Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 5, Problem 28ESP
A number of human–mouse somatic cell hybrid clones were examined for the expression of specific human genes and the presence of human chromosomes. The results are summarized in the following table. Assign each gene to the chromosome on which it is located.
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Interpreting the results of an RNA-Seq analysis is pivotal in understanding the underlying genetic mechanisms of diseases such as breast cancer. In this analysis, Figure 1 provides comprehensive data on differentially expressed genes associated with breast cancer. By delving into the provided information, we can gain valuable insights into the molecular landscape of this disease. First focus is on the gene with the highest fold change, EYA4, situated on chromosome 6. With a staggering fold change of 3604.4176, EYA4 exhibits an unprecedented level of overexpression in cancerous cells compared to normal cells. This profound alteration suggests a pivotal role for EYA4 in breast cancer pathogenesis. The log2 fold change of 11.81555 further emphasizes the magnitude of this difference in gene expression. Statistical significance is evident, with an exceptionally low p-value of…
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Chapter 5 Solutions
Concepts of Genetics (12th Edition)
Ch. 5 - In a family with one autistic child the risk for...Ch. 5 - Given that the prenatal test can provide only a...Ch. 5 - Consider two hypothetical recessive autosomal...Ch. 5 - With two pairs of genes involved (P/p and Z/z), a...Ch. 5 - In Drosophila, a heterozygous female for the...Ch. 5 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 5 - Review the Chapter Concepts list on page 94. Most...Ch. 5 - Describe the cytological observation that suggests...Ch. 5 - Why does more crossing over occur between two...Ch. 5 - Explain why a 50 percent recovery of...
Ch. 5 - Why are double-crossover events expected less...Ch. 5 - What is the proposed basis for positive...Ch. 5 - What two essential criteria must be met in order...Ch. 5 - The genes dumpy (dp), clot (cl), and apterous (ap)...Ch. 5 - Colored aleurone in the kernels of com is due to...Ch. 5 - In the cross shown here, involving two linked...Ch. 5 - In a series of two-point mapping crosses involving...Ch. 5 - Two different female Drosophila were isolated,...Ch. 5 - In Drosophila, a cross was made between femalesall...Ch. 5 - Another cross in Drosophila involved the...Ch. 5 - In Drosophila, Dichaete (D) is a mutation on...Ch. 5 - Drosophila females homozygous for the third...Ch. 5 - In Drosophila, two mutations, Stubble (Sb) and...Ch. 5 - If the cross described in Problem 18 were made,...Ch. 5 - Are mitotic recombinations and sister chromatid...Ch. 5 - What possible conclusions can be drawn from the...Ch. 5 - An organism of the genotype AaBbCc was testcrossed...Ch. 5 - Based on our discussion of the potential...Ch. 5 - Traditional gene mapping has been applied...Ch. 5 - DNA markers have greatly enhanced the mapping of...Ch. 5 - In a certain plant, fruit is either red or yellow,...Ch. 5 - Two plants in a cross were each heterozygous for...Ch. 5 - A number of humanmouse somatic cell hybrid clones...Ch. 5 - A female of genotype produces 100 meiotic tetrads....Ch. 5 - In laboratory class, a genetics student was...Ch. 5 - Drosophila melanogaster has one pair of sex...Ch. 5 - In Drosophila, a female fly is heterozygous for...Ch. 5 - The gene controlling the Xg blood group alleles...Ch. 5 - Prob. 34ESP
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- Not all inherited traits are determined by nuclear genes (i.e., genes located in the cell nucleus) that are expressed during the life of an individual. In particular, maternal effect genes and mitochondrial DNA are notable exceptions. With these ideas in mind, let’s consider the cloning of a sheep (e.g., Dolly). A. With regard to maternal effect genes, is the phenotype of such a cloned animal determined by the animal that donated the enucleatedegg or by the animal that donated the somatic cell nucleus? Explain.arrow_forwardIn the Ames test shown in Figure 16-17, what is the reason for adding the liver extract to each sample?arrow_forwardThe data below are from a DNAse-Seq experiment of chromosome 22. DNAse-seq is another method for measuring chromatin organization. In the experiment, regions of chromatin are digested with DNAse I and DNA is sequenced to determined to identify cleavage sites. See this short article for complete explanation. Which of the following statements are true based on the data below (select all that apply)? A. Region C appears to be an actively transcribed region in all tissue types except trophoblasts B. Region A could be an origin of replication for the chromosome C. Region B is highly associated with nucleosomes in BE2C and Natural KIller Cells D. The genes in the fetal kidney tissue are the least actively transcribed on chromosome 22. E. The data for each tissue must be from separate people to have such a variation in DNAse sensitivityarrow_forward
