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 27QP
Summary Introduction
To predict: The pedigree chart for the family based on the given information.
Introduction: The pedigree chart helps to depict the complete history of the pattern of inheritance of genetic disease from one generation to another generation. The inheritable genetic disease is either autosomal dominant or recessive or sex-linked based on which the probability of disease in a future generation can be predicted.
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Students have asked these similar questions
A mother who is heterozygous for Huntington’s disease (child from mother in #8) and a
father who is homozygous normal want to have a baby. What genotypes and phenotypes
are possible for their child?
A WOMAN IS HETEROZYGOUS FOR TWO
HARMFUL RECESSIVE ALLELES IN
DIFFERENT CHROMOSOMES, ONE FOR
PHENYLKETONURIA (PKU) AND THE
OTHER FOR CYSTIC FIBROSIS (CF). SHE
MARRIES AN UNAFFECTED MAN WHO IS
A CARRIER FOR NEITHER DISEASE.
IF SHE HAS A DAUGHTER, WHAT IS THE
PROBABILITY THAT THE CHILD WILL
CARRY NEITHER OF THE RECESSIVE
ALLELES?
EXACTLY ONE?
BOTH?
Mendel’s observation that two different traits could be inherited independently of each other can be explained by understanding that:
During meiosis, the process of chromosome assortment into daughter cells is random
Alleles of the genes reside on homologous chromosomes
All of the listed choices are correct
During meiosis, maternal and paternal members of homologous chromosomes are distributed separately into daughter cells
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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- A karyotype shows that a child has Klinefelter syndrome (47,XXY). If the child is also colorblind (due to a recessive X-linked allele), despite his parents having normal color vision, in which parent and stage of meiosis did nondisjunction occur? And explain why .arrow_forwardAn abnormality of the large intestine called intestinal polyposis (IP) is dependent on dominant gene A. A nervous disorder called Huntington’s chorea (HC) is determined by dominant gene H, a lethal gene the effects of which appear on a mean age of 40 years. A and H are on nonhomologous chromosomes. A man carrying the gene A marries a woman carrying the gene H. If this couple had 7 children, what is the chance that at least 2 of their children will survive up to old age? What are the genotypes of the man and his wife?arrow_forwardJim and Pam would like to have a baby, but they know that albinism (an autosomal recessive disease) runs in both of their families. They do a genetic test, and learn that they are both carriers of the disease allele. What is the chance that their baby will have albinism? hint: use a punnet square O 2/3 O 1/8 O 1/4 O 1/16 O 1/2arrow_forward
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