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In this question, you need to discuss the results you obtained from the BSA standards to generate your standard curve.
Are you confident in the absorbance values you measured from the BSA standards? Was there variability between your duplicates? Comment on the shape of your standard curve and note the highest absorbance you would consider for interpolating the concentration of an unknown sample.
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- True of False 9) Gels are read from left to right. Answer: 10) Functional segments of genes (introns) code for proteins, molecules that carry out most cellular functions. Answer: 11) The human genome contains small, repetitive DNA elements that have become randomly inserted into it over millions of years. Answer:18:20 O O OO0 • 4G+ * ll 7% Restriction enzyme tutorial 2021_1_ - Saved 1. Calculate the size of the resulting fragments as they will occur after digestion and write the sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (see figure 3A). %3D B BamHI 5305 22,346 27,972 34499 41,732 op) ECORI, 21,226 26,104 31,747 39,168 44,972 (bp) D Hindll 23,130 27,479 36,895 37,584 44141 25,157 (bp) 37,459 2. How many fragments would you expect to see for each of the maps indicated in question 1 above? Мар В: Мар С: Мар D: 3. Use the sequence indicated in figure 1 (also refer to figure 2 and 3) and indicate how many times does the sequence GAATTC occur in the A DNA sequence? What about AAGCTT and GGATCC? GGATC: GAATTC: АAGCTT: B IU e A = E12. When a person is determining the concentration of a sample, they will frequently repeat the samemeasurement of a sample by preparing three or more identical tubes (i.e. make replicates).a) Although you did not set up replicates today, what would be the purpose of setting upreplicates?b) If you had tested three replicates of test tube #2 in the determination of glucoseconcentration, what would you do with the data from the three replicates?
- Complete the following statement about conjugation. Bacterial confirmation is the equivalent of sexual reproduction in eukaryotes in that it necessitates cell to cell contact. The donor cell just contain a 1. Blank which is distinct from the larger bacterial chromosome. The 2. Blank carry genes in the 3. Blank which regular the conjugation processQUESTION 13 A 2000 nt plasmid with 10 negative supercoils has L= 210 200 190 400A media contains 2 E. coli cells which have a generation time and a replication time of 60 and 40 mins respectively. How many cells will be produced after 5 hours?
- We need to prepare a stock solution of medium for your culture cells, which usually includes liquid salt solution and bovine serum. Our liquid salt solution is supplied in a 50X concentration, and we need to dilute it to 1X for use. We also need to add 75% fetal bovine serum for a final concentration of 15%. How would we make up 0.80 liters of this culture media using water as our solvent?Arial BIUA 11 + .. | I 1 I 3 I 4 i.) Fill in the table for each of the E. coli: (0) = No Activity (+) = Basal Activity and (+++) = High Activity E. coli chromosome F' Plasmid B-gal activity? Permease activity? When When When When Glucose is Lactose is Glucose is Lactose is present present present present a.) I+ P+ O+ Z+ Y+ Inone +++ +++ b.) I^[S] P+ O+ Z+ Y+ none c.) I+ P+ O^[c] Z+ Y+ none d.) I+ P+ O- Z- Y+ none e.) I+ P+ O+ Z+ Y+ I^[S] P+ O+ Z+ Y+ f.) I^[S] P+ O+ Z- Y+ I+ P+ O^[c] Z+ Y- g.) I^[TB] P+ O+ Z+ Y 1+ P+ O^[c] Z- Y+ h.) I+ P+ O^[c] Z+ Y- I+ P+ O+ Z** Y+ i.) I^[TB] P+ O^[c] Z+ Y- 1+ P+ O+ Z- Y+ Z** is a polar mutation ii. ) If the lac operon in 'a' carried a mutation in the CAP binding site that rendered it nonfunctional, how would that affect the level of ß-galactosidase protein activity with and without lactose present, why? MacBook Air 000Following is the data and notice that it is a terrible idea to culture hMSCs longer than 10 days. You’re strongly Days # cells0 50001 75002 125003 125004 218005 287006 530007 1143008 1653009 19200010 19200011 11680012 8950013 8830014 78300 Part1 You are working for a start-up that is pursuing a clinical trial. The trial involves grafting hMSCs intopatients suffering from interveterbral disc disease using a degradable polymer scaffold. You are going to 3Dprint a porous cylindrical scaffold that is 2 cm in radius and 1 cm in height (matching the dimensions of adegenerated disc). Assume a porosity of 50%. You will fill available volume of the scaffold with hMSCs at adensity of 1 million cells per cm3. Based on the data above, what starting number of cells will you use andhow long will it take you to get enough cells for the trial? Part2The trial is a failure (patients did not report any reduction in back pain). Your team wants to try againusing 85% hMSCs and 15% nucleus pulposus cells .…
- 5 X O https://student.masteryconnect.com/?iv it41199tzPsQ_21hXG6OXABtokesw7jBW 912 Approved Sites SCS 21-22 SCI BIO District CFA2 Perez Jimenez, Dorian 00 4 of 20 The diagram shows the structure of DNA with complementary base pairing between strands. What is the benefit of this complementary nature of DNA? O It helps in controlling the amount of protein produced by the cells. O It helps in storing all the information required for the cell to function efficiently. It helps in the synthesizing of two new identical DNA strands from each parent strand O I t helps in the unwinding of DNA allowing for the formation of chromosomes during mitosis Type here to searchSeveral cell culture lines of epithelial cells, called “Cell Line A, B, or C,” are incubated with 104 infectious particles of influenza virus, and viral titers in the culture media are measured 2 days later. Looking at the results of this experiment, it is apparent that the three lines do not all show the same response to the virus. To investigate these differences, mixing experiments are performed, where cells from two different cell lines are mixed together at a 1:1 ratio before the Influenza infection. Based on the results shown in the figure below, propose an explanation for these data.A B Which of the bacterial colonies shown (A or B) was nonlethal when injected in mice? А В