Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
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
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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…
Concordance studies of twins for a neurodegenerative disorder show MZ=
46% and DZ= 15%. Further studies have shown a possible link to a gene
on chromosome 9, however, there are some individuals in the study who
have the allele but do not develop the disorder (group 1), and there are
other individuals who do not have the allele yet develop the disorder
(group 2). Amita's older sister and maternal uncle have this disorder.
Currently, Amita & her 2 younger brothers do not show symptoms. Amita's
paternal grandfather was rumored to have this disorder.
1. Draw the pedigree for Amita's family and determine the mode of
inheritance if any.
2. Explain how the 2 groups in the study could be possible?
3. What would you tell Amita about the heritability of this disorder?
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...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- stion 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_forwardFamilial 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 if the couple wanted prenatal testing so that a normal fetus could be aborted?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 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_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_forward
- A 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_forwardTay–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_forwardEhler-Danlos syndrome is a rare disorder caused by a mutation ina gene that encodes a protein called collagen (type 3 A1). Collagenis found in the extracellular matrix that plays an important role inthe formation of skin, joints, and other connective tissues. Peoplewith Ehler-Danlos syndrome have extraordinarily flexible skin and very loose joints. The pedigree below contains several individualsaffected with this syndrome, shown with black symbols. Based onthis pedigree, does the syndrome appear to follow autosomalrecessive, autosomal dominant, X-linked recessive, or X-linkeddominant inheritance? Explain your reasoning.arrow_forward
- 5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoomarrow_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_forwardI have seen that this was answered as C, Why is the answer C, how is that not evidence of it being genetic? Shouldnt it be none of the above? Question: Of the following, which supports the idea that alcoholism has no genetic or a low genetic component? a) Some strains of mice select alcohol over water 75% of the time, whereas others shun alcohol. b) The concordance value is 55% for MZ twins and 28% for DZ twins. c) Biological sons of alcoholic men who have been adopted have a rate of alcoholism more like that of their adoptive fathers. d) There is a 20% to 25% risk of alcoholism in the sons of alcoholic men. e) None of these.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY