Download Bio 102 Practice Problems

Bio 102 Practice Problems
Recombinant DNA and Biotechnology
Multiple choice: Unless otherwise directed, circle the one best answer:
1. Which of the following DNA sequences could be the recognition site for a restriction enzyme?
A.
B.
C.
D.
E.
TGCCGT
TGCGCA
TGCTGC
All of the above.
None of the above.
2. Which of the following is not needed for DNA sequencing by the method we discussed in class?
A.
B.
C.
D.
E.
Radioactive primer
DNA polymerase
Fluorescent dideoxy nucleotides
Ordinary nucleotides (dNTPs)
All of the above are required.
3. What is the key enzyme used in PCR?
A.
B.
C.
D.
E.
ATP synthase
Taq DNA polymerase
DNA ligase
Restriction enzymes
Sigma factor
4. The creation of a cDNA molecule relies upon
A. Rubisco.
B. PCR.
C. DNA ligase.
D. reverse transcriptase.
E. Taq DNA polymerase.
5. Which DNA sequence is a palindrome and thus might be a restriction enzyme site?
A. GCAGCA
B. CCGGCC
C. GGGGGG
D. TACGTA
E. CCTAGG
6. To change a simple cloning vector (or plasmid) into an expression vector, we would need to add
A. a site that can be cut by a restriction enzyme.
B. an origin of replication.
C. one or more introns.
D. a -10 and a -35 box.
E. a gene encoding resistance to an antibiotic.
Short answer (show your work or thinking to get partial credit):
1. Recombinant human insulin, produced by bacteria carrying a cloned insulin gene, is now the major form of
insulin used to treat diabetes. The human insulin gene encodes an mRNA only 333 nucleotides long, but the
entire gene spans more than 4000 nucleotides. There are three exons and two introns.
a. If we were to clone this gene directly from the nuclear DNA, bacteria would not be able to express the
insulin protein. Explain why this is true.
This gene has introns; bacteria don't. If we cloned the gene directly, the bacteria would produce mRNA
that includes the introns and be unable to splice it. Therefore, no functional protein could be made.
b. What technique should be used instead in order to get a functional insulin coding sequence cloned into
bacteria? Describe briefly how this technique works.
We should use cDNA cloning. In this technique, we isolate mRNA from the cytoplasm of a cell, so that it
has already been spliced. We then use reverse transcriptase enzyme to make a DNA copy (cDNA) of
the mRNA. This DNA can then be cloned, producing a bacteria-readable gene that lacks introns.
c. Every cell in the human body has the same DNA, so every cell has an insulin gene. However, in order to
use the technique you described in "b," you would have to start with cells from the pancreas--the only
body cells that actually produce the insulin protein. Why are these the only cells that would work?
We need insulin mRNA in order to do cDNA cloning. If the cells are the only ones that make insulin
protein, then most likely they are also the only ones that make significant amounts of insulin mRNA.
2. Human gene therapy remains a promising possibility but is still plagued by problems. In the table below are
listed two possible vectors and two problems. For each combination, please briefly explain if the specific
problem is expected to be encountered for the vector.
Immune
Response
Insertional
Mutagenesis
Retroviral Vector
Could be a problem; someone who’s
had a retroviral infection before could
be immune to the vector, or if multiple
treatments are necessary, development
of immunity could be a big drawback.
Could be a problem, because the
retrovirus inserts DNA randomly into
the cell’s genome; it could hit a gene
by chance.
Liposome
Should not be a problem. Proteins,
not lipids, usually trigger the immune
response.
Liposomes usually deliver plasmids
or other DNA that won’t integrate
into the genome. Shouldn’t be a
problem.
3. The diagram below represents a section of the human genome. The coding sequence of a gene, YFG, is shown
by an arrow, and boxes indicate the locations of some regulatory sequences. Locations of recognition
sequences (cut sites) for three common restriction enzymes (EcoRI, BamHI, and NcoI) are also marked.
You would like to clone this gene in E. coli for further study. You have available the expression vector
(plasmid) shown below:
a. Why is it important for this plasmid to be an expression vector?
