Biology: The Unity and Diversity of Life (MindTap Course List)
15th Edition
ISBN: 9781337408332
Author: Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr
Publisher: Cengage Learning
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Textbook Question
Chapter 13, Problem 4GP
For each genotype listed, what allele combinations will occur in gametes?
- a. AABBCC
- b. AaBBc
- c. AaBBcc
- d. AaBbCc
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Chapter 13 Solutions
Biology: The Unity and Diversity of Life (MindTap Course List)
Ch. 13 - Prob. 1DAACh. 13 - Prob. 2DAACh. 13 - Prob. 3DAACh. 13 - Prob. 1SQCh. 13 - An organisms observable traits constitute its...Ch. 13 - Prob. 3SQCh. 13 - Prob. 4SQCh. 13 - The offspring of the cross AA aa are ________. a....Ch. 13 - Refer to question 5. Assuming complete dominance,...Ch. 13 - A testcross is a way to determine ________. a....
Ch. 13 - Assuming complete dominance, a cross between...Ch. 13 - The probability of a crossover occurring between...Ch. 13 - True or false? All traits are inherited in a...Ch. 13 - One gene that affects three traits is an example...Ch. 13 - The phenotype of individuals heterozygous for...Ch. 13 - _______ in a trait is indicated by a bell curve....Ch. 13 - Match the terms with the best description. ______...Ch. 13 - Mendel crossed a true-breeding pea plant with...Ch. 13 - Assuming that independent assortment occurs during...Ch. 13 - Refer to problem 2. Determine the predicted...Ch. 13 - For each genotype listed, what allele combinations...Ch. 13 - Prob. 5GPCh. 13 - Suppose you identify a new gene in mice. One of...Ch. 13 - Mutations in the TYR gene may render its enzyme...Ch. 13 - In sweet pea plants, an allele for purple flowers....Ch. 13 - Red-flowering snapdragons are homozygous for...
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- Considering only the five genes A, B, C, D, and E, how many genetically distinct gametes (i.e. gametes withdifferent genotypes) could be produced through independent assortment by an individual with the genotypeAABbCcDDEe? a. 4b. 8c. 16d. 64arrow_forwardUse the product rule to determine the chance of obtaining an offspringwith the genotype Rr Yy from a dihybrid cross between parents withthe genotype Rr Yy.a. ½ c. 1/4b. ¼ d. 1⁄16arrow_forwardHemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is referred to as a sex-linked disease. The recessive allele causes the disease. A man with hemophilia (xhy) marries a woman who is homozygous dominant (XHXH. A. Illustrate using a Punnett square the probability that their children will have the disease. B. Will any of their children have the disease? C. Predict the probabilities of their children having the disease.arrow_forward
- Using the same information listed above for hair and eye color, (E = dominant brown eyes with e = recessive blue eyes; R = dominant brown hair and r = recessive blond hair), if the mom was heterozygous for both eye color and hair color what would her genotype be? a. EERR b. eerr c. EeRr What would be the 4 possible gametes the heterozygous mom could contribute? a. RE, Re, rE, re b. Rr, Ee, RE, re c. RR, rr, EE, eearrow_forwardThe pea plants used in Mendel’s genetic inheritance studies were diploid, with 14 chromosomes in somatic cells. Assuming no crossing over events occur, how many unique gametes could one pea plant produce? a. 28 b. 128 c. 196 d. 16,384arrow_forwardWhat are the possible allele combinations that could be formed in the gametes produced from a parent with the genotype RrWw? Select one: a. RW, RW, rw, and rw b. RW, Rw, rW, and rw c. RW and rw d. RW, Rw, and rwarrow_forward
- Three genes (A, B, and C) are found on three different chromosomes. For the following diploid genotypes, describe all of thepossible gamete combinations.A. Aa Bb Cc C. Aa BB CcB. AA Bb CC D. Aa bb ccarrow_forwardAssume that one of Merida's sons, who is heterozygous for orange hair color, married a girl that was also heterozygous. Create a Punnett square to show the possibilities that would result if they had children. a. List the possible genotypes and phenotypes for their children. b. What are the chances of a child with orange hair? c. What are the chantes of a child with yellow?arrow_forwardPhenylketonuria (PKU) is a disease that results from a recessive gene.Suppose that two unaffected parents produce a child with PKU. a. What is the probability that a sperm from the father will contain the PKU allele?b. What is the probability that an egg from the mother will contain the PKU allele?c. What is the probability that their next child will have PKU?d. What is the probability that their next child will be heterozygous for the PKU gene?arrow_forward
- A boy is color-blind (X-linked recessive) and has a straight hairline (autosomal recessive). Which could be the genotype of his mother? Select one: А. Вbxwxw B. Bbww C. Xbxbww D. Xbywwarrow_forwardAssuming no gene linkage, in a dihybrid cross ofAABB x aabb with AaBb F1 heterozygotes, what is the ratio of the F1 gametes (AB, aB,Ab, ab) that will give rise to the F2 offspring? a. 1:1:1:1 b. 1:3:3:1 c. 1:2:2:1 d. 4:3:2:1arrow_forwardPhenylketonuria (PKU) is a disease that results from a recessive gene. Suppose that two unaffected parents produce a child with PKU. a. What is the probability that a sperm from the father will contain the PKU allele? b. What is the probability that an egg from the mother will contain the PKU allele? c. What is the probability that their next child will have PKU? d. What is the probability that their next child will be heterozygous for the PKU gene?arrow_forward
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