Sample Lab Report

Goh Si Fan (U1020639J)

Experiment C7Vapour Liquid Equilibrium

NANYANG TECHNOLOGICALUNIVERSITYSCHOOL OF CHEMICAL AND BIOMEDICAL ENGINEERING

(Division of Chemical & Biomolecular Engineering)

CH 2701: Chemical/Biomolecular Engineering Laboratory 3

Experiment C4: Polymerase Chain Reaction (Formal Lab Report)

Year 2/Semester 1

Laboratory: N1.2 B2-41

Name: Goh Si Fan

Matriculation No: U1020639J

Lab Group Group 9

Date of Experiment: 13th September 2011

Lane 2

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Experiment C4Polymerase Chain

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Table of Content

1. Objectives 3

2. Abstract 4

3. Introduction 5 - 6

4. Equipments & Materials used 7

5. Procedure 8-12

6. Results 13-14

7. Questions 15-22

8. Discussion 23

9. Conclusion 24

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1 Objectives

PCR Amplification

To learn the technique of performing PCR to amplify a specific sequence of DNA.

To learn how DNA is extracted from hair follicles.

To learn the way a thermal cycler works

To understand the principles of PCR and how the same basic processes are used for

DNA replication.

Gel Electrophoresis of Amplified PCR Samples

To get hands on experience in setting up and perform gel electrophoresis to separate

DNA fragments.

To learn proper techniques on know how to handle the agarose gel.

To visualize DNA bands in the gel after performing staining techniques on it.

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2 Abstract

This experiment basically involves polymerase chain reaction (PCR) to amplify a specific

chain of DNA (PV92 region of chromosome 16). The objective is to determine whether Alu

sequence is presence within the region. PCR is used to amplify the target sequence of the

DNA extracted from the hair follicles.

We will also learn about the steps involving PCR, which includes the main 3 steps as

followed:

1. Denaturation Step

2. Annealing Step

3. Extension Step

We will then analyse the results by performing gel electrophoresis on the specific DNA

sequence. There will be 3 types of results depending on the DNA size of the PCR product.

Genotype DNA Size of PCR Product (bp)

Homozygons (+/+) 941

Homozygons (-/-) 641

Heterozygons (+/-) 941-641

Table 2.1- Classification of Genotype vs its DNA size of PCR product

ALU

Intron

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3 Introduction

3.1 Polymerase Chain Reaction Amplification

PCR is a technique used to amplify specific regions along a strand of DNA up to 10 kilo base

pairs. It is performed through thermo cycling where the following 3 processes will occur,

namely, denaturation, annealing of primers and extension of the new DNA strands.

For Denaturation process, the mixture is heated to a high temperature in order for the

parent strand DNA to separate.

During the Annealing process, the mixture is cooled down, and primers seek its

complementary sequences in the DNA strand and anneal them via hydrogen bond.

For the Extension process, the mixture is then heated up again to 72°C. This is the optimum

activity temperature for the Taq polymerase to synthesize new complementary DNA strands

starting from the 3’-OH ends of the primers. The three processes are performed 40 times in

cycles.

Details of PCR cycle is explained under Section 7 Question 4.

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3.2 Gel Electrophoresis of Amplified PCR Samples

Figure 3.1 – Gel Electrophoresis

In this experiment, we used Gel Electrophoresis to determine the target sequences. Gel

Electrophoresis works on the principle of size and charge based on the fragment separation.

When the samples are loaded into the wells of the gel, current is passed through the 2

electrodes at each end. The DNA fragments itself are negatively charged, thus it will move

across the gel and be attracted to the positive end of the gel.

Thus, the bigger the size of the fragment, the slower it will move across the gel matrix as it

will encounter more hindrance across the gel.

Thus, through such movement of DNA fragments, we are able to know the size of the DNA

fragments and whether it contains the Alu insertion.

