Concept explainers
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?
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.
Thus, mt+ strain is the donor of chloroplast DNA and the chloroplasts gain their characteristics from mt+. The mitochondria obtain their features from mt-.
Want to see more full solutions like this?
Chapter 9 Solutions
EBK CONCEPTS OF GENETICS
- Time mapping is performed in a cross involving the genes his,leu, mal, and xyl. The recipient cells are auxotrophic for all fourgenes. After 25 minutes, mating is interrupted, with the resultsin recipient cells shown below. Diagram the positions of thesegenes relative to the origin (O) of the F factor and to one another.(a) 90% are xyl+(b) 80% are mal+(c) 20% are his+(d) None are leu+arrow_forwardA 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_forwardIn com, male sterility is controlled by maternal cytoplasmic elements. However, the presence of a nuclear fertility restorer gene (F_) restores fertility to male sterile lines. a. What are the crosses male sterile female x FF male? Give the genotypes and phenotypes of the offspring in each cross. Explain.arrow_forward
- The following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.arrow_forwardFrom 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.arrow_forwardNeurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe listarrow_forward
- n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines sing the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of corn plants (i.e. the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male-fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Using the following color-coded circles, simulate the crosses indicated below. Put the illustrations of crosses in the spaces provided. Be sure to include in the labels the genotypes and phenotypes of the offspring in each cross. Big light green circle - male-sterile cytoplasm Big orange circle - male-fertile cytoplasm Small orange circle - FF nucleus Small half-light green-half-orange circle - Ff nucleus Small light-green circle - ff nucleusarrow_forward
- For 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_forwardAn individual is heterozygous for a reciprocal translocation, with the following chromosomes: A • B C D E F A • B C V W X R ST • U D E F R ST • U V W X Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.arrow_forwardAn ebony strain of flies was discovered to be sensitive to carbon dioxide. Crossing a female sensitive strain with male resistant strain gave all sensitive offspring. The offspring of an F1 female crossed with a resistant male were all sensitive. Cytoplasmic extracts from sensitive flies, when injected to resistant strains, can cause the latter to develop sensitivity to carbon dioxide. Provide an explanation for this observation.arrow_forward
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education