Below is a pedigree of a family with a history of Best disease (or vitelliform macular dystrophy). All members of this family underwent ASO testing using two different ligonucleotides (ASO 1 and ASO 2): one that hybridizes to the only known mutation that leads to Best disease and one that hybridizes to the wild-type gene sequence in the same region. From the information you have, how is Best disease most likely inherited? L II. III. autosomal dominant autosomal recessive x-linked recessive x-linked dominant you cannot determine ASO 1 ASO 2 1-1 1-2 11-1 11-2 11-3

Below is a pedigree of a family with a history of Best disease (or vitelliform macular dystrophy). All members of this family underwent ASO testing using two different ligonucleotides (ASO 1 and ASO 2): one that hybridizes to the only known mutation that leads to Best disease and one that hybridizes to the wild-type gene sequence in the same region. From the information you have, how is Best disease most likely inherited? L II. III. autosomal dominant autosomal recessive x-linked recessive x-linked dominant you cannot determine ASO 1 ASO 2 1-1 1-2 11-1 11-2 11-3

Name: Below is a pedigree of a family with a history of Best disease (or vi
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Description: Below is a pedigree of a family with a history of Best disease (or vitelliform macular dystrophy). All members of this family underwent ASO testing using two differentoligonucleotides (ASO 1 and ASO 2): one that hybridizes to the only known mutation

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 1QP: The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show...

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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.
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 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 DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.
The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●
The DNA of every individual in the pedigree shown below has been sequenced at the causative locus. All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this autosomal dominant condition, but they exhibit no symptoms. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12
DNA obtained from the indicated donor strains is used to TRANSFORM the indicated recipient strains. The resulting progeny are plated on minimal medium so that only wild-type recombinants are scored. The number of wild-types for each cross is given in the chart below. donor donor arg+ val- ser- val+ ser- arg- val- ser+ val- ser+ arg+ val- ser- O valine recipient recipient none of these. O arginine arg- val+ ser+ arg+ val- ser+ Which locus is closest to serine? arg+ val+ ser- wild-types wild-types 70 98 0 2 arg- arg-
Consider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.
In McCune-Albright syndrome, fibrous connective tissue replaces bone, tan patches (café-au-lait spots) dot the skin, and hormone abnormalities cause early puberty and malfunction of the thyroid, pituitary, and adrenal glands. The phenotype is highly variable, and all patients are somatic mosaics for the mutation, which is in the gene GNAS1. Why is the condition seen only in mosaics?
A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies.These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic,25% adenine-requiring auxotrophsa. Explain the difference between these two types ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?
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
Consider two maize plants:a. Genotype C/c m ; Ac/Ac+, where cm is an unstable allele caused by a Ds insertionb. Genotype C/c m, where cm is an unstable allele caused by Ac insertionWhat phenotypes would be produced and in what proportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a is crossed with the plant in part b? Assume that Ac and c are unlinked, that the chromosome-breakage frequency is negligible, and that mutant c /C is Ac+.