Application Information

This drug has been submitted to the FDA under the reference 016931/001.

Names and composition

"R-GENE 10" is the commercial name of a drug composed of ARGININE HYDROCHLORIDE.

Forms

ApplId/ProductId Drug name Active ingredient Form Strenght
016931/001 R-GENE 10 ARGININE HYDROCHLORIDE INJECTABLE/INJECTION 10GM per 100ML

Similar Active Ingredient

ApplId/ProductId Drug name Active ingredient Form Strenght
016931/001 R-GENE 10 ARGININE HYDROCHLORIDE INJECTABLE/INJECTION 10GM per 100ML

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Answered questions

10 points!! Do you think plants that possesses two recessive alleles of the R gene will be able to fight off a potential fungal pathogen?
why? or why not? Help Please! Asked by Curtis Luminati 1 year ago.

No, if the wild-type allele is "R" (disease resistant) and this plant has 2 recessive "r" (not resistant) then it will not be able to produce the enzymes needed to fight the pathogen without intervention. Answered by Sarina Barra 1 year ago.

If they are rr then there is no dominant resistant gene R and the plants will be sensitive to the fungus. Answered by Kenya Schurz 1 year ago.


Phenotypic ratio of linked genes in a dihybrid cross?
A dihybrid cross was conducted between PpRr x PpRr. Normally, this would make a 1:3:3:9 ratio. If the genes were linked, though, what would the ratio be? Asked by Shenita Gherardi 1 year ago.

The exact ratio depends on how tightly linked they are and how often recombination occurs. However the general pattern is that you get two classes of progeny that are like their parents and are the most common and then two recombinant classes which have much smaller numbers. If for example P and R genes were linked 10 map units apart, you would expect to get 90% parental classes and 10% recombinant classes. However, if you cross two heterozygotes it may be difficult to work out linkage properly, as some of the recombinants may be hidden in the parental classes - so you need to do a test cross with one double homozygote recessive to identify each parental and recombinant class individually. This last bit is hard to explain without a drawing. If your parents were Pr/pR then the non-recombinant gametes would be Pr and pR, while your recombinant ones would be PR and pr, if you were to sit down and work out the progeny from these different gametes you would see what I mean. Answered by Branden Merklin 1 year ago.

you will probable desire to permit me recognize what each and each and each and each pertain too, TtRr = i.e tall and dark, or tall and lightweight, its particularly relies upon on what they propose...you have a incorrect question spectacular here, and do basically a punnet sq. to return to a selection the genotype, and then jointly as u recognize what each and each and each and each pair stands for then you definitely particularly can be sure the phenotype ratio. this is common. yet upload greater infor on your question Answered by Carly Zarucki 1 year ago.

3:1. Gametes - PR and pr x PR and pr gives PPRR,PpRr,PpRr and pprr Answered by Nathaniel Shrieves 1 year ago.


9 of 10 children in a family are right handed. Right handedness is a dominant trait?
The rest of the question says: "Assume that this trait is controlled by a single pair of genes." Ok so im completely lost. This was an assignment I was given over vacation and my teacher never uttered a word of genetics.The following questions are:What is the genotype of the left handed... Asked by Renea Biederwolf 1 year ago.

The rest of the question says: "Assume that this trait is controlled by a single pair of genes." Ok so im completely lost. This was an assignment I was given over vacation and my teacher never uttered a word of genetics. The following questions are: What is the genotype of the left handed child? What are the possible genotypes of the right handed children? What are the possible genotypes of the parents? If these parents were to have produced only one child, would it be left or right handed? Thanks so much for all the help. Its alot of questions, but I honestly have no idea what to do. Answered by Bettye Lilley 1 year ago.

