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Flashcards in Exam 2 long questions Deck (14):
1

Describe two techniques that could be used to purify a target protein.

One technique is differential centrifugation. The cells are disrupted by being spun in a centrifuge at low speeds to yield a pellet consisting of nuclei and a supernatant. This is repeated a few times to yield a series of pelleted enriched in the cell materials and supernatant and form a homogenate. Another technique is dialysis. Separation in dialysis is based on the size of the molecules for buffer exchange. The solution is placed in a cellophane bag with pore that are too small to allow protein to diffuse but also big enough to let salt equilibrate.

2

Describe three techniques or measurement used to evaluate protein purification protocol for effective isolation of a pure enzyme.

One technique is measurement of specific activity. Total activity is divided by total protein, which enables us to measure the degree of purification by comparing specific activities after each purification step. Another technique is measurement of yield. Yield is a measure of the total activity retained after each purification step as a percentage of the activity in the crude extract. Another technique is measurement of purification level. It measures the increase in purity and is obtained by dividing the specific activity after each purification step by the specific activity if the initial extract.

3

Explain what the term delta G tells you about a chemical reaction.

DeltaG is the change in free energy when a reaction occurs. It provides information on the spontaneity of the reaction, but not the rate of the reaction. When delta G is negative, the reaction is exergonic or spontaneous. When delta G is positive, the reaction is endergonic or nonspontaneous. When delta G is neutral, the reaction is at equilibrium and there is no net change in the amount of reactants and products in the reaction.

4

Discuss the formation of the enzyme-substrate complex and how this facilitates catalysis of biochemical reactions.

Enzymes bring together substrates in enzyme-substrate complexes. The substrates bind to specific regions on the enzyme called the active sites. When the enzyme and substrate interact, it promotes the formation of the transition state. This facilitates catalysis of biochemical reactions because it lowers the delta G of the reaction, providing the rate-enhancement characteristic of enzyme action.

5

What is the parameter Km in Michaelis-Menten kinetics? What is its significance and how can it be used to compare different enzymatic reactions?

Km is the substrate concentration at half of Vmax, which is the fastest possible rate for an enzyme at fixed concentration and for a specific substrate. Km is significant because it is a measure of the affinity an enzyme possesses for a specific substrate. It can be used to compare different enzymatic reactions by determining which substrate an enzyme has a higher affinity for from looking at the Km value. If it is high, there is a greater amount of substrate required to reach half of Vmax, and vice versa.

6

Compare and contrast the kinetics (as a function of substrate concentration) of a Michaelis-Menten enzyme and an allosteric enzyme. Include the physical basis of each.

Allosteric enzymes have an initial lag in reaction velocity in response to the substrate, whereas Michaelis-Menten enzymes do not have that initial lag. Velocity increases instantly in response to the substrate. Allosteric enzymes display a sigmoidal curve on a graph comparing velocity to substrate concentration, and Michaelis-Menten enzymes increase rapidly and level off sooner. Allosteric enzymes are also more complex than Michaelis-Menten because they consists of a committed step and remaining steps, which are facilitated by Michaelis-Menten enzymes.

7

Describe how allosteric effectors influence enzyme function. Include a graph to illustrate your answer.

Positive effectors, also called activators, bind to the enzyme at the regulatory site and stabilize it by transforming it into the R form. This lowers the substrate threshold for activity. An example of this would be ATP. Negative effectors, also called inhibitors, bind to the enzyme at the regulatory site and stabilize it by transforming it into the T form. This increases the substrate threshold for activity. An example of this would be CTP.

8

What is the molecular basis of gout?

Gout is the result of consuming foods rich in purine nucleotides, for example, adenine and guanine. They are degraded into urate, which forms crystals in the fluid and lining of your joints. It can also be caused by loss of allosteric regulation by an important enzyme in purine synthesis, PRS. It is normally feedback inhibited by purine nucleotides, however, in certain people the regulatory site has undergone a mutation that renders PRS insensitive to feedback inhibition. This leads to a glut of purine nucleotides, which are then converted into urate. Excess urate will accumulate and cause gout

9

Describe the effects of the three different types of reversible inhibitors on enzymes.

Competitive inhibitors are structurally similar to substrates and bind to the active site, which prevents the actual substrate from binding. This causes the Km value to increase but does not have an effect on the Vmax value. Uncompetitive inhibitors will bind only to the enzyme-substrate complex somewhere other than the active site. It causes both the Km value and Vmax value to decrease. Noncompetitive inhibitors will bind to either the enzyme or the enzyme-substrate complex somewhere other than the active site. This causes the Vmax value to decrease but has no effect on the Km value.

10

Describe the chymotrypsin active site and how these residues contribute to its catalytic mechanism.

Chymotrypsin contains the highly reactive serine 195, histidine 75, and aspartic acid 102. Serine 195 produces a two-step hydrolysis in the enzyme. The first step is the burst phase because the covalent attachment is very rapid, and the second step is the steady phase because the hydrolysis release is much slower. Histidine 75 plays a role in the modification by TPCK. The reactive group in TPCK covalently modifies one of the ring nitrogens of histidine 75. It also serves to position the serine side chain and to polarize its hydroxyl group so that it is poised for deprotonation. Aspartic acid 102 helps orient the histidine residue and make it a better proton acceptor through hydrogen bonding and electrostatic effects.

11

Describe the general structure of adult hemoglobin, Highlight the structural differences between deoxy- and oxy-hemoglobin.

Hemoglobin has a protoporphyrin ring with a central iron atom coordinated by four nitrogen atoms of heme. It also contains four heme groups. Deoxyhemoglobin is hemoglobin before oxygen is found, and the iron atom is slightly out of the plane of the heme ring. Oxyhemoglobin is hemoglobin after oxygen is found, and the iron atom is pulled into the plane the heme ring is in.

12

What are the three allosteric effectors for hemoglobin? Describe their effects on the allosteric state of hemoglobin and illustrate how they change the oxygen-binding curve.

2a. ,3-BPG is an allosteric effectors that stabilizes the T state and facilitates the release of oxygen. CO2 and hydrogen are also allosteric effectors and are produced by actively respiring tissues to enhance oxygen release by hemoglobin. The stimulation of oxygen release by CO2 and H+ is called the Bohr effect.

13

Describe the molecular basis as to how membrane channels are selective for certain ions use the potassium channel as an example.

Membrane channels are selective for certain ions because the channel is just the right size for K+ to fit. Na+ cannot fit because it is too small, and other larger molecule physically cannot fit. The opening to the outside and the cavity are filled with water, and K+ can fit without losing its shell of bound water molecules. When K+ moves farther into the cavity, its gets more narrow and the ions must remove their water molecules and interact with groups from the protein. After the bound water molecules are removed, another K+ is moved into the holding site and the K+ is pushed to the next site. This process of pushing K+ ions further into the channel as others enter eventually kicks the K+ ions out and into the opposite side of the membrane.

14

Compare and contrast pumps and secondary transporters.

Pumps facilitate the movement of molecules against their concentration gradient and require energy to do so. This is called active transport. Secondary transporters use the power of one concentration gradient to drive the transportation of a different molecule against its concentration gradient. This does not require energy but does require two different molecules.