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Welcome to IB 201! Genetics and Evolution. Extra Credit Question: For extra credit question, please use the index cards provided. Print your name, TA,

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Presentation on theme: "Welcome to IB 201! Genetics and Evolution. Extra Credit Question: For extra credit question, please use the index cards provided. Print your name, TA,"— Presentation transcript:

1 Welcome to IB 201! Genetics and Evolution

2 Extra Credit Question: For extra credit question, please use the index cards provided. Print your name, TA, and section # at top of card. Thanks! A phenotype ratio of 9:3:3:1 in the offspring of a mating between two individuals that are heterozygous for two traits occurs when: A. the genes reside on the same chromosome B. each gene contains two mutations C. the gene pairs assort independently during meiosis D. only recessive traits are scored E. none of the above

3 Strategies for Learning Genetics Strategy Take notes during lecture and review before exam Summarize lecture in your own words after class Self-questioning during lecture Exam Grades Worst Intermediate Best

4 Why Genetics and Evolution? Evolution is unifying principle of biology. “Nothing in biology makes sense except in the light of evolution” Theodosius Dobzhansky (geneticist)

5 Darwin’s Postulates  Individuals within species are variable.  Some of these variations are passed on to offspring.  In every generation, more offspring are produced than can survive.  Survival & reproduction of individuals are not random. Those that survive and reproduce are those with the most favorable variations. They are naturally selected.

6 Darwin’s Success  He compiled massive amounts of evidence that all these postulates are true.  Showed conclusively that individuals within species are variable, and that some of these variations are passed on to the offspring.

7 Darwin’s Problems Because mechanisms of inheritance were not known, two objections were raised: 1. Inheritance acts like pigment in paint (“blending inheritance”). Any new variant would mix with existing traits and be swamped by them. Thus, even if new variation could be created, it would not persist. 2. Variation within species is limited. Once existing variation is exhausted, evolution by natural selection will grind to a halt. Discoveries in genetics solved both these problems.

8 Mendel solves one of Darwin’s Problems  When Origin was published, an unknown monk was cultivating peas in central Europe. Gregor Mendel would demonstrate that objection 1 is invalid by showing that inheritance is particulate, not blending.  Darwin had Mendel’s paper but did not read it,; he also did crosses but did not go beyond the F1!

9 Morgan solves the other  In 1900 DeVries, Von Tschermak, and Correns published experiments similar to Mendel’s, and attributed priority of discover to him.  Within a few years, T.H. Morgan & colleagues showed that new hereditary variation (mutation) occurs in every generation.  By beginning of the 20th century, both objections to Darwin’s principles had been conclusively eliminated by discoveries in genetics.

10 “Modern Synthesis”  Evolution by natural selection was not fully accepted until the hereditary mechanism was understood.  The “Modern Synthesis” is the synthesis of Mendelism and Darwinism that occurred in the early 20th century.  The Modern Synthesis is the foundation of modern evolutionary biology.

11 Modern genetics is also based on evolutionary biology and the Modern Synthesis  Medical population genetics: patterns of genetic disease and disease resistance  Genetic epidemiology:identification and spread of new disease vectors: Hanta virus, AIDS, SARS  Genome biology: Sequencing projects for non-human organisms can tell us where did human genes come from, which genes are conserved and which have diverged,and how did eukaryotic genomes come to be organized the way they are (introns, exons, repetitive DNA).

12 Course Outline Deviations from Mendelism: Epistasis; Unusual Modes of Inheritance Genetic Data Analysis: Probability & Statistics Chromosomal Inheritance: Chromosomal Abnormalities; Sex Determination Mapping: Gene and Genome Mapping Traits Affected by Genes & Environment Quantitative Traits Genes in Populations Genetic Mechanisms of Evolution; Population Genetics of Disease and Disease resistance Genomes and Genome Evolution

13 Additional Deviations from Mendelism Lethal Alleles Epistasis Unusual sex linkage Sex influenced inheritance Genetic Anticipation

