AOL 2 : Nucleic Acids

Discuss the central dogma of molecular biology.

• The central dogma describes the usual flow of genetic information in a cell: DNA → RNA → protein.

• Step 1: Transcription (DNA → RNA)

  • The sequence of one DNA strand, the template strand, is copied into a complementary RNA molecule by the enzyme RNA polymerase.

  • Transcription starts at a promoter and ends at a terminator.

  • Information is “transcribed” in the same language of nucleic acids.

• Step 2: Translation (RNA → protein)

  • The sequence of messenger RNA (mRNA) is read by ribosomes in nonoverlapping groups of three nucleotides called codons.

  • Each codon specifies one amino acid, according to the genetic code. tRNAs with matching anticodons deliver the amino acids.

  • The amino acids are joined by peptide bonds into a polypeptide.

  • Information is “translated” from the language of nucleotides to the language of amino acids.

• Replication (DNA → DNA) is also part of the picture.

  • DNA copies itself using each strand as a template, with DNA polymerase building the new strands by complementary base pairing.

  • This passes genetic information from cell to cell and from parent to offspring.

• Together, transcription and translation make up gene expression, the production of a functional gene product.

  • Gene expression is regulated: not every gene is expressed in every cell at all times.

  • Example: muscle cells express contraction proteins, but skin and liver cells do not, even though they contain the same DNA.

• Prokaryotes vs. eukaryotes

  • In prokaryotes, transcription and translation both occur in the cytoplasm and are coupled in space and time. The primary transcript is the mRNA.

  • In eukaryotes, transcription occurs in the nucleus and translation in the cytoplasm.

  • Eukaryotic primary transcripts first undergo RNA processing: a 5′ cap, a poly(A) tail, and splicing to remove introns.

• Exceptions to the usual flow

  • HIV transfers information from RNA → DNA (reverse transcription).

  • SARS-CoV-2 transfers information from RNA → RNA.

  • In most cells, the flow is still DNA → RNA → protein.

Describe the structure and organization of DNA.

• Building blocks: nucleotides

  • Each nucleotide has three components: a 5-carbon sugar (deoxyribose), a phosphate group, and a nitrogenous base.

  • In deoxyribose, the 2′ carbon carries a hydrogen (–H) rather than a hydroxyl (–OH).

  • The base attaches to the 1′ carbon, and the phosphate attaches to the 5′ carbon.

• Four bases

  • Purines (double-ring): adenine (A) and guanine (G).

  • Pyrimidines (single-ring): thymine (T) and cytosine (C).

• A single strand is a polymer of nucleotides

  • Nucleotides are linked by phosphodiester bonds between the 3′ carbon of one nucleotide and the 5′ phosphate of the next.

  • The repeating sugar-phosphate units form the sugar-phosphate backbone.

  • The strand has polarity: one end has a free 5′ phosphate and the other has a free 3′ hydroxyl.

  • Sequences are written 5′ → 3′ by convention.

• Double helix

  • Two strands wind around each other in a right-handed helix.

  • The sugar-phosphate backbones are on the outside and the bases point inward.

  • The strands are antiparallel: the 5′ end of one strand lies opposite the 3′ end of the other.

  • Dimensions: 10 base pairs per turn and a diameter of 2 nm.

  • The uneven twisting creates a major groove and a minor groove, where proteins can contact bases and recognize specific sequences.

• Base pairing (complementarity)

  • A pairs with T (2 hydrogen bonds) and G pairs with C (3 hydrogen bonds).

  • Every base pair contains one purine and one pyrimidine, which keeps the distance between the backbones constant.

  • The two strands are complementary, not identical.

• Forces that stabilize the helix

  • Hydrogen bonds between bases on opposite strands are individually weak, but millions of them add up to give stability.

  • Base stacking occurs between neighboring bases in the same strand. The nonpolar, flat bases cluster together away from water.

  • Phosphodiester bonds are strong covalent bonds that hold each strand together.

• Function from structure

  • Information is stored in the sequence of bases. There is no restriction on the order of bases, so the information capacity is enormous.

  • Complementary pairing explains how DNA is replicated.

Compare and contrast the structures of DNA and RNA.

• Similarities

  • Both are polymers of nucleotides linked by phosphodiester bonds.

  • Both have polarity, with a 5′ end and a 3′ end.

  • Both use the purines A and G and the pyrimidine C.

  • Both store information in their base sequence.

