Population Genetics: Correlating Genotypic Differences in Our Population with Different Phenotypic Outcomes [1] Jayanth (Jay) Krishnan T.A. Ms. Bianca Pier Lab Partner: Ms. Catherine Mahoney Section 1: Biology November 3rd, 2011

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Introduction/Purpose: Why Did We Study This Problem? There are roughly thirty genes for bitter taste receptors in humans. TAS2R38 is unique 1002 base-pairs human bitter taste receptor gene which aids in the tasting of phenylthiocarbamide (PTC). In 2003, it was discovered that are three unique nucleotide positions. These positions can vary and each of them is known as a single nucleotide polymorphism (SNPs – replacement of a single nucleotide). A combination of the three SNPs is known as a haplotype which strongly influences tasting ability. [2] TAS2R38 has two different alleles. A dominant allele denoted by, “T,” and a recessive allele denoted by, “t.” An individual’s genotype is comprised of two alleles (not necessarily one of each type). If even a single, “T,” is present in an individual’s genotype, his or her phenotype will signify that they are a taster for the bitter PTC. If both of the two alleles of an individual are the recessive “t,” then the individual is a non-taster. In order to do analysis on populations such as our Biology 1010 lab class, we each needed to amplify our DNA and identify our phenotypes. We used the technique of Polymerase Chain Reaction (PCR) to amplify DNA. DNA from cheek cells was amplified for each individual in our population. The amplified PCR product for both tasters and non-tasters is 221 base-pairs. To distinguish the PCR products of those who taste versus non-tasters we use the restriction enzyme HaeIII which can identify and cut the GGCC group. GGCC is a group of base-pairs that is apparent in the amplified sequences for exclusively tasters. [2] Homozygous recessive non-tasters have only a single uncut 221 base-pair sequence. The 221 base-pair sequence of homozygous dominant strong tasters are cut by HaeII where the SNPs

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are located and hence there are two bands at 177 base-pairs and 44 base-pairs. The DNA for homozygous dominant strong tasters is also cut by HaeII where the SNPs are located and hence there are two bands at 177 base-pairs and 44 base-pairs. Lastly, the DNA of heterozygous weak tasters are also cut by HaeII where the SNPs are located and hence there are three bands at 221 base-pairs, 177 base-pairs, and 44 base-pairs. Reiterating what we stated earlier, in this lab we wanted to see phenotypic diversity among our class using a tasting PTC paper test and the genotypic diversity using PCR. I, for example was found to be homozygous recessive and therefore predicted to be a non-taster of PTC. With the taste test, I realized that I was indeed a non-taster. We then used Chi square statistics (a test for two table categorical data) to compare the wet-lab results we obtained to the frequencies we calculated using Hardy-Weinberg equilibrium. If the results correlated, which is based on the p-value of the test, the distributions meet the conditions for Hardy-Weinberg equilibrium. The test was implemented not only for my biology lab class, but for all lab classes combined as well. The student population in my lab and the rest of the Biology 1010 labs was very close with the empirical frequencies obtained from the Hardy-Weinberg equilibrium and hence using further methods of analysis (Chi-squared tests) we were able to conclude that that both population distributions were in Hardy-Weinberg equilibrium.

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Materials and Methods I: How was our DNA collected and amplified: SNPs Lab Part A: [2] In order for us to amplify our DNA we first needed to collect it. In this experiment we used saline mouthwash to harvest our cheek cells. Each individual swished the mouthwash vigorously against their cheeks and gums for thirty seconds and spit the contents into a cup. Each of these cups then contained our respective DNA. One milliliter of this solution containing our DNA was suspended with a vortex and then transferred to a microfuge tube. The tube was then subsequently centrifuged forming pellet cells while the supernatant was dumped into a cup. By using the vortex and the centrifuge, the DNA was able to be extracted out of the cells. The pellet was then suspended in a tube with Chelex. Chelex is negatively charged that grabs ions such as Mg2+ which are required cofactors for enzymatic reactions. Essentially the Chelex is the reason behind why the DNA is not degraded by hydrolytic enzymes released when membranes are destroyed. The contents then around went through two ten minute cycles of boiling and sitting in an ice bath. Following this procedure, the DNA was vortex-ed and centrifuged which makes the cellular debris and the Chelex into a pellet. The DNA is now the supernatant which was extracted and stored separately. SNPs Lab Part B: [3] The next lab class we obtained our separated DNA and a vial with a ready-to-go-bead in it obtained from our lab TA to begin PCR of the 221 base-pairs of the TAS2R28 gene that may contain the SNPs if the individual is a taster. This bead contains all the necessary ingredients including Taq polymerase, dNTP, Mg2+, buffer. Each of these reagents is needed to amplify the 4 [Reference 1] Lab Report was written according to the guidelines of the rubric

