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Tutor profile: Eric J.

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Eric J.
Ph.D. in Molecular Biology
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Questions

Subject: Biomedical Science

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Question:

You are tasked with designing a new treatment against various forms of cancer. What are some cellular pathways or processes that would make good targets in your drug design?

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Eric J.
Answer:

The way to approach this problem is to try and understand the difference between cancer cells and normal cell tissues. You can then design therapies that will have greater effect on processes and properties cells in cancerous cells than healthy cells. One thing you can take advantage of is that cancer cells undergo more frequent cell divisions than healthy cells. This increased cell division is often due to a lack of “check point” and damage control genes. These genes sense DNA damage and then arrest processes like mitosis to allow for the DNA to be corrected before proceeding. As such, one therapy against cancer is to deliberately introduce DNA damage. This accrual of DNA damage will kill more cancer cells than normal cells, as the normal cells will not be duplicating their damaged DNA at the same rate. This is the basis behind chemo/radiation therapy in real therapies. Another thing that you can design drugs for is to attack “cellular markers” that are specific to cancer cells. Cancer cells often lose their original “tissue identity” and begin to express a different set of genes from their healthy counterpoints. One of the consequences in this different genetic program is that cancer cells begin to express different proteins on the surface of their cell membranes. A way to take advantage of this is to design viruses that recognize these surface proteins that are unique to cancer cells. The viruses will then selectively infect and kill cancer cells while leaving normal cells uninfected. These are called oncolytic viruses. Another target for anti-cancer therapies are ones that attack metabolic processes enriched in cancer cells. Cancer cells undergo a transition into Warburg metabolism. Which involves a decrease in cellular respiration and increase in utilizing glycolysis to generate ATP. During this shift, cancer tissues become extremely Hypoxic (Lacking oxygen). As such, a good way to hurt cancer cells more than healthy cells is to inhibit enzymes involved in glycolysis. Additionally, relieving the hypoxia will allow the healthy cells to out compete cancer cells since cellular respiration is much more efficient than glycolysis. Two other effective cancer therapies are in relieving the spread of cancer or metastasis. The way to do that is to either block the cancer cells from shedding its connections from its surrounding epithelial tissue and become isolated/migratory cells that will spread to other parts of the body. This process called Epithelial to Mesenchymal transitions (EMT), do not normally occur in healthy adult tissues, so inhibition of this can stop metastasis without harming the healthy tissue. Another thing that is important in metastasis is for the cancer cell to break through the extra-cellular matrix of the surrounding tissue so that it is free to migrate to other parts of the body. Cancer cells break down this network of collagen, laminins and fibronectin (Key proteins in the ECM) through use of actin rich protrusions called invadopodium. These structures, which induce changes in shape to the cancer cell’s membrane, drill into the extra-cellular matrix and then degrade it with proteases called matrix-metallo proteases (MMP). Thus inhibition of either invadopodium formation or MMP secretion can cause the cancer cells to remain in place and block the worst outcomes and prognoses that are associated with metastasis.

Subject: Biochemistry

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Question:

Use the genetic coding chart to answer the following questions: A small gene encoding Protein X has the following sequence on its template strand: 5’-GGTACGGAACGTTGCCAGGGGCACTCATT-3’ is mutated into one of 4 genetic variants: 5’-GGTACGGAACGTTGCCAGGAGCACTCATT-3’ 5’-GGTACGGAACGTTGCCAGGGGCACTAATT-3’ 5’-GGTACGGAACGATTGCCAGGGGCACTCATT-3’ 5’-GGTACGGAACCTTGCCAGGGGCACTCATT-3’ Which of the following mutations would be the least harmful (Mutations in bold), what type of mutation is this?

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Eric J.
Answer:

The first step is to find the start of the reading frame (ATG) and begin reading codons from there. Because this is template DNA, the RNA is the complementary sequence to this strand, with Thymine being replaced by Uracil (U). Thus, in RNA the sequence is as follows: DNA: 5’-GG TAC GGA ACG TTG CCA GGG GCA CTG ATT-3’ RNA: 5’-CC AUG CCU UGC AAC GGU CCC CGU GAG UAA-3’ Amino Acid:5’-_MET PRO CYS ASN GLY PRO SER GLU STOP-3’ since the reading frame first begins at methionine. Looking the above mutations, the one that is least harmful is the first one. That is because the change from GGG to GGU is a conservative substitution mutation in which the amino acid is unchanged. This means it is unlikely to affect the function of the protein and is thus the least harmful mutation. In the second mutation, the amino acid changes from Glutamate (GAG) to Aspartate (GAU). While these two AA have the same charge, they is still a chance the change can affect protein function. The third mutation is an insertion that causes a frame shift and introduces a premature stop codon by shifting the reading frame by 1 (UAA). The last one changes a cysteine (UGC) to a tryptophan (UGG) which could have profound effect on protein function. The type of mutation can be answered a number of ways. 1 way is that the mutation is a substitution mutation. Another correct answer is a Transversion mutation since the nucleotide being exchanged is a Guanine for a Uracil. The last correct answer is that this is a synonymous mutation. This is because while the nucleotide is altered in this mutation, the amino acid remains unchanged.

Subject: Biology

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Question:

If 2 parents with brown eyes give birth to 3 children, one of which is a daughter with blue eyes. What are the odds their next child will also have blue eyes? What type of inheritance is blue eyes? Bonus: Why would one trait be dominant over the other in this case?

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Eric J.
Answer:

The key to this question is seeing that despite the fact that both parents have brown eyes, they are able to give birth to a blue-eyed child. This lets you know that despite both of them having the phenotype for brown eyes, they both contain the alleles for blue eyes and are heterozygotes in their genotype. This allows you to construct a Punnett square in order to determine the probability of a blue-eyed child. Since both parents have brown eyes despite having the allele for both traits we know that brown eyes are the dominant trait (B) and since blue eyes do not appear in either parent, we know that it is the recessive trait (b). B b BB Bb B Bb bb b Any genotype that contains a dominant allele (B) will cause the child to have brown eyes and so only bb will give blue eyes. bb appears in 1 out of 4 of the cases and so the child has a 25% chance of having blue eyes. The key to understanding the type of inheritance here is to know that the person with the recessive trait is a daughter. We know from the previous problem that blue eyes are recessive. Since the daughter has 2X chromosomes, one of which comes from the father who has brown eyes. Since the father only has one (dominant) X-chromosome to give, there would be no way for a daughter to have blue eyes if the gene were X-linked. Since the gene is not X-linked, we know that the trait for blue eyes is autosomal recessive. BONUS: This is one to really get you thinking about how and why traits may be dominant or recessive. In this case, eye color is caused by the deposition of the melanin pigment. Brown eyes are dominant because even one copy of the allele is sufficient to produce this pigment. Blue eyes are caused by the complete absence of this pigment and are thus recessive. The reason why absence of the pigment yields blue eyes are for the same reason that clear skies and water is blue in our atmosphere; the diffraction of light.

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