The next three questions go together. Duchenne's muscular dystrophy (DMD) is a disease caused by a recessive allele at a gene on the X chromosome. Polydactyly is caused by a dominant allele at an autosomal gene. If a phenotypically normal woman who's father had DMD has a child with man who is heterozygous for the allele causing polydactyly but is otherwise normal, what is the probability that the child will have elther DMD or polydactyly, but not both? Part I: What is the cross? A ppxx x PpxDxD B- PPXPXX Ppx©y C- ppx"x x PpXdy D- ppxPx* x PpxDY Question 11 Part II: What is the probability the child will have polydactyly but not DMD? Enter your answer to two decimal places les. 0.88).
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- An allele responsible for Marfan syndrome Section 13.4 is inherited in an autosomal dominant pattern. What is the chance that a child will inherit the allele if one parent does not carry it and the other is heterozygous?A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
- A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?Analysis of X-Linked Dominant and Recessive Traits The following is a pedigree for a common genetic trait. Analyze the pedigree to determine whether the trait is inherited as: a. autosomal dominant b. autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linkedX-linked ichthyosis is an X-linked recessive trait that manifests in part as dry, scaly skin (“ichthy-” = fish or fish like). Suppose a couple are considering having a child together. Parent A is heterozygous for the ichthyosis allele while Parent B is hemizygous negative for the ichthyosis allele. What is the probability their child would be unafflicted with ichthyosis but be a carrier of the ichthyosis-causing allele? a.0% b.25% c.50% d.75% e.100%
- Geneticists are concerned about three different genetic conditions present within their family. Achondroplasia is an autosomal dominant genetic disorder that results in short-limbed dwarfism. (A = achondroplasia allele; a = average height allele). Red-Green Color blindness is an X-linked recessive genetic disorder (Xc = color blindness allele; XC = normal color vision allele). The 7-year-old son has an autosomal recessive form of deafness. (h = deafness allele; H = normal hearing allele.) All three family members were karyotyped, and additional genetic testing was performed to determine the chromosomal location of the genes being studied. The figures below show the karyotypes of the mother and father with respect to the chromosomes containing the three genes mentioned above. Question: The couple’s 7-year-old son has all three of these genetic conditions (achondroplasia, color blindness and deafness). Taking into consideration the genotypes of his parents, what is the complete…Neurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.autosomal recessive allele (not sex-linked). Omplete the following monohybrid crosses for different types neritance pattefh autosomal dominant, sex linked recessive, and dominant inheritance. Inheritance of autosomal recessive traits Female parent phenolype: Example: Albinism Albinism (lack of pigment in hair, eyes and skin) is inherited as an Male parent phenatype: Using the codes: PP Pp (normal) (albino) la) Enter the parent phenotypes and complete the Punnett square for a cross between two carrier genotypes. A Give the ratios for the phenotypes from this cross. Pp (carrier) eggs sperm Phenotype ratios: Inheritance of autosomal dominant traits Example: Woolly hair Woolly hair is inherited as an autosomal dominant allele. Each affected individual will have at least one affected parent. Using the codes: WW (woolly hair) Female parent phenotype: Male parent phenotype: Ww (woolly hair, heterozygous) W w (normal hair) (a) Enter the parent phenotypes and complete the Punnett square for a…
- This pedigree shows the segregation of a rare disease in a family. What is the most likely mode of inheritance of this disease? (assume complete penetrance) autosomal recessive autosomal dominant x-linked recessive x-linked dominant maternal imprinting paternal imprintingPedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12Shown above is a family pedigree tree in which family members afflictedwith the disease Haemophilia are shown with filled-in squares (male) or circles (females). A couple is trying to determine the likelihood of passingon the disease to their future children (represented by the ? symbolabove) because the hemophilia runs in the woman’s family. Turner syndrome is a disease in which an individual is bornwith only a single X chromosome. Suppose the woman in thecouple is a carrier for hemophilia and has a child with Turnersyndrome. Would this child have the disease?