Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 5, Problem 29P
A
dominant neuromuscular disorder spinocerebellar ataxiatype
Does either group of lod scores indicate statisticallysignificant odds in favor of genetic linkage? Explainyour answer.
What is the maximum value for each set of lod scores?
Based on the available information, is
Based on available information, is
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Concordance studies of twins for a neurodegenerative disorder show MZ=
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A man in his early 30s suddenly developed weakness in hishands and neck, followed weeks later by burning musclepain—all symptoms of late-onset muscular dystrophy. Hisinternist ordered genetic tests to determine whether he had oneof the most common adult-onset muscular dystrophies—myotonicdystrophy type 1 (DM1) or myotonic dystrophy type 2 (DM2). Thetests detect mutations in the DMPK and CNBP genes, the onlygenes known to be associated with DM1 and DM2. While awaitingthe results of the gene tests, the internist explained that thedisease-causing mutations in these genes do not result in changesto the coding sequence. Rather, myotonic dystrophies resultfrom increased, or expanded, numbers of tri- and tetranucleotiderepeats in the 3 untranslated region of the DMPK or CNBP genes.The doctor went on to explain that the presence of RNAs withexpanded numbers of repeats leads to aberrant alternative splicingof other mRNAs, causing widespread disruption of cellular pathways.This discussion…
A man in his early 30s suddenly developed weakness in hishands and neck, followed weeks later by burning musclepain—all symptoms of late-onset muscular dystrophy. Hisinternist ordered genetic tests to determine whether he had oneof the most common adult-onset muscular dystrophies—myotonicdystrophy type 1 (DM1) or myotonic dystrophy type 2 (DM2). Thetests detect mutations in the DMPK and CNBP genes, the onlygenes known to be associated with DM1 and DM2. While awaitingthe results of the gene tests, the internist explained that thedisease-causing mutations in these genes do not result in changesto the coding sequence. Rather, myotonic dystrophies resultfrom increased, or expanded, numbers of tri- and tetranucleotiderepeats in the 3 untranslated region of the DMPK or CNBP genes.The doctor went on to explain that the presence of RNAs withexpanded numbers of repeats leads to aberrant alternative splicingof other mRNAs, causing widespread disruption of cellular pathways.This discussion…
Chapter 5 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 5 - For parts a, b, and c, draw a diagram illustrating...Ch. 5 - 5.2 In a diploid species of plant, the genes for...Ch. 5 - A pure-breeding tall plant producing oval fruit as...Ch. 5 - 5.4 Genes E and H are syntenic in an experimental...Ch. 5 - In tomato plants, purple leaf color is controlled...Ch. 5 - 5.6 In Drosophila, the map positions of genes are...Ch. 5 - 5.7 Genes A, B, and C are linked on a chromosome...Ch. 5 - Gene G recombines with gene T at a frequency of...Ch. 5 - Genes A, B, C, D, and E are linked on a...Ch. 5 - Syntenic genes can assort independently. Explain...
Ch. 5 - 5.11 The recombination frequency between linked...Ch. 5 - On the DrosophilaX chromosome, the dominant allele...Ch. 5 - Researchers cross a corn plant that is pure -...Ch. 5 - 5.14 syndrome is an autosomal disorder affecting...Ch. 5 - 5.15 Three dominant traits of corn seedlings,...Ch. 5 - 5.16 In a diploid plant species, an with the...Ch. 5 - Prob. 17PCh. 5 - The Rh blood group in humans is determined by a...Ch. 5 - 5.19 Genetic linkage mapping for a large number of...Ch. 5 - 5.20 with the genotype form tetrads in the...Ch. 5 -
Gene and gene are genetically linked. Answer...Ch. 5 -
T. H. Morgan’s data on eye color and wing form,...Ch. 5 - Prob. 23PCh. 5 -
The boss in your laboratory has just heard of a...Ch. 5 - In rabbits, chocolate-colored fur (w+) is dominant...Ch. 5 - Prob. 26PCh. 5 - 5.27 In tomatoes, the allele for tall plant height...Ch. 5 - 5.28 Neurofibromatosis is an autosomal dominant...Ch. 5 - A 2006 genetic study of a large American family...Ch. 5 - 5.30 A experiment examining potential genetic...Ch. 5 - A genetic study of an early onset form of heart...Ch. 5 - In experiments published in 1918 that sought to...Ch. 5 - DNA sequence for 10 individuals are Identify the...Ch. 5 - 5.34 The accompanying pedigree below shows a...Ch. 5 - 5.35 Based on previous family studies, an...Ch. 5 - Divide a clean sheet of paper into four quadrant...Ch. 5 - 5.37 For six genes known to be linked on chromosme...
