Practical Techniques in Genomics & Proteomics PV92 practical

Practical Techniques in Genomics & Proteomics PV92 practical
DNA Template Preparation
1. Why is it necessary to chelate the metal ions from solution during the boiling/lysis step at 100°C?
What would happen if you did not use a chelating agent such as the InstaGene matrix?
2. What is needed from the cells for PCR?
3. What structures must be broken to release the DNA from a cell?
PCR Amplification
1. Why is it necessary to have a primer on each side of the DNA segment to be
amplified?
2. How did Taq DNA polymerase acquire its name?
3. Why are there nucleotides (A, T, G, and C) in the master mix? What are the other
components of the master mix, and what are their functions?
4. Describe the three main steps of each cycle of PCR amplification and what reactions occur at each temperature.
5. Explain why the precise length target DNA sequence doesn’t get amplified until the third cycle. (Tip: the use of a drawing on a separate piece of paper will clarify your answer)
Gel Electrophoresis of Amplified PCR Samples
1. Explain the difference between an intron and an exon.
2. Why do the two possible PCR products differ in size by 300 base pairs?
3. Explain how agarose electrophoresis separates DNA fragments. Why does a smaller DNA fragment move faster than a larger one?
4. What kind of controls are run in this experiment? Why are they important? Could others be used?
Analysis and Interpretation of Results
Remember that this Alu sequence is inserted into a noncoding region of the PV92 locus on chromosome 16 and is not related to a particular disease, nor does it code for any protein sequence. It is simply a sequence that can be used to study human genotypic frequencies.
Because Alu repeats appear in the general population at random, the Alu insert in chromosome 16 is very useful for the study of gene frequencies in localized human populations. Theoretically, in some small, geographically isolated populations, all individuals may be homozygous +/+. In others, the individuals may all be homozygous –/–. In a “melting-pot” population, the three genotypes (+/+, +/–, –/–) may exist in equilibrium.
The frequencies of genotypes and alleles are basic characteristics that population geneticists use to describe and analyze populations. The results you obtain in this exercise
provide a real-life opportunity to calculate genotypic and allelic frequencies of the Alu insert in your class and to use the Hardy-Weinberg equation. The results of the PCR reactions reveal your and your classmates’ genotypes: +/+, +/–, and –/–. Knowing your genotypes, you can count up the alleles of your class “population” and determine their frequencies. You can then compare the allelic and genotypic frequencies
of your class population to published reports of larger population sizes.
1. What is your genotype for the Alu insert in your PV92 region? Include a copy of the gel photograph.
2. What are the genotypic frequencies of +/+, +/–, and –/– in your class population? Fill in the table below with your class data obtained from allele server.
Table 1. Observed Genotypic Frequencies for the Class
Category
Number
Frequency (# of Genotypes/Total)
Homozygous (+/+)
Heterozygous (+/–)
Homozygous (–/–)
Total =
= 1.00
Allelic frequencies can be calculated from the numbers and frequencies of the genotypes
in the population. Population geneticists use the terms p and q to represent the frequencies of the (+) and (–) alleles, respectively. Allele frequencies can be calculated from either the numbers or the frequencies of the genotypes (since they are related to each other).
p = frequency of (+) allele = number of (+) alleles
total number of alleles (both + and –)
= 2(# of +/+ students) + 1(# of +/– students)
total number of alleles (both + and –)
= frequency of (+/+) students + ½ (frequency of (+/–) students)
q = frequency of (–) allele = number of (–) alleles
total number of alleles (both + and –)
= 2 (# of –/– students) + 1(# of +/– students)
total number of alleles (both + and –)
= frequency of (–/–) students + ½ (frequency of (+/–) students)
3. What is the frequency of each allele in your class sample? Fill in the table below with your class data. Remember, a class of 32 students (N) will have a total of 64 (2N) instances of each locus.
Table 2. Calculated Allelic Frequencies for the Class
Category
Number
Frequency (# of Genotypes/Total)
(+) alleles
p =
(–) alleles
q =
Total alleles =
= 1.00
4. The following table presents data from a USA-wide random population study.
Table 3. Genotypic Frequencies for Alu in a USA Sample
Category
Number
Frequency (# of Genotypes/Total)
Homozygous (+/+)
2,422
0.24
Heterozygous (+/–)
5,528
0.55
Homozygous (–/–)
2,050
0.21
Total =10,000
= 1.00
Now, using the data above, calculate the allelic frequencies for the USA data as you did
for your class population in Table 2.
Table 4. Calculated Allelic Frequencies for USA
Category
Number
Frequency (# of Genotypes/Total)
(+) alleles
p =
(–) alleles
q =
Total alleles =
= 1.00
5. How do your actual class data for genotypic and allelic frequencies compare with those
of the random sampling of the USA population?
Would you expect them to match?
What reasons can you think of to explain the differences or similarities?
The Hardy-Weinberg equation, p2 + 2pq + q2 = 1, is one of the foundations of population genetics. It is the algebraic expansion of (p + q)2 = 1, where p + q = 1. The equation describes the frequencies of genotypes in a population that is at “genetic equilibrium”, meaning that the frequencies are stable from generation to generation. The Hardy-Weinberg theory states that, for a population to achieve this equilibrium, the population must be quite large, the members must mate randomly and produce offspring with equal success, and there must be no migration of individuals into or out of the population, or an excessive mutation converting one allele to another. Given these conditions, and the allelic frequencies p and q, the Hardy-Weinberg equation says that
p2 = the expected frequency of the (+/+) genotype in the population
2pq = the expected frequency of the (+/–) genotype in the population
q2 = the expected frequency of the (–/–) genotype in the population
It is important to understand that p2, 2pq, and q2 are expected, theoretical genotype
frequencies of a population under Hardy-Weinberg equilibrium conditions, and they may not be realized in real-life population samples if one of the conditions is not met. These theoretical frequencies are calculated using the observed values for p and q; they may or may not be the same as the observed genotypic frequencies such as those shown in Table 1. If the observed and expected genotypic frequencies are the same, this indicates that the population is in Hardy-Weinberg genetic equilibrium.
