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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Chapter 9, Problem 3PDQ
Streptomycin resistance in Chlamydomonas may result from a mutation in either a chloroplast gene or a nuclear gene. What
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In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?
DNA from a strain of Bacillus subtilis with genotype a+ b+ c+ d+ e+ is used to transform a strain with genotype a− b− c− d− e−. Pairs of genes are checked for cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformation a+ and b+ No b+ and d+ No a+ and c+ No b+ and e+ Yes a+ and d+ Yes c+ and d+ No a+ and e+ Yes c+ and e+ Yes b+ and c+ Yes d+ and e+ No On the basis of these results, what is the order of the genes on the bacterial chromosome?
From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
Chapter 9 Solutions
Concepts of Genetics (11th Edition)
Ch. 9 - Chlamydomonas, a eukaryoric green alga, may be...Ch. 9 - In aerobically cultured yeast, a petite mutant is...Ch. 9 - DNA in human mitochondria encodes 22 different...Ch. 9 - Prob. 4NSTCh. 9 - In 2000, Rocco Baldelli was a first-round draft...Ch. 9 - In 2000, Rocco Baldelli was a first-round draft...Ch. 9 - Prob. 3CSCh. 9 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 9 - Review the Chapter Concepts list on page 196. The...Ch. 9 - Streptomycin resistance in Chlamydomonas may...
Ch. 9 - A plant may have green, white, or green-and-white...Ch. 9 - In diploid yeast strains, sporulation and...Ch. 9 - Predict the results of a cross between ascospores...Ch. 9 - In Lymnaea, what results would you expect in a...Ch. 9 - In a cross of Lymnaea, the snail contributing the...Ch. 9 - In Drosophila subobscura, the presence of a...Ch. 9 - A male mouse from a true-breeding strain of...Ch. 9 - Consider the case where a mutation occurs that...Ch. 9 - What is the endosymbiotic theory, and why is this...Ch. 9 - In an earlier Problems and Discussion section (see...Ch. 9 - The specification of the anteriorposterior axis in...Ch. 9 - The maternal-effect mutation bicoid (bcd) is...Ch. 9 - (a) In humans the mitochondrial genome encodes a...Ch. 9 - Mutations in mitochondrial DNA appear to be...Ch. 9 - Researchers examined a family with an interesting...Ch. 9 - Payne, B. A. et al. (2013) present evidence that a...
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- Three haploid fungal mutants that require compound W for growth were isolated. Each mutant contains a recessive allele in a single gene. Three compounds (A, B and C) in the biosynthetic pathway to W are known, but their order in the pathway is unknown. Each compound is tested for its ability to support the growth of each of the three mutants. Phenotypes of all of the three mutants are shown in the following table (“+" indicates growth, "-" indicates no growth). A C W Mutant 1 Mutant 2 Mutant 3 What would be the phenotype of a haploid mutant that contains both mutant alleles in mutant 2 and 3? Phenotype refers to growth or absence of growth on compounds A, B, C and WN. O Like mutant 1 O Like mutant 2 Like mutant 3 O Like wild typearrow_forwardBy conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the following table:Explain how these results are consistent with the idea that thebacterial chromosome is circular?arrow_forwardA series of Hfr strains that have genotype m+ n+ o+ p+ q+ r+ are mixed with an F− strain that has genotype m− n− o− p− q− r−. Conjugation is interrupted at regular intervals and the order of the appearance of genes from the Hfr strain is determined in the recipient cells. The order of gene transfer for each Hfr strain is What is the order of genes on the circular bacterial chromosome? For each Hfr strain, give the location of the F factor in the chromosome and its polarity. Hfr5 m+ q+ p+ n+ r+ o+ Hfr4 n+ r+ o+ m+ q+ p+ Hfr1 o+ m+ q+ p+ n+ r+ Hfr9 q+ m+ o+ r+ n+ p+arrow_forward
- A cross was performed between a yeast strain that requires methionine and lysine for growth (met− lys−)and another yeast strain, which is met+ lys+. One hundred asci were dissected, and colonies were grownfrom the four spores in each ascus. Cells from thesecolonies were tested for their ability to grow on petriplates containing either minimal medium (min), min+ lysine (lys), min + methionine (met), or min + lys+ met. The asci could be divided into two groupsbased on this analysis:Group 1: In 89 asci, cells from two of the four spore colonies couldgrow on all four kinds of media, while the other two spore coloniescould grow only on min + lys + met.Group 2: In 11 asci, cells from one of the four spore colonies couldgrow on all four kinds of petri plates. Cells from a second one ofthe four spore colonies could grow only on min + lys plates andon min + lys + met plates. Cells from a third of the four sporecolonies could only grow on min + met plates and on min + lys+ met. Cells from the…arrow_forwardFor a haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in and absorbed from the environment. The arginine biosynthetic pathway is: Enzyme A Enzyme B Enzyme Ç Compound X It is know that genes encoding enzymes A and C are on two different chromosomes. Compound Y Compound Z- Arginine A mutant strain of genotype a (lacking only enzyme A) is crossed to a mutant strain of genotype c (lacking only enzyme C) to generate a diploid strain. Sporulation (i.e. meiosis) is subsequently induced in the resulting diploid strain. What proportion of the spores (haploids formed by sporulation) is expected to grow on medium without arginine but supplemented with Compound Y? O 100% 50% 0% 25%arrow_forwardDetermine the coefficient of coincidence and the interferencearrow_forward
- DNA from a strain of Bacillus subtilis with genotype a* b* c* d* e* is used to transform a strain with genotype a b c d e. Pairs of genes are checked for cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformatic a* and b* No b* and d* No a* and c* No b* and e* Yes a* and d* Yes c* and d* No a* and e* Yes c* and e* Yes b* and c* Yes d* and e* No On the basis of these results, what is the order of the genes on the bacterial chromosome?arrow_forwardGiven the following complementation chart for holes in Monstera, give me the biochemical (phenotype) pathway. A В D E Holes Class 1 + + + - - Class 2 + + Class 3 + + + Class 4 Class 5 + + - - + +arrow_forwardSchizosaccharomyces pombe, also known as "fission yeast," is a powerful model organism in molecular and cell biology. While performing a genetic screen, you discover an auxotrophic S. pombe strain that is unable to synthesize one or more vitamins. The following table represents the key experiments you performed during your genetic screen. Fill in the table with the outcome of each experiment for your mutant strain (using + for growth and - for no growth). Medium Rich media Minimal media Minimal media + all vitamins Minimal media + all amino acids Growth Wild-type + + + + Mutant + + + > > >arrow_forward
- In E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What information do you know based onthe question and your understanding of the topic?arrow_forwardIn E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What topic in genetics does this question address?arrow_forwardDetermine the map distances between the genes.arrow_forward
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