Test 3: Essays Flashcards
**Discuss the origin and functional significance of gene families.**
Gene families consist of functionally related genes that have very similar sequences. Members of the same family share a common ancestral form. The members slightly differ as a result of mutations during duplication. Next, a transposition takes place, followed by a series of duplication and mutation sequences. This leads to the gene family. For example, globin genes are derived from the ancestral gene myoglobin. The series of duplication and mutations lead to the formation of α-globin and β-globin gene families. Myoglobin binds oxygen tightly and will not release it under normal conditions. Hemoglobin, however, will freely transfer oxygen under normal conditions. This evolution of the globin gene allowed for terrestrial animals to evolve from aquatic animals. All vertebrate globin genes are highly conserved with regarded to their exons. There is a high degree of sequence homology and the numbers of amino acids that are used are highly conserved. The introns contain a fair amount of variability. Pseudogenes are also present in the globin gene families, and a buffer to mutation exists in the pseudogenes and introns. Mutations in the pseudogene sequence or intron sequences have no effect on the function of the genes. Gene families also display adaptive value because different genes in the family may be expressed at different times in the life of the organism and play different roles in different environments. In the case of β-hemoglobin, embryonic form is much like myoglobin, due to the fact that the embryo is in an aquatic environment. As the fetus begins to use oxygen from the mothers circulatory system, the fetal form takes over. The adult form is expressed later, as the individual breathes air only, utilizing the evolved low affinity form of hemoglobin. Organization of gene families varies between species. The genes in a family may be clustered on one chromosome or dispersed among several chromosomes. There is also variability in the number of functional genes and pseudogenes in the family, as well as in their orientation in the DNA.
Classifying individual genes into families helps researchers describe how genes are related to each other. Researchers can use gene families to predict the function of newly identified genes based on their similarity to known genes. Similarities among genes in a family can also be used to predict where and when a specific gene is active (expressed). Additionally, gene families may provide clues for identifying genes that are involved in particular diseases.
**Discuss the details of prokaryotic translation.**
_Initiation: _
The small ribosomal sub unit binds to Initiation Factor 3 (helps keep ribosome in place b/c of its weak binding). The small sub unit/IF3 complex then binds to the Shine-Delgarno sequence of the mRNA. Meanwhile, the initiator tRNA (tRNA-fmet) binds to IF2, which promotes the binding of the tRNA to the start codon. The N-formyl methionine blocks the N-terminal so the polypeptide being produced can grow in only one direction. The small sub unit/IF3 complex & the tRNA/IF2 complex bind to the start codon to form the initiation complex. The large ribosomal unit then joins (aided by IF1) which establishes the 3 active sites. Peptidyl (P), Amino (A), & Exit (E) sites.
Elongation:
Begins when the ribosome is intact with mRNA & initiator tRNA, the P-site is filled with initiator tRNA, and the A & E sites are empty. Ef-Tu/GDP then shuttles a charged tRNA into the A-site where it is determined if the anticodon of the tRNA is a match. Peptidyl transferase (enzyme) then breaks the amino acyl bond between the fmet-tRNA & catalyzes the formation of a peptide bond between the two amino acids. This is followed by translocation, where the de-acylated tRNA moves from the P-site to the E-site and moves the new peptidyl tRNA from the A-site to the P-site. The A-site is now empty and ready for the next tRNA with the appropriate anticodon. This process continues until the final stop codon of the mRNA is reached.
Termination:
Translation is halted when the stop codon enters the A-site. This is because there is no tRNA with an anticodon that corresponds to any of the stop codons. A release factor then binds to the A-site, which triggers the peptidyl transferase to cleave the polypeptide from the final tRNA into the cytoplasm. The release of the polypeptide chain allows for IF3 to bind to the small ribosomal sub unit, dissociating the ribosomal subunits and releasing the tRNA/mRNA. IF3 stays bound to the small ribosomal sub unit, ready for the next round of initiation.
**Discuss the various methods by which UV damage to DNA is repaired.**
UV damage is the absorption of energy that cause covalent bonds to be formed between 2 adjacent bases. Most common form is T-T dimers.
_Photo reactivation: _ DNA photolyases recognizes and binds to the thymine dimers. The energy of blue light changes the conformation of the photolyases resulting in the dimers breaking apart. There is no loss of material and no substitutions used in this repair.
