Inquiry into Life
14th Edition
ISBN: 9780073525525
Author: Mader, Sylvia S./
Publisher: McGraw-Hill College
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Chapter 26, Problem 2TC
In a genomic comparison between humans and yeast, what genes would you except to be similar?
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In an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n = 7) moves slowly but eventually forms seven bands, which represent DNA fractions that are of different sizes and hence have moved at different speeds. These bands are presumed to be the seven chromosomes. How would you show which band corresponds to which chromosome?
It is possible to take the DNA of a gene from any source and place it on a chromosome in the nucleus of a yeast cell. When you take DNA of a human gene and put it into a yeast cell chromosome, the yeast cell can synthesize the human protein. However, when you remove the DNA for a gene normally present on yeast mitochondrial chromosomes and put it on a yeast chromosome in the nucleus, the yeast cell cannot synthesize the correct protein, even though the gene comes from the same organism. Explain. What would you need to do to ensure that such a yeast cell could make the correct protein?
Chapter 26 Solutions
Inquiry into Life
Ch. 26.1 - Describe the steps in forming Recombinant DNA.Ch. 26.1 - Discuss how the polymerase chain reaction works.Ch. 26.1 - Prob. 3LOCh. 26.1 - Prob. 1CYPCh. 26.1 - Prob. 2CYPCh. 26.1 - Would you be willing to provide your DNA for a...Ch. 26.1 - Prob. 2QTCCh. 26.1 - Prob. 3QTCCh. 26.2 - Prob. 1LOCh. 26.2 - Prob. 2LO
Ch. 26.2 - Prob. 1QTCCh. 26.2 - Prob. 2QTCCh. 26.2 - Prob. 3QTCCh. 26.2 - Prob. 1CYPCh. 26.2 - Prob. 2CYPCh. 26.3 - Compare and contrast in vivo and ex vivo gene...Ch. 26.3 - Prob. 1CYPCh. 26.3 - Prob. 2CYPCh. 26.3 - Prob. 1QTCCh. 26.3 - Prob. 2QTCCh. 26.3 - Prob. 3QTCCh. 26.4 - Prob. 1LOCh. 26.4 - Prob. 2LOCh. 26.4 - Prob. 3LOCh. 26.4 - Prob. 1CYPCh. 26.4 - Explain how comparative genomics can provide...Ch. 26.4 - Prob. 3CYPCh. 26 - Prob. F25.3BYBCh. 26 - Prob. F25.5BYBCh. 26 - Prob. S25.3BYBCh. 26 - Prob. 1CSCh. 26 - What is the difference between a genetically...Ch. 26 - Prob. 3CSCh. 26 - Prob. 1ACh. 26 - Prob. 2ACh. 26 - Prob. 3ACh. 26 - The polymerase chain reaction Use RNA polymerase...Ch. 26 - Prob. 5ACh. 26 - Prob. 6ACh. 26 - Prob. 7ACh. 26 - Prob. 8ACh. 26 - Prob. 9ACh. 26 - Prob. 10ACh. 26 - Prob. 11ACh. 26 - Prob. 12ACh. 26 - Prob. 13ACh. 26 - Prob. 1TCCh. 26 - In a genomic comparison between humans and yeast,...
