Phenotype frequencies in the ABO blood system are as follows.\nplease note, probability(Prob pi) are given in brackets \n\nPhenotype   O(0.441)  A(0.417)   B(0.104)  AB(0.039)\n \n\nIf an individual is chosen at random, what is the probability that his blood group will match the blood group of a crime stain?  (Use the law of total probability.)\n(PM = Probability of a match = Prob an innocent person would have the blood type which matches that at  a crime stain).
217311
P(Match) = P(Person's BT is the same as BT of stain)\n= P(Person is O and Stain is O OR Person is A and Stain is A OR Person is B and Stain is B OR Person is AB and Stain is AB)\nAll of these events are mutually exclusive or disjoint, so\n= P(Person is O and Stain is O) + P(Person is A and Stain is A) + P(Person is B and Stain is B) + P(Person is AB and Stain is AB)\nSince the person was chosen at random, his/her blood type is independent of the blood type of the stain.\n= P(Person is O)P(Stain is O) + P(Person is A)P(Stain is A) + P(Person is B)P(Stain is B) + P(Person is AB)P(Stain is AB)\n= 0.441^2 + 0.417^2 + 0.104^2 + 0.039^2\n= I'm sure you can do this.
633285
Can a man have, let's say, B+? Yes.\nCan a woman have, let's say, B+? Yes.\nCan they get married? Yes.\n\nExact same blood type will not be a problem. Only way for problem is if baby is + and mom is -. Only way for that to happen is if Dad is + and mom is -.
301933
P(selecting a person at random and he has the symptom and is detected by the procedure)\n\n= P(patient having it) and P(detecting it using the procedure)\n\n= (0.5/100)*(98/100)\n= 0.0049\n\nBut if the person is not randomly selected and you are sure s/he has the problem and its just a matter of detecting it with the procedure, then Probability = (98/100) = 0.98
93124
Codegirl gave a great answer, although she left out the possibility of your husband being AB and giving an A to make your child AA.  All I'd add is that you have to understand the difference between phenotype (which is the expressed form) versus the genotype (with is the combination of genes).  When you and your baby were typed, they only look at the phenotype, not the genotype.\n\nA person tested as A can be genotypically AA (A from both parents) or AO (A from one parent; O, which is recessive, from the other).\nA person tested as B can likewise be BB (B from both parents) or BO (B from one parent, O from the other).\nA person tested as AB must be genotypically AB, getting one from each parent.\nA person taht is tested as O must likewise be OO, getting O from each parent.\n\nA parent gives one of their two alleles to their child.  If you are AB, you could give your child either A or B.  Since their are two possible combinations of alleles that make A (AA and AO), the father might have contributed either an A or an O.  So the only thing known for certain about the father's genotype is that he is NOT BB.  He could be either AA, AB, or AO (and contributing an A to make your child genotypically AA), or he could be either BO, OO or AO (and contributed an O to make your child genotypically AO).  But phenotypically, the father could be anything: A, B, AB, or O.
6168
I'm not familiar with the algorithm, but I'll take a stab at the question. Reconstructing phylogenies genetically has to do with comparing DNA sequences. The mathmatical analysis has to do with determining what is the likelihood that two sequences of DNA can be the same by chance alone. If the odds are that they would be unlikely to be the same by random chance, they would be considered to have had a common ancestor.\n\nYou can think about the probabilities here. If you flip a coin twice, the chances of getting heads both times is pretty good. If you flip it a hundred times, the chance of getting heads everytime is diminishingly small. \n\nThe algorithm probably compares DNA sequences and determines how likely it is that they are related, and how closely related. I don't do this kind of work, so someone else may know specifically, but this would be the idea.
58434
first you will need the blood sample, anti-A antibody and anti-B antibody. \nplace a drop of anti-A and anti-B antibody on separate slides.\nsmall amount of blood is taken from the patient's 4th finger. the blood sample is then mixed with the droplets of the antibodies.\nafter a few minutes, observe any signs of agglutination on the slides. if agglutination occurs on slide with anti-A antibody, the blood contains antigen A (blood type A). if agglutination occurs on slide with anti-B antibody, the blood contains antigen B (blood type B). if there's agglutination on both slides, the person has blood type AB. if there's no agglutination, the person has blood type O.
