Campbell Biology: Custom Edition
18th Edition
ISBN: 9781323717271
Author: Urry, Cain, Wasserman, Minorsky, Reece
Publisher: PEARSON C
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Question
Chapter 14.4, Problem 3CC
Summary Introduction
To determine: The genotype of “Person J” who has polydactyly.
Concept introduction:
Extra digits can be present in either fingers or toes. It is a physical anomaly that is known as polydactyly. It is an autosomal dominant trait.
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Joan was born with 6 toes on each foot, a dominant trait called polydactyly. Two of her 5 siblings and her mother but not her father also have extra digits. What is Joan’s genotype for the number of digits character? Explain your choice. Use D and d symbolize the alleles for this character.
Ben, Marie, and Korra are siblings. Ben is blood type O, Ana is blood type B, and Korra is type A. What could be the possible genotypes of their parents?
You are trying to find a blood donor to help treat some of your patients.
The C-yonce and Kay-Z have 3 children with blood types A , B, and O. The youngest with blood type O needs a transfusion and you're trying to figure out if either of the parents are a match so you test the parent genotypes.
Note: Alleles for each blood type are expressed as follows:
blood type A= I^A
blood type B= I^B
blood type O= i
Hint: Remember that 2 alleles contribute to a persons blood type. Pay attention to dominant and recessive relationships among the alleles.
1-C-yonce is type B. What must her genotype be?
2-What must Kay-Z’s genotype be?
3-Which parent can give blood to their child ?
4-After running lab tests, it was discovered that the children's blood types were reported incorrectly.(This does not affect the genotypes entered for C-yonce and Kay-Z in the previous questions.) Their youngest actually has blood type AB, and still needs a blood transfusion. Which parent can donate blood to the child…
Chapter 14 Solutions
Campbell Biology: Custom Edition
Ch. 14.1 - DRAW IT Pea plants heterozygous for flower...Ch. 14.1 - WHAT IF? List all gametes that could be made by a...Ch. 14.1 - MAKE CONNECTIONS In some pea plant crosses, the...Ch. 14.2 - Prob. 1CCCh. 14.2 - Two organisms, with genotypcs BbDD and BBDd, are...Ch. 14.2 - WHAT IF? Three characters (flower color, seed...Ch. 14.3 - What two properties, one structural and one...Ch. 14.3 - If a man with type AB blood marries a woman with...Ch. 14.3 - WHAT IF? A rooster with gray feathers and a hen...Ch. 14.4 - Beth and Tom each have a sibling with cystic...
Ch. 14.4 - Prob. 2CCCh. 14.4 - Prob. 3CCCh. 14.4 - MAKE CONNECTIONS In Table 14.1, note the...Ch. 14 - When Mendel did crosses of true-breeding purple-...Ch. 14 - DRAW IT Redraw the Punnett Square on The right...Ch. 14 - Inheritance patterns are often more complex than...Ch. 14 - Both members of a couple know that they are...Ch. 14 - DRAW IT Two pea plants heterozygous for the...Ch. 14 - A man with type A blood marries a woman with type...Ch. 14 - A man has six fingers on each hand and six toes on...Ch. 14 - DRAW IT A pea plant heterozygous for inflated pods...Ch. 14 - Flower position, stem length, and seed shape are...Ch. 14 - Hemochromatosis is an inherited disease caused by...Ch. 14 - The genotype of F1, individuals in a tetrahybrid...Ch. 14 - What is the probability that each of thc following...Ch. 14 - Prob. 9TYUCh. 14 - Prob. 10TYUCh. 14 - In tigers, a recessive allele of a particular gene...Ch. 14 - In maize (com) plants,a dominant allele I inhibits...Ch. 14 - The pedigree belowtraces the inheritance of...Ch. 14 - Imagine that you are a genetic counselor, and a...Ch. 14 - EVOLUTION CONNECTION Over the past half century,...Ch. 14 - SCIENTIFIC INQUIRY You are handed a mystery pea...Ch. 14 - Prob. 17TYUCh. 14 - SYNTHESIZE YOUR KNOWLEDGE Just for fun, imagine...
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- α-thalassemia is another blood disorder in which abnormal hemoglobin molecules are formed, leading to anemia. One mutant allele known to cause α-thalassemia occurs in α-hemoglobin and is called Constant Spring. Normal α-hemoglobin is 141 amino acids long, while the Constant Spring protein is 172 amino acids long. Include answers to both A and B in your response. A maximum of 2 sentences per part. A) Explain how a frameshift mutation in the coding region of α-hemoglobin could result in the Constant Spring protein. B) Explain how a single base-substitution mutation in the coding region of α-hemoglobin could result in the Constant Spring protein.arrow_forwardProduce a Punnett square to illustrate the dihybrid cross described below (NOTE: It is the same story as in question 9, above so you may use the square that you produced in answer to question 9 to answer this question, you do not need to draw it twice). use it to fill in the blanks in the answer text below. NOTE: please type in whole numbers only, no symbols, no letters, no spaces. There are two common alleles for the TAS2R38 gene on Chromosome 7. This gene encodes a seven-transmembrane G-protein coupled receptor. This receptor controls the ability to taste glucosinolates. Phenylthiocarbamide (PTC) is a synthetic glucosinolate. The recessive TAS2R38 allele produces a non-functional receptor. The father in this dihybrid cross is heterozygous for these alleles, meaning that he can taste PTC. The mother is homozygous recessive, meaning that she cannot taste PTC The father has X-Linked Protoporphyria which means that he is very sensitive to sunlight exposure, he is hemizygous for the…arrow_forwardProduce a Punnett square to illustrate the dihybrid cross described below (NOTE: It is the same story as in question 9, above so you may use the square that you produced in answer to question 9 to answer this question, you do not need to draw it twice). use it to fill in the blanks in the answer text below. NOTE: please type in whole numbers only, no symbols, no letters, no spaces. There are two common alleles for the TAS2R38 gene on Chromosome 7. This gene encodes a seven-transmembrane G-protein coupled receptor. This receptor controls the ability to taste glucosinolates. Phenylthiocarbamide (PTC) is a synthetic glucosinolate. The recessive TAS2R38 allele produces a non-functional receptor. The father in this dihybrid cross is heterozygous for these alleles, meaning that he can taste PTC. The mother is homozygous recessive, meaning that she cannot taste PTC The father has X-Linked Protoporphyria which means that he is very sensitive to sunlight exposure, he is hemizygous for the…arrow_forward
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