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Sex Chromosomes & Probabilities and Chi Squared

Sex Chromosomes & Probabilities and Chi Squared. X-Linked Inheritance. Special chromosomes determine sex in many organisms X and Y chromosomes = sex chromosomes which are non-identical but share some genes Hemophilia is a classic example of human X-linked inheritance

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Sex Chromosomes & Probabilities and Chi Squared

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  1. Sex Chromosomes&Probabilities and Chi Squared

  2. X-Linked Inheritance • Special chromosomes determine sex in many organisms • X and Y chromosomes = sex chromosomes which are non-identical but share some genes • Hemophilia is a classic example of human X-linked inheritance • Males produce two different types of gametes: one containing X and another Y chromosome (heterogametic) • Females have two X chromosomes and produce only X-bearing gametes • In some organisms (birds, butterflies and some reptiles), females are heterogametic

  3. Morgan studied inheritance patterns in Drosophila melanogaster and observedthat with some traits Males X Females produced different ratios than when Females X Males

  4. Inheritence of Hemophilia A among the descendants of Queen Victoria.

  5. How is this trait most likely inherited?

  6. How is this trait most likely inherited?

  7. Predicting the liklihood of observing a specific outcome in the progeny of a given cross If a couple has three children, what’s the liklihood that they have exactly one girl? (the problem can be broken down using the multiplication and addition rules) They could either have (Boy AND Boy AND Girl) OR (Boy AND Girl AND Boy) OR (Girl AND Boy AND Boy) So the probability (P) would be P = (1/2 X 1/2 X 1/2) + (1/2 X 1/2 X 1/2) + (1/2 X 1/2 X 1/2) P = (1/8) + (1/8) + (1/8) P = 3(1/8) P = 3/8

  8. Probability of having exactly 1 girl If the probability of possibility A is p and the probability of the alternative possibility B is q, then the probability that, in n trials, A is realized s times and B is realized t times is 3! 2!1! = (1/2)2(1/2)1= n! psqt s!t! 3/8 In the previous example of a couple having three children, where the probability of having a boy is p and the probability of having a girl is q… the liklihood of observing each outcome is.. (p + q)3 = (p + q)(p2 + 2pq + q2) = p3 + 3p2q + 3pq2 + q3 The liklihood of observing specific outcomes in the progeny of a given cross is predicted by the binomial probability (p + q)n 3p2q = 3(1/2)2(1/2) = 3/8

  9. Chi-Square Analysis • The test of goodness of fit = test analyzes whether observed data agree with theoretical expectation • A conventional measure of goodness of fit is a value called chi-square, c2 • c2 = ∑( observed-expected)2 / expected • A value of c2 = 0 means that the observed numbers fit the expected numbers perfectly

  10. Chi-Square Analysis • Probability P that a worse fit (or one equally bad) would be obtained by chance, assuming that the genetic hypothesis is true • The critical values of P are conventionally chosen as 0.05 (the 5 percent level) and 0.01 (the 1 percent level) • Statistically significant refers to the magnitude of the difference between the observed and the expected numbers • To determine the P value corresponding to a calculated c2 we need the number of degrees of freedom of the particular chi-square test • The number of degrees of freedom equals the number of classes of data minus 1

  11. G g G g G g G g Calculation of chi-square for a suspected monohybrid cross … by example OBSERVED Experiment: We cross two pea plants that we suspect are monohybrids for pea color. When we sampled 144 peas from the progeny and got 99 yellow peas and 45 green peas 99 45 EXPECTED Hypothesis: (If) This is a monohybrid cross, (then) we should observe a 3:1 ratio of yellow:green peas in the progeny. This means, out of 144 peas, we should expect to observe 108 Yellow:36 Green. 108 36 c2 = ∑(observed-expected)2/expected = (observedYel-expectedYel)2/expectedYel + (observedGrn-expectedGrnl)2/expectedGrn = (99 - 108)2 / 108 + (45 - 36)2 / 36 = 81/108 + 81/36 = 3.0

  12. We have two classes of data (either yellow or green) so…our degrees of freedom = (# of classes) - 1 = 2 -1 = 1 Giving a P value of ~0.08. So the amount of fluctuation that we observed was not statistically different from what we expected to see for this sample. Indicating that our hypothesis could be correct… and we a dealing with a trait inherited as a a monohybrid cross.

  13. What would if the ratios stayed the same when we sampled 14,400 peas, instead of 144? So we observed 9,900 yellow and 3600 green peas. 1st sample 144 peas 2nd sample 14,400 peas ? ? 9900 10800 ? ? 4500 3600 14400 14400 ?

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