- A genetics student analyzes the sugar components of nucleic acids and notes that DNA contains deoxyribose while RNA contains ribose. This structural difference has several functional consequences. Which statement best explains why this sugar difference affects nucleic acid stability?
Answer: Deoxyribose has fewer hydroxyl groups, making DNA less reactive and more chemically stable than RNA
When you encounter questions about nucleic acid structure, focus on how small molecular differences create major functional consequences. The key here is understanding how sugar composition affects chemical reactivity and stability.
The structural difference between DNA and RNA sugars is crucial: deoxyribose (in DNA) lacks a hydroxyl (-OH) group at the 2' carbon that ribose (in RNA) possesses. This seemingly minor difference has profound implications for stability. Fewer hydroxyl groups mean fewer sites for chemical reactions to occur. The 2'-OH group in RNA makes it much more susceptible to hydrolysis reactions, where water molecules can attack and break the phosphodiester bonds in the RNA backbone. DNA, lacking this reactive group, is significantly more chemically stable and less prone to spontaneous degradation. This explains why DNA serves as long-term genetic storage while RNA typically functions in shorter-term processes.
Looking at the wrong answers: B incorrectly suggests ribose's extra hydroxyl groups create stronger hydrogen bonds - while RNA can form more hydrogen bonds, this actually increases reactivity rather than stability. C is factually wrong because deoxyribose doesn't contain extra methyl groups; it simply lacks a hydroxyl group. D reverses the truth - ribose's structure makes RNA more susceptible, not resistant, to enzymatic breakdown.
Study tip: Remember this pattern: fewer reactive groups = greater stability. This principle applies throughout biochemistry, from nucleic acids to proteins. When comparing DNA and RNA properties, always consider how that missing 2'-OH group in DNA affects the molecule's behavior.
- During RNA transcription, the enzyme RNA polymerase synthesizes RNA using DNA as a template. If the template DNA strand has the sequence 3'-TACGGAATC-5', what will be the sequence of the newly synthesized RNA molecule?
Answer: 5'-AUGCCUUAG-3' with proper RNA nucleotide incorporation and directionality
When you encounter RNA transcription questions, remember that RNA polymerase reads the DNA template strand in the 3' to 5' direction and synthesizes the new RNA strand in the 5' to 3' direction, following complementary base pairing rules.
Given the template DNA strand 3'-TACGGAATC-5', RNA polymerase will read this sequence and create a complementary RNA strand. The base pairing rules for transcription are: A (DNA) pairs with U (RNA), T (DNA) pairs with A (RNA), G pairs with C, and C pairs with G. Reading the template 3' to 5', the complementary RNA sequence becomes 5'-AUGCCUUAG-3'.
Answer A is correct because it shows the proper RNA nucleotides (U instead of T) and the correct 5' to 3' directionality of the newly synthesized RNA strand.
Answer B contains the correct RNA nucleotides but shows incorrect directionality (3' to 5'), which violates the fundamental rule that RNA synthesis always proceeds 5' to 3'.
Answer C maintains proper directionality but uses DNA nucleotides (T instead of U), forgetting that transcription produces RNA, not DNA.
Answer D represents the same sequence as the template strand rather than its complement, suggesting the RNA would be identical to the template instead of complementary to it.
Study tip: For transcription questions, always remember the acronym "5-3-U": RNA synthesis goes 5' to 3' and uses U instead of T. This will help you quickly eliminate answers with wrong directionality or incorrect nucleotides.
- A student examines the chemical structure of nucleotides and notes that the phosphate group has a negative charge at physiological pH. How does this property affect the overall structure and behavior of nucleic acid molecules in aqueous solution?
Answer: The negative charges make nucleic acids soluble in water and require cations for structural stability
When you encounter questions about nucleic acid structure, focus on how the phosphate groups create both opportunities and challenges for these massive molecules in cellular environments.
The phosphate groups in nucleotides carry negative charges at physiological pH because they lose protons from their hydroxyl groups. This creates a molecule that's inherently water-soluble due to favorable interactions between the charged phosphates and polar water molecules. However, this same negative charge creates a structural problem: the sugar-phosphate backbone becomes a chain of negative charges that strongly repel each other. Without some way to stabilize this structure, the DNA double helix would be electrostatically unstable. Cations like Mg²⁺ and histone proteins provide this stabilization by neutralizing or shielding the negative charges, allowing the compact structures we see in chromosomes.
Choice A is incorrect because like charges repel, not attract—negative phosphates would push nucleic acids apart, not bring them together. Choice B gets the solubility right but misses the crucial role of cations; also, DNA strands are held together by hydrogen bonds in base pairs, not separated by phosphate repulsion. Choice C incorrectly suggests that base pairing eliminates electrostatic effects, but base pairing occurs between bases, not phosphates—the backbone charges remain problematic.
Remember that nucleic acids face a fundamental challenge: they must be water-soluble to function in cells, but their highly charged nature requires additional stabilization mechanisms. This dual requirement shapes much of chromosome structure and DNA packaging.
- A molecular biologist studies the 3D structure of tRNA molecules and observes that they fold into an L-shaped tertiary structure. This folding is primarily stabilized by base pairing between different regions of the same RNA molecule. What term best describes this type of base pairing?
Answer: Intramolecular base pairing creating secondary and tertiary structure within single tRNA
When analyzing RNA structure, you need to understand the difference between intermolecular and intramolecular interactions. tRNA molecules are fascinating examples of how a single RNA strand can fold into complex 3D structures through strategic base pairing within itself.
The correct answer is B because tRNA folding involves intramolecular base pairing - interactions between different regions of the same RNA molecule. The single tRNA strand contains complementary sequences that pair with each other, creating the characteristic cloverleaf secondary structure that then folds into the L-shaped tertiary structure. This internal base pairing stabilizes the molecule's functional conformation.
Answer A is incorrect because it describes intermolecular interactions between separate tRNA molecules, not the folding of individual tRNA structures. While tRNA does interact with other molecules in the ribosome, the L-shaped folding occurs within each individual tRNA molecule.
Answer C misses the key distinction being tested. While tRNA does follow Watson-Crick base pairing rules (A-U, G-C), this answer doesn't address whether the pairing is intra- or intermolecular, which is the central concept here.
Answer D is wrong because it suggests pairing only occurs between the 5' and 3' ends. In reality, tRNA's internal base pairing involves multiple regions throughout the molecule - the acceptor stem, D arm, anticodon loop, and TψC arm all participate in creating the stable structure.
Remember: when you see questions about RNA folding, focus on whether the interactions occur within one molecule (intramolecular) or between different molecules (intermolecular). This distinction is crucial for understanding nucleic acid structure and function.
- In eukaryotic cells, newly transcribed mRNA undergoes several modifications before translation. If a researcher treats cells with an inhibitor that specifically blocks the addition of the 5' cap structure, what would be the most likely consequence for the affected mRNA molecules?
Answer: The mRNA molecules would be rapidly degraded by cellular exonuclease enzymes
When you encounter questions about mRNA processing in eukaryotes, focus on the three key modifications: 5' capping, 3' polyadenylation, and splicing. Each serves a specific protective or functional role.
The 5' cap is a modified guanosine nucleotide added to the 5' end of newly transcribed mRNA. This structure serves two critical functions: it protects the mRNA from degradation by 5' exonucleases (enzymes that digest RNA from the 5' end) and helps ribosomes recognize and bind to the mRNA for translation initiation. Without this cap, the mRNA becomes highly vulnerable to cellular degradation machinery.
When an inhibitor blocks 5' cap addition, the unprotected mRNA molecules would be rapidly degraded by cellular exonuclease enzymes, making choice B correct.
Choice A is wrong because uncapped mRNA is actually translated less efficiently, not more efficiently, since ribosomes have difficulty recognizing and binding to uncapped transcripts. Choice C incorrectly suggests that capping affects nuclear export - while the cap does play a role in export, mRNA molecules wouldn't remain "permanently attached" to the nuclear membrane; they would either be exported uncapped or degraded. Choice D confuses capping with splicing - these are separate processes, and blocking cap addition wouldn't cause excessive splicing of exons.
Remember that mRNA processing modifications work as a quality control and protection system. When you see questions about blocking any of these modifications, think about what protective function is lost and what cellular machinery might then act on the unprotected mRNA.
- A researcher studies RNA molecules and observes that some can fold into complex three-dimensional structures with catalytic activity. What property of RNA structure enables this folding capability that is generally not observed in DNA?
