Population Genetics: Correlating Genotypic Differences … · 1 [Reference 1] Lab Report was...

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[Reference 1] Lab Report was written according to the guidelines of the rubric

PopulationGenetics:

CorrelatingGenotypicDifferencesinOur

PopulationwithDifferentPhenotypic

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

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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

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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 base-

pairs. 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

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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

strong taster

Phenotype

weak taster

Phenotype

non-taster

Totals

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 Hardy-

Weinberg 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 0.113

d2/e for Tt (.40 -.3384)2/.3384 0.011

d2/e for tt (.33 - .1296)2/.1296 0.3098

Chi-Squared 0.434

P-Value ~.50

Reject hypothesis 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 0.0062

P-Value ~.90

Reject

hypothesis 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 Hardy-

Weinberg 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.

<http://rpilms.rpi.edu/webct/ContentPageServerServlet/Lab/2GeneticsAndMedicine/WetSNP_I/

1Pre/1.Lab_Exercise-SNPs_PtI.pdf>.

[3] Radlick. Using Single NucleotidePolymorphism (SNP) to Predict Bitter Tasting Ability -

Part II RPI LMS. Web.

<http://rpilms.rpi.edu/webct/ContentPageServerServlet/Lab/2GeneticsAndMedicine/WetSNP_I/

2/2.Lab_Exercise-SNPs_Pt2.pdf>.

[4] "Hardy-Weinberg." Kansas State University. Web. 02 Nov. 2011. <http://www.k-

state.edu/parasitology/biology198/hardwein.html>.

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