Jekyll-Hyde Afflicted Spider Curse Afflicted Jekyll-Hyde/Spider Curse Human 2. II 2 III 3. Jekyll-Hyde disease is characterized by transformation into an unfeeling, aggressive, alter ego at night. While the "Spider Curse" is a disease that causes affected individuals to grow extra arms and extra eyes during a full moon. The genes responsible for these diseases sort independently and both traits run in one family. Based on the above pedigree answer the following questions: What is the mode of inheritance for Jekyll-Hyde disease? (Hint: look at individuals IIl-1 & III-3) O a. Autosomal Dominant O b. Autosomal Recessive O c. X-Linked Dominant O d. X-linked Recessive What evidence supports your hypothesis? Give at least 2 pieces of information from the pedigree that support your answer to 3a. Answer:

Jekyll-Hyde Afflicted Spider Curse Afflicted Jekyll-Hyde/Spider Curse Human 2. II 2 III 3. Jekyll-Hyde disease is characterized by transformation into an unfeeling, aggressive, alter ego at night. While the "Spider Curse" is a disease that causes affected individuals to grow extra arms and extra eyes during a full moon. The genes responsible for these diseases sort independently and both traits run in one family. Based on the above pedigree answer the following questions: What is the mode of inheritance for Jekyll-Hyde disease? (Hint: look at individuals IIl-1 & III-3) O a. Autosomal Dominant O b. Autosomal Recessive O c. X-Linked Dominant O d. X-linked Recessive What evidence supports your hypothesis? Give at least 2 pieces of information from the pedigree that support your answer to 3a. Answer:

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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The Hemoglobin gene has two very common alleles: HbS and HbA. Bob’s professor asks him to draw the chromosomes for an individual that is HbSHbA and label the alleles. Bob draws this (below). Explain why Bob is wrong and then draw it correctly. FOXP2 is a gene in humans; humans that have mutated copies of FOXP2 often have trouble speaking. It is on Chromosome 7, which is about 160 million base pairs long. FOXP2 is at position 115 million base pairs. Which of the drawings below appropriately diagrams the location of FOXP2? Why?
Why would a mutation in a somatic cell of a multicellular organism not necessarily result in a detectable phenotype?
Ch. 13-7 Mutations in the TYR gene may render its enzyme product—tyrosinase—nonfunctional.Individuals homozygous for such mutations cannot make the pigment melanin. Albinism, the absence of melanin, results. Humans and many other organisms can have this phenotype (right). Mutated tyrosinase alleles are recessive when paired with the normal allele in heterozygous individuals.In the following situations, what are the probable genotypes of the father, the mother, and their children?a. Both parents have normal phenotypes; some of their children have the albino phenotype and others are unaffected.b. Both parents and children have the albino phenotype.c. The mother and three children are unaffected; the father and one child have the albino phenotype.
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 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.
How might a point mutation in DNA affect the phenotype. How the genome is expressed in the organism? (Please explain this at the molecular level).
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