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Chapter 3-Independent Assortment-September 23.docx

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Department
Biology
Course
BIOL 205
Professor
Kenton Ko
Semester
Fall

Description
CH3: INDEPENDENT ASSORTMENT OF GENES Sept 23/2011 – Pg 89-99, 102-103, 106-119 3.1 Mendel’s Law of - If two genes are on different chromosomes, the gene pairs are separated by semicolon – A/a; Independent Assortment B/b. - If they are on the same choromosome, the alleles on one homolog are written adjacently with no punctuation and are separated from those on the other homolog by a slash – AB/ab or Ab/aB. - A heterozygote for a single gene (A/a) is a monohybrid Dihybrid - A double heterozygote such as A/a . B/b is a dihybrid and by studying dihybrid crosses, Mendel came up with his second principle of heredity. - 9:3:3:1 ratio for dihybrid cross is a consistent ratio in dihybrid F2 – phenotypic ratio - 1:1:1:1 gametic ratio - The proportions of the 4 possible outcomes could be calculated using the product rule to multiply along the branches in the diagram. o Ex. ¾ of F2 is round, and ¾ of round seeds will be yellow – so ¾ x ¾ = 9/16 of peas are round and yellow. Mendel’s Second Law - Mendel’s Second Law : different gene pairs assort independently in gamete formation and applies to genes on different chromosomes. 3.2 Working with Independent Assortment Predicting Progeny Ratios - There is an easier way than drawing punnett squares to find probabilities of specific phenotypes or genotypes from a cross. - Product rule states that the probability of independent events both occurring together is the product of their individual probabilities. o Ex. Probability of rolling 2 dice because outcome on one die is independent of the other. o Probability of rolling a 4 on both dice is 1/6 x 1/6 = 1/36 - Sum rule states that the probability of either of 2 mutually exclusive events occurring is the sum of their individual probabilities. - Focuses on outcome A’ or A’’ o Ex. Probability of rolling either 2 4s or 2 5s are mutually exclusive outcomes o 1/36 +1/36 = 1/18 What proportion of progeny will be a specific genotype? - If you have 2 plants of genotypes: A/a; b/b; C/c; D/d; E/e and A/a; B/b; C/c; d/d; E/e - You want to recover a progeny plant of a/a; b/b; c/c; d/d; e/e - Assuming the gene pairs assort independently, then we can do this calculation easily using the product rule: o Ex. A/a x A/a, one fourth of the progeny will be a/a o ¼ x 1/2 x ¼ x ½ x ¼ = 1/256 How many progeny do we need to grow? - To estimate how many progeny plants needed to be grown, we need to examine at least 256 progeny to stand an average chance of obtaining one individual plant of the desired genotype. - We need to ask what sample size would we need to be 95% confident that we will obtain at least one success? - First need to consider probability of failure: 1- (1/256n = 255/256 - Probability of no success in a sample of n is (255/256) - Hence, probability of at least 1 success is 1- (255/256) =0.95 How many distinct genotypes does a cross produce - In a self of the tetra hybrid – A/a; B/b; C/b; D/d there will be three genotypes for each gene pair and 4 gene pairs in total - 3 =81 genotypes - In a testcross of a tetrahybrid, there will be 2 genotypes for each gene pair (A/a and a/a) and a 4 total of 2 = 16 genotypes in the progeny. - Because we’re assuming that all genes are on different chromosomes, all these testcross genotypes will occur at an equal frequency of 1/16. Using the chi-square test on monohybrid and dihybrid cross - When a researcher is confronted with results close but not identical to an expected ratio, they need to check ratios2against expectations. - Chi-square test (X ) is a way of quantifying the various deviations expected by chance if a hypothesis is true. - A hypothesis is rejected as false if there is a probability of less than 5% of observing a deviation from expectations. o Ex. If you predict 50% A/a and 50% a/a and in reality you obtain 120 progeny with 55 red – dominant an2 65 white –2recessive, these differ from predicted. o Use the formula X = ∑ (O-E) / E for all classes o E is expected number in a class, O is observed number. o You obtain 0.42 for each class and total X = 0.84, which you’ll compare with table 3-1 (pg 99) o You see that your 0.84 lies between 50% and 10%, this probability value is much greater than the cutoff of 5% therefore your data is good. - The probability value is a probability of observing a deviation from the expected results at least as large on the basis of chance if the hypothesis is correct
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