Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and their children when albino male and normal female
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- Look at the pedigree below and answer the following questions related to the human genetic trait depicted in this pedigree. 1. Indicate whether the pattern of inheritance associated with this human trait is most likely to be (i) rare X-linked recessive, (ii) sex-influenced, or (iii) sex-limited. You may assume that the gene is fully penetrant. Then, provide a specific reason that justifies your selection of this pattern of inheritance as the correct one, and also explain why each of the other two alternatives are not correct. As part of your answer, include the phenotypic ratio, including the sex of the offspring, that you would expect to find in each of the three possible scenarios.arrow_forwardTwo autosomal mutations include albinism and dwarfism. Albinism (a) is recessive, and dwarfism (D) is dominant. Complete a dihybrid cross of two people that are heterozygous for normal pigmented skin. One person is normal height and has no genetic trace of dwarfism in the family. The other person has dwarfism but has a mother who is normal height. 21. Complete a full dihybrid Punnett square (6pts).arrow_forwardAmelogenesis imperfecta (AI) is a disorder of faulty tooth enamel formation. It is inherited in an autosomal dominant and X-linked dominant pattern. The expression of AI disorder is determined by mutations in the autosomal alleles. One copy of the mutated allele (A) will cause the disorder. The severity of the disorder is determined by mutations in a gene carried on the X chromosome. Normal (or non-severe) abnormality (XN) is dominant over the abnormality (or severe) (Xn) allele. In the absence of the autosomal dominant allele, the abnormality gene on the X chromosome is notexpressed. Question:A woman with normal teeth had four children with a man with non-severe form of AI: A boy was born without amelogenesis imperfecta A girl was born without amelogenesis imperfecta A boy was born with severe amelogenesis imperfecta A boy was born with non severe amelogenesis imperfecta Identify the parental genotypes. Complete the Punnett square for the parental cross, and identify the possible…arrow_forward
- Acatalasia is caused by a rare autosomal recessive gene. In heterozygous condition catalase activity is decreased slightly. A woman with a normal catalase activity have a husband with low catalase activity. What is the probability of children birth without anomaly in the family, if grandparents from both sides have reduced activity of catalase? Diagram the cross and mention the genotypes of all individuals in the family?arrow_forwardGive typing answer with explanation and conclusionarrow_forwardPLEASE ANSWER PART C.arrow_forward
- The questions below all refer to the following pedigree. The locus for allele D/d determines how cilia function within the body. Mutations at this allele cause abnormal ciliary function, resulting in a clinical disorder characterized by frequent respiratory infections (including in the ears, sinuses, and lungs), as well as infertility. In the pedigree, black circles/squares represent individuals affected by the disorder. White circles/squares represent unaffected individuals. Remember, if a trait is rare in a population (such as this one), we assume individuals marrying into the family are NOT carriers unless the information provided indicates otherwise. A1 A2. || B1. B2. B3 B4. B5 B6. II C1_ C2. C3. C4. C5 IV D1 D2- D4 - D5_ 1) What is the probability that B3 is a carrier of this disorder? Show your work with a Punnett Square in addition to stating the probabilityarrow_forwardThere are six types of agglutinogen named C,D, E and c,d,e.the first three are dominant and last three are recessive.discussarrow_forwardage An extra finger in humans is rare but is due to a dominant gene. When one parent is normal and the other parent has an extra finger but is heterozygous for the trait, what is the probability that the first child will have an extra finger? - /mod/quiz/attempt.php?attempt=1173673&cmid=3837312&page=11# Select one: O a. 25% O b. 50% O c. 75% O d. 0% O e. 100% @ 2 W # 3 E с $ 4 R G Search or type URL % 5 T MacBook Pro 6 Y & 7 ☆ U * 8 ( 9 F ) 0 0 Next page P (3) BO 11arrow_forward
- Please DEFINE the alleles involved in this cross, using genetic notation consistent with the pattern of inheritance exhibited by this trait. Dwarf allele: __________ Normal sized allele: __________arrow_forwardF1 and F2 offspring with given genders and disease phenotypes. The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype. There are three possible modes of inheritance that underlie the generation of the data. All are monogenic. They are: (i) Autosomal Recessive. (ii) Homozygous Lethal Dominant. (iii) Autosomal Dominant. One's phenotype is determined by their genotype at the disease locus and the mode of inheritance, as we have seen with Punnett Square. on the F2 data and make a decision, based on your statistical analysis, as to whether you reject or do not reject the computer-generated model of inheritance as being consistent with the observed data. The mode of inheritance you are to test on the observed data is autosomal recessive. PARENTAL CROSS Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 DATA Phenotype…arrow_forwardPEDIGREE: Shaded individuals in the pedigree have a genetic disease. Individuals marrying into the family, that is individuals II–1, II–4 and II–6, have no history of the disease in their families.1. Determine the mode of inheritance ______________________________2. Give the genotypes of the following individualsarrow_forward
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