Answer: RNA is typically single-stranded, permitting intramolecular base pairing and complex folding patterns
When you encounter questions about RNA's unique structural capabilities, focus on the fundamental differences between RNA and DNA that allow RNA to perform diverse functions beyond just storing genetic information.
RNA's ability to fold into complex three-dimensional structures with catalytic activity (like ribozymes) stems primarily from its single-stranded nature. Unlike DNA's double helix, single-stranded RNA can fold back on itself, creating intramolecular base pairs between complementary regions within the same molecule. This intramolecular base pairing generates complex secondary structures like hairpins, loops, and bulges, which then fold further into intricate tertiary structures capable of catalysis.
Let's examine why the other options fall short: Option A incorrectly suggests that uracil provides more flexible base pairing than thymine. While RNA does contain uracil instead of thymine, this substitution doesn't significantly impact folding flexibility—both form similar hydrogen bonding patterns with adenine. Option C mentions ribose's additional hydroxyl group, and while this 2'-OH group does contribute to RNA stability and function, it's not the primary factor enabling complex folding. The single-stranded nature is far more critical. Option D assumes RNA length determines folding capability, but this is incorrect—many functional RNA molecules are quite long, and folding depends on sequence complementarity, not molecule size.
Remember this key principle: RNA's structural versatility comes from being single-stranded, allowing it to be both an information carrier and a functional molecule. When you see questions about RNA's catalytic or structural roles, think "single-stranded flexibility" first.
- During DNA replication, helicase enzymes unwind the double helix by breaking hydrogen bonds between base pairs. If a researcher adds excess helicase to a DNA solution containing equal amounts of the following sequences, which would be unwound most rapidly? Sequence A: 5'-ATATATATAT-3', Sequence B: 5'-GCGCGCGCGC-3', Sequence C: 5'-AATTCCGGAA-3'
Answer: Sequence A unwinds fastest due to weaker A-T base pairing requiring less energy input
When you encounter DNA replication questions involving helicase activity, focus on the fundamental differences in hydrogen bonding between base pairs. Adenine-thymine (A-T) pairs form 2 hydrogen bonds, while guanine-cytosine (G-C) pairs form 3 hydrogen bonds, making G-C pairs significantly stronger and requiring more energy to break.
Helicase must break these hydrogen bonds to unwind the double helix. Since A-T pairs have fewer hydrogen bonds, they require less energy input from helicase to separate. Sequence A (5'-ATATATATAT-3') contains only A-T base pairs, making it the easiest target for helicase unwinding activity.
Looking at the incorrect options: Option B suggests that Sequence B (rich in G-C pairs) would unwind fastest due to its alternating pattern, but this ignores the crucial fact that G-C pairs are inherently stronger regardless of their arrangement. The "regular pattern" doesn't overcome the fundamental energetic barrier. Option C proposes that Sequence C's mixed composition provides optimal unwinding, but having both strong G-C and weaker A-T pairs would actually create an intermediate unwinding rate. Option D incorrectly assumes helicase activity is independent of base composition, when in reality, the enzyme's effectiveness directly depends on the strength of bonds it must break.
For DNA structure questions on exams, always consider the molecular details first—hydrogen bond numbers, bond strengths, and energy requirements. Don't be distracted by seemingly sophisticated explanations about patterns or flexibility when basic chemistry provides the answer.
- A genetics researcher analyzes the thermal stability of various DNA sequences and creates a predictive model. According to this analysis, a 20-base pair DNA sequence with the composition 12 A-T pairs and 8 G-C pairs would have what approximate melting temperature compared to a sequence with 8 A-T pairs and 12 G-C pairs?
Answer: Approximately 15-20°C lower melting temperature due to reduced hydrogen bonding strength
When you encounter DNA melting temperature questions, focus on the fundamental difference in hydrogen bonding between base pairs. A-T pairs form 2 hydrogen bonds, while G-C pairs form 3 hydrogen bonds, making G-C pairs significantly more stable and harder to separate.
Let's compare these sequences: Sequence 1 has 12 A-T pairs and 8 G-C pairs, while Sequence 2 has 8 A-T pairs and 12 G-C pairs. Sequence 1 has 4 fewer of the stronger G-C pairs and 4 more of the weaker A-T pairs. This substantial shift in base composition creates a notable difference in overall stability. The sequence with more G-C pairs requires significantly more thermal energy to denature, resulting in a melting temperature difference of approximately 15-20°C.
