X-linked recessive, Unaffected. father Carrier mother XY |Unaffected ] Affected ICarrier XY Unaffected Unaffected Carrier daughter Affected daughter U.S. National Library of Medicine son son Color blindedness is a sex-linked trait. If we could see the pedigree chart for several more generations of the family illustrated here, we would expect A) more males to be color blind. B) móre females to be color blind. no females to ever be color blind. D) an equal number of males and females that are color blind.

X-linked recessive, Unaffected. father Carrier mother XY |Unaffected ] Affected ICarrier XY Unaffected Unaffected Carrier daughter Affected daughter U.S. National Library of Medicine son son Color blindedness is a sex-linked trait. If we could see the pedigree chart for several more generations of the family illustrated here, we would expect A) more males to be color blind. B) móre females to be color blind. no females to ever be color blind. D) an equal number of males and females that are color blind.

Name: X-linked recessive, carrier motherUnaffected.CarriermotherfatherX YXX
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Description: X-linked recessive, carrier motherUnaffected.CarriermotherfatherX YXX|Unaffected|AffectedI CarrierXYUnaffected UnaffectedCarrierdaughterAffectedsondaughterU.S. National Library of MedicinesonColor blindedness is a sex-linked trait. If we could see t

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter4: Pedigree Analysis In Human Genetics

Section: Chapter Questions

Problem 19QP: Analysis of X-Linked Dominant and Recessive Traits Suppose a couple, both phenotypically normal,...

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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 As a genetic counselor investigating a genetic disorder in a family, you are able to collect a four-generation pedigree that details the inheritance of the disorder in question. Analyze the information in 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-linked
Duchenne muscular dystrophy is a recessive disorder caused by a rare,loss-of-function allele that is located on the X chromosome in humans. Anunaffected woman (i.e., without disease symptoms) who is heterozygousfor the X-linked allele causing Duchenne muscular dystrophy has childrenwith a man with a functional (non-disease-causing) allele. What is theprobability that this couple will have an unaffected son?
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement?
Two mothers give birth to sons at the same time at a busy urbanhospital. The son of mother 1 has hemophilia, a disease causedby an X-linked recessive allele. Neither parent has the disease.Mother 2 has a son without hemophilia, despite the fact thatthe father has hemophilia. Several years later, couple 1 suesthe hospital, claiming that these two newborns were swappedin the nursery following their birth. As a genetic counselor, youare called to testify. What information can you provide the juryconcerning the allegation?
Attached are three pedigrees. For each trait, considerwhether it is or is not consistent with X-linked recessiveinheritance. In a sentence or two, indicate why or why not.
Classical hemophilia is a sex-linked disease caused by a recessive gene on the X chromosome. (Hemophilia refers to diseases that cause delays in blood clotting.) If a woman who is acarrierof classical hemophilia has children with a normal male, give the ratios of the possible offspring with respect to classical hemophilia. Be sure to state both the genotypes and the phenotypes of each offspring. For genotypes, use X for a normal X chromosome, Xh for an X chromosome with the hemophilia gene, and Y for a normal Y chromosome. For phenotypes, if the offspring is female, be sure to state if homozygous normal, a carrier, or has the disease. If the offspring is a male, be sure to state if normal or has the disease.
Color blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?
A man, Penoy, whose sister died in early childhood from a recessive lethal disease marries a woman Esmae, who has the same family history. Because Penoy has survived beyond childhood, he does not have the disease, but he may be a carrier (i.e. heterozygous, as may also be the case with Esmae). What is the probability that their first child will suffer from the disease? [Hint: first calculate the probability that Penoy is heterozygous; then determine the probability that both parents are carriers. Remember that he has survived to adulthood when calculating this probability].
X-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%