StompOnStep1.com a. Assume this pedigree represents inheritance of a human disease. What is the most likely mode of inheritance for this disease? b. What is the most likely genotype person III-IV if we use A and a as the alleles for this disease? a. Autosomal dominant, b. Aa a. Y linked, b. Aa a. Autosomal dominant, b. AA a. Autosomal recessive, b. aa

StompOnStep1.com a. Assume this pedigree represents inheritance of a human disease. What is the most likely mode of inheritance for this disease? b. What is the most likely genotype person III-IV if we use A and a as the alleles for this disease? a. Autosomal dominant, b. Aa a. Y linked, b. Aa a. Autosomal dominant, b. AA a. Autosomal recessive, b. aa

Name: O StompOnStep1.coma. Assume this pedigree represents inheritance of a
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Description: O StompOnStep1.coma. Assume this pedigree represents inheritance of a human disease.What is the most likely mode of inheritance for this disease?b. What is the most likely genotype person III-IV if we use A and a asthe alleles for this disease?a. Au

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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Homozygous dominant parent (PP), and a Homozygous recessive or just simply say recessive parent (pp): a. Fill in the Punnett square. Each box represents a genotype possibility for an offspring. b. Place the allele donated by each parent in the corresponding box. Now list the possible genotypes and their corresponding phenotype. c. If an individual's genotype is heterozygous, the dominant trait will be expressed in the phenotype. Give the percent possible for the phenotypes. P P p p Genotype: ______________ Phenotype: ______________ Phenotype % probable: __________ 2. Cross between a Homozygous dominant parent (PP) and a Heterozygous parent (Pp). Fill in as in step one. 3. Cross between a Heterozygous parent (Pp), and another Heterozygous parent (Pp). Fill-in as before. 4. Test cross: A test cross is between a recessive parent (pp), and a Heterozygous parent…
Retinitis pigmentosa, a form of blindness in man, maybe caused either by a dominant autosomal gene R, or a recessive autosomal gene a. An afflicted man whose parents are both normal marries a woman with genotype AaRr. a. What proportion of the children are expected to suffer from this affliction if R and A are inherited independently? b. If this couple want to have normal children only, what isthe probability of having normal children?
11. Below is a pedigree chart of an autosomal recessive disorder. Answer the following questions using the correct genetic terminology (do not just write letters like “Ee”). A. What is the genotype of individual 1 in generation II? B: What is the genotype of individual 2 in generation I?C: Is it true that individuals 6, 7, 8, 9, and 10 in generation III all have the same genotype? Why or why not?
Analysis of Autosomal Recessive and Dominant Traits The father of 12 children begins to show symptoms of Huntington disease. a. What is the probability that Sam, the mans second-oldest son (II-2), will suffer from the disease if he lives a normal life span? (Sams mother and her ancestors do not have the disease.) b. Can you infer anything about the presence of the disease in Sams paternal grandparents?
Clubfoot is a common congenital birth defect. This defect is caused by a number of genes but appears to be phenotypically distributed in a noncontinuous fashion. Geneticists use the threshold model to explain the occurrence of this defect. Explain this model. Explain predisposition to the defect in an individual who has a genotypic liability above the threshold versus an individual who has a liability below the threshold.
Analysis of Autosomal Recessive and Dominant Traits a. What pattern of inheritance is suggested by the following pedigree? b. For genotype assignment, assume that the pedigree is for an autosomal dominant trait and that the affected male in the first generation is heterozygous. Assign genotypes to all other individuals in the pedigree.
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 Autosomal Recessive and Dominant Traits In the following pedigree, assume that the father of the proband is homozygous for a rare trait. What pattern of inheritance is consistent with this pedigree? In particular, explain the phenotype of the proband.
Analysis of Autosomal Recessive and Dominant Traits Huntington disease is a rare, fatal disease that usually develops in the fourth or fifth decade of life. It is caused by a single autosomal dominant allele. A phenotypically normal man in his twenties who has a 2-year-old son of his own learns that his father has developed Huntington disease. What is the probability that he himself will develop the disease? What is the chance that his young son will eventually develop the disease?
Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12