Free Unit Notes

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I. Meiosis (Topic 5.1)

A. Chromosomes and chromosome sets

1. A diploid (2n) cell has two chromosome sets, one from each parent. A haploid (n) cell has one set.

2. Homologous chromosomes carry the same genes at corresponding locations, or loci. They may carry different versions of a gene, called alleles.

3. Sister chromatids are copies of one chromosome made during DNA replication. They remain joined until they separate during division.

4. Meiosis reduces the chromosome number from diploid to haploid. Fertilization joins haploid gametes to form a diploid zygote.

B. DNA replication and chromosome counting

1. DNA replicates once during S phase before meiosis I. Two divisions follow, with no DNA replication between them.

2. Count chromosomes by their centromeres. Replication doubles the DNA amount but does not immediately change the chromosome number.

3. A human cell has 46 chromosomes before and after replication. After replication, it has 92 chromatids; a normal gamete has 23 unduplicated chromosomes.

C. Meiosis I: separation of homologous chromosomes

1. In prophase I, homologous chromosomes pair through synapsis, forming tetrads with four chromatids. Crossing over can occur between non-sister chromatids at sites called chiasmata.

2. Chromosomes condense, the spindle forms, and the nuclear envelope breaks down. Spindle microtubules attach to chromosome kinetochores.

3. In metaphase I, homologous pairs line up at the cell’s equator. Each pair’s orientation is random relative to the other pairs.

4. In anaphase I, homologous chromosomes move to opposite poles. Sister chromatids remain joined, so each chromosome is still duplicated.

5. In telophase I, chromosomes reach the poles, and nuclei may reform. Cytokinesis usually produces two haploid cells with duplicated chromosomes.

D. Meiosis II: separation of sister chromatids

1. In prophase II, spindles form and attach to the duplicated chromosomes. No new round of DNA replication occurs.

2. In metaphase II, chromosomes line up individually at the equator of each cell, with sister chromatids attached to opposite poles.

3. In anaphase II, the connections holding sister chromatids together break. The chromatids separate and become individual chromosomes.

4. In telophase II, nuclei form and chromosomes decondense. Cytokinesis usually produces four haploid cells with unduplicated chromosomes.

E. Meiosis compared with mitosis

1. Both processes use spindle microtubules to move chromosomes after DNA replication.

2. Mitosis has one division and usually produces two genetically similar cells with the parent cell’s chromosome number. It supports growth, repair, and asexual reproduction.

3. Meiosis has two divisions and produces haploid products with new allele combinations. Homologous pairing and crossing over distinguish meiosis I from typical mitosis.

II. Meiosis and Genetic Diversity (Topic 5.2)

A. Sources of genetic variation

1. Crossing over exchanges corresponding DNA segments between non-sister chromatids during prophase I. It creates recombinant chromosomes with new combinations of existing alleles.

2. Independent assortment results from the random orientation and separation of homologous pairs. Each gamete receives a mixture of maternal and paternal chromosomes.

3. For n homologous pairs, independent assortment alone allows 2ⁿ chromosome combinations. With n = 23, humans can produce 8,388,608 combinations before crossing over is considered.

4. Random fertilization combines gametes independently, further increasing possible allele combinations in offspring.

5. Mutation creates new alleles. Meiosis and fertilization mainly rearrange existing alleles; they do not guarantee that every offspring is genetically unique.

B. Errors in chromosome separation

1. Nondisjunction occurs when homologous chromosomes fail to separate in meiosis I or sister chromatids fail to separate in meiosis II.

2. If one pair fails to separate in meiosis I, all four products have an abnormal number of that chromosome: two have an extra copy and two lack a copy.

3. If the error occurs in one cell during meiosis II, two products have the normal number, one has an extra copy, and one lacks a copy.

4. Fertilization involving an abnormal gamete may produce aneuploidy, an extra or missing individual chromosome. Trisomy means three copies; monosomy means one copy.

5. Trisomy 21 involves an extra chromosome 21. Chromosome imbalances can disrupt development by changing the amounts of many gene products.

III. Mendelian Genetics (Topic 5.3)

A. Genes, genotypes, and phenotypes

1. A gene is a DNA sequence that contributes to a functional product. Alleles are alternative versions of a gene.

2. Genotype is an organism’s allele combination. Phenotype is an observable or measurable characteristic influenced by genotype and environment.

3. A homozygous individual has matching alleles at a locus, such as AA or aa. A heterozygous individual has different alleles, such as Aa.