479141
The Landsteiner system works like this, there are A and B antigens. Blood group AB means anyone who has this group (not many do, but I'm one!) has both anti A and anti B on the surface of the red blood cells, but neither in the serum.\nGroup A will have anti A on the red blood cells and anti B in the serum, whereas group B will have anti B on the red blood cells and anti A in the serum. \nBlood group O (usually the commonest) has no antigens on the red blood cells , but both anti A and anti B in the serum.\nYou may have heard of rhesus negative and positive? That refers to the D antigen and there are Cs and Es and loads of other ways to group blood, but the ABO Landsteiner system along with anti D are the most relevant and best known.\n(hopes she's remembered her stuff from the path labs donkey's years ago!)
66035
They can't be the parents.  The baby is AB, so it has an A allele and a B allele.  Since bloodtypes are codominant, a person will have the B antigen if and only if they have the B allele (so they'll be either B or AB).  The parents are O and A, so neither one of them has a B allele.  The baby had to get it from someone else, so the father isn't the real dad.  Either they took home the wrong baby, or the mom was cheating with someone whose bloodtype was either B or AB.
6741
Human MN Blood Group\nThe human MN blood group shows co-dominant inheritance at a single gene with two alleles. This makes the system ideal for population genetic studies in humans since all three genotypes can be identified and allele frequencies can be calculated directly. \nSince specific anti-M and anti-N antibodies are available, an individual can be blood typed as indicated in the following table:\n\n\nBlood\nType Anti-M\nReaction Anti-N\nReaction Genotype \nM + - MM \nN - + NN \nMN + + MN \n\nThe calculation of the frequency of either the M or N allele in a particular population is done by the simple relationship as follows:\n\n\np(M) = (# or frequency of M types + half the # or frequency of MN types) / total\nThe table below lists the observed blood type frequencies in several, different populations along with the calculated values of the M and N alleles and the expected frequencies of the three blood groups. The agreement between observed and expected frequencies is quite good.\n\n\nPopulation Blood Type\nM Blood Type\nMN Blood Type\nN p(M) q(N)  * \nAustralian Aborigine (obs.) 0.0301 0.2959 0.674 0.178 0.822 \nAustralian Aborigine (calc.) 0.0317 0.2936 0.6757   1.03 \nUS Native (obs.) 0.600 0.3512 0.0488 0.776 0.224 \nUS Native (calc.) 0.6002 0.3476 0.0502   0.77 \nUS Black (obs.) 0.2842 0.4964 0.2194 0.532 0.468 \nUS Black (calc.) 0.2830 0.4980 0.2180   0.19 \nUS White (obs.) 0.2916 0.4938 0.2146 0.5385 0.4615 \nUS White (calc.) 0.290 0.497 0.213   0.41 \nEskimo (obs.) 0.4673 0.4314 0.1013 0.683 0.317 \nEskimo (calc.) 0.4665 0.4330 0.1005   0.14 \nLapp (obs.) 0.287 0.521 0.192 0.548 0.452 \nLapp (calc.) 0.300 0.495 0.204   2.63 \nGerman (obs.) 0.3062 0.4913 0.2025 0.552 0.448 \nGerman (calc.) 0.3047 0.4946 0.2007   0.46
545904
Okay, keep in mind that this problem is asking the probablility that AT LEAST 8 were married - that means the probability that 8 were married plus the probability that 9 were married plus the probablility that 10 were married.\n\nSo let's start with the probability that 8 were married:\n\nIf the probability that someone was never married is .15, that means the probability that they were married is 1 minus .15, or .85. And if 8 people in the sample were married, that means that 2 were not. So what you're going to do is multiply .85^8 (for married people) x .15^2 (for unmarried). \n\nThe principle is the same for 9 married people and ten married people: \n\nFor 9 married, you multiply .85^9 x .15^1\n\nFor 10 married, you multiply .85^10 x .15^0\n\nFinally, add all three answers together, which gives you the probability that either 8, 9, or 10 people in the sample were married, which is what the question was asking! Hope that helped!
418638
The women and the man would have a heterozygous geneotype\n\nThe woman's genotype would consist of a gene for B blood and for O blood.\nThe man's genotypes would consist of a gene for A blood and B blood.  \n\nSince O is a ressisve gene, A and B would dominiant the blood type for their children who could have an A, B or AB blood type.  \n\nFollowing a basic 2 x 2 Punnett square, the statistics of a child getting a homozygous B blood type would be 1/4; a AB blood type would be 1/4; for getting a heterzygous A or B blood type with a ressive O gene would be 2/4 or 1/2.
