The probability is 0 (but the daughter will be a carrier of the color blind gene).
This is because the gene dictating whether someone is color blind or not is linked to the X chromosome (and not the Y). The color blind gene is a recessive gene whilst the normal color vision gene is a dominant gene. Hence if a girl (XX) has one normal vision gene (from one parent) and one color blind gene (from the other parent), her normal vision gene will be dominant to the recessive color blind gene and hence she will have normal vision (but she will be a carrier of the color blind gene). If both her parents contribute the recessive color blind gene to her, then she will be color blind.
For a woman (XX) to be color blind, she needs to be have both genes to be recessive (ie where there is no dominant normal color vision gene to dominate). For a man (XY), as long as the X gene contributed by his mother is a color blind gene, he will be color blind because he has no other X chromosome where a dominant normal color gene could reside.
Hence, to answer the question, a man with normal color vision (XY, with a dominant normal color vision X gene since the gene can't be the recessive color blind gene otherwise he will be colorblind) and a colorblind woman (XX, both recessive color blind genes), will each contribute an X each the child. The man will contribute his only X chromosome which carries the normal color vision X gene and the woman can only contribute a recessive color blind gene. The man's normal color vision X gene will be dominant, and hence the daughter will definitely have normal vision (despite being a carrier).
50%. 1 of the two male offspring will definately be color blind. Do a punnett square with the father having normal vision and the mother being a carrier.
Probability is everywhere! If you are entering a raffle, wouldn't you like to know what the probability of you winning is? Or perhaps the lottery? If you need to pick a certain color marble out of a bag of marbles to win a prize, wouldn't you like to know the probability of picking the winning color? Probability is everywhere and can help you in life.
The theoretical probability of randomly picking each color marble is the number of color marbles you have for each color, divided by the total number of marbles. For example, the probability of selecting a red marble is 3/20.
In a standard deck the probability is 0. There are not two, but 26 cards of the same colour.
It is a certainty if you pick 5 cards.
The father has to be colorblind for the daughter to be colorblind because both X chromosomes must have the colorblindness gene in females because the colorblind gene is recessive. If only the mother is color deficient, then she merely passes on the gene to one of the X chromosomes in a female. If both the mother and father are both colorblind, then both X chromosomes in the female are effected and the female is colorblind. There are two scenarios in which a daughter may be born colorblind. 1. The father is colorblind and the mother is a carrier of the colorblind gene. The daughter will be either colorblind or a carrier of the colorblind gene. 2. The father and mother are both colorblind. If this is the case, then all of the children will be born colorblind.
50% probability that the sons produced from this union will be color blind. 50% probability that a son will not have the disease. 50% probability that a daughter will be a carrier of the allele for color blindness. 50% probability that a daughter will not be a carrier. Phenotypically this would be 1male color blind, 1 male not color blind, 1 female carrier and 1 female non-carrier. Assuming that the probability of male and female offspring is identical...this would be 25% of each genotype mentioned above. Based on phenotype, without regard to sex the percentages of normal to colorblind would be 75% to 25% with the 25% displaying the characteristic all being male.
Women can not be colorblind, only men. For questions like these a punnett square is useful. Men can not carry the colorblind trait, but women can. I know this is kind of confusing. When a carrier ( a woman with the color blind trait) has children with a man ( color blind or not) her kids will have 50% chance of having that trait. If its a girl, she will be the carrier. If its a boy, he will have the colorblind trait. SO TO ANSWER YOUR QUESTION: Theoreticaly, 1 of the daughters will be the carrier, and the son will have a 50% chance of being colorblind. Women can be colorblind, its just rare. About every 6400 women one is colour blind and with men, every 80 men 1 is colour blind.
It depends on a number of different factors. If colorblind is common in either family but the mother/father were lucky enough not to get it, then its a recessive gene. but if its not common in either family then its a very low chance the child will be color blind. Although, if there are numerous people on both sides of the family the probability is very high. It all depends on a lot of different factors. How dominant is the gene? How many relatives are colorblind? If there are any, and they have children, are they colorblind? The easiest way to figure out if the child will be colorblind is to have the kid and test it for colorblind-ness.
Yes. A person is color blind if all his or her X chromosomes have the defective gene. A man have one X chromosome, and a woman has two. Thus: If only the father is color blind - The probability that the son is color blind is 0% - The probability that the daughter is color blind is 0% - The probability that the daughter is a bearer of the defective gene is 100% If the mother is a bearer of the defective gene, but is not color blind, and the father is not color blind - The probability that the son is color blind is 50% - The probability that the daughter is color blind is 0% - The probability that the daughter is a bearer of the defective gene is 50% If the mother is a bearer of the defective gene but not color blind, and the father is color blind - The probability that the son is color blind is 50% - The probability that the daughter is color blind is 50% - The probability that the daughter is a bearer of the defective gene is 100% If the mother is color blind and the father is not - The probability that the son is color blind is 100% - The probability that the daughter is color blind is 0% - The probability that the daughter is a bearer of the defective gene is 100% If both parents are color blind - The probability that the child, regardless of gender, is color blind is 100%
The girl must have a copy of the gene on each X chromosome. This means that she must have a father who is colorblind and a mother who is either a carrier or is colorblind.
No it is only possible.
Every dog is colorblind
no color blind people are mainly red-green colorblind or less commonly blue-yellow colorblind
50%. 1 of the two male offspring will definately be color blind. Do a punnett square with the father having normal vision and the mother being a carrier.
100% of all male offspring will be colorblind. 0% of all femal offspring will be colorblind.
Highly unlikely. Eye color and color blindness are not determined the same way within your DNA. -Eye color depends on what color your eyes your parents have and the dominance of specific traits and alleles. -Color blindness is linked primarily to gender. It's a sex linked mutation and occurs only on the X chromosome. Males are much more likely to be color blind than females, because it is linked only to the x chromosome. Males only have one x chromosome and females have two. For females to have the colorblind mutation, their father has to be colorblind and their mother has to be a carrier or colorblind. For a male to be colorblind their mother has to either be a carrier or color blind. It is much more likely that mother is carrier than colorblind.