molecular bio tech

Question 1

What is Polymerase Chain Reaction? (PCR)

Answer 1

It amplifies DNA from a limited source of DNA so that there is sufficient amount for analysis

Question 2

What are the components of the PCR?

hint: x5

Answer 2

*Template DNA
➔DNA containing the target sequence
to be amplified

* Primers
➔synthetic single-stranded DNA
 fragment (20-30 nucleotides long)
➔needed to initiate DNA synthesis by
 providing a free 3’OH group for Taq
 polymerase to bind to and extend
➔ 2 different primers are required.
 Each is complementary to the
 sequence at 3’end of each single
 stranded target DNA sequence.

➔ are required in large excess to
 increase the likelihood of them
 binding to target DNA sequences
 (instead of target DNA sequences
 reannealing)
➔ become part of amplified sequences

• Taq polymerase
  ➔ thermostable DNA polymerase
  which is resistant to denaturation at
  high temperature
• Deoxyribonucleotides (dNTPs)
  ➔ substrates for DNA replication made
  up of dATP, dTTP, dCTP and dGTP
• Buffer
  ➔ contains cofactor, Mg2+
  , for proper DNA polymerase function

Question 3

What is the 3 step process of PCR?

Answer 3

Denaturation

1. Heating to 95C causes the weak hydrogen bonds* between complementary bases of each
   strand to break due to increased molecular vibrations;
2. The denaturation of double-stranded DNA into single-stranded DNA exposes the bases for
   complementary base pairing;

Primer Annealing
3. Lower temperature of 64C allows primers* to anneal* specifically to the regions flanking the
target DNA sequence via complementary base pairing*;
4. The primers determine the segment to be amplified and provides a free 3’-OH group for chain extension;

Extension
At the optimum temperature of Taq polymerase* , 72 oC, which performs the synthesis of complementary* DNA strand;
6. Chain extension/ elongation* occurs from 3’ end of primer which provides free 3’ OH required
by polymerase;

Question 4

What are the advantages of PCR?

Answer 4

1. Only a minute amount of DNA* is required to carry out PCR as with each round of PCR, the
   number of copies of target DNA is doubled. Thus the number of desired sequence increases
   exponentially and there will be sufficient DNA for analysis.**
2. Use of thermostable* (i.e. resistant to denaturation at high temperatures*) Taq polymerase
   allows PCR to be automated so DNA can be amplified very quickly.

Question 5

What are the limitations of PCR?

Answer 5

1. Taq polymerase lacks 3’ to 5’ proofreading ability**. Hence errors occurring early in the PCR
   reaction will get compounded* with each subsequent replication cycle.
2. Knowledge of sequences flanking* (i.e. at the 3’ ends of) the target sequence** is required in
   order to design appropriate primers.
3. Taq polymerase tends to ‘fall off’ the DNA template before chain extension is complete if the
   strand is too long. Hence there is a limit to the size*** of DNA fragment (~3kb) to be amplified.
4. Minute amounts* of contaminant DNA* can be exponentially amplified along with target DNA*
   and affect the reliability* of the results.

