Bio 1A03 Post-lab 4 Assignment F2021: Gel electrophoresis PDF

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Practicing gel electrophoresis and analyzing the correlation of the AMY1A gene copy number on the amylase concentration in an individual....

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Post-lab 4 Assignment Due the day of your lab 5A at 6:00 am to the drop box on Avenue. 1) Include a screenshot of your results from the virtual electrophoresis activity (your actual results and the ideal results must be visible in your screenshot). (1 mark)

2) Why did the bands of DNA in lanes 2, 3, 4, and 5 migrate approximately the same distances when comparing Student A and Student B? How does this relate to the primer sets and DNA templates that were used during the PCR reaction? (2 marks) The distance a DNA band travels is typically due to its fragment size, therefore, the bands in 2, 3, 4, 5 would have DNA fragment sizes that are very similar or identical, since they travelled approximately the same distance. This also means that the DNA templates and primer sets that were used during the PCR reaction needed to be the same in each lane because if they were different, the fragment sizes of the DNA in Student A and Student B would’ve been different, which would’ve resulted in bands that weren’t the same distance apart. 3) What would need to have been done during the gel electrophoresis procedure in order to separate all of the 100 bp ladder bands in the Figure 4A-7 and have them appear as individual bands in the 1.5% agarose gel? (1 mark) To separate all the 100bp ladder bands and have them appear as individual bands

would require time. The bands would’ve needed to be left in the gel electrophoresis for a longer duration of time so that the DNA fragments could separate even more distinctly towards the positive electrode, depending on their size. If the gel electrophoresis is run until the last band is about to fall off, this would provide the most distinct bands, since it has allowed the fragments to separate for the longest period of time, without losing any of the bands. Adjusting the voltage of the apparatus could also separate the bands more distinctly since there is a stronger pull on the negatively charged DNA, which would pull fragments to the bottom faster. 4) Imagine that you repeat the DNA extraction, PCR, and gel electrophoresis experiment using Unknown Student A and B's DNA. After imaging the resulting gel, only the band representing the amylase gene fragment is present in Lane 2 of the gel in Figure 4A-7. You do not observe a band for the actin gene fragment in Lane 2, but you do observe an actin band in the samples in Lanes 3, 4, and 5. All four lanes 2-5 have the amylase band. What could possibly explain why you do not see a band for actin in lane 2? (1 mark) Since the experiment is being repeated and there are no modifications to the procedure, it would be assumed that the results would be similar to the first trial. Therefore, if the actin band is missing in Lane 2, there was an error during the procedure. For example, one possible error would be during the pipetting stage. If the pipette was accidentally pushed to the second stop when pipetting Lane 2, this would result in bubbles formed in the gel, which could’ve disrupted the sample, causing Actin DNA fragments to be lost. 5) What are the diploid amylase (AMY1A) gene copy numbers for Unknown Student A and Unknown Student B as calculated from your ImageJ and Excel analysis? Include the table from the excel file that you used to calculate their gene copy number. (2 marks) The diploid Amylase (AMY1A) Gene Copy number for Unknown Student A is 3 The diploid Amylase (AMY1A) Gene Copy number for Unknown Student B is 6

Student Unknown Student A Unknown Student B

Amylase Area Results

Actin Area Results

Normalized Amylase Data

7211.803

6256.761

20864.68196

11270.137

5065.518

32605.96888

Amylase: Actin Ratio 3.33474172 4 6.43684789 5

Number of diploid Amylase (AMY1A) Gene Copies

6) Attach the scatter plot figure showing the relationship between gene copy number and amylase enzyme concentration that you created using the data in Table 4A-3. Follow the formatting instructions given in this lab manual and in the Excel Tutorial #1 to create a publication quality figure. Don't forget to include a caption with your graph title, sample size, and correlation coefficient (r). (5 marks)

3 6

Amylase Enzyme Concentration (mg/mL)

3.3 3 2.7 2.4 2.1

1.8 1.5 1.2

0.9 0.6 0.3

0 0

1

2

3

4

5

6

7

8

9

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11

12

AMY1A Gene Copy Number

Scatter plot of Amylase Concentration vs AMY1A Gene Copy Number. Amylase concentrations (mg/mL) and AMY1A gene copy numbers of 21 unknown students was recorded to help determine the correlation between Amylase concentration and the AMY1A gene copy number. The correlation value was calculated to be 0.663821. 7) What can you determine from the r value about the relationship between gene copy number and amylase enzyme concentration? (1 mark) Based on the correlation value (r value) calculated from the scatter plot, it can be determined that the correlation between the AMY1A gene copy number and Amylase enzyme concentration is moderate. This is because the r value was found to be 0.663821 which is between 0.5 and 0.8. A moderate correlation value means that the value of one’s AMY1A gene copy number has a moderate effect on one’s value of amylase enzyme concentration.