- You have isolated 8 mutants in yeast that fail to grow on minimal media plates but do grow when they are supplemented with Arginine. You know that Arginine is synthesized in a biochemical pathway within wild-type yeast, but you do not know how many gene products it takes for the pathway. You have all of the lines as both a and a cells and mate each strain to each other in pairwise crosses and plate them on minimal media to see if they grow. You obtain the following results with (+) representing growth, and (-) indicating no growth: a 1 5 1 a 4 5 6 7 8 How many genes are represented? O 1 3 7 O Cannot tell from the data a + + + + + • + + i 0 +, + + + • + + 7 + + + + + , . + + + + + m + + + + + + + 2 + + + + + i + -I + + . . + + +arrow_forwardPlease consider providing a detailed explanation. Do not copy from previously provided solutionsarrow_forwardGenes with highly similar sequence are often located adjacent one another in the genome. Gene duplication commonly arises from errors in replication. When the organization of such adjacent genes is in an inverted orientation, this can reduce the expression of other genes that have similar sequence and are located on other chromosomes. Explain the mechanism of how this generally occurs. Please state the answer in details: what is the mechanism? How it happens? Why this happens? When it happens? And every other necessary information.arrow_forward
- All the cells of one organism share the same genome. However, during development, some cells develop into skin cells while others develop into muscle cells. Briefly explain how the same genetic instructions can result in two different cell types in the same organism.arrow_forwardWhen considering the cause of the disease in a patient who is the child of an inbred couple, it was found that 4 enzymes in the lysosome were inactivated. When comparing the amino acid sequences of these four enzymes with the wild-type enzyme, no difference was found in the number and arrangement of amino acids. Given mutations that may be the cause of the patient's disease above.arrow_forwardLINEs and SINEs are repetitive sequences in humans that as retrotransposons, can also insert into genes and cause disease. Select one: True FalseWhat is the most reasonable explanation for the observation that transposons in many multicellular genomes are more often found in nongenic sequences (i.e.,sequences that do not code for genes) such as centromeric heterochromatin rather than in genic sequences (i.e,. gene sequences)? Select one: a. The transposons are “safe” from harm from the host when in nongenic regions. b. Reverse transcriptase promotes integration into nongenic DNA preferentially. c. Insertion into nongenic DNA is less likely to do harm to the host and thus would not be selected against. d. Genic DNA is protected from transposon insertion by all the transcription factors bound to the region. e. The AT-rich nature of centromere and other nongenic sequences makes it easier for transposon insertion.arrow_forward
- A 19 year old female patient is diagnosed with chronic myelogenous leukemia. Karyotype analysis shows that the leukemic cells of this patient are heterozygous for a reciprocal translocation involving chromosomes 9 and 22. However, none of the normal, nonleukemic cells of this patient contain the translocation. a) Describe a molecular test to determine if chemotherapy given to the patient described would be completely succesful. (That is, devise a method to make sure that the patient's blood would be free of leukemic cells.) Be as specific as possible.arrow_forwardThe technique of fluorescence in situ hybridization (FISH) is described. This is another method for examining sequence complexity within a genome. In this method, a DNA sequence, such as a particular gene sequence, can be detected within an intact chromosome by using a DNA probe that is complementary to the sequence.For example, let’s consider the β-globin gene, which isfound on human chromosome 11. A probe complementary to theβ-globin gene binds to that gene and shows up as a brightly colored spot on human chromosome 11. In this way, researchers can detectwhere the β-globin gene is located within a set of chromosomes. Becausethe β-globin gene is unique and because human cells are diploid(i.e., have two copies of each chromosome), a FISH experimentshows two bright spots per cell; the probe binds to each copy ofchromosome 11. What would you expect to see if you used thefollowing types of probes?A. A probe complementary to the Alu sequenceB. A probe complementary to a tandem array near…arrow_forwardThe DNA sequence of one strand of a gene from threeindependently isolated mutants is given here (5′ endsare at left). Using this information, what is the sequence of the wild-type gene in this region?mutant 1 ACCGTAATCGACTGGTAAACTTTGCGCGmutant 2 ACCGTAGTCGACCGGTAAACTTTGCGCGmutant 3 ACCGTAGTCGACTGGTTAACTTTGCGCGarrow_forward
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