More Genetics Help Please!?
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.
221778
Homozygous red-eyed female: X^W X^W\nWhite-eyed male: X^w Y\n\nF1:\n\n------ X^W --------- X^W\n--------------------------\nX^w | X^W X^w --- X^W X^w\n----|\nY --| X^W Y ------ X^W Y\n\nGenotype: 2 X^W X^w : 2 X^W Y (1:1)\nPhenotype: 100% of offspring have red eyes.\n\nF2:\n\n------ X^W --------- X^w\n--------------------------\nX^W | X^W X^W --- X^W X^w\n----|\nY --| X^W Y ------ X^w Y\n\nGenotype: 1 X^W X^W : 1 X^W X^w : 1 X^W Y : 1 X^w Y\n\n1 of each.
405198
i would suggest consulting with a genetic counselor to discuss your particular situation. see info below...
591863
The answer is "b"\n\nAssuming the dwarf characteristic is recessive, since about half of the offspring were dwarf, that means that half the offspring have to by homozygous for the dwarf genotype (tt).  The only way to get that is to cross the Tt and tt.  Thus you get half the offspring being Tt (tall) and half being tt (dwarf).  If you cross a homozygous dominant (TT) with anything else, all the offspring will be tall, because all will carry at least 1 "T" gene.  If you cross "Tt" and Tt" then you get 1/4 TT (tall), 1/2 Tt (tall), and 1/4 tt (dwarf).
427517
Something to do w/ genetics............\n\n-Mr. Happy-\n:o)
125133
Genetic predisposition among other things.
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.
278052
You need another generation to figure much out.  The ratio there is about 50:50.  What I'd assume is that you're looking at one plant being homozygous for a recessive gene (aa) and another that is heterozygous for the dominant gene (Aa).  You can't know what color causes which gene from first principles, but the 50/50 split suggests this aa/Aa.  The punnett square for that cross would give half aa and half Aa.  Those with one copy of the dominant phenotype will express the dominant phenotype, and those with the double recessive will express the recessive trait.  It is the same genotype in the plants of a given color in each summer.\n\nSince the likely cause is that the recessive gene is one that fails to make the red pigment, the odds are that white is recessive and red is dominant, but from first principles, you can't know that.  What I'd suggest was necessary is to then do two sets of crosses, independent of one another, using the plants from the second generation.  In one, cross two white plants, and in the other, cross two red plants.  If white is recessive and red dominant, we'd expect the white/white crosses to yield 100% white flowers, while the red/red crosses should yield a quarter of the plants white and 3/4 red.  If red is recessive, then we should see the white/white cross giving 3/4 white and 1/4 red, while the red/red cross should give 100% red flowers.
707484
A)\n\nThe white mouse shares the phenotype and genotype of parents\n\nMother white mouse: aa\nFather white mouse: aa\nWhite mouse: aa\n\nThe Brown mouse might be 2 things, \neither heterozygous: Aa\nor homozygous dominant: AA\n\n(in these type of exercises, when you mate something with an animal from which you have no info, he's usually homozygous. And if he wasn't the next step would not work.)\n\nTherefore, the mating of the white mouse (aa) with the brown mouse (AA) give only heterozygous (Aa) brown mice.\n\nWhy not B?\nBecause if the white mouse was heterozygous (Bb), that meant that "White" was the dominant gene, and the brown mouse had to be homozygous recessive (bb). But now you can say that they still could have a brown mice offspring, because they both carry a recessive allele (bb). Well, yes, but eventually white mice would appear in the be born (BB, Bb). This way it would never be all brown.\n\nWhy not C?\nBecause all the offspring would then be all white and never brown. It's the inverse of example A).\n\nWhy not D?\nBecause the white mouse is alive and fertile.
1124743
its a genetic problem
98654
Do refer to websites, such as Wikipedia, Vivisimo or Google; you will find countless articles, links and so on. Start with any scientific term, such as 'chromosome' and it will get you somewhere. Kimballs Biology Pages are a wonderful bet too.
251702
Genetics. And Darwins theory.
904389
First, note the word "true-breeding".  That indicates each plant is homozygous for all those traits, which means that you don't have to worry about that plant being het for any of those genes.  And it means that they only make a single kind of gamete with regard to thsoe 4 genes.\n\nTechnically, these loci might all be linked to each other, you would need F2 data to know for sure.  But you are supposed to assume that four traits are controlled by 4 genes.\n\nNow, since the parents were homozygous for all their genotypes, the offspring will all be identical, they will all be het for those genotypes.  And which ever phenotype they are, that's the dominant one for that trait.  So purple, green, terminal, and yellow are the dominant phenotypes.\n\nAnd there are 16 different gametes to be made, because there are 4 genes (presumably) and each is het in the F1.\n\nThe only way for an F2 to show all 4 recessive phenotypes would be for it to get a gamete from each parent taht also had all four recessive alleles.  So 1 in16 gametes will fit the bill, so the odds of getting that gamete from on parent is 1 in 16, the odds of getting it from the other parent is 1/16, so the odds of getting it from both is 1/256.
207195
Genetics and Environment
877803
you need to speak to a genetics counselor usually located at a medical university hospital\nthen you would need a family history, then finally a sanple of tissue for the dna to be read of each but thats not needed once a genetics counsel takes place
824039
Not sure if this will help but try the link below\n\nhttp://www.answers.com/topic/gene
116986
Sorry No!Depends on your genetics...