Biochem Midterm 2: Unit 4 Nucleic Acids

4.1 | What are nucleic acids made of, and where is their information stored?

They are nucleotide polymers. Information is stored in base order; the sugar-phosphate backbone provides the supporting chain.

4.1 | Nucleoside versus nucleotide?

Nucleoside = base + sugar. Nucleotide = nucleoside + one or more phosphates. Nucleoside triphosphates are activated substrates for synthesis.

4.1 | What distinguishes DNA from RNA chemically?

DNA has 2-deoxyribose with 2′-H and bases A/G/C/T. RNA has ribose with 2′-OH and A/G/C/U.

4.1 | Which bases are purines and which are pyrimidines?

Purines: A and G, two fused rings. Pyrimidines: C, T, and U, one ring.

4.1 | How is a base attached to its sugar?

An N-glycosidic bond joins sugar C1′ to purine N9 or pyrimidine N1. Prime marks distinguish sugar carbons from base positions.

4.1 | What are the usual nucleoside names?

RNA: adenosine, guanosine, cytidine, uridine. DNA: deoxyadenosine, deoxyguanosine, deoxycytidine, and thymidine.

4.1 | What do AMP, ADP, and ATP mean, and which bonds connect their phosphates?

Adenosine with one, two, or three phosphates. Sugar-to-phosphate is a phosphate ester; adjacent phosphates in ADP/ATP are joined by phosphoanhydride connections.

4.1 | What connects neighboring nucleotides within a strand?

A 3′–5′ phosphodiester linkage: one sugar's 3′ oxygen connects through phosphate to the next sugar's 5′ oxygen. Phosphates make the backbone negative.

4.1 | How do you identify the 5′ and 3′ ends?

A typical 3′ end has a free 3′-OH; the terminal 5′ position can bear phosphate or OH. The p in pAGCT denotes a 5′ phosphate.

4.1 | What does antiparallel mean, and how are sequences reported?

Aligned duplex strands run in opposite directions. Sequences are conventionally written 5′→3′ unless labeled otherwise.

4.1 | How do you rewrite a strand versus find its reverse complement?

Rewriting the SAME strand from the opposite end only reverses order. Finding a partner also complements bases. Reverse complement of 5′-AGTCGA-3′ = 5′-TCGACT-3′.

4.1 | What are the standard DNA and RNA base pairs?

DNA: A–T and G–C. RNA: A–U and G–C. A–T has two hydrogen bonds; G–C has three.

4.1 | What stabilizes a DNA duplex?

Base pairing, base stacking, hydrophobic effects, and other noncovalent interactions. Backbones lie outside; bases stack inside, approximately perpendicular to the helix axis.

4.1 | Which DNA forms and grooves should you recognize?

B-DNA is the common right-handed form; the notes also mention A and Z forms. Major/minor grooves expose sequence-dependent chemical patterns for protein recognition.

4.1 | How does DNA structure support storage, copying, and repair?

A stable backbone preserves sequence. Complementary strands template replication and can provide information for repair of a damaged partner.

4.1 | Why is RNA more susceptible to alkaline backbone cleavage than DNA?

RNA's 2′-OH can attack an adjacent phosphate. DNA lacks that OH and is relatively more stable, although it can still be damaged.

4.1 | What are DNA melting, Tm, and reannealing?

Melting separates paired strands without normally cutting backbones. Tm is the melting midpoint. Reannealing is complementary-strand reassociation under suitable conditions.

4.1 | What generally raises or lowers DNA melting temperature?

At matched conditions, higher GC and moderate increases in salt generally raise Tm; mismatches lower it. Longer duplexes generally have higher Tm, especially compared with short duplexes.

4.1 | Which has higher expected Tm: 100 bp with 21% A or 100 bp with 23% A?

The 21% A duplex: its GC content is 58%, versus 54% for the 23% A duplex. Assume ordinary dsDNA and matched conditions.

4.1 | What are Chargaff's rules and their limitation?

For an ordinary complete duplex, A = T and G = C; all percentages total 100%. These equalities need not hold within one isolated strand.

4.1 | How do you solve base-composition and base-pair counting problems?

If A = 18%, T = 18% and C = G = 32%. A 200-bp duplex contains 400 nucleotides. GC% = G% + C%.

4.1 | What RNA structural features should you recognize?

