Color blindness is an X-linked recessive disease. A female and a male with normal vision have 4 children. The phenotypes of the children are listed below: 2 Colorblind males, 1 normal vision female, one normal vision male. Given these results, the female must be a carrier for colorblindness. True False
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- Analysis of X-Linked Dominant and Recessive Traits A young boy is color-blind. His one brother and five sisters are not. The boy has three maternal uncles and four maternal aunts. None of his uncles children or grandchildren is color-blind. One of the maternal aunts married a color-blind man, and half of her children, both male and female, are color-blind. The other aunts married men who have normal color vision. All their daughters have normal vision, but half of their sons are color-blind. a. Which of the boys four grandparents transmitted the gene for color blindness? b. Are any of the boys aunts or uncles color-blind? c. Is either of the boys parents color-blind?The young woman shown at right has albinismvery pale skin, white hair, and pale blue eyes. This phenotype is due to the absence of melanin, which imparts color to the skin, hair, and eyes. It typically is caused by a recessive allele. In the following situations, what are the probable genotypes of the father, the mother, and their children? a. Both parents have normal phenotypes; some of their children are albino and others are not. b. Both parents and all their children are albino. c. The mother is not albino, the father is albino, and one of their four children is albino.Redgreen color blindness is an X-linked recessive disorder in humans. Your friend is the daughter of a color-blind father. Her mother had normal color vision, but her maternal grandfather was color-blind. What is the probability that your friend is color-blind? (a) 1 (b) (c) (d) (e) 0
- 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?Can a male be a carrier of red-green color blindness?Cystic fibrosis is an autosomal disease that mainly affects the white population, and 1 in 20 whites are heterozygotes. Genetic testing can diagnose heterozygotes. Should a genetic screening program for cystic fibrosis be instituted? Should the federal government fund it? Should the program be voluntary or mandatory, and why?
- a. The ability to taste the chemical phenylthiocarbamideis an autosomal dominant phenotype, and the inabilityto taste it is recessive. If a taster woman with a nontasterfather marries a taster man who in a previous marriagehad a nontaster daughter, what is the probability thattheir first child will be(1) A nontaster girl(2) A taster girl(3) A taster boyb. What is the probability that their first two childrenwill be tasters of either sex?(a) Enter the parent phenotypes and complete the Punnett square Inheritance of sex linked recessive tralts Example: Hemophilia Female paront phenotype Inheritance of hemophilia is sex linked. Males with the recessive (hemophilia) allele, are affected. Females can be carriers. (normal female) Xxh (carrier female) ben-N werxhxh (hemophiliac female) leizobXY (normal male) xhy (hemophiliac male) Male parent phenotype: Using the codes: XX eggs (a) Enter the parent phenotypes and complete the Punnett square for a cross between a normal male and a carrier female. sperm (b) Give the ratios for the phenotypes from this cross. Phenotype ratios: Inheritance of sex linked dominant traits Example: Sex linked form of rickets A rare form of rickets is inherited on the X chromosome. Using the codes: XX Female parent phenotype: alvo eo (normal female); XY (normal male) (affected heterozygote female) XRXR (affected female) XRY Male parent phenotype: XRX eggs (affected male) Nor a cross between an…Red-green color blindness is an X-linked recessive disorder. If Allison is heterozygous (a carrier), and her husband, Michael, is NOT colorblind. What is the chance that their male children will be colorblind? (Note: you are calculating the probability for their MALE children only, in other words if they have 1 male child, what is the probability that he will be born with the disease?) 0% 25% or 1/4 50% or 2/4 75% or 4/4 100% or 4/4