Retinoblastoma can be seen as a familial cancer, inherited in an autosomal recessive manner (RB-/RB-), individuals heterozygous for the RB+ and RB- alleles can develop tumor as a result of…A mitotic crossover that leads to homozygosity for RB+ in some cells and RB- in other cellsA meiotic mutation in the RB+ allele that leads to homozygosity for RB+A somatic mutation in the RB- allele that leads to homozygosity for RB+The fact that RB- is dominant to RB+

Name: Retinoblastoma can be seen as a familial cancer, inherited in an autos
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Description: Retinoblastoma can be seen as a familial cancer, inherited in an autosomal recessive manner (RB-/RB-), individuals heterozygous for the RB+ and RB- alleles can develop tumor as a result of…A mitotic crossover that leads to homozygosity for RB+ in som

The Rb gene is an archetypal tumor suppressor gene. Mutations in this gene result into…

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Biology

Retinoblastoma can be seen as a familial cancer, inherited in an autosomal recessive manner (RB-/RB-), individuals heterozygous for the RB+ and RB- alleles can develop tumor as a result of… A mitotic crossover that leads to homozygosity for RB

Retinoblastoma can be seen as a familial cancer, inherited in an autosomal recessive manner (RB-/RB-), individuals heterozygous for the RB+ and RB- alleles can develop tumor as a result of… A mitotic crossover that leads to homozygosity for RB

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter12: Genes And Cancer

Section: Chapter Questions

Problem 12QP

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The propensity to develop retinal cancer can run in pedigrees of certain families. You have learned about the role of the retinoblastoma (Rb) protein in the cell cycle. A heterozygous individual (genotype = Rbm / Rb+) that inherits a mutant allele of the Rb gene (Rbm) as well as a normal allele of the Rb gene (Rb+) is at significant risk for developing cancer of the retina. The best explanation for this is that… A. the Rb(m) allele is dominant to the Rb(+) allele. B. a mutation in the Rb(+) allele can result in a Rb(m) / Rb(m) genotype in certain cells. C. the retina is frequently exposed to UV rays D. the Rb(m) allele directly converts the Rb(+) allele to another Rb(m) allele. E. the cells of the retina carry more than two copies of the Rb gene
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?
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?
The Xg cell-surface antigen is coded for by a gene located on the X chromosome. No equivalent gene exists on the Y chromosome. Two codominant alleles of this gene have been identified: Xg1 and Xg2. A woman of genotype Xg2/Xg2 bears children with a man of genotype Xg1/Y, and they produce a son with Klinefelter syndrome of genotype Xg1/Xg2Y. Using proper genetic terminology, briefly explain how this individual was generated. In which parent and in which meiotic division did the mistake occur?
Explain why familial breast cancer shows a dominant pattern of inheritance in a pedigree even though it is recessive at the cellular level.
The mother of a family with 10 children has blood typeRh+. She also has a very rare condition (elliptocytosis,phenotype E) that causes red blood cells to be oval rather than round in shape but that produces no adverseclinical effects. The father is Rh− (lacks the Rh+ antigen)and has normal red blood cells (phenotype e). The children are 1 Rh+ e, 4 Rh+ E, and 5 Rh− e. Information isavailable on the mother’s parents, who are Rh+ E andRh− e. One of the 10 children (who is Rh+ E) marriessomeone who is Rh+ e, and they have an Rh+ E child.a. Draw the pedigree of this whole family.b. Is the pedigree in agreement with the hypothesisthat the Rh+ allele is dominant and Rh− is recessive?c. What is the mechanism of transmission ofelliptocytosis?d. Could the genes governing the E and Rh phenotypesbe on the same chromosome? If so, estimate the mapdistance between them, and comment on your result
For the following diseases with their potential pedigree, mode of inheritance and the responsible gene: (Pedigrees A, B, and C) -> Do Pedigree B Pedigree B, Autosomal dominant, Huntingtin gene Pedigree B, Autosomal dominant, CFTR gene Pedigree B, Autosomal dominant, HexA gene Pedigree B, Autosomal dominant, FGFR3 gene Pedigree A, Autosomal recessive, CFTR gene Pedigree A, Autosomal recessive, Beta-globin gene Pedigree A, Autosomal dominant, FGFR3 gene Pedigree B, X-linked dominant. Factor VIII gene Pedigree A, Autosomal dominant, Beta-globin gene Pedigree A. Autosomal recessive, Huntingtin gene Pedigree C, X-linked recessive, Factor VIII gene Pedigree A, Autosomal recessive, HexA gene
For the following diseases with their potential pedigree, mode of inheritance and the responsible gene: (Pedigrees A, B, and C) -> Do Pedigree C Pedigree B, Autosomal dominant, Huntingtin gene Pedigree B, Autosomal dominant, CFTR gene Pedigree B, Autosomal dominant, HexA gene Pedigree B, Autosomal dominant, FGFR3 gene Pedigree A, Autosomal recessive, CFTR gene Pedigree A, Autosomal recessive, Beta-globin gene Pedigree A, Autosomal dominant, FGFR3 gene Pedigree B, X-linked dominant. Factor VIII gene Pedigree A, Autosomal dominant, Beta-globin gene Pedigree A. Autosomal recessive, Huntingtin gene Pedigree C, X-linked recessive, Factor VIII gene Pedigree A, Autosomal recessive, HexA gene
Hemophilia A is caused by a recessive X-linked allele that encodes a defective form of a clotting protein. If a affected father and a mother who is known to not be a carrier have children, what percentage of female offspring will have hemophilia?
For the following diseases with their potential pedigree, mode of inheritance and the responsible gene: (Pedigrees A, B, and C) -> Do Pedigree A Pedigree B, Autosomal dominant, Huntingtin gene Pedigree B, Autosomal dominant, CFTR gene Pedigree B, Autosomal dominant, HexA gene Pedigree B, Autosomal dominant, FGFR3 gene Pedigree A, Autosomal recessive, CFTR gene Pedigree A, Autosomal recessive, Beta-globin gene Pedigree A, Autosomal dominant, FGFR3 gene Pedigree B, X-linked dominant. Factor VIII gene Pedigree A, Autosomal dominant, Beta-globin gene Pedigree A. Autosomal recessive, Huntingtin gene Pedigree C, X-linked recessive, Factor VIII gene Pedigree A, Autosomal recessive, HexA gene
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