Foundation Phase: Core Science Fundamentals

Build explanations from cells, atoms, and bonds

By the end of this foundation lesson, you should be able to locate the stages of respiration, predict a simple genetic cross, calculate a limiting-reactant yield and pH, and connect an organic reaction to its mechanism. You need fractions, scientific notation, and basic logarithms. This is an introduction, not the complete science syllabus.

The U.S. DAT Survey of Natural Sciences contains 40 biology, 30 general chemistry, and 30 organic chemistry questions in one 90-minute section. Use the ADA's current candidate guide and updates to check the specifications for your test date.

1. Locate the process before memorizing the product

Bacteria and archaea are prokaryotes: they lack a membrane-bound nucleus. Animals, plants, fungi, and protists are eukaryotes. Both groups have DNA, ribosomes, cytoplasm, and a plasma membrane. A bacterium can make ATP without mitochondria.

For aerobic oxidation of one glucose in a typical eukaryotic accounting model:

Total: about 30–32 ATP, not a fixed yield for every cell. Pyruvate oxidation releases 2 CO₂; the cycle releases 4 more. The electron transport chain transfers electrons to oxygen and helps build a proton gradient. ATP synthase uses proton flow back toward the matrix to make ATP. The chain and ATP synthase have distinct roles. Oxidative phosphorylation.

Check the accounting: using approximate yields of 2.5 ATP per mitochondrial NADH and 1.5 per FADH₂ gives 10(2.5) + 2(1.5) + 4 = 32. If the two glycolytic NADH transfer their electrons through a lower-yield shuttle, the estimate falls by 2 ATP to 30. NADH itself does not simply cross the inner membrane. Mitochondrial energy conversion.

Try it: blocking electron transfer to oxygen would reduce oxidative phosphorylation. Glycolysis does not directly consume oxygen, but sustained glycolysis still requires regeneration of NAD⁺.

2. Separate DNA copying from inheritance

DNA strands are antiparallel. In ordinary DNA base pairs, A pairs with T through two hydrogen bonds; G pairs with C through three. During replication, each original strand templates a new complementary strand: replication is semiconservative.

Helicase separates the strands, primase provides primers, DNA polymerase extends a new strand 5′ → 3′, and ligase seals remaining backbone nicks after primer replacement. Both new strands grow 5′ → 3′; the lagging strand does so in Okazaki fragments. “Lagging” does not mean synthesis runs backward.

For an autosomal locus with complete dominance, an Aa individual has the same modeled phenotype as AA. Dominance describes the heterozygote, not which allele is common, beneficial, or stronger. A recessive X-linked allele can be expressed in a hemizygous individual, so “recessive means expressed only in homozygotes” is too broad.

The expected genotype ratio is 1 AA : 2 Aa : 1 aa. With complete dominance, the phenotype ratio is 3 dominant : 1 recessive. These are probabilities across offspring, not a promise about a family of four. Incomplete dominance produces an intermediate heterozygous phenotype; codominance produces distinguishable expression of both alleles, as in AB blood type. Alleles do not blend into a new allele.

Worked cross: RrYy × RrYy, with complete dominance at each locus and independent assortment. Round, green seeds require R_ and yy. Thus P(round and green) = (3/4)(1/4) = 3/16. The familiar 9:3:3:1 ratio depends on those assumptions; linked loci may not produce it.

3. Use atoms and bonds to predict properties

Atomic number Z counts protons; mass number A counts protons plus neutrons. A neutral atom has equal numbers of electrons and protons. Losing electrons produces a cation without changing the element.

Across a main-group period, atomic radius generally decreases while first ionization energy and electronegativity generally increase. Down a group, added shells generally increase radius. These are trends with exceptions; they are not a substitute for electron configurations.

Ionic bonding is electrostatic attraction between ions; NaCl is a useful example. Covalent bonding involves shared electron density. Polar bonds share unequally, but bond polarity and molecular polarity are different: CO₂ has polar C=O bonds whose dipoles cancel in its linear geometry. Electronegativity-difference cutoffs are rough classroom guides, not universal boundaries between bond types.

4. Track units through stoichiometry

Use mass → moles → mole ratio → requested units. Coefficients compare numbers of molecules or moles, not masses.

