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 41P
Table 11.2 and Fig. 11.27 together portray the search for the mutation causing Nic Volker’s severe inflammatory bowel disease. Neither of Nic’s parents had the condition, so geneticists narrowed their investigation by focusing on rare variants that showed a recessive pattern and those on the X chromosome.
| a. | For candidate variants on an autosome, would the researchers have looked only for variants for which Nic is homozygous? Explain. |
| b. | Apart from the recessive and X-linked hypotheses, do any other possible explanations exist for Nic’s condition? |
| c. | The causative mutation was pinpointed by analyzing only Nic’s exome, because at the time of these investigations, whole-genome or whole-exome sequencing was too expensive to perform on his parents. How could you determine inexpensively whether or not this mutation occurred de novo in the germ line of one of his parents (that is, during the formation of the particular egg or sperm that produced Nic)? Your answer should not involve whole-genome or whole-exome sequencing. |
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Susan’s grandfather was deaf, and passed down a hereditary form of deafness within Susan’s family as shown in Figure Q19–12.A. Is this mutation most likely to be dominant or recessive?B. Is it carried on an autosome or a sex chromosome? Why?C. A complete SNP analysis has been done for all of the 11 grandchildren (4 affected, and 7 unaffected). In comparing these 11 SNP results, how long a haplotype block would you expect to find around the critical gene? How might you detect it?
Friedreich ataxia (FRDA) is an autosomal recessive, neurodegenerative disease that causes a lack of voluntary coordination of muscle movements. Affected individuals are homozygous for an unusually large number (expansion) of repeats of a trinucleotide sequence (GAA) in the first intron of the X25 gene. Unaffected individuals typically have between 7 and 38 repeats of the trinucleotide (GAAGAAGAAGAA…). FRDA patients have anywhere from 66 to over 1,700 repeats.
To understand how the GAA trinucleotide expansion leads to FRDA, researchers looked at X25 gene expression by extracting RNA from affected and unaffected patients and doing a northern blot analysis (see the figure below):
In panel “a,” the researchers used a probe to detect X25 mRNA.
In panel “b,” the researchers used a probe on a duplicate of the original blot to detect human GAPDH mRNA (GAPDH is an enzyme involved in glycolysis).
The sample labeled “YR” is mRNA from yeast cells that was used as a control.
Explain…
Consider two hypothetical recessive autosomal genes a and b, where a heterozygote is testcrossed to a double- homozygous mutant. Predict the phenotypic ratios under the following conditions: (a) a and b are located on separate autosomes. (b) a and b are linked on the same autosome but are so far apart that a crossover always occurs between them. (c) a and b are linked on the same autosome but are so close together that a crossover almost never occurs.
Chapter 11 Solutions
Genetics: From Genes to Genomes
Ch. 11 - Choose the phrase from the right column that best...Ch. 11 - Would you characterize the pattern of inheritance...Ch. 11 - Would you be more likely to find single nucleotide...Ch. 11 - A recent estimate of the rate of base...Ch. 11 - If you examine Fig. 11.5 closely, you will note...Ch. 11 - Approximately 50 million SNPs have thus far been...Ch. 11 - Mutations at simple sequence repeat SSR loci occur...Ch. 11 - Humans and gorillas last shared a common ancestor...Ch. 11 - In 2015, an international team of scientists...Ch. 11 - Using PCR, you want to amplify an approximately 1...
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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