Molecular Biology 1st Exam - spring 1997

ALL EXAMS ARE DUE AT 9:30 am ON MONDAY February 3, 1997.

There is no time limit on this test. You may find it easier to take this test in two hour blocks over several days, though if you are confident in your molecular skills, you could wait until Sunday night. However, I predict it may take you a bit longer to think of all the answers (just some friendly advice). You are not allowed to use your notes, any books or journals, nor are you allowed to discuss the test with anyone until all exams are turned in at 9:30 am on Monday, Feb. 3. While you are taking the exam, you may not study for any science classes that will help you with this exam (e.g. Organic Chem.) I do not know which classes may present problems so talk to me if you are unsure. You may use a calculator. The answers to the questions must be typed, though you may want to supplement your text with hand drawn figures (write neatly for any labels in your figures).

Please do not write 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)?

Average for 1997 = 85%

You might find it more fun to take this test while humming (Dunt, dunt, dunt, dunt, dunt, dunt...) the tune to Mission Impossible.

20 pts.
1) (Good news, bad news.) You bought a ticket to London for spring break and the plane is over booked (bad news). You get bumped to first class (good news) and the man sitting next to you looks a lot like Tom Cruise (really good news). He is reading what looks like a secret file. On this file is the following:
A) This not-so-smart-agent needs help. Write down the DNA sequence shown here in the proper 5' to 3' orientation.

B) He tells you that the sequence you have written down is the template strand for an mRNA. There are no introns, but he needs you to write down the mRNA sequence in the proper 5' to 3' orientation.

C) Now he asks you to translate the mRNA using the table provided to you at the bottom of this page. Start with the 5' most mRNA nucleotide and then write down the amino acid sequence beginning with the N-terminus.

D) Now he asks you to see if you can find an open reading frame. You must tell him yes or no and explain why you have answered this way.

15 pts.
2) You are having lunch at a cafe, and you see the following on your menu. Please convert these letters to the full names of the amino acids:
glycine threonine serine aspartic acid isoleucine proline serine
isolucine tryptophan glutamate glutamate asparagine histidine glutamate
asparagine leucine serine isoleucine cysteine
glycine phenylalanine threonine valine
arginine glutamate

14 pts.
3) A woman sits down and tells you that a new monoclonal antibody has been generated and the CIA thinks that it will bind to acetylcholinesterase. This agent asks you to design one or more experiments (including controls) that would:

A) find out if the antibody's epitope is also present on acetylcholinesterase isolated from flies, mice, worms, and humans and if so, how big each version of acetylcholinesterase is.
+ control: known and purified acetylcholinesterase in a lane by itself-
control: a tissue known to lack acetylcholinesterase, such as yeast cells
a lane for each of the species with total protein loaded. I would choose muscle tissue for mice and humans.
a lane of MW markers
Run these samples on an SDS-PAGE, transfer to nitrocellulose, probe with the alleged monoclonal antibody. Detect the antibody.
You should see a band in the + control lane and none in the - control lane. The + control lane will tell you the approximate size to expect in the other species. Note where the bands are in each lane and calculate the MW for each species' form of acetylcholinesterase, using the MW markers.

B) since all of these species do express acetylcholinesterase, clone the cDNA that encodes the fly version of acetylcholinesterase. (It has never been cloned from flies, though it has from the other species.)
Two choices, either use the other species' cDNA as a probe for a cDNA library (lambda gt10), or use the antibody to probe an expression library (lambda gt11).
But either way will involve these steps:
isolate purified fly mRNA
make cDNA
make cDNA library in lambda library (see above)
make your probe (see above)
screen the appropriate library
isolate lambda phage DNA
sequence the cDNA insert to confirm you have the right clone

9 pts.
4) Below, you see two northern blots. One for Free Bacterial Intake protein, and one for actin, the universally expressed protein. What can you deduce from these data? Explain how your made your deduction.
Actin controls: they worked in all lanes; it appears that each lane has about the same amount of total RNA loaded except human which appears to have more RNA in that lane since the actin band is more intense.
Flies: the FBI mRNA has a molecular weight of about 1.6 kb. It appears to be the least abundant of all FBI mRNA.
Worms: appears to have two alleles of FBI, one about 1.1 kb and one about 1.3 kb in length. These two forms are collectively about twice as abundant as the fly FBI mRNA.
Mice: the molecular weight is about 0.9 kb and appears to be the most abundant FBI mRNA due to the thickness of the band being largest.
Humans: molecular weight is about 1.3-1.4 kb. Its relative abundance is difficult to judge since there is more total RNA in this lane than in any of the others. But it is not the most abundant form since this band is the same thickness as the mouse FBI band.
12 pts.

5) When you return to your hotel room, you find a balance, a pH meter, some chemical bottles, various pipets, a calculator and the following directions:

A) make solution with a final volume that is 225 ml of a 0.35 M NaCl, 5% v/v acetic acid. Tell me all the volumes and all reagents. Formula weights: NaCl = 58.5, Tris = 121; acetic acid = 58; EDTA = 372; BamH I and Hind III = 103 each.
4.6 g NaCl + 11.25 ml acetic acid and water up to 225 ml

B) A stock solution is 2.59 mg/ ml of DNA and you want to make 100 µl that is 1 µg/ µl. How do you do this?
use 38.6 µl of stock DNA and 61.4 µl water

C) Set up a restriction digestion of 1 µg the DNA above and cut it in 15 µl final volume with the enzymes
Bam HI and Hind III.
1 µl DNA
1.5 µl 10X buffer
< 0.75 µl Bam
< 0.75 µl Hind III
11 µl water

D) Make 250 ml of 10X TE which is 100 mM Tris (pH 8.0) and 10 mM EDTA.
3.025 g Tris in about 200 ml
0.93 g EDTA
pH with HCl
water up to 250 ml
10 pts.

6) How could you subclone the following 2 fragments (no need for exact volumes; just the strategy)? Clone each of them into separate plasmids and convince me that you know which is which. The boxes represent segments of DNA that terminate with the indicated restriction sites. There are two segments of DNA that are nonfunctional (blank) but the rest of the DNA segments have been labeled. Your task is to clone two fragments into a typical cloning plasmid. The two fragments you should clone are: fragment 1 is from 400 to 1600, and fragment 2 is from 4400 to 5200.
Digest plasmid with Bam and gel purify the 1200 bp fragment. This will be a 50/50 mixture of DNA. Ligate this into a new plasmid (that does not contain lac Z) cut with Bam, transform into bacteria, and plate on medium that would allow for the detection of ß-galactosidase (colonies turn blue). Digest DNA to confirm the size of insert and that there are not two inserts.
Digest plasmid with Sal and purify the 1800 bp fragment. Then digest this with Bam and purify the only 800 bp fragment. Ligate into Bam cut plasmid (that does not contain tetR), transform into bacteria and plate onto tetracylcine containing media. Select colonies that grow.

10 pts.
7) John Majors (#2 in pedigree) is heterozygous for a rare genetic disease of compulsive secret telling. You can see a family pedigree and a series of RFLPs below.

A) Which band is associated with the diseased allele? How
do you know this?
The upper band since the two suffering with this disease (4 and 6) are homozygous for the disease allele and homozygous for the upper band.

B) Which grandchild should not be trusted with secrets?
I would not trust child number 8. He is homozygous for the upper band. However, he may not suffer from this disease since he is not shaded in. This would be due to recombination between the RFLP marker and the disease locus.

10 pts.
8) A top covert operative almost lost his life when he refused to drink the customary milk drink that all international spies drink. Design a method to generate a hypoallergenic cow's milk. Since many people develop allergies to a milk protein called casein, how could you produce casein-free milk using transgenic technology?
I would make a transgenic cow that expresses casein anti-sense RNA. To do this, I would take the wild-type casein promoter and ligate it onto the anti-sense strand of DNA for casein (see diagram).

10 pts.
9) Finally, you return home. You are glad to be away from international spies and the doorbell rings. (Dunt, dunt, dunt, dunt, dunt, dunt...) The secrete service has been following your vacation moves, and wants some advice. It appears that a biological weapon has been discovered in the White House. It is a normal protein except a single amino acid has be changed from cysteine to methionine. After looking at the structures below, tell the agent what possible effect this substitution might have on the protein.
Two major possible disruptions: 1) the loss of a cystein may lead to a disrupted disulfide double bond. This would alter secondary, tertiary, and quaternary structure of the wild-type protein and since form gives function, the mutation could have an altered property. 2) There is the loss of a hydrophilic R group and the addition of a hydrophobic R group. This could also disrupt secondary, tertiary, and quaternary structure of the wild-type protein.

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