Eukaryotic genes have regulatory signals like enhancers and TATA boxes that are not recognized by
bacteria and lack the -10 and -35 sequences that bacterial RNA polymerase needs to start
transcription. In addition, eukaryotic ribosomes find the correct AUG codon by scanning from the cap
to the first AUG, while bacteria rely on a Shine-Dalgarno sequence in the mRNA. These bacterial
regulatory signals are provided by the expression vector. If we use a plasmid that did not have these
signals, it's very unlikely that the bacteria would be able to transcribe and translate our gene.
b. Why is it important for this plasmid to have an antibiotic-resistance gene?
This gives a way to select for bacteria that acquire the plasmid. The frequency of successful
transformation is small, so we need a way to know that we got the clone into a cell. Only cells that
acquire this plasmid will be able to grow in the presence of this antibiotic.
c. What restriction enzyme would you use to clone this gene? Explain your choice.
The plasmid has sites for all three restriction enzymes. However, EcoRI cuts the plasmid twice, so we
would wind up chopping out a piece of the plasmid, including the needed Shine-Dalgarno region.
This is a bad choice.
BamHI is also a bad choice, because it cuts in the middle of the gene we want to clone. We want an
enzyme that will leave our gene intact.
NotI is the best choice. It will cut our gene out of the genome and makes a single cut for inserting it into
the plasmid. Notice that it's not important that the enhancer and TATA sequences will be left behind;
the plasmid provides the needed regulatory signals.
4. You would like to use PCR to amplify (make many copies of) the boxed section of the DNA sequence below:
5´ ACGACCGATAGACGACGTAGGACTTACTTACTTACGTAGGCA 3´
3´ TGCTGGCTATCTGCTGCATCCTGAATGAATGAATGCATCCGT 5´
You ask your lab partner to order a pair of primers that can be used in the PCR reaction. The sequences of the
primers he orders are:
Primer #1: 5´ ATAGAC 3´
Primer #2: 5´ ACTTAC 3´
a. Oops! Looks like you shouldn’t have trusted your lab partner on this one. Which of the two primers is
wrong, and why won’t it work?
Primer #1 is OK: it will bind to the bottom strand, and its 3’ end will then be pointed in the right direction
for Taq DNA polymerase to synthesize the desired DNA.
Primer #2 won’t work, because it binds to other end of the same strand and so its 3’ end is pointed out
away from the sequence to be copied:
5’ATAGAC->
5´ ACTTAC->
3´ TGCTGGCTATCTGCTGCATCCTGAATGAATGAATGCATCCGT 5´
b. Give the sequence of a primer that will work and could be used instead of the wrong one. Be sure to
indicate the 5´ and 3´ ends.
We need a primer that will bind to the top strand, with its 3’ end pointed toward the left so that the other
strand can be copied. For example, 5’ GTAAGT 3’:
5´ ACGACCGATAGACGACGTAGGACTTACTTACTTACGTAGGCA 3´
<-TGAATG 5’
5’ATAGAC->
3´ TGCTGGCTATCTGCTGCATCCTGAATGAATGAATGCATCCGT 5´
5. You are trying to find the gene responsible for a human genetic disorder. You have mapped the gene to a
particular chromosome region, and examining the human genome sequence for that region gives you the
nucleotide sequence below:
5’ CATACTTACTACTAGATTACGATTAGACGATTAGGATG|GCC|GAC|TCG|TGC|AGT|AAC|
AGC|ATG|ACC|GAG|GCC|TAGACCAGATTAGGAGCCGGACCAGGACGGACCAGCGACT 3’
a. Assuming you are reading the non-coding strand and that there are no introns, find an open reading frame
(ORF) in this region. Circle the point where translation will start, and put a box around the point where
translation will stop. Then give the number of amino acids in the protein this gene would encode:
12 amino acids (start codon encodes an amino acid; stop codon doesn’t)
b. If you wanted to express this gene in E. coli, what would need to be present in your cloning vector to
ensure that it will be transcribed and translated?
The gene’s promoter sequence won’t be recognized by E. coli’s sigma factor, so you’ll need a prokaryotic
promoter (–10 and –35 sequences). E. coli ribosomes can only find the correct start codon by first
binding the Shine-Dalgarno sequence, so the vector should also have this sequence positioned just
before the desired start codon.
c. How might the protein produced by E. coli differ from the protein produced from the same gene in a
human cell?
The amino-acid sequence (primary structure) will be the same. However, eukaryotic proteins are often
modified in the ER or Golgi: carbohydrates added, phosphate groups added, etc. These modifications
probably won’t happen in E. coli. Also, it’s possible that the protein won’t be correctly folded, if
some specific protein or condition in the eukaryotic cell is needed for folding.
6. You have cloned a cDNA encoding a human hormone, and you hope to produce the hormone in bacteria in
order to treat a severe genetic disorder. Unfortunately, when you insert this DNA into a plasmid and transform
it into the bacteria, you get no hormone production. Give two valid reasons for your failure, and suggest a
possible solution in each case.
(1) Hormone must be modified after synthesis; bacteria lack ER and Golgi. Solution: determine the needed
modification and try to reproduce it chemically, clone needed enzymes or use a eukaryotic host.
(2) cDNA has no promoter and can’t be transcribed. Solution: insert it into an expression vector that includes
a promoter.
(3) cDNA from a eukaryotic cell doesn’t have a Shine-Dalgarno sequence needed for translation initiation in
bacteria. Solution: insert it into an expression vector that includes a promoter.
(4) cDNA is made from mRNA, and the mRNA for the hormone will only be present in a cell that normally
produces the hormone. Could be that the source of the DNA needs to be changed.
(5) The hormone might be toxic to bacteria. Solution: could try a different cell as the host.
7. Circle the DNA sequence(s) below which could potentially be a recognition site for a restriction enzyme.
5´ ATTTTA 3´
3´ TAAAAT 5´
5´ CGCG 3´
3´ GCGC 5´
5´ GGATCC 3´
3´ CCTAGG 5´
5´ AGGAGG 3´
3´ TCCTCC 5´
8. A Bio 102 student gets so excited about the tyrosinase experiment that she decides to try
to clone the tyrosinase gene. She grinds up some potato, extracts the DNA from it and
digests the DNA with two different restriction enzymes (separately, not together): EcoRI
and BamHI. She then obtains a cloning vector and digests it with the same two enzymes.
She then runs a gel, which is shown at the right.
a. Which enzyme would she want to use for cloning the potato DNA: EcoRI, or
BamHI? Explain why you made your choice.
BamHI, because it only cuts the plasmid once—if you cut the plasmid twice, then
both pieces must go back together along with your insert in order to get a
functional recombinant plasmid.
b. Notice that the cloning vector made nice, tight bands on the gel, but the potato DNA
just looks like a smear with no distinct bands. However, this is just what the student
expected, so she’s not worried about it at all. Explain why this is the expected result.
Plasmids are small and might have only one or two cut sites for a particular enzyme.
The potato genome is huge and will have hundreds or thousands of sites for that
enzyme. So we expect many, many more fragments, leading to the smeary
appearance. That’s OK, though, because after cloning all the fragments (to make
a library), we’ll have a way to identify the one correct clone.
c. The student now mixes the potato DNA (digested with the enzyme you specified in part A) with cloning
vector DNA (digested with the same enzyme). She then adds the mixture to E. coli cells that have been
treated with CaCl2, heats briefly to 42°C, adds growth medium and incubates for an hour. What would be
her next step? Be as specific as possible.
Now the cells would be put on an antibiotic-containing plate. This will kill any cell that didn’t get a
plasmid and allow those that did to grow into colonies.
d. Unfortunately, after doing the next step as you specified, she doesn’t get a single bacterial colony. Not
even one! When she reviews her procedure, she realizes she left out a critical step. What did she forget,
and why would this be necessary?
She forgot to add the ligase! Ligase enzyme is needed to join the potato DNA with the cloning vector to
make a single, circular recombinant DNA molecule.
9. Suppose you want to clone the gene for human Hexokinase, so that you can use bacteria to produce the
protein and obtain it in pure form for further study of its activity.
a. Your first task is to isolate the hexokinase gene from the human genome. Assuming you have some
human DNA on hand and access to the genomic databases on the Web, what technique might you use to
obtain pure hexokinase DNA?