Question 6

What is the purpose of agarose gel

Answer 6

Separate DNA based on fragment size**

Question 7

Describe the steps of agarose gel electrophoresis

Answer 7

1. A slab of agarose gel is placed in a buffer solution* contains ions* which allows the conduction of electricity** when the current is turned on
2. The DNA sample is mixed with a dense loading dye* containing glycerol* & 2 coloured dyes. Glycerol makes the DNA sample denser than the buffer solution so that the DNA sample can sink to the bottom of the well
3. Since DNA is invisible, the dyes colour the DNA sample and will indicate if the DNA has been loaded correctly into the well***
4. The 2 coloured dyes thus act as visual markers* which help to monitor the progress of the migration* of the invisible DNA fragments in the gel during electrophoresis
5. One dye (corresponds to a 100bp DNA fragment) and often runs ahead of the DNA sample* and gives an indication of when gel electrophoresis must be stopped so that the samples do not run out of the gel. The other dye (corresponds to a 1100bp DNA fragment) and gives an indication of the position of the larger fragments on the gel
6. DNA samples are pipetted into the wells* in the gel near the negative electrode*
7. A DNA ladder (i.e. DNA molecular weigh markers) which contains DNA fragments of known sizes, is run in one of the lanes and acts as a standard for which to compare fragments of unknown size* in the sample
8. Negatively charged DNA* is attracted towards the positive electrode (anode) when subjected to an electric current**
9. The agarose gel** matrix made of a meshwork of polymer fibres** which impedes movement* of longer fragments more than shorter fragments. The longer fragments thus migrate more slowly* compared to shorter fragments, leading to a banding pattern observed on the gel
10. Before the loading dye reaches the end of the gel, the current is turned off
11. To visualize the bands***, the gel can be treated with a a staining dye that binds DNA (e.g. ethidium bromide, a carcinogen) and fluoresces under UV light.
12. Thus
    a) the fragment size* can be estimated (based on position* of the band relative to bands in the molecular weight marker) and
    b) the amount of DNA** can possibly be estimated (based on intensity and thickness* of the band)

Question 8

What is southern blotting

Answer 8

Tool to detect specific nucleotide sequences within a sample of DNA

Question 9

Describe the procedures of Southern blotting and nucleic acid hybridisation

Answer 9

(Continued from Gel electrophoresis)
1. Gel slab is placed on top of the sponge and under a nitrocellulose membrane. A stack
of paper towels placed on top of nitrocellulose membrane. These are placed in a tray of
alkaline solution. A heavy weight is placed above the paper towels.

2. Absorbent paper towels draw the solution towards themselves and the alkaline solution**
   denatures* double-stranded DNA* into single-stranded DNA*
3. Single stranded DNA on the gel is then drawn upwards** onto the nitrocellulose
   membrane* and binds to the membrane* (in exactly the same position as they were in
   the gel).
4. Nitrocellulose membrane is removed and incubated with single-stranded, radioactive*
   DNA probe which hybridises* via complementary base pairing* to part of the target
   sequence.
5. The excess unhybridised probes are washed off.
6. Autoradiography* is performed placing X-ray film** over membrane. Radioactive regions
   exposes (& hence blackens) the film* forming an image that correspond to the bands
   that have base-paired with probe.

Question 10

What are Restriction Fragment Length Polymorphisms (RFLPs)?

Answer 10

RFLPs are unique banding patterns** among individuals when their DNA is digested by restriction enzymes** and separated by gel electrophoresis ( and subjected to Southern hybridisation)

Question 11

What variations are there in RFLPs

Answer 11

• RFLPs arise due to the polymorphic nature of DNA in the different individuals, there will be variations* in the
  1. number/location of restriction sites*
  or
  2. number of tandemly repeated nucleotide sequences* (e.g. short tandem repeats)
  This will result in unique banding pattern*

Question 12

What type of DNA polymorphism is present in sickle cell anaemia?

Answer 12

• a single nucleotide polymorphism*
• difference in a single base pair* due to a point mutation
• in single cell anaemia, this SNP is within the coding region

e. g. in sickle-cell anaemia, the disease-causing mutation occurs at restriction site for Mst II within the beta-globin gene.
1. In the disease causing allele, the Mst II restriction site is eliminated.
2. In the normal allele, Mst II restriction site is retained.
- However the majority of SNPs* used for RFLP analysis are found in non-coding regions

Question 13

Describe the use of RFLP analysis in determining the genotype of a person for B-globin gene?

Answer 13

Digest genomic* DNA from both samples with MstII to obtain different restriction fragments will arise. (i.e. cut both DNA samples with the
SAME* restriction enzyme, Mst II)

2. Perform gel electrophoresis* (which will produce smears on the gel) followed by southern blotting, probing* and autoradiography*
   (which will allow the visualisation of distinct bands).
   The same single-stranded radioactive probe* (which is complementary to part of the target sequence) will be used to detect both the
   presence of A-type and S-type DNA. Hence the site that the probe binds on the DNA fragments should give different banding patterns on
   the autoradiogram

Question 14

Explain the use of RFLP in DNA fingerprinting

Answer 14

• As no individuals (exception of twins) have the same genome, therefore they will not have the same DNA profile.
• The DNA profile is the unique banding pattern that identifies individuals.

