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1.5: Chapter 1 Questions

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    Section 1 Questions

    Question \(\PageIndex{1}\)

    In Figure 1.2, two examples of types of enzyme-substrate binding are shown: the Lock-and-Key model and Induced-Fit. What are some situations in which one style of the enzyme would be favored over the other?

    Answer

    Lock and key enzymes are highly specific for their substrate and therefore do not need a transition state to undergo the catalytic reaction. This could be used for substrate channels like Na+/K+ pumps in which a reaction doesn’t need to occur.

    Induced fit enzymes utilize a transition state, to convert a substrate into a product. The transition state is able to cause a conformational change in the active site of the enzyme and facilitate high-energy reactions such as breaking or forming chemical bonds

    Question \(\PageIndex{2}\)

    Label the following type of import/export mechanisms as passive, active, or facilitated and explain why: endocytosis, ion channels, pores, transporters/permeases. Some may have more than one answer.

    Answer

    Endocytosis: Active, facilitated. Endocytosis or “cell eating” is a multi-enzyme mediated process that allows the cell to uptake large particles from its environment. This involves membrane modification, protein receptors, and digestive enzymes and organelles working across gradients.

    Pores: Passive, facilitated. Once porins establish pores, such as in the nuclear envelope, small molecules like DNA and RNA can passively diffuse in and out of the membrane without the need for carrier proteins.

    Ion Channels: Active, facilitated OR passive, facilitated. Active ion channels pump small molecules across a gradient and are typically considered to be “gated,” meaning that the enzymes can open and close in a regulated manner to control what is being moved across the membrane. Passive ion channels are permanently open to facilitate transfer and rely on a constantly established concentration gradient to allow for transport to occur.

    Transporters/Permeases: Active, facilitated. Transporters move larger molecules across a concentration gradient and assist in the movement of soluble proteins and molecules through the hydrophobic membrane

    Section 2 Questions

    Label the functional groups present in the chemicals shown below:

    Screen Shot 2022-03-16 at 9.36.05 PM.pngScreen Shot 2022-03-16 at 9.37.14 PM.pngScreen Shot 2022-03-16 at 9.38.36 PM.pngScreen Shot 2022-03-16 at 9.40.50 PM.png

    Screen Shot 2022-03-16 at 9.41.35 PM.pngScreen Shot 2022-03-16 at 9.42.43 PM.pngScreen Shot 2022-03-16 at 9.43.28 PM.pngScreen Shot 2022-03-16 at 9.44.30 PM.pngScreen Shot 2022-03-16 at 9.45.01 PM.png

    Answers:

    Section 3 Questions

    1) a. Consider a subset of reactions of glycolysis given below. ΔG'°, substrates, and products are given from colon cancer cells (nmol/g tissue). After examining the conditions of the cell for each enzymatic reaction, predict if the ΔG of the reaction will increase or decrease. (Data from Hirayama A et al. 2009 Cancer Research. The ratio of NAD+/NADH is 10:1 and the concentrations of the cofactors are ATP (110) and ADP (300).

    Reaction ΔG'° [Substrate] [Product]

    #1 Hexokinase
    Glucose → Glucose-6-phosphate

    		 -16.6 123 75
    #2 Phosphoglucose Isomerase
    Glucose-6-phosphate → Fructose-6-phosphate    		 1.67 75 50
    #3 Phosphofructokinase
    Fructose-6-phosphate → Fructose-1,6-bisphosphate    		 -14.2 50 50
    #10 Pyruvate Kinase
    Phosphoenolpyruvate → Pyruvate    		 -31.4 5 25
    Lactate Dehydrogenase -25.1 25 25,000

    Answer:
    Reaction #1 - Increase.
    Reaction #2 - Decrease.
    Reaction #3 - Increase.
    Reaction #10 - Increase.
    Lactate Dehydrogenase - Increase.

    2) Consider the reaction below along with the thermodynamic properties: ΔH° = -760 kJ/mol, ΔS° = -0.185 kJ/mol K, and ΔG = -705 kJ/mol
    Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq)

    At what temperature would this reaction have an equilibrium constant of 1?

    Answer: ΔG° = RTln(Keq)

    Because we want know know the temperature at which Keq = 1, and we know that the ln(1) = 0, ΔG° = 0 when Keq = 1.

    ΔG° = ΔH° - TΔS°
    0 = -760 kJ/mol - T(-0.185 kJ/mol K); Rearrange and solve for T = 4108.1 °K

    This page titled 1.5: Chapter 1 Questions is shared under a not declared license and was authored, remixed, and/or curated by Henry Jakubowski and Patricia Flatt.