4. Under complete dominance, Aa and AA have the same phenotype. The recessive phenotype appears in aa; dominant does not mean more common, stronger, or beneficial.

B. Mendel’s laws and meiosis

1. The law of segregation states that the two alleles at a locus separate during gamete formation. Each normal gamete receives one allele.

2. Separation of homologous chromosomes in meiosis I provides the basis for segregation. Chromatids separate in meiosis II, completing allele distribution.

3. The law of independent assortment states that alleles of different unlinked genes enter gametes independently. Random orientation of homologous pairs explains this pattern for genes on different chromosomes.

4. Genes far apart on the same chromosome can behave as unlinked because recombination separates their alleles frequently.

C. Monohybrid, dihybrid, and test crosses

1. A Punnett square combines possible parental gametes to predict offspring genotypes and phenotypes. A monohybrid cross follows one gene.

2. In Aa × Aa, expected genotypes are 1/4 AA, 1/2 Aa, and 1/4 aa. Complete dominance gives a 3:1 dominant-to-recessive phenotype ratio.

3. A dihybrid cross follows two genes. An AaBb parent produces AB, Ab, aB, and ab gametes in equal proportions if the genes assort independently.

4. AaBb × AaBb gives a 9:3:3:1 phenotype ratio when the genes assort independently, both show complete dominance, and their effects do not alter one another.

5. A test cross pairs an individual with a dominant phenotype and unknown genotype with a homozygous recessive individual.

   a. AA × aa produces all Aa offspring. Aa × aa predicts equal proportions of Aa and aa.
   b. Recessive offspring reveal the unknown parent’s recessive allele. A small sample with no recessive offspring does not prove that parent is AA.

D. Probability rules

1. For independent events, multiply: P(A and B) = P(A) × P(B). In AaBb × AaBb, P(aabb) = 1/4 × 1/4 = 1/16 if the genes assort independently.

2. For mutually exclusive outcomes, add: P(A or B) = P(A) + P(B). In Aa × Aa, Aa can arise as A from one parent and a from the other in either order: 1/4 + 1/4 = 1/2.

3. Predicted ratios describe probabilities over many offspring. Each fertilization is a new event; earlier offspring do not determine the next offspring’s genotype.

E. Pedigrees and autosomal inheritance

1. A pedigree traces a trait through generations. Squares commonly represent males, circles represent females, and filled symbols mark the studied phenotype.

2. Autosomal genes lie on chromosomes other than sex chromosomes. Autosomal dominant and recessive traits can affect either sex.

3. An autosomal recessive trait can occur in children of two unaffected heterozygous carriers. Two carriers have a 1/4 affected-offspring probability for each pregnancy.

4. An autosomal dominant trait often appears in successive generations. Under a simple model with full expression, an affected individual has an affected parent unless a new mutation occurred.

5. Test possible genotypes against all family relationships. A small pedigree may fit more than one inheritance model.

F. Chi-square goodness-of-fit testing

1. A null hypothesis states that observed counts fit a specified inheritance model; differences from expected counts result from chance sampling.

2. Calculate each expected count as total offspring × expected proportion. Use counts, rather than percentages, in χ² = Σ[(O − E)²/E], where O is observed and E is expected.

3. Add the calculated terms for all categories. For a standard genetic goodness-of-fit test with fixed predicted proportions, degrees of freedom = number of categories − 1.

4. Compare χ² with the critical value for the chosen significance level and degrees of freedom. At α = 0.05 and df = 1, the critical value is about 3.84.

5. Reject the null hypothesis if χ² exceeds the critical value. Otherwise, fail to reject it; this does not prove the model is true.

6. For 80 offspring under a 3:1 model, expected counts are 60 and 20. Observed counts of 54 and 26 give χ² = 36/60 + 36/20 = 2.40, so the model is not rejected at α = 0.05.

7. Use independent observations and sufficiently large expected counts, typically at least five per category. A significant result suggests the proposed model needs further investigation.

IV. Non-Mendelian Genetics (Topic 5.4)

A. Incomplete dominance, codominance, and multiple alleles

1. In incomplete dominance, the heterozygote has an intermediate phenotype. Red and white flower alleles may produce pink heterozygotes; the alleles remain distinct.

2. In codominance, both alleles are expressed in a heterozygote. AB blood cells display both A and B antigens.

3. A cross between two heterozygotes can give a 1:2:1 phenotype ratio under incomplete dominance or codominance. These patterns change phenotype relationships without violating allele segregation.

