Genetics Final Reveiw

is a set of coordinately regulated genes that

• 1. usually encode proteins involved in a common function, in this case lactose utilization,
• 2. are linked on the chromosome,
• 3. are transcribed as a unit,
• 4. are coordinately regulated.

a disaccharide of glucose and galactose linked together

an enzyme that cleaves the B-galactoside bond to generate glucose and galactose. In a minor side reaction it also produces allolactose. This enzyme is encoded by the lacZ gene

an enzyme that transports lactose into the cell. This enzyme is encoded by the lacY gene.

an enzyme that may detoxify toxic by products. This enzyme is encoded by the lacA gene.

• 1. Lactose must be present, therefore these genes are inducible by lactose.
• 2. Glucose must be absent from the media, therefore these genes are repressible by glucose.

mutations lead to the synthesis of the three structural enzymes even in the absence of lactose

mutations lead to the inability of the cell to synthesize the structural enzymes at all, even when lactose is present.

• They hypothesized that the lacZ-lacY-lacA genes are transcribed as a single, polycistronic, mRNA, beginning just before the start of the B-galactosidase coding sequence and proceeding through the three structural genes.
• This arrangement of genes transcribed as a unit is also called an operon. The lacI gene is a separate gene that encodes the lac repressor protein.
• The repressor binds to lacO, or the operator. This binding prevents RNA polymerase from transcribing the structural genes.
• When lactose is present, it binds to the repressor, preventing it from binding to the operator DNA. Thus, RNA polymerase can transcribe the genes.

An important part of the Jacob and Monod model is that the three structural genes comprise an operon, that is they are transcribed as a polycistronic mRNA.

are mutations that only affect the genes immediately adjacent to the mutation.

Mutations in elements like promoters, transcriptional termination sites, origins of DNA replication, centromeres of chromosomes, would be cis-dominant.

• 1. The genes must be clustered, that is map to a single locus.
• 2. Since genes in an operon are transcribed as a single mRNA, there must be only a single promoter at the 5' end. Thus promoter up and down mutations:
• a. Map to one end of the operon.
• b. Affect expression of all the genes in the operon in a cis-dominant fashion. 3. Nonsense mutations often have a polar effect on expression of genes in the operon.

Thus, the CAP-cAMP complex is required for transcription of the lac operon.

• Glucose repression, or catabolite repression, is a global regulatory mechanism; it controls not only the lac operon, but operons required for the utilization of a large number of sugars.
• There are two components to glucose regulation: the small molecule cyclic AMP, or cAMP, and the catabolite activator protein or CAP.
• The CAP protein binds cAMP to form a CAP-cAMP complex that binds to a specific sequence of DNA.
• Such a sequence is located in the lac operon immediately upstream from the -35 region of the lac promoter.
• The -35 region of the lac promoter is a poor -35 sequence, and normally RNA polymerase will not recognize it.
• However, the CAP-cAMP complex helps RNA polymerase bind to the lac promoter, overcoming the lack of a good -35 sequence.
• Mutations in the promoter that generate a better -35 region, eliminate the requirement for CAP- cAMP and allow induction in the presence of glucose plus lactose.Thus, the CAP-cAMP complex is required for transcription of the lac operon.When glucose is present in the medium, cAMP levels are low. When glucose is absent, cAMP levels are high.

Generally the activity of these proteins is regulated by some external signal, such as lactose for the lac repressor, or indirectly by glucose for CAP.

With respect to lactose, the lac operon is an inducible-negative system, whereas with respect to glucose, it is a repressible-positive system.

• E. coli requires five genes to encode the enzymes required for the biosynthesis of the amino acid
• tryptophan. They constitute an operon with the genes arranged in the order: trpE-trpD-trpC-trpB-trpA.
• The cell regulates the synthesis of these enzymes according to the availability of tryptophan in the media; when tryptophan levels are low, the enzymes are synthesized, when tryptophan levels are high, the enzymes are not synthesized.

transcriptional repression and attenuation, the prevention of transcriptional termination.

