Concepts of Genetics (12th Edition)
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 9, Problem 1NST

Chlamydomonas, a eukaryoric green alga, may be sensitive to the antibiotic erythromycin, which inhibits protein synthesis in bacteria. There are two mating types in this alga, mt+ and mt. If an mt+ cell sensitive to the antibiotic is crossed with an mt cell that is resistant, all progeny cells are sensitive. The reciprocal cross (mt+ resistant and mt sensitive) yields all resistant progeny cells. Assuming that the mutation for resistance is in the chloroplast DNA, what can you conclude from the results of these crosses?

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Summary Introduction

To determine: The conclusion that can be drawn from the results of the given crosses.

Introduction: Chlamydomonas is a eukaryotic green alga and shows two mating types mt+ and mt-. The cross between mt+ cell (sensitive to antibiotic) and mt- cell (resistant to antibiotic) produces sensitive progeny cells.

Explanation of Solution

The following conclusions can be drawn from the results of the given crosses:

  • The mt+ strain of Chlamydomonas is the donor of the chloroplast DNA. This is because the inheritance of resistance or sensitivity to the antibiotic relies on the status of mt+ gene.
  • In Chlamydomonas, the chloroplasts obtain their features from mt+, whereas mitochondria obtain their features from the mt- strain.
Conclusion

Thus, mt+ strain is the donor of chloroplast DNA and the chloroplasts gain their characteristics from mt+. The mitochondria obtain their features from mt-.

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Wild-type strains of the haploid fungus Neurospora canmake their own tryptophan. An abnormal allele td renders the fungus incapable of making its own tryptophan.An individual of genotype td grows only when its medium supplies tryptophan. The allele su assorts independently of td; its only known effect is to suppress the tdphenotype. Therefore, strains carrying both td and su donot require tryptophan for growth.a. If a td ; su strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progenyand in what proportions?b. What will be the ratio of tryptophan-dependent totryptophan-independent progeny in the cross of part a?
A Neurospora colony at the edge of a plate seemed to be sparse (low density) in comparison with the other colonies on the plate. This colony was thought to be a possible mutant, and so it was removed and crossed with a wild type of the opposite mating type. From this cross, 100 ascospore progeny were obtained. None of the colonies from these ascospores was sparse, all appearing to be normal. What is the simplest explanation of this result? How would you test your explanation? (Note: Neurospora is haploid.)
The genotypes below that give viable animals for an mdm2+/- p53+/- cross are the following (p53 on chr 17 and mdm2 on chr 12) are:

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Concepts of Genetics (12th Edition)

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