AP Biology : DNA, RNA, and Proteins

Study concepts, example questions & explanations for AP Biology

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Example Questions

Example Question #1 : Understand Regulation Of Transcription

What is the function of RNA polymerase in prokaryotes? 

Possible Answers:

None of the other answers 

Transcribe RNA from a DNA template

Replicate DNA during cell division

Translate RNA to protein

Repress transcription by binding operators

Correct answer:

Transcribe RNA from a DNA template

Explanation:

The correct answer is transcribe RNA from a DNA template. RNA polymerases are DNA-dependent, meaning that they require a DNA template; however, the new daughter strand that they create is composed of RNA. This RNA will then be translated into a functional protein by prokaryotic ribosomes. 

Example Question #1 : Understanding Prokaryotic Genes

Which of the following is true of Rho-dependent transcription termination in prokaryotes?

Possible Answers:

Rho uses GTP as a source of energy

Rho attaches to a Rho recognition site on the DNA template 

None of the other answers

Rho unwinds the DNA-RNA helix at the transcription bubble

Rho is a nuclease that terminates transcription by cutting template DNA

Correct answer:

Rho unwinds the DNA-RNA helix at the transcription bubble

Explanation:

Rho attaches to a Rho recognition site on the mRNA strand and uses ATP to move along the mRNA strand towards RNA polymerase. When RNA polymerase pauses at the terminator, Rho unwinds the DNA-RNA hybrid. RNA polymerase, Rho, and the newly synthesized mRNA are released. 

Example Question #1 : Understanding Prokaryotic Genes

Prokaryotic genes are often grouped together based on function. What are these groups of genes called?

Possible Answers:

Promoter sequences

Operator sequences

Linked genes

Operons

Clusters

Correct answer:

Operons

Explanation:

Bacteria organize some of their genes into operons. Operons contain genes of a similar function grouped together, and these genes are all transcribed together. For example, the lac operon involves the three genes required for breaking down lactose. There is no point in only transcribing one or two of the three genes since they are all required to break down lactose. Thus, they are under the control of a single operator and are all transcribed when the operator is active.

Example Question #1 : Understanding Prokaryotic Genes

Bacterial plasmids can __________.

Possible Answers:

be present in multiple copies

All of these

transfer between bacterial cells

integrate into the bacterial genome

carry antibiotic resistance genes

Correct answer:

All of these

Explanation:

Plasmids are small pieces of DNA that are not part of a bacteria's genome. The genes contained on plasmids are not necessary for proper function of the bacteria. However, bacterial plasmids can carry genes to confer antibiotic resistance, and commonly do. Plasmids can be transferred between bacteria via conjugation, and can be integrated into their genomic DNA.. Plasmids are usually present in more than one copy per cell.

Example Question #161 : Dna, Rna, And Proteins

Which of the following technological advances allows for the creation of a “recombinant plasmid”? 

Possible Answers:

Viral vectors

cDNA libraries

Restriction endonucleases

Watson-Crick base pairing

Sanger sequencing

Correct answer:

Restriction endonucleases

Explanation:

Restriction Endonucleases, or otherwise known as restriction enzymes, allows biologist to “cut and paste” different DNA sequences together. The use of restriction endonucleases is critical for the creation of recombinant plasmids. Viral vectors is incorrect, as viral vectors are useful in the application of recombinant DNA plasmids, delivery to host cell, but not in forming.

Example Question #1 : Understand Regulation Of Transcription

A mutation in the bacterial cell escherichia coli makes it unable to break down the sugar lactose. Which of the following is not a potential mutation that could result in this phenotype?

Possible Answers:

A frameshift mutation in the LacZ coding region

A mutation in the operator that blocks the activator from binding

A mutation in the region DNA polymerase binds to in the promoter

A point mutation in the LacZ gene that abolishes enzyme activity

Correct answer:

A mutation in the region DNA polymerase binds to in the promoter

Explanation:

DNA polymerase is not involved in the process of transcription. The incorrect answer could be corrected if it was rewritten as "a mutation in the promoter region where the RNA polymerase binds to."

Example Question #1 : Understanding Prokaryotic Genes

When a repressor binds to the __________, the LacZ gene does not get expressed. 

Possible Answers:

operator

gene

terminator

promoter

Correct answer:

operator

Explanation:

The region of a gene where activator/repressors bind in prokaryotic organisms is called an operator. The protein that's bound to the operon strongly influences the level of gene expression.

Example Question #9 : Understand Regulation Of Transcription

What is an important distinction between the process of translation in prokaryotes versus eukaryotes?

Possible Answers:

Prokaryotes produce a different set of amino acids from translation than eukaryotes

Eukaryotes perform translation before transcription, while prokaryotes perform transcription first

Only eukaryotes perform translation

Translation may occur at the same time as transcription in prokaryotes; they always occur separately in eukaryotes

Prokaryotes do not use RNA; eukaryotes do

Correct answer:

Translation may occur at the same time as transcription in prokaryotes; they always occur separately in eukaryotes

Explanation:

The lack of a nuclear membrane in prokaryotes has the advantage of allowing the cell to translate RNA as it is transcribed from DNA. This means that even before the full RNA is produced, the protein coded by that RNA can start being made. Eukaryotes produce RNA inside the nucleus, so it must first be fully transcribed and undergo modifications before it can be moved to the cytoplasm, where translation occurs.

Example Question #11 : Transcription

Hydrogen bonds form between RNA nucleotide residues and DNA nucleotide residues, forming a temporary DNA-RNA hybrid. This process is a part of which phase of prokaryotic transcription?  

Possible Answers:

Termination

Template recognition

Elongation

Initiation

Correct answer:

Elongation

Explanation:

Prokaryotic transcription has three essential steps: initiation, elongation, and termination. The initiation process involves the binding of RNA polymerase to the correct region of DNA, and is characterized by the binding of the sigma factor to the RNA polymerase. Elongation occurs as the RNA strand is synthesized from the DNA template. Termination occurs when the RNA polymerase enzyme encounters a rho factor or particular DNA structure that causes it to release the DNA strand and cease RNA synthesis.

It is during the elongation process that RNA nucleotides are matched to the DNA template. The temporary DNA-RNA hybrid exists only briefly at the point of transcription before phosphodiester bonds form between adjacent ribonucleotides.

Example Question #11 : Understand Regulation Of Transcription

Which of these does not contribute to the termination of RNA transcription in bacteria?

Possible Answers:

Reattachment of the sigma factor

Disruption of uracil-adenine RNA-DNA hybrid

Rho factor

Hairpin loops

Correct answer:

Reattachment of the sigma factor

Explanation:

The sigma factor is an important part of initiation for prokaryotic transcription. Once transcription has been initiated, however, the sigma factor is released during elongation. RNA polymerase synthesizes the RNA product until it is interrupted in one of two ways. In rho-dependent termination, a rho factor protein interferes with RNA polymerase binding and causes it to release the DNA strand. In rho-independent termination, structural features of the DNA cause RNA polymerase to become detached. The structures include hairpin loops, which generate steric hindrance, and adenine-rich sequences, which lead to weak binding of the RNA product to the DNA template.

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