Hardy Weinberg lab report PDF

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Introduction: Population genetics is the study of the frequency of alleles within a population of the same species. All the people within the group share the same gene pool and are able to interbreed amongst one another. One main principle of principle genetics is the Hardy Weinberg Law. This law describes how the genetic frequencies of a population are affected by reproduction and Mendelian principles. If a population meets the Hardy Weinberg Equilibrium (HWE), then the allele and genotype frequencies will remain constant from generation to generation. In order for that to occur, “the population must be large, randomly mating, and not affected by mutation, migration, or natural selection”. However, this is rare in present day due to the constant changes in the environment and the mutations present throughout generations. The Hardy Weinberg Equilibrium is used to calculate the allele frequencies in a population using p as the frequency for the dominant allele and q as the frequency for the recessive allele. Allele frequency is the frequency of the alleles of a particular gene within the population. Genotypic frequency is the frequency of the genotypes within the population. In order for the population to be in equilibrium, p + q must equal 1. The same rule applies for genotypic frequency, which is calculated using the equation: p² + 2pq + q² = 1. The frequency of individuals with genotype AA is represented by p², individuals with genotype Aa is represented by 2pq, and individuals with the genotype aa is represented by q². The allele frequencies of the population is utilized to calculate the frequencies of genotypes within the population. With all this data, it can be further used in the Chi-Square test to validate whether the population is or is not in Hardy Weinberg Equilibrium (HWE). To test out HWE, we will use the TAS2R38 gene from our DNA in our cheek cells and amplify it using the PCR method. PCR is a tool used to amplify DNA and make billions of copies from a segment of DNA. After it gets amplified, the PCR product will get digested with the restriction enzyme

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HaeIII, which includes a SNP (single-nucleotide polymorphism) and will cut one allele and not cut the other to produce a RFLP (Restriction Fragment Length Polymorphism). The RFLP can be easily separated on a 2% agarose gel and will help us analyze which individuals in the class have the dominant gene and which have the recessive gene. The observed values from the RFLP test and the expected values from the hardy weinberg equilibrium will be utilized to complete the Chi-square test. The value of the Chi-square test will determine if the class population is or is not in Hardy Weinberg Equilibrium. The purpose of the first half of the experiment is to perform PCR on our cheek cell DNA and digest it with the restriction enzyme HaeIII to produce RFLPs. The purpose of the second half of the experiment is to analyze the RFLPs by separating it onto agarose gel and identifying the genotypes. Then to take the info to generate T & t allele frequencies within the class gene pool and use that to conduct the chi-square test to test whether the class population is in Hardy-Weinberg equilibrium. Materials used in Week 1: ● Saline mouthwash with a spitting cup ● Eppendorf tube ● Centrifuge ● P200 Micropipetter ● Waterbath ● PCR reaction tube ● 5 uL of our cheek cell DNA ● 5 uL 10x Buffer ● 5 uL of 2 mM dNTPs ● 1 uL RFLP Primer 1

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● 1 uL RFLP Primer 2 ● 2 uL Taq polymerase ● 31 uL H2O ● 10% chelex Materials used in Week 2: ● Digested DNA sample from week 1 ● 2% agarose gel ● Loading dye ● TAE (tris-acetate-EDTA) gel running buffer ● Ethidium Bromide (CAUTION: Carcinogen) ● UV gel box (CAUTION: causes skin damage) ● Gel running apparatus Procedure for Week 1: 1. Rinse your mouth with the saline mouthwash for 30 seconds. This will dislodge the epithelial cells from the mouth. 2. Spit into the cup to allow easy removal. 3. Pipette 1.5 mL of the cheek cells from the cup into the labeled eppendorf tube. 4. Bring the tube to the centrifuge and spin for 2 minutes at max speed. 5. Retain the pellet of cells by discarding 100 uL of supernatant. a. 0.1 mL should remain in the tube which contains cells. 6. Resuspend the cells by pipetting up and down with a p200 pipetman set to 80 uL. 7. Transfer 80 uL of resuspended cells to the tube with 10% chelex (100 uL). 8. Bring the tube to the water bath and boil for 10 minutes.

