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Comprehensive Study Notes: Mendelian Inheritance, Gene Expression, Linkage, and Molecular Biology Concepts

Genes, alleles, and genotype vs phenotype

  • Genes and alleles
    • A gene is a unit that determines a trait (e.g., flower color).
    • Alleles are two versions of a gene, e.g., A and a. Each organism has two alleles per gene (one from each parent).
    • Dominant allele typically masks the recessive allele in the phenotype; genotype can be homozygous dominant (AA), homozygous recessive (aa), or heterozygous (Aa).
  • Genotype vs phenotype
    • Genotype: the alleles an individual carries for a gene (e.g., AA, Aa, aa).
    • Phenotype: the observable trait expressed by the genotype (e.g., purple flower color).
    • If two dominant alleles (AA) → dominant phenotype.
    • If two recessive alleles (aa) → recessive phenotype.
    • If heterozygous (Aa) → typically shows the dominant phenotype.
  • Example interpretations
    • A flower with purple phenotype but genotype Aa is heterozygous yet purple due to dominance.
    • A flower with purple phenotype and genotype AA is homozygous dominant.
  • Terminology related to sex chromosomes
    • Genotypes can be homozygous (same alleles) or heterozygous (different alleles).
    • Hemizygous refers to having only one copy of a gene, as in many X-linked traits in males (males have one X; thus they can’t be carriers like females do).

Monohybrid vs dihybrid crosses; test crosses

  • Test cross
    • Use a dominant phenotype with unknown genotype (e.g., purple flower) and cross it with a homozygous recessive tester (e.g., recessive for the trait).
    • Offspring phenotypes reveal whether the dominant parent is homozygous dominant or heterozygous.
  • Dihybrid crosses
    • Cross involving two traits with two genes (e.g., A and B) where each gene has two alleles.
    • If both parents are dihybrids (Aa Bb × Aa Bb), the classic phenotypic ratio in offspring is
    • 9:3:3:1
    • This arises because each gene assort independently and each trait has a dominant/recessive pattern.
  • Two-trait Punnett square expectations
    • When two genes are unlinked and assort independently, the phenotypic classes appear in the order: dominant for both, dominant for A but recessive for B, recessive for A but dominant for B, recessive for both.
    • The corresponding ratio is 9:3:3:1 across the four phenotypic categories, out of 16 equally likely gamete combinations.
  • Probability reminders
    • If crossing Aa × Aa for a single trait (monohybrid), the probability of a dominant phenotype in offspring is P( ext{dominant}) = rac{3}{4} (3/4 for Aa × Aa).
  • Practice questions approach
    • For a cross like AaBb × AaBb, you can memorize the 9:3:3:1 ratio or derive it via a full 16-square Punnett square.
    • If you know the parental genotypes for each trait, you can compute specific phenotype probabilities by multiplying independent outcomes (e.g., dominant for A and recessive for B → rac{3}{4} imes rac{1}{4} = rac{3}{16}).

Linkage, crossing over, and gene distance

  • Linked genes
    • If two genes are on the same chromosome (linked), they do not assort independently and you may not observe the 9:3:3:1 ratio.
    • The parental (non-recombinant) combinations tend to be more frequent than recombinant combinations.
  • Crossing over and distance
    • Crossing over between genes during meiosis can separate linked alleles, producing recombinants.
    • The frequency of crossing over indicates how far apart genes are on the chromosome.
    • A greater distance yields more opportunity for crossing over.
    • Example distance: 25 centiMorgans (cM) corresponds to a recombination frequency of 25% → 0.25 as a proportion.
    • Therefore, distance in Morgans equals recombination frequency: ext{distance} = r, ext{ where } r = ext{recombination frequency} \ (0 o 1); here r = 0.25 \Rightarrow ext{distance} = 25 ext{ cM}
  • Practical implication
    • If observed offspring show fewer parental types than expected, consider linkage and recombination between linked genes.

Human and classical genetic patterns and examples

  • X-linked (hemizygous) traits
    • Males have only one copy of many X-linked genes, so they cannot be carriers in the same sense as females.
    • A pattern example: carrier females (heterozygous) can pass the allele to offspring; affected males pass the allele to all daughters but not sons.
    • In autosomal traits you may see carriers in both sexes; in X-linked, male phenotypes directly reflect their single allele.
  • Albinism (autosomal recessive example)
    • Gene on chromosome 11; recessive allele causes albinism when present in two copies (aa).
    • Normal melanin production requires at least one functional copy: AA or Aa produce melanin; aa produces lack of pigment.
    • Gametes after meiosis: possible combinations include normal allele-containing gametes vs mutant allele-containing gametes; fertilization determines child phenotype.
  • Hemophilia and color-blindness (sex-linked) patterns
    • Schematic examples illustrate how a mutation on the X chromosome can produce affected males and carrier females.
  • Inheritance patterns summaries
    • Autosomal dominant vs recessive, autosomal inheritance shows both sexes affected roughly equally.
    • X-linked recessive patterns often show affected males with carrier or unaffected females; daughters can be carriers.

Grafical concepts: pleiotropy, polygenic traits, and codominance/incomplete dominance

  • Pleiotropy
    • A single gene influencing multiple traits or phenotypes (e.g., a gene affecting several bodily systems).
    • Example mention: a gene associated with cancer risk where more dominant risk alleles increase overall risk.
  • Polygenic traits
    • Traits controlled by multiple genes, often leading to continuous variation (e.g., height, skin color) — complex inheritance beyond simple Mendelian ratios.
  • Incomplete dominance
    • Heterozygotes display an intermediate phenotype rather than a dominant phenotype masking the other allele.
    • Classic example: red (RR) and white (rr) cross yields pink (Rr).
    • Incomplete dominance produces distinct phenotypes for each genotype (RR, Rr, rr).
    • Example discussed: pink flowers between red and white alleles; red and white are not blending in a single intermediate phenotype unless a system is described as such.
  • Codominance
    • Both alleles are fully expressed in heterozygotes.
    • Classic example: ABO blood types on chromosome 9.
    • Alleles A and B are codominant to each other and both are dominant over O (which is recessive).
    • Genotypes and phenotypes:
    • IAIA → phenotype A
    • IBIB → phenotype B
    • IAIB → phenotype AB (both A and B antigens expressed)
    • IOIO → phenotype O
    • Why O is recessive here: O allele does not produce a functional glycoprotein antigen.
  • ABO blood group as a paradigmatic codominance example
    • Alleles A and B code for glycoproteins that determine blood antigen; O allele is a null allele.
    • Crosses yield phenotypes following codominance rather than simple dominance relationships.
  • Worked example leave-behind questions
    • Cross AB parent with A parent: possible offspring phenotypes and genotypes include A, B, AB, with specific combinations depending on the second parent’s genotype.
    • A father color-blind (X-linked) and mother normal: explore inheritance outcomes for sons and daughters.

Gene-to-protein connections and example diseases

  • Central concept
    • A gene codes for a protein; the protein’s function contributes to the phenotype.
    • Example: cystic fibrosis involves a defective transmembrane protein, leading to mucus buildup and related symptoms (cough, mucus issues).
  • Labrador retriever coat color example
    • A gene determining whether a dog is black or chocolate, illustrating how a single gene can influence phenotype through a specific protein or pathway.

DNA structure, replication, and transcription basics

  • Semi-conservative replication
    • Each new DNA double helix contains one parental strand and one newly synthesized strand.
    • This was demonstrated by classic experiments and is summarized as semi-conservative replication.
  • Enzymes involved in DNA replication
    • Helicase: unwinds the DNA double helix.
    • Topoisomerase: helps relieve torsional strain as helicase unwinds DNA.
    • Primase: lays down RNA primers to provide a starting point for DNA synthesis.
    • DNA polymerase I: removes RNA primers and fills in with DNA.
    • DNA ligase: seals the nicks between Okazaki fragments on the lagging strand.
  • Leading vs lagging strand
    • Leading strand: synthesized continuously in the 5' to 3' direction as the fork opens.
    • Lagging strand: synthesized in short fragments (Okazaki fragments) in a discontinuous manner because it runs opposite to fork movement; each fragment begins with a primer.
    • Okazaki fragments are joined by ligase after primers are replaced by DNA.
  • Primer direction and anti-parallelism
    • DNA strands are antiparallel; one runs 5' to 3' and the other 3' to 5'.
    • RNA primers are oriented to provide a 3' end for DNA polymerase to extend.
  • Mutations and chromosomal changes
    • Deletion: missing segment of DNA.
    • Duplication: segment repeated.
    • Inversion: a segment of DNA flips orientation within the chromosome.
  • PCR primer design and product sizing
    • PCR amplifies a targeted region between two primers facing toward each other.
    • Product size equals the sum of primer lengths plus the intervening sequence: if primer A is 25 bp, primer B is 25 bp, and the intervening sequence is 40 bp, then the product length is 25 + 40 + 25 = 90 ext{ bp}.
    • Gel electrophoresis separates products by size; a 90 bp product would appear between the markers that bracket 90 bp.
  • Primer orientation and overlap
    • Primers should anneal to opposite strands so that extension proceeds toward the target region and with sufficient overlap to ensure specificity.
  • Transcription basics (brief recap from lecture)
    • RNA synthesis uses a DNA template; RNA polymerase reads from a 3' to 5' DNA template, synthesizing RNA in a 5' to 3' direction.
    • Base pairing rules: G pairs with C; A pairs with U in RNA (T in DNA).
  • RNA processing: introns and exons
    • Introns: non-coding segments that must be removed from the primary transcript (pre-mRNA) via splicing.
    • Exons: coding segments that remain in the mature mRNA.
    • Mature mRNA contains only exons after splicing and is then translated into protein.

Quick reference: key formulas and numerical ideas

  • Mendelian dihybrid phenotype ratio (unlinked genes):
    • 9:3:3:1. For a AaBb × AaBb cross, corresponding phenotypic proportions are rac{9}{16}, rac{3}{16}, rac{3}{16}, rac{1}{16}.
  • Monohybrid dominant phenotype probability (Aa × Aa):
    • P( ext{dominant}) = rac{3}{4}.
  • Recombination frequency and distance between genes
    • r = rac{ ext{recombinant offspring}}{ ext{total offspring}}.
    • If r = 0.25, distance = 25 \, \text{cM}.
  • Product size in PCR when primers are of known lengths
    • If primer lengths are l1, l2 and the intervening fragment length is n, then product length L = l1 + n + l2.
    • Example: L = 25 + 40 + 25 = 90\text{ bp}.