DNA containing the genes gin+ phe+ lys+ was used to transform bacterial cells of genotype gln- phe- lys-. The following cell types were obtained with the number of each cell type shown below: Genotype Number Igin+phe+lys+ 25 gln+|phe-lys- gin phe-lys+ 4. gin-phe+ lys- 14 gin-phe+lys+ gin-phe iys+ gin-phe-tys- 19 Which gene is in the center? Which two genes are the closest together? 2. 00

DNA containing the genes gin+ phe+ lys+ was used to transform bacterial cells of genotype gln- phe- lys-. The following cell types were obtained with the number of each cell type shown below: Genotype Number Igin+phe+lys+ 25 gln+|phe-lys- gin phe-lys+ 4. gin-phe+ lys- 14 gin-phe+lys+ gin-phe iys+ gin-phe-tys- 19 Which gene is in the center? Which two genes are the closest together? 2. 00

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Linkage maps in an Hfr bacterial strain are calculated in units of minutes (the number of minutes between genes indicates the length of time that it takes for the second gene to follow the first in conjugation). In making such maps, microbial geneticists assume that the bacterial chromosome is transferred from Hfr to F − at a constant rate. Thus, two genes separated by 10 minutes near the origin end are assumed to be the same physical distance apart as two genes separated by 10 minutes near the F −attachment end. Suggest a critical experiment to test the validity of this assumption.
By conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the table:Draw a map that shows the order of genes and the locations ofthe origins of transfer among these different Hfr strains?
You are using the restriction enzyme HAEIII to digest different samples of the taster gene isolated from cheek cells of different people and amplified by PCR. When viewing the bands on the electrophoresis gel, one would expect that a taster (homozygote) would have---------band(s), whereas a carrier (heterozygote) would show--------band(s), and a non-taster would show------band(s).
DNA from a strain of Bacillus subtilis with the genotype trp" tyr* was used to transform a recipient strain with the genotype trp tyr". The following numbers of transformed cells were recovered: Number of transformed Genotype cells trp* tyr 154 trp tyr* trp" tyr* 312 354 What do these results suggest about the linkage of the trp and tyr genes?
DNA 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 co-transformation and results shown below are obtained. Based on these results, what is the order of the genes on the bacterial chromosome? please show how you worked out the order by diagraming, thank you! Pair of genes Co-transformation a+ and b+ no a+ and c+ no a+ and d+ yes a+ and e+ yes b+ and c+ yes b+ and d+ no b+ and e+ no c+ and d+ no c+ and e+ yes d+ and e+ no
13) What is the purpose of the negative selectable marker in a mouse knock out cassette? a) Why don't we need negative selectable marker on a yeast knock out cassette? b) Draw the target chromosome in the mouse if illegitimate recombination has occurred with the knock out cassette.
In an analysis of five rII mutants, complementation testing yielded the following results:
In E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What information do you know based onthe question and your understanding of the topic?
What is a contig?
In a generalized-transduction experiment, phages arecollected from an E. coli donor strain of genotype cys+leu+ thr+ and used to transduce a recipient of genotypecys- leu- thr-. Initially, the treated recipient populationis plated on a minimal medium supplemented with leucine and threonine. Many colonies are obtained.a. What are the possible genotypes of these colonies?b. These colonies are then replica plated onto threedifferent media: (1) minimal plus threonine only, (2)minimal plus leucine only, and (3) minimal. Whatgenotypes could, in theory, grow on these three media?c. Of the original colonies, 56 percent are observed togrow on medium 1, 5 percent on medium 2, and nocolonies on medium 3. What are the actual genotypes ofthe colonies on media 1, 2, and 3?d. Draw a map showing the order of the three genes andwhich of the two outer genes is closer to the middle gene
Knock out mice that are mutant for a gene product X die. Variants of the X gene that were reintroduced through homologous recombination indicate that variant proteins that lack the C-terminal region of the protein cannot rescue the lethality (did not allow them to live), and always localized to the cytoplasm. Those that retained the C-terminal region rescued the lethality (conferred viability!) and were consistently localized within the nucleus. RNA-seq analysis of the mutant cells vs the wild-type cells indicated that the expression of many genes that are essential for neural function was reduced in the knock-out mutant cells. The N-terminal region of Protein X is 100% conserved between mouse and humans at both the amino acid and the nucleotide level. The predicted mouse mRNA sequence is shown below where the AUG corresponds to the translational start site (AUG). 5’-AUGUUUACAGAGGGGAAU... -3’ d) What motif could be present to direct this protein to its correct destination? e)…
Knock out mice that are mutant for a gene product X die. Variants of the X gene that were reintroduced through homologous recombination indicate that variant proteins that lack the C-terminal region of the protein cannot rescue the lethality (did not allow them to live), and always localized to the cytoplasm. Those that retained the C-terminal region rescued the lethality (conferred viability!) and were consistently localized within the nucleus. RNA-seq analysis of the mutant cells vs the wild-type cells indicated that the expression of many genes that are essential for neural function was reduced in the knock-out mutant cells. The N-terminal region of Protein X is 100% conserved between mouse and humans at both the amino acid and the nucleotide level. The predicted mouse mRNA sequence is shown below where the AUG corresponds to the translational start site (AUG). 5’-AUGUUUACAGAGGGGAAU... -3’ Q3a) Using a total RNA sample obtained from human cells, what is the first primer you would…