For 2H₂ + O₂ → 2H₂O, assume oxygen is in excess. With the supplied rounded molar masses H₂ = 2.0 g/mol and H₂O = 18.0 g/mol:

5.0 g H₂ × (1 mol / 2.0 g) × (2 mol H₂O / 2 mol H₂) × (18.0 g / 1 mol) = 45 g H₂O.

The product can weigh more than the starting hydrogen because oxygen contributes mass. Without enough oxygen, this is only a maximum.

Worked limiting-reactant example: 10.0 g Al and 15.0 g Cl₂ react by 2Al + 3Cl₂ → 2AlCl₃. Use Al = 27.0, Cl₂ = 71.0, and AlCl₃ = 133.5 g/mol throughout.

Carry extra digits until the final result. Comparing reactant masses directly would incorrectly identify the limiting reactant.

5. State the conditions for pH

A Brønsted–Lowry acid donates a proton; a base accepts one. For dilute aqueous solutions, use pH ≈ −log₁₀[H⁺] and pOH ≈ −log₁₀[OH⁻], with numerical concentrations in mol/L. The exact definitions use activities.

At 25 °C, K_w ≈ 1.0 × 10⁻¹⁴, so pH + pOH ≈ 14. Neutral means [H⁺] = [OH⁻]; neutral pH is about 7 at this temperature. Values below 0 and above 14 are possible, and neutral pH changes with temperature. pH and pOH.

Worked example at 25 °C: [OH⁻] = 2.5 × 10⁻⁴ M.

1. pOH = −log₁₀(2.5 × 10⁻⁴) = 3.60.
2. pH = 14.00 − 3.60 = 10.40, basic.
3. Cross-check: [H⁺] = 10⁻¹⁴/(2.5 × 10⁻⁴) = 4.0 × 10⁻¹¹ M, giving the same pH.

Try it: if [H⁺] = 1.0 × 10⁻⁹ M at 25 °C, pH is 9.00 and the solution is basic. A tenfold rise in [H⁺] lowers pH by one unit.

6. Recognize functional groups and three-dimensional structure

Primary, secondary, and tertiary amines exist; NH₂ alone is not a definition of every amine. For simple naming problems, identify the principal characteristic group before choosing and numbering the parent. Do not use “nearest functional group” as an unrestricted numbering rule. In 2-methylbutane, the lower substituent locant is 2 rather than 3.

Constitutional isomers differ in connectivity. Stereoisomers share connectivity but differ in spatial arrangement. Enantiomers are nonsuperimposable mirror images; diastereomers are stereoisomers that are not mirror images. A tetrahedral carbon bonded to four different groups is a common source of chirality, but chirality can also arise without such a center. Enantiomers.

Alanine, CH₃–CH(NH₂)–COOH: C1 is the carboxyl carbon, C2 is attached to H, NH₂, CH₃, and COOH, and C3 is methyl. C2 is the sole stereogenic center. An ordinary flat connectivity drawing identifies the center but does not specify its configuration. L/D labels describe configuration, not the sign of optical rotation; glycine is achiral.

7. Explain reactions by electron movement

Substrate, solvent, nucleophile/base, and leaving group all matter. A tertiary substrate alone does not uniquely select SN1 over elimination.

For propene + HBr without peroxides, the major product is:

CH3–CH=CH2 + HBr  →  CH3–CH(Br)–CH3
propene                2-bromopropane

Protonation that leads to the more stable secondary carbocation is favored, then Br⁻ attacks. In this example H goes to the terminal carbon, which already has more hydrogens. With radical-initiating peroxide conditions, HBr can instead give anti-Markovnikov addition; do not apply that exception indiscriminately to all HX reagents. Markovnikov orientation.

For comparable simple aldehydes and ketones, aldehydes are generally more reactive toward nucleophilic addition because the carbonyl carbon is less crowded and less stabilized by electron-donating alkyl groups. NaBH₄ reduction followed by workup turns an aldehyde into a primary alcohol and a ketone into a secondary alcohol. Methanal has no alkyl groups, so “every aldehyde has one alkyl group” is false.

Resonance contributors describe one delocalized electronic structure, not molecules switching back and forth. Stabilization depends on contributor quality and equivalence, not just the number of drawings.

Before moving on, solve the limiting-reactant and pH examples without looking, then explain one wrong approach. Being able to name the assumption that failed is more useful than remembering the numerical answer.