1. Study the given alleles. Write the correct phenotype for each genotype.X – normalGen otypexC - Color-blindPhenotypeXXXYXXCxCY2. Study the given alleles. Write the correct genotype for each phenotype.xH - HemophiliacPhenotypeX- normalGen otypeHemophiliac femaleHemophiliac maleNormal female carrier of the geneNormal maleNormal female283. 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 areb. There arec. There ared. There are% normal sons.% normal daughters.% color-blind sons.% color-blind daughters.% normal female, carrier of the disorder.orororore. There areor

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Description: 1. Study the given alleles. Write the correct phenotype for each genotype.X – normalGen otypexC - Color-blindPhenotypeXXXYXXCxCY2. Study the given alleles. Write the correct genotype for each phenotype.xH - HemophiliacPhenotypeX- normalGen otypeHemo

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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 Xxc XXC xCY

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 11GP: A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the...

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A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
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?
1. A woman who is heterozygous for hemophilia marries a man with hemophilia. What are the genotypes and phenotypes of the children? a) What is the genotypic ratio of the above cross (Write number next to genotype)? ___________________________________________________________________ b) What is the phenotypic ratio of the above cross? (Write number next to phenotype)? ___________________________________________________________________
1. Hemophilia is due to a sex-linked gene. It is recessive and found on the X chromosome. A woman who is a carrier for hemophilia marries a normal man. What will be the possible phenotypes of their children? Use a punnet square. 2. A phenotypically normal man, who has a hemophiliac brother, marries a normal woman, who is not a carrier. What is the probability that any of their children will be hemophiliac? Use a punnet square. 3. Color-blindness is an X-linked, recessive trait. If a normal-sighted woman, whose father was color-blind, marries a colorblind man, what is the probability that they will have a son who is color-blind? Use a punnet square. 4. A man and woman, both of normal vision, have: 1) a color-blind son (#1) who has a daughter of normal vision 2) a daughter (#1) of normal vision who has one color-blind son and one normal vision son 3) another daughter (#2) of normal vision who has five sons, all with normal vision What are the probable genotypes of the…
7. The ability to roll your tongue (R) is dominant to lack of this ability (r). A. What is the genotype of a man who can roll his tongue if his father couldn’t? B. If a man who is heterozygous for tongue rolling marries a woman who is heterozygous, what would you predict for the genotypes and phenotypes of their children?
8. Huntington’s disease is a degenerative disease of the nervous system that strikes in middle age. The allele that causes the disease (H) is dominant to the allele that results in the normal condition (h). Answer the following questions about the inheritance of this disease. A. What is the genotype of a man who is normal but whose father had Huntington’s disease? B. What is the genotype of a woman who has Huntington’s disease if both of her parents had Huntington’s disease? C. If a man who is heterozygous for Huntington’s disease marries a woman who is normal, what would you expect for the genotypes and phenotypes of their children? D. If a normal man marries a woman who is homozygous for Huntington’s disease, what do you expect for the genotypes and phenotypes of their children?
1. A woman is colorblind. Construct a Punnett square and using versions of the letter “C” determine if she marries a man with normal vision: What are the chances her sons will be colorblind? What are the chances her daughters will be colorblind? What are the chances of having a carrier daughter? 2. Both the husband and wife have normal vision. Their son is colorblind. What can you conclude about the father’s genotype? What about the mother’s genotype? Father: Mother: 3. If a young girl has fragile X syndrome, a recessive trait (f), what is her genotype? What are the possible genotypes of her father and mother? Mother: _________ or __________ Father: _________________ If her brother also developed this condition, which parent (father, mother, or both) contributed a disease allele? Explain your answer as well as constructing Punnett squares to support your answer. 4. Both the mother and father of a hemophiliac son have normal blood clotting. What are the genotypes of the…
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?
identify if it is True or False 1. ABO Blood Group show codominance. 2. Alleles are different forms of genes. 3. Sex-linked inheritance means that only the X-chromosomes are affected. 4. The carriers of hereditary materials are the genes. 5. Hemophilia is a sex-linked trait.
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.