What is the probability that a color-blind woman and a man with normal vision will have a color-blind daughter
493494
x x -> color-blind female\nX Y -> normal male\n----\nx Y ; x X -> offsprings\n\ngiven the conditions, the chances are:\n0%  to have a color-blind daughter\n\n50% to have normal but carrier children\n50% to have color-blind children\n\n0% normal sons\n100% color-blind sons\n0% color-blind daughters\n100% normal but carrier daughters\n\nsince red-green color-blindness is a recessive trait, the mother would have the trait in both her x chromosomes. the father had normal vision so he would have the normal gene which is dominant for the trait. \n\nsince males get their x chromosome from their mother and the y chromosomes from their father, all sons borne would be color-blind because whatever the x chromosome they receive from their mother would have the gene for color-blindness. though the trait is recessive, this one gene would be enough to manifest itself because males only have one x chromosome. \n\nsince females get an x chromosome from each parent, all daughters would get a normal x chromosome from their father and an affected x chromosome from their mother. and because females have two x chromosomes and color-blindness is recessive, they would need to have two affected genes for the trait to be manifested thus all daughters borne would have normal vision. the dominant gene they received from their father would mask or overpower the affected gene they inherited from their mother. however, these daughters are also called carriers because the affected x chromosome they received from their mother can be passed on to their own sons wherein the color-blindness trait can then be manifested.
759196
On your side, the probability to have brown hair is one. Your fiancee have 0.75 chance of both parent had brown hair, I guessed.\nSo your baby will have >.75 probability to have brown hair.
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.
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
799354
In animals? I don't know, how that works with animals.\nI suppose it depends on the species.\nWith people, you are more likely to get a blue eyed child\nbecause the blue eyes are on a dominant gene.\nI could make you a chart, but I won't.\nThe order of dominance goes as such:\nBrown\nBlue\nhazel\ngreen\ngray\nviolet.\nI'm not positive on the order of the last two though.\nBut....you could have brown eyes, and your wife might have\nblue, and you could have a child with hazel, because you\nboth carry the gene for hazel eyes. Get it?\n\nGenes don't mix up like paints, either.\nSo you if neither of you carry the gene for hazel eyes,\nyou will not have a child with hazel eyes, if say, you have\nbrown eyes and your spouse has green. You would\nmost likely get a child with brown eyes.\nBut there are other factors to consider.\nFor example:\nSay you have blue eyes and your wife has green eyes.\nYour genes are Blue/violet and hers are green/violet.\nYou don't carry a gene for green eyes and she doesn't carry a \ngene for blue eyes, so your child would have violet eyes.\nThat is how genetics basically works. More or less.\nAnd that is why we still have recessive genes.
264930
The answer to a) is exactly the same as the answer to b), since all children should be equal. The answer to a) & b) is 0.25*0.25=0.125\n\nc) The answer is 0.25*0.25*0.75*0.75*0.75 * 10 = 135/512 We multiply 0.25*0.25*0.75*0.75*0.75 to get the probability of first 2 being blood type O and the others not. We multiply by 10 because there are 10 possible combinations of this. (Such as the last 2 being blood type 0 and the first 3 not...etc)\n\nd) Exactly none: 0.75^5 = 243/1024\nExactly one: 0.25 * 0.75^4 * 5 = 405/1024\nExactly two: 135/512 as above\nExactly three: 0.25^3 * 0.75^2 * 10 = 45/512\nExactly four: 0.25^4 * 0.75 * 5 = 15/1024\nExactly five: 0.25^5 = 1/1024
709683
Color blindness is a sex-linked thing (giggles from the 5th grader)\nThe gene is on the X chromosome.  Men get one X and one Y chromosome.  If their X has the gene for color blindness, there is no matching gene on the Y chromosome to "cover" for the defective gene.  The color blindness is expressed.\n\nWomen, however, are blessed with 2 X chromosomes.  If they get a gene for color blindness, the other X usually has a normal gene that can "do the work" that the abnormal gene cannot.  Color blindness is not expressed, and she is a carrier of the gene.\n\nWomen can be color blind if their color blind Dad marries a carrier Mom, and she gets 2 defective genes.\n\nHow's that?
18336
Try Wikipedia.\n\nrizzblizz:\n\nCan it be proven that only men can be color blind? Do you know any women who are color blind?
1115447
Most certainly not! Where did you hear that?!\n\nColor-blindness is a genetic trait that is carried on the X chromosome. It is purely hereditary.
355647
It depends on what genes are passed down.  I have brown eyes and my husband has blue eyes...our daughter has green eyes.  We both must carry a green eye gene that she got.
