Nucleotides, Base Pairing, and DNA Stability for the MCAT

MCAT content library / Biology & biochemistry / Nucleotides, Base Pairing, and DNA Stability

Pair the strands in the correct direction, read what a melting curve measures, and distinguish DNA entry from evidence of hereditary function.

About 15 minutes, including the worked examples and checks.

From a gene to a functional result · Lesson 1 of 5 · Continue in this path

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Follow the sugar, the backbone, and the sequence separately
Build the paired strand before rewriting its direction
Read a melting comparison under its actual conditions
Use pairing to explain reannealing and hybridization
Connect molecular structure to hereditary evidence
Check your reasoning
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Follow the sugar, the backbone, and the sequence separately

A DNA strand carries information in the order of its bases. Its sugar–phosphate backbone provides a directional chain on which that order can be read. Base pairing connects one sequence to another, but it does not make the two strands run in the same direction. AAMC places nucleic-acid structure, pairing, denaturation, reannealing, and hybridization within the MCAT’s gene-expression foundation.¹

A nucleoside contains a base attached to a sugar. Adding one or more phosphate groups makes a nucleotide. Adenosine is therefore a nucleoside; AMP, ADP, and ATP are nucleotides. In DNA the sugar is deoxyribose, with H at the 2′ carbon; RNA has ribose, with a 2′ OH. Adenine and guanine are two-ring purines. Cytosine, thymine, and uracil are one-ring pyrimidines. Standard DNA uses A, G, C, and T; standard RNA uses A, G, C, and U.²

Within a strand, 3′–5′ phosphodiester linkages join neighboring sugars through phosphate. The prime marks identify sugar carbons, and the 5′ and 3′ labels identify opposite ends of the chain. Do not confuse this covalent backbone with the interactions between bases on different strands. Separating an ordinary DNA duplex by heating need not cut its backbone or change its base sequence.³

Build the paired strand before rewriting its direction

In a standard Watson–Crick DNA duplex, A pairs with T and G pairs with C. The strands are antiparallel: opposite ends face each other. A–T has two interbase hydrogen bonds and G–C has three, but duplex stability also depends strongly on stacking between adjacent bases. Counting hydrogen bonds alone is not a complete thermodynamic explanation.³⁴

For the illustrative strand 5′-AGTCGA-3′, first place each complementary base directly underneath its partner. The aligned strand is 3′-TCAGCT-5′. If a question asks for that partner written 5′ to 3′, read the lower strand from right to left: 5′-TCGACT-3′. Complementing changes the bases; reversing changes the written order. Both operations are needed when the original and its partner must each be reported 5′ to 3′.

The same orientation rule applies when an RNA strand pairs with DNA, with U replacing T in the RNA sequence. A single RNA can also fold so that complementary stretches within it pair. “Single-stranded RNA” does not mean every base is permanently unpaired.⁹

Read a melting comparison under its actual conditions

Melting is loss of duplex structure as strands separate. For a simple two-state duplex transition, the melting midpoint, T_m, is the temperature at which half the strands are duplex-associated and half are separated. A higher midpoint indicates greater thermal stability under that assay’s conditions. Length, base sequence, mismatches, ionic conditions, and strand concentration can affect the result; T_m is not a number determined by GC percentage alone.⁵

The original illustrative data below compare two fully complementary, 30-base-pair DNA duplexes: L contains 40% GC pairs and H contains 60%. Each complementary strand is present at 1.0 µM. Both samples use the same pH 7.2 buffer and 100 mM NaCl, equilibrate at each temperature, and show reversible separation without backbone cleavage, aggregation, or competing hairpins. A calibrated signal reports the fraction of strands separated; no conversion from raw absorbance is required.

Original illustrative melting data under the matched conditions described above. Lines connect the table values.

Temperature (°C)	L: fraction separated	H: fraction separated
48	0.05	0.00
56	0.27	0.02
60	0.50	0.05
64	0.73	0.12
72	0.95	0.50
80	0.99	0.88
88	1.00	0.98

Illustrative equilibrium melting data for two matched duplex samples

Read each curve at a separated fraction of 0.50: L has T_m = 60°C and H has T_m = 72°C, a 12°C difference. At 64°C, the duplex-associated fractions are 1 − 0.73 = 0.27 for L and 1 − 0.12 = 0.88 for H. H therefore remains more extensively paired at that temperature. This is consistent with the usual greater thermal stability of GC-rich DNA under comparable conditions, without establishing a universal conversion from GC percentage to degrees Celsius.

The controls are part of the result. Raising sodium concentration can stabilize duplex DNA by reducing the electrostatic cost associated with its charged backbones. If H had been tested at higher salt than L, sequence and solution conditions would be mixed together. Neighboring-base arrangements also matter: equal length and equal GC percentage do not guarantee equal melting behavior.⁴⁵

Use pairing to explain reannealing and hybridization

After separation, complementary strands can reassociate when temperature, salt, concentration, and available time permit. Reannealing describes restoring paired structure; hybridization uses complementary pairing to form a duplex between selected nucleic-acid strands, such as a probe and its target. The strands may come from different preparations. The information needed for recognition lies in their compatible sequences, not in whether they were previously neighbors.¹⁵

A probe result consequently needs conditions. A mismatched duplex may dissociate under conditions where a fully matched duplex remains paired. Failure to detect binding at one temperature does not establish that a target sequence is absent: the probe could be poorly matched or the assay too stringent. Conversely, permissive binding does not establish a perfect sequence match. These are applications of the same balance between sequence-specific pairing and the assay environment.