Question 15

How to carry out DNA fingerprinting?

what are the 2 diff steps

Answer 15

1. Restriction digestion** of genomic DNA** by restriction enzymes** ( cut DNA from different individuals with SAME restriction enzyme)
   followed by gel electrophoresis to separate DNA fragments***
2. Southern blotting** using of nitrocellulose membrane.**
3. Nucleic acid hybridisation* using radioactive probes complementary to the STRs
4. Visualisation of bands** via autoradiography* using X-ray films**

OR

1. PCR using primers that flank target DNA sequence
2. Restriction Digest
3. Gel electrophoresis
4. Bands visualised by staining gel with ethidium bromide

Question 16

What is meant by a restriction enzyme [3]

Answer 16

1. Enzyme that recognizes and bind** a specific* 4-6 base pair DNA sequence* called a
   restriction site* as its active site is complementary to the DNA sequence; Note:
   Binding must be mentioned.
2. Enzyme cuts/breaks phosphodiester bonds** on specific positions on both DNA
   strands;
3. Giving rise to either blunt or sticky ends* depending on type of restriction enzyme;
4. Used as a defence mechanism by bacteria against bacteriophage by cutting up
   foreign DNA*, hence restricts multiplication of viruses;

Question 17

What are sticky ends

and what are blunt ends?

Answer 17

Sticky ends:
- produced when restriction enzymes leave a staggered cut resulting in single-stranded overhangs/ sticky ends

• short overhangs will form H bonds and anneal by complementary base pairing with complementary singe-stranded stretches on other DNA molecules cleaved on the same site

Blunt ends: produced when restriction enzymes make a simple cut across both strands at a single point

Question 18

How gel electrophoresis is used to separate fragments of DNA. [5]

Answer 18

1. Dense* loading buffer is mixed with DNA sample to help it sink* to the bottom of
   wells located nearest the negative electrode/cathode**;
2. Loading dyes also added to DNA sample to allow visualisation** of progress of
   electrophoresis;
3. Negatively-charged** DNA migrates out of well towards direction of positive
   electrode/anode when subjected to an electric field / current;
4. Meshwork* of agarose polysaccharides impede movement of longer fragments
   more*** than shorter fragments
5. causing them to migrate slower** than shorter fragments and end up nearer to the well**;
6. A method of visualization is suggested. Either stain gel with ethidium bromide
   followed by visualization under UV light. Or Southern Blot *followed by use
   radioactive DNA probe** and autoradiography** with an X-Ray film**;

Question 19

Q1/ tutorial: With reference to Fig. 1.2, explain the banding pattern of individual C. [4]

C - contains 3 bands
B- contains 1 band
A- contains 2 bands

Answer 19

1. C is heterozygous for sickle cell
2. Normal and mutant alleles are on the same gene loci on a pair of homologous chromosomes
3. Normal allele gives rise to an intermediate and short fragment* upon restriction digestion with MstII and these fragments correspond to the middle and bottommost bands respectively
4. Mutant allele* gives rise to a single long fragment** upon restriction digestion with MstII and this corresponds with the uppermost band*

Question 20

IV6 was also found to suffer from the disease, even though genetic screening using the RFLP
marker in question identified him as being only a carrier. Suggest a possible explanation for
this. [2]

Answer 20

1, During formation of gametes in III4 *;

2. Crossing over* could have occurred at a locus between disease allele and RFLP
   marker, such that disease allele became linked to RFLP allele 3;
   IV6 therefore inherited 2 copies of disease allele.
   NB: This is one limitation when studying RFLPs flanking the gene responsible for the disease.
   However, such RFLPs are still used in genetic screening because they lie very close to the
   gene of interest, and therefore crossover frequencies tend to be very low.

