GRE Subject Test: Biology : DNA Analysis

Study concepts, example questions & explanations for GRE Subject Test: Biology

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Example Questions

Example Question #1 : Dna Analysis

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

Possible Answers:

I and II

I only

II and III

I, II, and III

Correct answer:

I and II

Explanation:

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 #2 : Dna Analysis

Which of the following DNA sequences could most likely be cleaved by an endonuclease?

Possible Answers:

5'-CCTAGG-3'

3'-GGATCC-5'

5'-TATACC-3'

3'-ATATGG-5'

5'-GGAACC-3'

3'-CCTTGG-5'

5'-CGATTA-3'

3'-GCTAAT-5'

Correct answer:

5'-CCTAGG-3'

3'-GGATCC-5'

Explanation:

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 #3 : 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?

Possible Answers:

Fungi

Plants 

Animals

They are synthetic

Bacteria

Correct answer:

Bacteria

Explanation:

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 #4 : 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?

Possible Answers:

Five

Six

Three

Four

It cannot be determined from the given information

Correct answer:

Five

Explanation:

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 #5 : Dna Analysis

The determination of the order of the four bases—adenine, thymine, cytosine, and guanine—in a strand of DNA is termed __________.

Possible Answers:

DNA amplification

genetic recombination

DNA sequencing

gene flow

DNA replication

Correct answer:

DNA sequencing

Explanation:

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 #6 : Dna Analysis

Which of the following accounts for the inability of dideoxynucleotide triphosphates to further polymerize in Sanger DNA sequencing?

Possible Answers:

Lack of a triphosphate moiety on the 5' carbon

Lack of a hydroxyl group on the 2' carbon

Presence of a hydroxyl group on the 2' carbon

Lack of a hydroxyl group on the 3' carbon

Presence of a hydroxyl group on the 3' carbon

Correct answer:

Lack of a hydroxyl group on the 3' carbon

Explanation:

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 #7 : Dna Analysis

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?

Possible Answers:

Cyclic nucleotides

Deoxyadenosine

Deoxyribonucleotides

Dideoxynucleotides

Purines

Correct answer:

Dideoxynucleotides

Explanation:

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 #8 : Dna Analysis

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?

Possible Answers:

The reaction will work, but at a significantly slower rate

The reaction will work, but the product will contain many undesired mutations

The reaction will be completely unsuccessful

The reaction will work, but amplify a region that was not his target

Correct answer:

The reaction will be completely unsuccessful

Explanation:

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 #1 : Dna Analysis

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?

Possible Answers:

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

None of the other answers

1. Use a restriction enzyme to digest HBB

2. Amplify the expression vector via PCR 

3. Ligate the digested HBB and vector

1. Amplify HBB via PCR

2. Use a restriction enzyme to digest the expression vector

3. Ligate HBB and the digested vector

Correct answer:

1. Amplify HBB via PCR

2. Use a restriction enzyme to digest the expression vector

3. Ligate HBB and the digested vector

Explanation:

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 #9 : Dna Analysis

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? 

Possible Answers:

PCR mutagenesis 

Electrophoretic mobility shift assay (ESMA)

DNA ligation 

Restriction enzyme digestion

None of the other answers

Correct answer:

PCR mutagenesis 

Explanation:

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. 

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