OK--first, when one writes the symbol for genes, a capital letter is used to specify the dominant or wild-type version, and a lower-case letter is used to indicate the recessive, or mutant trait. Thus, "R" would be the gene which normally gives right-handedness, and "r" is for left-handedness. "Dominant" means that as long as someone has at least one copy of the R version on their chromosomes, then that is the version of the trait that will show up. So--A left-handed child must be "rr". Why? Because the other possibilities (RR and Rr) are right-handed, since the R version is dominant. Similarly, a right-handed child could be "Rr" or "RR", since only *one* copy of the "good" version is needed. In order for a child to be left-handed, both parents must carry at least one copy of the r gene. If both parents were RR, then neither child could get a copy of r. Since each child gets one copy of the gene from each parent, then *neither* parent could be RR. Thus, both parents must be at least Rr, and one could be rr. Now with the father Rr and the mother Rr, there are four possibilities from each sperm/egg combination: RR, R from father/r from mother; r from father/R from mother; and rr. If you understand probabilities, this means the genotypes of the kids will be 1RR:2Rr:1rr. So there will be on average 1 left-handed kid per 3 right-handed ones. If the father were Rr and the mother rr (left-handed herself), then the sperm/egg combinations would be R father/r mother, R father r mother (the other copy of the mother's r gene), r father/r mother and r and r father/r mother (the other copy of the mother's r gene), so the ratio would be 2Rr;2rr, or 1:1. With 1 of ten kids left-handed, the Rr/Rr combination is more likely, so that's probably what the parents are--however, Rr/rr is still a possibility. If you want to earn extra credit, you could say that *both* parents in theory *could* be RR, but a spontaneous mutation arose in the process of sperm or egg formation that converted an "R" copy into an "r" copy. Anyway--if the parents produced only one child, the Rr/Rr combination could be right or left-handed; same for the other combination. Answered by Ellsworth Matkins 1 year ago.


Year 9/10 genetics questions....?
why some characteristics seem to skip a generation. Asked by Lucienne Gostlin 1 year ago.

Suppose you're an utter retard. We shall represent this with a little r, and normal intelligence with a big R. You have two copies of every gene. One comes from mom, one from dad. Mommy was Rr and daddy was Rr. As long as they have at least one good copy of the intelligence gene they're perfectly fine; the working one just does double duty. You were unlucky. Instead of getting two Rs or an R and an r, you got two rs and now you're stupid. You marry a normal functional RR man because you're cute. Brainless, but cute. Like Britney Spears. Anyways, all your children come out Rr because you can only contribute an r gene and your husband can only contribute an R. Remember that Rr is an okay combination; it still works and your children really are smarter than you are. In fact, they seem perfectly normal. This is the generation the family curse is skipping. Suppose one of your Rr children marries another person who is Rr. The result is a 25% chance of RR, 50% chance of Rr and a 25% chance of rr. This means that one in four of your grandkids will be retarded. If your children marry an rr they have a 50% chance of Rr grandkids and a 50% chance of rr. Don't let them marry a retard. Try to get them to find someone with RR genes so there's a 50% chance of RR and a 50% chance of Rr, both of which are good. Traits where you need two bad genes for the effect to appear are called recessive. Answered by Heather Rider 1 year ago.


Genetics problem. Can anyone help? Thanks!?
here is an image link of the question that is better formatted (my table kinda got smashed) http://media.cheggcdn.com/media/15c/15cab72f-59e9-4cd9-8f4a-ce40a3ad7a34/phpDAyXAC.png Asked by Hyo Marine 1 year ago.

Bacillus subtilis, can grow on mannose, metabolized by the products of the man operon. Expression of the operon is controlled by manR. Depending on conditions the regulatory protein may bind at one of two sites in the operon, as follows: (i) When mannose is absent from the cell, the regulatory protein R1 can bind specifically at an operator site manO. Binding of R1 at manO reduces transcription of the operon four-fold from a basal level of 20 units. (ii) When mannose is present in the cell, it binds to the regulatory protein, causing it to undergo an allosteric transition from conformation R1 to a new conformation R2 that cannot bind at manO but it can bind specifically at a different site called the initiator, manI. Binding of R2 at manI increases transcription of the operon two-fold from the basal level Mutations, m1-m3, which affect expression of this operon, were identified. Each mutation affects only 1 component of the operon. Levels of operon activity were measured in haploids. They were also measured in partial diploids with an F’ carrying the wild-type alleles of all genes and regulatory elements described above Operon activity Haploids Partial Diploids (-) mannose(+) mannose(-) mannose (+) mannose wild type 5 40 10 80 m1 20 2010 80 m2 5 20 10 60 m3 20 40 25 80 For each mutation, describe which component is affected. In addition, explain the observed activity in the haploid and partial diploid in each case. Answered by Pamila Giovanni 1 year ago.

