Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 25, Problem 1Q
Summary Introduction
To determine: The likelihood of a child having polydactyly who has a normal father and a mother with polydactyly.
Introduction: Alteration of the
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A gene is composed of two alleles. An allele can be either dominant or recessive. Suppose that a husband and wife, who are both carriers of the sickle-cell anemia
allele but do not have the disease, decide to have a child. Because both parents are carriers of the disease, each has one dominant normal-cell allele (S) and one
recessive sickle-cell allele (s). Therefore, the genotype of each parent is Ss. Each parent contributes one allele to his or her offspring with each allele being equally
likely. Complete parts a) through c) below.
a) Genes are always written with the dominant gene first. Therefore, there are two instances the offspring could have genotype Ss (one if the mother contributes the
dominant allele and the father contributes the non-dominant allele; and one if the father contributes the dominant allele and the mother contributes the non-dominant
allele). List the other two possible genotypes of the offspring.
(Use a comma to separate answers as needed.)
For a recessive condition, two normal heterozygous individuals have children.
What is the likelihood of their children being affected by this condition?
What is the likelihood of their children being carriers without the condition?
What is the likelihood of their asymptomatic children being carriers?
Suppose that an individual with the condition has children with a heterozygous individual, what is the likelihood of their children being carriers?
Albinism is caused by an autosomal recessive allele that interferes with skin pigmentation in mammals. Two normally pigmented human parents already have an albino boy. They plan to continue to have children until they get a girl. Some or all of this information is important for each of the questions below.
a) What is the probability that their next child (currently unborn) will be a girl with albinism? Explain your reasoning.
b) What is the probability their first female child will be albino? Explain your reasoning.
c) The answer to part (b) is different (and, yes, the answer is different) from the answer to part (a). Explain why. (Hint: it has something to do with the underlined words.)
Chapter 25 Solutions
Becker's World of the Cell (9th Edition)
Ch. 25 - Cloning can be done by somatic cell nuclear...Ch. 25 - If the DNA content of a diploid cell in the G1...Ch. 25 - Prob. 25.3CCCh. 25 - Prob. 1QCh. 25 - Prob. 25.4CCCh. 25 - What do you think would happen to a pathogenic...Ch. 25 - Prob. 25.6CCCh. 25 - The Truth About Sex. For each of the following...Ch. 25 - Ordering the Phases of Meiosis. Drawings of...Ch. 25 - Telling Them Apart. Briefly describe how you might...
Ch. 25 - Prob. 25.4PSCh. 25 - More about DNA. Let X be the amount of DNA present...Ch. 25 - Meiotic Mistakes. Infants born with Patau syndrome...Ch. 25 - Prob. 25.7PSCh. 25 - QUANTITATIVE Punnett Squares as Genetic Tools. The...Ch. 25 - QUANTITATIVE Genetic Mapping. The following table...Ch. 25 - Prob. 25.10PSCh. 25 - Prob. 25.11PS
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- Red-green color blindness is inherited as a sex-linked recessive trait. The gene is found on the X chromosome. How can a man with normal color vision father a daughter who is red-green color-blind? Group of answer choices The man is heterozygous for red-green color blindness. The woman with whom he mates is red-green color-blind. He can't (unless there is a mutation). The man's mother carries an allele for red-green color blindness, and the expression of the trait skipped a generation.arrow_forwardIn humans, the gene for brachydactyly (abnormally short fingers) is a dominant gene. The normal finger condition is the recessive trait. Both parents are heterozygous at this location. What is the phenotype of the father? Of the mother? What percentage of their kids should be expected to have brachydactyly?arrow_forwardA couple who are both heterozygous for an autosomal recessive mutation that is narrowly expressed and fully penetrant are planning on having three children. What is the probability that one their children will be normal (unaffected) and two children will have the recessive mutant phenotype? Show your work. Please answer this question using the Bayes’ Theoremarrow_forward
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