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 11 III IV V 1/4 1/2 1/2 3 ₂0₁ 1/2 1/2 Wild Type Female 1/4 1/2 B Affected Known carrier Affected female Normal female Affected male Normal male D ...

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 11 III IV V 1/4 1/2 1/2 3 ₂0₁ 1/2 1/2 Wild Type Female 1/4 1/2 B Affected Known carrier Affected female Normal female Affected male Normal male D ...

Name: The DNA of every individual in the pedigree shown in image B (below)
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Description: 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 alle

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 16QP: Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior...

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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 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
You have identified a SNP marker that in one largefamily shows no recombination with the locus causinga rare hereditary autosomal dominant disease.Furthermore, you discover that all afflicted individuals in the family have a G base at this SNP on theirmutant chromosomes, while all wild-type chromosomes have a T base at this SNP. You would like tothink that you have discovered the disease locus andthe causative mutation but realize you need to consider other possibilities.a. What is another possible interpretation of the results?b. How would you go about obtaining additional genetic information that could support or eliminateyour hypothesis that the base-pair difference is responsible for the disease?
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.What is the percentage level of penetrance for the condition in the diagram?
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?
Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6
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?
An experimental assay for the blood-clotting protein called factorIX is available. A blood sample was obtained from each individual in the following pedigree. The amount of factor IX protein, shown within each symbol on the pedigree, is expressed as a percentage of the average amount observed in individuals who do not carry a mutant copy of the gene.
In Neurospora, the mutant stp exhibits erratic stop-andstart growth. The mutant site is known to be in the mtDNA. If an stp strain is used as the female parent in a crosswith a normal strain acting as the male, what type ofprogeny can be expected? What about the progeny fromthe reciprocal cross?
This pedigree traces the inheritance of a rare disease in humans. a. Based on this pedigree, is the allele for this disease dominant or recessive? Explain. b. What genotypes are possible for the individuals labeled 1, 2, and 3?
11). This pedigree illustrates a family in which some members have a completely penetrant disease caused by a dominant mutation. This mutation is linked at a distance of 10 map units from a SNP marker with three different alleles (1, 2 and 3). The SNP alleles found in each family member are indicated below each pedigree symbol. It is not yet evident whether the very young individuals labeled A and B will develop the disease. a. What is the probability that individual A will develop the disease? b. What is the probability that individual B will develop the disease? 1,3 2,2 1,2 3,2 O 1,2 3,2 A B
A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1and another strain that carried the mating-type allele aand the wild-type allele for arg-1 (+). Four hundred linear octads were isolated, and they fell into the sevenclasses given in the table below. (For simplicity, they areshown as tetrads.)a. Deduce the linkage arrangement of the mating-typelocus and the arg-1 locus. Include the centromere orcentromeres on any map that you draw. Label all intervalsin map units.b. Diagram the meiotic divisions that led to class 6. Labelclearly