Additional Questions 2 Flashcards
One of a set of alternative forms of a gene. In a diploid cell, each gene will have two of these, each occupying the same position (locus) on homologous chromosomes.
Allele
Process by which DNA sequence information can be transferred from one DNA helix (which remains unchanged) to another DNA helix whose sequence is altered.
Gene conversion
Experimental process in which two complementary nucleic acid strands form a double helix; a powerful technique for detecting specific nucleotide sequences.
Hybridization
X-shaped structure observed in DNA undergoing recombination, in which the two DNA molecules are held together at the site of crossing-
over, also called a cross-strand exchange.
Holliday junction
T/F: Homologous recombination requires relatively long regions of homologous DNA on both partners in the exchange.
True. Homologous recombination requires long stretches of nearly identical DNA to initiate and complete recombination. This ensures that recombination occurs at corresponding points along the chromosome rather than between similar but non-identical repeated DNA elements found throughout eukaryotic genomes.
T/F: Gene conversion requires a limited amount of DNA synthesis.
True. Conversion refers to a change in the frequency of nucleotide markers during recombination. Initially, each marker is equally present in the starting duplexes. However, after recombination, this balance shifts, resulting in altered frequencies such as 2:0, 4:0, or 3:1 instead of the original 1:1 (or 2:2). This change is primarily driven by mismatch repair and DNA synthesis, both of which involve some degree of DNA replication.
Discuss the following statement: “The Holliday junction contains two
distinct pairs of strands (crossing strands and noncrossing strands),
which cannot be interconverted without breaking the phosphodiester
backbone of at least one strand.”
This statement is incorrect. Crossing and noncrossing pairs of strands can be interconverted by rotation movements that do not require strand breakage.
In addition to correcting DNA mismatches, the mismatch repair system functions to prevent homologous recombination from taking place between similar but not identical sequences. Why would recombination between similar, but nonidentical sequences pose a problem for human cells?
A significant portion of the human genome consists of repetitive elements like Alu sequences, which are dispersed across chromosomes. If recombination were to occur between such sequences on different chromosomes, it could lead to translocations, potentially destabilizing the genome. Unrestricted recombination between these repeats would rapidly rearrange the genome, resulting in nonviable progeny and endangering the species. To prevent this, the mismatch repair system acts as a safeguard. Since repeated sequences vary slightly, recombination between them creates heteroduplex regions with numerous mismatches. If the mismatch repair system detects an excessive number of mismatches, it aborts the recombination process, ensuring that only nearly identical sequences at the same locus undergo successful recombination.
Length of DNA that moves from a donor site to a target site either by cut-and-paste transposition or by replicative transposition.
DNA-only transposon
Enzyme that makes a double-strand DNA copy from a single-strand RNA
template molecule.
Reverse transcriptase
RNA-containing virus that replicates in a cell by first making a double-strand DNA intermediate.
retrovirus
Rearrangement of DNA that depends on the breakage and rejoining of two DNA helices at specific sequences on each DNA molecule.
conservative site-specific recombination
T/F: When transposable elements move around the genome, they rarely integrate into the middle of a gene because gene disruption—a potentially lethal event to the cell and the transposon—is selected against by evolution.
False. Transposable elements integrate almost randomly, often disrupting or altering genes in the process. While some of these integration events are lethal to the cell and the transposable element itself, most are not. By spreading throughout the genome, even at the expense of a few cells, transposable elements ensure their persistence within the species.
Which one of the following functions best describes the role of DNA polymerase I in bacterial cells?
A. Modifies Okazaki fragments for joining into intact strands.
B. Repairs any DNA that is damaged during transcription.
C. Synthesizes most of the cellular DNA during replication.
D. Synthesizes RNA primers to initiate Okazaki fragments.
A. Modifies Okazaki fragments for joining into intact strands.
DNA polymerase I processes Okazaki fragments by removing RNA primers with its 5’-to-3’ exonuclease activity and filling in the resulting gaps with DNA. DNA polymerase III primarily handles genome replication, while DNA primase synthesizes RNA primers. DNA polymerase is not responsible for repairing damaged bases during transcription.
