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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A panel of cell lines was created by human–mouse somatic-cell hybridization. Each cell line was examined for the presence of human chromosomes and for the production of three enzymes. The following results were obtained:On the basis of these results, give the chromosomal locations of the genes encoding enzyme 1, enzyme 2, and enzyme 3.
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?
A yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells 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 of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?
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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- 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_forwardConsider the following types of cells: F+, F-, Hfr, and F’ cells. Which of these four types of cells are capable of acting as a donor during conjugation? What genes does each cell that is capable of acting as a donor donate to the recipient cell?arrow_forwardA normal appearing female infant was identified with a positive newborn screen, linked to chromosome 12. Few years ago, her older sibling had developed profound hypoglycemia, liver failure leading to coma, and subsequent irreparable brain damage, following a viral illness. The sibling was subsequently shown by clinical testing to have the same disorder that this female infant is screened positive for. a. What is the most likely diagnosis? b. What biomarkers would confirm this on the newborn screening process? Describe the mechanism that causes this metabolic defect.arrow_forward
- The 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_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_forwardNot 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_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_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_forwardIn the Holliday model for homologous recombination shown, the resolution steps can produce recombinant or nonrecombinantchromosomes. Explain how this can occur.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_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_forwardNorthern blotting, RT-PCR, and microarrays can be used to analyze gene expression. A lab studies yeast cells, comparing their growth in two different sugars, glucose and galactose. One student is comparing expression of the gene HMG2 under these two conditions. Which technique(s) could he use and why? Another student wants to compare expression of all the genes on chromosome 4, of which there are approximately 800. What technique(s) could she use and why?arrow_forward
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