8) Do Unknown Student A and Unknown Student B’s self-reported starch levels, AMY1A gene copy numbers, and salivary amylase concentrations support an association between gene evolution, gene copy number, and amylase enzyme production? Please use Unknown Student A and Unknown Student B’s data to explain and support your choice for the presence or absence of an association. (2 marks) Student A and Student B’s self-reported starch levels, AMY1A gene copy numbers, and salivary amylase concentrations do support an association between gene evolution, gene copy number and amylase enzyme production. According to the data, Student A only has 3 AMY1A gene copy numbers, while Student B has 6. This means that Student B has more gene copies, allowing Student B to produce more salivary amylase (about 0.591 mg/mL), while Student A will produce less than Student B (only about 1.947 mg/mL).

Inferring from the data, it can be concluded that Student B’s ancestral diet would be high in starch since the high starch diet would explain the higher amount of AMY1A gene copy numbers and higher concentration of salivary amylase. On the other hand, Student A’s ancestral diet would have a lower amount of starch which corresponds to the Student A’s data having lower concentrations of salivary amylase and lower AMY1A gene copy numbers. Overall, it can be concluded that students who have a high amount of starch in their ancestral diet will have a larger AMY1A gene copy number, and as a result, will lead to a higher production/concentration of salivary amylase. 9) Submit a labelled picture (similar to Figure 4A-7) of your restriction digest gel image from Lab 4B. Number the lanes on the gel and provide a brief description of what is present in each lane (similar to the descriptions in Table 4A-2) (5 marks).

Lane 1: Contains the 1 kb ladder, used to size the DNA fragments in other lanes Bp Values of the 1 kb ladder 10000 bp

8000 bp 6000 bp 5000 bp 4000 bp 3500 bp 3000 bp 2500 bp 2000 bp 1500 bp 1000 bp 750 bp 500 bp 250 bp Lane 2: Contains the control sample, which has no restriction enzymes, only the phage DNA. No restriction enzymes are present, therefore no cuts to the DNA have been made, making all the fragments the same size. Lane 3: DNA fragments of the phage DNA after being cut by the HindIII restriction enzyme Lane 4: DNA fragments of the phage DNA after being cut by the EcoRI restriction enzyme Lane 5: DNA fragments of the phage DNA after being cut by the BamHI restriction enzyme Lane 6: DNA fragments of the phage DNA after being cut by the BglII restriction enzyme Each band in lanes 3-6 represents a DNA fragment that’s been cut by the respective restriction enzymes into a different base pair size. The larger base pair length fragments stay closer to the top of the well while the smaller fragments move down the well, creating multiple bands. Some of the restriction enzymes created more bands of fragments, while others created less, which means that each restriction enzyme had different amounts of recognition sites.

10) Submit your correctly formatted 100 bp ladder standard curve that you created in Excel. Follow the instructions in the lab 4B manual and Excel Tutorial #2 to complete your publication quality figure. (5 marks)

100 bp Ladder Standard Curve 10000

DNA Fragment Size (bp)

y = 33917e-0.086x R² = 0.9977 1000

100

10

1 0

5

10

15

20

25

30

35

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45

50

55

60

65

70

Distance Migrated (mm)

Yellow: Unknown Student A Actin Gene (~300 bp) Green: Unknown Student A Amylase Gene (~105 bp) Red: Unknown Student B Actin Gene (~275 bp) Purple: Unknown Student B Amylase Gene (~130 bp) 11) What are the lengths of the DNA fragments, in base pairs, for the two bands that are seen in all of lanes 2 to 5 of Figure 4A-7? Show your calculations for both bands using the equation of the line generated from your 100bp standard curve. Which base pair value represents the amplified DNA that was targeted by the primers that were specific to a segment of an actin gene? Which base pair value represents the amplified DNA that was targeted by the primers that were specific to a segment of an amylase ( AMY1A) gene? (4 marks) Trendline Equation: y=33917e-0.086x Unknown Student A: Actin Gene (55 mm) y=33917e-0.086x y=33917e-0.086(55) y=299.37 bp Unknown Student A: Amylase (AMY1A) Gene (67 mm) y=33917e-0.086x y=33917e-0.086(67) y=106.67 bp Unknown Student B: Actin Gene (56 mm) y=33917e-0.086x

y=33917e-0.086(56) y=274.70 bp Unknown Student B: Amylase (AMY1A) Gene (65 mm) y=33917e-0.086x y=33917e-0.086(65) y=126.68 bp

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