A woman diagnosed with early-onset Alzheimer's due to a mutation of the APP genehas children with a man that has no family history of familial Alzheimer's. Give the probability of each possible genotype with corresponding phenotype. (The woman has two possible allele combinations. You must show both possibilities)
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- A woman diagnosed with early-onset Alzheimer's due to a mutation of the APP genehas children with a man that has no family history of familial Alzheimer's. Give the probability of each possible genotype with corresponding
phenotype . (The woman has two possible allele combinations. You must show both possibilities)
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- Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known concordance statistics, would his MZ twin be at high risk for the disease? Would the twins risk decrease if he were raised in an environment different from that of his schizophrenic brother?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?Figure 1-15 shows the family tree, or pedigree, for LouiseBenge (Individual VI-1) who suffers from the diseaseACDC because she has two mutant copies of the CD73gene. She has four siblings (VI-2, VI-3, VI-4, and VI-5)who have this disease for the same reason. Do all of the10 children of Louise and her siblings have the samenumber of mutant copies of the CD73 gene, or mightthis number be different for some of the 10 children?While sitting at home during Movement Control Order (MCO) because of pandemic covid19, observe two different traits of a couple in your family (eg. your mom & dad or your sister & her husband or your brother & his wife, etc). Draw a genetic cross that involves cross of the parents with the chosen 2 pairs of their contracting traits. Imagine that the cross obeys the Mendelian Laws, show the cross and gametes production for each generation (P, F1 and F2). By Using a Punnet square as symbolic representation of the results for the cross, determine the phenotypes, genotypes, phenotypic ratio and genotypic ratio of F2 generation in the family.
- Huntington's disease is a very rare, debilitating disease that affects approximately 1 in 10,000 people. The disease is caused by a particular type of genetic mutation (large number of CAG repeats) in a gene called huntingtin (HTT). Designate the disease-causing allele HTTP and the non-disease- causing allele HTTN. We know that individuals with genotype HTTN will not develop Huntington's Disease, but people with either HTTD/HTTP or HTTN/HTTD will develop the disease. Which allele is dominant--HTTN or HTTP? Explain your reasoning thoroughly, using the correct definition of dominance.Duchenne muscular dystrophy is sex linked and usuallyaffects only males. Victims of the disease become progressively weaker, starting early in life.a. What is the probability that a woman whose brotherhas Duchenne’s disease will have an affected child?b. If your mother’s brother (your uncle) had Duchenne’sdisease, what is the probability that you have receivedthe allele?c. If your father’s brother had the disease, what is theprobability that you have received the allele?Huntington disease is a rare dominant condition in humans that results in a slow but inexorable deterioration of the nervous system. The disease shows what might be called age-dependent penetrance, which is to say that the probability that a person with the Huntington genotype will express the phenotype varies with age. Assume that 50% of those inheriting the HD allele will express the symptoms by age 40. Susan is a 35-yearold woman whose father has Huntington disease. She currently shows no symptoms. What is the probabbility that Susan will show symptoms in five years?
- A polydactylous, normal-visioned, brown-eyed man with wavy hair has a nonpolydactylous, blue-eyed mother. He proposed marriage to a nonpolydactylous, astigmatic, blue-eyed, and curly-haired lady whose mother has normal vision. However, the lady is worried about the proposal since sickle-cell anemia has been known to run in both their families. Is there a genetic basis for her worry? Explain.Hemophilia is an X-linked disease associated with the inability to produce specific proteins in the blood-clotting pathway. Shown above is a family pedigree tree in which family members afflicted with the disease are shown with filled-in squares (male) or circles (females). A couple is trying to determine the likelihood of passing on the disease to their future children (represented by the ? symbol above) because the hemophilia runs in the woman’s family 4.Assuming that the woman in the couple is a carrier, what is the probability that the couple’s first son will have hemophilia?In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman? HR/hr Hr/hr hr/hR Hr/hR HR/Hr