pick a physical trait that is determinedby two alleles?
823120
Is it a human, plant, or animal?  For plant I would have to say that traits of pea plants would work.  Either tall or short, purple or white flowers, yellow or green seeds, and wrinkled or smooth seeds.  This is one of the reasons Gregor Mendel (the father of modern genetics) used pea plants, because they have only two outcomes for any given trait.
33435
A condition in which both alleles of a gene pair in a heterozygote are fully expressed, with neither one being dominant or recessive to the other.
188037
get different genetics
149630
1:2:1\n1 homozygous dominant\n2 heterozygous (the hybrid)\n1 homozygous recessive
515090
dominant e.g. Aa, AA - Recessive -aa
490031
Ok, here's how it works. It seems that you need genotypes. A homozygous trait is a trait that is both dominant and recessive. Therefore, the organism possesses the dominant trait. A heterozygous genotype is always dominant (dominant is dominant over recessive). For example, brown hair (B) is dominant over blond hair (b). a heterozygous person will be Hh, and will have brown hair even though both alleles are present.
507695
Heredity. You got the genes for that trait from your ancestors.
694878
1) Homozygous at that locus.\n\n2) The human tailbone.  It is what remains of a tail that we no longer need.
78968
A Punnett Square, for those unaware, is the box used for determing allel (trait) combination in genetics.  A 2x2 box is the most commonly seen.  I'll use this as my example:\nDraw a square. Inside, draw a plus sign, so that you get 4 identical squares inside the larger square.\nNow, we need two 'breeders.'  Let's say Dad has brown hair, with resessive blond; Bb.  Mom is the same way; Bb.\nAt the top of the square, write Dad's alleles (the big and little 'b')...one per column.  On the left side, place Mom's alleles...one per row.\nNow, multiply.\nIn the four squares, you should end up with the following:\n1 BB\n2 Bb\n1 bb\n\nWhat does this mean?\nIt means that if Mom and Dad have 4 kids, the odds are they will have 3 with brown hair and one blond.  (Hopefully Dad understands recessive alleles, or Mom might have some explaning to do.)\n\nWill every 4th kid be blond?  No.  But, each kid will have a 1 in 4 chance of being blond, and 3 in 4 chance of being brunette.
796794
Do I only pick one or can I choose all of the above?
903177
Take each of the three traits seperately and compare to expected mendalian ratios. (Testcross means the male was completely recessive. If the female is a heterozygote then the expected ratio is 1:1, if homozygous dominant 1:0, if homozygous recessive 0:1). \n\nFor example\n243+241+15+10=509 black (1000/509=2)\n235+226+12+18=491 not black (2)\n==>approximately 1:1\nThe female had to be Bb because both types are visable.\n\nscarlet =491\nnot scarlet=509\n==>1:1, STst\n\nhooked=514\nnot hooked=486\n==>1:1, HKhk\n\nFemale is B b, ST st, HK hk\n\nFor linkage/mapping distance, take the total progeny that do not look like either parent and had to come from a crossover event and divide by the total number of progeny. Assume that the combinations you see most often are parental  (243+241+235+226). Work with two traits at a time. Significant deviation from mendalian ratios indicates linkage.\n\nb and st\nBST(parental)=235+18=253\nbst(parental)=243+10=253\nBst=226+12=238\nbST=241+15=256\nparental=506, recombination=494...approximately1:1 no linkage \n\nb and hk\nBHK=12+18\nbhk=15+10\nBhk(parental)=235+226\nbHK(parental)=243+241\nparental=945, recombination=55   \n55/1000=.055(100)=5.5cM between b and hk\n\nYour map distance is cM.
293753
Your friend is correct. I just studied this in school. Dark hair always dominates light hair. And the allele for green eyes overpowers the allele for blue eyes.
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
52176
Genetic predisposition.
633663
the portion of genes that determines traits are called alleles (sp).  If you want to find out which ones I am sure they are mapped out if you use a search engine\n\nAt one time scientists thought that a single gene pair, in a dominant/recessive inheritance pattern, controlled human eye color. The allele for brown eyes was considered dominant over the allele for blue eyes. The genetic basis for eye color is actually far more complex. At the present, three gene pairs controlling human eye color are known. Two of the gene pairs occur on chromosome pair 15 and one occurs on chromosome pair 19. The bey 2 gene, on chromosome 15, has a brown and a blue allele. A second gene, located on chromosome 19 (the gey gene) has a blue and a green allele. A third gene, bey 1, located on chromosome 15, is a central brown eye color gene. \n\nsaved you the trouble
1115185
25% chance of being O\n75% chance of being A\nonly if both parents are heterozygous\n\neven if one is homozygous no chance of getting "O"
633573
an essential and distinguishing attribute of something or someone