Genetics: From Genes to Genomes
Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 11, Problem 34P

Approximately 3% of the population carries a mutant allele at the CFTR gene responsible for the recessive disease cystic fibrosis. A genetic counselor is examining a family in which both parents are known to be carriers for a CFTR mutation. Their first child was born with the disease, and the parents have come to the counselor to assess whether the new fetus inside the mother’s womb is also diseased, is a carrier, or is homozygous wild type at the CF locus. DNA samples from each family member and the fetus are tested by PCR and gel electrophoresis for an SSR marker within one of the CFTR gene’s introns. The following results are obtained:

 Chapter 11, Problem 34P, Approximately 3 of the population carries a mutant allele at the CFTR gene responsible for the

a. What is the probability that the child who will develop from this fetus will exhibit the disease?
b. When this child grows up, what is the probability that any one of her own children will be afflicted with the disease?
c. The cystic fibrosis gene displays extensive allelic heterogeneity: More than 1500 different mutations of the CFTR gene have been shown to be associated with cystic fibrosis worldwide. With this fact in mind, why might human geneticists choose to test the fetus in the indirect manner described in this problem rather than focusing directly on the mutations that actually caused the disease in the first child?
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Cystic fibrosis (CF) is an inherited chronic disease that affects the lungs and digestive system of about 30,000 children and adults in the United States. It is caused by a recessive allele that encodes a defective chloride channel. Genetic screening has determined that two parents are carriers (heterozygous). What is the probability that they have a child with CF? Rabbits with the genotype CC° are black, heterozygous rabbits CCW appear brown, and rabbits that are CWC" have white fur. VWhat is the phenotypic and genotypic ratio of the offspring of a cross between a brown and a white rabbit? Using your book, determine the probability of each blood type in the offspring of a heterozygous man who has type B blood, and a heterozygous woman who has type A blood. A woman who is type O gives birth to a baby who is also type 0. In a paternity case, a man claims that he cannot be the father because he has type A blood. Is he right? Explain.
X‑linked, recessive diseases, such as hemophilia, are extremely rare in the population. However, many women are carriers and show no sign of the disease. The pedigree illustrates the inheritance of an X‑linked, recessive disease. Determine whether the unknown individuals are affected by the disease, unaffected by the disease, or carriers of the X‑linked recessive allele. Unaffected individuals are not carriers of the X‑linked recessive allele.
Cystic fibrosis (CF) is an autosomal recessive trait. A three-generation pedigree is shown below for a family that carries the mutant allele for cystic fibrosis. Note that carriers are not colored in to allow you to figure out their genotypes.      Normal allele =  F      CF mutant allele =  f        What is the genotype of individual #13? A) ff B) FF C) Ff  D) it is impossible to tell

Chapter 11 Solutions

Genetics: From Genes to Genomes

Ch. 11 - Prob. 11PCh. 11 - The previous problem raises several interesting...Ch. 11 - You want to make a recombinant DNA in which a PCR...Ch. 11 - You sequence a PCR product amplified from a...Ch. 11 - Prob. 15PCh. 11 - The trinucleotide repeat region of the Huntington...Ch. 11 - Sperm samples were taken from two men just...Ch. 11 - Prob. 18PCh. 11 - a. It is possible to perform DNA fingerprinting...Ch. 11 - On July 17, 1918, Tsar Nicholas II; his wife the...Ch. 11 - The figure that follows shows DNA fingerprint...Ch. 11 - Microarrays were used to determine the genotypes...Ch. 11 - A partial sequence of the wild-type HbA allele is...Ch. 11 - a. In Fig. 11.17b, PCR is performed to amplify...Ch. 11 - The following figure shows a partial microarray...Ch. 11 - Scientists were surprised to discover recently...Ch. 11 - The microarray shown in Problem 25 analyzes...Ch. 11 - The figure that follows shows the pedigree of a...Ch. 11 - One of the difficulties faced by human geneticists...Ch. 11 - Now consider a mating between consanguineous...Ch. 11 - The pedigree shown in Fig. 11.22 was crucial to...Ch. 11 - You have identified a SNP marker that in one large...Ch. 11 - The pedigrees indicated here were obtained with...Ch. 11 - Approximately 3 of the population carries a mutant...Ch. 11 - The drug ivacaftor has recently been developed to...Ch. 11 - In the high-throughput DNA sequencing protocol...Ch. 11 - A researcher sequences the whole exome of a...Ch. 11 - As explained in the text, the cause of many...Ch. 11 - Figure 11.26 portrayed the analysis of Miller...Ch. 11 - A research paper published in the summer of 2012...Ch. 11 - Table 11.2 and Fig. 11.27 together portray the...Ch. 11 - The human RefSeq of the entire first exon of a...Ch. 11 - Mutations in the HPRT1 gene in humans result in at...Ch. 11 - Prob. 44P
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