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4 Equipments and Materials Used

PCR Amplification

1. 200 μL InstaGeneTM matrix plus protease

2. 1 x tube of complete master mix (with primers) on ice

3. 2 x strands of Hair from the same person

4. Foam micro test tube holder

5. Vortex mixer

6. Stop watch timer

7. PCR tube

8. Capless micro test tube

9. P-20 Micropipet

10. Pipet Tips

11. Screwcap tube

12. Water bath at 56 °C

13. Water bath at 100°C

14. Centrifuge

15. Mycycler Thermal Cycler

Gel Electrophoresis

1. Student's PCR sample

2. 10 μL PV92 XC DNA loading dye

3. 275 mL of TAE electrophoresis buffer

4. 10 μL MMR (DNA standard)

5. 10 μL PV92 homozygous (+/+) control sample

6. 10 μL PV92 homozygous (−/−) control sample

7. 10 μL PV92 heterozygous (+/−) control sample

8. 120 mL Fast BlastTM DNA stain (100x)

9. 1.5 - 2 L warm tap water

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5 Procedures

5.1 Extraction of DNA template from Hair Follicle

For this part of the experiment, we are supposed to collect 2 strands of hair for each group in order

to prepare the DNA template for the next session of lab group. Thus, using the hair bulb and the

InstaGene matrix plus protease, we are able to extract the DNA template from the hair follicle and

conduct PCR amplification.

Figure 5.1 – Place base of hair into screwcap tube

Figure 5.2- Place it into 56°C water bath for 5 minutes

Figure 5.3- Vortex for several times

Figure 5.4- Place tube back into water bath again for 5 minutes

Figure 5.5- Then place tube into 100°C water bath for 5 minutes

Figure 5.6- Then place tubes centrifuge for 5 minutes at 6000x g

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5.2 PCR Amplification

Step 2a: Obtain a PCR tube containing 10 μL of yellow PCR master mix.

Step 2b: Using a micropipette, pipette 10 μL of the DNA template from the supernatant in

the screwcap tube into the PCR tube containing the master mix.

Step 2c: Ensure mixture mixed well by pipetting up and down 2 to 3 times. In addition,

make sure there are no air bubbles were avoided, especially at the bottom of the PCR tube.

Step 2d: Place PCR tube in MyCycler thermal cycler.

Step 2e: 40 cycles of amplification will take place within the 3 hours.

Figure 5.7- 10 μL of the DNA template from the supernatant

Figure 5.8- MyCycler thermal cycler readings

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5.3 Gel Electrophoresis of Amplified PCR Samples

In order to conduct the Gel Electrophoresis, we used the PCR samples obtained from the previous

laboratory group. This is due to time constraint to carry out the entire PCR reaction.

Step 3a: Place the PCR tube in a capless micro test tube and then into the centrifuge. Pulse-

spin it for 3 seconds at 200 x g.

Step 3b: 10 μL of PV92 XC loading dye added to the PCR tube and mixed thoroughly with

the micropipette.

Step 3c: Agarose gel obtained and placed with the casting tray onto the gel box platform.

Ensure the wells are connected correctly.

Step 3d: The samples were loaded into the 8 wells of the gel using a clean tip for each

sample in this order:

Step 3e: The power supply was switched on and the gel electrophoresis apparatus set to

100V and left running for 30 minutes.

Figure 5.9- Loading samples into well of gel

Figure 5.10- Gel Electrophoresis in process

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Lane Sample Load Volume

1 MMR (DNA Standard) 10 μ

2 PV92 Homozygous (+/+) control 10 μ

3 PV92 Homozygous (-/-) control 10 μ

4 PV92 Heterozygous (+/-) control 10 μ

5 Group 1 20 μ

6 Group 2 (My Group) 20 μ

7 Group 3 20 μ

8 Group 4 20 μ

Table 5.1- Amount of load samples in the gel

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5.4 Quick Staining of Agarose Gels in 100x Fast Blast DNA Stain

