Transcript Slide 1
Chapter 14 – Mendelian Genetics Blending vs. Particulate Theory of Inheritance Over time, populations do not Traitsofover offspring Infer become uniformtime, were a “blend” of Mendel His observations observed lead that populations begin thelook parental genes Often to what’s traits retain now that their to uniform traits. separate seemed accepted to identities ashave the and look alike “disappeared” Particulate Theory of would reappear in inheritance subsequent generations Vocabulary Review Genotype = Genetic Homozygous ==When Heterozygous When make-up of of both alleles the both alleles of the of Trait: Characteristic Allele: Different forms characteristics genotype are the same Phenotype = expressed genotype are different an organism of a gene represented by the 2 alleles trait, basedresponsible on for different traits (Hybrid) in organisms Ex.diploid Plant Height genotype Ex. T = Tall t = short TT TT or Tt Tall Tt tt tt Short Single-Factor Crosses P = Parental Generation F1 = 1st Filial Generation X t t T T Tt Tt Tt Tt Memorize these ratios! F2 = 2nd Filial Generation X T t T t TT Tt Tt tt Genotypic Phenoytypic Ratio Ratio of of monohybrid monohybrid cross: cross 3:1 1:2:1 Two Factor Crosses (2 traits) Y = yellow, y = green R = Round, r = wrinkled X Dihybrid Cross X RrYy RrYy Memorize these ratios! Test Crosses If a plant has a dominant phenotype, (for example yellow seeds) and we are unsure of its genotype (YY, or Yy), you can determine it’s genotype by crossing it with another with a recessive phenotype (green seeds) with the genotype yy. Yy If F1 = 100% Yellow Then P must be = YY X YY or Yy? yy Yy yy If F1 = 50% Yellow, 50% green Then P = Yy (hybrid) Pedigree Aa ? A = tongue roller a = can not roll tongue aa ? ? ? ? AA aa Aa aa Can you figure out the rest of the genotypes on your own? male Mating couple female Children/Siblings Shaded = trait being followed Other Pedigree Symbols Examples of connected symbols: • Fraternal twins • Identical twins Unique Degrees of Dominance (exceptions to the rule) Incomplete Dominance: Dominant trait “blends” when combined with a recessive allele Notice how the genotypes are written… X CRCR CRCW CWCW Unique Degrees of Dominance Co-Dominance: When there are multiple alleles that are dominant and What’s are of equal strength expressed - then both dominant alleles will be expressed when combined when we’re allele for I’m the dominant - a dominant allele will always mask a recessive allele. together? Type A blood! Example: Blood Type Don’t count me out just I’m the dominant allele ‘cuz I’m recesssive. I’m for Type blood !B blood B or B i =B type IiIBAi III= I BA = type O allele! type blood ortype IA O i= AB type blood A blood Notice how the genotypes are written… Unique Gene Interactions Pleiotrophy: The ability of a gene to affect an organism in many phenoytypic ways Ex. Sickle Cell Anemia Blood Clumping Physical Weakness or Brain Damage Unique Gene Interactions Polygenic Inheritance: when multiple genes have an added effect on a single phenotype (Opposite of Pleiotrophy) – ex. Skin color, height Notice the range in genotypes… aabbcc AABBCC Unique Gene Interactions Epistasis: when a gene at one locus alters the expression of a gene at another locus Alleles for Fur Color: B = Black Fur b = brown fur C = Color c = albino BBCC, BBCc, BbCC, BbCc bbCC, bbCc BBcc, Bbcc, bbcc Since cc genotype is albino, the alleles for fur color (B or b) are not expressed Genetics is cool! But wait…sample probability problems to come! How do you calculate probability? Rule of Multiplication (Alternatives to Punnett Squares) Sample CheckProblem your work! Brown eyes are dominant over blue eyes. Parent A X eyes, b while b has blue ParentBB isBb heterozygous Bb for brown eyes. What is b bb bb the probability that they will have a child with blue eyes? 50% probability for blue eyes Parent A (blue eyes) = bb Parent B (brown eyes) = Bb Process: Blue-eyed Child has to be bb Probability of parent A donating one “b” allele= 1 Probability of parent B donating the other “b” allele = ½ 1 X ½ = ½ (50% probability) Rule of Multiplication and Addition Steps for solving: 1. Write out the genotypic possibilities 2. Use rule of X (multiply probabilities of each genotypic combination) 3. Use rule of + Sample Problem #2: In a cross between AaBbCc x Aabbcc, what is the probability that at least two of the three recessive traits is present in the offspring? AAbbcc AA (1/2) x bb (1/2) x cc (1/2) = 1/8 Aabbcc Aa (1/2) x bb (1/2) x cc (1/2) = 1/8 aaBbcc Aa (1/4) x Bb (1/2) x cc (1/2) = 1/16 aabbCc Aa (1/4) x bb (1/2) x Cc (1/2) = 1/16 aabbcc Aa (1/4) x bb (1/2) x cc (1/2) = 1/16 Sum of the fractions: 6/16 = 3/8