14 Manx Cats

15 Lethal alleles F1: MmxMm F1: 1 MM 2 Mm 1 mm F2: 1 Lethal: 2 Manx: 1 Normal F2 phenotypic ratio: 2:1 instead of 3:1

16 Other lethal mutations Achondroplasia (humans) Yellow body color (domestic mice) Curly wings (Drosophila)

17 Agouti: wild type

18 F2 Phen. ratio: 9 agouti : 3 black : 4 albino agouti F1: agouti P: agoutialbino F2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb cc Simple dominant phenotype? novel phenotype BB CCxbb cc black albino Bb Cc

19 Epistasis Locus 1 Locus 2 BB Bb bbCC Cc cc agouti agoutiblack no effect no effect albino

20 Biochemical model CC or Cc: tyrosinase is produced (involved in production of melanin) BB or Bb: controls distribution of the pigment

21 Epistasis Normal dihybrid ratio is altered from 9:3:3:1 to 9:3:4 C and B gene have an epistatic interaction

22 YellowBrown X F 2 generation F 1 generation Parental generation Red 9/16 Yellow 3/16 Brown 3/16 Green 1/16 Red Self-fertilization Figure 10.18b Crosses between pure lines produce novel colors. R-Y- rrY- R-yy rryy R-Y- rrYY RRyy Codominance?

23 GenotypeColorExplanation of color R-Y- rrY- R-yy Red Yellow Brown Red pigment + no chlorophyll Yellow pigment + no chlorophyll Red pigment + chlorophyll rryyGreenYellow pigment + chlorophyll Gene 1Gene 2 R = Red r = Yellow (-) = R or r Y = Absence of green (no chlorophyll) y = Presence of green (+ chlorophyll) (-) = Y or y Figure 10.18c Model to explain 9 : 3 : 3 : 1 pattern observed above: Two genes interact to produce pepper color.

24 Practice Problem In Labrador retrievers, coat color is controlled by two loci each with two alleles B,b and E,e respectively. When pure breeding Black labs with genotype BB EE are crossed with pure breeding yellow labs of genotype bb ee the resulting F1 offspring are black. F1 offspring are crossed (Bb Ee x Bb Ee). Puppies appear in the ratio: 9/16 black; 3/16 chocolate; 9/16 black; 3/16 chocolate; 4/16=1/4 yellow. What genotypes correspond to these three phenotypes? 9/16 B- E- 3/16 B- ee3/16 bb E-1/16 eebb B- E-B- ee bb E- and bb ee

25 Other kinds of epistasis 1/16 aabb 9/16 A-B- 3/16 A-bb3/16 aaB- Hint: usually given numbers, not fractions 27 agouti; 12 albino; 9 black 28 agoute; 11 albino; 4 black

26 Practice Problem In the summer squash (Cucurbita pepo) fruit shape is determined by two genes. Two different true-breeding spherical types were crossed. The F1's were all disk, and the F2's segregated 35 disk, 25 spherical and 4 long. Explain these results. What’s the first step? Notice novel phenotype: disk, long. What’s the next step? Notice there are three F2 phenotypes. What kind of inheritance will give three F2 phenotypes? Incomplete, codominance Epistasis Expected F2 ratio? 1:2:1 Variation on 9:3:3:1

27 Practice Problem, cont. In the summer squash (Cucurbita pepo) spherical fruit is recessive to disk, True-breeding spherical types from different geographic regions were crossed. The F1's were disk, and the F2's segregated 35 disk, 25 spherical and 4 long. Explain these results. Are the phenotypic ratios closer 1:2:1 or to a variant of 9:3:3:1 ? If phenotypic ratios closer to a variant of 9:3:3:1, then what variant is it? Total # of individuals = = 64 64/16 = 4 9*4 = 36 6*4 = 24 1*4 = 4 Phenotypic ratio close to 9:6:1

28 Practice Problem, cont. In the summer squash (Cucurbita pepo) spherical fruit is recessive to disk, True-breeding spherical types from different geographic regions were crossed. The F1's were disk, and the F2's segregated 35 disk, 25 spherical and 4 long. Explain these results. If phenotypic ratios are close to 9:6:1, then what are the genotypes associated with each phenotype? 35 disk 25 spherical 4 long 9/16 A- B- 3/16 A- bb + 3/16 aa B- 1/16 aa bb What were the genotypes of the original spherical parents? AA bb aaBB


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