One problem with Taq polymerase is that it incorporates the wrong base
approximately once every 8000 bases, a frequency of mistakes that is far higher than the error rates for DNA polymerases that carry out DNA replication in cells. Which one of the following statements best explains the
high error rate of Taq polymerase in PCR?
A. Taq polymerase cannot make Okazaki fragments efficiently.
B. Taq polymerase lacks a 3’-to-5’ proofreading exonuclease.
C. Taq polymerase often falls off the DNA, interrupting synthesis.
D. Taq polymerase synthesizes DNA in the 3’-to-5’ direction.
A. Taq polymerase cannot make Okazaki fragments efficiently.
Highly accurate DNA polymerases possess 3’-to-5’ proofreading exonuclease activity, allowing them to detect and remove mismatched bases at the growing strand’s 3’ end. Okazaki fragments (Choice A) are not involved in PCR. Premature polymerase dissociation (Choice C) reduces efficiency but does not introduce errors. All DNA polymerases, including Taq polymerase, synthesize DNA in the 5’-to-3’ direction, making Choice D incorrect.
Another problem with Taq polymerase is that it is only capable of synthesizing relatively small pieces of DNA, with an upper limit of around 4000 bases. Which one of the following is a likely explanation for the inability of Taq polymerase to synthesize long pieces of DNA?
A. Taq polymerase can make only relatively short Okazaki fragments.
B. Taq polymerase cannot remove mismatched bases from the 3’ end.
C. Taq polymerase lacks the helicase required for strand separation.
D. Taq polymerase reactions lack the primase needed for new primers.
B. Taq polymerase cannot remove mismatched bases from the 3’ end.
All DNA polymerases prefer a matched base at the 3’ end of the growing strand. Without 3’-to-5’ proofreading exonuclease activity, Taq polymerase stalls at a mismatch, limiting fragment length. Choice A is incorrect as Okazaki fragments are not involved in PCR. Choice C is incorrect because PCR does not require helicase activity. Choice D is incorrect since primers are included in the PCR mix from the start, not synthesized during the reaction.
Helps to position the RNA polymerase correctly at the promoter, to aid in pulling apart the two strands of DNA to allow transcription to begin, and to release RNA polymerase from the promoter into the elongation mode once transcription has begun.
General transcription factor
Small RNA molecules that are complexed with proteins to form the ribonucleoprotein particles involved in RNA splicing.
snRNA (small nuclear RNA)
Nucleotide sequence in DNA to which RNA polymerase binds to begin transcription.
Promotor
A large protein complex containing multiple 3’-to-5’ RNA exonucleases that degrade improperly processed mRNAs, introns, and other RNA debris retained in the nucleus.
Exosome
The enzyme that carries out transcription.
RNA polymerase
RNA molecule that specifies the amino acid sequence of a protein.
mRNA (messenger RNA)
Process in which intron sequences are excised from RNA transcripts in the nucleus during the formation of messenger and other RNAs.
RNA splicing
Signal in bacterial DNA that halts transcription.
Terminator
Segment of a eukaryotic gene consisting of a sequence of nucleotides
that will be represented in mRNA or other functional RNAs.
Exon
Large multiprotein structure forming a channel through the nuclear envelope that allows selected molecules to move between nucleus and
cytoplasm.
Nuclear pore complex
T/F: The consequences of errors in transcription are less severe than those of
errors in DNA replication.
True. DNA replication errors are heritable and affect all subsequent RNA and protein products, while transcription errors are temporary and limited to a subset of RNA molecules. This explains why DNA polymerases have much lower error rates than RNA polymerases—natural selection strongly favors high-fidelity DNA replication to maintain genetic integrity across generations.
T/F: Since introns are largely genetic “junk,” they do not have to be removed
precisely from the primary transcript during RNA splicing.
False. Although intron sequences are largely dispensable, their precise removal is essential. Even a single nucleotide error during splicing could shift the reading frame, leading to an abnormal protein. The splicing machinery ensures accuracy by recognizing conserved splice site sequences, preventing potentially harmful mutations.
