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- 1. Study the given alleles. Write the correct phenotype for each genotype. X– normal Genotype XC – Color-blind Phenotype XX XY XCXC www m XX www w ww w 2. Study the given alleles. Write the correct genotype for each phenotype. X- normal Genotype хн- Hemophiliaс Phenotype Hemophiliac female Hemophiliac male Normal female carrier of the gene Normal male Normal female1. Study the given alleles. Write the correct phenotype for each genotype. X – normal Gen otype xC - Color-blind Phenotype XX XY XXC xCY 2. Study the given alleles. Write the correct genotype for each phenotype. xH - Hemophiliac Phenotype X- normal Gen otype Hemophiliac female Hemophiliac male Normal female carrier of the gene Normal male Normal female 28 3. Determine the genotype and phenotype of the offspring. A color-blind mother (XCx) married a normal sighted (XY) father. Genotype: Phenotype: Genotype: Phenotype: Genotype: Genotype: Phenotype: Phenotype: a. There are b. There are c. There are d. There are % normal sons. % normal daughters. % color-blind sons. % color-blind daughters. % normal female, carrier of the disorder. or or or or e. There are or12. A. B. C. D. 9:24 A. .5G prairiestate.desire2learn.com does that prove ne is the father of the baby or the person who committed the crime? Explain your answer. Hemophilia is an x-linked disease in which the blood does not clot normally; it is sometimes called "bleeder's disease." Hemophilia is caused by a recessive allele (h). The dominant allele (H) produces blood that clots normally. What is genotype of a man who is a hemophiliac? What is the genotype of a man with normal blood clotting? What is the genotype of a woman with normal clotting blood if her father was a hemophiliac? If a man with normal clotting blood and a heterozygous woman have children together, what would you expect for the genotypes and phenotypes of the children? 13. A dominant x-linked allele (B) gives normal color vision but the recessive allele (b) causes red-green color blindness. What is the genotype of a man with normal color vision whose father was color- blind?
- y 301 Amelogenesis imperfecta is X-linked dominant. Affected XY individuals have extremely thin enamel on the teeth while XX carriers have grooved teeth from uneven deposition of enamel. If an unaffected XY individual were to produce children with a XX carrier partner, a. what would be the expected chance of a XY child being affected with the disease? b. what would be the expected chance of a XY child being affected with the disease?the offspring ès) can often be autosomal I 1.1 III ad amalg od no IV 9 OTO 5 0000 anivellor and goje If individual 2 were to marry a woman with no family history of the disease, which of the following would most likely be true of their children? a. All of the children would have the disease. b. None of the children would have the disease. c. Only the sons would have the disease. d. All of the sons would be carriers of the disease. e. None of the daughters would be carriers of the disease.97. This family is (picture above) affected with blindness. Individual (IV.1) is clinically unaffected. What is the chance that he is homozygous for the normal allele? A-½ B-% C- almost 0 D-1/3 E- almost 100% ANS: IV TO O 560-556
- II. Write TRUE if the statement is correct and FALSE if otherwise. -7. A temale parent possessing an X-linked dominant mutation is considered a carrier and will not manifest clinical symptoms of the disorder. _8. Y-linked traits are passed from the father to son, without the occurrence of genetic recombination. 9. Somatic mosaicism results to abnormalities based on the amount and distribution of normal cells while gonadal mosaicism affects the germline tissues leading to a new dominant mutation. 10. A test cross is done to determine which allele is dominant and which is recessive.4. In this pedigree of another family with hemophilia A. If the woman Il-2 has a son, what is the chance that he is affected? II B. If the woman Il-2 has a daughter, what is the chance 1 2 3 4 III that she is affected? -1 C. If the woman Il-2 has 2 sons, what is the chance that neither will be affected? D. What are the chances that the child of II-4 and Il-5 will be affected?1 II 2 5 7 III 1 2 1. Determine the mode of inheritance 3 4 5 6 7 8 9 2. Give the genotype of the following individuals 1-1: I-3: II-1: 1-2: Il-6: II-3: Il-1: :7-ןו I-6: Il-2: Il-8: II-7: 3.
- 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?
