MCB problem 6 quiz

1)

a) A normal cell with no genetic defects was exposed to large amounts of a DNA-damaging chemical (chemical X) for 5 minutes. After a few hours, the cell died. Based on what you had learnt today, account for the death of the cell. (5 marks)


b) Some environmental pollutants (e.g. cigarette smoke) can cause adducts to form on DNA. These are organic molecules that bind to DNA, causing a bulge in the molecule. Suggest how these adducts can interfere with cell replication. (5 marks)






a) The reason why the cell will die is because the DNA has been denatured. In today's lesson, we learned that there can be many possible ways of DNA denaturing. Firstly, it could be due to a foreign molecule attaching to the bases, and not allowing the DNA polymerase or helicase to pass through the DNA for transcription or for replication. Secondly, it could be that there is a cross link, which causes pyrimidine dimers to occur. These cross link can be caused by the UV rays emitting photons to the DNA. When this occurs, the sugar-phosphate backbone will also get distorted too. In this case, it is able to repair when the photolyase are able to remove the pyrimidine dimers. Thirdly, it will cause the sugar-phosphate backbone to break. In this case, the DNA ligase IV is able to join the sugar-phosphate backbone back. Forthly, it is the Base mismatch. The DNA glycosylase can recognize the mismatch base site and remove its base. Then, the endonuclease removes neighboring nucleotides, and the gap is filled by DNA polymerase I and DNA ligase.
The death of the cells could be due to the fact that the cell do not have enough immunity. Like for example maybe it does not have enough of the repairing mechanism. Like for example, in the problem statement, we could see that the one with a lack of DNA ligase IV cannot survive very long. Therefore, it does not have the ability to do faster repairing. Also, the damage rate is more than the repair rate, therefore, the rate of survival will keep dropping.
b) In cell replication, helicase will firstly go through the DNA, and it will split the double strand into a single strand, and therefore it will start reading a single strand coding, and therefore it will then replicate mRNA. In order to replicate mRNA, the helicase needs to go though the DNA, and with the molecule obstructing the pathway, the helicase might not be able to split the double strand.
Also, in the replication process, where the DNA will replicate from a single strand to a double strand by the DNA polymerase III. The DNA Polymerase III will not be able to read the complete sequence of the DNA, and with the molecule obstructing the reading, it will not be able to go on to complete the full strand coding, therefore the DNA sequence will not be complete.

MHC

a) In the MHC Class I, the antigenic proteins which enters the Antigen Presenting Cell will firstly undergo degradation by a protein called protease. After degrading, it will be transported by TAP, and then the antigenic peptide will be loaded onto the Class I MHC Molecule. The MHC then move to the cell surface and display the antigenic peptide. The T killer cell bind onto the MHC molecule, and the killer T cell will then be activated, and then it will be able to kill the infectious foreign particles directly.

b) In the MHC Class II, when the antigens enter the Antigen Presenting Cell, phagosomes are formed, and the phagosomes containing the antigens are merged with the lysosome. Then, the MHC class II molecules migrate into the phagolysosome where the antigen is being degraded, and the invariant chain is removed for peptide binding. The Class II MHC molecule then moves to the cell surface, and present the antigenic peptide to the helper T-cell, and it will help to activate immune cells to fight against the foreign infectious particles.

DNA Folding (School revision)

1) Firstly, the long strand double helix DNA coil a few rounds around the octomer histones. The octomer histones are positively charged.

2) In order to stabilise the DNA being uncoiled from the octomer histone, another protein called the H1 histone will attached onto the DNA and the octomer like a scotch tape, to stabilise it. The ending product is called nucleosome.

3) DNA continued to form more and more nucleosomes. After that, these nucleosomes in a long chain will coil itself like a spring shape. It will then be known as solenoid.

4) The solenoid then attach to another protein called scaffold to form loops. That is called the loop solenoid.

5) It then made up into chromosomes.

(WARNING: Information may be incorrect. If there are any incorrect information, or anything more to add, please give comments. )

DNA Replication (MCB school revision)

Okazaki fragment - short fragment of DNA.