Complementary RNA regions form stems joined by unpaired loops. RNA also forms complex folds and noncanonical base pairs; paired appearance alone does not identify DNA.

4.1 | How are eukaryotic chromosomes packaged?

DNA associates with proteins as chromatin; DNA wrapped around histone cores forms nucleosomes. A typical human diploid cell has about 2 m DNA in 46 chromosomes.

4.1 | Why do histones bind DNA, and how do they differ from transcription factors?

Basic Lys/Arg-rich histones attract negative DNA phosphates and package DNA. Many transcription factors instead recognize particular DNA sequence patterns.

4.1 | How does genome packaging affect gene expression?

Tightly packed chromatin generally reduces access to DNA. Bacterial DNA is compacted through supercoiling and associated proteins rather than ordinary eukaryotic nucleosome organization.

4.2 | What is semiconservative replication?

Each daughter duplex has one original template strand and one newly synthesized strand.

4.2 | What does DNA polymerase require?

A template, paired primer with free 3′-OH, dATP/dGTP/dCTP/dTTP, suitable divalent ions such as Mg²⁺, and the enzyme. Ordinary DNA polymerase cannot start unprimed.

4.2 | What are the directions of DNA synthesis and template reading?

New DNA grows 5′→3′ by adding to its 3′ end; polymerase reads the template 3′→5′.

4.2 | What is the chemistry of nucleotide addition?

The growing 3′-OH attacks the incoming dNTP's alpha phosphate, forming a phosphodiester bond and releasing PPi. PPi hydrolysis helps drive synthesis forward.

4.2 | What establishes replication fidelity?

Complementary base-pair geometry, polymerase discrimination, and proofreading. Incorrect nucleotides can sometimes bind and be incorporated; selection is not perfect.

4.2 | What is proofreading, and what happens if it is lost?

3′→5′ exonuclease removes mismatched terminal nucleotides. Losing it while retaining synthesis allows more errors to escape correction.

4.2 | How do E. coli DNA polymerases I and III differ?

III performs most chromosome replication and proofreads. I removes primers, fills with DNA, and participates in repair; I has both 5′→3′ and 3′→5′ exonuclease activities.

4.2 | What roles do polymerases II, IV, and V and repair pathways serve?

Repair or damage tolerance. Lesions illustrated include bulky base adducts, missing bases, and covalently joined bases.

4.2 | What is an origin of replication?

A DNA site where replication begins; E. coli's origin is OriC. AT-rich regions can facilitate opening because they are generally less stable than matched GC-rich regions.

4.2 | What do helicase, SSB, and topoisomerase do?

Helicase uses ATP to separate strands; SSB stabilizes exposed ssDNA; topoisomerase relieves torsional strain ahead of replication.

4.2 | What does primase make, and which daughter strands need it?

Primase makes short RNA primers, about 10 nucleotides in the bacterial example. Leading synthesis needs initial priming; each lagging-strand fragment needs a primer.

4.2 | Leading versus lagging strand, and why are both needed?

Antiparallel templates require continuous leading growth toward the fork and discontinuous lagging growth as Okazaki fragments away from it. Both grow 5′→3′; the notes illustrate bacterial fragments around 1000 nucleotides.

4.2 | What is the trombone model?

Changing loops in the lagging-strand template let leading and lagging polymerases work together in a coordinated replication complex.

4.2 | How are Okazaki fragments matured?

Remove RNA primers, replace them with DNA, then seal remaining nicks. Losing polymerase I's 5′→3′ exonuclease impairs primer removal, not proofreading directly.

4.2 | What exactly does DNA ligase do?

Seals an adjacent 3′-OH and 5′-phosphate into a phosphodiester connection. It cannot fill a missing stretch of nucleotides.

4.2 | How do you label a replication-fork diagram?

Identify template 5′/3′ ends and fork motion. Leading growth follows fork motion; individual lagging fragments grow oppositely. Find short primers at fragment 5′ ends.

4.2 | How do bacterial and eukaryotic replication compare?

E. coli: circular chromosome, one origin. Eukaryotic nuclear DNA: multiple linear chromosomes, many origins. Both use antiparallel templates and leading/lagging synthesis.

4.2 | What are replicons and licensing?

A replicon is DNA replicated from an origin. Licensing helps ensure each origin initiates only once per replication cycle; many origins allow large chromosomes to be copied in parallel.

4.2 | What are polymerase alpha–primase and polymerase switching?

