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27 Cards in this Set
- Front
- Back
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What was the purpose of observing the effects of MnO2? |
To compare the effects of an inorganic catalyst (MnO2) with an organic one (catalase). |
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Is MnO2 a catalyst? Is MnO2 an enzyme? Why or why not? |
It is not an enzyme (because it’s inorganic), but it is a catalyst because it brought about a chemical rxn |
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What effect did fractionating cells (grinding in the mortar) have on the catalytic activity of catalase? Why? |
Fractionation increased the catalytic activity because it opened the cells, thereby releasing more catalase enabling it to catalyze H2O2 at a much faster rate. |
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What effect did boiling and HCl have on the catalytic activity of catalase? |
Boiling and HCl eliminated the catalytic activity because it denatured the enzymes. (boiling due to high temp; HCl due to low pH) |
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How do you think the reaction would be altered if the concentration of peroxide used was greater than the 3% we used? Explain |
The rxn would increase because the concentration of the substrate (H2O2) would be greater. |
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What is the purpose of performing the splint test? What does a positive result indicate to you about the catalytic activity of the enzyme? |
The splint test determined if a flammable gas was produced (in this case, oxygen). The reignition of the splint indicates that the catalase was functioning, i.e., breaking down H2O2 into O2 and H2O |
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In what type of cell is catalase abundantly found? |
Red blood cells |
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What was the purpose of observing the sand in H2O2? |
To prove that the sand had no effect in causing the increased catalytic activity in the bloody pulp test |
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Number of electrons needed for a stable carbon atom |
8 |
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Explain what a hydrocarbon is. |
It’s the simplest organic compound made of only C and H. Many of them are combustible |
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-ane = |
single bond |
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-ene = |
double bond |
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-yne = |
Triple Bond |
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Substrate in toothpick lab |
Toothpicks |
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Competitive inhibitor in toothpick lab |
Pins |
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Active sight in toothpick lab |
fingertips |
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product in toothpick lab |
broken toothpicks |
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what color is iodine normally |
amber |
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Explain what was happening to the iodine (at the molecular level) after it was added to the potato slice. |
The iodine entered the potato cells as its natural amber color. Eventually it came into contact with the starch molecules in the leucoplasts and turned black. The iodine “slid” into the helical structure of amylose and amylopectin and reflected bluish-black. |
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What was the color of the iodine after you added it to the sugar? WHY? |
Amber, because sugar does not contain starch |
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What was the color of the iodine after you added it to the crushed crackers? WHY? |
Black, because crackers contain starch |
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What was the purpose of the sugar and crushed cracker? |
The sugar was the negative control (no color change; received no amylase) and the cracker was the positive control (color change; received no amylase). BTW, amylase was the manipulated variable. |
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Why did one partner chew for one minute and the other chew for four? |
So that we could see the difference between the digestion of the cracker after one minute and the digestion after four minutes. The iodine was a darker color after one minute because the starches had not been broken down as much as after four minutes by the amylase. |
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Why did you and your partner spit the cracker out at the same time? |
So that there wasn’t one cracker mush being digested in the Petri dish while the other one was still being chewed |
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Explain what was happening in the iodine (at the molecular level) after it was added to the cracker mush. |
Same thing as what was happening in the potato |
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Why did you get different results between the cracker mushes |
The one-minute cracker contained more starches because it had not been exposed to the amylase as long as the four-minute cracker |
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Is this investigation qualitative or quantitative? Explain your answer |
It’s qualitative because the results were observed (color changes) and no numerical data were collected.
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