Figure 5.10- Removing of agarose gel

Figure 5.11- Stain the gel with Fast Blast stain for 2-3 minutes

Figure 5.12- Rinse the stained gel with warm water

Figure 5.13- Rock the gel on the rocking platform for 5 minutes

Figure 5.14- Now we are able to see the bands appearing on the gel

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Lane 4Lane 1 Lane 2 Lane 3 Lane 5 Lane 6 Lane7 Lane 8



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6 Results

Figure 6.1- Result of Experiment

Group 1

Group 2

(My Group)

Group 3 Group 4MMR

(DNA Standard)

Homozygous

(+/+) controlHomozygous

(-/-) control

Heterozygous

(+/-) control

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Lane Explanation

1 Lane 1 contains of the DNA standard and on the gel, it shows up as several bands which corresponds to 1000, 700, 500, 200 and 100 base pairs (bp) fragments. However, from Figure 6.1, we are unable to clearly see the individual fragments due to the face of improper loading of dye in the gel well before the start of the gel electrophoresis.

2 In Lane 2 it contains PV92 homozygous (+/+) control sample. In Figure 6.1, a strong single band can be observed in Lane 2 near the 1000 bp mark (when compared to the DNA standard).This clearly indicates that the Alu sequence is presence in both chromosomes that are 941 base pairs long.

3 For Lane 3, it contains PV92 homozygous (-/-) control sample. In Figure 6.1, a single band can be observed, and it is close to the 700 bp mark when compared to Lane 1. This means that Alu sequence is missing from both of the chromosomes which are 641 base paris long.

4 In Lane 4, it contains PV92 heterozygous (+/-). We are able to view the presence of 2 single bands in Lane 4. 1 is close to the 1000bp mark, and the other is close to the 700 bp mark. This proves that Alu sequence is present in only 1 of the either chromosome, with the one with Alu inserted is 941 base pairs long, and the one without Alu insert is 641 base pairs long. Thus, the presence of 2 bands being observed.

5 For Lane 5, it is the DNA sample of Group 1. Based on Figure 6.1, 2 bands can be observed. We can conclude that has the genotype of heterozygous (+/-).

6 For Lane 6 (My Group), from the results 1 band is observed. Thus, we can deduce that the DNA sample for my group has the genotype of Homozygous (+/+) and the Alu insert is presence for both chromosomes.

7 As for Lane 7, it contains result from Group 3. It has the same band appearing as Land 6, and we can also safely deduce that Group 3 has the DNA sample of genotype of a homozygous (+/+) and the Alu sequence is present in both chromosomes as the 1 single band appear near the 1000 bp mark.

8 Lastly, for Lane 8 which contains the DNA sample for Group 4, we deduce that it has the genotype of a homozygous (-/-) as it is near the 700 bp mark. This means that the Alu sequence is absent in both chromosomes.

Table 6.1-Exaplanation of Results

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7 Questions

PCR Amplification

1. Why is it necessary to have a primer on each side of the DNA segment to be

amplified?

Single-stranded DNA circle cannot serve as a template for DNA polymerase because the

enzyme cannot initiate the formation of DNA strand. It can only add nucleotides to the 3'

hydroxyl terminus of an existing strand. The strand that provides the necessary 3' OH

terminus is called the primer. The primers are pieces of DNA complementary to the DNA

template and it defines the starting point for replication. They are needed to start DNA

replication and are needed on both sides of the strands. Thus, primers will attach to the

beginning of the template before the polymerase can attach itself to the free 3' end of

the primer.

2. How did Taq DNA polymerase acquire its name?

Taq DNA polymerase is known as Thermophilus acuaticus. It is a polymerase purified

from a thermostable bacterium. Thermus aquaticus is a thermophilic bacteria which can

be found in hot springs at high temperature. This suggest that the high heat resistance of

the polymerase are able to withstand high temperatures and thus can be used in the

denaturation step of PCR.

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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?

Nucleotides are the building blocks of DNA and they are present in the master mix. They are

added one at a time to the primer in order for DNA polymerases to synthesize new

complementary strand from the DNA template. Thus, without them, DNA replication will

not occur.