4. Multiple alleles means a population has more than two alleles at a locus. A diploid individual generally carries only two.

5. The ABO alleles are Iᴬ, Iᴮ, and i. Iᴬ and Iᴮ are codominant, and both are dominant to i.

   a. Type A is IᴬIᴬ or Iᴬi; type B is IᴮIᴮ or Iᴮi.
   b. Type AB is IᴬIᴮ; type O is ii.

B. Linked genes and genetic mapping

1. Linked genes lie on the same chromosome. Closely spaced genes tend to be inherited together because crossing over is less likely between them.

2. Parental combinations match the allele arrangements on the original parental chromosomes. Recombinant combinations differ because crossing over rearranged those alleles.

3. In a test cross, offspring phenotypes can reveal gamete combinations from the heterozygous parent. An unlinked dihybrid test cross predicts a 1:1:1:1 ratio; excess parental combinations suggest linkage.

4. Recombination frequency = (recombinant offspring ÷ total offspring) × 100%. A frequency of 1% corresponds to about one map unit, or centimorgan (cM), over short intervals.

5. If 84 of 600 offspring are recombinant, the frequency is 14%, giving an estimated distance of 14 cM.

6. Observed two-gene recombination frequency cannot exceed 50%. Multiple crossovers can hide recombination, so distant genes on one chromosome may appear unlinked.

C. Sex-linked inheritance

1. Sex-linked genes lie on sex chromosomes. In a typical XX/XY system, an XY parent passes X to daughters and Y to sons.

2. For most X-linked genes, an XY individual has one allele. An X-linked recessive allele can therefore be expressed without a second recessive copy.

3. An X-linked recessive carrier mother and unaffected father predict affected sons in half of their sons and carrier daughters in half of their daughters.

4. Fathers do not pass X-linked alleles to sons. An affected father passes his X-linked allele to every daughter, whose phenotype also depends on the maternal allele.

5. A Y-linked allele passes from father to son in the usual XY system. A trait being more common in one sex does not, by itself, prove sex linkage.

6. Sex determination differs among species. Female birds are typically ZW and males ZZ; in honeybees, unfertilized haploid eggs usually develop into males and fertilized diploid eggs into females.

D. Relationships among genes and traits

1. Pleiotropy occurs when one gene influences several traits. Those effects share a genetic cause rather than segregating as independent genes.

2. Polygenic inheritance occurs when multiple genes affect one trait. Their combined effects can produce continuous variation, such as a range of heights.

3. Epistasis occurs when one gene changes or masks another gene’s effect. For example, a gene that prevents pigment production can hide a second gene’s effect on pigment color.

E. Non-nuclear inheritance

1. Mitochondria and chloroplasts contain DNA. Traits encoded by this DNA can follow inheritance patterns different from those of nuclear genes.

2. In animals, mitochondria usually come from the egg. Mitochondrial alleles are therefore typically maternally inherited by offspring of either sex.

3. In many plants, mitochondria and chloroplasts are inherited through the ovule. Organelle inheritance is usually maternal, although exceptions occur.

4. Organelles are distributed among gametes and daughter cells. When organelles carry different DNA variants, their proportions can differ among cells and offspring, affecting phenotype.

V. Environmental Effects on Phenotype (Topic 5.5)

A. Genotype and environment

1. Environmental conditions can affect gene expression, protein activity, and development. The same genotype can therefore produce different phenotypes without a DNA sequence change.

2. Phenotypic plasticity is a genotype’s ability to produce different phenotypes under different environmental conditions.

3. Nutrition can affect height and growth. Temperature can affect pigment-producing enzymes in some animals, changing coat color.

4. Soil conditions can affect flower color in some hydrangeas; soil pH influences aluminum availability, which affects pigment color.

5. A reaction norm describes the phenotype produced by a genotype across an environmental range. Different genotypes may respond differently to the same environmental change.

B. Studying environmental effects

1. Compare genetically similar organisms under different controlled conditions. Change the environmental factor of interest while keeping other relevant conditions consistent.

2. Measure the same phenotype, use adequate replication, and compare results with a control group. Correlation alone does not establish cause and effect.

3. Heritability describes how much trait variation within a particular population and environment is associated with genetic variation. It does not describe the genetic share of one individual’s trait or mean the environment cannot change it.

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