6. Using the values for p and q that you calculated in Table 2 for your class population, calculate p2, 2pq, and q2.
Do they come out to be the same as the genotype frequencies that you found in Table 1? If they do, your class resembles a Hardy-Weinberg genetic equilibrium. If your observed (actual) genotype frequencies are not the same as the expected values, what might be some of the reason(s) for the difference?
7. Using the values for p and q that you calculated in Table 4 for the USA population sample, calculate p2, 2pq, and q2.
Do they come out to be the same as the genotype frequencies that you found in Table 3? Does this USA-wide sample suggest that the population of the USA is in Hardy-Weinberg equilibrium?
Analysis of Classroom Data Using Bioinformatics
Bioinformatics is a discipline that integrates mathematical, statistical, and computer tools to collect and process biological data. Bioinformatics has become an important tool in recent years for analyzing the extraordinarily large amount of biological information that is being generated by researchers around the world. You will perform a bioinformatics exercise to investigate the genotypic frequencies for the Alu polymorphism in your class population and compare them with the genotypic frequencies of other populations.
Following PCR amplification and electrophoresis of your samples, you will analyze your experimental data to determine your genotypes for the Alu insertion within the PV92 locus on chromosome 16. The classroom genotype data can then be entered into the Allele Server database located at Cold Spring Harbor Laboratory’s Dolan DNA Learning Center. Allele Server is a Web-based database that contains genotype data from a range of populations around the world as well as other classrooms and teacher training workshops. It also provides a collection of statistical analysis tools to examine the Alu insertion polymorphism at the population level. You can either analyze your classroom data as an individual population or compare your population with other populations in the database. Once you enter classroom data into Allele Server, you can perform a Chi-square analysis to compare the Alu genotype frequencies within the class population with those predicted by the Hardy-Weinberg equation. The genotypic frequencies of the class population can also be compared with the genotypic frequencies of another population in the database, using the Chi-square analysis.
Using Allele Server
Note: The Dolan DNA Learning Center web site is continually updated. Some of the
following information may change.
1. On your Web browser,go to Bioservers at Dolan Learning Center (http://www.bioservers.org/bioserver/) and register as a student.
2. Then go to and log in to Allele server
3. Click on Add data. If your PCR did not work, then go straight to step 6.
4. In pop up box, select NTU MSc 2012-2013, password rugby , User no. is given in the excel spreadsheet on NOW.
5. Follow instructions provided by clicking help. Add your data
6. Recheck at a later date that whole class data has been entered. To carry out analyses, you can select groups by clinking on manage groups. Further details on using Allele Server are available here http://www.dnalc.org/help/sad/index.html
7. Perform Chi squared test on class data, and include result here
Where in the world is pv92?
One of the reasons pv92 is interesting is that it is polymorphic — some humans have it, some do not. Moreover, the humans that do have it tend to come from a particular part of the world. In this simple exercise, you will use a map to find out where this area is.
You will need a world map to write on. If you don’t have one, you can print out this map (map courtesy of ITA’s Quick Maps).
Step 1
Go to manage groups and choose ten of the reference populations that seem interesting. Try to get a range of populations from around the world. If you are not sure where a population is from, you can look it up in an encyclopedia, or ignore it. Click OK to select the populations
Step 2
Click the open button next to a population. Note the + allele frequency for this population (second row of information in the window). Plot this frequency on the map where the population is.
Step 3
After repeating step 2 for all your populations, look at the map. Where in the world do you see a clear concentration of the + allele? If there is no clear concentration, pick some more populations and plot them on the map. Once you have identified an area of concentration, you may want to pick a few more populations to verify your hypothesis.
1. Is there any particular reason that pv92 is concentrated here? (hint: does pv92 have a selective advantage?)
2. If the answer to 1 is no, why is there such a difference in allele frequencies around the world (hint: natural selection is not the only form of evolution)
3. Think of a hypothesis about the origin and dispersal of the pv92 allele that acocunts for the observations.
Population Similarities and Differences
Alu insertion frequencies can be used to study relationships between human populations. In this exercise, we will be using the Chi Square comparison to explore these relationships.
Step 1
Choose four or five reference groups to examine, and also your class data.
Then click OK.
Step 2
Select two of the populations (start with your own). Choose “Chi Square” for your comparison, and click the Compare button. Take a look at the pie charts. How do the populations seem to compare? Record the p-value for this comparison in the table below (replace number on grid with name of population used.)
Repeat this step for all combinations of populations you’ve selected
Population
1
2
3
4
5
1
2
3
4
5
Step 3
To interpret the results, you need to understand what the test is doing. for the test, we formulate a “null hypothesis” — that the two groups are indistinguishable. Our test will attempt to disprove the hypothesis. If the test returns a p-value of 0.05 or below (5%), we can say the null hypothesis is disproved, and therefore that the two groups are statistically different. If the p-value is over 0.05, we can reach no conclusions about the two groups. This is an important point — p-values of over 0.05 do not mean that the two groups are the same, just that the test was inconclusive.
Note which groups are statistically different.
1. Which groups are different, and which are basically the same?
2. Are there some groups that you would assume would seem different, but aren’t? What does this imply about examining populations based on a single allele?

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