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- What percentage of the DNA in the genome actually corresponds to genes? How much is actually protein-coding exons? What makes up the rest?arrow_forwardA yeast strain with a mutant spo11- allele has been isolated. The mutant allele is nonfunctional; it makes no spo11 protein. What do you suppose is the phenotype of this mutant strain?arrow_forwardWhy The Arrangement of Genesin the Genome Is Not Uniform?arrow_forward
- Which one of the following options most accurately describes how the ~3000 megabases (Mb; ~3,000,000,000 nucleotides) of the haploid human genome are distributed on chromosomes? The haploid human genome is distributed on autosomes 1-22 and X and Y (so, 24 chromosomes). The haploid human genome is distributed on the homologous pairs of autosomes 1-22 and X and Y (so, 46 chromosomes). The haploid human genome is distributed on a single chromosome (so, 1 chromosome). The haploid human genome is distributed on one pair of homologous chromosomes (so, 2 chromosomes).arrow_forwardYou are working with two different yeast cultures to study their genetics. But, you are not sure what the mating type of either of the yeast cultures is. (As a reminder, yeast mating types are either a or alpha). As a test, you mix the two cultures together and then observe them under a microscope. If this is the image that you see, what can you conclude about the mating types of the yeast? Explain your answer, including what type of cell morphology you see on this microscope slide.arrow_forwardPart A: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be AB? Enter your answer as a decimal to three places (for example: 0.120). Part B: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be Ab? Part C: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be aB? Part D: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a signal plant will be ab?arrow_forward
- In yeast, LYS5, ADE1, and URA2 are genes required to synthesize lysine, adenine and uracil, respectively. A MATa lys5 ADE1 URA2 haploid was mated to a MATA LYS5 ade1 ura2 haploid to make a triply heterozygous diploid. This diploid was put through meiosis, 100 tetrads were dissected and the spore colonies transferred to medium lacking either lysine, adenine or uracil to determine which spores were prototrophic or auxotrophic for those nutrients. The following results were obtained: LYS5-ADE1 ADE1-URA2 LYS5-URA2 PD NPD Ι 20 22 58 18 20 62 50 8 42 Based on this information, select the statements below that are TRUE. Select 4 correct answer(s) A) The LYS5 and URA2 genes are linked. B) The genotypes of the four spores in the ADE1-URA2 tetratypes is ade1 ura2 ade1 URA2 ADE1 ura2 ADE1 URA2 C) The map distance between ADE1 and URA2 is 51 CM. D) ADE1 and URA2 are likely located on the same chromosome. E) The map distance between LYS5 and URA2 is 29 CM. ☐ F) LYS5 and ADE1 independently assort…arrow_forwardBased on the data in Table 1, which individual(s) is/are heterozygous? Select all that apply a) B IV-8 b) C IV-3 c) B IV-9 d) A IV-3arrow_forwardThe image shows the genetic code of an organism before and after the occurrence of a spontaneous error during chromosomal crossover. G-T-G-C-C-A-T-C-A → G-T-G-T-C-A Based on the information provided, which type of mutation most likely occurred? O deletion O nondisjunction O substitution O translocationarrow_forward
- A diploid (ie, contains TWO sets of chromosomes) organisms with a 45,000-kb haploid (counts only one set of its chromosomes) genome contains 21% G residues. Calculate the number of A, C, G< and T residues in the DNA of each cell in this organism. Can you help explain why this is the answer, thank you! Answer: Since the haploid genome contains 21% G, it must contain 21% C (Because G=C) and 58% A + T, or 29% A and 29% T. Each cell is a diploid, containing 90,000 kb or 9x10^7 bases. Therefore, A=T = (0.29)(9x10^7) = 2.61 x 10^7 bases and G=C=(0.21)(9x10^7) = 1.89x10^7 bases.arrow_forwardHumans and many other complex organisms are diploid, possessing two sets of genes, one inherited from the mother and one from the father. However, a number of eukaryotic organisms spend most of their life cycles in a haploid state. Many of these eukaryotes, such as Neurospora and yeast, still undergo meiosis and sexual reproduction, but most of the cells that make up the organism are haploid. Considering that haploid organisms are fully capable of sexual reproduction and generating genetic variation, why are most complex eukaryotes diploid? In other words, what might be the evolutionary advantage of existing in a diploid state instead of a haploid state? And why might a few organisms, such as Neurospora and yeast, exist as haploids?arrow_forwardWhen comparing evolutionary similarities between different genes within a gene family, it is usually more straightforward to compare genes by using the protein sequences of gene products rather than DNA sequences of the genes themselves. Explain why this is the case. (Write 5 points)arrow_forward
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