• The trp repressor protein is a dimer that can't bind DNA unless tryptophan is bound to it, exactly the opposite relationship that we saw between the lac repressor and allolactose.
• In this case, tryptophan is the co-repressor.
• Thus, when tryptophan is plentiful, transcription is repressed, and the structural genes are not expressed.

• 1. As in the case of regulation of the lac genes, energy is wasted if unneeded enzymes are synthesized.
• 2. Even if tryptophan is not supplied in the medium, the rate of tryptophan synthesis varies depending upon what other things are present. The synthesis of tryptophan uses carbon skeletal compounds generated in other metabolic pathways. If tryptophan is synthesized at too high a rate, it will be at the cost of something else cell requires.

• In this case, what the cell really wants to monitor is not how much tryptophan is present, but how much charged tryptophanyl-tRNA is present relative to the protein synthetic needs of the cell.
• This parameter is determined in a process termed attenuation

• In such a state, sequence 1 is covered by ribosomes, and sequence 2 and 3 are free to pair. The 2 and 3 pairing forms rather than 3:4 because these two sequences are transcribed before 4.
• Since 3:4 does not form, transcription is not terminated.

• In this state, sequences 1 and 2 are covered with ribosomes, and 3 is free to pair with 4.
• The 3:4 pair results in termination before the five trp structural genes are transcribed.

Again 3:4 causes termination and the genes are not transcribed.

• This regulatory mechanism allows the cells to directly assess whether there is sufficient tryptophan for protein synthesis.
• Many other amino acid biosynthetic pathways are regulated in a similar manner, involving both the gross control by specific repressors and fine tuning by attenuation.

• 1. Early gene expression- the phage infects the cell and mobilizes the cell machinery to replicate its DNA.
• 2. Middle gene expression- DNA replication occurs.
• 3. Late gene expression- Phage coat proteins are synthesized and cell lysis functions are expressed.

• Clearly, if the late genes were expressed too soon, the number of phage produced would be greatly reduced because the limited DNA molecules would all be packaged before replication could proceed. Also, if lysis functions were expressed early, infection couldn't proceed.
• Thus, lambda provides us with a model system for temporal gene expression.

The lambda genetic map is represented as a linear chromosome. The DNA is linear in the phage head, but when it is injected into the cell, it circularizes which is necessary for the integration event and the initial rounds of DNA replication.

• The coat proteins A-J are encoded on the left arm.
• Genes encoding proteins required for establishing and maintaining lysogeny are on the right arm.
• Genes required for DNA replication, O and P, are farther out on the right arm.
• Genes encoding cell lysis functions, S and R, are on the extreme right arm.

When the chromosome circularizes, the S and R genes are brought close to the coat protein genes so that all the late functions are clustered together.

is a regulator important for the lysogenic/lytic choice and repression of the early genes as we will see later and is transcribed from pR (promoter for rightward transcription).

is an anti-terminator protein transcribed from pL (promoter for leftward transcription). N binds to the DNA at the nut (N utilization) sites and jumps on the RNA polymerase as it transcribes through that region. It prevents termination at the terminator sites tL and tR1 to allow the second round of transcription.

is a DNA binding protein that activates transcription. It binds to the region of pRE and pI to promote the transcription of CI and int. These proteins are important for lysogeny as we will discuss later.


CII activates two promoters that can not be transcribed by E. coli RNA polymerase alone

are required for DNA replication, so as N accumulates and allows expression of O and P, DNA replication will begin.

is a regulator of late gene expression, genes S and R encode lysis functions and A-J, the coat proteins.is an

anti-terminator like N except that it binds at qut (Q utilization) and prevents termination of transcription at tR3.

accumulates in the cell, it will turn on the late genes.

the accumulation of Q and the expression of the late genes is increased dramatically as the lambda DNA is replicated.

lambda chromosome must integrate, and second, the genes of the lytic cycle must be repressed.

critical gene for the integration

critical gene for repression

• we can induce lambda out of lysogeny by shining uv light on the cells.
• uv light damages DNA, so when the cell is subject to such damage, repair functions are induced.
• One of the repair proteins is RecA which has a protease activity; it cleaves proteins at a specific sequence.
• The lambda repressor has evolved to have such a sequence so that it is cleaved and, consequently inactivated, by RecA.
• Once the repressor is inactive, the initial round of transcription that will lead to the lytic cycle is begun.
• Included in genes transcribed in the second round is xis which excises lambda from the chromosome.

