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Question 1
Below is a photograph of gel electrophoresis result obtained from the experiment.
Figure 1
L G-NTC G-DNA G-P H-NTC H-DNA H-P
The gel electrophoresis result showed 14 visible bands from the lane loaded with the ladder. No visible band was observed on the lane containing G-NTC sample. Visible bands of G-DNA, G-P, H-DNA, and H-P samples were also produced. The bands of G-DNA and G-P samples produced coincide with eleventh band on the lane loaded with the ladder. The size of this band is 400bp. Consequently, the size of G-DNA and G-P products is 400bp. On the other hand, bands of H-DNA and H-P coincide with the twelfth band on the ladder lane. The size of this band is 300bp. Therefore, the size of H-DNA and H-P is 300bp. There was no visible band on the lane loaded with H-NTC sample.
Question 2
Figure 2
- Below are two gels: the one on the left is the gel provided while the other one is my gel.
Band 2817 is visible for all the reactions in my gel and the gel provided.
Six bands are visible on the lane loaded with the ladder. The number of bands in the other lanes is as shown below:
- Cs: 3 visible bands
- S1: 4 visible bands
- S2: 4 visible bands
- S3: 4 visible bands
- S4: 5 visible bands
- S5: 4 visible bands
The number of bands formed from any given lane depends on the extent to which the restriction enzyme added digests the sample loaded on the lane. This implies that S4, which forms the highest number of bands on the gel, is the most completely digested sample among the six samples.
In each lane, the band furthest from the loading well contains the shortest DNA strands. On the other hand, the bands nearest to the loading wells on the gel contain the longest DNA strands. This is due to the fact that short DNA strands more faster and more easily through the gel than long ones.
Electrophoresis result shown above reveals that the restriction fragment profile of suspect 3 matches the profile obtained from the crime scene. This implies that the sample of blood collected from the crime scene belonged to the suspect. Consequently, the suspect qualifies for prosecution.
Question 3
Principles of DNA gel electrophoresis
Gel electrophoresis is a technique widely employed in molecular biology and biochemistry to separate macromolecules. Besides, the technique can also be applied in purifying macromolecules (Garrett & Grisham, 2013). The macromolecules separated by this technique are mainly nucleic acids and proteins. A solid support medium is needed for separation of fragments of DNA. The solid support medium commonly used is agarose gel. Polyacrylamide can also be used. However, agarose gel is preferred for being non-toxic and for its ability to separate wider range of DNA fragments than Polyacrylamide. An electrophoresis buffer is also needed to provide ions for transfer of charges. Besides, the buffer helps in maintaining PH of the gel.
Macromolecules contain charges (Hochachka & Mommsen, 1995). Whereas protein can have either positive or negative charge, nucleic acids have only negative charges. During electropholerisis, the molecules are placed in an electric field. Given that they contain charges, they tend to move towards charges negative to the ones they contain. Therefore, nucleic acid molecules move towards the anode. This causes the molecules to be separated along the gel based on the size of each. Shorter strands of DNA are move further from the wells on the gel where samples are loaded. On the other hand, longer strands cover short distance from the loading wells.
Question 4 Analysis of results from Practical 4
Suppose a sample whose concentration is 0.25 Mm produces an absorbance of 0.320 as shown in the results for practical 4, the product concentration and velocity can be calculated as shown below:
- Product concentration.
Using Beer Lambert Equation and assuming that path length is 0.1cm;
A = Ԑ*C*d
C = A/( Ԑ*d)
C = 0.320/(3600*0.1)
C = 0.320/360
C= 0.00089mmolL-1
- Velocity
Velocity = product concentration(C)/ reaction time in minutes.
Velocity = 0.00089mmolL-1/7min
Velocity = 0.0001271mmolL-1min-1
The Vmaximum is likely to be 0.00025714mmolL-1min-1. On the other hand, KM can be estimated as 0.75mM
Question 5
Question 6
- Results obtained from practical 6
- Vmax of the reaction with enzyme inhibitor is lower than that of reaction without inhibitor. This implies that the inhibitors involved are competitive inhibitors.
References
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