Arabidopsis thaliana is a diploid plant model organism with 2n 10. Please select the number options to match the followingnumber of copies of each gene an Arabidopsis thaliana1 leaf cell hasnumber of chromosomes an Arabidopsis thaliana leaf cellcontains1. two2. fivenumber of chromosomes an Arabidopsis thalianagamete cell contains3. tenpairs of homologous chromosomes an Arabidopsis thaliana2leaf cell contains

Diploid number of chromosomes means that chromosomes are present in pairs. Haploid number of…

number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana - gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana -- leaf cell contains

number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana - gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana -- leaf cell contains

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?
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown? (b) Using a drawing, demonstrate how these chromosomeswould pair during meiosis. Be sure to label the differentsegments of the chromosomes.(c) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement? The woman in above problem has had two miscarriages. Shehas come to you, an established genetic counselor, with thesequestions:(a) Is there a genetic explanation of her frequent miscarriages?
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown? (b) Using a drawing, demonstrate how these chromosomeswould pair during meiosis. Be sure to label the differentsegments of the chromosomes.(c) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement? The woman in above problem has had two miscarriages. Shehas come to you, an established genetic counselor, with thesequestions:(a) If not, what is the chance that she could have a normalchild? Provide an informed response to her concerns
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown?
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement?
A tomato geneticist attempts to assign five recessivemutations to specific chromosomes by using trisomics.She crosses each homozygous mutant (2n) with each ofthree trisomics, in which chromosomes 1, 7, and 10 takepart. From these crosses, the geneticist selects trisomicprogeny (which are less vigorous) and backcrosses themto the appropriate homozygous recessive. The diploidprogeny from these crosses are examined. Her results, inwhich the ratios are wild type:mutant, are as follows:Which of the mutations can the geneticist assign towhich chromosomes? (Explain your answer fully.)
The accompanying pedigree shows a family in which one child (II-1) has an autosomal recessive condition. On the basis of this fact alone, provide the following information. 1) What is the chance that among the three children in generation II who have the dominant phenotype, one of them is AAAA and two of them are AaAa? (Hint: Consider all possible orders of genotypes.) Express your answer to two decimal places.
10 cM separates two hypothetical autosomal human genes. The dominant alleles are have complete penetrance and will result to Crossed eyes (e+) and short thumbs (th+). Four children are born to a normal guy and cross-eyed, small-thumbed woman. Two of the children have short thumbs and the other two have crossed eyes. She is carrying her fifth child. What is the probability that this fifth child will be cross-eyed and have short thumbs?
Campomelic dysplasia (CMD1) is a congenital humansyndrome featuring malformation of bone and cartilage.It is caused by an autosomal dominant mutation of agene located on chromosome 17. Consider the followingobservations in sequence, and in each case, draw whateverappropriate conclusions are warranted.(a) Of those with the syndrome who are karyotypically46,XY, approximately 75 percent are sex reversed,exhibiting a wide range of female characteristics.(b) The nonmutant form of the gene, called SOX9, isexpressed in the developing gonad of the XY male,but not the XX female.(c) The SOX9 gene shares 71 percent amino acid codingsequence homology with the Y-linked SRY gene.(d) CMD1 patients who exhibit a 46,XX karyotypedevelop as females, with no gonadal abnormalities.
A rare blinding disease that has a relation to dengenerative factors is partially penetrant. In the following pedigrees for two families, the affected symptomatic individuals (black circles and squares) have been diagnosed with this disease due to the mutation in mitochondrial DNA m.14484T>C. If III.4 is homoplasmic for m.14484T>C in hair, blood, urine and other tissues examined. What will occur with IV.7 then?
Shown (the picture) is a partial gene map for Drosophila melanogaster. In Drosophila, vermillion eyes are recessive (v) to red eyes (V) and the gene resides on chromosome 1; dumpy wings (d) are recessive to normal wings (D); and speckled body (s) is recessive to normal body (S). The dumpy wing gene and the speckle body gene are linked on chromosome 2 and have 7 map units separating them. Assume that chromosome 1 and 2 are autosomal chromosomes and that there are no known interactions between the genes. In an experiment, an undergraduate student crossed pure breeding males that have red eyes, normal wings, and normal body with pure breeding females who have vermillion eyes, dumpy wings, and speckle body to obtain an F1. Using the Drosophila gene map, answer the following questions: a. What fraction of the F1 gametes will contain the alleles for red eyes, normal wings and speckle body? b. In a cross between an F1 flies and a tester, what fraction of the offspring do you expect to have…