encodes a galactose permease that brings galactose into the cell.

encode enzymes required for the conversion of galactose to glucose-6-phosphate which can then enter the glycolytic pathway for energy generation.

encodes an enzyme that is secreted from the cell to break down the disaccharide melibiose which is composed of a galactose plus a glucose molecule. This gene is also regulated as part of the gal regulon.

repressed (in the presence of glucose)-1X,

derepressed (in the presence of a non-glucose energy source and in the absence of galactose)-10X,

induced (in the presence of galactose and the absence of glucose)-1,000X.

mutations are recessive and uninducible for the gal genes

mutations are dominant and express the gal genes constitutively except on glucose where the genes are still repressed.

mutations are recessive and express the gal genes constitutively except on glucose where the genes are still repressed.

Model 1. In the presence of galactose, Gal4 binds to DNA to activate expression of the gal genes. In the absence of galactose, Gal80 binds to Gal4 to prevent it from activating gene expression.

Model 2. In the absence of galactose, Gal80 binds to DNA to repress expression of the gal genes. In the presence of galactose, Gal4 binds to Gal80 to prevent it from repressing, thereby allowing induction of gene expression.

Model 3. In the absence of galactose, Gal80 binds to DNA to repress gal gene expression. In the presence of galactose Gal80 is removed from the DNA, and Gal4 binds to activate transcription.

• Model 1. The double mutant would be uninducible. Since Gal4 is absent, no activation of gene expression can occur. It does not matter whether Gal80 is present or not, since its only function is to prevent the function of Gal4 in the absence of galactose.
• Model 2. The double mutant would be constitutive. Since Gal80 is absent, the genes are not repressed and high levels of expression would occur. The presence or absence of Gal4 does not matter if its only function is to prevent Gal80 repression in the presence of galactose.
• Model 3. The double mutant would be partially induced in the absence of galactose and expressed at lower than wild type levels in the presence of galactose. The absence of Gal80 would cause loss of repression, but the absence of Gal4 would not allow full induction.

One gene masks the expression of a different gene for a different trait.

An activator is a DNA-binding protein that regulates one or more genes by increasing the rate of transcription.

is a cytoplasmic protein that binds galactose when it enters the cell.

The ability of Gal80 to bind to Gal4 is regulated by a third protein, Gal3

Thus, Gal80 shuttles back and forth between the nucleus and the cytoplasm and is trapped in the cytoplasm by binding to Gal3 when galactose is present, and is trapped in the nucleus by binding to Gal4 when galactose is not present.

• Glucose represses the gal structural genes by repressing GAL4 transcription.
• The presence of glucose in the cell activates a repression complex comprised of several proteins: Mig1 and Mig2 (DNA binding proteins that bind to a region upstream of the GAL4 gene) and a general repression complex composed of two proteins Tup1 and Ssn6.
• Repression of GAL4 is achieved by either Mig1 or Mig2 binding to the DNA and bringing the general repressors to the gene. The Tup1/Ssn6 complex then represses through a complex mechanism involving altering nucleosome structure and contacting the TATA box complex to prevent transcriptional activation.

• 1. Repressed. In the presence of glucose, the repressors Mig1 and Mig2 are activated and prevent GAL4 transcription. As a result, there is no transcriptional activator for the gal genes in the cell and very little transcription occurs.
• 2. Derepressed. In the absence of glucose or galactose, the GAL4 gene is transcribed, but Gal80 prevents transcriptional activation of the gal genes. Under these conditions, some transcription does occur because Gal4 is present.
• 3. Induced. In the presence of galactose (and the absence of glucose), GAL4 is transcribed, and Gal4 activity is not inhibited by Gal80. This condition results in strong transcriptional activation of the gal genes.

• These haploids vary in the genes they express.
• a cells produce a pheromone which causes a cells to arrest at the beginning of the cell cycle and grow an protuberance towards the source of the pheromone. a cells produce the a pheromone that has similar effects on a cells.
• a cells produce a pheromone receptor that resides in the cell membrane, binds a pheromone, and signals the cell of its presence. a cells produce receptor that binds a pheromone and signals the cell of its presence.