T/F: RNA polymerase II generates the end of a pre-mRNA transcript when
it ceases transcription and releases the transcript; a poly-A tail is then quickly added to the free 3’ end.
False. The 3’ ends of most pre-mRNA transcripts produced by RNA polymerase II are defined not by the termination point of transcription, but by cleavage of the RNA chain 10-30 nucleotides downstream of the
sequence AAUAAA.
Match the following list of RNAs with their functions.
1. blocks translation of selected mRNAs
2. modification and processing of rRNA
3. protects germ line from transposable elements
4. components of ribosome
5. splicing of RNA transcripts
6. directs degradation of selected mRNAs
7. codes for proteins
8. adaptor for protein synthesis
A. mRNA
B. rRNA
C. snoRNA
D. snRNA
E. tRNA
F. piRNA
G. miRNA
H. siRNA
A : 7
B : 4
C : 2
D : 5
E : 8
F : 3
G : 1
H : 6
Why doesn’t transcription cause a hopeless tangle? If the RNA polymerase does not revolve around the DNA as it moves, it will induce two DNA supercoils—one in front and one behind—for every 10 nucleotides it transcribes. If, instead, RNA polymerase revolves around the DNA—avoiding DNA supercoiling—then it will coil the RNA around the DNA
duplex, once for every 10 nucleotides it transcribes. Thus, for any reasonable-size gene, the act of transcription should result in hundreds of coils or supercoils…and that’s for every single RNA polymerase! So why
doesn’t transcription lead to a complete snarl?
RNA polymerase does not rotate around DNA as it transcribes, preventing RNA from wrapping around the template strand. Instead, transcription induces positive supercoiling ahead and negative supercoiling behind the polymerase. However, topoisomerases regulate this supercoiling, preventing excessive tension buildup and maintaining optimal levels for efficient transcription.
Which of the following mutational changes would you predict to be the
most deleterious to gene function? Explain your answers.
1. Insertion of a single nucleotide near the end of the coding sequence.
2. Removal of a single nucleotide near the beginning of the coding sequence.
3. Deletion of three consecutive nucleotides in the middle of the coding
sequence.
4. Deletion of four consecutive nucleotides in the middle of the coding
sequence.
5. Substitution of one nucleotide for another in the middle of the coding
sequence.
Mutations that cause frameshifts early or in the middle of a coding sequence tend to be the most harmful, as they result in a severely altered and often truncated protein. In contrast, frameshifts near the end of the sequence may have minimal effects. Deletion of three nucleotides does not disrupt the reading frame but may remove or alter an amino acid, sometimes with little impact. Single-nucleotide substitutions can be harmless if they do not change the amino acid, but if they introduce a stop codon, they can lead to a nonfunctional, truncated protein.
Consider the properties of two hypothetical genetic codes constructed
with the four common nucleotides: A, G, C, and T.
A. Imagine that one genetic code is constructed so that pairs of nucleotides
are used as codons. How many different amino acids could such a code
specify?
B. Imagine that the other genetic code is a triplet code; that is, it uses three
nucleotides to specify each amino acid. In this code, the amino acid specified by each codon depends only on the composition of the codon—not the sequence. Thus, for example, CCA, CAC, and ACC, which all have the
composition C₂A, would encode the same amino acid. How many different amino acids could such a code specify?
C. Would you expect the genetic codes in A and B to lead to difficulties in
the process of translation, using mechanisms analogous to those used in
translating the standard genetic code?
A. A genetic code based on nucleotide pairs would generate only 16 possible codons (4×4), which is insufficient to encode all 20 standard amino acids. This limitation explains why the genetic code evolved to use triplets, allowing for 64 codons—more than enough to specify all amino acids while also providing redundancy.
B. A triplet genetic code based solely on codon composition, rather than sequence order, would generate 20 unique codons: four with identical bases, twelve with two identical and one different, and four with three different bases. This system could specify a maximum of 20 amino acids but lacks the redundancy of the actual genetic code, which enhances error tolerance.
