7006005004003002001003010MW|||E1 E2 E3 E4 E5 E6What type oftrinucleotide repeatdisease is Adamssyndrome most likely tobe?If implanted, whichembryo is most likely toproduce a blind baby?1. autosomal dominant3.2. X-linked dominantMW= molecularweight markers inbasepairsdotted lines indicatethe different lanes ofthe gelY-linked dominant4. autosomal recessive6. E15. X-linked recessive7. E2Which embryo ishomozygous for the wild- 8. E3type adams gene and9. E4should be used forimplantation?What is the mode ofinheritance for Adamssyndrome?10. E513.11. E612. A polyQ trinucleotide repeatdiseaseA non-polyQ trinucleotide repeatdisease Shown below is a pedigree for a completely penetrant trait called Adamssyndrome in which babies are born blind. This trait occurs when an allele ofthe adams gene is associated with ≥200 tandem trinucleotide repeats (thenormal number is 10). First cousins, III-1 and III-2 married and their first child(IV-1) was blind. For their next child, they decided to do in vitro fertilization withIII-1's sperm and III-2's eggs to generate six embryos (labeled E1-6). Wheneach embryo contained eight cells, a single cell was removed and genomicDNA was isolated. PCR reactions using primers that flank the trinucleotiderepeat region were then performed and the resulting fragments werefractionated on an agarose gel. PCR reactions using genomic DNA from III-1,III-2 and IV-1 were included as controls. The DNA was visualized using afluorescent dye and the gel is shown below.Based on this information, select the best answer from the list to the questionsbelow.||E1 = embryo 1IVE1E2 E3 E4 E5E6Agarose gel of PCR fragments generated using genomic DNA andprimers flanking the trinucleotide repeat region

Name: 7006005004003002001003010MW|||E1 E2 E3 E4 E5 E6What type oftrinucleot
Uploaded: 2024-03-20T06:42:55.000Z
Channel: Bartleby
Description: 7006005004003002001003010MW|||E1 E2 E3 E4 E5 E6What type oftrinucleotide repeatdisease is Adamssyndrome most likely tobe?If implanted, whichembryo is most likely toproduce a blind baby?1. autosomal dominant3.2. X-linked dominantMW= molecularweight m

Inheritance of genetic disease could be as follows,Autosomal dominant(if a single parent is…

Science

Biology

700 600 500 400 300 200 100 30 10 ||| MW III-1 III-2 IV-1 E1 E2 E3 E4 E5 E6 MW= molecular weight markers in basepairs dotted lines indicate the different lanes of the gel 1. autosomal dominant 2. X-linked dominant What type of trinucleotide repeat disease is Adams syndrome most likely to be? If implanted, which embryo is most likely to produce a blind baby? 3. Y-linked dominant 4. autosomal recessive 5. X-linked recessive 6. E1 7. E2 Which embryo is homozygous for the wild- 8. E3 type adams gene and 9. E4 should be used for implantation? 10. E5 What is the mode of inheritance for Adams syndrome? 11. E6 12. A polyQ trinucleotide repeat 13. disease A non-polyQ trinucleotide repeat disease

700 600 500 400 300 200 100 30 10 ||| MW III-1 III-2 IV-1 E1 E2 E3 E4 E5 E6 MW= molecular weight markers in basepairs dotted lines indicate the different lanes of the gel 1. autosomal dominant 2. X-linked dominant What type of trinucleotide repeat disease is Adams syndrome most likely to be? If implanted, which embryo is most likely to produce a blind baby? 3. Y-linked dominant 4. autosomal recessive 5. X-linked recessive 6. E1 7. E2 Which embryo is homozygous for the wild- 8. E3 type adams gene and 9. E4 should be used for implantation? 10. E5 What is the mode of inheritance for Adams syndrome? 11. E6 12. A polyQ trinucleotide repeat 13. disease A non-polyQ trinucleotide repeat disease

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...

See similar textbooks

Similar questions

The following pedigree shows the pattern of inheritance of red-green color blindness in a family. Females are shown as circles and males as squares; the squares or circles of individuals affected by the trait are filled in black. What is the chance that a son of the third-generation female indicated by the arrow will be color blind if the father is not color blind? If he is color blind?
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?
The following pedigree illustrates the inheritance of Nance–Horan syndrome, a rare genetic condition in which affected people have cataracts and abnormally shaped teeth.
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?
In the pedigree below, male II-1 has Klinefelter syndrome, which is the result of an XXY karyotype. On the X chromosome, a gene called G6PD has two codominant alleles, G6PDA and G6PDB. In this pedigree, A, B, and AB refer to the phenotypes associated with the alleles of this gene. (Note: In this family, no individuals have the AB version of the phenotype.) A A B Based on the information in the pedigree, when could nondisjunction have occurred? Select all correct answers. In Il-1's father, during meiosis I In II-1's mother, during meiosis I In II-1's mother, during meiosis II In Il-1's father, during meiosis II
This is a pedigree for a dominant trait caused by gene A in humans. Shaded symbols show individuals affected with the trait; non-shaded individuals are normal (aa). Among the progeny arising from the marriage of individual III-1, what proportion would be expected to show the trait? Among the progeny arising from the marriage of individual III-6, what proportion would be expected to show the trait?
Duchenne muscular dystrophy is a recessive disorder caused by a rare,loss-of-function allele that is located on the X chromosome in humans. Anunaffected woman (i.e., without disease symptoms) who is heterozygousfor the X-linked allele causing Duchenne muscular dystrophy has childrenwith a man with a functional (non-disease-causing) allele. What is theprobability that this couple will have an unaffected son?
Help me create a pedigree of this information: Pedigree analysis: Generation 1: Normal parents (AA x AA) Generation 2: Carrier parents (AA x AS) Generation 3: Affected child (AS x AS) Generation 4: Affected grandchild (SS) This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generation
The pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family. I (B+ AB- 1 2 3 4 II O- A+ В- B- AB+ A+ 1 2 4 5 6 a. What are the ABO blood group genotypes of individuals I-1 and I-2? b. Which child/ren of individual I-4 can donate blood to him? c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?
The image shows a pair of homologous chromosomes from a single parent before gamete production. M1 and M2 are maternal chromosomes, while P1 and P2 are paternal chromosomes. Two traits are shown: D represents seed color (D – green, d – yellow), while F represents flower color (F – purple, f – white). These two traits follow the patterns of basic Mendelian genetics. During crossing-over between the M2 F allele and the P1 f allele, a mutation occurred and the portion of P1 did not reattach to the chromosome. Which of the following explains what would happen to the proportion of white flowers in a population resulting from this mutation? A - There would be an increase in the proportion of white flowers because the f allele is distributed to more gametes. B - There would be a decrease in the proportion of white flowers because the f allele is not distributed to as many gametes. C - There would be an increase in the proportion of white flowers because the f allele would not be masked by the…
The following pedigree describes the inheritance of Lesch-Nyhan syndrome, an x-linked recessive disease. Affected individuals are shaded. what is the probability, that the indicated child (IV.1) will be affected by Lesch-Nyhan syndrome? show solution