Nucleic Acids and Strand Direction

Reading CAT from the opposite end gives TAC. Direction also matters when you read a DNA sequence, but DNA’s two ends are chemically different: one is called 5′ and the other 3′. Labeling those ends lets you identify how a strand is oriented and write its complementary strand correctly.

Watch the process

Nucleic Acids

A nucleotide has exactly three parts: a five-carbon sugar, one or more phosphate groups attached to the sugar’s 5’ carbon, and a nitrogenous base attached to the 1’ carbon. This is the standard nucleotide structure used in nucleic-acid synthesis; phosphate can occupy other positions in some cellular nucleotides. Read those small marks before going further. The five carbons of the sugar are numbered 1 through 5, and each number carries a prime mark, read aloud as “five prime” or “three prime,” only so that the sugar’s carbons are never confused with the separately numbered atoms of the base. The labels 5’ phosphate, 3’ hydroxyl, and 2’ hydroxyl identify positions on the sugar. In DNA the sugar is deoxyribose, which lacks the hydroxyl at the 2’ position that ribose carries in RNA. The 2’ hydroxyl makes RNA chemically more reactive and less stable, which contributes to differences in stability between RNA and DNA.

The bases divide into two-ring purines, adenine and guanine, and one-ring pyrimidines, cytosine plus thymine in DNA or uracil in RNA. In double-stranded DNA, adenine pairs with thymine through two hydrogen bonds and guanine pairs with cytosine through three. In standard Watson-Crick pairing, a purine pairs with a pyrimidine, which keeps the width of the double helix constant along its length. The two strands are antiparallel: they run in opposite directions.

The pairing rules have a measurable consequence that Erwin Chargaff found before anyone knew the structure. In any double-stranded DNA sample, the amount of adenine equals the amount of thymine and the amount of guanine equals the amount of cytosine. Those equalities are now called Chargaff’s rules, and they are not arbitrary bookkeeping: every A on one strand is opposite a T on the other, so the totals must match. Notice what the rules do not say. They do not require A to equal G, and the ratio of A-T pairs to G-C pairs varies widely between species. Use the pairing relationships to calculate an unknown base percentage from a supplied one.

Notice also the energy difference built into the pairing. An A-T pair holds with two hydrogen bonds; a G-C pair holds with three. A region of DNA rich in G and C is generally harder to separate under otherwise matched conditions. Base stacking also contributes to stability, which is why such regions have higher melting temperatures and why the origins of replication in many genomes are A-T rich. Compare melting temperatures only under otherwise matched conditions.

The Backbone Has a Direction, and That Is the Point

Polymerases join activated nucleotides through a phosphodiester bond that links the phosphate on the 5’ carbon of one nucleotide to the hydroxyl on the 3’ carbon of the next. Run that linkage down a strand and a typical unmodified strand has a 5’ end and a 3’ hydroxyl end. A 5’ end may carry a phosphate or hydroxyl, and some RNAs have additional end modifications; the sugar-carbon numbering still defines the direction. That asymmetry is 5’ to 3’ directionality, and it is the structural fact underneath the entire Information Storage and Transmission big idea.

Think about why an information molecule needs it. A sequence is only information if it can be read the same way every time, and reading the same way every time requires a defined starting end and a defined direction of travel. The sugar- phosphate backbone supplies both. The DNA and RNA polymerases studied here add new nucleotides only to a free 3’ hydroxyl, so every nucleic acid strand is synthesized in the 5’ to 3’ direction while the template is read 3’ to 5’. This directionality helps explain leading and lagging strands, Okazaki fragments, and the direction of transcription along a template.

RNA differs from DNA on four counts worth holding together: ribose instead of deoxyribose, uracil instead of thymine, typically single-stranded instead of double-stranded, and often shorter-lived, although lifetimes vary among RNA types. Being single-stranded lets an RNA fold back on itself through internal base pairing, which is how transfer RNA gets its shape and how ribozymes get catalytic activity. A single-stranded molecule that folds into a working shape is behaving the way a protein does, and that overlap is why RNA can both carry information and do chemical work.

Reading a Strand Correctly

A single DNA strand is written \(5’\text{-ATGCCG-}3’\). Give the sequence of the complementary strand, written in the conventional direction.

Pair each base first, keeping the strands aligned: A pairs with T, T with A, G with C, C with G, C with G, G with C. Reading the complement in the same left-to-right order as the original gives TACGGC, but that reading runs \(3’\) to \(5’\) because the strands are antiparallel. Convention writes sequences \(5’\) to \(3’\), so reverse it.

Answer

\(5’\text{-CGGCAT-}3’\). Pairing alone is not enough; the antiparallel arrangement means the complement must be reversed before it is written.

Using Chargaff’s Rules on Real Numbers

A double-stranded DNA sample is 22 percent guanine by base count. A second sample, single-stranded, is 22 percent guanine and 31 percent cytosine. For each sample, determine what can be concluded about the adenine content, and explain why the two samples must be handled differently.

