Concept explainers
a.
To determine:
The
Introduction:
The bread mold or the Saccharomyces cerevisiae is extensively studied in the field of genetics. The fusion of the gametes gives rise to the spores that are in a specific structure called ascus. The spores formed are eight in number and follow
b.
To determine:
The difference between the uniparental and biparental inheritance as seen in the bread mold and the validation of the paradoxical statement.
Introduction:
Microorganisms are small microscopic organisms that are present almost everywhere. They inhabit nearly all the areas from the air, water, soil, hot springs, snow, and so on. The microbes can be infectious or non-infectious. Microorganisms can be unicellular (single-celled) or multicellular.
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ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
- A mutant haploid strain of Saccharomyces cerevisiae(yeast) called cox2-1 was found that was unable togrow on media containing glycerol as the sole sourceof carbon and energy. (Glycerol is a nonfermentablesubstrate for yeast.) This strain could, however, growon the fermentable substrate glucose. Researchers discovered that cox2-1 cells lack a mitochondrial proteincalled cytochrome c oxidase.a. Explain why cox2-1 cells can grow on mediumcontaining glucose but not on glycerol medium.arrow_forwardA mutant haploid strain of Saccharomyces cerevisiae(yeast) called cox2-1 was found that was unable togrow on media containing glycerol as the sole sourceof carbon and energy. (Glycerol is a nonfermentablesubstrate for yeast.) This strain could, however, growon the fermentable substrate glucose. Researchers discovered that cox2-1 cells lack a mitochondrial proteincalled cytochrome c oxidase.a. Explain why cox2-1 cells can grow on mediumcontaining glucose but not on glycerol medium.b. When cox2-1 was crossed with a wild-type yeaststrain and the resultant diploid cells were allowedto grow mitotically, it was found that about half thediploid clones were able to grow on glycerol, whilethe other half could not. The diploid clones thatcould grow on glycerol were induced to sporulate,and they yielded tetrads with four spores that wereall able to grow on glycerol medium. In all of thesetetrads, two of the haploid progeny were of matingtype a and two of mating type α. The diploids thatcould not…arrow_forwardThe intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.arrow_forward
- Yeast cells are eukaryotes, and they require a functional secretory pathway to grow and to maintain cellular organization. But surprisingly, when SRP is removed from yeast cells by deleting the relevant genes, the cells are still alive (although they grow slowly). a) How can yeast survive without SRP? Propose two alternative hypotheses. b) How might electron microscopy of normal and SRP-deficient yeast cells help you to distinguish between these two hypotheses?arrow_forwardResearchers isolated a yeast phosphofructokinase (PFK) mutant in which a serine at the fructose-2,6-bisphosphate binding site was replaced with an aspartate residue. The amino acid substitution completely abolished the binding of fructose-2,6-bisphosphate to PFK. There was a dramatic decline in glucose consumption and ethanol production in the mutant compared to control yeast. Why can't the mutant PFK bind with fructose-2,6-bisphosphate?arrow_forwardwhat is the nature and likely location(s) of a mutant that would, 1)allow constitutive expression of the lac gene? 2)prevent the cell from responding to lactose ( genes are not induced when exposed to lactose)? 3) not allow the cell to utilize lactose even when the genes are inducedarrow_forward
- The synthesis of flower pigments is known to be dependent on enzymatically controlled biosynthetic pathways. For the crosses shown here, postulate the role of mutant genes and their products in producing the observed phenotypes: (a) P1: white strain A * white strain B F1: all purple F2: 9/16 purple: 7/16 white (b) P1: white * pink F1: all purple F2: 9/16 purple: 3/16 pink: 4/16 whitearrow_forwardThere are three different chitin synthase genes that control the chitin synthesis in Saccharomyces cerevisiae, namely CHS I, CHSII, and CHSIII. If a mutation occurs in each of these genes, what will happen to the S. cerevisiae?arrow_forwardA mutation creates a dominant negative allele of a particular gene. The gene encodes a protein that forms a trimer within the cell. If one or more of the subunits has the mutant structure, the entire trimeric protein is inactive. In a heterozygous cell, if the proteins of both alleles are present at the same levels, what percent of the trimers present in the cell will be active? A) 100% B) 5% C) 50% D) 33% E) 5%arrow_forward
- Clary Fray used the pET vector system to express her prokaryotic amylase enzyme. She added peptone into her culture broth of BL21(DE3) Escherichia coli strain to induce protein expression. At the end of the experiment, she discovered that her protein was not expressed. She repeated three more times but her protein of interest was still not produced. (i) (ii) (iii) (iv) (v) Explain the reason why Clary failed to obtain her protein of interest and suggest a solution to troubleshoot this problem. Clary plans to express her protein along with a polyhistidine-tag. Explain the importance of His-tag in protein work. Is DH5a Escherichia coli suitable to propagate the plasmid before protein expression? Besides heat shock method, elaborate another procedure Clary could utilize to transform the recombinant pET vector into the host cell. If her supervisor instructs her to express a gene from gold fish (Carassius auratus), is the expression system she is using now suitable for this experiment?…arrow_forwardLeber's Hereditary Optic Neuropathy (LHON) is a disease that involves degeneration of neural cells in the retina and results in loss of central vision. The disease is caused by mutations in any one of three genes in the mitochondrial genome that encode proteins involved in oxidative phosphorylation. In a genetic counseling clinic, a woman and her husband seek advice on the potential that any of their children would be afflicted with LHON. The husband's mother and father, both exhibit symptoms of the disease, but the woman does not. What is a reasonable advising statement to make? There is not enough information to advise this couple. a. The couple should be advised that all of their children are likely to display symptoms of LHON. b. The couple should be advised that all their female children will display symptoms of LHON, but male children will be unaffected. C. The couple should be advised that all their male children will display symptoms of LHON, but female children will be…arrow_forwardLeber’s Hereditary Optic Neuropathy (LHON) is a disease that involves degeneration of neural cells in the retina and results in loss of central vision. The disease is caused by mutations in any one of three genes in the mitochondrial genome that encode proteins involved in oxidative phosphorylation. In a genetic counseling clinic, a woman and her husband seek advice on the potential that any of their children would be afflicted with LHON. The husband's mother and father, both exhibit symptoms of the disease, but the woman does not. What is a reasonable advising statement to make? a. The couple should be advised that all of their children are likely to display symptoms of LHON. b. There is not enough information to advise this couple. c. The couple should be advised that none of their children will be affected. d. The couple should be advised that all their male children will display symptoms of LHON, but female children will be unaffected. e. The couple should be…arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning