All GRE Subject Test: Biology Resources
Example Questions
Example Question #1 : Understanding Restriction Endonucleases And Rfl Ps
Which of the following is a reason that cDNA clones of eukaryotic genes are capable of being expressed in prokaryotic cells?
I. cDNA clones do not contain any of the introns present in the genomic DNA
II. Prokaryotes have the basic translational machinery needed to express these genes
III. Prokaryotes are capable of making similar post-transcriptional modifications as eukaryotes
I, II, and III
I only
II and III
I and II
I and II
cDNA clones are incredibly powerful tools that can be used to express eukaryotic proteins in prokaryotic cells. One of the primary reasons that this is possible is because both eukaryotes and prokaryotes contain very similar systems of translating mRNA transcripts. Another reason is because the cDNA clone does not contain any of the information that would normally be spliced (introns). Prokaryotes do not make the same post-transcriptional modifications that eukaryotic cells do, and must have introns removed by external means in order to create viable protein products.
Example Question #1 : Dna Analysis
Which of the following DNA sequences could most likely be cleaved by an endonuclease?
5'-GGAACC-3'
3'-CCTTGG-5'
5'-TATACC-3'
3'-ATATGG-5'
5'-CCTAGG-3'
3'-GGATCC-5'
5'-CGATTA-3'
3'-GCTAAT-5'
5'-CCTAGG-3'
3'-GGATCC-5'
Endonucleases cut DNA at restriction sites. These restriction sites are palindromic, meaning that they read the same forward as they do backwards.
The sequence 5'-CCTAGG-3' will have the complementary strand 3'-GGATCC-5'. Since this sequence is read the same forward as it is backwards, it is a palindromic sequence that can be potentially cleaved by an endonuclease.
5'-CCT|AGG-3'
3'-GGA|TCC-5'
Example Question #2 : Dna Analysis
When performing recombinant DNA techniques, it is often necessary to cut desired DNA sequences with restriction enzymes. Where are these enzymes typically isolated from?
Animals
Bacteria
Fungi
They are synthetic
Plants
Bacteria
The correct answer is bacteria. Both bacteria and archae utilize restriction enzymes as a defense mechanism to cut foreign intracellular DNA (specifically DNA of bacteriophage). Restriction enzymes are named based on the organism in which they were discovered - for example EcoRI is named after E.coli, and was the first restriction enzyme, found in the R strain.
Example Question #2 : Dna Analysis
After a double digest with EcoRI and HindIII, a gel electrophoresis shows that you have several restriction fragment bands. You see bands of lengths 3kb, 4kb and 5kb. From previous tests, you know that there are two EcoRI restriction sites and two HindIII restriction sites in the DNA fragment you are studying. How many restriction fragments are really in your reaction?
Six
Four
Three
It cannot be determined from the given information
Five
Five
The correct answer is five. Although you can only see three bands in your gel, a DNA fragment with four different restriction sites will be cut into five distinct pieces (if you're having trouble picturing this, drawing it out helps). In this case, some of the bands must be matching sizes, so they do not appear to be separate on your gel. For instance, there may be two 3kb bands, two 4kb bands, and one 5kb band.
Example Question #1 : Understanding Dna Sequencing
The determination of the order of the four bases—adenine, thymine, cytosine, and guanine—in a strand of DNA is termed __________.
DNA amplification
DNA sequencing
gene flow
DNA replication
genetic recombination
DNA sequencing
DNA sequencing is used to determine the exact order of nucleotides that compose a given DNA molecule. This process can refer to any method that is used to determine the order of the nitrogenous bases in a strand of DNA.
DNA amplification is the duplication of regions of DNA to form multiple copies of a specific portion of the original region. DNA replication is the process of producing two identical replicas from one original DNA molecule. Gene flow is the movement of alleles from one population to another, owing to migration of individual organisms. Genetic recombination is the generation of new combinations of alleles on homologous chromosomes due to the exchange of DNA during crossing over.
Example Question #1 : Understanding Dna Sequencing
Which of the following accounts for the inability of dideoxynucleotide triphosphates to further polymerize in Sanger DNA sequencing?
