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Published byKenneth Simpson Modified over 4 years ago

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How Much Do You Remember???

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Character A heritable feature

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Trait A variant for a character

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True-breeding Plants that produce offspring of the same variety when they self-pollinate

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Hybridization Crossing of two true-breeding varieties

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P Generation Parental generation; the true- breeding parents

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F 1 Generation First filial generation; hybrid offspring from the P generation

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F 2 Generation Second filial generation; offspring of F 1 hybrids that self- pollinate

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Allele Alternative versions of a gene

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Dominant allele When 2 alleles at a locus are different, it determines the organism’s appearance

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Recessive allele When two alleles at a locus are different, it has no noticeable effect on the organism’s appearance *both alleles must be recessive to see this trait

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Law of Segregation The two alleles for a heritable character separate during gamete formation and end up in different gametes *an egg or sperm only gets one of the two alleles that are present in the somatic cells

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Punnett Square A diagram for predicting the allele composition of offspring from a cross between individuals of known genetic makeup *Practice…

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Homozygous An organism having a pair of identical alleles for a character

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Heterozygous An organism that has two different alleles for a gene

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Phenotype An organism’s traits

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Genotype And organism’s genetic makeup

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Test Cross Breeding of a recessive homozygote with an organism of dominant phenotype but unknown genotype to determine the unknown genotype

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Monohybrid Organism that is heterozygous for one character

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Dihybrid Organism that is heterozygous for two characters

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Law of independent assortment Each pair of alleles segregates independently of other pairs of alleles during gamete formation

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Laws of Probability and Inheritance Patterns

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Laws of Probability Probability: 1 = will occur, 0 = will NOT occur Probabilities of all possible outcomes must add up to 1. For a coin, If the sides are both “heads,” the probability of landing on that side is 1; and the probability of landing on “tails” is 0. If the coin has two different sides, there is a ½ chance of landing on a particular side. For a stack of 52 different cards, there is a 1/52 chance that you will select any given card, and there is a 51/52 chance of selecting a card other than the one you want. Outcome is not affected by previous trials. Probability: 1 = will occur, 0 = will NOT occur Probabilities of all possible outcomes must add up to 1. For a coin, If the sides are both “heads,” the probability of landing on that side is 1; and the probability of landing on “tails” is 0. If the coin has two different sides, there is a ½ chance of landing on a particular side. For a stack of 52 different cards, there is a 1/52 chance that you will select any given card, and there is a 51/52 chance of selecting a card other than the one you want. Outcome is not affected by previous trials.

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Laws of Probability Just as each coin toss’s outcome is independent of the others, so the alleles of a gene segregate into gametes independently of another gene’s alleles. (law of independent assortment) Two rules will help predict the outcome of the fusion of gametes: Multiplication Rule Addition Rule Just as each coin toss’s outcome is independent of the others, so the alleles of a gene segregate into gametes independently of another gene’s alleles. (law of independent assortment) Two rules will help predict the outcome of the fusion of gametes: Multiplication Rule Addition Rule

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Multiplication Rule Take the individual probabilities of the given outcome and multiply them together Example: For a monohybrid cross Rr x Rr (R is dominant and r is recessive) the possibility of each allele for a particular gamete being given to the offspring is ½. The probability of both gametes giving the same allele to the offspring is ½ x ½ = ¼. Take the individual probabilities of the given outcome and multiply them together Example: For a monohybrid cross Rr x Rr (R is dominant and r is recessive) the possibility of each allele for a particular gamete being given to the offspring is ½. The probability of both gametes giving the same allele to the offspring is ½ x ½ = ¼.

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Addition Rule Add the individual possibilities together when determining if one of two or more mutually exclusive events is going to occur. For example: In a monohybrid cross (Rr x Rr), the probability of the dominant allele being passed on by one of the gametes is ½ x ½ = ¼, and the probability of the dominant allele being passed on by the other gamete ½ x ½ = ¼. Probability of a heterozygote (Rr): ¼ + ¼ = ½. Add the individual possibilities together when determining if one of two or more mutually exclusive events is going to occur. For example: In a monohybrid cross (Rr x Rr), the probability of the dominant allele being passed on by one of the gametes is ½ x ½ = ¼, and the probability of the dominant allele being passed on by the other gamete ½ x ½ = ¼. Probability of a heterozygote (Rr): ¼ + ¼ = ½.

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Solving Complex Genetics Problems with the Rules of Probability Extending Mendelian Genetics for a Single Gene: the Spectrum of Dominance Complete Dominance – heterozygote and dominant homozygote are indistinguishable Mendel’s pea crosses (white OR purple, round OR wrinkled) Codominance – both phenotypes are exhibited at the same time Human blood surface molecules (MN has M AND N molecules) Incomplete Dominance – phenotype is between the phenotypes of the parents Snapdragons with red and white parents have pink offspring Extending Mendelian Genetics for a Single Gene: the Spectrum of Dominance Complete Dominance – heterozygote and dominant homozygote are indistinguishable Mendel’s pea crosses (white OR purple, round OR wrinkled) Codominance – both phenotypes are exhibited at the same time Human blood surface molecules (MN has M AND N molecules) Incomplete Dominance – phenotype is between the phenotypes of the parents Snapdragons with red and white parents have pink offspring

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Multiple Alleles Human Blood Type: A, B, AB, O Determined by which of two carbohydrates (A or B) are found of the surface of a person’s red blood cells An enzyme (I) attaches the carbohydrates I adds A, I adds B, i adds neither Each person has 2 alleles, so there are 6 possible genotypes and 4 phenotypes Human Blood Type: A, B, AB, O Determined by which of two carbohydrates (A or B) are found of the surface of a person’s red blood cells An enzyme (I) attaches the carbohydrates I adds A, I adds B, i adds neither Each person has 2 alleles, so there are 6 possible genotypes and 4 phenotypes A B Genotype Phenotype

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The End

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