The allele for an albino child in humans is "a". The normal skin color allele is "A''. What is the Probability
113474
9/64 because\n\nparents - Aa - heterozygous \n\nbelow is a punnet square\n\n___A a\nA AA Aa\na Aa aa\n\n1/4 chance of having albino child.\n3/4 chance of not having albino child\n\nall possible outcomes are:\n\n1/4 * 1/4 * 3/4 = 3/64\n1/4 * 3/4 * 1/4 = 3/64\n3/4 * 1/4 * 1/4 = 3/64\n\nadd them\n\n3/64 + 3/64 + 3/64 = 9/64
848151
Its actually more complex than a typical simple black and white mix (referring to color)\n\nAlbinism is a genetic disorder that is quite recessive.  The possibility of the resulting child having the color of black, white or "blended" would depend if the child inherits the albinism. \n\nInheriting the afican parent's albinism depends on the caucasion's popsibilty of passing along albanism as well. \n\nTaking the albinism factor out of the equation and it is safe to say that the offspring of a person with black skin and a person with white skin will result in a blending of the skin colors. \n\nIt is also safe to assume that the offspring will appear to have more of an african-american (black) look as opposed to a caucasion (white) look and would appear to be a child of mixed heritage.
802960
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.
692190
Only if a genetically recessive trait runs in the family and both the guy and his sister have the defective chromosome, only then does the baby risk having a birth defect or mental retardation - even then(ie, with the defective recessive chromosome in both), the chance would be 1 abnormal child out of 4. If the guy and the sister have no such defective chromosomes, then the question of an abnormal/retarded child does not arise.\n\nIf, however, a genetically dominant trait runs in the family and either the guy or his sister have the defective chromosome, then the risk increases to 1 in 2  and if both have it, then the risk is 3 out of 4.\n\nIn any case, whether the baby will be normal or abnormal cannot be determined until a karyotyping of both the guy and his sis is done. Having said that, the rate of such disorders is quite low and in all probability, the child will be normal.
639406
When Mendel began to scientifically analyze "heredity", he concluded that each parent (mother + father) donates an "allele" for a characteristic (trait) in question.\n\nA lettering system was used: Capital letters indicated that the allele is DOMINANT; lower-case letters indicate that the allele is RECESSIVE.  This is the case for 22 of the 23 chromosome pairs in humans.  The 23rd pair are called the sex chromosomes.  Females (mother) always donate an "X" allele; fathers can donate either an "X" or a "Y" allele.\n\nIf a person's genotype for their sex is "XX", this means two things.  1: Both X's are the same (in other words, X and X are identical).  Identical alleles are called HOMOZYGOUS.  2: A human who is "XX" is female.\n\nA human who is "XY" is male.  "X" allele and "Y" allele are DIFFERENT.  Thus, they called HETEROZYGOUS.  We refer to men as being "heterozygous" in that a male's SEX alleles, 'X' and 'Y' are different.\n\nLet us say "B" is the dominant allele for brown eyes, and "b" is the recessive allele that can result in blue eyes.  A person who is "Bb" is heterozygous (B is a different allele than b).  A person's who genotype is Bb is likely to have brown eyes.  A person who is "BB" is HOMOZYGOUS (1st B is identical to the 2nd B) and will definitely have brown eyes.  A person who is "bb" is HOMOZYGOUS (1st allele 'b' is identical to the 2nd allele, 'b') for the recessive allele; this person will have blue eyes.
909716
Mother's genotype is X*X and dad's genotype is XY.\nX*X -------> 1/2 X* & 1/2 X\nXY --------> 1/2 X & 1/2 Y\n\nKids genotype will be: 1/4X*X, 1/4XX, 1/4X*Y, 1/4XY\nTherefore, there is a 25% chance of having a daughter who is a carrier, 25% chance of having a daughter who is not a carrier, 25% chance of having a hemophilic son and 25% chance of having a healthy son.
