He marries a woman who is also heterozygous for dimples and a carrier for PKU disease.\nDetermine the following\nGenotype of man\nGenotype of woman\nPossible gametes of the man\nPossible gametes of the woman\nPossible phenotypes of their children and the ratio of each
215010
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.
390191
The majority of miscarriages is spontaneous abortion to which her body is aware that there is a genetical defect to the fetus and whcih case it desides to remove it from the body.   What she and her husband  need to do is go to a genetic counslor and they will run genetical tests.  The fact that she had a child that died of a rare heart disease is a HUGE signal.  Probably the doctors aren't thinking logially about his situation and not really seeing the overall picture.
413968
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...
537183
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
631235
Yes, He´s married to a brunette model. Together have a foundation for helping poor children.\nThey been married about 4 years.\nThey have no kids.
477039
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.
364748
It's called genetics.\n\nCoach
436163
As I recall, the map unit is a 1% chance that two genes will sort independently. So, 13% will be al/+ phenotype, 13% dp/+, 37% are al/dp, and the other 37% are wildtype. I think... definitely check this before you accept it as true.
49203
100% of sons will be colorblind\n100% of daughters will be carriers\n\nYes this could stand up in court because there is no chance that the daughter recieved one of the recessive alleles from her father (because he would have to express the trait if he did).\n\nHowever, the man can still be her father, if a random mutation occured, x inactivation occured on the chromosome containing the dominant allele, chimerism, or perhaps an error in another gene caused the color-blindness...\n\nMost likely though, the man would not be her father. A deduction would not be made only on the fact that the daughter is color blind and the father isn't. A paternity test would need to be done.\n\nAnd I don't know what the hell kittybriton is talking about. People shouldn't attempt questions they don't know the answer to.
470301
he IS different,and a good start towards a unique expression of love in friendship.if he can prove every day that his love for her is strong he does it because he wants to.if he got married to her to learn the same lesson he would be obligated to love her.there is a difference.he can date my daughter
297269
These website explains it all...I hope it helps.\n\nhttp://home.earthlink.net/~rickpen16/catholicmarriages/index.html
295197
http://www.law.cornell.edu/topics/Table_Marriage.htm
398609
A)\n\nAs far as pod size, at least one parent must be homozygous for H, since all the offspring have one pod.  The other parent coudl have any genotype for that gene\n\nThe leaf shape is showing the 3:1 ratio characteristic of a dihybrind cross.  So both parents must be het at the leaf locus.\n\nB) is similar, since all the offspring have 3 pods, the parents must be homozygous for that recessive allele.  The other parent coudl have any genotype for that gene.\n\nThe 3:1 ratio of leaf shape shows that the parents must again both be het for the leaf shape locus.\n\nC)  If all the offspring show the dominant pheontype, one parent must be homozygous for the pod number, and one must be homozygous for the leaf shape.  Either the same parent is homozygous for both, or each is homozygous for one trait.  Again, if one parent is homozygous for a gene, the other parent coudl have any genotype at that locus.\n\nD)  The ratio of one-pod/3-pod is again, 3:1, so both parents must be het for that gene.  Since half the offspring have the dominant leaf shape, and half have the recessive leaf shape, that means that one parent must be homozygous for the recessive allel, and the other must be het.\n\ne) The ratio of one-pod/3-pod is closest to 3:1, so both parents must be het for that gene.  again, the ration is leaf shapes is closest to 1:1, so one parent must be homozygous recessive, the other het.\n\nBut the ratios aren't quite right, there's something non-Mendelian affecting the offspring here.
48180
bud_leaf_... answered it for me, but generally, if you put all of the generation on a pedigree (a family tree that lists similar traits), the carrier always passes the genes on to the offsspring. because the carrier gences are both recessive, if the person mates with another carrier, or a victim the offspring WILL get the disease, unless if it is genetically altered. if the carrier mates it a non carrier with  dominant gene (example, Tt) the offspring will be a carrier. If the mate does not have the disease with both dominant(TT) the offspring will either not be a carrier, or be a carrier. but with the case of hemophelia,it is a sex-linked disease, so the males will be more likely to get it because they have ONLY ONE X CHROMOSOME, WHILE FEMALES HAVE TWO, WHERE THE SECOND  X ACTS AS A BACKUP. THe females will be carriers, while, sadly, the males will be victims. That is a example of a SEX-LINKED DISEASE. It really depends on what type the disease is, though (sex-linked, or just genetic)
49061
It has autosomal dominant inheritance pattern.  So, assuming your spouse does not have neurofibromatosis and is recessive (aa), and you are either homozygous or heterozygous (AA or Aa).  Your child has a 100% chance of receiving it if you are homozygous and a 50% chance of receiving it if you are heterozygous.
550386
If the P phenotype is\nPPSS x ppss\nThen all of F1 is PpSs\nAll would be Purple Smooth\n\nb F1 crossed with ppss: PpSs x ppss\nthe ratio would be 1:1 for PpSs and ppss\nor, Purple Smooth to Yellow Wrinkled.\n\nYou need to learn to use the Punnet square!
495006
B. a heterozygous female\n\nHeterozygous means that she has both a dominant and recessive copy of the gene, so the recessive gene isn't expressed. Being a carrier means that she can have a daughter that is a carrier(or a daughter with the recessive gene expressed if the father has the recessive gene), a daughter that doesn't have the gene at all (if the father doesn't have the gene), or a son that may or may not have that recessive gene expressed(50% to 50%), depending onn of they get the dominant or recessive gene.\nOnce again, feel free to contact me if you need help.