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 14, Problem 6SQ
Choose the statement that is incorrect.
- a. A son can inherit a recessive allele on an X chromosome from either parent.
- b. An individual may inherit three or more of each type of chromosome characteristic or the species, a condition called polyploidy.
- c. A female child inherits one X chromosome from her mother and one from her father.
- d. Pedigree analysis can be used to determine a future child’s chance of being born with achcondroplasia.
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Webbed fingers is inherited as an X-linked disease An unaffected male marries an
affected female.
a. Draw a Punnett square of the possible offspring.
b. List the phenotypes of the possible children
c. Draw a pedigree that displays the inheritance in you Punnett square
Choose correct option and do explain.
Considering an X-linked dominant trait, if an affected woman and an unaffected man decide to have children, which of the answer choices is possible for their children?
a.
All of their sons are expected to show the dominant trait.
b.
Their daughters are expected be heterozygous for the gene.
c.
Their daughters are not expected to show the dominant trait.
d.
Their sons are expected to be heterozygous for the gene.
e.
All their children, whether male or female, are expected to show the dominant trait.
Hemophilia 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.
Chapter 14 Solutions
Biology: The Unity and Diversity of Life (MindTap Course List)
Ch. 14 - Prob. 1DAACh. 14 - Prob. 2DAACh. 14 - Prob. 3DAACh. 14 - Prob. 4DAACh. 14 - Prob. 5DAACh. 14 - Prob. 1SQCh. 14 - Pedigree analysis is necessary when studying human...Ch. 14 - A recognized set of symptoms that characterize a...Ch. 14 - Prob. 4SQCh. 14 - A trait that is present in a male child but not in...
Ch. 14 - Choose the statement that is incorrect. a. A son...Ch. 14 - Prob. 7SQCh. 14 - Prob. 8SQCh. 14 - Klinefelter syndrome (XXY) can most be easily...Ch. 14 - Prob. 10SQCh. 14 - Does the phenotype indicated by the red circles...Ch. 14 - G6PD deficiency is an X-Linked recessive disorder....Ch. 14 - Marian syndrome (Section 13.5) is inherited in an...Ch. 14 - Duchenne muscular dystrophy, which is inherited in...Ch. 14 - Human females have two X chromosomes (XX); males...Ch. 14 - A mutation on an autosome causes a particular...Ch. 14 - Expression of the SRY gene on the Y chromosome...Ch. 14 - The somatic cells of most individuals with Down...Ch. 14 - Mutations in the genes for clotting factor VIII...
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- A type of baldness is sex –limited such that the allele that causes it is dominant in men and recessive in women. Thus baldness is much more common in men than women even though the locus is on an autosome. How is the pattern of inheritance in this example different from X-linked inheritance of a recessive allele? Group of answer choices A. the trait can be passed from fathers to sons B. the trait can be passed from fathers to daughters C. the trait can be passed from mothers to sons D. the trait can be passed from mothers to daughtersarrow_forwardA man who has color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The woman’s father had color blindness. Color blindness is determined by an X-linked gene, and blood type is determined by an autosomal gene. a. What are the genotypes of the man and the woman? b. What proportion of their children will have color blindness and type B blood? c. What proportion of their children will have color blindness and type A blood? d. What proportion of their children will be color blind and have type AB blood?arrow_forwardTay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.arrow_forward
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