Unit 1 Test Review Package Answers 2024F PDF

Unit 1 Test Review Package [Answers] Deets Knowledge and Understanding Thinking Application 20 multiple choice questions 4-5 short answer questions 4-5 short answer questions 2 require written explanation 2 req...

Unit 1 Test Review Package [Answers] Deets Knowledge and Understanding Thinking Application 20 multiple choice questions 4-5 short answer questions 4-5 short answer questions 2 require written explanation 2 require written explanation /20 marks ~ /15 marks ~ /15 marks Practice Questions Lessons 1-7 1. Complete the following chart - Which molecule would have a Which molecule would have a higher boiling point? Why? lower freezing point? Why? ~ both molecules are nonpolar - so we The greater the intermolecular force are dealing with London dispersion (see previous box), the easier it would forces. The larger the molecule, the be to freeze (lock in a relatively fixed greater the london dispersion force shape). Therefore, pentane will have a between the molecules (greater higher freezing point; propane molecular weight = more electrons = (molecule #1) has a LOWER freezing more opportunities for instantaneous point - needs to remove more kinetic dipoles to take place).. And in turn, the energy for the change in state to be greater the force of attraction, higher achieved. the temperature needed to boil / separate the substance. So, it’s molecule #2 (pentane). 2. A protein has a tertiary structure formed by interactions between the side chains. For each pair below, identify the strongest type of interaction between the two amino acids - Aspartic acid + lysine Phenylalanine and alanine Serine and lysine ~ dipole-dipole (NH2 and OH) Dipole / dipole London dispersion force ~ ionic-dipole possible if NH2 is ionized Ionic possible (if ionized in aqueous solution - NH3+ COO-) to NH3+ 3. Draw the basic molecular structure (monomer) of each of the three macromolecules and complete the rest of the table - carbohydrates lipids proteins  Monomers diagram (draw it if you can - not just copy and paste) Polarity of monomer ~ 5 OH groups (polar) ~ glycerol - 3 hydroxyl ~ amino group (polar) ~ ether in carbon ring (polar) groups (polar) ~ carboxyl group (polar) ~ fatty acids - non-polar CH bonds + carboxyl group at the end for bonding How the monomers join together (reaction type + drawing the resulting dimer (monomer + monomer) would be nice!) 4. The breakdown (hint: lysis) of carbohydrates and proteins cannot take place in the absence of water. Please explain why. The process is called hydrolysis. Opposite of condensation, water (H and OH) is added to break a bond - so from a larger molecule to two smaller units. 1. First state the importance of a protein’s 3-D shape. Then, describe the four structural levels of a protein, focusing on the bonds formed in each level. Why is the 3-D shape primary secondary Tertiary Quaternary important? Shape determines Beads on a string - the The folding of the All the intermolecular All the intermoelcular function; for example, sequence of amino ‘beads on a string’ due forces of interaction forces of interaction the protein’s specific acids; to hydrogen bonding of among the R-groups of between polypeptides shape is specific to the the backbone - the amino acids in the (most proteins are substrate it catalyzes Main bond - peptide specifically between polypeptide, including made of multiple (as an enzyme); or the bond between each the amino hydrogen dipole-dipole, polypeptides together) protein’s specific shape (partially positive) and H-bonding, ionic, and again, this includes amino acid allows for specific the carbonyl oxygen hydrophobic dipole-dipole, receptors to bind (if the (partially negative) (london-dispersion H-bonding, ionic, protein is a surface forces); hydrophobic receptor) (london-dispersion forces); Lesson 8 5. How does enzymatic activity affected by the following - Substrate temperature pH Feedback regulation Competitive inhibition concentration ~ the greater the ~ the higher the ~ optimal pH around ~ could be positive or ~ the inhibitor molecule [substrate], the higher temperature to around 7ish; negative regulation competes with the o the rate of reaction 37 C, the greater the ~ activity dramatically ~ the product of the substrate for the enzyme activity; decreases in acidic / enzyme reaction (most enzyme’s active site ~ the enzyme activity basic solution likely in a series of ~ lower intended dramatically decreases reactions) acts to enzyme activity; afterwards, especially further promote / o after 50 C. deactivate the initial enzyme activity; 6. What type of regulation is each diagram depicting? Allosteric activation Allosteric inhibition / non-competitive Competitive inhibition 7. Distinguish between the following terms - co-factors co-enzymes Prosthetic groups non-A.A. components of a functional ~ organic cofactors ~ permanently attached protein  8. Is it likely that an enzyme that can catalyze the breakdown of the glycosidic bond of a sugar molecule could also catalyze the breakdown of the peptide bond of protein? Please explain by referring to the bonds and molecular structure of both types of molecules. ~ glycosidic bond - ether bond between two monosaccharides (two carbons) ~ peptide bond - amide bond (O=C-N) between two A.A. Completely different bonds - would need completely different active binding sites; therefore different enzymes; 9. Examine the graph. Based on the line, how do we know that enzymes are involved? And what conditions on [enzyme] and [substrate] can we conclude? By referring to the graph line, how do what conditions on [enzyme] and we know enzymes are involved in the [substrate] can we conclude graph ~ deactivation of enzyme efficacy above ~ before deactivation, it’s a simple linear o 37 C. line suggesting no limiting reagent / saturation, etc. - so likely plenty of available substrates and functional enzymes; 10. 👁️‍🗨️The red line (without a letter) represents normal enzyme reaction rate. Which line (A, B, or C) is an example of non-competitive inhibition? Please support answer. ~ line A actually shows increased enzyme activity - so not inhibition [likely some sort of activator, or an increase in enzyme concentration, or increase in temperature without denaturing the protein] ~ line B shows very little change from normal enzyme activity; there’s an initial decrease, but then reaches the same plateau - that when [substrate] increases, the enzyme activity also increases; this suggests competitive inhibition, since both are competing for the active site, the higher the concentration of substrate, the more likely the substrates will be able to overwhelm the active site (more collision rate relative to this competing inhibitor) - such that when we approach maximum substrate concentration (x-axis), there is almost no difference in reaction rate between line B and normal enzyme activity (red line); ~ line C shows a clear decrease in enzyme activity regardless of substrate level - suggesting an alternative place of inhibition that remains consistent independent of [substrate] - so non-competitive inhibition Lesson 10 11. 