• Diploid cells do not express the pheromones or the receptors. They do not express the
• mating functions that haploid cells do.
• Diploid cells can undergo meiosis and sporulate when starved for glucose and nitrogen, while haploid cells can not.

determines the mating type of haploid cells.

• asg (a-specific genes) are genes expressed uniquely in a cells.
• asg (a-specific genes) are genes expressed uniquely in a cells.
• hsg (haploid-specific genes) are genes expressed in both a and " cells, but not in diploid cells.

• a1 is a transcriptional activator that binds upstream of the a and activates their transcription.
• a2 is a repressor that binds upstream of the asg and brings the Tup1/Ssn6 repressor complex to these genes to achieve repression.
• a1 has no function in a cells, but in diploid cells a1 forms a dimer with a2 to repress hsg. One other protein is important in this regulatory scheme. Mcm1 is a protein produced in all cells
• that acts as a transcriptional activator for the asg, asg, and hsg.

• Salmonella is genus of bacteria closely related to E. coli. A number of species cause diseases,
• like typhoid fever.
• Any successful parasite of mammals must evolve mechanisms to avoid the host immune system,
• and phase variation is one used by Salmonella.To counter the immune response,
• Salmonella can switch between two distinct forms of the flagella that are completely different as far as the immune system is concerned.
• The two antigenic distinct forms are composed to two different proteins designated H1 and H2.

• Cells switch from one from one form to the other at a frequency of 1/1,000. Thus, a culture of cells with the H1 flagella will spontaneously give rise to a small number of cells with the H2 type. Similarly, a culture of cells with the H2 flagella will spontaneously give rise to a small number of cells with the H1 type.
• In an infection, this means that when the host mounts an immune response to and kills off one form, which takes several days, the second form grows up to take its place and a new response must be mounted.

Laboratory strains of Saccharomyces cerevisiae have been bred to be stable in the haploid state. However, in the wild, yeast are homothallic meaning that haploid cells can switch mating type.

If an a cell from a homothallic strain is allowed to grow in pure culture, some of the cells will switch mating type and mate with those that have not, generating a culture of diploid cells. Similarly " cells can switch to the a mating type.

• A single a cell will bud to generate a mother (original) cell and a daughter (bud1) cell.
• The mother cell will switch its mating type to a, but the bud1 will remain a. The mother cell will bud again, giving rise to bud2. The mother cell will then switch
• back to an a cell, while the bud2 will remain a. Meanwhile, bud1 (an a cell) grows and buds, giving rise to a bud1-1. Bud1 will then
• switch to an a cell while bud1-1 remains an a.

• HMLa (homothallic left a) and HMRa (homothallic right a) are silent loci, silent because the
• information is not expressed. However, HMLa and HMRa contain exactly the same DNA sequences as the MATa and
• MATa, respectively.
• The reason they are not expressed is due to the E and I sequences surrounding each locus.

• direct the cell to condense the nucleosomes covering the two silent loci into heterochromatin which can not be transcribed. This is analogous the packaging of the mammalian X chromosome in X inactivation.
• It is the silent loci that serve as donors to provide the information when cells switch to the opposite mating type.

• is initiated by an enzyme called HO, an endonuclease like the restriction enzymes. It recognizes and cleaves a specific sequence neat the MAT locus indicated as ho on the chromosome III diagram.
• After cleavage, other nucleases chew up the DNA in the MAT locus destroying the information there.
• To repair this gap in the DNA, the cell copies the information from one of the silent loci, leading to the gene conversion event.
• The silent locus that is chosen is the one that has the opposite information that resided in the MAT locus, thus switching the mating type.
• Some important features of this mechanism are:
• As a gene conversion event, the information at the silent locus is copied, but remains intact at the silent locus. Thus the information is preserved if further switching events are required.
• The HO endonuclease is only expressed in mother cells, thus only mother cells switch mating type.

