Mendel’s 2nd Law Law of Independent Assortment Probability and Chi-Square Test Rules of Probability • Product or multiplication rule – The probability that two or more independent events will occur together is the product of their individual probabilities • Sum rule – The probability of any one event occurring out of two or more mutually exclusive events is calculated by adding the probabilities of these events (key indicators are either and or) What is the probability of rolling a “2” on a pair of dice? We will use the product rule. What is the probability of rolling a “7” on a pair of dice? Use both product rule and sum rule. What is the probability of rolling a “7” on a pair of dice? Use both product rule and sum rule. Probability in Genetics F1 X F1 Gamete Formation D/d (Tall) X ½D ½D ½d D/d (Tall) ½d Probability in Genetics F1 X F1 Gamete Formation D/d (Tall) X D/d (Tall) ½D ½d ½D ¼ D/D ¼ D/d ½d ¼ d/D ¼ d/d Probability in Genetics F1 X F1 Gamete Formation D/d (Tall) ½D ½d X D/d (Tall) ½D ½d ¼ D/D Tall ¼ d/D Tall ¼ D/d Tall ¼ d/d dwarf Mendel’s 2nd Law Law of Independent Assortment Gene pairs on different chromosomes assort independently during gamete formation Seven traits Mendel studied Pure lines Mendel’s Dihybrid Cross • What information can you conclude from these results? – Yellow is dominant over green – Round is dominant over wrinkled True-Breeding Yellow is dominant over green Round is dominant over wrinkled For the Y or R locus What type of gametes are produced? Y R/r ½ R ½ Y R ¼ y ½ R/r r ½ R ½ Y r ¼ r ½ y R ¼ y r ¼ Resulting gametes Yellow is dominant over green Round is dominant over wrinkled Mendel’s Dihybrid Cross Mendel’s Dihybrid Cross 12/16 Yellow= 3/4 Yellow 4/16 green= 1/4 green Mendel’s Dihybrid Cross 12/16 round 4/16 wrinkled Independent Assortment • Two genes are inherited independently • Frequency of F2 phenotypes can be obtained by the product law of probabilities Of all offspring of all offspring ¾ Round combined probabilities 9/16 Yellow, Round ¾ Yellow ¼ Wrinkled 3/16 Yellow, Wrinkled ¾ Round 3/16 Green, Round ¼ Wrinkled 1/16 Green, Wrinkled ¼ Green Concepts • Genes-DNA – Inherited – Expressed • Producing a gene product (usually a protein polypeptide) • The expression of the genes produces the physical traits or phenotype of the organism. • Genes are located on chromosomes at defined locations – These position on the chromosome are called a locus • A change in the DNA sequence of a gene (mutation) can result in a change in the phenotype of the organism • Diploids – Contain two copies of each chromosome. – One copy is inherited from each parent • Genes located on different chromosomes will independently assort from each other during meiosis Chi-square test Goodness of Fit Evaluating your experimental data • If you were testing the hypothesis that a trait is recessive, you would predict that mating two heterozygous parents would yield ¼ of the progeny showing the mutant phenotype. • If you found 14 mutant phenotypes in a group of 50 offspring, you wouldn’t be surprised that you didn’t get exactly 12.5 mutants. Because the amount of deviation from what you expected is fairly small. • If you found only 3 mutant phenotypes, the deviation from what you expected is very large and your intuition would suggest that the original hypothesis may not be correct. Genetic Analysis • Outcomes of genetic crosses are subject to random fluctuations by chance deviation Genetic Analysis • Outcomes of genetic crosses are subject to random fluctuations by chance deviation – A/a (brown) X a/a (white) – expect ½ A/a (brown) and ½ a/a (white) – observed results 22 brown to 18 white • Ratio 1:1 – observed results 25 brown to 15 white • Is this really a 1:1 ratio, or is there other genetic explanation for this result? – To evaluate the role of chance in producing these deviations from our expected results, the statistical test Chi-square test is used. Chi-square test “Goodness-of-fit” • This test provides information about how well the observed values fit the expected values – This test will indicate the probability that the difference between the observed results and the expected results is due to chance – Chi-square test is always applied to the progeny observed, raw data, NOT the ratios • Task of Analyzing genetic ratios is two-fold – Form a hypothesis- Determine the expected ratios • Form the HYPOTHESIS – Observed results never exactly match prediction results – Determine how closely the actual results correspond to the expected values • Do these results ‘FIT’ our hypothesis • Asking the ‘GOODNESS OF FIT’ • Chi-squared test asks: – What is the probability that the data obtained would diverge this far from the theoretical (predicted), if the null hypothesis were correct? In coat color in mice, black is dominant over white. In a cross of two predicted heterozygote black mice, 50 progeny were produced: 30 black mice and 20 white mice. • B/b X B/b – What F2 result is expected? • 3:1 ratio of black to white mice – Out of 50 progeny, how many should be black and how many white? • ¾ (50) black and ¼ (50) white • 37.5 black and 12.5 white – The cross yielded 30 black and 20 white – Could this difference be due to chance? Coat color in mice, where black is dominant over white. In a cross of two heterozygote (predicted) black mice, 50 progeny were produced: 30 black mice and 20 white mice. • B/b X B/b – What F2 result is expected? • 3:1 ratio of black to white mice • Hypothesis – Out of 50 progeny, how many should be black and how many white? • ¾ (50) black and ¼ (50) white • 37.5 black and 12.5 white • Expected values – The cross yielded 30 black and 20 white • Observed values – Could this difference be due to chance? • Chi-square test – goodness of fit Chi-square Analysis--- X2 • X2 = sum (observed – expected)2 expected • Χ2 = ∑ (o – e)2 e + ∑ (o – e)2 e o = observed value e = expected value Σ = sum of calculated value for each category in the ratio Chi-square Analysis--- X2 • X2 = sum (observed – expected)2 expected black white • Χ2 = ∑ (30 – 37.5)2 + ∑ (20 – 12.5)2 37.5 12.5 • X2 = ∑ 56.25 + 56.25 37.5 12.5 • X2 = ∑ 1.5 + 4.5 • X2 = 6.0 Interpretation of X2 Value • Calculate (df) degrees of freedom = n-1 where n is number of categories • For 3:1 ratio; n = 2, so 2 – 1 = 1 df • Compare the calculated X2 value to theX2 Table of calculated probabilities values (p value) to X2 values correlated with the different degrees of freedom. Chi-squared probability table Interpretation of p value • Hypothesis is never proven nor disproven absolutely • p value of 0.05 is arbitrary point for rejection of null hypothesis • In our case, null hypothesis is rejected 
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