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Chapter 3 – Basic Principles of Heredity
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Johann Gregor Mendel (1822 – 1884) Pisum sativum Rapid growth; lots of offspring Self fertilize with a single plant; cross fertilize between two plants
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Pisum sativum 7 characteristics –Each had only 2 forms –True-breeding varieties When allowed to self-fertilize, all offspring had same parental trait
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Modern Genetic Terminology Gene – inherited factor that codes for a specific characteristic Locus – physical location of a gene on a chromosome Allele – alternate forms of a gene –what specifically the gene codes for (black hair, blond hair)
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Modern Genetic Terminology Genotype – set al individual’s alleles; its genetic makeup –Homozygous – 2 of the same allele for a gene –Heterozygous – 2 different alleles for a gene Phenotype – outward expression of a gene –An allele may be present but not expressed in the phenotype
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Monohybrid cross Cross between plants that differ in a single characteristic P (paternal) generation –True-breeding for trait
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Monohybrid cross F 1 (filial) generation –All have trait of one parent –Reciprocal cross – sex of parent with trait made no difference
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Monohybrid cross F 2 generation –Phenotype ratio 3:1 3 = trait in F1 1 = trait not seen in F 1 ; seen in P generation –“lost” phenotype reappeared
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Monohybrid cross conclusions Each plant has two “factors” (modern terms - genes) In heterozygotes, one allele will be expressed; other will be masked, but can be passed on and expressed in offspring –Dominant allele – expressed Capital letter –Recessive allele – masked Lowercase letter 2 alleles separate with equal probability
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Principle of Segregation Each diploid organism has 2 alleles for each gene Alleles segregate from each other randomly in gamete formation
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Punnett square Illustrates possible gametes formed and possible fertilization combinations
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Probability Likelihood of the occurrence of a particular event expressed as a fraction or a decimal Multiplication rule “AND” –Probability of two or more independent events occurring together
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Multiplication rule Probability of rolling a four – 1/6 Probability of rolling a four AND then a 3 –1/6 x 1/6 = 1/36
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Multiplication rule Cross between two heterozygous purple flowered plants (Pp x Pp) ? Probability of having a purple offspring, AND then a white ? Probability of having two white offspring
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Addition rule “either/or” Probability of having 2 or more mutually exclusive events occur together Probability of rolling a three OR a four –1/6 + 1/6 = 2/6 (1/3)
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Albinism – autosomal recessive disorder 2 carriers mate (Aa x Aa) ? Probability of having three children with albinism –¼ x ¼ x ¼ = 1/64 ? Probability of having 2 “normal” and 1 albino (order not important) –1 st affected = ¼ x ¾ x ¾ = 9/64 –2 nd affected = ¾ x ¼ x ¾ = 9/64 –3 rd affected = ¾ x ¾ x ¼ = 9/64 –Add all possible combinations = 27/64
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Binomial expansion a = probability of albinism (1/4) b = probability of “normal” pigmentation (3/4) 5 children (a + b) 5 a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + b 5
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Binomial expansion a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + b 5 Term number = n +1 First term has a n, second term a n-1 b, etc –a always loses 1; b gains 1 For coefficient (number in the front) –1 st term is always 1 –2 nd term – same power as binomial (5) –3 rd term – multiply preceding coefficient (5 from 2 nd term) by exponent of a in the 2 nd term (4), then divide by term # (2) = (5x4)/2 = 10 Coefficient for 3 rd term is 10
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Binomial expansion ? probability of 2 carriers of albinism having 5 albino children and 1 “normally” pigmented child (a + b) 6
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Test Cross Determination of genotype of a dominant phenotype individual Cross with homozygous recessive individual –If any offspring demonstrate recessive phenotype, unknown must be heterozygous
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Types of genetic crosses Reciprocal –Sex of parents with a specific trait is switched Test –Cross of unknown dominant with recessive Back –Cross of individual with a parent
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Dihybrid cross 2 different traits are examined at the same time P generation – true breeding for both traits
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Dihybrid cross F 1 exhibits both dominant forms of the traits Heterozygous for both – can form 4 different types of gametes
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Dihybrid cross F 2 generation 9:3:3:1 ratio 9 – both dominant traits 3 – dominant for color; recessive for shape 3 – recessive for color; dominant for shape 1 – both recessive traits Principle of Independent Assortment –Alleles at different loci segregate independent from one another
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Branch diagram Uses probability rules 1 st column lists proportions of phenotypes of 1 st trait 2 nd column lists proportions of phenotypes of 2 nd trait, etc Faster than a Punnett square when dealing with multiple loci –Specifically when you need one particular phenotype
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Ratios Punnett squares and Branch diagrams deal with probability Observed ratio is rarely EXACTLY the expected ratio Goodness of fit Chi-Square test –Indicates probability that deviation between observed and expected ratio is due to chance alone
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Chi-Square example
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Chi-Square example cont Number is squared, so it’s always a positive number X 2 = 2.0 Need Table Degrees of freedom = n – 1, where n = possible phenotypes
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Chi-Square example cont Df = 1 X 2 = 2.1< p <.5 10% < p < 50% that variability is due to chance – hypothesis is accepted Cut off is usually p = 0.05 (5% variation due to chance)
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Cat example – Chi-square Assuming black is dominant to gray, a cross between Bb x Bb yields an expected ratio is 3:1 Offspring = 30 black cats and 20 gray Accept or reject hypothesis?
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Cat example calculations
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