Ahhh Genetics Help Please!?
830750
1. heterozygous: Punnett square is Pp x pp\n2. dominant b/c if it was recessive that would be the only type of allele the parents would pass on and all their children would have to be intensely freckled\n3. the Punnett square would be a 4x4 box with the B's and C's independant of each other. so on one side you have BC, Bc, bC, and bc with all of those showing up on the other side as well b/c the female was the same genotype as the male.  Obviously the black color is dominant b'c the male is heterozygous and black so all heterozygous offspring and all homozygous dominant offspring will be black as well.  the question gets a little fuzzy here though, i think you may have left some out b/c we don't know which of the two genes controls color or if they both do how the interaction between them works.  I'd be glad to help further if you have any more information to supply.
867443
OK people...here's the lesson for the day...this is an actual bonified case of a genetic trait that is actually inherited through birth....you will notice that it's not some sort of sickness or illness that we all like to pass off as genetic, but it's an actual physiological trait that can be attributed to a genealogical factor...ooooooohhh...amazing isn't it?.......it's just  pigmentation...nothing to worry about...actually kind of unique and cool...a genetic trait stamp like a birthmark, high intelligence or big feet etc...this is what we inherit genetically from our lineage...v cool...
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).
1147347
Not possible. Genetics give you what you have, be thankful for it.
1024331
I apologize but all that i can come up with is one.\n\nYou could raise fruit flies (Drosphila) and relate them to Gentics. i.e. punnet sqaures, Gregory Mendel, common characteristics, principle of dominance and recessiveness, etc.\n\nif you would like me to go further into detail about this project, please e-mail me at \n\nthe_lonley_kender@yahoo.com or send me an IM on Yahoo! at the_lonley_kender or on AIM at  thelonleykender
50708
This is really all about genetics. Each person has differing genes and inherit family traits!
292443
Your genetics..dna..you know..the whole human biology thing..
1096018
There is a WONDERFUL site that you can input both parents eye colors (and their parents) and it will tell you the percentages of what you offsprings eyes could be!\n\nhttp://www.thetech.org/genetics/\n\nDown near the bottom left it will have the link for you to launch the application to do the test.
217979
heredity and makeup!!!
844582
Genetics and environment and stuff... oh and gravity! Unless if the tree is on a steep hill.  :-)
410584
Are the genes linked? I don't have my book with me here.\n\nIf not, mate a male with one of them. Take the male progeny from that mating and cross with the other mutant. Select hermaphrodites from the period of egg laying when the sex ratio is 1:1 (say the second 8-hour period after they're together). Allow these hermaphrodites to self and select double mutants from the progeny. I think this should do it, but you'd better draw it out.\n\nIf they're linked I think it will just be harder to find double mutant progeny in the final stage, again draw it out to check.
779422
heterozygous - two different genes\nR=round\nW=wrinkled\nA)round - RW, wrinkled - WW\nB)RW, RW, WW, WW\nC)50% round with RW genotype, 50% wrinkled with WW genotype\nD)25% round with RR, 50% round with RW, 25% wrinkled with WW
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.
735604
the cross is going to be Hhdd x hhDd you can get this far because we know the parent of the ones having kids.  the male's mother has to be hh because the straight is recessive and so the male could only get an h from her. since he is wavy his genotype must be Hh and dd because red is recessive.  the female's father is dd because he is red headed so she could only get a d from him and being dark herself must be Dd and straight is hh.\n\nhope that wasn't to confusing.\n\np.s. it is 1:1:1:1 dark wavy: red wavy: dark straight: red straight
99271
genetics and hormones
842615
Your question requires more knowledge of science and genetics than I have.  However, since you haven't received much help with it I thought I would give you a couple of sites.  Try these - they may be of some help to you.  Sorry I couldn't help more.
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.