The afflicted individuals in the pedigree have dysphasia-like condition. The gel below shows resultsof PCR amplification of a molecular marker that may be linked to the condition trait.I.|-1 1-2 Il-1 |I-2 |I-3 ||-4 l-1 III-2 III-3 IV-1ВI.14IIE123A. Based on the evidence, is the molecular marker linked to the condition? In your own words, explainyour reasoning.B. What is the mode of inheritance? Explain your reasoningC. Suppose III-3 had a child IV-1 with an afflicted female (no molecular data is available for her).(IV-1 is not indicated in the pedigree, but the molecular data is shown). Based on the data, what isthe risk (the chance) that the child IV will have the condition? Explain your reasoning.

Introduction Dysphasia, commonly known as aphasia, is a communication impairment. It has an impact…

The afflicted individuals in the pedigree have dysphasia-like condition. The gel below shows results of PCR amplification of a molecular marker that may be linked to the condition trait. I. I-1 1-2 Il-1 |I-2 |I-3 ||-4 |Il-1 III-2 III-3 IV-1 1 A П. 3 4 D III. E 2 3 A. Based on the evidence, is the molecular marker linked to the condition? In your own words, explain your reasoning.

The afflicted individuals in the pedigree have dysphasia-like condition. The gel below shows results of PCR amplification of a molecular marker that may be linked to the condition trait. I. I-1 1-2 Il-1 |I-2 |I-3 ||-4 |Il-1 III-2 III-3 IV-1 1 A П. 3 4 D III. E 2 3 A. Based on the evidence, is the molecular marker linked to the condition? In your own words, explain your reasoning.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter16: Reproductive Technology, Genetic Testing, And Gene Therapy

Section: Chapter Questions

Problem 18QP

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A RFLP is discovered that is linked to the gene for Duchenne’s muscular dystrophy (DMD). DMD is an X-linked, recessive trait. The RFLP is 2 map units from the gene for DMD. Consider the following pedigree and Southern blot using a probe that hybridizes to the RFLP. Which band/s is/are associated with DMD? What is the genotype for individuals 3 and 4? (Remember, this is an X linked disease, so use X’s and Y’s to denote). Individual 9 married a man who does NOT have muscular dystrophy, and she is pregnant. DMD is an X-linked trait. What is the probability for their child to have DMD? An amniocentesis is performed and it is determined that 9’s child in utero has only a 10 kb band that hybridizes to the same probe used above. What can you say about the child now?
The figure below represents a two-generation pedigree representing the inheritance of hereditary deafness (HD), which, has been shown to be autosomal recessive. Below the pedigree (and aligned with each person in each generation) are Southern blot gels for two RFLP loci. The BamH1 RFLP locus has alleles of either 7 kb or 5 kb and the EcoRI RFLP locus gives either a 4 kb or 3 kb EcoRI fragment. a. Determine if HD is linked to either of these markers. b. If linkage is detected between HD and one of the RFLPs, are there any recombinant offspring that you can detect? c. If linkage is detected between HD and one of the RFLPs, what is the approximate map distance between the two? Do you think that this distance will efficiently and effectively allow the RFLP to track the HD gene?
A. Look at the pedigree, and DISREGARD individual II-8 for the moment. Is the pattern of inheritance of Unetan syndrome dominant or recessive? You may assume that the gene is FULLY-PENETRANT in this family. Please give two specific reasons that support your conclusion. B. Now, looking at BOTH the pedigree AND at the Southern blot, is this trait autosomal, X-linked, or Y-linked? Please give two specific reasons that support your conclusion. Once again, disregard II-8 for the moment. One of your two reasons must refer specifically to evidence present in the Southern blot. C. Define the gene alleles associated with Unetan syndrome. Your alleles MUST be consistent with the pattern of inheritance, AND your genetic notation must be consistent with that used throughout the course. Unetan syndrome allele: ________ Normal allele: ________
The Figueroa family has a genetically inherited trait called ocular albinism—lack of pigment in the eye—which is caused by a 516-bp deletion in an exon of a single gene. DNA for this gene was amplified from each member of the family and run on the gel shown below. The thickness of the band indicates the relative amount of DNA. Use the pedigree and the gel to determine the most likely mode of inheritance for this disease. a) x-linked dominant b)x-linked recessive c)autosomal dominant d) autosomal recessive e) cant be determined
A cross is made between two E. coli strains: Hfr arg + bio + leu+ × F− arg − bio − leu−. Interrupted mating studies show that arg+ enters the recipient last, and so arg+ recombinants are selected on a medium containing bio and leu only. These recombinants are tested for the presence of bio + and leu+. The following numbers of individuals are found for each genotype: arg+ bio+ leu+ 320 arg+ bio- leu+ 0 arg+ bio+ leu- 8 arg+ bio- leu- 48a. What is the gene order? b. What are the map distances in recombination percentages?
A three point cross is carried out and the following 1000 offspring are recovered. PHENOTYPES NUMBER OBSERVED CP+ 5 CPR 330 C++ 115 C+R 70 +++ 385 ++R 10 +PR 70 +P+ 15 - A. B. C. D. E. What is the correct gene map for these genes? (A-E) F. G. H. I. J. What is the interference value? (F-J) A. P----20----C----10----R B. P----20----R----10----C C. C----20----R----10----P D. R----20----C----10----P E. C----20----P----10----R…
PLEASE ANSWER THE FOLLOWING LETTERS: a,b,c, and d Examine the pedigree of the McGraw family shown below. Certain individuals in this family are affected by a brain condition that makes them more susceptible to vertigo. As a genetic counselor, you interview the family and draw DNA samples. You discover that the condition is caused by a mutation that changes the sequence 5’GCATTC3’ to 5’GAATTC3’ introducing an EcoRI cut site. You decide to amplify a 1200bp fragment from the DNA that spans this mutation and then digest it with EcoRI. You run the results on a gel next to a marker that shows bands at 2000bp, 1200bp, 900bp, 800bp, and 400bp. Some individuals from the pedigree are identified on the gel.
Compared to the normal A allele, the disease-causing allele in sickle cell anemia (S allele) is missing an MstII restriction site. On a Southern blot of genomic DNA cut with MstII and hybridized with the probe shown on the diagram below, a person with sickle anemia, carrying two S alleles, will show Choose an answer below: a single band at 1.1 kb. a single band at 1.3 kb. a single band at 0.2 kb. one band at 0.2 and one at 1.3 kb. one band at 1.1 and one at 1.3 kb.
Aliens with orang eye color allele (o) is recessive to the dominant black eye color allele (O). The locus of the orange gene from 10 pure breed orange eyes aliens and 10 pure breed black eyes aliens. You notice a difference in the DNA sequence linked to each allele of the orange gene and you decide to use it as a physical marker to follow recombination between this sequence linked to the orange gene. These sequences consist of short tandem repeat (STR) with two different number of repeats, each associated with one of the two orange gene alleles. a PCR test distinguishes the 10 repeat STR and the 6 repeat STR associated with the O and o alleles respectively. Using primers on each side of the STRs you can amplify by PCR this sequence and visualizing the size of the 10 repeat and 6 repeat STRs in an electrophoresis gel. You can follow the two STR sequences (10 and 6 repeats) linkage and recombination frequencies with the orange gene locus. The black eye aliens yield a PCR fragment that is…
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?
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?