There are also other components in the master mix and they are as followed:

- Deoxnucleoside triphosphates (dNPs): These are namely, Adenine, Guanine,

Thymine and Cytosine which are the raw materials and building blocks that are

required for DNA polymerases to synthesize new complementary strand from

the DNA templates.

- DNA polymerase: It is an enzyme that reads the DNA template strand assembles

the nucleotides into a new DNA chain via complementary base pairing. Taq DNA

polymerase is the DNA polymerase used in this experiment.

- Magnesium ions: Magnesium ions cofactors required by DNA polymerase to

sustain its the function in order to create the DNA chain.

- Oligonucleotide primers: These are fragments of DNA which are complementary

to the 3’ ends of the DNA template. They will anneal to the template strands and

serve as starting point for Taq DNA polymerase to begin replication.

- Salt buffer: Buffer solution is able to provide an ionic environment and pH so as

to optimize the reaction and to maintain the stability of the DNA polymerase.

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4. Describe the three main steps of each cycle of PCR amplification and what reactions

occur at each temperature.

Steps Description

Denaturation DNA template heated to approximately 94°C for 1 minute. This will lead

to the double-stranded DNA to separate as the high temperature will

cause the hydrogen bonds which keep the DNA strands together will

break. Thus, we used Taq DNA polymerase that can withstand the high

temperature during the denaturation step, thus it will not denature.

Annealing The annealing step is conducted at 60°C, and during this stage, the

primers are attached to their complementary sequences on each strand

of the double stranded DNA template. Ionic bonds between the primers

and the templates are constantly formed and broken. Thus, once the

polymerase attach to the template, the ionic bonds will be strong and

not break. The temperature of 60°C is the optimum temperature to

allow primers to bind on the template. When the temperature is below

the optimum non specific annealing of primers will occur and the

amplification of the undesired DNA sequence will occur. When the

temperature conducted is too high, annealing will also not occur.

Extension For extension step, the temperature is raised from 60°C to 72°C. This is

to attain optimal polymerase activity. Taq DNA polymerase adds

nucleotides one at a time to the primers attached at the starting point

to create the complementary copy of the DNA template.

The Taq DNA polymerase reads the DNA template from

the 3’ to 5’ end and adds nucleotides to the primer from the 5’ to 3’ end

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to create a complementary duplicate of the template..

A complete PCR amplification undergoes 40 cycles

Table 7.1- PCR Amplification Cycle Explanation

5. Explain why the precise length target DNA sequence doesn’t get amplified until the

third cycle. You may need to use additional paper and a drawing to explain your

answer.

Heat Denaturation

Primer Annealing

Primer Extension 1st

Cycle

2nd Cycle

3rd Cycle

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Figure 7.1-PCR Cycle

As seen in Figure 7.1, we are able to observe that amplification only starts at Cycle 3.

Cycle 1 & 2:

During the first 2 cycle, the DNA template undergoes heat denaturation. The previously

synthesized strand is used as the template. The primer binds to it and extension proceeds.

This time, the extra replicated region on the previously synthesized strand is not replicated.

Hence, the strand being synthesized will contain only the target sequence.

Cycle 3:

In Cycle 3, the target DNA sequence is being amplified from this cycle onwards. During the

first 2 cycles, it only involves the replication of parent strain, which includes the segment of

no interest.

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Gel Electrophoresis of Amplified PCR Samples

1. Explain the difference between an intron and an exon.

Introns do not code for translated protein and do not contribute to mRNA, thus its

known as the non-coding region. Introns will be eventually spliced out during RNA

splicing and the snRNPs is responsible for forming splicesome and to splice the introns

out. As for exon, it is the actual protein coding sequence and part of it will contribute to

mRNA and it determine how the proteins will look like afterwards. Once splicing occurs, it

will leave behind exons and join together to form a mature RNA.