DNA for the several different reactions that are performed during the three stages of PCR. We then added PTC primer mix/loading dye to the tube with the bead to dissolve the bead. Our DNA was finally added, along with the restriction enzyme HaeIII, and the mixture was spun. The TA also stained our agarose gel with ethidium bromide which slips between stacked base pair of the DNA so that florescence is produced when UV light is shown on the gel. We then implement PCR through denaturing, annealing and extending steps to amplify our DNA. The first step, denaturing takes a total of thirty seconds in 94 degrees Celsius where chromosomal DNA is denatured into single strands. The second step, annealing takes forty-five seconds at 64 degrees Celsius. In this step primers for hydrogen bonds combine with complementary sequences on either sides of the target region on the TAS2R38 receptor. Lastly, extending takes forty-five seconds at 72 degrees Celsius where Taq Polymerase extends the primers and makes complementary DNA begin with each primer. This process was repeated thirty-five times in a thermal cycler. II: Describe what you did to distinguish between taster and non-taster alleles: Regarding what we wanted from PCR is only to amplify a region of 221 base-pairs of the TAS2R38 gene. The enzyme HaeIII was added to distinguish between taster and non-taster alleles. HaeIII finds the flanked GGCC nucleotide patterns for DNA of trace receptors and perform a digestion or cut that result in a band in the gel. In addition to loading our digested solution and see where bands are created we also load our undigested solution (without HaeIII) to use as a control. Reiterating concepts from the introduction, homozygous recessive non-tasters only have a single uncut 221 base-pair sequence. The 221 base-pair sequence of homozygous dominant 5 [Reference 1] Lab Report was written according to the guidelines of the rubric

strong tasters are cut by HaeII where the SNPs are located, and hence there are two bands at 177 base-pairs and 44 base-pairs. The 221 base-pair sequence of homozygous dominant strong tasters are cut by HaeII where the SNPs are located and hence there are two bands at 177 base-pairs and 44 base-pairs. The 221 base-pair sequence of heterozygous weak tasters are also cut by HaeII where the SNPs are located and hence there are three bands at 221 base-pairs, 177 base-pairs, and 44 base-pairs. For my PCR, I used 5 microliters of a 100 base-pairs ladder and 10 microliters of undigested sample to use as controls. I then loaded the gel with 16 microliters of digested sample. When UV light was shined over my gel, because of the ethidium bromide, we can use fluorescence to compare the lane filled with 16 microliters of digested sample and the controls and thus predict my genotype. [3] Thus, using this experiment, I can conclude if the individual is a strong taster, a weak taster or a no taster. Non-tasters with genotype tt, like me, have no cuts and therefore only a band at 200 base-pairs. Strong tasters with the genotype TT have bands at 177 base-pairs and 44 basepairs. Lastly weak tasters with genotype Tt have bands at 221, 177, and 44 base-pairs. In order to visualize these bands it is crucial that we compare the results to the 100 base-pair ladder. Sometimes one can inadvertently see a primer dimer in the PCR of those who are heterozygous. This band results as an artifact of PCR where primers overlapping one another amplify themselves. When our PCR was done and all members of the lab identified their genotype the lab TAs helped us take a picture of our gel using a flash – which is a type of camera used to take pictures of agarose gels.

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III: Describe how you determined your phenotype Everyone in the class determined their phenotype experimentally by first applying a blank slit of paper that had no chemicals on it. This piece of paper was essentially used as the control to make sure students know what blank paper tasted like. Students were then given papers with PTC on it. Those who were non-tasters tasted nothing and were therefore predicted to be homozygous recessive. Those who were weak tasters only slightly tasted the bitterness and were perceived to be heterozygous. Lastly those who were homozygous dominant experienced a very strong bitter taste. All of these predictions were confirmed with PCR evidence which is present at the end of step II of the methodology above.

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Results: The Aftermath of Experimentation My genotype as predicted by the gel was that I am tt indicating that I am homozygous recessive and thus a non-taster. This is indeed true as I could not taste the PTC during the empirical demonstrations or step III of the materials and methods. Hence my phenotype is consistent with my genotype.