The alpha–primase complex makes RNA primer plus short DNA. More processive polymerases then take over; the notes emphasize alpha→delta switching. Epsilon mainly supports leading synthesis and delta lagging synthesis in modern descriptions.

4.2 | What are telomeres and the end-replication problem?

Telomeres protect linear chromosome ends. Removing the terminal primer can leave an end that ordinary synthesis cannot fully fill because no suitable upstream 3′-OH is available.

4.2 | What does telomerase do, and why does it matter?

It uses its own RNA template to extend the G-rich 3′ DNA end: an RNA-dependent DNA polymerase. Protective end structures and telomere maintenance support chromosome stability; many cancers maintain telomerase activity.

4.2 | How do polymerase names identify template and product?

DNA-dependent DNA polymerase: DNA→DNA. DNA-dependent RNA polymerase: DNA→RNA. RNA-dependent DNA polymerase: RNA→DNA. RNA-dependent RNA polymerase: RNA→RNA.

4.2 | What are the requirements and stages of PCR?

Template DNA, two DNA primers, thermostable polymerase, dNTPs, and suitable Mg²⁺/buffer. Repeated stages: heat denaturation, primer annealing, and extension.

4.2 | How are PCR primers oriented, and which cellular enzymes are unnecessary?

Primers bind opposite strands with 3′ ends pointing inward. Heat replaces helicase-mediated separation; supplied primers remove the need for primase.

4.2 | How is ideal PCR amplification calculated?

N = N₀ × 2ⁿ. Three starting molecules after five ideal cycles give 3 × 32 = 96; actual efficiency may be lower.

4.3 | What are a gene, genome, and gene expression?

A gene produces a functional RNA, sometimes directing protein production. A genome includes all genetic material. Gene expression makes functional RNA or protein from genetic information.

4.3 | What does transcription require, and how does it differ from DNA synthesis?

RNA polymerase uses DNA, ATP/GTP/CTP/UTP, and suitable Mg²⁺ or Mn²⁺. It can initiate without a primer; it makes RNA 5′→3′ while reading template 3′→5′.

4.3 | Coding versus template strand?

Template/antisense is read by polymerase. Coding/sense matches the 5′→3′ RNA except T replaces U. Different genes can use different DNA strands as templates.

4.3 | How do you obtain RNA from coding or template DNA?

Coding DNA: orient 5′→3′ and replace T with U. Template DNA: make its antiparallel RNA complement, then report RNA 5′→3′.

4.3 | What is transcribed from template 3′-TACCTTGGAATT-5′?

RNA 5′-AUGGAACCUUAA-3′. The corresponding coding DNA is 5′-ATGGAACCTTAA-3′.

4.3 | What is a promoter, and what do bacterial sigma and consensus sequences mean?

A promoter is initiation-directing DNA; sigma helps bacterial RNA polymerase recognize it. A consensus shows common bases among related promoters; -10/-35 elements are upstream of initiation.

4.3 | What happens during transcription initiation?

Polymerase binds promoter DNA in a closed complex, then opens DNA to form an open complex. Sigma mainly assists initiation and can dissociate after promoter escape.

4.3 | What happens during elongation and termination?

A moving bubble unwinds DNA ahead and rewinds it behind; RNA emerges with its free 5′ end, while its growing 3′ end stays at the active site. A termination signal ends synthesis.

4.3 | What structural and chemical details characterize transcription elongation?

The notes show about 17 open DNA bases and an approximately eight-base DNA–RNA hybrid. RNA 3′-OH attacks incoming NTP alpha phosphate; PPi is released.

4.3 | Intrinsic versus Rho-dependent termination?

Intrinsic termination uses an RNA hairpin followed by a U-rich region. Rho-dependent termination requires the bacterial Rho protein.

4.3 | How do rifampicin and actinomycin D inhibit transcription?

Rifampicin blocks very early bacterial RNA extension. Actinomycin D intercalates into DNA and interferes with template use, including eukaryotic transcription. Blocking transcription can help measure decay of existing mRNA.

4.3 | Where do transcription and translation occur in bacteria versus eukaryotes?

Bacteria can couple them without a nuclear barrier. Eukaryotic nuclear transcription and processing occur in the nucleus; exported mRNA is translated in the cytoplasm or on rough-ER-associated ribosomes.

