Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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Chapter 3, Problem 36QP
a.
Summary Introduction
To determine: The possible
Introduction: The human blood group is an example of codominance where the alleles for the two antigens A and B are equally dominant. It means that when the two antigens A and B are present, the blood group is AB whereas in the absence of both the antigens, the blood group is O.
b.
Summary Introduction
To determine: The genotype of the individuals with blood group O.
Introduction: The individuals with the blood group O lack the surface antigens for both the blood groups A and B. Therefore, the expression of type O blood group occurs when the individual inherits both the recessive alleles. It makes people with type O blood group to become universal donors.
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Blood types in humans are Codominant.
a. What are the possible genotypes of someone with type O blood?
b. What are the possible genotypes of someone with type B blood?
c. What is the phenotype of someone’s whose genotype is AO?
d. A man with AB blood is married to a woman with AB blood. What blood types will their children be and in what proportion?
. Assuming no involvement of the Bombay phenotype(in case you’ve already read ahead to Section 3.2):a. If a girl has blood type O, what could be the genotypes and corresponding phenotypes of her parents?b. If a girl has blood type B and her mother has bloodtype A, what genotype(s) and correspondingphenotype(s) could the other parent have?c. If a girl has blood type AB and her mother is alsoAB, what are the genotype(s) and correspondingphenotype(s) of any male who could not be thegirl’s father?
In bears fur colour can be either yellow or tan, and is determined by two alleles of the C gene. A heterozygous bear is observed to present a fur pattern of yellow and tan patches. Explain why this can occur.
Chapter 3 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
Ch. 3.4 - Why do scientists design experiments to disprove...Ch. 3.4 - Should Ockhams razor be considered an irrefutable...Ch. 3.7 - Prob. 1EGCh. 3.7 - For most cases, a p value of 0.05 is used to...Ch. 3 - Prob. 1CSCh. 3 - Prob. 2CSCh. 3 - Prob. 3CSCh. 3 - Prob. 1QPCh. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - Crossing Pea Plants: Mendels Study of Single...
Ch. 3 - Prob. 4QPCh. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - Prob. 6QPCh. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - Crossing Pea Plants: Mendels Study of Single...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - Prob. 14QPCh. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - Prob. 17QPCh. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - More Crosses with Pea Plants: The Principle of...Ch. 3 - Meiosis Explains Mendels Results: Genes Are on...Ch. 3 - Meiosis Explains Mendels Results: Genes Are on...Ch. 3 - Meiosis Explains Mendels Results: Genes Are on...Ch. 3 - Prob. 26QPCh. 3 - Prob. 27QPCh. 3 - Variations on a Theme by Mendel A characteristic...Ch. 3 - Prob. 29QPCh. 3 - Variations on a Theme by Mendel Pea plants usually...Ch. 3 - Prob. 31QPCh. 3 - Prob. 32QPCh. 3 - Prob. 33QPCh. 3 - Prob. 34QPCh. 3 - Prob. 35QPCh. 3 - Prob. 36QP
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