Question 21

Explain why (QV: 1. As number of repeats in STR increases from 7- 12， temperature needed to
separate the double strands increases from 53OC (below 55 OC ) - 63.5OC
2. temperature was increasing more slowly as the number of repeats increased )

Answer 21

3. With an increasing number of STRs** there would be an increase in length** of
   polynucleotides
4. means larger number of H-bonds** between bases of the two strands to be broken
5. Hence, more heat energy** needed to break more bonds** and separate strands

Question 22

Outline the process of DNA hybridization that allows the RFLP for a particular gene to be visualised. [5]

Answer 22

1. ds DNA is denatured / made single-stranded and by alkaline / NaOH solution and
   transferred to a nitrocellulose membrane - exactly the same position as they were in the
   gel;
2. Nitrocellulose membrane incubated with a radioactive probe, that is complementary in
   sequence* to part of target sequence / gene.
3. DNA fragments containing this part of target sequence will hybridise to probe* by
   complementary base-pairing**;
   4(!!) . After hybridisation, membrane is washed to remove any unhybridised probes.
4. Using autoradiography**/X-ray film over membrane, banding pattern can be visualised.
   (Radioactivity of bound probes exposes film to form an image corresponding to bands that
   have base-paired to probe.)

Question 23

Describe the role of the buffer solution in the gel electrophoresis protocol. [2]

[2] - 2 points

Answer 23

1. Buffers contain ions which allows conduction of electric current
2. Thus allowing negatively-charged DNA** molecules to move from negative
   electrode to positive electrode

TAKE NOTE OF POINT 2

Question 24

In June 2002, over 500 ivory tusks and 42 000 other ivory items were seized from a ship
arriving in Singapore. Evidence seemed to link this ivory to elephants from Malawi in Africa.
(i) Outline the process of genetic fingerprinting using RFLP that could be used to test this
seized ivory. [4]

Answer 24

1. Genomic DNA is extracted* from soft tissue/dried blood from seized samples and
   Malawian elephants, and cut with same restriction enzyme**;
2. DNA is separated according to size in gel electrophoresis where negatively-charged DNA** migrates towards positive electrode/anode when subjected to an
   electric field / current;
3. Meshwork of agarose impedes movement of longer fragments more than shorter
   fragments* resulting in longer fragments migrating slower than shorter fragments* and
   end up nearer to the well*;
4. ds DNA is denatured* by alkaline / NaOH solution and
   transferred to a nitrocellulose membrane*;
5. Carry out Southern blotting/nucleic acid hybridization by incubating membrane
   with single strand radioactive probe* which will hybridise with DNA fragment
   through complementary base pairing***;
6. Using autoradiography/X-ray film*** over the membrane, the banding pattern can visualised

Question 25

(ii) Explain how the genetic fingerprints of the seized ivory could be used to confirm that it
originated from elephants in Malawi. [4]

Answer 25

1. Genetic fingerprint is due to different alleles* producing different bands* in gel resulting in unique banding pattern in individuals;
2. Different bands arise due to polymorphic* nature of DNA in different individuals, there
   will be variations in number and location of restriction sites* and number of tandemly
   repeated nucleotide sequence* among individuals.
3. Genetic fingerprint of animals that provided ivory can be compared against*
   fingerprint of elephants from Malawi to see how closely related they are.
4. If fingerprint pattern is similar to pattern to that of Malawi elephants, then ivory haul
   is from Malawi.

Question 26

Explain the role of primers in the PCR [2]

Answer 26

1. Primers are short, single stranded DNA sequences that can anneal to the target DNA sequences;
2. they are complementary* to the regions flanking the gene of interest and thus determine the
   segment to be amplified;
3. they provide a free 3’-OH group for chain extension by DNA polymerase so that the gene of
   interest can be amplified;

Question 27

Explain why marker DNA was included in this experiment? [2]

Answer 27

1. The marker DNA consists of a mixture of DNA fragments of known sizes***;
2. It acts as a standard that the fragments of unknown sizes from the DNA samples can be compared with to estimate their fragment size**

Question 28

TnR 11 q2(e)/
The above analysis used a single VNTR locus. Suggest why DNA profiling labs normally compare 13 such loci when conducting investigations?

Answer 28

1. When using a single locus, it is possible for individuals to have the same banding pattern, especially if they are related
2. using 13 loci will give rise to unique DNA banding patterns that enable investigator
3. probability of 2 individuals with identical DNA banding profile is extremely low
4. able to distinguish between individuals thus allowing accurate identification of the rapist
   - does not increase reliability!!