Excision Repair: Repair that is characterized by the dimers being physically removed from the strand. Endonuclease (UVRA & UVRB) binds to dimer and applies two cuts in the damaged strand (8 nuc. upstream & 4-5 nuc. downstream from the dimer). DNA poly. I synthesize a new strand to replace the excised segment, beginning at the free 3’ OH. The old strand that contains the dimer is degraded and the last phosphodiester bond is closed by DNA ligase.
Recombinational Repair: A short-term fix so that replication does not shut down. DNA poly. skips over the dimer on the template and is reinitiated downstream, leaving a gap in the daughter strand. This activates sister strand salvage where the single strand region from the undamaged template is excised and inserted into the gap in the daughter strand. DNA poly. I then uses the new strand as a template and fills the gap in the undamaged template. The dimers remain but the normal daughter strand can go about replication.
_SOS Repair: _ Error prone repair that only occurs when the damage level is high. During replication if dimers are in the template, random bases are inserted in the daughter strand. Increased damage in SS DNA activates RecA which activates other parts of the SOS. This will alter the editing system to allow for lower fidelity and high mutation
Discuss the details of regulation for inducible and repressible operons in bacterial systems.
Inducible operons are generally in the “off” position by default. The negative effector trans-acting element binds to the operator of the promoter sequence. This binding blocks RNA polymerase, preventing it from binding to the promoter and inhibiting transcription. This is the default condition. The operon can be potentially turned “on” with the introduction of the pathways substrate into the environment. The substrate typically acts as the inducer. The inducer will bind allosterically with the negative effector, preventing the negative effector from binding the operator at the promoter. If enough of the inducer is present, all of the negative effector will be blocked from the promoter, and polymerase cant be blocked from transcribing the operons genes. This opens the potential for transcription with the removal of the negative effector. In the case of the lac operon, the cell must now “decide” what its optimal efficiency will be, which is to ignore the lactose or to use the lactose if glucose is scarce. The decision is based on relative amounts of ATP, ADP, & AMP. A high level of ATP means a high level of glucose present, but a high level of ADP/AMP means a low level of glucose. CAP normally unbound, helps the cell “decide” what to do. AMP is converted to cAMP by adenylate cyclase, and if glucose levels are low in the cell, cAMP accumulates. This causes CAP to become active and bind to cAMP. The CAP-cAMP complex acts as a positive effector, and enhances transcription by enhancing the binding of RNA polymerase to the promoter. For genes to be expressed in an inducible operon, induction must remove the negative effector, and CAP-cAMP must bind the promoter, enhancing RNA polymerase binding and enhancing transcription.
Repressible operons are usually in the “on” position by default. The repressor is constitutively produced but does not bind DNA under normal conditions. The product of the repressible system acts as the co-repressor. A sudden availability of product changes the system, and the co-repressor binds to the repressor, forming a complex that binds to the promoter, which inhibits transcription and turns the operon “off”. However, repressible operons are not expressed as “all or none”. This is referred to as attenuation, which results in off, on or partially on, depending on the concentration of product available in the ell. Attenuation is linked with simultaneous transcription and translation. In the case of the tryptophan operon, the leader polypeptide (containing a sequence of 2 tryptophan codons in a row) is transcribed and translated first. The rate of translation of the leader determines whether the operon will be transcribed fully or if termination will occur at the end of the leader. If a higher than normal amount of tryptophan is present, the ribosome keeps pace closely with RNA polymerase. This allows the formation of the terminator stem/loops (1-2 &3-4 stem/loops). If a low amount of tryptophan is present, tryptophan-charged tRNA molecules will be in short supply and the ribosome will stall. This stalling and separation of the ribosome from the polymerase allows the formation of the anti-terminator stem/loop (2-3 stem/loop). The formation of this stem/loop allows the polymerase to transcribe the entire operon. So the amount of accumulation of product determines the expression of repressible operons. No product accumulation results in fully “on”, where a high accumulation of product results in fully “off”. In between accumulation results in a partially “on” state that varies based on the level of product concentration.
Discuss the details of regulation of the lac operon.
Regulation of the lac operon is an example of an inducible regulator. It is always turned off until the repressor is inhibited then the lac operon is activated. The repressor is allosteric and when glucose is available the repressor will bind to the operator, resulting in the hindering of the operon. If glucose is not available and lactose is, the lactose will bind to the allosteric repressor, changing its shape and the repressor will no longer bind to the operator. Without the repressor binding, the lac operon can break lactose down so that the glucose can be used. When glucose and lactose are present, it is still more energy efficient to use glucose instead of lactose.