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9. Vigorously shake the tube for 5 seconds after boiling. 10. Bring the tube to the centrifuge to spin for 2 minutes at max speed. 11. Carefully transfer only the supernatant to the new labelled tube. DNA will be within the supernatant. 12. Bring the tube to the instructor’s table and transfer 5 uL of DNA to PCR reaction tube. Make sure to record the number of reaction tube. PCR Program done by the lab professor: 1. Make the PCR reaction tube consisting of: 5 uL of our cheek cell DNA, 5 uL of 10X buffer, 5 uL of 2mM each dNTPs, 1 uL RFLP primer 1, 1 uL RFLP primer 2, 2 uL Taq, and 31uL of H2O. The total solution should add up to 50 uL. 2. The PCR program consists of 6 steps: a. Denature the tube at 95* C for 5 minutes. b. Denature the tube at 94* C for 30 seconds. c. Anneal the tube at 65* C for 30 seconds. d. Extend at 72*C for 40 seconds. i.

Do Steps B, C, & D for 30 cycles.

e. Extend at 72*C for 7 minutes. f. Hold at 4*C, indefinitely. 3. Restriction digest will occur next: a. Add 1 uL of HaeIII restriction enzyme to each PCR reaction. b. Place the reaction tube in a 37*C incubator for 2 hours to ensure the DNA is fully digested. Procedure for Week 2:

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1. Obtain the digested DNA sample from week 1. 2. Add the TAE gel running buffer into the gel box and stop once the gel is fully covered. 3. Load 2% agarose gel with all the DNA samples mixed with loading dye in this order: 1. DNA ladder 2. Negative Control (uncut): mix one person’s uncut DNA with 6x loading dye 3. Positive Control (TT-cut) 4. DNA of student 1 (our group member was #1) 5. DNA of student 2 (our group member was #2) 6. DNA of student 3 (our group member was #5) 7. DNA of student 4 (our group member was #6) [THIS IS MY DNA] 8. DNA ladder 4. Load 20 uL of each part from #2 into all the wells. 5. Place a protective top on the gel box and connect the box to the power supply. 6. Start the gel to run at 20V for 5 minutes. After 5 minutes, turn the power up to 100V and run until the lowest dye is ¾ down the gel. 7. While the gel electrophoresis is running its course, conduct the ptc paper test by placing one paper strip into your mouth to see whether there is a bitter taste or no taste. 8. Bring the gel to the ethidium bromide station at the back of the lab room a. CAUTION: Ethidium Bromide is a known carcinogen. 9. Gel will be incubated in ethidium bromide for 20 minutes. 10. Once the incubation is complete, then view the gel on the UV gel box. a. CAUTION: UV exposure causes skin damage. 11. The gel images will be printed for each group member.

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Data: RFLP Test Results

Paper Test Results

# Tasters (TT)

9

# Tasters (Tt)

3

# Non-Tasters (tt)

3

5

Total Pop.

15

15

10

Figure 1: RFLP test vs the Paper test for the class population ✗²

Obs #

Freq. of Genotypes

Exp. #

O-E

(O - E)²

(O - E)²/ E

Taster (TT)

9

p²

7.35

1.65

2.7225

.37

Taster (Tt)

3

2pq

6.3

-3.3

10.89

1.729

Non-taster (tt)

3

q²

1.35

1.65

2.7225

2.0167

Total

15

100%

15

----

-----

✗² = 4.1157

Figure 2: Chi-square of the RFLP test result

Figure 3: Chi-square Table

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Figure 4: Our picture of the gel electrophoresis. Calculations: F (T) = (2 x 9) + 3 = 0.7 = p -------------(2 x 15) F (t) = (2 x 3) + 3 = 0.3 = q -----------(2 x 15)

p+q=1 0.7 + 0.3 = 1

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p² + 2pq + q² = 1 (0.7)² + 2 (0.7)(0.3) + (0.3)² = 1 0.49 + 0.42 + 0.09 = 1 0.49 x 15 = 7.35 0.42 x 15 = 6.3 0.09 x 15 = 1.35 Hypothesis: Our hypothesis for the population if it is in Hardy Weinberg Equilibrium will be: 7.35 people will be taster TT, 6.3 people will be taster Tt, and 1.35 people will be a non-taster tt. Results: In order to conduct our data and results, we began by performing the paper PTC test in our class and 10 people were tasters, and 5 people were non-tasters as shown in figure 1. These values determined our 2 phenotypes of taster and nontaster. The genotypes for tasters were TT and Tt, and the genotype for non-tasters was tt. Then we conducted the gel electrophoresis of our DNA to obtain our RFLP analysis and figure out the genotypes of the class. The genotypes of my groupmates from figure 4 were as follows: student 1 was TT in the 300 bp lane, student 2 was tt in the 200 bp lane, student 5’s DNA did not show up, and student 6 was tt in the 200 bp lane as seen in figure 4. We established these based off of our negative control in well 2 that was uncut and positive control in well 3 that was cut (TT). Student 3 must have made a human error when she conducted the procedure, which led to no DNA showing up on the gel electrophoresis. In Figure 1, the overall class population data from the RFLP analysis was as follows: 9 tasters TT, 3 tasters Tt, and 3 non-tasters tt. These numbers were our observed data. We plugged them into the Hardy-Weinberg equation to find the allele frequency and the genotypic frequencies. For allele T, we did (( 2 x 9) +3) / 30 = 0.7. This is the p frequency in the Hardy-