Figure 1: The PCR gel when shined with UV Student A is my lab partner Catherine Mahoney who is a taster which can be seen from the number of bands in her gel. Student C is alike me - a non-taster with no breaks and only a single band 8 [Reference 1] Lab Report was written according to the guidelines of the rubric

Table 1 shows my lab class’s count for each type of genotype based on gel results. It also presents our phenotypes based on the PTC paper test. 15 out of a class of 55 were homozygous dominant who should be strong tasters. 22 out of a class of 55 were heterozygous and should be weak tasters. Lastly 18 out of 55 were homozygous recessive and should be no tasters. The genotypic outcomes did indeed jive with the phenotypic results for the most part.

Table 1: Section 1 Data Results – The table describes how well the data obtained from an individual PCRs correlated with the empirical/field test when he or she guessed his or her phenotype based on the aftermath of when the strip of PTC was licked. As one would notice for the most part, excluding whether heterozygous individuals are strong or weak tasters, the results are quite consistent. Genotype Phenotype Phenotype Phenotype

Totals

strong taster

weak taster

non-taster

By Genotype

TT

13

2

0

15

Tt

10

12

0

22

tt

0

3

15

18

Table 2 is a very similar representation that presents the type of genotype for all the Biology lab classes combined. 43 out of a class of 172 were homozygous dominant who should be strong tasters. 79 out of a class of 172 were heterozygous and should be weak tasters. Lastly 50 out of 172 were homozygous recessive and should be no tasters. The genotypic outcomes did indeed jive with the phenotypic results for the most part.

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Table 2: The Genotype distribution for all BIOL-1010 Students – The information for each individual’s genotype was recorded based on the results of the number of bands observed when PCR was done on an individual basis

Genotype

Totals By Genotype

TT

43

Tt

79

tt

50

Now that we had our results, we wanted to confirm whether or not they were in HardyWeinberg equilibrium: Hardy-Weinberg states two equations: a) p + q = 1 b) p2 + q2 + 2pq = 1 So first we calculated the allelic frequency of our class and the allelic frequency of all the classes combined.

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I: For my Biology lab Class – Refer to Table 1 Allelic Frequency of T: [(2*15) + 22]/110 = 52/110 = .47 Allelic Frequency of t: [(9*2) + 22]/110 = 40/110 = .36 Now we wish to use the Allelic frequency we calculated to determine the expected genotype frequencies. Expected Genotypic Frequency for TT: p2 = (.47 * .47) = .22 Expected Genotypic Frequency for Tt: 2pq = 2(.47 * .36) = .3384 Expected Genotypic Frequency for tt: q2 = (.36 * .36) = .1296 We then calculate observed frequency using Table 1. Observed Genotypic Frequency for TT: 15/55 = .27 Observed Genotypic Frequency for Tt: 22/55 = .40 Observed Genotypic Frequency for tt: 18/55 = .33 Now in order to determine if our class population is in Hardy-Weinberg Equilibrium we must calculate the X2 value: X2 = SUM[(Observed Genotypic Frequency – Expected Genotypic Frequency)2/Expected Genotypic Frequency] d2/e for TT = (.27-.22)2/.22 = .113 d2/e for Tt = (.40 -.3384)2/.3384 = .011

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d2/e for tt = (.33 - .1296)2/.1296 = .3098 X2 = .434 This value falls between the intervals of .016 and .46 signifying that the p-value is slightly above .50. Since I took a level of significance of .05, the value was not significant and I should not reject the hypothesis that the class population is in Hardy-Weinberg equilibrium. Hence with about 50% significance there is no difference between the observed numbers of students of each genotype and expected ones.