Start with the double-stranded sample, where the pairing rules apply. Guanine pairs with cytosine, so cytosine is also 22 percent, and the two together account for \[22 + 22 = 44 \text{ percent}.\] The remaining bases must be adenine and thymine: \[100-44 = 56 \text{ percent}.\] Because adenine pairs with thymine, that 56 percent divides equally, so adenine is 28 percent and thymine is 28 percent.

Now the single-stranded sample, and here the reasoning has to stop short. Chargaff’s rules follow from base pairing between two strands. A single strand has no partner, so nothing forces its adenine content to match its thymine content — and the given numbers confirm it, since guanine at 22 percent and cytosine at 31 percent are already unequal. All you can say is that adenine plus thymine together make up \(100-22-31 = 47\) percent, split in some unknown ratio.

Answer

The double-stranded sample is 28 percent adenine. The single-stranded sample allows only the combined figure of 47 percent adenine plus thymine, because the pairing rules require a complementary partner strand to hold.

Review: Nucleic Acids and Directionality

A nucleic acid is an information molecule because its backbone has a defined beginning and a defined direction, not merely because it contains four kinds of base.

Change the sugar and you change stability and lifetime; change the base and you change the message; reverse the direction and you change the meaning of the same sequence entirely.

When you write a complement, complement and then reverse. Check both steps: complement the bases, then reverse their order to write the answer in the requested direction.

The Confusion Worth Clearing Up Now

A nucleotide, a nucleoside, and a base contain different components. A nitrogenous base is one of adenine, guanine, cytosine, thymine, or uracil. A base plus a sugar is a nucleoside. A base plus a sugar plus at least one phosphate is a nucleotide, and that is the monomer. When a question says the monomer of a nucleic acid, it means all three parts.

Distinguish the bonds between paired bases from the bonds along the backbone. Within one strand, neighboring nucleotides are held by phosphodiester bonds, which are covalent and strong. Between the two strands, paired bases are held by hydrogen bonds, which are weak and individually easy to break. That difference is exactly why the double helix can be unzipped for replication and transcription without the strands themselves falling apart. If a question describes heating DNA until the strands separate while each strand stays intact, it is describing hydrogen bonds breaking and phosphodiester bonds surviving — the same logic as protein denaturation, applied to a different molecule.

Nucleotides and directionality

Practice question 1

The phosphodiester bond in a nucleic acid strand joins

1. the nitrogenous base of one nucleotide to the base of the next

2. the 5’ phosphate of one nucleotide to the 5’ phosphate of the next

3. the phosphate group on the 5’ carbon of one nucleotide to the hydroxyl on the 3’ carbon of the adjacent sugar

4. two sugars directly with no phosphate between them

Practice question 2

Which set of differences correctly separates RNA from DNA?

1. RNA uses deoxyribose and thymine and is typically double-stranded; DNA uses ribose and uracil and is typically single-stranded

2. RNA uses ribose and uracil and is typically single-stranded; DNA uses deoxyribose and thymine and is typically double-stranded

3. RNA uses ribose and thymine and is typically single-stranded; DNA uses deoxyribose and uracil and is typically double-stranded

4. RNA pairs adenine with guanine and DNA pairs adenine with thymine, though both are built on the same sugar

Practice question 3

A DNA polymerase adds nucleotides only to a free 3’ hydroxyl. It follows that

1. the new strand is built 5’ to 3’ while its template is read 3’ to 5’

2. the new strand and its template are built in the same direction

3. the enzyme can begin synthesis at either end of the new strand

4. directionality applies to DNA replication but not to transcription

Practice answer key

1. C; 2. B; 3. A.

Practice answer explanations

1. Nucleotides and directionality, Question 1. Choice C is correct. A phosphodiester bond joins the 5’ phosphate of one nucleotide to the 3’ hydroxyl of the neighboring sugar, which is what makes the backbone directional. Choice A describes base pairing between strands rather than the covalent backbone. Choice B names a phosphate-to-phosphate linkage rather than the sugar-phosphate-sugar connection of the nucleic-acid backbone. Choice D removes the phosphate that gives the backbone its name and its charge.

2. Nucleotides and directionality, Question 2. Choice B is correct. RNA carries ribose and uracil and is typically single-stranded, while DNA carries deoxyribose and thymine and is typically double-stranded. Choice A swaps every feature between the two molecules. Choice C swaps only the bases, keeping the sugars right, which is the halfway error to watch for: uracil belongs to RNA and thymine to DNA. Choice D misstates standard complementary pairing, in which A pairs with T or U rather than G, and also denies the sugar difference that names the two molecules.

3. Nucleotides and directionality, Question 3. Choice A is correct. Adding only to a free 3’ hydroxyl forces synthesis in the 5’ to 3’ direction, and because the strands are antiparallel the template is necessarily read 3’ to 5’. Choice B ignores the antiparallel arrangement. Choice C contradicts the stated restriction on where nucleotides can be added. Choice D limits directionality to replication, but RNA polymerase obeys the same 3’ hydroxyl requirement.

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