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Tutor profile: Brady P.

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Brady P.
Chemistry tutor and Biology TA for 4 years, future Medical Student
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Questions

Subject: Organic Chemistry

TutorMe
Question:

How many electrons occupy the bonding molecular orbitals of a CN triple bond?

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Brady P.
Answer:

a. 2 b. 4 c. 6 d. 8

Subject: Biochemistry

TutorMe
Question:

Why is the phosphorylation of glucose to glucose-6-phosphate an important first step of glycolysis?

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Brady P.
Answer:

After glucose enters the cell through a cell transporter, it is possible for it to escape similarly to the way it arrived. Phosphorylation of the 6th carbon on glucose gives the molecule a charge that inhibits glucose from leaving the cell due to hydrophilicity and repelling charges, therefore trapping it in the cell. The phosphorylated carbon is now a reactive site important for cell intermediate reactions moving forward throughout glycolysis.

Subject: Biology

TutorMe
Question:

Why would a non-polar molecule such as benzene diffuse passively across a cellular membrane, but a large polar molecule like glucose would not? What does this tell us about the characteristics of cell membranes? How would glucose get into the cell?

Inactive
Brady P.
Answer:

Benzene has a carbon skeleton structure with little to no polarization, it is therefore hydrophobic and will not react strongly with water through hydrogen bonding. The cell membrane is composed of a lipid bi-layer, where the middle portion of the membrane consists of long non polar carbon chain tails of lipids. Because these tails are non polar and hydrophobic, it allows for a molecule like benzene to passively diffuse through, due to the "like dissolves like" principle. Glucose on the other hand is a large polarized molecule, making it hydrophilic, and unable to diffuse passively across the membrane. Because of this phenomenon, glucose will require transmembrane channels with similar hydrophilic characteristics lining the channel in order to enter a cell.

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