8.3 Genetics

8.3 Genetics on NMAT Biology

Genetics on CEM NMAT Biology (part of the 30-item / ~30-minute academic subtest) links inheritance patterns to molecular mechanisms. You must compute transmission probabilities, read simple pedigrees, and follow DNA replication, transcription, and translation — then connect mutations and chromosome errors to phenotype.

Quick frame: Every inheritance item is a story about alleles in gametes combining. Every molecular item is a story about information copied or expressed. Keep those stories separate until a stem explicitly joins them (e.g., mutation causing a new allele).

Mendelian laws (core transmission logic)

Gregor Mendel’s pea experiments yield two laws still foundational:

1. Law of Segregation: The two alleles of a gene separate during gamete formation so each gamete carries one. Mechanistically this is homolog separation in meiosis I (with sister chromatids still linked until meiosis II).
2. Law of Independent Assortment: Alleles of different genes on different chromosomes (or far apart on one chromosome) assort independently. Physically this is random orientation of homologous pairs at metaphase I.

Vocabulary:

Classic monohybrid F2 ratio (complete dominance): cross Aa × Aa → genotypes 1 AA : 2 Aa : 1 aa; phenotypes 3 dominant : 1 recessive.

Classic dihybrid F2 (independent assortment, complete dominance both loci): 9:3:3:1 phenotypic ratio from AaBb × AaBb.

Punnett squares (method)

1. Write parental genotypes.
2. List unique gametes for each parent (apply segregation).
3. Fill grid with combinations.
4. Tally genotypes → map to phenotypes using dominance rules.
5. Convert counts to fractions or percents; check sum = 1.

Worked problem 1 — monohybrid
Trait: freckles dominant (F) over none (f). Parents: heterozygous freckled × freckled heterozygote → Ff × Ff.
Gametes: F, f each. Offspring: 1 FF : 2 Ff : 1 ff. Phenotypes: 3/4 freckled, 1/4 non-freckled. Probability a child is homozygous freckled: 1/4.

Worked problem 2 — test cross
Dominant phenotype, unknown genotype × recessive aa.

• If all offspring dominant → parent likely AA.
• If 1:1 dominant:recessive → parent Aa.
  Test crosses reveal genotype by recessive probe.

Worked problem 3 — dihybrid probability
AaBb × AaBb, independent genes. Probability of aabb?
aa = 1/4, bb = 1/4 → (1/4)×(1/4) = 1/16 (matches 9:3:3:1 corner).

Incomplete dominance, codominance, and related extensions

ABO blood type table (high-yield)

Worked problem 4 — incomplete dominance
Red (CᴿCᴿ) × white (CᵂCᵂ) → all CᴿCᵂ pink. F2 from pink × pink: 1 red : 2 pink : 1 white. Note: genotypic and phenotypic ratios match because heterozygote is distinct.

Worked problem 5 — blood type
Type A parent (could be IᴬIᴬ or Iᴬi) × type B (IᴮIᴮ or Iᴮi). Child type O (ii) is possible only if both parents carry i → parents Iᴬi and Iᴮi. Child AB is also possible from those genotypes.

Sex-linkage and pedigrees

In humans, XX female, XY male; many genes on X have no counterpart on Y. X-linked recessive traits (e.g., classic hemophilia, red–green color blindness teaching examples):

• Males express whatever allele is on their single X (hemizygous)
• Females need two copies to express recessive phenotype
• Affected males transmit the allele to all daughters (as carriers if recessive) and no sons (sons get dad’s Y)
• Carrier females transmit to half of sons (affected) and half of daughters (carriers), on average

Pedigree symbols (basics): squares = male, circles = female; shaded = affected; horizontal line = mating; vertical line down to offspring.

Worked problem 6 — X-linked
Carrier mother XᴴXʰ × normal father XᴴY. Sons: 1/2 XᴴY normal, 1/2 XʰY affected. Daughters: 1/2 XᴴXᴴ, 1/2 XᴴXʰ carriers — none affected if recessive. Probability a son is affected = 1/2; probability a random child is an affected son = (1/2 child is male)×(1/2) = 1/4 if sex ratio equal.

