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What are the possible genotypes of a parent that has hemophilia type B

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A:A parent with hemophilia type B has one dominant gene and one recessive gene. ChaCha again! [ Source: http://www.chacha.com/question/what-are-the-possible-genotypes-of-a-parent-that-has-hemophilia-type-b ]
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What are the possible genotypes of a parent that has hemophilia t...?
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A parent with hemophilia type B has one dominant gene and one recessive gene.

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Inheritance Problems (Help)?
Q: 1.What would be the genotype of a female carrier for hemophilia? What would be the genotype of a male with hemophilia?2.A woman’s uncle has hemophilia. Her father does not have the trait, but she worries that she might have a son with hemophilia. Should she worry? Assume that she is heterozygous. Explain your answer.3.A woman (whose father is colorblind) marries a colorblind man. Can they have an affected son? Can they have an affected daughter? Use a Punnett square to help answer the question.Multiple Allelelic traitsThe following traits are controlled by more than two alleles. One example of a multiple allelic trait is the blood group ABO. The following represents the phenotypes and genotypes of the ABO blood group. Type A blood has the genotypes IAIA or IAi, type B blood has the genotypes IBIB or IBi, type AB blood has the genotype IAIB, and type O blood has the genotype ii.4.If one parent has type O blood and the other has type AB blood, what is the chance that their first child will have type O blood? Type A blood? Type B blood? Or type AB blood? Use a Punnett square to help answer the question.5.If a child has type O blood and its mother has type A blood, could a type B man be the father? Explain your answer. For the following problems, you are not given the inheritance patterns. You must determine the inheritance pattern and complete the problems.6.A woman is heterozygous for Cystic fibrosis and her husband does not have the trait. What is the likelihood of their children having the trait?7.A woman is heterozygous for Huntington’s disease and her husband is heterozygous for it. What are the chances of their children having it?8.A 15 year old boy finds out that he has muscular dystrophy. Neither one of his parents have the disorder. How is this possible? Use a Punnett square to explain your answer.sorry my teacher is sooooo bad and i need to do this worksheet in order to pass... problem is i have NO idea what im doing.
A: This is TAXATION .
Homework for Genetics?
Q: 1.Ear lobe attachment is an autosomal trait. Free earlobes are completely dominant to attached earlobes. A woman with attached earlobes, both of whose parents have free earlobes, has children with a man who has free earlobes and whose mother has attached earlobes. What are the expected genotypes of their children?2.Hair texture in humans is an autosomal trait with curly hair partially (incompletely) dominant to straight hair, the recessive condition. Wavy hair is the heterozygous condition. A curly haired woman had children with a man with wavy hair and they had 16 children. How many of the children would be expected to be curly haired? 3.Narcolepsy is an autosomal recessive disorder that can affect Doberman pinschers (a breed of dog). Two normal Doberman pinschers when bred together produced 9 normal puppies and 3 that had narcolepsy. To the extent you can discern, what are the genotypes of all the dogs? What proportion of the normal puppies can be expected to be homozygous dominant? 4.Normal blood clotting is dominant to blood clotting deficiency (hemophilia) and alleles for this trait are on the X chromosome. A man with normal blood clotting has children with a woman who has hemophilia. If they had a son, what would be the chance that he would have hemophilia? Daughter? Any child?6.Jose has O- blood and Helen has B+ blood. Helen’s mother has O+ blood and her father has AB- blood. Jose and Helen have a son, Jake. What are the possible blood types (i.e. phenotypes) for Jake? 7.Assume the brown eye allele is dominant to the blue eye allele. Kiyoshi, a man with blue eyes, and his partner Aneesa, who has brown eyes, want to have a daughter who has brown eyes. Their first child, a son, had blue eyes. What is the probability that this couple will have a daughter with brown eyes? HINT: this is a dihybrid problem. 8.Achondroplasic dwarfism is a dominant trait. In the homozygous dominant condition embryos do not survive. Heterozygotes are dwarfs. Homozygous recessive is normal height. Michael and Michelle are both dwarfs. What are the chances that if this couple was to have a child, the child would be a dwarf?GENETICS QUESTIONS: 5 POINTS1.Are dominant characteristics always more frequent in a population than recessive characteristics? Why or why not? What determines the frequency of a phenotype in a population? 2.How is it possible to determine the genotype of an individual animal or plant having a dominant phenotype without using pedigrees or family histories? 3.Why is hybrid seed (heterozygotes created from parents that are homozygous dominant and homozygous recessive) so expensive to produce? 4.What blood types are not possible for children to have if their parents have AB blood? 0 blood? 5.Organisms heterozygous for a trait are often called carriers of that trait. What does this mean? Please help me! Thanks
A: 1.free - dominant = Eattached - recessive = eif she has attached earlobe then she must be 'ee' no-matter who her parent are, because if she has even one 'E' then because that is dominant it will be expressed.the man must be 'Ee' because his mother was 'ee' meaning he must have inherited at least 1 'e' and he has free earlobe meaning he must have recieved at least 1 'E'.now if we cross these: E ee| Ee| ee|e| Ee| ee|so there are 2 possibilities Ee 2/4 of the time and ee 2/4 of the time hence the genotype is:Ee - 50%ee - 50%Now thats all im going to show you because all the other questions are similar and if you just apply the same principles you can do them all
4 genetic problems answer ASAP Thank YOu very much?
Q: 1. In humans, brown eyes(B) are dominant over blue(b). A brown-eyed man marries a bluue-yed woman and they have three children, tow of whom are brown-eyed and one of whom is blue-eyed. Draw the punnett square that illustrates this marriage. What is the man's genotype? What are the genotypes of the children?2. Females have two x chromosomes, which can be represented by xx. Males have one x and one y chromosome, so we say they are xy. Draw the punnett square for the previous problem also keeping track of the sex of the various individuals. The two brown-eyed children are one girl andf one boy; the blue-eyed child is a boy. If these parents would hav ea bunch omre children so that they had 12 in all, how many of those 12 would you expect to be blue-eyed girls? If the first brown-eyed daughter marries a heterozygous brown-eyed man, draw a PUnnett square that predicts what their children will be.3.In Guinea pigs. black hair(B) is dominant over the white(b), rough coat texture(R) is dominant over smooth(r), and short hair(S) is dominant over long hair(s). Assuming these genes are on separate chormosomes, draw the PUNnett square for a cross between a homozygous black, rough, short-haired Guinea pig adn a white, smooth, long-haired one. What wouldt he phenotype(s) of the the offspring be? If two of hte F1 offspring were crossed, draw the punnett square for this cross. Hint: First make a list of a possible gametes, making sure each ahs exactly one copy of each of the genes(one allele for each gene) What would the genotype and phenotype rations be for the F2 generation?4. In humans, the genes for colorblindness adn hemophilia are both located on the X chromosome wiht no coresponding gene on the Y. HTese are both recessive genes. If a man and a woman, both wiht normal vision, marry and have a colorblind son, draw the punnett square that illustrates this. If the man dies and the woman remarries to a colorblind man, draw a Punnett square showing the type(s) of children that could be expected form this marriage. HOw many /what precentage of each could be expected?
A: 1) Genotype of Male is Bb and children's are Bb, Bb, bb. (Sorry I can't draw the punnett square here but show the cross between Bb and bb in that square.2)Drawing punnett square is impossible here.
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