Choice A correctly identifies this substantial temperature difference due to the reduced hydrogen bonding strength in the A-T rich sequence. Choice B underestimates the magnitude—a 4 base-pair shift from G-C to A-T creates more than just a "slight" difference given the 50% increase in hydrogen bonds per G-C pair. Choice C reflects the misconception that total base pair number determines melting temperature, ignoring composition entirely. Choice D incorrectly suggests A-T pairs somehow increase stability, when they actually decrease it due to weaker hydrogen bonding.
Remember this rule: each G-C pair contributes roughly 3-4°C more to melting temperature than an A-T pair. When calculating melting temperature differences, count the net change in G-C content and multiply by this factor for quick estimates.
- During DNA replication, the enzyme DNA polymerase can only synthesize new DNA in the 5' to 3' direction. Given this constraint, what structural feature of the DNA double helix creates the need for discontinuous synthesis on one strand?
Answer: The antiparallel orientation of the two DNA strands in the double helix
DNA replication questions often test your understanding of how the structural properties of DNA create functional challenges during synthesis. The key insight here is recognizing how DNA's architecture directly impacts the replication process.
DNA polymerase's 5' to 3' synthesis constraint creates a fundamental problem because of the antiparallel nature of the DNA double helix. In the double helix, one strand runs 5' to 3' while its complement runs 3' to 5' in the opposite direction. When the replication fork opens, DNA polymerase can synthesize continuously on the leading strand (following the fork in the 5' to 3' direction). However, on the lagging strand, the enzyme must work "backwards" relative to fork movement, forcing it to synthesize in short segments called Okazaki fragments. This creates the discontinuous synthesis pattern.
Option A is incorrect because groove width differences affect protein binding and recognition, not replication directionality. Option B misrepresents DNA structure—phosphodiester bonds have identical orientations in both strands, just running in opposite directions. Option D focuses on hydrogen bonding strength between base pairs, which affects stability and melting temperature but doesn't influence synthesis direction or continuity.
When studying DNA replication, always connect structure to function. Remember that antiparallel orientation is the root cause of asymmetric replication—it's why we have leading and lagging strands, Okazaki fragments, and the need for different enzymatic machinery on each strand. This structural constraint shapes the entire replication process.
- A researcher analyzes the nucleotide composition of a double-stranded DNA sample and finds that adenine comprises 28% of all bases. If this DNA sample contains 1200 total nucleotides, how many cytosine nucleotides are present?
Answer: 264 cytosine nucleotides
When you encounter DNA composition problems, remember Chargaff's rules: in double-stranded DNA, adenine equals thymine, and guanine equals cytosine. This base-pairing principle is your key to solving these questions.
Given that adenine comprises 28% of all bases, thymine must also be 28% due to A-T base pairing. Together, adenine and thymine account for 56% of the total bases (28% + 28% = 56%).
This leaves 44% for the remaining bases (100% - 56% = 44%). Since guanine and cytosine must be present in equal amounts due to G-C base pairing, each comprises 22% of the total bases (44% ÷ 2 = 22%).
With 1200 total nucleotides and cytosine representing 22% of all bases: 1200×0.22=264 cytosine nucleotides.
Looking at the wrong answers: B (336 nucleotides) represents 28% of 1200, which would be correct if you mistakenly thought cytosine equaled adenine rather than applying Chargaff's rules. C (528 nucleotides) represents 44% of 1200 – this would be the combined percentage of guanine and cytosine together, not cytosine alone. D (672 nucleotides) represents 56% of 1200, which is the combined percentage of adenine and thymine.
Study tip: Always start DNA composition problems by applying Chargaff's rules to find the percentage of each base, then multiply by the total number of nucleotides. Remember that complementary bases are always equal in double-stranded DNA. - A research team investigates DNA packaging in eukaryotic cells and discovers that the negative charges on DNA interact with positively charged histone proteins. If DNA were uncharged, what would be the most likely consequence for chromosome structure?
Answer: Chromosomes would be less organized due to weaker DNA-histone interactions and reduced packaging
When you encounter questions about DNA packaging, focus on the electrostatic interactions that drive chromatin organization. DNA's negative phosphate backbone creates strong attractions with positively charged histone proteins, forming the foundation of chromosome structure.