Lack of a hydroxyl group on the 3' carbon
Presence of a hydroxyl group on the 2' carbon
Lack of a hydroxyl group on the 2' carbon
Lack of a triphosphate moiety on the 5' carbon
Presence of a hydroxyl group on the 3' carbon
Lack of a hydroxyl group on the 3' carbon
Dideoxynucleotide triphosphates (ddNTPs) lack a 3' hydroxyl group, which is necessary for further polymerization past the ddNTP. In the absence of this 3' free hydroxyl group, ddNTPs cannot be linked to the next incoming nucleotide's 5' end. This halts DNA synthesis, and is the basis of Sanger (chain-termination) sequencing. Ordinary nucleotides do possess a 3' hydroxyl group, allowing them to polymerize further. Sanger sequencing takes advantage of this difference to terminate synthesis at various points.
Example Question #1 : Understanding Dna Sequencing
Traditional Sanger style DNA sequencing relies on a method called chain termination. What type of molecule is used to terminate DNA chains to create molecules of all possible lengths covering the fragment to be sequenced?
Purines
Cyclic nucleotides
Deoxyribonucleotides
Deoxyadenosine
Dideoxynucleotides
Dideoxynucleotides
The key to the chain termination method used in Sanger sequencing is the dideoxynucleotide. These modified bases, designated as ddNTP's rather than dNTP's, are missing their 3' hydroxyl group. This prevents further elongation of DNA strands being synthesized by a polymerase since DNA polymerases all require a 3' hydroxyl as their substrate. In the context of a whole sequencing reaction done before actual DNA sequencing, this creates strands of all possible lengths along your desired sequence.
Example Question #1 : Understanding Cloning And Pcr
A researcher is performing PCR to amplify a sample of DNA. Unfortunately, he forgot to add the DNA primer prior to starting the experiment. Which of the following results is he most likely to observe?
The reaction will be completely unsuccessful
The reaction will work, but amplify a region that was not his target
The reaction will work, but the product will contain many undesired mutations
The reaction will work, but at a significantly slower rate
The reaction will be completely unsuccessful
The primers are absolutely essential for the functionality of a PCR reaction. The heat-resistant DNA polymerases need the RNA primers to anneal before they can begin replicating DNA. Without the primers, there is no way for the DNA polymerase to function and, therefore, no product will be obtained.
Primers are essential for any DNA replication process and are synthesized by primase during in vivo replication. For PCR, primers targeted to specific genes are artificially created and introduced to the sample. Only when the primers are bound will DNA polymerase be recruited to the replication site.
Example Question #2 : Understanding Cloning And Pcr
A student researcher wants to clone the human hemoglobin gene (HBB) into an expression vector so that he can express HBB in mouse cells and observe the resulting phenotype.
Which sequence of techniques will allow the student researcher to successfully clone the HBB gene?
None of the other answers
1. Amplify HBB via PCR
2. Use a restriction enzyme to digest the expression vector
3. Ligate HBB and the digested vector
1. Use a restriction enzyme to digest HBB
2. Amplify the expression vector via PCR
3. Ligate the digested HBB and vector
1. Ligate HBB
2. Use a restriction enzyme to digest the expression vector
3. Amplify HBB and the digested expression vector via PCR
1. Use a restriction enzyme to digest the expression vector
2. Ligate HBB and the digested vector
3. Amplify HBB via PCR
1. Amplify HBB via PCR
2. Use a restriction enzyme to digest the expression vector
3. Ligate HBB and the digested vector
The most logical procedure for the student is to first use PCR to amplify the HBB gene from human genomic DNA with primers that have restriction sites at the ends (this will faciliate ligation to a restriction enzyme digested vector). Next, the expression vector should be digested using a restriction enzyme, and then the digested vector and the amplified HBB gene can be ligated together. The final product will be two segments of the original vector with the HBB gene spliced between them.
Example Question #2 : Understanding Cloning And Pcr
A student researcher wants to change five consecutive base pairs in the middle of a PCR amplified fragment of DNA. What technique is best suited for this experiment?
None of the other answers
PCR mutagenesis
Restriction enzyme digestion
DNA ligation
Electrophoretic mobility shift assay (ESMA)
PCR mutagenesis
The most suitable technique is PCR mutagenesis. The student will design primers that anneal imperfectly at the desired site of mutagenesis. These primers will contain the new 5-base pair sequence to be inserted in the new strand. One PCR will amplify a fragment containing the new mutant sequence and all upstream sequences. A second PCR will amplify a fragment containing the mutant sequence and all downstream sequences. Finally, a PCR will amplify one sequence from the two templates, using the original primers used to amplify the full-length fragment.
Certified Tutor
Certified Tutor