probability of have a child with these four parental genotyoes: FFxFF?
405198
i would suggest consulting with a genetic counselor to discuss your particular situation. see info below...
544769
yes, but the chances are low. The defective gene for colorblindness is on the X chromosome. A man has one copy of the X chromosome, so if he has that copy he'll be colorblind\n\nHowever a girl has 2 copies, and as long as they have one functioning copy, they won't be.\nHowever, if your father is colorblind, and your mother is a carrier for colorblindness, your chances as a girl for being colorblind is 50%
250838
Good God man you are on the net, you have more information than you know what to do with right at your finger tips. Go to Yahoo's main page and type genetics into the search bar.
692724
it is possible that half of the chromosomes in an ovum are always paternal, but that is not always the case... \n\nbecause in an organism with 56 chromosomes, even if crossing-overs are not taken in consideration, there are still 268,435,456 possible combinations of the gametes. there's a chance that all chromosomes in the ovum would be paternal. there's also a chance that all chromosomes in the ovum would be maternal. and there's also chances of any other combinations in between like more maternal than paternal chromosomes, or more paternal than maternal chromosomes, or equal numbers of paternal and maternal chromosomes.
568189
Well, I'm kind of worried that you mentioned your job as an Assistant Professor in Veterinary College, but you don't know how to calculate an inbreeding coefficient.\n\nAnyway, second cousin is usually unrelated enough. But go back in your family trees, and I'll bet you have many more relatives in common. Should that be the case, your children have a slightly higher chance of having birth defects. \n\nSupposedly, if two unrelated people in a population have an X percent chance of having a defective child, then first cousins have a 8/5X (1.6X) chance of having a defective child. Such defects occur in the low single digits anyway.
590754
6/16=37.5%\n\nSixteen different ways to have 4 children (excluding hermaphrodites.)\n\n1. b b b g  \n2. b b g b\n3. b g b b\n4. g b b b\n5. g g g b\n6. g g b g\n7. g b g g\n8. b g g g\n9. b b b b\n10. g g g g\n11. b b g g *\n12. b g b g *\n13. b g g b * \n14. g g b b *\n15. g b g b * \n16. g b b g *
334742
The normal form of dimples is a dominant trait controlled by a single allele.  Your dad clearly is heterozygous for dimples and one of your grandparents had them (and was also heterozygous).  When one parent is heterozygous for dimples, the probability of each child having dimples is 50%.  The probabilities that out of six kids the numbers of kids with dimples are as follows:\n\n0 kids with dimples: 1.6%\n1 kid with dimples: 9.4%\n2 kids with dimples: 23.4%\n3 kids with dimples: 31.2%\n4 kids with dimples: 23.4%\n5 kids with dimples: 9.4%\n6 kids with dimples: 1.6%
168198
Actually, the odds are not 50:50.  The chances are 51.4 male to 48.6 female.  So the odds are slighly in favor of a boy.\n\nhttp://www.ourstolenfuture.org/NewScience/reproduction/sexratio/sexratio.htm\n\nAs people have said, the births are not connected so outcome of each event is independent.  So each birth event would seem to have the same likelihood as any other birth event.\n\nHowever, there may be individual factors which are not accounted for in this study.  We have several friends who have all girls.  3 girls in one family, 4 in another. While statistically plausible, and I haven't found any specific studies on this, I'm betting that there are heredity and environmental factors that are involved. Certainly their's some interest in why the birth rates of certain countries deviate significantly from the norm.\n\nHere's another website that talks about ways of modifying your behavior that might increase the odds of a boy or a girl.  It's not scientific but in the quote: "Dr. Moore also notes in his book that selecting a child's gender shouldn?t be taken too seriously, but he thinks using his methods make trying that much more fun."\n\nhttp://preconception.com/resources/articles/tochance.htm\n\nOf course, if you choose to adopt and make a proactive choice of gender (boy or girl), you have a much strongest choice of influencing the odds.
740256
Since only one of you carries the Thalassemia gene, your child has a 25% chance of getting thalassemia minor, regardless of its sex. It's usually unnoticed at birth, but becomes apparent in the first year of life (child becomes anemic).\n\nThe hemophilia gene is carried on the X chromosome. For a father who has hemophilia, all sons will be unaffected and all daughters will have the condition (since the daughters will inherit the father's X chromosome for sure).
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.
488747
Well, I have this friend that had 4 kids and lost custody cause she was a crack head & also had 50lbs too much crack, anyway she might be a good choice and she would split and leave u the kid. I would have to say there would be no gaurantee on what the babies health would be though. But thats the chance you take , along with the mixture of your DNA ! She is a nice Religious girl , just a little ate up and out there ( Somewhere, Just don't know where). Oh by the way , watch those blond jokes u leave ladies cause u look like a natural platinum blond now !
