The Different Types of Spontaneous Mutation - Essay Example

Question 2 (50 marks) (a) Briefly describe the different types of spontaneous mutation that can change the nucleotide sequence of DNA, and the significance of these mutations. Spontaneous mutation refers to the changes in nucleotide sequence of a gene without any effort or it can be said that in spontaneous mutation changes in nucleotide sequence of gene suddenly occurs. There are three types of spontaneous mutation: 1) Substitution Mutation (Transition Mutation, Transversion Mutation) 2) Spontaneous deamination 3) Frameshift Mutation Substitution Mutation: Substitution mutation is the most common mutation due to substitution of base pair in DNA. During replication of DNA repair, wrong base pairs are incorporated. Base pair substitution is of two types, Transition & Transversion. Transition base pair substitution This type of substitution takes place due to Tautomerism. Tautomerism is the relationship between two structural isomers that are in chemical equilibrium and readily change into one another. Transition substitution resulted in substitution of one pyrimidine by another (Cytosine to Uracil) or one of the purine (Adenine or Guanine) by another. Transversion Base Pair Substitution Transversion mutation involves the substitution of purine by pyrimidine and pyrimidine by purine. This type of mutation is rare due to steric problems of pairing purines with purines, and pyrimidine with pyrimidines. Spontaneous Deamination: This type of mutation takes place due to chemical modification in a base. The chemical modification is caused due to two reactions one is deamination and second is depurination. In deamination Cytosine occasionally loose an amino group and then it is converted to Uracil. This resulted in pairing of Uracil-Guanin instead of Cytosine-Guanin. It is well known that Uracil pairs with Adenin. After Uracil-Adenin pair in first round of replication, thymine takes the place of Uracil in second round of replication. And finally Cytosine - Guanin pair is converted to Adenin - thymine Pair. In depurination loss of purine group (Adenine or Guanin) takes place. Frameshift Mutation: Due to deletion and insertion of one or a few nucleotides in DNA molecule, the shifting of reading frame of nucleotide takes place either in backward or in forward position, this results into a mutation. This is known as frameshift mutation. Generally this mutation occurs where there is a short repeated nucleotide sequence. Though mutations are harmful to an organism, but it is necessary to generate variability in organisms and contribute to the process of evaluation in nature. (b) Using examples, outline three ways in which mutagens can act. (6 marks) Any agent that directly cause damage to the DNA alters the base sequence or interferes with repairing system is known as mutagens. Mutagens may be some chemical or radiation. Here we are discussing how the mutagen can act: 1) Base Analogues: A base analogue is a chemical compound similar to the one or four bases of DNA. During normal replication process this base analogue can be incorporated into growing polynucleotide chain. For Example 5-Bromouracil (5-BU) is an analogue of thymine. It can pair with adenine in polynucleotide chain by replacing thymine. During the replication keto-form of 5-BU Substitutes for thymine and the replication of an initial Adenin:Thymine (A:T) pair becomes an A:BU pair. The rare enol form of 5-BU that pairs with Guanin is the first mutagenic round of replication. In the next round of replication Guanin pairs with cytosine. So that transition is completed from A:T to G:C pair. 2) Chemical Changing the Specificity of Hydrogen Bonding. There are many chemicals that after incorporation in DNA change the specificity of hydrogen bonding. For example Nitrous Oxide (HNO2) converts the amino group of the bases into keto group through oxidative deamination. Deamination of adenine results in formation of hypoxanthine, the pairing behavior of which is like guanine. Hence it pairs with cytosine instead of thymine replacing AT pair by GC Pairing. 3) Radiations Radiation is a physical mutagens. Radiations damaging the DNA molecule fall in the wave length range below 340nm. Due to absorption of ultraviolet radiations, DNA leads to cross linking, single strand breaks and base damage. Some other types of radiation like X-ray cause breaking of phosphate ester linkages of DNA helix. (c) If any of the proteins E1-E5 were able to recognize and cleave the DNA at the mismatch site, in what ways would the resultant fragments differ The protein used for detection of mutation in DNA is known as endonuclease Enzyme. This enzyme can also be used for cutting DNA. This enzyme finds a Specific sequence on heteroduplex DNA, where there is a mismatch. Here we are discussing the mutation detection in hetero duplex DNA, in which one strand appears green due to IRD 700 and the other strand appears red due to IRD 800. This heteroduplex DNA had a T/G mismatch. The heteroduplex DNA was incubated with endonuclease for 20 Hours at 42C. After incubation DNA gel clearly showing the mismatch point. Protein E1 and E5 were able to recognize and cleave the mismatch sites. The resultant fragment is differing in the pairing due to thymine and guanine. While the pairing should be in between Cytosine & Guanin. (d) The lanes for proteins 1 and 5 show two bands (labelled with white circles) that are not seen in the lanes for proteins 2, 3 and 4. What does this result indicate about the properties of proteins 1 and 5 compared with proteins 2, 3 and 4 From the gel, estimate the size of the larger product (z in Figure 1a). The lanes for protein 1 and 5 shows two bands that are not seen in the lanes for protein 2,3, and 4, because protein 1 and 5 is having such a chemical property which determines the mismatching sequence in heteroduplex DNA. The larger product size (z in figure 1a) is (228+444)-204=468bp. (e) (i) Why were the T/A and C/G combinations at position 229 included in this experiment Answer : To compare the results of mismatch pairs like A/A, T/T, C/C, G/G, A/C, T/C, A/G, T/G with the matched pair A/T and C/G, these match pair will serve as control. If we see the result in figure 2b, we found that each mismatch pair is showing reaction against endonuclease E1 and E5 as black circles, while match pair is not showing reaction against E1 and E5. (ii) In the course of DNA replication, what type/s of mutation event would result in the anomalous pairings of A/A and A/G In the course of DNA replication frameshift mutation may result in the anomalous pairing of Adenine:Adenine (A/A) and Adenine: Guanine(A/G). (iii) 'describe in detail the specificity of endonucleases in E1 and E5 shown in the results of this experiment. Endonucleases E1 and E5 has the specificity to recognize mismatch pair in a heteroduplex DNA. These enzymes have a specificity to cut the double stranded DNA at the mismatch pair position. During breaking of DNA endonucleases enzyme breaks the sugar -phosphate bond. While recognizing the mismatch pair, endonucleases enzyme check for at least 4 to 8 nucleotides in one go. Some times the sequence of nucleotide is palindromic as follows: 5' ---- C A T A T G ---- 3' I I I I I I 3' ----- G T A T A C ---- 5' The above sequence is palindrome sequence. There are five types of endonuclease E1 to E5. Type E1and E5 are generally large, multisubunit complexes containing both the endonuclease and methylase activities. (iv) What reservations might you have about the data shown The worries about the data shown into the figure, is only the visual inspection. During visual inspection of DNA gel, analyzer has to take care of the small tiny spots which are due to mismatches. If such spots are left then it may leads to incorrect data gathering. (v) Are there any other observations that could be made about the results shown in this figure (2 marks) Yes the observation of controls make a great impact on the result. Good & right Result of control conforms the reading of Mismatch pairs. (f) What do the results shown in Figure 3 show about the properties of E1 and E5 Why are three bands present What control experiment should the experimenters have included, and what would you have expected it to show Figure 3 demonstrate the reliability on endonuclease E1 and E5, which are responsible for cleavage of DNA. In this experiment it is try to prove that E1 and E5 can easily detect insertion mismatches. Control experiments are also carried out along with the sample experiment to rely on the integrity of the experiment. In this case control experiment provides the expected results, because control experiments are done for comparison purpose only. Read More