KEY

Fall 1998 Biology 111 Exam #3 - end of genetics + Bioenergetics


There is no time limit on this test, though I have tried to design one that you should be able to complete within 2.5 hours, except for typing. You are not allowed to use your notes, old tests, or any books, nor are you allowed to discuss the test with anyone until all exams are turned in at 9 am on Tuesday November 23. EXAMS ARE DUE AT CLASS TIME ON MONDAY November 23. You may use a calculator and/or ruler. The answers to the questions must be typed on a separate sheet of paper unless the question specifically says to write the answer in the space provided. If you do not write your answers on the appropriate pages, I may not find them unless you have indicated where the answers are.


Please do not write or type your name on any page other than this cover page.
Staple all your pages (INCLUDING THE TEST PAGES) together when finished with the exam.
Name (please print):



Write out the full pledge and sign:






How long did this exam take you to complete (excluding typing)?

 

 


Lab Questions:

6 pts.
1) Design an experiment using the Ames test to determine whether DDT is mutagenic or not. You must use two strains of cells and you also want to determine what affect 30 seconds of UV light might have on DDT's ability to be mutagenic. Make sure you tell me all the controls you would perform as well.

Plate 1 = strain one and nothing else (control)
Plate 2 = stain one with 10 µl DDT
Plate 3 = strain one with 50 µl DDT
Plate 4 = strain one with 100 µl DDT

Plate 1 = strain one and nothing else (control)
Plate 2 = strain one and 30 seconds of UV (control)
Plate 3 = stain one and 30 seconds of UV with 10 µl DDT
Plate 4 = strain one and 30 seconds of UV with 50 µl DDT
Plate 5 = strain one and 30 seconds of UV with 100 µl DDT

Plate 1 = strain two and nothing else (control)
Plate 2 = stain two with 10 µl DDT
Plate 3 = strain two with 50 µl DDT
Plate 4 = strain two with 100 µl DDT

Plate 1 = strain two and nothing else (control)
Plate 2 = strain two and 30 seconds of UV (control)
Plate 3 = stain two and 30 seconds of UV with 10 µl DDT
Plate 4 = strain two and 30 seconds of UV with 50 µl DDT
Plate 5 = strain two and 30 seconds of UV with 100 µl DDT

4 pts.
2) Why did some strains have different rates of spontaneous reversion if they all shared the exact same DNA polymerase alleles? For example, the spontaneous reversion rate for 1535 was about 20 while 102 was closer to 250.
Because each strain had a different kind of mutation (e.g. insertion v. deletion v. base substitution), the same DNA polymerase must be more prone to make the type of mutation in strain 102 than the type in strain 1535. From this, we should be able to determine the relative tendency of DNA polymerase to make different types of mutations.

Lecture Questions:

6 pts.
3) a) What is a restriction enzyme made of?
It is a protein, so it is made of amino acids.

b) What is its substrate?
Its substrate is DNA, and a particular sequence of bases, depending on the enzyme.

c) What is the product?
The product is restricted, or cut, DNA so that the phosphodiester backbone is cut on both strands. This results in one piece of DNA becoming two pieces.

6 pts.
4) What is the molecular cause for HD?
There is a trinucleotide repeat that causes too many copies of an amino acid in a row. This causes the huntingtin protein to bind too tightly to HAP-1 but why that causes the disease is unclear. We do know that the more repeats a person has, the sooner the disease will be evident.

10 pts.
5) Generate a genetic linkage map from these data about a dominant disease:

 

142 individuals were AA and had only the upper band

9 individuals were AA but had both bands

11 individuals were aa but had both bands

138 individuals were aa and had only the lower band

a) How many people had the disease?
149 have the disease plus any heterozygotes not listed above.

b) What is the map distance between the disease locus and the RFLP locus?
6.7 map units

c) What would you tell a couple whose fetus was tested and had only the lower band?
I would tell them that there is only a 0.4% (6.7% times 6.7%) chance that their child will not have this disease.

4 pts.
6) Why did investigators need to do chromosomal walking if they could use cDNA as a probe for CF or exon amplification for HD? Why not skip over chromosomal walking since it takes so long?
They had to narrow down the number of possible restriction fragments that they wanted to probe. Since they could not probe or do exon amplification on every restriction fragment in a genomic library, they needed a way to eliminate most of the DNA that was not between the RFLPs that were known to flank the disease locus.