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I: For All of the Biology 1010 Lab Classes - Refer to Table 2 Allelic Frequency of T: [(2*43) + 79]/344 = 165/344 = .48 Allelic Frequency of t: [(2*50) + 79]/344 = 179/344 = .52 Now we wish to use the Allelic frequency we calculated to determine the expected genotype frequencies Expected Genotypic Frequency for TT: p2 = (.48 * .48) = .2304 Expected Genotypic Frequency for Tt: 2pq = 2(.48 * .52) = .50 Expected Genotypic Frequency for tt: q2 = (.52 * .52) = .27 We then calculate observed frequency using Table 2. Observed Genotypic Frequency for TT: 43/172 = .25 Observed Genotypic Frequency for Tt: 79/172 = .46 Observed Genotypic Frequency for tt: 50/172 = .29 Now in order to determine if our class population is in Hardy-Weinberg Equilibrium we must calculate the X2 value: X2 = SUM[(Observed Genotypic Frequency – Expected Genotypic Frequency)2/Expected Genotypic Frequency]

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d2/e for TT = (.25-.2304)2/.2304 = .0016 d2/e for Tt = (.46 -.50)2/.50 = .0032 d2/e for tt = (.29 - .27)2/.27 = .0014 X2 = .0062 This value falls between the intervals of .0038 and .0016 signifying that the p-value is slightly above .90. Since I took a level of significance of .05, the value was definitely not significant and I should not reject the hypothesis that the class population is in Hardy-Weinberg equilibrium. Hence with about 90% significance I can assert that there is no difference between the observed numbers of students of each genotype and expected ones.

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Table 3: Calculations for my Lab Class – The table displays how I calculated the X2 value of my class in deciding whether the results jived with the standard Hardy-Weinberg Equilibrium d2/e for TT (.27-.22)2/.22 d2/e for Tt (.40 -.3384)2/.3384 d2/e for tt (.33 - .1296)2/.1296 Chi-Squared P-Value Reject hypothesis

0.113 0.011 0.3098 0.434 ~.50 No

Table 4: Calculations for All BIOL-1010 Lab Classes – The table displays how I calculated the X2 value of results from all classes combined in deciding the results jived with the standard Hardy-Weinberg Equilibrium: d2/e for TT

(.25-.2304)2/.2304

0.0016

d2/e for Tt

(.46 -.50)2/.50

0.0032

d2/e for tt

(.29 - .27)2/.27

0.0014

Chi-Squared P-Value Reject hypothesis

0.0062 ~.90 No

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Discussion: Why did we get such results? Reiterating my genotypic results which are present in Figure 1 of the Results section, I am homozygous recessive as I have only a single band. There is no difference between my digested and undigested samples as the effect of HaeIII was the same. This is consistent with my PTC results where I empirically tested as a non-taster. Interestingly enough you can notice a discrepancy in genotypic and phenotypic results for most heterozygotes in my class in Tables 1 and 2. This discrepancy probably resulted as it is difficult to distinguish between whether one is a strong taster or a weak taster. Additionally some members of the class may have classified their genotype incorrectly as they may have made a mistake by seeing a primer dimer or a band that was faint. Tables 3 and 4, along with my data provided in my results sections display two X2 tests I performed to assess whether my class and total biology classes were consistent with the HardyWeinberg equilibrium for the TAS2R38 taste receptors. As shown in the tables provided by the p-value of my tests, both distributions were indeed in Hardy-Weinberg equilibrium and thus not evolving. The first test which had a small population size had a p-value of slightly above .50. The second which was contained all labs pooled together, a larger population size, had a p-value of slightly above .90. Both results are consistent with what was predicted, but the second is even more accurate due to the law of large numbers. Lastly, I would definitely expect the bitter taste receptor allele to be in Hardy-Weinberg equilibrium as it is a trait that follows classic Mendelian laws, more specifically the law of dominance. The gene also seems to meet most of the conditions for a population that has a

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Hardy-Weinberg equilibrium including a large population (usually greater than 40), random mating (not satisfied), no mutation, no natural selection, and no emigration/immigration. Although, I am unfortunately not a taster, it would be advantageous to be a taster as many poisonous foods are bitter. Regarding gustation, bitterness is a taste that humans tend to naturally avoid. Perhaps some of our ancestors would have lived longer if they also possessed this trait. Other primates probably do not have this ability as we have not discovered that they had the gene for TAS2R38. However, regarding the prevalence of the gene in other species, mice are also known to have the same bitter taste receptor. [4]

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Citations: [1] Radlick, L. (2011). Grading Rubric: Population Genetics Lab Report [2] Radlick. Part I: Using A Single Nucleotide Polymorphism (SNP) to Predict Bitter Tasting Ability. RPI LMS. Web. . [3] Radlick. Using Single NucleotidePolymorphism (SNP) to Predict Bitter Tasting Ability Part II RPI LMS. Web. . [4] "Hardy-Weinberg." Kansas State University. Web. 02 Nov. 2011. .

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