Connect molecular structure to hereditary evidence

Evidence identifying DNA as genetic material combined physical tracking with functional experiments. Labeling experiments followed DNA and protein into infected cells, while selective degradation experiments asked which component was required for a transferable trait. A label locates material; persistence of a trait and selective loss of transfer test a different claim.⁶

For example, imagine a cell-free donor preparation that transfers a pigment trait to recipient cells, which retain it through many divisions. A label verified to remain attached to donor DNA is detected inside the recipients. That observation supports DNA entry, but entry alone does not establish that DNA caused the trait. If selective DNA digestion abolishes transfer while RNA and protein digestion do not—and recipient survival and enzyme specificity are controlled—the functional comparison specifically implicates intact DNA. The check below asks how far that conclusion can go.

Nucleotides also work outside long information-bearing chains. ATP can couple a favorable net hydrolysis reaction to cellular work; cAMP is a nucleotide-derived second messenger formed from ATP. The same molecular family therefore participates in information storage, energy transfer, and signaling. Recognizing the sugar, base, and phosphate arrangement helps identify which role a question is testing.⁷⁸

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A DNA strand is written 5′-GATCCA-3′. Which sequence is its fully complementary DNA partner, also written 5′ to 3′?

A. 5′-ACCTAG-3′

Review this choice. This reverses the original sequence without complementing its bases. The partner must satisfy both opposite strand direction and A–T/G–C pairing.

B. 5′-CTAGGT-3′

Review this choice. These are the aligned complementary bases, but their aligned direction is 3′ to 5′. Reversing that string gives the requested 5′-to-3′ partner.

C. 5′-GATCCA-3′

Review this choice. This reproduces the given strand. The supplied sequence is not its own reverse complement.

D. 5′-TGGATC-3′

Correct. Align 3′-CTAGGT-5′ beneath the original strand, then read it from its 5′ end. This produces 5′-TGGATC-3′.

The principle to carry forward

Write both ends before choosing a sequence. A correct set of complementary letters can still be reported in the wrong direction.

Check 2

Two different, fully paired 30-base-pair DNA duplexes are tested at equal strand concentration and pH. X has 40% GC and is tested in 20 mM NaCl; Y has 60% GC and is tested in 100 mM NaCl. Y has the higher equilibrium melting midpoint. Which comparison would best separate the sequence difference from the salt difference?

A. Measure both original samples with a faster heating ramp.

Review this choice. A faster ramp changes equilibration and may introduce a kinetic difference. It leaves the original salt confound in place.

B. Repeat both curves at 100 mM NaCl with their other conditions matched.

Correct. Matching the salt concentration removes that solution difference while preserving the two sequences. Any remaining difference can be interpreted as sequence-dependent under the common conditions, although GC percentage alone does not describe every sequence feature.

C. Repeat the original salt conditions after doubling both strand concentrations.

Review this choice. Equal changes in strand concentration do not remove the unequal salt concentrations. Concentration can also affect duplex melting.

D. Use a longer version of X while keeping Y and both buffers unchanged.

Review this choice. Changing X’s length introduces another determinant of duplex stability while the salt difference remains.

The principle to carry forward

A melting comparison requires matched solution and concentration conditions before attributing a difference to sequence.

Check 3

A cell-free donor extract transfers a pigment trait that recipients retain through many divisions. Per 10,000 viable recipients, untreated extract yields 40 stable pigment-positive colonies, RNase-treated extract yields 39, protease-treated extract yields 41, and DNase-treated extract yields 0; no-extract controls also yield 0. Complete selective digestion is verified, enzymes are removed before exposure, and recipient survival is unchanged. Purified donor DNA alone has not been tested. Which conclusion is best supported?

A. DNA degradation prevents transfer here, but DNA alone has not been shown sufficient.

Correct. Loss of transfer after selective DNA degradation makes intact DNA necessary for the observed activity under these conditions. Retained activity after the other treatments supports that interpretation. A DNA-only preparation would address sufficiency.

B. DNA degradation shows that recipient proteins are unnecessary for expressing the trait.

Review this choice. Only proteins in the donor preparation were digested. Recipient proteins and cellular machinery remain present in every condition.

C. DNA entry proves that the trait cannot depend on any recipient-cell process.

Review this choice. Entry and a selective requirement for donor DNA do not eliminate recipient mechanisms needed to maintain or express the trait.

D. DNA is sufficient because all extracts began with the same number of viable recipients.

Review this choice. Matching recipient viability excludes a survival explanation for the comparison. It does not isolate DNA as the only supplied donor material.

The principle to carry forward

Separate material entering a cell, material required for a transfer, and material sufficient by itself to produce that transfer.