Concept explainers
a.
To determine:
The genotypes of the spores within the two asci groups.
Introduction:
A cross performed in between yeast strain requiring methionine as well as lysine for growth, which is met-lys- and the strain that was met+lys+.
b.
To determine:
Whether the lys and met genes are linked and the map distance between these genes.
Introduction:
A cross between two yeast strains can be depicted as:
c.
To determine:
The phenotype of some discovered asci that have a different pattern and list the
Introduction:
The building blocks of the proteins are amino acids. Lysine and methionine come under the category of amino acids.
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Genetics: From Genes to Genomes
- Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3?arrow_forwardTwo auxotrophic triple Escherichia coli strains (A: met- phe- ade- val+ bio+ thr+ and B: met+ phe+ ade+ val- bio- thr-) are mixed in LB liquid medium, diluted and then spread on LB solid rich medium. Six colonies are observed: Then, replicates are performed on 6 different media (minimum medium + glucose + indicated substances). The results are shown below. Determine the genotype of the 6 colonies observed. Which ones are from strain A? From strain B? Which hypotheses can explain these results and which one do you prefer? met phe val bio Abbreviations: met ade val thr phe ade bio thr Met: methionine; Phe: phenylalanine; Ade: adenine; Val: valine; Bio: biotin; Thr: threonine.arrow_forwardAn Hfr strain that is leuA+ and thiL+ was mixed with a strain thatis leuA− and thiL−. In the data points shown in the following graph,the conjugation was interrupted at different time points, and thepercentage of recombinants for each gene was determined bystreaking on a medium that lacked either leucine or thiamine.What information do you know based on the question and your understanding of the topic?arrow_forward
- An Hfr strain that is leuA+ and thiL+ was mixed with a strain thatis leuA− and thiL−. In the data points shown in the following graph,the conjugation was interrupted at different time points, and thepercentage of recombinants for each gene was determined bystreaking on a medium that lacked either leucine or thiamine.Make a calculation.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_forwardDNA 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?arrow_forward
- 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_forwardIn 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+?arrow_forwardA fluctuation test was carried out to determine the rate of mutation to Azetidine resistance (a toxic proline analog) in S. typhimurium. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures were grown to 10° cells / ml. A 0.1 ml sample of each culture was then plated on each plate (a total of 20 plates) to detect AztR mutants. The results are shown in the following table. Calculate the mutation rate of S. typhimurium to AztR. Culture # # AztR mutants Culture # # AztR mutants 11 12 3 4. 13 14 15 4. 30 303 97 69 14 16 17 18 19 20 10 19arrow_forward
- The following two strains of E. coli are crossed with each other: Hfr pan* thi* ala* and F¯pan¯ thi¯ ala¯ It was shown that the pan marker entered last in interrupted conjugation experiments. By spreading the bacteria from these experiments on different selection media, the following results (in number of colonies) were obtained: MM + Gluc IM + Gluc + ala MM + Gluc + thi MM + Gluc + thi + ala 280 281 286 339 Give the number of each phenotypic class and justify your answer. Determine the order of the genes. What are the genetic distances between the different loci? Abbreviations: Gluc: glucose; Ala: alanine; Thi: thiamine; Pan: pantothenic acid.arrow_forwardWild-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?arrow_forwardThe figure below shows the life cycle of the fungus Neurospora. The adult stage of the Neurospora is a multicellular haploid. b) Neurospora has an arginine amino acid synthesis pathway shown below. Suppose I take the strain above that only grows with arginine supplements and cross it to a different mutant Neurospora strain that grows with arginine and citrulline supplements but not with ornithine supplements. Assuming gens A, B, and C are unlinked and there is only one mutation per stain: What percentage of the progeny will grow on ornithine? What percentage on citrulline? What percentage on arginine?arrow_forward
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