A male has a particular X-linked recessive genetic disorder. His partner is normal,but her father had the disorder. What is the chance that their sons will have thedisorder?Select one:Oa. 50%O b. 0%O c. 100%O d. impossible to determine from the information given.O e.25%LENACTranscribed Image Text: A male has a particular X-linked recessive genetic disorder. His partner is normal, but her father had the disorder.What is the chance that their sons will have the disorder? Select one: O a. 50% O b. 0% O c. 100% O d. impossible to determine from theinformation given. O e. 25%Po11W*****

Alleles are the alternative forms of a gene that are located on the same locus of a homologous…

A male has a particular X-linked recessive genetic disorder. His partner is normal, but her father had the disorder. What is the chance that their sons will have the disorder? Select one: O a. 50% O b. 0% O c. O d. impossible to determine from the information given. О e. 100% 25%

A male has a particular X-linked recessive genetic disorder. His partner is normal, but her father had the disorder. What is the chance that their sons will have the disorder? Select one: O a. 50% O b. 0% O c. O d. impossible to determine from the information given. О e. 100% 25%

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 6TYU: Redgreen color blindness is an X-linked recessive disorder in humans. Your friend is the daughter of...

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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
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?
One type of chronic porphyria is X-linked dominant. A woman with chronic porphyria whose father was normal has children with a normal man. What is the phenotypic ratio of their children? Oa. 1 female with porphyria: 1 normal male Ob. all of their children have porphyria O c 2 normal females: 1 normal male: 1 male with porphyria O d. 1 nomal female: 1 female with porphyria: 1 normal male: 1 male with porphyria O e. 2 females with porphyria: 1 male with porphyria: 1 normal male Of. all children are normal Next page Follow LIO MacBook Air DII F9 FB F7 80 F5 F3 esc F2 % 2# $ 8 6 4 この

D. If a man is color-blind, which parent did he inherit the gene from? Explain your answer. E. Which parent does a man get all his x-linked genes from? F. If a woman is color-blind, what do you know about the genotypes and phenotypes of her parents? G. Are men more likely than women to get a genetic disease which is x-linked and dominant? Are they more likely to get a disease that is x-linked and recessive? Explain your answers. H. Can a man with normal color vision have a colorblind daughter? Explain your answer. I. Which parent determines the sex of a child?
A male has a particular X-linked recessive genetic disorder. His partner is normal, but her father had the disorder. What is the chance that their sons will have the disorder? Select one: O a. 50% O b. 0% O c. 100% O d. impossible to determine from the information given. O e. 25%
An individual with 46, XX genotype is diagnosed with Duchenne-type Muscular Dystrophy, a recessive X-linked disorder. Genetic tests confirm that this individual is a heterozygote for this disorder. Briefly, but specifically, explain how it’s possible that they are showing symptoms of this disorder.
In this part, you will work out on X-linked traits. Remember that males have only one X chromosome and females have two. For X-linked recessive, use the following designations: XA= normal Xa= the trait (a genetic disease or abnormality) For X-linked dominant, use the following designations: XA= the trait (a genetic disease or abnormality) Xa= normal Determine if the pedigrees below can be for a trait that is X-linked recessive. Write the genotype next to the symbol for each person in the pedigree below. A. Is it possible that this pedigree is for an X-linked recessive trait? B. What can you conclude about the children of mothers affected with an X-linked recessive characteristic? C. What can you conclude about the father of an affected female?
A man who has color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The woman’s father had color blindness. Color blindness is determined by an X-linked gene, and blood type is determined by an autosomal gene. a. What are the genotypes of the man and the woman? b. What proportion of their children will have color blindness and type B blood? c. What proportion of their children will have color blindness and type A blood? d. What proportion of their children will be color blind and have type AB blood?
I. White eye cross White eye color is a recessive trait found on the X chromosome. Also called Sex-linked. This is called sex-linked. Gene symbols Xw White eye color X normal red eye color Y it’s a boy The parents were White eye color female XwXw and red eye color male XY Diagram of P1 cross: You are crossing XwY and XXw (figure out the phenotype) Diagram of F1 cross: Give expected F2 phenotypic ratio: Give expected F2 genotypic ratio:
Match the chromosome disorder to its descriptionin the key. Jacobs syndrome a. female with undeveloped ovaries and uterus, unable to undergopuberty, normal intelligence, can live normally with hormonereplacementb. XXY male, can inherit more than two X chromosomesc. male or female, mentally impaired, short stature, flat face, stubbyfingers, large tongue, simian palm creased. XXX or XXXX femalee. caused by nondisjunction during spermatogenesis