If 35% of da bases in a region of da mouse genome r cytosine, wat percentage in da same region r adenine?
468892
as cytosine pairs with guanine, that accounts for 70% of the bases leaving 30% to split between adenine and thymine \n\nso, 15%
778145
http://www.pathlights.com/ce_encyclopedia/15sim03.htm\n\nsmall monkey (Macaca rhesus): 42 \n\n48 chimp chromosomes.\n\nhttp://www.ridgecrest.ca.us/~do_while/sage/v7i4f.htm\n\nBut the human DNA sequence was just recently decoded. (It was published in the 16 February, 2001, issue of the journal Science.) Work on decoding the DNA sequence of chimps is just beginning. (See Monkey Business, in the Evolution in the News column in this issue.) How could they have been able to calculate the similarity of two sequences “decades” ago when neither had been decoded? How can they calculate it now when only one of them has been decoded? Did they just pull 98% out of the air?\n\nPerhaps they compared the 46 human chromosomes with the 48 chimp chromosomes. They could not possibly have done that and come up with a figure exceeding 97.8% Here’s why:\n\nLet’s compare the chromosomes of a boy and his mother. A boy gets 23 of his chromosomes from his mother and 23 from his father. Clearly, at least 23 of the chromosomes will be identical to his mother. Therefore, the similarity will be at least 50%.\n\nBut one of the chromosomes he gets from his father must be different because his mother has two X chromosomes (if she didn’t, she wouldn’t be female), and the son has one X chromosome and one Y chromosome (because he is male). Therefore, at most, 45 of the 46 (97.8%) chromosomes will be identical.\n\nSo, the boy’s Y chromosome is guaranteed to be different from his mother. What about the other 22 chromosomes he gets from his father? Societies generally try to discourage incestuous relationships, and scientists have discovered that there is a good reason for this. It reduces the chance that a baby will inherit the same defective gene from both mother and father. If the boy’s father and mother come from unrelated families, it is almost certain that most of those 22 chromosomes will have at least one different gene.\n\nSo, if you determine genetic similarity by counting identical chromosomes, the genetic similarity of a boy to his mother is likely to be closer to 50% than 100%. “Scientists” certainly could not have found that humans and chimpanzees are 98% alike by counting chromosomes.\n\nOf course we know that they didn’t count chromosomes when making the comparison. We only mention it to make the point that what you count will affect the result you get. You can use a method of counting that gives close to 50%, or you can use a different method that gives a much higher number........\n\n\nclick on link to read more.
771405
40.\nhttp://www.research.att.com/~njas/sequences/b000602.txt
951611
There are 4 different bases for DNA: adenine (A), guanine (G), cytosine (C), and thymine (T).  They can be broken up into two groups: purines (A and G) and pyrimidines (T and C).  A will base pair with T via 2 hydrogen bonds, while G will base pair with C via 3 hydrogen bonds.  This is the Watson-Crick method of DNA base pairing...thus, you get a purine pairing with its complementary pyrimidine.  You can find this information easily by searching Wikipedia for each base individually.
52398
All bases are present in DNA except URACIL and in RNA it is THYMINE. Adenine is joined with Thymine in DNA, as well as Cytosine is with Guanine. In RNA, The only difference is that Adenine Joins with Uracil. Guanine and Cytosine still Join together.
797775
although all the answers represent components of eukaryotic genomes, e is the best answer\n\nhere is a link with some additonal info\n\nhttp://www.biologynews.net/archives/2005/05/02/scientists_track_stealth_dna_elements_in_primate_evolution.html
901637
Let B mean the genotype for dominant black fur and b mean the genotype for brown fur. \n\nRecall that when a gene is dominant, it is always expressed when  it is present. When a gene is recessive then it is ONLY expressed  when there is no alternative.\n\na) Pure black mice would have B1B2 and brown mice must be b1b2 because it is a recessive gene. (I've put in the numbers to act as subscripts so it makes more sense.) Crossing these, you would get offspring that inherit one gene from each parent.\nB1b1 ==> black\nB1b2 ==> black\nB2b1 ==> black\nB2b2 ==> black\n\nSo false because all the offspring from pure black and brown rats would be black. \n\nb) True. Pure black mice have BB. As both parents have BB then there's not a chance for bb to occur!\n\nc) False. Brown mice are always bb (recessive) so bb x bb can never result in Bb or BB. So offspring of 2 brown mice are always brown.\n\nd) Heterozygous black mice are Bb. So crossed with brown mouse bb (b1b2):\nBb1 ==> black\nbb1 ==> brown\nBb2 ==> black\nbb2 ==> brown\nSo you get equal numbers of black and brown offspring. True.
