BIO 311D 1st EditionFinal Exam Study Guide: Lectures 1-31Mitosis and MeiosisLecture 1The Stages of Meiosis:1. Interphase: Chromosomes duplicate2. Meiosis I: Homologous chromosomes separatea. Prophase I: Chromosomes condense, homologous chromosomes pair up (synapsis), and crossing over occursb. Metaphase I: Tetrads line at metaphase plate, microtubules attach to kinetochoresc. Anaphase I: Pairs of homologous chromosomes separate and move toward pole by spindle apparatusd. Telophase I/Cytokinesis: Cleavage furrow divides the cell into two3. Meiosis II: Sister chromatids separatea. Prophase II: Spindle apparatus formsb. Metaphase II: Sister chromatids (due to crossing over, sister chromatids are no longer identical) align at the metaphase plate and microtubules attach to kinetochoresc. Anaphase II: Sister chromatids separate and move towards opposite polesd. Telophase II/Cytokinesis: Nuclei form and chromosomes decondense*Each daughter cell is genetically different from the othersComparing Mitosis and Meiosis:Property Mitosis MeiosisDNA Replication Interphase InterphaseNumber of Divisions One TwoSynapsis of Chromosomes None Prophase INumber of Daughter Cells Two (Identical) Four (Unique)Role in the Animal Body Growth, Repair Produces GametesCharacteristics Unique to Meiosis (all occur in Meiosis I):1. Prophase I: Synapsis and crossing over occurs2. Metaphase I: Tetrads (paired homologous chromosomes) line up at the metaphase plate3. Anaphase I: Homologous chromosomes separate instead of sister chromatidsMendel and the Gene IdeaLectures 2-3Advantages of pea plants:1. Many varieties with distinct heritable features2. Mating can be controlled3. Each flower has sperm producing organs (stamens) and egg producing organs (carpel)4. Cross pollination (fertilization between differentplants) involves dusting one plant with pollen fromanotherThe Law of Segregation:- Two alleles for a heritable character segregate duringgamete formation and end up in different gametes- Discovered by using monohybrids (crosses betweenone character)- The white color trait did not disappear but rather itwas recessive- Mendel called the purple flower the dominant trait- The white colored flower became the recessive traitThe Law of Independent Assortment:- Dihybrid cross determines whether two characters are transmitted to offspring as a package or independently- Dihybrid cross was used to discover the law of independent assortment- Found that each pair of alleles segregates independently of each other during gamete formation- However, genes located near each other on the same chromosome tend to be inherited togetherDEFINITIONS YOU SHOULD KNOW:- Monohybrid cross: A cross between individuals for one trait (ratio of 3:1)- Dihybrid Cross: A cross between individuals for two traits (ratio of 9:3:3:1)- Complete Dominance: Occurs when phenotype of the heterozygote and dominant homozygote are identical- Incomplete Dominance: Phenotype of F1 hybrids is somewhere between the phenotype of two parental varieties- Codominance: Two dominant alleles affect the phenotype in separate, distinguishable ways- Multiple Alleles: Most genes exist in populations in more than two allelic forms- Pleiotropy: Most genes have multiple phenotypic effects- Epistasis: A gene at one locus alters the phenotypic expression of a gene at a second locus- Polygenic Inheritance: An additive effect of two or more genes on a single phenotypeChromosomal Basis of InheritanceLectures 4-5Morgan’s Experimental Evidence:- Thomas Hunt Morgan (an embryologist) found evidence associating a specific gene with a specific chromosome- He performed an experiment with fruit flies which provided evidence that chromosomesare the location of Mendel’s heritable factors- Advantages of breeding fruit flies:o Fruit flies produce many offspringo A generation can be bred every two weekso They only have four pairs of chromosomesX and Y chromosome/Important information over Chromosomes:- X is larger than Y, although they have the same amount of genes- Karyotype is a complete set of chromosomes in the cell- X-linked disorders tend to be more common in males than females- Given a fly with a gray body and normal wings and another fly with a black body and vestigial wings, what will the F1 generation look like?- Test cross this F1 generation, what would the offspring look like?X Inactivation in Female Mammals:- Because females have XX, one X becomesinactive- Inactive x condenses into a barr body- One of the two x chromosomes in eachcell is randomly inactivated- Example: Female cats can have certainspots where she is brown and other spotswhere she is orange. This is due to theinactivation of random X chromosomesMapping the Distance Between Genes UsingRecombination Data:- Genetic map is an ordered list of geneticloci along a particular chromosome- He predicted that the farther apart twogenes are, the higher probability that acrossover will occur between them, therefore a higher recombination frequency- Linkage map: Genetic map of a chromosome based on recombination frequencies- Map Units: Distances between genes o One centimorgan represents a 1%recombination frequencyo Map units indicate relativedistance/order, NOT the preciselocation of genesThe Evolution of PopulationsLectures 6-7Microevolution: A change in allele frequencies in a population over generationsThe Hardy-Weinberg equation:- Can be used to test whether a population has evolved- The frequency of an allele in a population can be calculatedo For diploid organisms the total number of alleles at a locus is the total number of individuals times 2- By convention, if there are 2 alleles at a locus, p and q are used to represent their frequencieso P is in reference to dominanceo Q is used in reference to the recessive alleleo P + Q will always equal 1- For example, consider a population of wildflowers that is incompletely dominant for color:o 320 red flowers (CRCR)o 160 pink flowers (CRCW)o 20 white flowers (CWCW) - Calculate the number of copies of each allele:o CR = (320 ´ 2) + 160 = 800o CW = (20 ´ 2) + 160 = 200- To calculate the frequency of each allele:o p = freq CR = 800 / (800 + 200) = 0.8o q = freq CW = 200 / (800 + 200) = 0.2- The sum of alleles is always 1o 0.8 + 0.2 = 1Example: 1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis.1. When counting the phenotypes in a population