Worked problem 7 — pedigree reasoning
Two unaffected parents have an affected daughter. For a rare autosomal recessive trait this is possible (Aa × Aa). For a rare X-linked recessive, an affected daughter implies father is affected (she must get his X with the allele) — if father is unaffected, X-linked recessive is unlikely. That discrimination is classic exam logic.

DNA structure, replication, transcription, translation

DNA structure: Double helix; antiparallel strands (5′→3′ opposite 3′→5′); backbone sugar–phosphate; bases A–T (2 H-bonds) and G–C (3 H-bonds); complementarity enables accurate copying.

Replication (semi-conservative): Each daughter duplex keeps one parental strand. Key players (intro set):

• Helicase unwinds; SSB proteins stabilize single strands
• Primase lays RNA primers
• DNA polymerase elongates 5′→3′ only
• Leading strand continuous; lagging strand Okazaki fragments
• DNA ligase seals nicks; proofreading reduces error rate

Central dogma (intro): DNA → (transcription) RNA → (translation) protein. (Reverse transcriptase in some viruses is a known exception students should recognize by name.)

Transcription: RNA polymerase reads DNA template, synthesizes mRNA (coding sequence in eukaryotes is processed: 5′ cap, poly-A tail, splicing out introns).

Translation: Ribosome reads mRNA codons (triplet code, nearly universal). tRNA anticodons bring amino acids. Start codon AUG (Met); stop codons release chain. Stages: initiation, elongation, termination.

Worked problem 8 — complementary strand
Template DNA 3′–TAC GGA–5′. RNA polymerase reads the template 3′→5′ and builds mRNA 5′→3′, so the mRNA is complementary to the template and matches the coding strand with U in place of T. Template 3′–TAC GGA–5′ → mRNA 5′–AUG CCU–3′. AUG is the start codon (Met) and CCU codes for Pro, so translation begins Met–Pro…

Worked problem 9 — mutation type
Coding DNA change that substitutes one base may yield: silent (same amino acid), missense (different amino acid), nonsense (early stop). Frameshift (insertion/deletion not multiple of 3) scrambles downstream codons — usually severe.

Mutations and chromosomal abnormalities

Mutations: heritable DNA sequence changes (gene level). Causes: replication errors, mutagens, radiation, etc. Somatic vs germline determines organism-only vs offspring transmission.

Chromosomal abnormalities (intro):

Nondisjunction: Homologs (meiosis I) or sister chromatids (meiosis II/mitosis) fail to separate → gametes n+1 or n−1 → zygotes with trisomy or monosomy after fertilization.

Worked problem 10 — nondisjunction reasoning
If sister chromatids fail to separate in meiosis II in one cell lineage, two gametes may be normal, one n+1, one n−1 from that meiosis’s products — exam stems often ask which division error fits a karyotype pattern.

Integrated inheritance drills

Worked problem 11 — combine assortment
Autosomal Aa and independent Bb. Parent AaBb. Fraction of gametes ab? 1/4 (independent 1/2 × 1/2).

Worked problem 12 — phenotype probability with sex
X-linked recessive color blindness: carrier woman × normal man. Probability that a daughter is color blind? 0 (father contributes normal X). Probability a son is color blind? 1/2.

Worked problem 13 — incomplete dominance × ratio check
Pink × red (CᴿCᵂ × CᴿCᴿ) → 1/2 red, 1/2 pink; no white. If white appears, parentage or dominance model is wrong.

Error traps

Study protocol

1. Daily Punnett: one mono, one dihybrid, one sex-linked (5 minutes).
2. Explain segregation using meiosis vocabulary once without notes.
3. Trace a polypeptide from gene to RER secretory path (links to Section 8.2).
4. Classify five mutation descriptions into silent/missense/nonsense/frameshift.

Section checkpoint

You are ready for NMAT genetics when you can: (1) derive 3:1 and 9:3:3:1 from first principles, (2) solve incomplete dominance and ABO problems without mixing models, (3) predict X-linked recessive risks for sons vs daughters, (4) order replication → transcription → translation with correct products and sites, and (5) connect nondisjunction to aneuploid gametes in one causal sentence.