279794
figure out the efgfect of the enzyme. Or you can test DNA structure. Inhibitary enzymes have a blacked actionsite whereas activation enzymes have activation sites open to substrates
584877
It's not expected to be the same. \n\nIn double stranded DNA, A=T and C=G, therefore (A+C)/(G+T) = 1 always. The number of A is not necessarily equal as C. Therefore, (A+T)/(C+G) can be any number, different than 1.
691341
A DNA nucleotide ia made up of:\n1- phosphate \n2- deoxyribose sugar\n3- one of the 4 nitrogeneous bases: adenine, thymine, cytosine or guanine.
37783
The bases pair up like this: adenine/thymine and guanine/cytosine.  A/T form 2 hydrogen bonds, and G/C form 3 hydrogen bonds with eachother, so it would be impossible for A to pair with C or G.  The sequence of the base pairs codes for the genes of the organism.
614020
Here it is! With no hustle!\n\n\nFor pure DNA sample at 260nm wavelength measurements,\n\nOD of 1 = 50 micrograms/mL\n\nFor pure RNA, at the same wavelength\n\nOD of 1 = 40 micrograms/mL\n\nPeace!
677350
Wow, this answer is gonna be a long one...   There are numerous studies of the charting of the positions of genes on a DNA molecule or chromosome and the distance between genes of Escherisia Coli. It basically comes down to good old conjugation and transduction.  Having sequenced the E. coli K-12 genome, scienctists are creating a knockout of every gene, a clone of every ORF, and an extensive gene expression data set under a variety of physiological conditions. These clones and data sets are made available freely to the community.  Scienctists have also sequenced six additional enterobacterial strains which are pathogens closely related to E. coli K-12, and many plasmids. The comparative genomics aim of this effort is to characterize the gene pool of horizontally transferred elements and virulence determinants known as the pathosphere.\n\nI put a reference on it so be sure to check it out yourself. \n\nA mutator gene, mutD5, whose phenotype is conditional, has been identified in Escherichia coli. By P1 transduction it has been shown to lie at about 5.7 min on the chromosome, being co-transduced with proA and argF. In rich medium, streptomycin- and nalidixic acid-resistant mutation frequencies are 50 to 100 times higher than those in minimal medium. In minimal medium, the mutD5-induced mutation frequencies are still 50 to 100 times above co-isogenic wild-type (mut+) levels. Similar results were obtained with all markers tested. Mutant frequencies can be raised by thymidine in the medium at concentrations as low as 0.04 μM, or by the endogenous generation of thymidine from thymine plus a deoxyribosyl donor. Deoxyadenosine, various ribonucleosides, thymine, and 2-deoxyribose do not stimulate mutation. None of these effects are related to growth rate, since growth rate and mutation rate can be decoupled completely.\n\nThe gntP gene, located between the fim and uxu loci in Escherichia coli K-12, has been cloned and characterized. Nucleotide sequencing of a region encompassing the gntP gene revealed an open reading frame of 447 codons with significant homology to the Bacillus subtilis gluconate permease. Northern (RNA) blotting indicated that the gntP gene was monocistronic and was transcribed as an mRNA with an apparent molecular size of 1.54 kb. The transcriptional start point was determined by primer extension analysis. The gntP gene was found to be under catabolite repression and was not induced by gluconate. Also, expression seemed to be stringently controlled. Several observations indicated that the GntP protein is an inner membrane protein; it contains characteristic membrane-spanning regions and was isolated predominantly from the inner-membrane fraction of fractionated host cells. A topology analysis predicted a protein with 14 membrane-spanning segments. The inability of a mutant strain to grow on gluconate minimal medium could be relieved by introduction of a plasmid encoding the gntP gene. Finally, the kinetics of GntP-mediated gluconate uptake were investigated, indicating an apparent Km for gluconate of 25 microM.
837041
it's not 1 or 2 because they're nucleotide bases\n\n3 is the only chioce left
222657
25% normal non-carrier\n50% normal carrier\n25% albino
1068360
dna and genetic coding of chromosomes.
1094130
try this\n\nhttp://en.wikipedia.org/wiki/Epigenetics