1)What type of chromosome abnormality does the father have?a)Duplicationb)Inversionc)Deletiond)Translocation 2)Why does the father have a normal phenotype?a) Because the translocation was reciprocal.b) Because he has one wild-type copy of each chromosome.c) Because the translocation occurred only in the father's gametes.d) Because no genes are missing. A healthy couple with a history of previous spontaneous abortions just had a child with cri du chat syndrome-a disease caused by one copy ofa deletion of the end of chromosome 5. Their physician orders karyotype analysis of both parents and the child. The karyotype results forchromosome 5 and 12 are shown below.MotherFatherChildChromosome5 125 125 12I II .

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1) What type of chromosome abnormality does the father have? a)Duplication b)Inversion c)Deletion d)Translocation

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. 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?
2 3 4 5 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ху Which chromosome set has an extra What diagnosis would you give this patient?. Finish the notation for this patient's karyotype: What causes a dark band on the chromosome? What is a centromere?
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During a practical, a group of biology students are tasked to study the cell cycle of an insect. This insect has the karyotype 2n=6, with genotype Ff Gg Hh KK. The students must draw a cell at the end of telophase of mitosis, using red lines for maternal chromosomes, blue lines for paternal chromosomes and black dots to represent centromeres. The drawing provided by the students is shown below. kt a) Identify three mistakes the students made when producing this diagram. In your answer only refer to elements already present in the figure. b) This fly is used in a testcross. What is the probability of producing a fly with genotype ff Gg hh Kk?
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A women is heterozygous, Aa, for classic albinism and this gene maps to chromosome 11, and is a carrier for haemophilia, an X-linked trait (H/h). Provide a diagram of chromosomes 11 and the sexchromosomes (show the alleles associated with albinism and haemophilia) in the cells involved with normal oogenesis for this female. Start with the oogonium and end with the mature ovum; you can ignore all polar bodies. Your diagrams should show the chromatids clearly and all cells must be labelled. Indicatethe total number of chromosomes and chromatids for each cell stage.
n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines sing the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspring
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