4.3 | What do eukaryotic RNA polymerases I, II, and III make?

I: precursor for 18S/5.8S/28S rRNA in nucleolus. II: pre-mRNA and some other RNAs in nucleoplasm. III: tRNA and 5S rRNA. The notes show alpha-amanitin sensitivity: I insensitive, II very sensitive, III moderate.

4.3 | What do mRNA, tRNA, and rRNA do?

mRNA carries codons; tRNA adapts codons to amino acids; rRNA supplies structural and catalytic ribosome functions. Functional RNA does not have to become protein.

4.3 | What other functional RNAs are listed?

snRNA: splicing; snoRNA: rRNA processing/modification; miRNA: regulates mRNA use; siRNA: targeted RNA silencing/antiviral defense; piRNA: gene/transposon regulation; lncRNA: diverse regulation.

4.3 | What approximate lengths are listed for small regulatory RNAs?

snRNA/snoRNA: below 300 nt; miRNA/siRNA: 20–25 nt; piRNA: 29–30 nt; lncRNA: over 200 nt. These are reference ranges, not universal exact lengths.

4.3 | What are enhancers and cis-acting regulation?

An enhancer is regulatory DNA acting on the same DNA molecule; bound proteins can stimulate a distant promoter through looping. The enhancer is neither the protein nor the promoter itself.

4.3 | What does TATA-binding protein do?

Recognizes appropriate TATA-containing promoters and strongly bends/distorts DNA during transcription-initiation-complex formation; not every promoter has a TATA box.

4.3 | What three processes produce a typical mature eukaryotic protein-coding mRNA?

5′ capping, 3′ cleavage/polyadenylation, and intron removal/exon joining. The initial transcript is pre-mRNA.

4.3 | What is the 5′ cap and its function?

A modified guanosine attached by an unusual 5′–5′ triphosphate linkage. It supports RNA protection, nuclear export, and translation initiation.

4.3 | How is the poly(A) tail made and what does it do?

After 3′ cleavage, poly(A) polymerase adds adenylates without a DNA template for the tail. It supports stability, export, and translation; the notes quote about 250 residues initially, with variable actual length.

4.3 | What are introns, exons, and common splice-site signals?

Introns are removed; exons remain and are joined. Exons may include untranslated regions. Many introns have 5′ GU and 3′ AG boundaries, but sequence context is also necessary.

4.3 | What is alternative splicing?

Different exon combinations from one pre-mRNA produce different mature RNAs/proteins. The notes illustrate membrane-bound versus secreted antibodies.

4.3 | How are rRNA and tRNA processed?

Both bacteria and eukaryotes process these RNAs. rRNA undergoes cleavage and modifications such as methylation/pseudouridine formation; tRNA undergoes trimming, modification, and sometimes splicing, with a functional 3′-CCA acceptor end.

4.3 | How can an intronic mutation cause disease without changing an existing coding codon?

It can alter splice recognition, retain intron sequence or skip exons, introduce a premature stop, or trigger RNA decay. Examples include some beta-thalassemias and splicing defects involving BRCA1, CFTR, factor VIII, or SMN genes.

4.4 | What are the key properties of the genetic code?

Three-base codons, nonoverlapping and comma-free within a frame, read 5′→3′, degenerate but normally unambiguous, and nearly universal. There are 64 codons: 61 sense and three stops.

4.4 | What are the start and stop codons?

Usual start: AUG, methionine; bacterial initiation uses fMet. Stops: UAA/UAG/UGA; coding DNA equivalents TAA/TAG/TGA. Stops do not add amino acids.

4.4 | How do you translate a sequence fragment correctly?

Identify coding/template status and ends, obtain 5′→3′ mRNA, establish its frame, and read the codon table until an in-frame stop. A fragment's first displayed base need not start a codon.

4.4 | What are an anticodon and a charged tRNA?

An anticodon pairs antiparallel with a codon; 5′-GAA-3′ pairs with 3′-CUU-5′. A charged tRNA carries an amino acid at its 3′-CCA acceptor end.

4.4 | What does aminoacyl-tRNA synthetase do, and why does charging accuracy matter?

It uses ATP to attach the appropriate amino acid to tRNA. A mischarged tRNA can deliver the wrong amino acid despite correct codon–anticodon pairing.

4.4 | What are translation's stages and initiation requirements?