640801
These types of DNA are used to create a much more accurate diagnosis of DNA samples lef tfrom teh crime\n\nemail me a kkstrong0@yahoo.com if you have further questions and ill try to tell you what i know
266219
EEEEH. I'm not really good with this mumbojumbo stuff but here is a site on genotypes and stuff. -> http://janis7hepc.com/Genotypes.htm <-
5093
Blood types are not evenly distributed throughout the human population. O+ is the most common, AB- is the rarest. There are also variations in blood-type distribution within human subpopulations. The figures given here are for people of European descent.\n\nType Frequency \nO+ 38% \nA+ 34% \nB+ 9% \nO- 7% \nA- 6% \nAB+ 3% \nB- 2% \nAB- 1%
471764
Note: they aren't asking just for the chance that at least 2 children have blue eyes.  They are asking for the probability, given that you know at least one already does.\n\nThis is conditional probability.\n\nP(B | A) = P(A and B) / P(A)\n\nSo the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.\n\nFor that part though, you would use binomal distribution.\n\nFirst, what is the probability that at least one child has blue eyes?\nThis would be 1 minus the probability that nobody has blue eyes.  \nP ( blue >= 1 ) = 1 - (0.75)^5 \nP ( blue >= 1 ) = 0.762695312 \n\nNow the probability that at least *two* children have blue eyes:\nP( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ).  Again you can figure the negative statement and subtract from 1.  So this would be equivalent to:\n1 - P ( blue = 0 ) - P ( blue = 1 ).\n\nThis would be:\nP ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)\nThe C(5,1) means the ways to choose 1 person from 5, which is just 5.\nP ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5 \nP ( blue >= 2) = 0.3671875\n\nSo putting it all together:\nP ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )\n= 0.3671875 / 0.762695312\n= 0.481434059\n≈ 48.1%
88174
This is a fairly long answer... first you need to get the initial allele frequencies. equation is p2 + 2pq + q2 (p2 and q2 from here on in is p squared and q squared)\nSince I am left to assume that counter clockwise is recessive then .36 is the initial q2. Taking the square root of that we get .6 as the initial allele frequency for q. 1-q=p so p=1-.6=.4. In other words, the island initally has 80p (clockwise)alleles and 120q (counterclockwise) alleles since humans are diploid and have two copies of each gene.\nNow, the 25 sailors have have added 50 copies of the q allele thereby changing the allele frequencies. The new allele frequencies are (120+50)/250 for q and 80/250 for p. when you do the math you get .32 for p and .68 for q. This will be the allele frequencies for the next generation too. \nNow plugging those numbers into the equation that I sited earlier you get .1024 for p2, .4352 for 2pq and .4624 for q2. You can give those a dominant/recessive notation such as WW for p2, Ww for 2pq, and ww for q2 to show genotype frequencies. \n\nThe evolutionary force at work is migration. \nhope this is what you were looking for.
447711
Mendel's laws predict that a monohybrid cross between any two heterozygotes would produce offspring in a 3:1 phenotypic ratio.  That is:\nAa  X  Aa = \n1 AA - Dominant Phenotype\n2 Aa - Dominant Phenotype\n1 aa - Recessive Phenotype\n\nSo, 3 Dominant : 1 Recessive\n\nNow for your two instances:\n\nA.\nIn the case of codominance we have two different dominant alleles and one recessive allele.  When both dominant alleles are present, the phenotype is different from that seen when either of the alleles is present alone (homozygous or heterozygous).  A classic example of codominance is that of blood types.  Type A and type B are both dominant phenotypes, type O is the recessive phenotype, and type AB is the codominant phenotype.\n\nBack to your question, lets say:\nA = dominant allele 1\nA* = dominant allele 2\na = recessive allele\n\nCrossing two heterozygotes: Aa X A*a results in\n1 AA* - codominant phenotype\n1 Aa - dominant allele 1 phenotype\n1 A*a - dominant allele 2 phenotype\n1 aa - recessive phenotype\n\nSo we have a 1:1:1:1 phenotype ratio, not Mendelian expected 3:1\n\nB.\nRecessive lethal alleles are a situation where the homozygous recessive genotype is lethal to the organism.  So if we cross two heterozygotes:\n\nAa X Aa \n\nWe get:\n\n1 AA - dominant phenotype\n2 Aa - dominant phenotype\n1 aa - lethal, so this offspring is never seen\n\nSo here all the offspring seen are of the dominant phenotype, again defying Mendel's predicted 3:1 ratio.\n\nHope this helps!