EBK CONCEPTS OF GENETICS
12th Edition
ISBN: 9780134818979
Author: Killian
Publisher: YUZU
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Chapter 9, Problem 22ESP
Because offspring inherit the mitochondrial genome only from the mother, evolutionarily the mitochondrial genome in males encounters a dead end. The mitochondrial genome in males has no significant impact on the genetic information of future generations. Scientists have proposed that this can result in an accumulation of mutations that have a negative impact on genetic fitness of males but not females. Experiments with Drosophila support this possibility. What experimental data or evidence would you want to evaluate or consider to determine if an accumulation of mtDNA mutations negatively impacts the fitness of males of any species?
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Chapter 9 Solutions
EBK CONCEPTS OF GENETICS
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 - Why did Marcia choose mitochondrial testing to...Ch. 9 - Marcia saw an ad on television for ancestry DNA...Ch. 9 - How much importance should we place on the results...Ch. 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 - Mitochondrial replacement therapy (MRT) offers a...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...Ch. 9 - As mentioned in Section 9.3, mtDNA accumulates...Ch. 9 - Because offspring inherit the mitochondrial genome...
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- Following a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. What is the genotype of Males 4? Select only one answer. 01. C/C D/D O 2. C/C D/d O3. C/C d/d O 4. C/C D/D O 5. c/c D/d O 6. C/cd/d 07. c/c D/D O 8. C/C D/d O 9. c/cd/d Karrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call C and d. The two homozygous C/C and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/C and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for C and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Using the number of flies expected for the Parental and the Recombinant genotypes and the formula that defines recombination frequency, what is the distance between loci C…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. What is the genotype of the F1 (progeny of Females 3 x Males 4). Select only one answer. 1. C/c D/d 2. C/C d/d 3. C/C D/D 4. c/c D/d…arrow_forward
- Following a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Which is the genotype of the rhythmic flies resulting from the testcross? Select only one answer 1. C/C d/d 2. C/c D/d 3. C/c d/d 4. c/c…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Comparing the number of flies expected for the Parental and the Recombinant genotypes, what can be concluded about the C and D loci? Select only one answer. 1. The…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Using the number of flies expected for the Parental and the Recombinant genotypes and the formula that defines recombination frequency, what is the recombination…arrow_forward
- The mitochondrial genome of Chlamydomonas contains the lowest number of genes, 12, of all the species of green algae. However, only 7 proteins are produced inside these mitochondria. Which of the following statements is the best explanation for why only 7 proteins are made from the 12 genes In Chlamydomonas mitochondria? OA Five of the genes are redundant; another copy is expressed from the nuclear genome.. OB. Five of the genes are nonfunctional "pseudogenes". OC Five of the genes are transcribed, but not translated. OD. Five of the genes are coding for proteins that are translated in the cytoplasm. Reset Selectionarrow_forwardResearchers have exploited Minute mutations in orderto study the phenotypes associated with recessive lethal mutations (l−) that decrease the rate of cell divisionand thus make only very tiny homozygous mutant clones that are difficult to analyze. Many differentstrains of Drosophila carry dominant loss-of-functionMinute (M) mutations in a variety of genes encodingribosomal protein subunits. The M genes are haploinsufficient; flies with only one wild-type M+ gene copyhave a slower pace of cell division, and thus prolongeddevelopment and subtle morphological abnormalities.To circumvent the tiny clone problem, researchersgenerate GFP-marked homozygous l−/ l− clones thatare also M+/ M+, in flies that are l−/ l+ and M−/ M+.The loss of the Minute mutation only in cells withinthe clone gives the l−/ l− cells a growth advantageover their neighbors, enabling the mutant clone togrow large enough to study. Diagram chromosomesthat could be used to generate such clonesarrow_forwardTo understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and “w” means that the hybrid wiggled.a. Explain what this experiment was designed to test. b. Use this reasoning to assign genotypes to all 12 mutants. c. Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5arrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY