Bio Class 1

Question 1

Sucrose (Monosaccharides)

Answer 1

Glucose + Fructose

Question 2

Lactose (Monosaccharides)

Answer 2

Glucose + Galactose

Question 3

Maltose (Monosaccharides)

Answer 3

Glucose + Glucose

Question 4

3 Polysaccharides

Answer 4

1. Glycogen
2. Starch
3. Cellulose

Question 5

Glycogen

Answer 5

High degree of branching, every 8-10 residues. Chain length in the thousands. Alpha linkages. Stored in the liver for energy.

Question 6

Starch (2 Types)

Answer 6

Branching every 20 to 30 residues. Two types are Amylose (15-30%) and Amylopectin (70-85%)

Question 7

Amylose vs Amylopectin Branching

Answer 7

Amylose is unbranched
Amylopectin is branched

Question 8

How many isoprene units do you need to make a terpene?

Answer 8

2 isoprene units

Question 9

Squalene has how many isoprene units? How many terpenes does it have?

Answer 9

6 isoprene units
3 terpene units

Question 10

Cholesterol

Answer 10

3 six-carbon rings + five-carbon ring. In the cell membrane (50%) it is used to stabilize the membrane.

low temps - increase fluidity
high temps -decreases fluidity

Question 11

Exergonic vs Endergonic

Answer 11

Ex- negative Gibbs free energy
En- positive Gibbs free energy

Question 12

Competitive Inhibition

Answer 12

Binds at: active site
effect on Vmax: same
effect on Km: increase

Question 13

Lineweavor-Burk Plot (LBP)

Answer 13

Y-axis - Vmax = 1/Vmax
X-axis - Km = 1/ Km

Question 14

Competitive Inhibitor (LBP)

Answer 14

1/ Vmax doesn’t change. No change in the Y- intercept. There is a change in 1/Km which is the X- intercept. Reversible

Question 15

Non-competitive Inhibitors (LBP)

Answer 15

Would show a change in the Y-intercept but not in the X- intercept. Irreversible

Question 16

Uncompetitive Inhibitor (LBP)

Answer 16

Would change in the Y-intercept and the X-intercept but not in the slope of the line

Acts by binding the enzyme-substrate complex. Both Km and Vmax are decreased with increasing inhibitor concentration.

Question 17

Mixed Inhibitor (LBP)

Answer 17

Results in changes in slope, X- and Y- intercept of the line

Question 18

Blood Brain Barrier

Answer 18

Hydrophobic molecules pass easily and would have no difficulty passing through the blood brain barrier.

Carbonyls and tertiary amines are polar and would be a problem for the blood brain barrier. Charged groups like ammonium would cause more of a problem and is thus the main problem in crossing the barrier

Question 19

In which situation would an acetylcholinesterase inhibiting drug be used therapeutically.

Answer 19

An acetylcholinesterase inhibitor would be used therapeutically when the post-synaptic acetylcholine receptors are blocked by another molecule. Increased levels of acetylcholine in the synapse with make up for the loss of post-synaptic acetylcholine receptors.