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Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 4, Problem 40P
Males have hemophilia when they are hemizygous for a nonfunctional recessive mutant allele of the X-linked gene for clotting factor VIII. Factor VIII is normally secreted into the blood serum by cells in the bone marrow that produce it.
a. | Do you think that females heterozygous for the hemophilia disease allele could have hemophilia in some parts of their bodies and not others? |
b. | If such a female “carrier” of hemophilia suffered a cut, would her blood coagulate (form clots) faster, slower, or in about the same time as that of an individual homozygous for a normal allele of the factor VIII gene? Would the rate of clotting vary significantly among heterozygous females? |
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Students have asked these similar questions
A 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?
Color blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?
Hemophilia is a disease inherited as a X-linked recessive trait while pattern baldness is controlled by an autosomal gene that is dominant in males and recessive in females. A hemophilic man who is also homozygous for baldness has children with a woman who carries normal genes for both traits. What is the probability that any of their child will be:
a. Hemophilic, normal-haired male
b. Hemophilic, bald male
c Female with normal blood clotting and bald
Note: Show how you derived your answers.
Chapter 4 Solutions
Genetics: From Genes to Genomes, 5th edition
Ch. 4 - Choose the best matching phrase in the right...Ch. 4 - Humans have 46 chromosomes in each somatic cell....Ch. 4 - The figure that follows shows the metaphase...Ch. 4 - XX males who are sex-reversed because they have a...Ch. 4 - Researchers discovered recently that the sole...Ch. 4 - One oak tree cell with 14 chromosomes undergoes...Ch. 4 - Indicate which of the cells numbered iv matches...Ch. 4 - a. What are the four major stages of the cell...Ch. 4 - Answer the questions that follow for each stage of...Ch. 4 - Does any reason exist that would prevent mitosis...
Ch. 4 - One oak tree cell with 14 chromosomes undergoes...Ch. 4 - Which types of cell division mitosis, meiosis I,...Ch. 4 - Complete the following statements using as many of...Ch. 4 - The five cells shown in figure a e on the next...Ch. 4 - One of the first microscopic observations of...Ch. 4 - A person is simultaneously heterozygous for two...Ch. 4 - Assuming i that the two chromosomes in homologous...Ch. 4 - In the moss Polytrichum commune, the haploid...Ch. 4 - Does any reason exist that would prevent meiosis...Ch. 4 - Sister chromatids are held together through...Ch. 4 - The pseudoautosomal regions PARs of the X and Y...Ch. 4 - Somatic cells of chimpanzees contain 48...Ch. 4 - In humans: a. How many sperm develop from 100...Ch. 4 - Women sometimes develop benign tumors called...Ch. 4 - In a certain strain of turkeys, unfertilized eggs...Ch. 4 - Imagine you have two pure-breeding lines of...Ch. 4 - A system of sex determination known as...Ch. 4 - In Drosophila, the autosomal recessive brown eye...Ch. 4 - Barred feather pattern is a Z-linked dominant...Ch. 4 - When Calvin Bridges observed a large number of...Ch. 4 - In a vial of Drosophila, a research student...Ch. 4 - In 1919, Calvin Bridges began studying an X-linked...Ch. 4 - In Drosophila, a cross was made between a...Ch. 4 - As we learned in this chapter, the white mutation...Ch. 4 - The following is a pedigree of a family in which a...Ch. 4 - Each of the four pedigrees that follow represents...Ch. 4 - The pedigree that follows indicates the occurrence...Ch. 4 - Duchenne muscular dystrophy DMD is caused by a...Ch. 4 - The X-linked gene responsible for DMD encodes a...Ch. 4 - Males have hemophilia when they are hemizygous for...Ch. 4 - Consider the following pedigrees from human...Ch. 4 - Several different antigens can be detected in...Ch. 4 - The ancestry of a white female tiger bred in a...Ch. 4 - The pedigree at the bottom of the page shows the...Ch. 4 - In 1995, doctors reported a Chinese family in...Ch. 4 - In cats, the dominant 0 allele of the X-linked...Ch. 4 - In marsupials like the opposum or kangaroo, X...Ch. 4 - The pedigree diagram below shows a family in which...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Hemophilia A is caused by a recessive X-linked allele that encodes a defective form of a clotting protein. If a affected father and a mother who is known to not be a carrier have children, what percentage of female offspring will have hemophilia?arrow_forwardChoose 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.arrow_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
- A mother is heterozygous for the X-linked gene for colorblindness and also heterozygous for the autosomal inherited sickle cell anemia. She is married to a man who can see color normally and who is heterozygous for sickle cell trait. Using b (colorblind), B (normal color), S (normal hemoglobin), s (sickle cell), answer the following: a. What are the genotypes of the parents? b: What is the probability of having a child who is both color blind and has sickle cell anemia?arrow_forwardAnhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). a. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands. b. Why does the distribution of the patches of skin lacking sweat glands differ among the females depicted in the illustration, even between the identical twins?arrow_forwardDuchenne muscular dystrophy is sex linked and usuallyaffects only males. Victims of the disease become progressively weaker, starting early in life.a. What is the probability that a woman whose brotherhas Duchenne’s disease will have an affected child?b. If your mother’s brother (your uncle) had Duchenne’sdisease, what is the probability that you have receivedthe allele?c. If your father’s brother had the disease, what is theprobability that you have received the allele?arrow_forward
- In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman? HR/hr Hr/hr hr/hR Hr/hR HR/Hrarrow_forwardHemophilia is another example of a X-linked disease caused when a recessive allele (Xh) is expressed. If a normal male reproduces with a heterozygous normal female, what are the expected genotypes and phenotypes? Will any of their daughters develop hemophilia? As in the previous question, you must also give the gender of the child in your genotype and phenotype descriptions here.arrow_forwardA woman with a rare autosomal recessive disorder was told that it was unlikely that her children would have the disorderas her husband did not have it. However, her first child has the disorder. a. What is the most likely explanation? b. Diagram the cross between the woman and her husband using a Punnett square, give the genotypic ratio (GR) and phenotypic ratio (PR) from the Punnett square. c. Based on the Punnett square results, what is the chance that her next child will have the disorder?arrow_forward
- In humans, the genes for coloblindedness and hemophilia re both located on the X chromosome with no corresponding gene in the Y. These are both recessive alleles. a. If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnet square that illustrates this. b. If the man dies and the woman remarries to a colorblind man, draw a Punnet Square showing the type of children could be expected from hre second marriage. How many/what percentages of each could ne expectedarrow_forward15. The following pedigree shows inheritance of Huntington's disease, a fatal genetic disorder that causes neurodegeneration. Since signs and symptoms usually do not appear until adulthood, many who are carriers may not realize their risk of passing on the disease-causing allele. The following pedigree represents a family in which some people are affected by Huntington's disease. Reeessive Trit er btmnt be Mec yplicalty Hinheteearrow_forward. 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?arrow_forward
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