976375
AA = p(squared)\nAa = 2pq\naa = q(squared)\np + q = 1\n\nFirst you need to interpret "9% have the disease". This means that 9% of the population have both recessive alleles, which means the frequency of aa is 0.09. You are asked to calculate the percentage (or frequency) of the resistant (or dominant allele), which is denoted A. So by using the above equations:\n\naa = 0.09, which can also be expressed as a(squared) = 0.09\n\na = square root of a(squared) = square root of 0.09 = 0.3\n\nThe recessive allele (a in this case) is always donated by q, and the dominant allele (A in this case) is always denoted by p. So therefore; a = q.\n\np + q = 1\np + 0.3 = 1\np = 1 - 0.3\np = 0.7\n\nSince p = A, the frequency of the dominant allele (A) equals 0.7
867087
Yes.  An albino is just like anyone else except they're skin is lacking pigment and their eyes and hair are lighter. They are normal in every other way, basically. He is one of the coolest guys I've ever known. :)
716160
The "sickle cell" gene (technically, allele) is a result of a mutation that modifies the shape of red blood cells. It's what is known as an incompletely recessive trait, meaning that a person with one copy of the sickle cell allele and one copy of the normal allele has mostly normal red blood cells with a few sickled ones. This state has been shown to give the person carrying it resistance to malaria, a common and often fatal disease in tropical regions, particularly Africa. \nThus in Africa, the sickle-cell allele would have been rendered more common by natural selection, since being resistant to malaria was a huge advantage. The children of two people carrying the sickle-cell trait have a 25% chance of receiving both sickle-cell genes from their parents, which is known as sickle-cell anemia. \nIt's not that only people of African descent can have sickle-cell anemia, but through selection the responsible gene was made more common because of malaria resistance. The same would not be the case in areas of the world where malaria was not common; which is why people whose ancestors came from other places are not as likely to have two parents carrying the sickle-cell trait.
862829
Gene Allele Description\nB=black, b=brown\nC=color deposition, c= no color deposition\n\nParents cross: BbCc  x  BbCc\nGametes: BC; Bc; bC; bc  x  BC; Bc; bC; bc\n\nF1 Punnnet Square\n\n.........BC....Bc....bC....bc\nBC\n\nBc\n\nbC\n\nbc\n\nthe result of the punnet square is as follow:\n\n9/16 B_C_  Black (color deposition)\n3/16 bbCC  Brown (color deposition)\n3/16 BBcc  Yellow (no color deposition)\n1/16 bbcc   Yellow (no color deposition)\n\nratio: 9:3:4\n\ngene c is epistatic on gene B/b
780996
well the child would have one part of the blood type of mother and one of father like if father is OO and mother is  AO then the child would be type A.
799112
The children would have bloodtype A or bloodtype B.\n\nThe "O" which they get from one parent is merely an absence of the information necessary to have specific antibodies against other bloodtypes.
874576
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%
39991
Alleles are different 'forms' of the same gene.\n\nFor example, there are three alleles of the gene for blood groups; alleles A, B and O.\n\nIn diploid organisms like ourselves we have two complete sets of genes, so there are various combinations of alleles. One allele is inherited from the mother and one from the father.\nAA - blood group A\nAO - blood group A (because A is dominant to O)\nBB - group B\nBO - group B (B dominant to O)\nAB - group AB (alleles are codominant - equal dominance)\nOO - group O\n\nGeneologically, we can follow the inheritance of various characteristics (determined by alleles) and similarly blood groups can be used to eliminate or confirm parentage.\n\nThe 'classic' geneological example is the family tree of Queen Victoria of Great Britain. This follows the inheritance of the sex-linked gene for haemophilia.
1067528
Sickle cell anemia is inherited as an autosomal recessive trait.  This means that in order for a child to have sickle cell in its full blown form, (s)he must inherit the trait from BOTH parents. If two carriers of the trait marry, these are the odds:\n\nBoth parents are Ss, so Ss crossed with Ss yields a 50% chance that a child will be a carrier (have sickle cell trait), a 25% chance that a child will be normal (not have any sickle cell at all in their genes), and a 25% chance that a child will have sickle cell disease. \n\nIf a person with sickle cell marries a normal person, then all of their children will be carriers (have the trait).\n\nIf a person who is a carrier marries a normal person, then 50% of the children will have the trait and 50% will be normal.\n\nIf a person who is a carrier marries a person with sickle cell,  50% of the children will be carriers, and 50% will have the disease.\n\nIn the unlikely case that two people with sickle cell disease marry, then ALL of their children will have the disease.\n\n\nNote that these numbers apply to EACH pregnancy: no matter what the outcome of each child, the chances are still the same for any siblings. So the only way a child can inherit the actual sickle cell disease is if the trait is carried by both parents or at least one of them actually has the disease.
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.
1150347
dark pigments in their skin...a genetic trait