small subunit matches tRNAs to the codons. The large subunit catalyzes the formation of peptide bonds. There are 3 sites for tRNAs on the ribosome: A site, aminoacyl tRNA site; P site, peptidyl tRNA; E site, exit site. Two of these are occupied at any one time. The mRNA is bound in proximity to the A and P sites. mRNA is decoded in a 5' to 3' direction, one codon at a time. The Process of Translation (Figure 7-33) Assume that the ribosome is already in the process of translating an mRNA. A tRNA in the P site is linked to the growing polypeptide. An aminoacyl tRNA accesses the A site following basepair rules. The energy of the aminoacyl-tRNA bond in the P site is used to form a peptide bond between the amino group of the amino acid in the A site with the carboxyl group of the amino acid residue in the P site. The reaction is catalyzed by a peptidyl transferase activity in the ribosome. The reaction is coupled to a conformational change in the ribosome that effectively results in a shift of the small subunit relative to the large subunit. In turn this results in a shift of the tRNAs to the E and P sites from the P and A sites respectively. In the final step of the cycle, the small subunit moves downstream precisely 1 codon (3 bases), placing a new codon in the A site and resulting in the release of the tRNA from the E-site. Initiation of Translation (Figure 7-35, 7-36) Initiation is the key step in deciding whether an mRNA is to be translated. Translation begins at an AUG codon and with a special initiator tRNA that carries methionine (Met). In eucaryotes the initiator tRNA is loaded onto the small subunit with initiation factors (proteins). The loaded small subunit recognizes the 5' Cap (not to be confused with the bacterial transcription factor) of the mRNA and moves in a 5' to 3' direction until it finds an AUG codon. The initiation factors then dissociate, allowing the large subunit to bind the small subunit. In this process the initiator tRNA is positioned at the P site. In bacteria multiple open reading frames (ORFs) are often found in a single message. The result is that more than one protein must be translated from the RNA. The ribosome can initiate translation at internal AUGs. Specificity comes from the fact that 5' of each functional AUG is a ribosome binding site which is required for efficient recruitment of the ribosome. Termination of Translation (Fig. 7-37) Stop codons signal the end of translation. Release factors associate with the ribosome when any one of the three stop codons reaches the A site of the ribosome. These factors cause the peptidyl transferase activity to catalyze the addition of a water molecule to the end of the chain. Antibiotics and Translation (see Table 7-3) The majority of known antibiotics used to treat bacterial infections block translation. The complexity and importance of translation makes it a prime target for disruption. The one notable exception to this is the penicillin/ampicillin family of antibiotics. Topic 25 Review Question 25-1. At what steps in translation is the specificity of the reaction determined? 10 Topic 26: RECOMBINANT DNA TECHNOLOGY GENETIC ENGINEERING Required Readings Chapter 10 Readings from Genome 3; p 63 – 65, 112 – 115, 133 – 143 Definition: The techniques by which DNA fragments from different sources (different chromosomes, different organisms, or man-made) are recombined to make a new DNA molecule. Introduction People have manipulated genes for thousands of years using selective breeding. Dog breeds and present day crops are prime examples. While obviously productive these efforts are slow and limited to breeding species. Until the early 1970s, DNA was hard to work with at the molecular level. Individual genes are contained amongst other genes in a long linear polymer. The composition of most DNAs is similar as they are composed of only 4 bases, therefore, traditional biochemical approaches were not well suited for separating and analyzing individual genes. Now, the manipulation of genes in the lab is easy using the tools of recombinant DNA technology. The implications are enormous! In Research. Recombinant DNA Technology (RDT) has allowed the rapid expansion of scientific knowledge. This has not only been in areas of gene structure (promoter and origin mapping, intron-exon structure, etc) and function (transcription, splicing, transposition, replication, translation) but in all areas of biochemistry and cell biology. The power of RDT directly reflects the link between genes and proteins as defined in the Central Dogma. RDT has thus introduced a novel way to study protein structure and function most notably in the form of site directed mutagenesis, protein overexpression, and protein engineering. The combination of RDT and computational science has also enabled the analysis of whole genomes (Bioinformatics). The complete genome sequences of many hundreds of organisms are now known. This has dramatically changed the approaches we use to understand cell function. In Biotechnology (defn. the use of organisms to do work for man). Recombinant DNA technology is the driving force behind the modern biotechnology industry. The impact on human health is evident in diverse areas, but most notably in the pharmaceutical industry and in agriculture. RDT has had a major impact in biotechnology because:  It has enabled the production of large amounts of otherwise rare proteins.  The techniques to introduce genes across species barriers have been developed.  RDT has allowed the creation of novel proteins, modified organisms, and recently ―novel‖ organisms (www.nytimes.com/2010/05/21/science/21cell.html). Areas where RDT has allowed dramatic changes in biotechnology include: Medicine: Drug production and creation (e.g. vaccines, human insulin, growth hormone, Factor VIII), diagnosis, genetic counseling and screening, as well as the potential for gene therapy. Agriculture: The production of crops with unique features e.g. vitamin enhanced crops (golden rice). The potential for crops with increased yield e.g. canola Environment: e.g. PCB and oil eating microbes Forensics and Law: DNA fingerprinting The importance of RDT in biotechnology and its major role in the creation, design and product
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