If DNA were uncharged, the critical DNA-histone interactions would be dramatically weakened. Histones are rich in positively charged lysine and arginine residues specifically because they need to neutralize DNA's negative charges and wrap DNA tightly around histone octamers to form nucleosomes. Without these electrostatic attractions, DNA would have much weaker affinity for histones, leading to looser, more disorganized chromatin structure. This makes B correct—chromosomes would indeed be less organized due to weaker DNA-histone interactions and reduced packaging efficiency.
A is incorrect because it ignores that DNA strands in chromosomes aren't primarily held apart by electrostatic repulsion between different DNA molecules, but rather organized through histone interactions. C misses the point entirely—while base pairing is crucial for DNA structure, histone-mediated packaging is what organizes DNA into chromosomes. Base pairing alone cannot compact meters of DNA into microscopic nuclei. D incorrectly assumes the primary consequence would be enzyme accessibility, when the fundamental issue is the loss of basic organizational structure.
Remember that chromatin organization depends heavily on charge complementarity. When you see questions about DNA packaging modifications, always consider how changes would affect the electrostatic interactions between DNA's phosphate backbone and histone proteins—this relationship drives chromosome condensation and organization.
- A biochemistry student learns that DNA polymerase requires a 3'-OH group to add new nucleotides during synthesis. RNA polymerase, however, can initiate synthesis without a primer. What structural difference between these enzymes' products explains this functional distinction?
Answer: DNA polymerase can only extend existing strands while RNA polymerase can begin new chains de novo
When you encounter questions about DNA and RNA polymerases, focus on their fundamental functional differences rather than getting distracted by structural details of their products.
The key distinction lies in how these enzymes initiate synthesis. DNA polymerase has a critical limitation: it can only add nucleotides to an existing 3'-OH group, meaning it must extend a pre-existing strand (either a primer or another DNA strand). This enzyme simply cannot start synthesis from scratch. RNA polymerase, however, possesses the unique ability to initiate synthesis de novo—it can begin creating a new RNA chain without requiring any pre-existing 3'-OH group to build upon.
This explains why DNA replication requires primers (short RNA sequences synthesized by primase) to get started, while RNA transcription can begin immediately when RNA polymerase binds to a promoter region.
Answer C correctly identifies this fundamental functional difference: DNA polymerase can only extend existing strands while RNA polymerase can begin new chains de novo.
Answer A focuses on sugar differences, but ribose versus deoxyribose doesn't explain the primer requirement—both sugars have 3'-OH groups available for extension. Answer B mentions strand number (double versus single), but this describes the final products, not why primer requirements differ. Answer D discusses base differences (thymine versus uracil), but base identity doesn't affect the enzymes' initiation mechanisms.
Remember: DNA polymerase needs a "starting point" (primer), while RNA polymerase can create its own starting point. This functional distinction is crucial for understanding replication versus transcription mechanisms.
- A student examines two nucleic acid samples under identical conditions. Sample X has a melting temperature (Tm) of 87°C while Sample Y has a Tm of 72°C. Both samples have the same length and concentration. What can be concluded about the base composition of these samples?
Answer: Sample X contains more guanine and cytosine nucleotides than Sample Y does
When you encounter questions about DNA melting temperature (Tm), you're dealing with the stability of hydrogen bonds between complementary base pairs. The melting temperature is the point where 50% of the DNA strands separate due to broken hydrogen bonds.
The key insight is that guanine-cytosine (G-C) base pairs form three hydrogen bonds, while adenine-thymine (A-T) base pairs form only two hydrogen bonds. This means G-C pairs are stronger and require more energy (higher temperature) to break apart. Therefore, DNA with a higher percentage of G-C content will have a higher melting temperature.
Since Sample X has a Tm of 87°C compared to Sample Y's 72°C, Sample X must contain more guanine and cytosine nucleotides than Sample Y. This makes option B correct.
Looking at the wrong answers: Option A is backwards—more A-T content would actually lower the melting temperature, not raise it. Option C misses the point entirely; purines (A and G) and pyrimidines (T and C) don't determine melting temperature—the specific base pairing does. Both samples would have equal amounts of purines and pyrimidines in double-stranded DNA anyway. Option D confuses DNA structure with melting behavior; the sugar backbone doesn't significantly affect hydrogen bonding between bases.