👁️‍🗨️Predict how each of the following changes will affect the cell membrane - The phospholipid’s fatty acids Cholesterol is removed from the cell The cell membrane is all of sudden become more saturated membrane placed in a fat-soluble solution ~ denser - more fatty acids are able to ~ removal of stabiliser; risk - ~ the cell membrane will turn inside out be packed into a given space; closer ~ low temperature - over-condensing of - the phosphate heads (hydrophilic) will together = more VDW forces bilayer - becomes too dense for turn inwards toward each other (Away Result: less fluid / more rigid cell substances to diffuse through; from fat-soluble solution) membrane ~ high temperature - enough molecular ~ while the fatty acid chains will turn motion to potentially break from VDW outsides in both directions facing forces (phospholipids not in bilayer outwards; formation - cell membrane breaks) 12. Contrast the chemical makeup (R-groups of A.A.) and the arrangement of A.A. between integral vs. peripheral membranes. Integral membrane Peripheral membrane ~ large portions of hydrophobic A.A. needed to be able ~ depending on how far the protein extends into the to embed in the cellular membrane; phospholipid bilayer ~ it may be just hydrophilic A.A.s - so that it’s attached / connected to the phosphate heads; 13. Label the following structures of the cell membrane - label name A Membrane protein (integral) B Membrane protein (peripheral) C Phosphate head (hydrophilic) D Fatty acid tails (hydrophobic) E Phospholipid F Cytosol G Cholesterol Lesson 11 14. List and explain two reasons why starch cannot move through a plasma membrane via simple diffusion - Way too large of a molecule - won’t be able to physically fit Polar - lots of hydroxyl groups that further attract water in through the phospholipid bilayer molecules - won’t be able to get through the phospholipid bilayer  15. 👁️‍🗨State and explain what type of transport each substance below would most likely use for entry into a cell. O2 glucose H2O Triglyceride Na+ ions Glycerol Simple passive Facilitated Simple passive Facilitated Facilitated Simple passive diffusion passive diffusion (slow) and transport transport diffusion (but Facilitated relatively slow) Non-polar + small Secondary active diffusion transport (aquaporins - faster) Larger non-polar Ions cannot [depending on molecule simply pass Small + polar concentration; but in Polar + relatively through cell all cases, cannot simply pass through small membrane cell membrane] 16. 👁️‍🗨️What happens when you place a freshwater fish (used to living in low-solute conditions) into the ocean (high-solute conditions)? A high-solute environment = hypertonic; going to draw water OUT 17. The solutes cannot move across the membrane. (dotted red line) Which way would water (the solvent) move - right or left? Please explain. A - more solutes = higher solute concentration = lower water concentration B - fewer solutes = lower solute concentration = higher water concentration Osmosis - water move from higher water concentration to lower water concentration therefore from B to A. Water level in A part of the tube will rise 18. 👁️‍🗨️The cell has a 20% glucose cytosol (assume the rest is water). The extracellular fluid (ECF) is 10% glucose - also assume rest is water. The cell is in what type of environment (hypotonic, hypertonic, isotonic)? Assuming the membrane is semipermeable - allowing only water (but not glucose) to move. Will the cell swell up, stay the same size or shrink? Please explain. What environment is Will the cell swell up, stay the same or shrink? the cell in? Please explain. Hypotonic (lower solute Water will move INTO the cell (higher solute concentration than inside concentration at 20% will draw water in) the cell) Inside the cell = higher solute concentration = lower water concentration - water moves from high (outside of cell) to low (inside of cell) 19. The left pump shows the first step; middle pump the second step; the right pump the final step. What type of transport is this? Simple passive? Facilitated passive? Primary active? Secondary active? Please explain.  This is moving Na+ ions from cytoplasm into ECF - this is going against concentration gradient - from an area of lower concentration to higher - therefore, it’s active; This is primary active - to set up an electrochemical gradient that will allow for symport / antiport (in secondary active transport) 20. Explain how transporting ions against gradient has the possibility of generating more potential energy than if it were just neutral molecules being transported against gradient. Two forces at play here - 1. Chemical gradient - when there’s a difference in concentration of particles, there’s a tendency / force for particles to move from an area of higher concentration to an area of lower concentration (diffusion) - so if there’s a gradient set up, there’s a potential force to move from higher concentration to lower concentration 2. Electrical gradient - if the substance also is charged - there’s an electrical gradient that will also force / drive ions to move from an area of similar charges (like charges repel) to an area of opposite charge (or less concentration of the same charge) So when dealing with ions - it’s called electrochemical gradient Lesson 12 21. Under what circumstances would endocytosis / exocytosis most likely to occur? Small or large molecules? Polar or nonpolar molecules? Small quantities or larger quantities? Larger molecules Polar molecules Larger quantities (bulk!) 22. Of the three types of endocytosis, identify and explain the type of endocytosis most likely to occur for each scenario below - Immune cell engulfing a foreign The intake of nutrients from The intake of specific hormones / invader surrounding by an unicellular cell messengers phagocytosis pinocytosis Receptor-mediated endocytosis 23. COVID-19’s spike protein docks onto the receptor cells of our lungs’ epithelial cells to enter our body. Which type of endocytosis is this describing? Receptor-mediated endocytosis