4 pts.
7) If chlorophyl a (the pigment in the reaction center) can only absorb blue and red light, then why don't leaves look orangy-yellow?
There are other pigments that absrob in the yellow orange range so the only light that bounces back is the green light which is not absorbed.

6 pts.
8) Define (as positive or negative) the DG, DH and DS of the light reaction.
DG = positive
DH = positive
DS = negative

6 pts.
9) Tell me what the 3 final products of the light reaction are and how each is produced.
ATP is produced when the H+ gradient (produced when electrons are transported in the first half of non-cyclic electon flow and the cyclic electron flow) is used as an energy source. The H+ ions pass through the ATP synthase which produces ATP.
NADPH is produced when NADP+ is the final electron acceptor for the second half of non-cyclic electron flow.
Oxygen is produced as a waste product when water is split to provide two electrons to fill the hole in PSII after chlorophyll a has been photooxidized.

8 pts.
10) What is consumed in the dark reaction and what is produced?
.ATP, NADPH, and CO2 are consumed.
Fixed carbon, in the form of glyceraldehyde-3-phosphate is produced.

4 pts.
11) In this reaction, is this molecule being oxidized or reduced? To get credit, you must explain your answer.

The molecule is being oxidized. You can tell this because the carbons have less hydrogen associated with them and more oxygen.

4 pts.
12) What role does Rhizobium play in the life of a soy bean plant?
It fixes nitrogen so that the plant can absorb it in a reduced molecule.

Is nitrogen fixation reduction or oxidation when ammonium is produced? Explain.
Ammonia has more hydrogens associated with it which is indicative of reduction.

8 pts.
13) List all the different reducing agents produced during cellular respiration and how much is produced from one glucose molecule.
One glucose produces 10 NADH molecules and 2 FADH2 molecules.

List the processes within cellular respiration which produce these reducing agents.
Glycolysis produces 2 NADHs.
"Pre-citric acid cycle" produces 2 NADHs.
Citric acid cycle produces 6 NADHs and 2 FADH2.

6 pts.
14) Plants consume water, CO2, and O2. Where and why are each of these consumed?
Water is consumed in the light reaction, as explained in number 9 above. It is also consumed in the citric acid cycle of cellular respiration.
CO2 is consumed in the dark reaction when the carbon is reduced into glyceraldehyde-3-phosphate.
O2
is consumed during oxidative phosphorylation, during cellular respiration.

We do not consume one of these three, which one and why?
We cannot fix carbon, therefore we do not consume CO2. We do need water.

6 pts.
15) What is cyclical about the citric acid cycle? Explain in general terms how carbons cycle through this pathway.
The fact that a four carbon sugar joins with the acetyl group from acetyl-CoA and the two carbons on the acetyl group are processed into 2 CO2s, leaves the same four carbons back in their original form to be recycled again. What goes around comes around (i.e. 4 carbons).

6 pts.
16) List the three carbon-based products of fermentation, excluding ATP and NAD+.
CO2 and ethanol are produced by fermentation.
Lactic acid is also produced by fermentation.

Which one or ones do you think Clostridium tetanii produces in us and why?
Lactic acid, since we do not tend to produce bubbly alcohol in our tissue when infected with the bacterium.

4 pts.
17) Facultative anaerobes need a control mechanism that repsonds to the presence or absence of oxygen. Develop a reasonable hypothesis which describes how such an organism can switch from anaerobic metabolism to aerobic. There are several correct answers to this question but keep your answer short and limited to only one hypothesis. Note that it is worth only 4 points.
Many possible answers here.

2 pts.
18) Why would anaerobic bacteria not be harmed by either cyanide or paraquat?
Since they do not have an electron transport systems, there would not be any harm or loss of energy.

 

+2 pts possible but no points lost
Bonus Question:
Why do people get dizzy when their blood glucose level drops even though there is plenty of protein and lipids available for catabolism in the brain?
Brain cells cannot metabolize anything but glucose, so they run out of energy when glucose levels drop. They cannot convert the other molecules into products that can enter cellular respiration (glycolysis and Krebs).



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