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- A man in his early 30s suddenly developed weakness in hishands and neck, followed weeks later by burning musclepain—all symptoms of late-onset muscular dystrophy. Hisinternist ordered genetic tests to determine whether he had oneof the most common adult-onset muscular dystrophies—myotonicdystrophy type 1 (DM1) or myotonic dystrophy type 2 (DM2). Thetests detect mutations in the DMPK and CNBP genes, the onlygenes known to be associated with DM1 and DM2. While awaitingthe results of the gene tests, the internist explained that thedisease-causing mutations in these genes do not result in changesto the coding sequence. Rather, myotonic dystrophies resultfrom increased, or expanded, numbers of tri- and tetranucleotiderepeats in the 3 untranslated region of the DMPK or CNBP genes.The doctor went on to explain that the presence of RNAs withexpanded numbers of repeats leads to aberrant alternative splicingof other mRNAs, causing widespread disruption of cellular pathways.This discussion…arrow_forwardstion 6 of 18 Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F, females in a testcross. She obtains the flies from this testcross. Phenotype Number of flies vg* qu+ 230 vg qu 224 vg qut vg* qu 97 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for this hypothesis. Provide the X2 to one decimal place. X2 = Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? the partial table of critical values for X2 calculations to test this hypothesis.arrow_forwardNiemann Pick Type C disease is a recessive disorder that causes the accumulation of cholesterol and other lipids in lysosomes, ultimately affecting both the liver and the nervous system. Below are the genotypes and phenotypes of offspring of a family with a history of Niemann Pick. 7 NN ( all normal phenotype) 3 Nn (all normal phenotype) 4 nn (1 early onset dementia, 1 mid-life onset dementia, 2 late-onset dementia). From this information, Niemann-Pick disease is an example of: A) variable expressivity B) incomplete dominance C) incomplete penetrance D) variable expressivity and incomplete penetrance E) multiple allelesarrow_forward
- Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?arrow_forwardA 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?arrow_forwardA 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?arrow_forward
- 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?arrow_forwardAchondroplasia is an autosomal dominant form of dwarfism caused by a single gene mutation. Calculate the mutation rate of this gene given the following data: 10 achondroplastic births to unaffected parents in 245,000 births.arrow_forwardA pedigree analysis was performed on the family of a man with schizophrenia. Based on the known concordance statistics, would his MZ twin be at high risk for the disease? Would the twins risk decrease if he were raised in an environment different from that of his schizophrenic brother?arrow_forward
- Tay–Sachs disease is caused by loss-of-function mutations ina gene on chromosome 15 that encodes a lysosomal enzyme.Tay–Sachs is inherited as an autosomal recessive condition.Among Ashkenazi Jews of Central European ancestry, about1 in 3600 children is born with the disease. What fraction ofthe individuals in this population are carriers?arrow_forwardPlease draw it out so I understand how it's suppose to be drawnarrow_forwardFollowing the analysis of a pedigree, a genetic link at 4cM is considered between a mutation leading to a pathology and the molecular marker HUMTH01. The study counts 14 "parental" and 3 "recombinant" individuals. We call p(theta=0.04) is the probability of obtaining such a pedigree in case of a 4cM genetic linkage, p(theta=0.5) is the probability of obtaining such a pedigree in case of independence between the mutation and the marker, Z(theta=0.04) the value of the Lod-score under the assumption of 4cM genetic linkage. Tick all the correct answers: p(theta=0,04)=1,79.10E-9 and Z(theta=0,04)=0,47 p(theta=0,5)=7,18.10E-12 and Z(theta=0,04)=0,77 and Z(theta=0,04)=0,67 p(theta=0,04)=2,75.10E-10 p(theta=0,5)=6,04.10E-10 and Z(theta=0,04)=0,47 p(theta=0,5)=9,36.10E-12 and Z(theta=0,04)=1,33 p(theta =0,5)=5,82.10E-11 and Z(theta=0,04)=0,67 no correct answer p(theta=0,04)=4,31.10E-11 p(theta=0,04)=2,01.10E-10 and Z(theta=0,04)=0,77 and Z(theta=0,04)=1,33arrow_forward
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