Helicase - separates the double strand.

Single-strand binding protein – prevents single strand coiling back to original shape.

DNA Polymerase synthesize only 5’ to 3’.

5’ to 3’ (direction of DNA polymerase) = Leading strand (3’ to 5’ old strand, 5’ to 3’ new strand, direction towards helicase)

5’ to 3’ (direction of DNA polymerase) = Lagging strand (5’ to 3’ old strand, 3’ to 5’ new strand, direction away from helicase)

Leading strand –

1) RNA Primase - attach a RNA Primer onto the strand
2) DNA Polymerase 3 - adds deoxyribonucleotides to synthesize the new complementary strands of DNA.

Lagging Strand –

1) RNA Primase – attaches a RNA Primer
2) DNA Polymerase III - adds deoxyribonucleotides
3) DNA Polymerase I – removes the RNA and replace it with DNA
4) DNA ligase – Forms phosphodiester bonds

Molecular and Cell Biology (School revision) - Transcription and translation

In order to undergo the central dogma, the DNA has to undergo transcription and translation.

In transcription, the RNA Polymerase splits the double strand into a single strand. The RNA Polymerase will then bring the nucleotides, and then the nucleotides will firstly find the start codon. After finding the start codon, the RNA Polymerase will then begin to bind the nucleotides to form a single strand. As the RNA Polymerase moves along to form a longer RNA strand, the front part of the mRNA will detached from the single strand, and after the RNA Polymerase reaches the stop codon, the mRNA strand will be formed.

The mRNA strand will then move out of the nucleus to the cytoplasm, and then translation will occur.

For translation, firstly, the ribosome will bind onto the mRNA strand, and then it will also find the start codon. After it found the start codon AUG, it will then start to read.

As it moves along, the tRNA with the anti-codon on the other side will then bind onto the codon. On the other end, it has the amino acid. As the ribosome moves along, there will be more tRNA binding onto the codon.

As the ribosome moves along, the tRNA on the front part will started to detached with the mRNA strand, and leaving behind the amino acid. The ribosome will also help to form the peptite bond of the amino acid.

After the ribosome reaches the end codon, the amino acid will then be formed.

In order to form protein, there will be more amino acids formed, which will then made up proteins.

Immunology - Complement System C1 to C9

In Complement System, there are different types of proteins in the complement system. These proteins are named from C1 to C9.

Firstly, the antibody will bind onto the infectious particle. After that, C1 will attached onto the antibody.

After attaching onto the antibody, C1 will then be activated, and then it will split to form C2a and C4b fragments.

When these 2 fragments joined together, it will then form the C3 convertase.

C3 convertase is able to cleave C3 into C3a and C3b.

C3a is able to cleave C5 into C5a and C5b.

C5b, C6, C7, C8, and C9 are able to attach itself onto the infectious cell membrane, and therefore opening a hole on the membrane, and allowing the infectious cell content to lysate, and it leads to self-destruct. This process is known as MAC, which is known as Membrane Attacking Complex.

C3b is able to attach itself to the infectious cell, and therefore it will help in opsonisation, and it will then undergo phagocytosis.

Clonal Selection (School Revision) Immunology

Clonal Selection is the process of forming new antibodies. When there is a new antigen(foreign infectious particles) invading our body, it will trigger the immune system. The T-cell will then activate the B-cell. After the B-cells are being activated, it will start to undergo a few cell divisions. After going through many cell divisions, it will form many memory cells and plasma cells.

The functions of the memory cell is to store the information of the specific antibody that has the immunity against the new antigens. If there are memory cells, the immune system will then be able to produce the specific antibody that has immunity to the new antigens.

The functions of the plasma cells is to produce antibodies to fight against the new foreign antigens.

After we recovered from the disease, the memory cell still remains in our body. Therefore if we infected the same disease again, we still have the memory cell which are present in our body, therefore, the plasma cell will produce the same antibody to fight against the disease. Therefore we can have immunity on the second time we infected the same disease.