The A gene is autosomal. A man with genotype Aa has a child with a woman whose genotype is AA. Their child's genotype is Aaa and this child is viable. What can be concluded? Choose all that apply. The child has Downs syndrome The child's genotype is aneuploid The child has Klinefelters Syndrome Non-disjunction occurred during O meiosis I or meiosis II in the mother (cannot determine which stage) Non-disjunction occurred in either O meiosis I or meiosis II in the father (cannot determine which stage) The child's genotype is euploid Non-disjunction occurred during meiosis II in the father

The A gene is autosomal. A man with genotype Aa has a child with a woman whose genotype is AA. Their child's genotype is Aaa and this child is viable. What can be concluded? Choose all that apply. The child has Downs syndrome The child's genotype is aneuploid The child has Klinefelters Syndrome Non-disjunction occurred during O meiosis I or meiosis II in the mother (cannot determine which stage) Non-disjunction occurred in either O meiosis I or meiosis II in the father (cannot determine which stage) The child's genotype is euploid Non-disjunction occurred during meiosis II in the father

Name: answer The A gene is autosomal. A man with genotype Aahas a child with
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Description: answer The A gene is autosomal. A man with genotype Aahas a child with a woman whose genotype is AA.Their child's genotype is Aaa and this child isviable. What can be concluded? Choose all thatapply.The child has Downs syndromeThe child's genotype is

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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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. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
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?
Shown above is a family pedigree tree in which family members afflictedwith the disease Haemophilia are shown with filled-in squares (male) or circles (females). A couple is trying to determine the likelihood of passingon the disease to their future children (represented by the ? symbolabove) because the hemophilia runs in the woman’s family. Turner syndrome is a disease in which an individual is bornwith only a single X chromosome. Suppose the woman in thecouple is a carrier for hemophilia and has a child with Turnersyndrome. Would this child have the disease?
In which parent the chromosomal mutation occurred. Father or Mother At what stage in meiosis (I or II) the chromosomal mutation occurred see image attached
Consider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9X
d) If the proband (III-2) marries a heterozygous woman, what is the probability that they will have a heterozygous daughter? If this same couple has a son, what is the probability that he will be affected? Show complete solutions and box or highlight your final answers.
The gene asSociated with the recessive disorder hemophilia is on the X chromosome, where h is the recessive copy and H is the dominant copy. The picture ("starting cell") shows a diploid cell from a male who has hemophilia, showing his X and Y chromosomes. Starting cell Xh Y Which of these pictures A-E will NOT occur at any time in meiosis of this starting cell? Select ALL that apply. D Xh Xh C YIY В X YI YY E Xh Y A X Xh ¡YY
identify and label the chromosomes number (1-22). also, identify if it is male or female by labeling the X and Y chromosomes (this may be normal or abnormal)
Sickle cell anemia is inherited as an autosomal recessive condition. It also exhibits incomplete dominance in that the heterozygous genotype displays a mild form of the disease known as sickle cell trait while individuals with the homozygous recessive genotype have a severe form of SCA. A man who has severe sickle cell anemia marries a woman who suffers from a mild trait. What is the probabilitu they will have a child with severe SCA?What is the probability they will have a child with mild SCA? What is the probability they will have a normal child? Show ALL work using punnett squares.
Under normal conditions, the members of a pair of homologous chromosomes separate from each other __________________. during anaphase of meiosis I and meiosis II during anaphase of mitosis, meiosis I, and meiosis II only during anaphase of meiosis I only during anaphase of meiosis II only during anaphase of mitosis
Assuming sexual reproduction and that no mutations have occurred. Please draw one possible genotype for offspring of these parents including two chromosomes with 3 alleles each.
Mutations of the CFTR gene causes ( Cerebral Plasy or Cystic Fibrosis or Craniofacial timbrosis or Fragile X syndrome). ? Which is (a/an) (Autosomal Dominant or X-linked ressessive or X-linked dominat or Recessive Deleterious)