796672
OK, draw yourself up a nice big checkerboard with 16 squares! Down the left hand side, write beside each square the 4 possible gametes produced by 1 parent - for parents with the genotypes you have given above, the gametes produced are: AB, Ab, aB, ab - this is because each egg or sperm produced will have only half the number of genes of the parent (so ONE copy instead of 2) so that when the egg and sperm meet, they will have 2 copies of each and not more!!! So the gametes will have ONE copy of the 'A' gene (which may be either the dominant 'A' allele or recessive 'a' allele - half the gametes will get one type of allele and half the other) and the same goes for the 'B' gene, to produce the possible gamete types above. Then, along the top of your little checkerboard diagram, write above each of the 4 squares the 4 possible gamete types for the SECOND parent - which in your case, are the same as the first!\n\nNow - you want to fuse your eggs and sperm to produce offspring! Do this by combining the gamete genotype of 1 parent with that of the other in the box where they meet up - eg if your very top left-hand square has the genotype 'AB' above it, and 'AB' to the left of it, then the genotype of the offspring produced will be AABB - homozygous for both! And the square 2 down from the top, in most left column, might be combining the gametes 'Ab' with AB - producing offspring AABb, which is homozygous for gene 'A' but not for gene 'B'.... and so on and so forth for the rest of the squares!\nI WOULD RECOMMEND DRAWING UP THIS SQUARE TO SHOW YOUR WORKING - YOUR TEACHER WILL BE WANTING TO SEE THAT!\ndouble homozygotes are AABB, AAbb, aaBB, or aabb...
331049
yes.  If both parents are heterozygous for brown eyes (i.e., the genotype would be Bb) then the chances of having a blue eyed child are 1 in 4.  The chances of a brown eyed child are 3 in 4, with 2 of these 3 being heterozygous for brown eyes and the other 1 being homozygous for brown eyes.
962212
1. Pp (heterozygous)\n2. Both parents are heterozygous (Ff)\nSince both parents show intense freckles, it must be a dominant trait. In order to have non-freck. children, they both must be heterozygous. Ff x Ff ----> 3:1 ratio expected; frec. to non-freck.\n3a) All four blood groups are possible\n  b)25% possibility for each\n\nFor#3, do a Punnett square-   A i   x   B i
968185
First the probability that a family of 4 is all girls is:\n1/16 (and the chance that they are not is 15/16).\n\nRather than figuring the chance of at least one family having this arrangement, figure the probability that *none* of the families do.  Then subtract from 1.\n\nThe answer will be 1 - (15/16)^10\nor about 47.6%\n\nIt's not a simple multiplication by n... if it were, then with 16 families you should have a 100% chance.  And with 32 families you would have a 200% chance.  That logic would be correct if you couldn't repeat a combination, but you can.  Just because family A has four boys, for example, doesn't mean family B can't also have four boys.\n\nLabel the 10 families as A, B, C, D, ... , I, J.  The probability that A has four girls is 1/16.  The probability that B has four girls is 1/16, etc.\n\nThe chance that *none* of them have four girls is 15/16 x 15/16 x 15/16 x ... x 15/16.  This is (15/16)^10 or about 52.4%.  Therefore the chance that at least one of them has four girls is 100 - 52.4 or 47.6%
849665
The first two answers are the same answer ... 1/16 = 0.0625 ... and both are incorrect.  (no_life computed the odds that the first child only is the male ... but forgot about the cases where the second child is the only male, or the third child, or the fourth.  And Steph computed the odds that *all four* are boys ... in which case, yes it is 0.5*0.5*0.5*0.5.)\n\nThe correct answer is 1/4  = 0.25 = 25%.\n\nHow many possible combinations can you have?\n2x2x2x2 = 16\n\nHow many of them are "one boy + 3 girls"?\n4 ... the first child is a boy (and the rest girls), or the second child, or the third, or the fourth.\n\nSo the odds are 4/16 = 1/4 or "1 in 4" (or "4 to 1 against").
401681
to what? have four children with four different blood types?\n\nYes, theoretically. Depending on the full genotype of the mother and father, but if they are both heterozygous (i.e., if Michelle's genotype is AO, and Mike's genotype is BO), then the possibilities for children include:\nAB\nAO\nOB\nOO\n\nIf they are both homozygous, however (ie, AA and BB), all their children will be AB. \n\nAnd then of course there are various combinations if Michelle is AO and Mike is BB or Michelle is AA while Mike is BO. A punnet square will help you out there.