Trypanosomas is a genus of protozoan parasites

• One disease caused by these parasites is sleeping sickness. Sleeping sickness is spread through an insect vector, the tse-tse fly. It is characterized by a persistent, cyclical manifestation of the disease.
• The reason for the persistence and cyclical disease is that the infecting organism changes its antigenic pattern as we discussed for Salmonella.Each time the host appears to recover through the immune system killing the parasite, a few cells that switched their antigenic determinant mount a new round of infection.
• This cycle repeats until the host becomes too weak to keep mounting new immune responses and dies.

• a protein that this organism uses to change its antigenic determinant
• All the VSGs have a similar carboxy-terminus, but vary greatly in the rest of the protein.

The ELC contains a promoter and the C terminal region. A basic copy can be placed into the ELC, replacing the basic copy, through a gene conversion event.

• The antibody molecules, or immunoglobulins, are composed of two copies of each of two proteins, the light chain and heavy chain designated such because of their relative sizes.
• Both the light and heavy chains are composed of two domains, the N-terminal variable region and the C-terminal constant region.
• Each immunoglobulin can react with two antigen molecules.
• Each immunoglobulin recognizes a unique surface of the foreign molecule or antigen.
• When the variable regions complex with the antigens, they form a complex that is recognized and digested by other cells of the immune system.

• The immunoglobulins are produced by B lymphocytes.
• Each B lymphocyte can only produce one type of immunoglobulin, only one type of light and one type of heavy chain, that will recognize a single antigenic determinant.
• The immunoglobulin is bound to the surface of the lymphocyte as it circulates in the blood.
• Lymphocytes have a life-span of only a few days.

• 1. Begin dividing,
• 2. Produce more immunoglobulin, which is not bound to the cell surface but secreted into the blood.
• 3. Generate long-lived (years) memory cells.

lambda and kappa

are required for activation of the B lymphocytes.display antibody-like molecules in their surfaces that are generated through DNA rearrangements. They are produced in the thymus (thus T cells).

• HIV-1 and HIV-2
• HIV infects a subclass of T-cells, the helper or CD4 cells of the immune system.
• CD4 cells have the CD4 receptor on their cell surface, a protein that is involved in activating the T-cell.

• The HIV virion is comprised of a protein coat (made up of gag proteins) and a membrane envelope (comprised of the env proteins).
• One of the HIV envelope proteins, gp120, specifically interacts with the CD4 receptor to allow the virus entry into the cell. This is why the virus is specific for CD4 T-cells.
• HIV infection results in the elimination of these T-cells and, consequently, the host immune response is compromised.
• The HIV genome, as for all retro viruses, is a single-stranded RNA molecule that encodes the proteins required for infection. There are two copies of the genome in the viral envelope.

• viral-encoded enzyme
• Reverse transcriptase uses a single stranded RNA template to synthesize a single-stranded DNA, then uses that DNA as template to produce a double-stranded DNA copy of the original RNA called a cDNA.
• Like DNA polymerases, reverse transcriptase requires a primer. The HIV virus uses a cellular lysyl-tRNA as a primer. The HIV RNA genome has a region complementary to this tRNA, which hybridizes to the RNA to serve as the primer.
• Upon entry into the cell, the virion is made permeable to dNTPs and other cellular components to allow the packaged reverse transcriptase to synthesize the cDNA.

• promotes a recombination event that integrates the cDNA into the genome.
• Integration occurs at random sequences, that is, integration occurs at a different place in the genome in each infected cell

• 1. The major translation product produced is the gag protein that makes up the virion envelope. The gag protein is actually a polyprotein that is cleaved by a protease into four proteins. Translation of the gag region terminates with a normal termination codon.
• 2. The pol polyprotein is encoded in a region that overlaps slightly with the end of the gag protein coding region.

• 1. HIV protease-this protein is responsible for cleavage of the gag and pol proteins into the mature proteins. It recognizes a specific string of amino acids and cleaves at those sites. The protease inhibitor class of AIDS drugs inhibit this protease which is not found in uninfected cells.
• 2. Reverse transcriptase
• 3. Integrase-This protein is responsible for the integration of the cDNA into the genome.

are genetic elements share many of the features of retroviruses, but never leave the cell.