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Ok, colour blindness and blood type are independently assorting traits, that means that having one does not affect having the other, the question even tells you that the traits are on different chromosomes.Β
To answer your question we'll have to calculate the probability of both traits in the child and then multiply them together to get the overall probability of both being present in the child.Β
Let's start with blood type: there are four types of blood type: AB, A, B and O. Blood type is determined by what protein each chromosome codes for: so if one chromosome is I^A and codes for A type protein, while the other is I^B and codes for B type protein, then the overall blood type will be AB. O blood type is recessive, it doesn't code for any protein at all and is given the symbol i.Β
Since both parents are type A and their first child is type O, you know that both parents are heterozygous for blood type A, meaning that one chromosome they have is I^A and the other is type or or i. For a child to be blood type O, each parent has to give their type O chromosome to the child. There is a 1/2 chance that one parent will give the recessive O chromosome, so for both parents to give their recessive allele the probability is 1/2 * 1/2 = 1/4Β
Colour blindness is an X-linked condition, that means that women can be carriers but not manifest the colour blindness because they have one "good" X chromosome which masks the recessive colourblindness gene on the other X. The father cannot have a colourblindness gene or else he'd be colourblind himself, since he's only got 1 X-chromosome. Because dad has healthy eyesight, any girls he has will have healthy eyesight too, since his X will always be healthy and mask the X from mom if she happens to give the recessive colourblind gene. So the probability that the girl will have normal vision is 1/1.Β
However, it is not a given that the baby born will be a girl, there's a 50/50 chance that it'll be one or the other sex, so we need to consider the probability involved with the child being a girl. this probability is 1/2.Β
So now we multiply all our probabilities together to calculate what the chances of a normal visioned, type O blood typed girl is:Β
Normal vision (1) * blood type O (1/4) * girl baby (1/2) = 1/8Β
So your probability is 1/8!Β
Hope this helps :)
To answer your question we'll have to calculate the probability of both traits in the child and then multiply them together to get the overall probability of both being present in the child.Β
Let's start with blood type: there are four types of blood type: AB, A, B and O. Blood type is determined by what protein each chromosome codes for: so if one chromosome is I^A and codes for A type protein, while the other is I^B and codes for B type protein, then the overall blood type will be AB. O blood type is recessive, it doesn't code for any protein at all and is given the symbol i.Β
Since both parents are type A and their first child is type O, you know that both parents are heterozygous for blood type A, meaning that one chromosome they have is I^A and the other is type or or i. For a child to be blood type O, each parent has to give their type O chromosome to the child. There is a 1/2 chance that one parent will give the recessive O chromosome, so for both parents to give their recessive allele the probability is 1/2 * 1/2 = 1/4Β
Colour blindness is an X-linked condition, that means that women can be carriers but not manifest the colour blindness because they have one "good" X chromosome which masks the recessive colourblindness gene on the other X. The father cannot have a colourblindness gene or else he'd be colourblind himself, since he's only got 1 X-chromosome. Because dad has healthy eyesight, any girls he has will have healthy eyesight too, since his X will always be healthy and mask the X from mom if she happens to give the recessive colourblind gene. So the probability that the girl will have normal vision is 1/1.Β
However, it is not a given that the baby born will be a girl, there's a 50/50 chance that it'll be one or the other sex, so we need to consider the probability involved with the child being a girl. this probability is 1/2.Β
So now we multiply all our probabilities together to calculate what the chances of a normal visioned, type O blood typed girl is:Β
Normal vision (1) * blood type O (1/4) * girl baby (1/2) = 1/8Β
So your probability is 1/8!Β
Hope this helps :)
Β Ok, colour blindness and blood type are independently assorting traits, that means that having one does not affect having the other, the question even tells you that the traits are on different chromosomes.Β
To answer your question we'll have to calculate the probability of both traits in the child and then multiply them together to get the overall probability of both being present in the child.Β
Let's start with blood type: there are four types of blood type: AB, A, B and O. Blood type is determined by what protein each chromosome codes for: so if one chromosome is I^A and codes for A type protein, while the other is I^B and codes for B type protein, then the overall blood type will be AB. O blood type is recessive, it doesn't code for any protein at all and is given the symbol i.Β
Since both parents are type A and their first child is type O, you know that both parents are heterozygous for blood type A, meaning that one chromosome they have is I^A and the other is type or or i. For a child to be blood type O, each parent has to give their type O chromosome to the child. There is a 1/2 chance that one parent will give the recessive O chromosome, so for both parents to give their recessive allele the probability is 1/2 * 1/2 = 1/4
Colour blindness is an X-linked condition, that means that women can be carriers but not manifest the colour blindness because they have one "good" X chromosome which masks the recessive colourblindness gene on the other X. The father cannot have a colourblindness gene or else he'd be colourblind himself, since he's only got 1 X-chromosome. Because dad has healthy eyesight, any girls he has will have healthy eyesight too, since his X will always be healthy and mask the X from mom if she happens to give the recessive colourblind gene. So the probability that the girl will have normal vision is 1/1.Β
However, it is not a given that the baby born will be a girl, there's a 50/50 chance that it'll be one or the other sex, so we need to consider the probability involved with the child being a girl. this probability is 1/2.Β
So now we multiply all our probabilities together to calculate what the chances of a normal visioned, type O blood typed girl is:Β
Normal vision (1) * blood type O (1/4) * girl baby (1/2) = 1/8Β
So your probability is 1/8!
Source: Yahoo Answers
________________________________________________
I hope the answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
To answer your question we'll have to calculate the probability of both traits in the child and then multiply them together to get the overall probability of both being present in the child.Β
Let's start with blood type: there are four types of blood type: AB, A, B and O. Blood type is determined by what protein each chromosome codes for: so if one chromosome is I^A and codes for A type protein, while the other is I^B and codes for B type protein, then the overall blood type will be AB. O blood type is recessive, it doesn't code for any protein at all and is given the symbol i.Β
Since both parents are type A and their first child is type O, you know that both parents are heterozygous for blood type A, meaning that one chromosome they have is I^A and the other is type or or i. For a child to be blood type O, each parent has to give their type O chromosome to the child. There is a 1/2 chance that one parent will give the recessive O chromosome, so for both parents to give their recessive allele the probability is 1/2 * 1/2 = 1/4
Colour blindness is an X-linked condition, that means that women can be carriers but not manifest the colour blindness because they have one "good" X chromosome which masks the recessive colourblindness gene on the other X. The father cannot have a colourblindness gene or else he'd be colourblind himself, since he's only got 1 X-chromosome. Because dad has healthy eyesight, any girls he has will have healthy eyesight too, since his X will always be healthy and mask the X from mom if she happens to give the recessive colourblind gene. So the probability that the girl will have normal vision is 1/1.Β
However, it is not a given that the baby born will be a girl, there's a 50/50 chance that it'll be one or the other sex, so we need to consider the probability involved with the child being a girl. this probability is 1/2.Β
So now we multiply all our probabilities together to calculate what the chances of a normal visioned, type O blood typed girl is:Β
Normal vision (1) * blood type O (1/4) * girl baby (1/2) = 1/8Β
So your probability is 1/8!
Source: Yahoo Answers
________________________________________________
I hope the answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