2. Why do the two possible PCR products differ in size by 300 base pairs?

This may be due to the fact that there is Alu insertion. The PV92 region will be 941 base

pairs long if there is the presence of Alu insertion. However, when there is an absence of

Alu insertion, it will be 641 base pair long.

This clearly indicates a difference of 300 base pairs. Thus we can conclude that Alu insert

is only present in some individuals only.

3. Explain how agarose electrophoresis separates DNA fragments. Why does a smaller

DNA fragment move faster than a larger one?

Gel electrophoresis works by separating DNA fragments based on the size and charge of

the DNA fragments. When we first load the DNA samples into the agarose gel wall, and

they are at the negative ends of the gel box. DNA fragments, being negatively charged,

will move across the gel due to the release of hydrogen ions from their phosphate group.

When current is passed through the electrodes, the DNA fragments will then move across

the positively charged anode. Smaller size DNA fragments are able to migrate faster

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because it can move through the gel faster as it faced less resistance from the gel matrix.

As the gel matrix is like a molecular sieve, larger fragments will face more resistance and

move at a slower rate.

4. What kind of controls are run in this experiment? Why are they important? Could

others be used?

The controls used in this experiment are:

- PV92 Homozygous (+/+)

- PV92 Homozygous (-/-)

- PV92 Heterozygous (+/-)

These above controls are important in the experiment as they allowed us to make

comparisons for the DNA samples to determine the presence of Alu insert. We are thus able

to deduce the type of genotype of our sample. We are unable to use other types of control

as our main aim is to find out whether the DNA samples contains Alu insert. .

5. What is your genotype for the Alu insert in your PV92 region?

From the results in Figure 6.1, the genotype for the Alu insert for the DNA sample of my

group is a Homozygous (+/+). Lane 6 contains the DNA sample of my group, and 1 thick

single band can be observed. Hence, the possibility of genotype of Heterozygous (+/−)

can be eliminated. When the position of the band is compared with the DNA standard in

Lane 1, it can be seen that the band is near the 1000 base pairs mark. The PV92 region

with Alu sequence present is 941 base pairs which is approximately 1000 base pairs.

Thus, it can be concluded that the genotype is a Homozygous (+/+).

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6. What are the genotypic frequencies of +/+, +/- and -/- in your previous group

population? Fill in the table below with your previous group data.

Catergory Number Frequency (# of Genotypes/Total)

Homozygous (+/+) 2 2/4

Heterozygous (+/-) 1 ¼

Homozygous (-/-) 1 ¼

Total = 4 1.00

Table 7.2- Observed Genotypic Frequencies for the Class

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8 Discussion

Sources of ErrorsErrors Ways to improve

Concentration of Magnesium (Mg2+) ions

Mg2+ ions are the cofactors for the Taq polymerase.When the concentration of Mg2+ is high, it will not stabalise primer to the correct binding sites. This will lead to the wrong extension of DNA sequence. However, through trial and error method to determine the optimum concentration of Mg2+ used for PCR beforehand.

Contamination of DNA Samples

Contamination may occur when we are collecting the DNA samples from our hair. Another way to prevent contamination is to conduct the collection and harvesting of DNA under the fumehood.

Table 8- Sources of errors

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9 Conclusion

After conducting this experiment, we are able to understand the principles of PCR

amplification and gel electrophoresis.

For PCR, 1 main emphasis is on how specific segments of DNA samples can replicated and

multiplied from just a trace amount. And this technique is indeed useful for forensic and

biological research. After observing how to extract DNA and how the amplification process

takes place allows me to further appreciate the beauty of PCR.

As for gel electrophoresis, it allows us to learn how different sizes of DNA fragments can be

separated in this setup. In addition, it also enables us to identify the types of genotype for

the DNA, which can be done by using the correct types of controls. Moreover, we also

gained hands on experience by performing gel staining in order to view the bands on the

gel, also also proper loading method of loading the DNA samples and dyes into the gel well.

Lastly, based on our experiment results, we have also concluded that Alu element is not

present in all individuals after performing PCR followed by gel electrophoresis.

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Sample Lab Report