The following data for the genotypes at the alcohol dehydrogenase locus wereobserved from a sample of Drosophila melanogaster.Sample size1000FF550FF=0.55; FS=0.34; SS=0.11O F-0.72; S=0.28FF=0.078; FS=0.40; SS=0.52FSWhat are the Hardy-Weinberg equilibrium genotype frequencies for each genotypebased upon this sample?OFF=0.52; FS=0.40; SS=0.078340SS110

The Hardy-Weinberg law states that when a population is in equilibrium, the genotypic frequencies in…

Answered: The following data for the genotypes at…

Biomedical Instrumentation Systems

1st Edition

ISBN:9781133478294

Author:Chatterjee

Publisher:Chatterjee

Chapter13: Artifacts And Noise In Medical Instrumentation

Section: Chapter Questions

Problem 7P

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Population genetics of Prairie blazingstar in Kansas A researcher surveys genetic variation in the Skdh (shikimate dehydrogenase) gene from a Kansas population of prairie blazingstar plants (Liatris pychnostachya). She obtains the data in the table below. There are two alleles, Fast (F) and Slow (S), named for their position in the electrophoresis gel. Values in the table are number of individuals with each genotype. FF 50 FS 200 SS 150 What is the frequency of the SS genotype in this population? (The observed genotype from the data.) Report your answer to the nearest 0.001. Type your answer.... - Using these data, calculate the frequency of the Fallele. Report your answer to the nearest 0.001. Type your answer... D -000 DO What is the expected genotype frequency for the FF genotype, assuming this population is in Hardy-Weinberg equilibrium? Report your answer to the nearest 0.001. Type your answer...
Assume that a series of compounds has been discovered in Neurospora. Compounds A–F appear to be intermediates in a biochemical pathway. Conversion of one intermediate to the next is controlled by enzymes that are encoded by genes. Several mutations in these genes have been identified and Neurospora strains 1–4 each contain a single mutation. Strains 1–4 are grown on minimal media supplemented with one of the compounds A–F. The ability of each strain to grow when supplemented with different compounds is shown in the table (+ = growth; o = no growth). Which biochemical pathway fits the data presented? Media Supplement Strain A B C D E F 1 o o o + + + 2 o o o o + + 3 o o o o + o 4 o o + + + + A) A → B → C → D → E → F B) A → B → C → F → D → E C) F → B → C → D → A → E D) A → B → C → D → F → E E) A → B → F → E → C → D
You are using nitrosoguanidine to “revert” mutant nic-2(nicotinamide-requiring) alleles in Neurospora.You treat cells, plate them on a medium without nicotinamide, and look for prototrophic colonies. You obtainthe following results for two mutant alleles. Explainthese results at the molecular level, and indicate how youwould test your hypotheses.a. With nic-2 allele 1, you obtain no prototrophs at all.b. With nic-2 allele 2, you obtain three prototrophiccolonies A, B, and C, and you cross each separately with awild-type strain. From the cross prototroph A × wildtype, you obtain 100 progeny, all of which are prototrophic.From the cross prototroph B × wild type, you obtain100 progeny, of which 78 are prototrophic and 22 arenicotinamide requiring. From the cross prototrophC × wild type, you obtain 1000 progeny, of which 996 areprototrophic and 4 are nicotinamide requiring.
A classic way to isolate thymidylate synthase–negative mutants of bacteriais to treat a growing culture with thymidine and trimethoprim. Most ofthe cells are killed, and the survivors are greatly enriched in thymidylatesynthase–negative mutants.(a) What phenotype would allow you to identify these mutants?(b) What is the biochemical rationale for the selection? (That is, why are themutants not killed under these conditions?)(c) How would the procedure need to be modified to select mammalian cellmutants defective in thymidylate synthase?
The major phenotype of deficiency of protolithic enzyme alpha-antitrypsin is the destruction of pulmonary alveoli resulting in chronic obstructive pulmonary disease or emphysema. The gene for this enzyme is highly polymorphic, with more than 70 different alleles described. 90% of white Europeans have MM genotype (normal). Two mutant alleles, S and Z, account for most of the disease associated with alpha-antitrypsin deficiency. The flowing table tabulates frequencies of different clinically important genotypes in the French population. Genotype Frequency MM 0.90 MZ 0.04 SS 0.001 SZ 0.0012 ZZ 0.0004 Using the data presented, calculate the allele frequencies for Z and S.
The Saccharomyces cerevisiae nuclear gene ARG8encodes an enzyme that catalyzes a key step in biosynthesis of the amino acid arginine. This protein isnormally synthesized on cytoplasmic ribosomes, butthen is transported into mitochondria, where the enzyme conducts its functions. In 1996, T. D. Fox andhis colleagues constructed a strain of yeast in which agene encoding the Arg8 protein was itself moved intomitochondria, where functional protein could besynthesized on mitochondrial ribosomes.a. How could these investigators move the ARG8gene from the nucleus into the mitochondria, whilepermitting the synthesis of active enzyme? In whatways would the investigators need to alter theARG8 gene to allow it to function in the mitochondria instead of in the nucleus?b. Why might these researchers have wished to movethe ARG8 gene into mitochondria in the firstplace?
The amino acid asparagine is shown below in its zwitterionic form. It possesses two Ka values: Ka1 = 9.5 x 10-³ and Ka2 = 1.6 x 10-⁹. ne NH3 asparagine (a) Sketch the pH speciation diagram for asparagine for the range of 1 < pH < 12, annotate with the position of the two pKa values and label the appropriate curves with the chemical structures of the three forms of asparagine in the pH range stated. Fraction 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 H₂N. 2 3 4 5 6 pH 7 00 8 9 10 11 12
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. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…
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? After crossing the F1 generation of the cross between mutant strains 1…
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?
The pathway for arginine biosynthesis in Neurospora crassa involves several enzymes that produce a series of intermediates as shown. O O O O ornithine citrulline ARG-E arginosuccinate arginine N-acetylornithine arginine You did a cross between ARG-E ARG-H* and ARG-E* ARG-H¯¯ Neurospora strains and identified an Arg- strain from an NPD tetrad. (Assume that Neurospora forms tetrads in the same way yeast do.) Which compound would rescue growth of this Arg- spore? N-aceltylornithine ARG-F ornithine citrulline ARG-G ARG-H → argininosuccinate
Considerable effort has been directed toward determining the genes in which sequence variation contributes to the development of type 2 diabetes. Approximately 800 genes have been implicated. Propose an explanation for this observation.

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