Since you can use the human genome database to determine the sequence of the enzyme, you can design
primers that could be used to amplify only the hexokinase gene from a human DNA sample by PCR.
b. In your initial attempt, you succeed in obtaining hexokinase DNA and ligating it into a plasmid vector,
but when you transform the recombinant plasmid into bacteria, you get no hexokinase protein produced at
all. When you discuss your problem with your friend, she suggests that you might want to start with
mRNA instead of with DNA. What problem can be overcome by starting with mRNA instead?
The hexokinase gene, like most human genes, probably contains introns. Bacteria can’t splice out introns,
so the bacteria couldn’t make the correctly processed mRNA (with an uninterrupted coding
sequence), so they can’t make the hexokinase protein starting with the complete human gene.
c. What will you need to do with your mRNA before you can make a new recombinant plasmid?
Use reverse transcriptase to make a DNA copy of it (convert it to cDNA).
d. What else would you need to provide in order for the bacteria to correctly transcribe and translate the
human hexokinase gene?
The cDNA won’t include a promoter (-10 and -35 sequences, for bacteria) or a Shine-Dalgarno sequence,
so the bacteria may not correctly recognize the gene unless these are added. An expression vector is
commonly used for cloning cDNA for this reason.
e. After consulting genome databases, you find that the human
hexokinase amino-acid sequence begins with Met-Trp-Lys-TrpTrp-Met. But the protein made from your plasmid begins with
Met-Trp-Met-Trp-Trp-Met! A mutation must have occurred!
Below, write the DNA sequence of the non-template strand for
the beginning of the actual (unmutated) hexokinase gene. Don’t
forget 5′ and 3′ ends.
There is only one codon for Met, AUG, and only one codon for
Trp, UGG. So the mRNA sequence for the un-mutated gene
must be: 5′-AUG-UGG-(Lys)-UGG-UGG-AUG.
But, there are two possible codons for Lys: AAA and AAG.
How could we know which one it was? Well, in the mutated
version, Lys is changed to Met (AUG). AAG could change
to AUG with only one mutation, while AAA would need
two mutations to change to AUG. So it’s most likely that the
Lys codon was AAG. Now we have 5′- AUG-UGG-AAGUGG-UGG-AUG for the mRNA.
Now, the question asks us for the non-template strand of the DNA. The non-template strand looks like the
mRNA, but of course it would have T’s, not U’s. So: 5′-ATGTGGAAGTGGTGGATG.
f.
Perhaps you could salvage your recombinant protein by treating it with a mutagen in order to get a
reversion mutation that restores the correct amino-acid sequence! Which of the following mutagens might
work? Circle all that apply.
1) ethidium bromide (makes one-base insertions or deletions)
2) aminopurine (causes A→G substitutions)
3) imidazolecarboxamide (causes A→T substitutions)
This could work because it could change the Met codon (TAC on the template strand) to TTC, which
would be AAG (Lys) in the mRNA.
4) aflatoxin B1 (allows any nucleotide to base-pair with a T during replication)
This could work because it could affect the T in the ATG Met codon; if T paired with T, we would get
TTG on the template strand, producing an AAC Lys codon.
10. Explain one significant problem with using retroviruses for gene therapy.
(1) Because they insert DNA randomly into host DNA, they could produce a mutation
(2) Because they contain a promoter, they could affect expression of other genes near the insertion
Matching:
1. In our example of cloning insulin, we used several different enzymes. Match each enzyme below with its
function in the cloning process.
F
Reverse transcriptase
G
Restriction enzyme
D
DNA ligase
C
Taq DNA polymerase
J
DNA helicase
a. Separating strands of DNA for PCR
b. Causing “sticky ends” to hydrogen bond
c. Copying a specific region of DNA
d. Covalently joining a DNA fragment to a cloning vector
e. Transcribing the recombinant gene
f. Making a DNA copy of RNA
g. Cutting DNA backbone at a specific sequence
h. Replicating plasmid
j. Not used in cloning
2. Matching. Match the enzymes on the left with their roles in gene cloning on the right. You may use the letters
once, more than once or not at all, but only one letter per blank.
E
F
A
B
F
DNA ligase
DNA polymerase
Restriction enzyme
Reverse transcriptase
Helicase
a. Cut expression vector to produce “sticky ends”
b. Produce cDNA from mRNA
c. Separate strands of double-stranded DNA
d. Produce many copies of a specific DNA sequence
e. Make covalent bonds between two DNA molecules
f. Not used in gene cloning