How do i find the genotyped for the 4 gametes from---?
823256
A. RY,rY\nB.Ry\nc.Ry,RY,ry,rY\nd. r,Y,ry\njust the differerent possiblities of rs and ys
1151978
Try this\nhttp://www.globalonlinesvcs.com/DNA_Testing.html
1075136
Assuming your species is diploid, it goes like this:\n\nSupposing there is a field of red, pink and white flowers. Red flowers have genotype RR, pink Rr or rR, and white rr. \n\nYou count the number of each kind of flower. It turns out there are 150 red-flowered plants, 480 pink, and 170 white ones. \n\nTherefore there are 2*(150+480+170) = 1600 alleles out there. There are 300 Rs in the red plants, 480 in the pink ones and 0 in the white. Added together there are 780. 780/1600 = 0.4875, so the frequency of the R allele is 48.75%\n\nAs for the r allele, there are 0 in the red-flowered plants, 480 in the pink, and 340 in the white. Added together, there are 820. 820/1600 is 51.25%.\n\nAs a double check, 48.75% + 51.25% = 100%
151482
Try here\nhttp://www.genomesize.com/search.php\nThey have a good searchable database, see the column labeled "Ch" for chromosome diploid number.
17652
http://www.ncbi.nlm.nih.gov/About/primer/genetics_cell.html
113474
9/64 because\n\nparents - Aa - heterozygous \n\nbelow is a punnet square\n\n___A a\nA AA Aa\na Aa aa\n\n1/4 chance of having albino child.\n3/4 chance of not having albino child\n\nall possible outcomes are:\n\n1/4 * 1/4 * 3/4 = 3/64\n1/4 * 3/4 * 1/4 = 3/64\n3/4 * 1/4 * 1/4 = 3/64\n\nadd them\n\n3/64 + 3/64 + 3/64 = 9/64
122666
http://www.seedbiology.de/structure.asp
371698
meiosis steps\n1) Meiosis I\nProphase I\nMetaphase I\nAnaphase I\nTelophase I\n2)meiosis II\nProphase II\nMetaphase II\nAnaphase II\nTelophase II
972775
In short Meiosis and Fertilisation.\n\nFertilisation brings each haploid gamete cell together, that is, a sperm and an egg. Each of these gametes carries a single copy of each chromosome. It is not until fertilisation that the chromosomes from each parent come together to form the diploid embryo.\n\nBut it's important to note that although we are walking around with 2 copies of our chromosomes, our gametes only have one. This is because the gametes are formed through the process of Meiosis, which halves our DNA. This allows the offspring to have a copy of each chromosome from each parent.
473864
Genotypes of man and woman : DdPp x DdPp\nPossible gametes of man and woman: DP Dp dP dp X DP Dp dP dp\nChildren: I guess its a dihybrid cross with 9:3:3:1 ratio of phenotypes since you have not mentioned the kind of inheritance.
1149823
A paternity test is the only conclusive way. Try the following website as an example
641740
its triploid,that means three sets of chromosomes.one from the haploid sperm of the pollen grain and two from the diploid secondary nucleus present in the embryo sac of the ovule,in the ovary.got it now!
876744
go www.learn.genetic.com i hope u can slove ur problem.
881626
Construct a punnet square:\n\nMom=  I^A & i\nDad= I^B & i\nOffspring will have a 1/4 chance of being: I^AI^B, I^Bi, I^Ai, ii\n\nIf the mother is I^Ai and the father is I^Bi, the possible blood phenotypes in the offspring are AB, B, A, and O.  There is a 25% chance of each blood type.
667403
Go to this website, you will find all the answers you need:\n\nhttp://plato.stanford.edu/entries/genotype-phenotype/
956460
To start with, You need more information as to the Parental genotype for each question, \nFor example in question (a) since you have only mentioned full dominance, we can either assume both parents will be A/A x A/A. However this cross is useless because you know there will not be variation in the F1 progeny because both are dominant. thus the ratio is 100% all dominant A/A.\nThe second assumption is to cross a dominant parent A/A with a recessive parent a/a. If that's the case then you choose one gamete from each parent to form the F1 as follow\n\nParental Genotype        A/A    x    a/a\nGametes                          A     x    a\n\nF1                                         A/a\n\nfrom here, if you need to get genotype and phenotype ratio for your F2, you need information as to what kind of cross you want do. You can self the F1 or you can testcross with a recessive parent. In either case the ratio is different\n\nSelf-cross                  A/a    x      A/a\nGametes              1/2 A , 1/2 a    x   1/2 A , 1/2 a\n\nF2  punnet square\n..............1/2  A..........1/2 a\n\n1/2 A......1/4 A/A........1/4 A/a\n\n1/2 a.......1/4 A/a........1/4 a/a\n\nPhenotypic ratio of F2:  1/4 A/A; 2/4 A/a; 1/4 a/a = 1:2:1\nGenotypic ratio of F2:  (1/4 A/A + 2/4 A/a); 1/4 a/a = 3:1\n\n\n\nTest-cross                  A/a    x      a/a\nGametes              1/2 A , 1/2 a    x   1/2 a , 1/2 a\n\nF2  punnet square\n..............1/2  A..........1/2 a\n\n1/2 a......1/4 A/a........1/4 a/a\n\n1/2 a.......1/4 A/a.......1/4 a/a\n\nPhenotypic ratio of F2:  1/2 A/a; 1/2 a/a = 1:1\nGenotypic ratio of F2: same as genotypic ratio\n\nquestion (a) is not formulated in a clear way. for example. question (a) should have mentioned something like this:\n\na) The F2 of monohybrid cross with both parents full dominance, or 1 parent full dominance, 1 parent recessive.\n\nIs hard to work in genetics problems when the problem lacks crucial information. let me know if there is more information that you did not write down on your question.
910758
sorry im a high schooler who doesnt know exactly how to do it but i think to approach it:\n1) find genotype thats just a little bit of guess and check although you know which ones are dominate since they give capital letters and you know they're heterozygous\n2) order youll have to find again with logic and guess and check\n3) genetic distance will be the times that it screws up from what would theoretically happen. Each gene locus in distance is about 1% so the number of percents that accidently cross will be your distance of alleles in map units\n4) this sounds like you would have to do this inorder to get #3... I really dont know too much about this but thats my stab at process of how to solve\nIf there is a pro here, I'd like to know how to do this too