Please answer fast The pedigree below shows the inheritance of red-green colorblindness, an X-linked recessivetrait (XD = normal vision, Xd = colorblindness allele). For all of the questions below, assumethat there is no abnormal event or new mutation; also, ignore X-inactivation for this problem.I+ GenotypeII123?IIWhat are the genotypes of the following individuals for color vision. Use "?" to indicate anyallele about which you are uncertain.1-1 =1-2 =Il-1 =Il-2=Il-3 =(a) What is the probability that III-1 will be colorblind? Explain your reasoning to show howyou arrived at you answer.

X chromosome of son is always comes from the mother. Only Y chromosome comes from the father . Any…

The pedigree below shows the inheritance of red-green colorblindness, an X-linked recessive trait (XD = normal vision, Xd = colorblindness allele). For all of the questions below, assume that there is no abnormal event or new mutation; also, ignore X-inactivation for this problem. - Genotype II 1 2 3 II What are the genotypes of the following individuals for color vision. Use "?" to indicate any allele about which you are uncertain. 1-1 = 1-2 = Il-1 = Il-2= Il-3 = (a) What is the probability that III-1 will be colorblind? Explain your reasoning to show how

The pedigree below shows the inheritance of red-green colorblindness, an X-linked recessive trait (XD = normal vision, Xd = colorblindness allele). For all of the questions below, assume that there is no abnormal event or new mutation; also, ignore X-inactivation for this problem. - Genotype II 1 2 3 II What are the genotypes of the following individuals for color vision. Use "?" to indicate any allele about which you are uncertain. 1-1 = 1-2 = Il-1 = Il-2= Il-3 = (a) What is the probability that III-1 will be colorblind? Explain your reasoning to show how

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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In 1995, doctors reported a Chinese family in whichretinitis pigmentosa (progressive degeneration of theretina leading to blindness) affected only males. Allsix sons of affected males were affected, but all of thefive daughters of affected males (and all of thechildren of these daughters) were unaffected.a. What is the likelihood that this form of retinitispigmentosa is due to an autosomal mutationshowing complete dominance?b. What other possibilities could explain the inheritance of retinitis pigmentosa in this family? Whichof these possibilities do you think is most likely?
This is a typical pedigree for a family that carries neurofibromatosis. Is individual I-2 most likely homozygous or heterozygous? Why? Explain.
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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
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The mutant FMR-1 allele that causes fragile X syndrome is considered to be X-linked dominant withincomplete penetrance and variable expressivity.Why do most females heterozygous for one mutantand one normal allele have at least some symptomsof the disease?
IN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.