Report the slope and the y-intercept below. Slope () = = y-intercept () =Can this standard curve be used for E.coli cells (why/why not?) 0.80.70.60.50.40.30.20.100200000004000000060000000Concentrationy = -4E-09x + 0.4021R² = 0.169180000000100000000C

In the given graph, an equation is provided. y=-4E-09x + 0.4021R2= 0.1691 The graph represents the…

Answered: Report the slope and the y-intercept…

Report the slope and the y-intercept below. Slope () = = y-intercept () = Can this standard curve be used for E.coli cells (why/why not?)

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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Five different HFR strains were isolated from a single F+ strain of E. coli. Timed matings showed that they transferred the following loci during the first hour of mating. The earliest loci are on the left, the slowest on the right for each order. Map these loci to a circle. strain early →→late 1. QZ ARL What markers are between B and X (shortest route)? SNWT O EMQ 3. 5. RBLE XAR BEMQZ EBLRA ZSNWT strain early →late 2. 4. TXARL XTWNS
Both the isolation buffer and assay buffer that was used in isolation of mitochondria contained 0.3M mannitol. What was the purpose of including mannitol at this concentration?
Using the Dynamic Programming algorithm for pairwise local alignment we covered in class, construct the dynamic programming score table for a local alignment of the following two sequences, using the following scoring parameters: match score = +5, mismatch score = -3, gap penalty = -2.: ACGTATCGCGTATA GATGCTCTCGGAAAWhat is score of the best local alignment between these two sequences? Show the alignment of these sequences. asap
At higher amounts of protein, the Bradford assay is not linear. Consider the plot to the right: what is the maximum amount of protein a sample could contain and still fall within a standard curve? Briefly explain your reasoning.
for this question please help find pairwise differences and segregating sites
The following chromatogram is obtained from the sequencing reaction. Using the key below, what message did you integrate into the yeast genome? AATATTAAAGTATAGAACGACTGTGAA A B с D E F G H I AAA AAT AAG AAC ATT ATG ATC AGT AGG J K L M N O P Q R Key GGG GGA GGT GGC GAA GAC GAT GTT GTA S T U V W X Y Z (SPACE) TTT TTC TTA TTG TGT TGA TGG TAA TAG
Answer part C of the following: A) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? See graph B. B) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC and allowed them to reanneal, you would find that approximately 15% of the…
Bacterial mutants can be identified in screens or selections.In a screen, individual colonies are tested for a particularphenotype. In a selection, only bacteria with the phenotypein question are recovered as colonies, enabling theidentification of extremely rare _________.
Select all that apply from the following to each transformation process : The following are White non-glowing colonies White glowing colonies Blue non-glowing colonies Blue glowing colonies None of the above The transformation are: a. LB + kanamycin plates that have been spread with the E. coli cells transformed with your ligations, what phenotype of the colonies do you expect to obtain . b. LB + kanamycin plates that have been spread with the E. coli cells transformed with water, what phenotype of the colonies do you expect to obtain. c. LB plates that have been spread with the E. coli cells transformed with your pHSG298, what phenotype of the colonies do you expect to obtain. d. LB + kanamycin plates that have been spread with the E. coli cells transformed with PHSG298, what phenotype of the colonies do you expect to obtain. e. LB plates that have been spread with the E. coli cells transformed with your ligations, what phenotype of the colonies do you expect to obtain. f. LB plates…
Second, Question 2
You have four strains, and you test for growth on the following medias: Rich, Minimal, Minimal + Lysine, Minimal + Histidine, and Minimal +Lysine and Histidine. The table below shows your results. Strain Rich 1 2 3 4 Minimal Min + Lysine Strain 2 Growth No Growth Growth Growth No Growth No Growth Strain 4 Growth Growth Strain 1 Lysine auxotroph only Growth Histidine auxotroph only Strain 3 is not an auxotroph, but a pro Min + Histidine No Growth Is an auxotroph, but not for ly ✓ No Growth Growth Determine whether each strain is auxotroph or a prototroph, and if an auxotroph what sort of auxotroph? Min + Lysine and Histidine Growth No Growth No Growth No Growth No Growth Growth Growth Growth
Energy-carrying compound produced during the Krebs cycle. (a) Oxygen (b) NADH+ (c) ETC (d) ATP Synthase