• Differences

Feature	DNA	RNA
Sugar	Deoxyribose (H on 2′ carbon)	Ribose (OH on 2′ carbon)
Pyrimidine base	Thymine (T)	Uracil (U)
5′ end	Typically a monophosphate	Typically a triphosphate
Strands	Usually double stranded (double helix)	Usually single stranded, though it can fold into hairpins and 3D shapes
Length	Very long (millions of nucleotides)	Usually much shorter (e.g., a few thousand nucleotides)
Stability	More stable	Less stable
Main role	Long-term information storage	mRNA, rRNA, tRNA, snRNA, and regulatory RNAs

• Key reasoning

  • The extra 2′ hydroxyl on ribose is a reactive group, which is part of why RNA is less stable than DNA.

  • Uracil lacks the methyl group that thymine has on its 5 carbon. The groups that form hydrogen bonds are identical, so U pairs with A just as T does.

  • Folded RNAs can have complex 3D structures, and some act as catalysts. This supports the RNA world hypothesis.

Provide the complementary sequences of a given sequence of DNA.

• Rules

  • A pairs with T, and G pairs with C.

  • The strands are antiparallel, so the complementary strand runs in the opposite direction.

• Method, step by step

  1. Write the original strand and label its ends (5′ → 3′).

  2. Replace each base with its partner: A↔T and G↔C.

  3. Label the new strand with the opposite orientation (3′ → 5′).

  4. To write it in the standard 5′ → 3′ direction, reverse the complementary sequence.

• Example 1

  • Original: 5′-ATGC-3′

  • Complement: 3′-TACG-5′

  • Written 5′ → 3′: 5′-GCAT-3′

• Example 2

  • Original: 5′-ACTTTCAGCGAT-3′

  • Complement: 3′-TGAAAGTCGCTA-5′

  • Written 5′ → 3′: 5′-ATCGCTGAAAGT-3′

• If the question asks for the RNA transcript from a DNA template

  • Use the same rules but put U in place of T.

  • For the template 5′-ACTTTCAGCGAT-3′, the RNA is 5′-AUCGCUGAAAGU-3′.

• Common mistakes to avoid

  • Forgetting that the complementary strand is antiparallel. Reading 3′-TACG-5′ left to right as if it were 5′ → 3′ is incorrect.

  • Using T in RNA or U in DNA.

Describe how gel electrophoresis separates nucleic acid fragments based on size.

• Purpose: to separate DNA or RNA fragments by length.

• Why it works

  • The phosphate groups in the backbone are ionized at cellular pH, so nucleic acids carry a uniform negative charge.

  • Charge per unit length is roughly constant, so size, not charge, determines migration speed.

• Procedure

  1. Prepare a gel (usually agarose; polyacrylamide for small fragments) with wells at one end. The gel is a porous mesh.

  2. Submerge the gel in a conducting buffer between two electrodes.

  3. Load the samples into the wells, along with a size ladder (fragments of known lengths).

  4. Apply an electric field. Nucleic acids move toward the positive electrode (anode).

  5. Stain the gel (e.g., ethidium bromide or a fluorescent dye) to visualize the bands.

• How separation occurs

  • Smaller fragments move through the pores more easily and travel farther.

  • Larger fragments are slowed by the mesh and stay closer to the wells.

  • Each band is a group of fragments of the same length.

• Reading the gel

  • Compare band positions with the ladder to estimate fragment size.

  • Distance traveled is roughly proportional to the log of fragment length.

  • Brighter bands generally indicate more DNA.

• Example: DNA cut with a restriction enzyme produces fragments of different lengths, which appear as a distinct pattern of bands.

Describe how Southern blots allow specific genes or DNA sequences to be visualized on a gel.

• Problem: after gel electrophoresis, a genome digest yields thousands of fragments that form a smear, and you cannot tell which fragment holds your gene of interest.

• Solution: a Southern blot uses a labeled probe to detect only the fragment(s) with a specific sequence.

• Steps

  1. Cut genomic DNA with restriction enzymes to produce fragments.

  2. Separate the fragments by gel electrophoresis.

  3. Denature the DNA in the gel (e.g., with alkaline solution) to make it single stranded.

  4. Transfer (blot) the single-stranded DNA from the gel onto a membrane (nitrocellulose or nylon). The fragments keep the same pattern as on the gel.

  5. Hybridize the membrane with a labeled probe: a single-stranded DNA or RNA complementary to the target sequence, labeled with a radioactive or fluorescent tag.

  6. Wash away unbound and weakly bound probe.

  7. Detect the label (e.g., autoradiography or fluorescence). Bands appear only where the probe hybridized.

• Key principle: the probe binds by complementary base pairing to its matching sequence on the membrane.