442637
Yes, if they are both carriers of a recessive light eye color gene, then there is a 25% chance (1 in four) that a child will have light eyes.
940033
Let's take the Blue eyes - brown eyes example.  Brown is the dominant trait, so we'll use the capital B letter.  Blue is the recessive trait, so we'll use the b letter.  A person with blue eyes has the genotype bb (got a b from dad and a b from mom).  A person with brown eyes can be homozygous BB (got B from dad and B from mom) or heterozygous Bb (got B from one parent and b from the other parent).\nA first generation usually starts with pure breeds, so both homozygous.  We'll have BB and bb as the starting parents.  Those parents produce gametes in their gonads.  Let's say that mom has blue eyes and dad has brown eyes.  All her eggs will have the b gene as all dad's spermatozoids will have the B gene.  If you want to make the Punnett square, you'll find that all babies will have 100% probability to be Bb\n____|__B__|__B__\n__b_|_Bb__|_Bb__\n__b_|_Bb__|_Bb__\nThe second generation is made by taking two of those offspring and crossing them.  First, you have to find out what alleles will be in the gametes.  Half of the eggs will be b and half will be B.  Also, half of the spermatozoids will be B and half will be b.  In the Punnett square you'll have.\n____|__B__|__b__\n__B_|_BB_|_Bb_\n__b_|_Bb__|_bb__\nHere, the babies will have 25% chance to have blue eyes (bb) and 75% chance to have brown eyes (25% homozygotous BB and 50% heterozygotous Bb).
806831
All your responders so far have assumed that the next child's gender is independent of the prior children's genders. That is a correct assumption.\n\nThey have also assumed that the probability of having a boy is 50%. This is not quite true. Population-wide, the ratio of baby boys to baby girls is approximately 105 to 100, so the percentage is actually a bit over 51%. \n\nThis is also based on the assumption that the probabilities do not vary from one couple to another. You may have read descriptions of techniques for increasing the probability of having a girl, or of having a boy. And if so, you know that the things you have to do (I won't get too graphic here) are things that some couples are likely to be doing already (because of the geometry of their bodies, or the chemistry of their bodies, or because of their technique). So I suspect (and this is only my opinion) that the probability of having a child of one gender or the other varies from couple to couple.\n\nThat last-mentioned suspicion could probably be tested by determining the relative frequency of families with different combinations of boys and girls. Do siblings of the same sex tend to occur more frequently than if each birth were a 51/49 proposition? If so, then there is likely a variation from couple to couple. And if they occur LESS frequently, then we REALLY have a conundrum, and I have NO idea what it would mean.
124648
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.
755941
4/16 or 25%\n\nThe follow table are the possibilities\n\nGirl    0   1   2   3   4\nBoy  4   3   2   1   0\n\nThe probability of having all boys is easy to figure out; it is 1/16. That is calculated by multiplying 1/2x1/2x1/2x1/2. The probability of having all girls is the same. 1/16.  \n\nThe probability of 2 boys and 2 girls is calculated by noting that 2 circumstances exist. In one the first 2 children are the same sex; the last 2 must be the opposite sex. Therefore you multiply  1/2x1/2x1/2 and get 1/8 (or 2/16) of this occurring. In the chance that  first 2 are equally split one boy one girl then the third child can be either sex and only the last one matters leading to another 1/2 x 1/2 = 4/16. For  a total of 6/16\n\nTherefore the remaining probably of 3 boys and 1 girl  or the alternative of 1 boy and 3 girls is the difference namely 8/16. \n\nSince it is equally probable to get 3B/1G or 1B/3G, the probability of 1 girl is half of the 8/16, namely 4/16
164659
Red/green and blue color blindness appear to be located on at least two different gene locations. The majority of affected individuals are males. Females are carriers but are not normally affected. This indicates that the X chromosome is one of the locations for color blindness. Male offspring of females who carry the altered gene have a 50 percent chance of being color-blind. The rare female that has red/green color blindness, or rarer still, blue color blindness, indicates there is an involvement of another gene. As of 2004, the location of this gene was not yet identified.\n\nHe's right about the 7th chromosome,though.
821644
Is sex-linked. Sex linkage by definition involves all traits located on the sex chromosomes. Sex linkage doesn't mean that only males can be colour-blind, but it means that the sex of a child influences its chance to have that trait.\n\n(BTW "the" allele for colour blindness is crap... a normal human actually has 3 colour vision genes, two of these - "red" and "green" - on the X chromosome; meaning several types of colour blindness. A red or green can simply be left out when X chromosomes exchange DNA in egg cell formation, or you can have defective versions of either gene. As the third gene, that which codes for the blue receptor pigment, is located on another chromosome, there exists non-sex-linked colour blindness as well.)