If I may offer my opinion... 1. in wildtype In haploid state of wild type, If mannose is absent, then the repressor will be in R1 state. The R1 will bind to manO, then the expression of man operon will be repressed to 4-folds of basal state, hence it will the level will be at 5. If mannose is present then, the the repressor will undergo transition into R2 which is then attached to manI, causing the expression to increase in 2-folds of basal state; 2 x 20 = 40. In the partial diploid state, all the regulators and the structural genes of man operon will be doubled (there is 2 copies or strands of man operon regulators and structural genes). It will be simply like in the eukaryotic cells, that the genes will have 2 alleles, except that in this case of partial diploid, it is only applied for only man operon. Therefore, whether mannose is present or absent, the overall expression on man operon in partial diploid state of wild type, will be 2 times higher than in haploid state. 2. m1 mutation From the data, it can be concluded that m1 mutation occur in the man R gene, causing defective repressor protein R productiob. The mutation causes repressor protein to function improperly, incapable of to transform either to R1 or R2. Therefore, it can not bind to either manO or manI. In the haploid state, since there is no funtional repressor protein, the expression will be at basal level, 20. But, in the partial diploid, since the partial wild type alleles are introduced, the overall expression will be like 'rescued' or normalized. Since, the introduced alleles bring wild type manR, hence the production of wild type repressor protein will be carried on by the manR from introduced alelle. The normal or wildtype repressor protein R from the introduced allele, is capable to bind to manO or manI on the mutated allele and wildtype, introduced allele. Therefore in partial diploids, the expression will be detected as the same level as in wild type. 3. m2 mutation The m2 mutation is occured in the manI. If you look at the first glance in the haploid state, you will also assume that the m2 mutation is occured at manR, causing defective repressor protein production which is unable to transform into R2 conformation, thus unable to bind to manI. It will be incapable to elevate the level of expression by the present of mannose, hence the level of expression is at 20 while the mannose is present. But later, if you observe at the partial diploid state, the first assumption that the m2 mutation is occured at manR will be unsupported by the data. By first assumption, the profile of expression level will be also 'rescued' since in partial diploid state, there will be wild type manR from the introduced allele, thus the profile will be observed as in wild type. Therefore, the first assumption is incorrect. If you observed a little more, then you will find that bythe present of mannose, the expression of man operon in partial diploid state is at 60. The only possible explanation is that the m2 mutation is occured in manI, causing manI to be defective; the R2 can not bind to manI, thus incapable to elevate the expression by 2-folds if mannose is present. The mutated allele or man operon will have defective manI, hence its expression is detected at 5, while mannose is absent and at 20, while mannose is present. The introduced allele, will bring wild type man operon, hence its expression is detected at 5, while mannose is absent and at 40, while mannose is present. Therefore, the overall expression will be at 5+5 = 10 if mannose is absent and 20+40 = 60, if mannose is present. 4. m3 mutation The m3 mutation is occured in manO. The overall explanations will be in 'similar' way as in m2 mutation (manI), but in this case, instead of incapabality to elevate expression by 2-folds, it will be incapability to alleviate the expression by 4-folds in the absent of mannose. The m3 mutation will causinf manO to be defective; the R1 can not bind to manO, thus the expression will be at 20 even if mannose is absent. Since only manO is disturbed, while manI is relatively normal, the expression will still be elevated by 2-fold if mannose is present. In the partial diploid state, the mutated allele or man operon will have expression level at 20 while mannose is absent, and at 40 while mannose is present. Meanhile, the wildtype allele will have expression level at 5 while mannose is absent, and at 40 while mannose is present. Therefore, the overall expression will be at 20+5 = 25 if mannose is absent and 40+40 = 80, if mannose is present. Hope it is not too complicated...and good luck. Answered by Joyce Jeudy 1 year ago.


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