Initiation, elongation, termination. Initiation assembles mRNA, ribosomal subunits, initiation factors, and initiator tRNA; the initiator enters the P site.

4.4 | What do the ribosome's A, P, and E sites do?

A receives incoming charged tRNA during elongation; P holds peptidyl-tRNA before transfer; E holds departing uncharged tRNA.

4.4 | How are peptide bonds formed and tRNAs moved?

Large-subunit rRNA catalyzes transfer of the peptide from P-site tRNA to the amino acid on A-site tRNA. Translocation advances one codon and shifts tRNAs toward P and E.

4.4 | How do translation termination and protein directionality work?

Release factors recognize stop codons and hydrolyze the peptide–tRNA connection. Proteins grow N→C: the emerging free end is N-terminal, and the growing tRNA-attached end is C-terminal.

4.4 | What bacterial ribosome features help label a diagram?

30S binds/decodes mRNA; 50S contains the peptidyl-transferase center and peptide exit tunnel. Together they form 70S, not 80S; sedimentation values are not simply additive.

4.4 | Which translation inhibitors affect decoding or aminoacyl-tRNA entry?

Streptomycin: bacterial initiation/misreading. Tetracycline: blocks bacterial aminoacyl-tRNA binding at the 30S-associated A-site region.

4.4 | Which translation inhibitors affect peptide formation, movement, or release?

Chloramphenicol: bacterial 50S peptidyl transferase. Erythromycin: bacterial 50S elongation/translocation. Cycloheximide: eukaryotic elongation/translocation. Puromycin: tRNA analogue causing premature termination in both systems.

4.4 | What is an operon and a polycistronic mRNA?

An operon groups genes under shared transcriptional control, often producing one mRNA encoding several proteins. Most eukaryotic nuclear protein-coding mRNAs encode one main protein.

4.4 | What do lacZ, lacY, lacA, and lacI encode?

lacZ: beta-galactosidase; lacY: lactose permease; lacA: transacetylase. Separately controlled lacI encodes the repressor.

4.4 | How does lactose relieve lac repression?

The repressor binds operator DNA and inhibits transcription. Lactose-derived allolactose binds the repressor and reduces operator binding: relief of negative regulation.

4.4 | How does glucose control positive lac regulation?

Low glucose favors higher cAMP, which activates CAP/CRP to stimulate transcription. Lactose removes repression; low glucose supplies strong activation.

4.4 | What are the four standard lac operon states?

Glucose+/lactose−: very low/off. Glucose−/lactose−: very low/off. Glucose+/lactose+: low/basal. Glucose−/lactose+: high.

4.4 | Synonymous versus missense versus nonsense mutation?

Synonymous: same amino acid. Missense: different amino acid. Nonsense: premature stop. Synonymous changes can still sometimes affect RNA processing, stability, or translation.

4.4 | Frameshift versus in-frame insertion/deletion?

A coding insertion/deletion not divisible by three shifts downstream codons. A multiple of three preserves the downstream frame but adds/removes residues and can still damage function.

4.4 | How can mutations change phenotype through protein sequence or amount?

Coding changes can alter folding, catalysis, or interactions; promoter changes can alter transcript/protein abundance; splice changes can alter mature RNA. Severity depends on position and context.

4.4 | What does 5′-AUG GAA CCU UAA-3′ encode?

N–Met–Glu–Pro–C. Read from AUG through the three sense codons; UAA terminates without adding a residue.

4.4 | Coding DNA is displayed 3′-…AGTTCC…-5′ with a stop. What is the final residue?

Reverse to 5′-CCT TGA-3′, then RNA CCU UGA: proline precedes the stop. Reverse a backward coding strand; do not complement it.

4.4 | What are the coding strand and RNA for worksheet template 3′-ATGGATTACCACGTGGACTGAGGACTC-5′?

Coding DNA: 5′-TACCTAATGGTGCACCTGACTCCTGAG-3′. RNA: 5′-UACCUAAUGGUGCACCUGACUCCUGAG-3′.

4.4 | What peptide does the worksheet RNA encode from its internal AUG?

N–Met–Val–His–Leu–Thr–Pro–Glu… . The displayed remaining fragment contains no in-frame stop; earlier RNA bases are upstream of the selected start.

4.4 | What does the worksheet's template CTC→CAC substitution do?

Coding GAG→GTG and RNA GAG→GUG change glutamate to valine: a missense mutation.