distiguish between positive and negative gene control?
970315
As one example, lactose leads to the expression of the genes of the lac operon. So, the presence of lactose brings about the expression of the genes.  (You can find in your text the details of HOW that all happens.)  Histidine is an example of negative gene control.  The presence of histidine prevents the expression of the genes involved in synthesizing histidine.(Again, I'm sure you'll need the deatils which you can find in your text.)
153903
Epistasis refers to the phenomenon where one or more genes will mask the effects of another gene in a different position on a chromosome (or locus). This shouldn't be confused with dominance though where the interaction takes place in the same locus.\n\nSimple Example: The gene that causes Albinism hides the gene controlling the color of a person's hair.\n\nAnother Example: Bombay Phenotype - This involves the ABO Blood group system. They lack a protein called the H antigen which is used to form the A and B antigens. So even though they may have A or B genes, they appear to be blood group O because of the lack of H antigens.
971621
"A" is the answer.  It happens in prophase of meiosis 1 and it does exactly what choice "A" states.  The rest of the choices happen during meiosis as well, but do not explain the importance of crossing over.
355307
It isn't a very good control if your cells die before the experiment is complete.  More info is needed to get a more specific answer.
430234
mostly determined by the genotype and mode of inheritance. Sometimes it can be epigenetic. some rules like mandelian inheritance, random seggregation, imcomplete penetrance, codomminance and give different phenotypes
197178
Erase, cancel out all the introns, promoter and tata box, and you are left with those sequence you want. Then use the dna translation table to figure out the amino acid sequence. Then use those amino acid sequence to figure out what phenotype they correspond to via doll codes.
980113
Do you want to know if it is a negative or positive feedback control mechanism?  If so, it is a negative feedback control mechanism.\n\nNot sure if that is what you are looking for.
872183
You have 10 to 100 Trillion cells in your body.  Every cell has the same genetic code.  How would you change that code in so many cells?\n\nAdditionally, the genome is turning out to be much more complicated than was thought just a few years ago.  Much of a gene's 'performance' is determined by 'epigenetic' regulation of the gene's expression.  You can have 'normal' genes, but if they are not being expressed properly, that is the same as having defective genes.\n\nThis is a very, very complex issue.
371069
Well, at the least conjugative plasmids are known to occasionally integrate into bacterial chromosomes. I think some others are, but if I had to pick just one, that would be it.
188037
get different genetics
594843
Recombination (I presume this is what you're talking about) can do both. It creates new combinations of alleles at different gene loci that might be especially advantageous in a particular environment. The down side is that it can separate them also, so a parent that has a good combination will produce some offspring that don't have the combination because recombination has separated them in the gametes.\n\nI hope this is what you were asking about.
190009
yeah like the first person says. it all depends because there are dominant and recesive genes or chromosomes or whatever, and i think green is dominant, but if you get get 2 blues they dominate over 1 green, i think that is how it works.
62552
DNA is in every cell, but not every gene is expressed in every cell.  The way that the cell controls which genes are activated and which aren't is through DNA-binding molecules called repressors and promoters.  A promoter binds to DNA near a gene and causes transcription of the gene to make mRNA that will go on to serve as a template for protein synthesis.  Conversely, a repressor can bind to a gene and prevent the transcription of that gene into an RNA template (usually depending on conditions in the cell at the time, like chemical messengers or availability of nutrients, etc.).
153580
there are nonsense parts in our DNA called transposable elements, they can move around our DNA on their own and insert themselves into genes and promoters drastically changeing there function
476517
Complementation Testing\nOccasionally, multiple mutations of a single wild type phenotype are observed. The appropriate genetic question to ask is whether any of the mutations are in a single gene, or whether each mutations represents one of the several genes necessary for a phenotype to be expressed. The simplest test to distinguish between the two possibilities is the complementation test. The test is simple to perform --- two mutants are crossed, and the F1 is analyzed. If th e F1 expresses the wild type phenotype, we conclude each mutation is in one of two possible genes necessary for the wild type phenotype. When it is shown that shown genetically that two (or more) genes control a phenotype, the genes are said to form a complementation group. Alternatively, if the F1 does not express the wild type phenotype, but rather a mutant phenotype, we conclude that both mutations occur in the same gene.\nThese two results can be explained by considering the importance of genes to phenotypic function. If two separate genes are involved, each mutant will have a lesion in one gene while maintaining a wild type copy of the second gene. When the F1 is produc ed, it will expresses the mutant allele of gene A and the wild type allele of gene B (each contributed by one of the mutant parents). The F1 will also express the wild type allele for gene A and the mutant allele for gene B (contributed by the other muta nt parent). Because the F1 is expressing both of the necessary wild type alleles, the wild type phenotype is observed. \n\nConversely, if the mutations are in the same gene, each homolog will express a mutant version of the gene in the F1. Without a normal functioning gene product in the individual, a mutant phenotype occurs. \n\nEye color in Drosphila is a good model to demonstrate the complementation test. A wide array of spontaneous mutations have been studied. These experiments demonstrated that five genes (white, ruby, vermillion, garnet and carnation) controlling ey e color reside within 60 cM of each other on the X chromosome. The dominant wild type allele for each gene produces the deep red eyes. The mutant alleles produce a different color. If mutants from any of these five genes are crossed, the F1 would expre ss deep red eye color (wild type phenotype). \n\nFive different alleles (buff, coral, apricot, white and cherry) are also known to exist for the white gene, each representing a mutation at a different position in the gene. If mutant flies for any of the five white alleles are crossed, the F1 offspring would have a mutant eye color. Therefore, two genes involved in the expression of a phenotype complement each other. But complementation bet ween two alleles of the same gene does not occur
576339
Defination of epistasis - A double mutant where one mutation masks the phenotype of another mutation.\n\nNote that epistasis is not the same thing as dominance. With epistasis a mutation in one gene masks the expression of a different gene. With dominance, one allele of a gene masks the expression of another allele of the same gene.\n\nFor instance, a mutant gene that causes complete baldness would be epistatic to a mutant gene that determines hair color. Likewise, a mutation preventing an early step in a biochemical pathway will be epistatic to a mutation that prevents a step later in the same pathway (that is, an "upstream" mutant phenotype will obscure a "downstream" mutant phenotype). Two examples of how tests of epistasis can be used to analyze biochemical reactions are shown below. \n\n\n\n--------------------------------------------------------------------------------\n\nQUESTION: Two mutations were isolated with different effects on proline utilization: the mutation put-1020 makes cells constitutively express the putP gene at high levels, while the mutation put-1222 prevents expression of the putP gene. To determine if these two mutations affect different "regulatory pathways" or if they both affect the same "regulatory pathway", the double mutant was constructed. The phenotype of the put-1020 put-1222 double mutant is constitutive expression of the putP gene. Suggest a simple explanation for these results. What results would you expect if the two mutations affected different pathways? \n\nANSWER: The put-1020 is epistatic to put-1222. The simplest explanation of these results is that both the put-1020 mutation and the put-1222 mutation affect the same regulatory pathway. If the two mutations affected independent regulatory pathways then the phenotype of the double mutant would probably be intermediate between the two extremes observed in the single mutants. \ncontinues here:\nhttp://www.sci.sdsu.edu/~smaloy/MicrobialGenetics/topics/genetic-analysis/epistasis.html
1001272
no becuase the combinations of the genes