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Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Textbook Question
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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A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies.These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic,25% adenine-requiring auxotrophsa. Explain the difference between these two types ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?
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Chapter 5 Solutions
Concepts of Genetics (11th 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 - Prob. 3CSCh. 5 - Prob. 4CSCh. 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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- this is what i have said about this image so far, what else can be said aswell including the raw count column. " 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…arrow_forwardIn humans, dosage compensation is accomplished by: inactivating one X chromosome in female somatic cells inactivating one homolog from each homologous pair of chromosomes in female somatic cells inactivating the Y chromosome in male somatic cells increasing gene expression from the X chromosome in male somatic cellarrow_forwardKnockout mice are mice in which a functional gene or a group of functional genes are rendered nonfunctional by a special technique involving homologous recombination. Predict what happen to knockout mice which have their RAG-1 and RAG-2 genes “removed".arrow_forward
- 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_forwardYou 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_forward
- Please 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_forward3) You have identified an interesting mutant in gene P. Using a Punnett square, demonstrate the cross you perform to determine if it is a dominant or recessive mutation compared to the WT gene P allele. Write the expected ratios for either scenario. 4) You have determined the mutation is dominant when compared to the WT allele. Briefly describe a technique you could use to determine if expression levels of gene P have been altered in this mutant. 5) You have determined the expression level of gene P has increased. What class/type of mutation would cause this?arrow_forward
- The figure below shows differential methylation patterns for various genes in samples of 2 embryonic stem cells (ES), 5 induced pluripotent stem cells (iPSC), and 4 somatic cells. Some genes were not successfully reprogrammed. Use this figure to answer the following question. which of the following is the most accurate statement about the figure? - The genes in regions A, B, and C, all seem to be expressed in embryonic stem cells, and are silenced primarily in adult cells. - The genes in regions A, B ,and C are active in embryonic stem cells. - HDACs may be acting on genes in regions A, B , and C in the ES cells. - All of the iPSC cells are safe to use in therapy.arrow_forwardSuppose, you want to detect the CAG repeat expansion within a particular gene (30 repeats in normal changes to 250 repeats in disease) in a certain disease. How will you diagnose this disease condition? Can you identify Y chromosome microdeletion (which involves the deletion of AZF locus) using conventional karyotyping? If not, then why? How will you diagnose a chromosomal translocation event? (Discuss any one of the processes)arrow_forwardTo detect the CAG repeat expansion with a particular gene where 30 repeats in Normal changes to 250 repeats in a certain disease, how can we diagnose the condition. How To identify Y chromosome microdeletion ( which involves the deletion of AZF locus) using conventional karyotyping? If not then why. How will you diagnose a chromosomal translocation event?arrow_forward
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