C. A doublet code could be translated using a mechanism similar to the standard genetic code. However, a triplet code based solely on nucleotide composition, without considering sequence order, would be challenging to translate since base-pairing would no longer function as a recognition mechanism. For example, an AUG codon would not pair with the same anticodon as a UGA, making accurate translation difficult.
Polycistronic mRNAs are common in prokaryotes but extremely rare in eukaryotes. Describe the key differences in protein synthesis that underlie this observation.
In eukaryotic cells, protein synthesis begins by scanning from the 5’ end of the mRNA until the first AUG codon is found, though sometimes a later AUG is used in a process called leaky scanning. This ensures ribosomes start translating near the 5’ end and must reinitiate scanning after completing translation. In contrast, prokaryotic cells initiate translation through base-pairing between mRNA sequences near the start codon and the 16S rRNA of the ribosome, allowing multiple start sites within the same mRNA. This difference enables prokaryotes to produce multiple proteins from a single polycistronic mRNA.
A hypothetical state of evolution that existed on Earth before modern
cells arose, in which RNA both stored genetic information and catalyzed
chemical reactions in primitive cells.
RNA world
What is so special about RNA that it is hypothesized to be an evolutionary precursor to DNA and protein?
RNA can store genetic information like DNA and catalyze reactions like proteins, making it a strong candidate for the origin of life. Its role in fundamental cellular reactions supports this idea. However, a clear pathway from nonliving matter to an RNA world remains unknown. Due to RNA’s instability, some speculate an earlier, more stable RNA-like molecule existed before RNA itself.
What is it about DNA that makes it a better material than RNA for the storage of genetic information?
DNA is more stable than RNA due to the absence of the 2’ hydroxyl group, which prevents self-catalyzed breakage. Its double-helix structure allows accurate repair by referencing the complementary strand. Additionally, using thymine (T) instead of uracil (U) helps detect and correct deamination damage, making DNA more resilient.
T/F: The differences in the patterns of proteins produced in different specialized cell types are accurately reflected in the patterns of expressed
mRNAs.
False. Gene expression differences between cell types are underestimated when only mRNA patterns are compared, as regulation occurs at multiple post-transcriptional levels, affecting the final protein profile.
A protein that binds to a specific DNA sequence and influences the rate at which a gene is transcribed.
Transcription regulator
Many transcription regulators form dimers of identical or slightly different subunits on the DNA. Suggest two advantages of dimerization.
Dimerization enhances DNA-binding affinity by doubling potential contacts and expands DNA-binding specificity through combinatorial pairing of subunits.
Natural defense mechanism in many organisms that is directed against foreign RNA molecules, especially those that occur in double stranded form.
RNA interference (RNAi)
A class of short noncoding RNAs that regulate gene expression; roughly one-third of human genes are thought to be regulated in this way.
MicroRNA (miRNA)
Small RNAs that transcriptionally silence intact transposon genes and
destroy any mRNA produced by them.
piRNA (piwi-interacting RNA)
A defense mechanism in bacteria that allows them to destroy viral invaders they have seen before.
CRISPR
An RNA longer than 200 nucleotides that does not encode a protein.
Long noncoding RNA (lncRNA)
T/F: Because siRNAs are so widespread among species, they are believed to be the most ancient form of RNA interference, with miRNAs being a later refinement.
True. siRNA-mediated defense mechanisms are crucial for plants, worms, and insects, although in mammals, a protein-based system has largely taken over the task of fighting off viruses.
T/F: piRNAs and crRNAs serve analogous functions; they defend against foreign invaders.
True. The CRISPR system in bacteria and the piRNA system in eukaryotes both incorporate invader sequences into genomic regions to create RNA guides for defense. CRISPR targets double-stranded DNA from external invaders, while piRNAs target single-stranded RNA from transposable elements already in the genome.
List and briefly discuss three features that make miRNAs especially useful regulators of gene expression.
First, a single miRNA can regulate a whole set of different mRNAs, so long as the mRNAs carry a common sequence in their untranslated regions. Second, regulation by miRNAs can be combinatorial, allowing
incremental changes in translation of mRNAs, which can fine-tune gene expression. Third, an miRNA occupies little space in the genome compared with a protein.