Remember this pattern: Higher Tm = more G-C content = stronger hydrogen bonding. This relationship appears frequently in molecular biology questions, so always associate melting temperature with base pair stability.
- A molecular biologist discovers a new RNA molecule that contains both guanine and uracil but completely lacks adenine and cytosine. If this RNA can form stable secondary structures through base pairing, which statement best describes the base pairing pattern?
Answer: Only wobble base pairing occurs between guanine and uracil with single hydrogen bonds
When you encounter questions about unusual nucleic acid structures, focus on the fundamental chemistry of hydrogen bonding between bases. Standard Watson-Crick base pairing (A-T and G-C) relies on complementary hydrogen bond donors and acceptors, but when forced to pair with non-complementary bases, only weaker interactions are possible.
In this RNA molecule containing only guanine and uracil, these bases must pair with each other despite being non-complementary. Guanine and uracil can form hydrogen bonds, but not in the stable, multiple-bond arrangements seen in normal base pairs. Instead, they form "wobble" base pairs—a type of non-Watson-Crick pairing that involves a single hydrogen bond and requires the bases to adopt slightly altered geometries. This wobble pairing is weaker than standard base pairing but still allows for secondary structure formation.
Option A is incorrect because G-U pairing cannot achieve the three-hydrogen-bond stability of G-C pairs—the chemical structures simply don't align properly for this. Option B is wrong because G-U pairing doesn't mimic A-T pairing; it's a distinct interaction with different geometry and bond strength. Option D is incorrect because guanine and uracil do form hydrogen bonds, not just van der Waals forces.
Remember that wobble base pairs occur naturally at the third position of codons during translation, where non-standard pairings like G-U are tolerated. When you see questions about unusual base compositions, think about the hierarchy of interactions: Watson-Crick pairs are strongest, wobble pairs are intermediate, and van der Waals forces are weakest.
- Use the table above to answer the question. A researcher measures the base composition of DNA from three different bacterial species under controlled conditions. Based on these data, which conclusion about DNA structure is most strongly supported?
Answer: Species C has the highest melting temperature due to its base composition pattern
Species C has 45% G and 45% C (90% total G+C content), which would give the highest melting temperature due to stronger G-C base pairs (3 hydrogen bonds vs 2 for A-T). All species follow Chargaff's rules (A≈T, G≈C). Choice A incorrectly focuses on purine/pyrimidine ratios rather than specific base pairing. Choice B incorrectly suggests Species B violates Chargaff's rules. Choice D is correct about Chargaff's rules but doesn't address the question focus. Choice E incorrectly suggests equal stability despite different G+C content.
- In prokaryotic cells, transcription and translation can occur simultaneously because both processes happen in the cytoplasm. However, in eukaryotes, these processes are separated spatially and temporally. What structural feature of eukaryotic mRNA allows this temporal separation to occur effectively?
Answer: The poly-A tail that provides stability during transport from nucleus to cytoplasm
When you encounter questions about prokaryotic versus eukaryotic gene expression, focus on the key difference: spatial separation. In prokaryotes, ribosomes can attach to mRNA while it's still being transcribed because everything happens in the cytoplasm. Eukaryotes evolved a different system where transcription occurs in the nucleus and translation in the cytoplasm.
The poly-A tail is crucial for making this separation work effectively. This stretch of adenine nucleotides added to the 3' end of eukaryotic mRNA serves as a protective cap that prevents degradation by cellular enzymes. Without this stability feature, mRNA would be broken down during its journey from nucleus to cytoplasm, making translation impossible. The poly-A tail essentially gives the mRNA molecule enough "shelf life" to survive transport and remain functional for protein synthesis.
Looking at the incorrect choices: Choice A describes introns, which are indeed removed during processing, but their removal happens quickly during transcription—they don't create the temporal gap between transcription and translation. Choice C is incorrect because ribosome binding sites aren't significantly modified during nuclear processing in eukaryotes. Choice D refers to a non-existent phenomenon; eukaryotes don't have different genetic codes for nuclear versus cytoplasmic processes.
Remember this pattern: when questions ask about temporal or spatial separation in eukaryotic gene expression, look for answers involving mRNA stability and transport. The poly-A tail, along with the 5' cap, are the key modifications that allow eukaryotic mRNA to function across compartments.