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Weinberg equation, and we got the number 30 from multiplying the population of 15 by 2 because there are 2 alleles in each individual. We did the same for allele t by doing ((2 x 3) + 3) / 30 = 0.3. This is the q frequency in HWE. We then took these allele frequencies and entered them into the equation p² + 2pq + q² = 1 and our equation looked like this: (0.7)² + 2 (0.7)(0.3) + (0.3)² = 1. We took each individual answers and multiplied it by the population number as follows to determine the number of expected people for each genotype: 0.49 x 15 = 7.35, 0.42 x 15 = 6.3, 0.09 x 15 = 1.35. These values became our expected values with 7.35 as our TT taster, 6.3 as Tt taster, and 1.35 as tt non-taster as shown in figure 2. Once we established the observed and expected values, we conducted the chi-square test and found our values for all 3 genotypes. The chi-square formula we used was (O-E)² / E. The degree of freedom used in the test was 1 because 2 alleles -1 = 1. The chi-square result from adding up the 3 genotypes was ✗² = 4.1157. We used 1 degree of freedom and the chisquare value of 4.1157 to find the p-value (probability) from figure 3. The answer we got was p = 0.05 to 0.01. Since the critical value was 3.84 and anything above it is rejected, our hypothesis was rejected. So we reject the hypothesis and the population is not in Hardy-Weinberg equilibrium. Discussion: The purpose of the experiment was to test the Hardy-Weinberg equilibrium at the Tas2R38 gene locus using methods such as PCR, RFLP analysis, and the Chi-square test. Our hypothesis that we were testing was that the expected value of tasters TT would be 7.35, tasters Tt would be 6.3 and non-taster tt would be 1.35. Our observed results from the RFLP analysis on the gel electrophoresis showed us that 9 people were tasters TT, 2 were tasters Tt, and 3 were

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non-tasters tt according to figure 1. To find the validity of these results and to test for HardyWeinberg equilibrium of the class population, we performed the chi-square test. The chi-square result we got was = 4.1157. Since the critical value from figure 3 is 3.84 and the p-value = 0.050.01, which was below the significant value of 0.05, we rejected the hypothesis. This means that the population is not in Hardy-Weinberg equilibrium (HWE) at the Tas2R38 gene locus. For the population to be in HWE would mean that there is no mutation, migration, or natural selection, which is unrealistic in this scenario. In our class in Farmingdale State College there are people from various backgrounds, towns, cities, and countries. This creates diversity and constant migration and mutations to occur, which leads to it being harder for the population to maintain Hardy-Weinberg equilibrium. Along with that, one student from our population was eliminated from the class count because her DNA band did not show up in the gel electrophoresis. The bands that showed up were based off of the digested PCR products that were cut and the ones that were uncut (like the positive control). If there was a single band in the same position as the uncut control at 221 bp, then it was a tt nontaster (homozygous recessive). If there were two bands at 177 bp and 44 bp, then it was a TT taster (homozygous dominant). If there were three bands present at 221 bp, 177 bp, and 44 bp, then it was a Tt taster (heterozygous). The eliminated student was a part of my group, student number 3 on the DNA ladder and class member #5. The entire group reading from figure 4 is as follows: student 1 was TT in the 300 bp lane, student 2 was tt in the 200 bp lane, student 5’s DNA did not show up, and student 6 was tt in the 200 bp lane as seen in figure 4. A multitude of human errors could have occurred when preparing her DNA for RFLP analysis. When she first rinsed her mouth with the saline, she might have not rinsed it for long enough; she may have removed a part of her pellet when discarding the supernatant from step 5

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of procedure 1; she may have not resuspended her cells properly from step 6 of procedure 1; she may have not transferred her DNA properly in steps 11 and 12 from procedure 1. These are just a few of the errors that may have occurred throughout the whole procedure. These errors can be avoided in future experiments by being more cautious when transferring DNA, and using a better handling technique when using a micropipette. Overall, the expected values and the observed values were not too far apart and 13 out of 15 people had the same genotype as their phenotype....