How can the frequecies of crossing-over be used to map chromosomes on genes?
439187
Genes that are close together will tend to stay together after a cross-over. The distance between the genes will be roughly proportional to the frequency at which they are separated by cross-overs.
732538
1RR, 2 Rr, and 1 rr   (a genotypic ratio of 1:2:1)\n\nwe just learned this in my bio class too!\nyou can use a punnett square to solve problems like this. you can look up punnett squares online, but you would set one up with an R and an r on both sides.\n\ni hope that helps!
801991
Independent Assortment:\nThe chromosomes of each parent sort independently in the production of gametes, so any combination of the chromosomes is possible in the offspring.\n\nCrossing Over:\nThe swapping of segments of a chromosomes between each other.  Crossing Over creates novel combinations of alleles for various gene loci.  These combination of alleles were not present in either parent.
1017859
Well, the equation is : (p^2) + (2pq)+(q^2)= 1....where (p^2) and (q^2) represent the frequency of homozygous alleles and (pq+qp=2pq) represent the frequency of heterozygous alleles.  \n\nThe way to use this formula is for example : " if you have a plant population that consists of 84% plants with Red flowers (RR, Rr -alleles) and 16% White flowers ( rr-alleles)," you use the formula above as follows:\n\n(q^2) = .16= White flowered plants (rr trait)\n(p^2)+ (2pq)= .84= Red flowered plants (RR, or Rr traits)\n\n\n\nNow there is another equation : p+q=1\n\nSo, take the square root of (q^2)=.16, q=.4 and then plug that into p+q=1. Now solve for p,  p should be .6.  Now  note that q=.4 and p=.6 and with that said plug it into the original equation in order to find the frequency. \n\n(p^2) +(2pq)+(q^2)=1\n\n(p^2) = (.6) (.6) = .36  or 36%  for the homozygous dominant trait (RR)\n\n(q^2)= (.4)(.4)= .16 or 16% for the homozygous recessive trait- (rr)\n\n(2pq) = (2)(.6)(.4)= .48 or 48% for the heterozygous trait - (Rr).\n\nAll of this should equall 100%.  So 36%+16%+48%=100%.\n\nSo, basically the Hardy-Weinberg equation is to see the frequency of the alleles (RR,Rr, and rr) in a population.
380018
From what I have read, it appears that it is autosomal dominant and is not X-linked as the gene most likely responsible lies on chromosome 1.\nHope this helps
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...
997040
Eye color is a polyploid trait.  the actual genes occur on a couple of different chromosomes.  Sorry.  The human genome project has it mapped out which chromosomes they are my poster is at work.
854413
to solve this, the easiest way is to find which two distances, when added together, equals the distance between another two loci (sounds confusing? wait till you see my example)\n\nfor example, 35 + 13 = 48, right? ok, now find the common locus in these two distances--> 35 = dp to b, 13 = al to dp --> common locus = dp --> plus 48 = al to b ==> dp MUST be between loci al and b\n\nsame logic applies when you look at 35 + 6 = 41, from which you can determine that locus b must be between loci dp and pr\n\n==> putting the two findings together, you then can draw out the linkage map:\n\nal <--13mu--> dp <-----35mu-----> b <-6mu-> pr
152438
#1)  Primarily by mutation.  A mutation must occur a germ cell (cells that produce sperm and eggs).  This can produce progeny with novel traits.  This is fairly rare event (one out of every billion nucleotides copied during meiosis).  Our body has ways of fixing these errors, but this is the biological basis of evolution.\n#2)  By crossing over.  During meiosis, parental chromosomes will swap pieces with each other.  For example, I used pieces of my mother's chromosome #1 and my father's chromosome #1 when I made the chromosome #1 that I passed to my daughter.  I passed a different combination of pieces to construct a chromosome #1 for my second daughter.\n#3)  By independent assortment during meiosis.  So not only did I make novel chromosomes out of pieces of my parents chromosomes, but I created a gamete using a unique combination of my mother and father's chromosomes.  For example I might have used chromosomes 1-11 of my father's (with pieces of my mother's chromosomes 1-11) and numbers 12-22 of my mother's (with pieces of my father's chromosomes 12-22).  For my second daughter it could have been totally different.  It is a unique situation each time!  No two gametes are the same.
398876
it's 50 50 every time.  genetics states that you get one chromosome from your mom (always X) and one from the dad, X or Y.  XX means girl, XY means boy. (assuming dad has even amounts of both).\n\nHence you use condition probablity assuming each occurance is a 50% chance.\n\n50% x 50% x 50% = 12.5%
1068360
dna and genetic coding of chromosomes.
450329
are you talking about genes? the complex arrangement of genes in numerous chains make up the characteristics of primates and most of all duel-reproductive organisms. out of the thousands, only a few have been mapped or pulled apart into single strands. the UCLA medical school has links to genetic mapping web sites. very cool!
718759
the frequency of a given phenotype. ie, the incidence of blonde hair, the incidence of brown hair, and the incidence of red hair. genotype frquency would be similar - incidence of BB (for brown hair) Bb (for brown hair) and bb (for blonde). Obviously, hair color is more than just one gene, but i think you get the idea.
826482
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.
696673
well certain genes on the chromosomes will be transfered to different ones which means the gamits produced in meiosis will have different genes than the parent and it will be a mixed half not a normal half
96823
1. Changes in allelic frequencies over time\n2. Change in a lineage of populations between generations
327006
You're trying to say you have a synthetic chromosome. I would check all the experiments for contamination, and check the cells for endogenous expression before jumping to that conclusion. There are a lot of hurdles to leap, as mentioned in the other posts.