• Related blots

  • Northern blot: detects RNA (gene expression).

  • Western blot: detects protein (using antibodies).

Assess the chemical composition of a DNA molecule given information about some of its constituents.

• Chargaff’s rules (for double-stranded DNA)

  • %A = %T and %G = %C.

  • Therefore %A + %G = %T + %C = 50% (purines = pyrimidines).

  • The ratio of A+T to G+C varies between species, but the pairing rules always hold.

• Method

  1. Identify the known base percentage or count.

  2. Use the partner rule (A = T, G = C).

  3. Subtract from 100% to find the remaining bases and split them equally between the pair.

• Worked example

  • A DNA sample is 20% adenine.

  • T = 20%, so A + T = 40%.

  • The remaining 60% is G + C, so G = 30% and C = 30%.

• Other composition facts

  • Each nucleotide has one sugar, one base, and at least one phosphate.

  • A double-stranded DNA of n base pairs has 2n nucleotides, so 2n sugars, 2n bases, and 2n phosphates.

  • Each strand of n nucleotides has n − 1 phosphodiester bonds.

  • A–T pairs have 2 hydrogen bonds and G–C pairs have 3, so total hydrogen bonds = 2(#A–T) + 3(#G–C).

  • Higher GC content means a more stable, harder-to-melt helix.

• Warning: Chargaff’s rules do not apply to single-stranded DNA or to RNA (which has U and is usually single stranded).

Predict whether two nucleic acid molecules will hybridize based on their sequences.

• Hybridization is the formation of a double-stranded molecule by base pairing between two single-stranded nucleic acids.

• Requirements

  • The sequences must be complementary: A–T (or A–U with RNA) and G–C.

  • They must be antiparallel: one strand 5′ → 3′ pairs with the other 3′ → 5′.

• Method for predicting

  1. Write the first strand 5′ → 3′.

  2. Write its complement, then reverse it to get the sequence as 5′ → 3′.

  3. If the second molecule matches this sequence, they will hybridize.

• Example

  • Strand 1: 5′-ATGC-3′. Its complement written 5′ → 3′ is 5′-GCAT-3′.

  • 5′-GCAT-3′ will hybridize.

  • 5′-TACG-3′ will not (it is the complement written in the wrong direction).

  • 5′-ATGC-3′ will not hybridize with itself, because the strands would be parallel.

• Strength of hybridization

  • More matching base pairs make a more stable hybrid.

  • Mismatches weaken the hybrid.

  • Higher GC content makes a stronger hybrid because G–C pairs have 3 hydrogen bonds.

  • Higher temperature or lower salt (higher stringency) favors only near-perfect matches.

• Other combinations

  • DNA–RNA hybrids can form (as in transcription, with U pairing with A).

  • RNA–RNA hybrids and RNA hairpins form by the same rules.

• Applications: probes in Southern and Northern blots, and DNA replication and transcription.

Interpret results from experiments using gel electrophoresis or Southern/Northern blots as a method.

• Gel electrophoresis: how to read a gel

  • Position of a band shows size. Bands closer to the wells are larger and bands farther from the wells are smaller. Compare with the ladder.

  • Brightness of a band shows relative amount of nucleic acid.

  • Presence or absence of a band shows whether a fragment of that size exists.

  • Lane-to-lane differences point to differences between samples.

• Southern blot (DNA)

  • A band means a fragment contains sequence complementary to the probe.

  • The number of bands can indicate how many copies of the sequence are present.

  • The band size shows the fragment length after restriction digestion.

  • A mutation that creates or destroys a restriction site changes the band sizes. This is the basis of restriction fragment length polymorphism (RFLP) analysis, used to compare individuals or detect disease alleles.

  • No band means the sequence is absent, such as a deletion.

• Northern blot (RNA)

  • A band means the gene is being expressed (transcribed) in that sample.

  • A darker band means more transcript, so higher expression.

  • No band means the gene is not expressed in that tissue or condition.

  • Band size shows the length of the mRNA. Multiple bands from one probe can suggest alternative splicing.

  • A smear often means the RNA was degraded.

• Example interpretation

  • A probe for gene X gives a band in muscle RNA but not liver RNA. Gene X is expressed in muscle but not liver.

• Check the controls

  • Always look at the ladder and any loading control to make sure differences reflect real biology and not unequal loading.

• Limit: a probe only detects sequences complementary to it. Anything that does not hybridize stays invisible.