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since 26 Sept 1999

1. In a cross of two AaBbCcDdEeFfGg parents, what is the probability of an AABBEE offspring? Of an AaBbCcFfGG offspring?
   1. Prob(AABBEE) = (1/4)^3
   2. Prob(AaBbCcFfGG) = (1/2)(1/2)(1/2)(1/2)(1/4)

2. If we expect, on average, two mutations in our experiment, what is the chance that we will find none? One? Two?
   1. Use Possion, wih parameter 2.
      1. Prob(0) = exp(-2) = 0.135,
      2. Prob(1) = (2)exp(-2)/1! = 0.271,
      3. Prob(2) = (2)^2 exp(-2) /2! = 0.271

3. Consider the experiment in problem (2) above. What is the probability of seeing no mutations after two such experiments? Three such experiments? Five such experiments?
   1. Prob(see none) on any given experiment is 0.135.
   2. Prob(none in two experiments) = (0.135)^2 = 0.0183
   3. Prob(none in three experiments) = (0.135)^3 = 0.0025
   4. Prob(none in five experiments) = (0.135)^5 = 0.000045

4. If the probability that a random individual displays a disease is 0.01, how many individuals will we have to sample to have at least a 75% chance of seeing at least one such diseased individual?
   • Geometric (waiting-time) problem. Prob(no success after N individuals) = (1-p)^N where p = success probability (here showing the disease, with p = 0.01).
   • Hence, Prob(at least 1 success) = 1- Prob(no successes) = 1 - (1-p)^N.
   • We wish N such that 1-(1-0.01)^N = 0.75, or (0.99)^N = 0.25, Takings logs, N* ln(0.99) = ln (0.25), or N = ln (0.25) / ln (0.99) = 138

5. Consider the offspring from two AaBb parents where the two loci are unlinked. If 20 offspring are examined, what is the chance that no aabb offspring are seen? What is the chance that at least one AABB offspring is seen?
   • Binomial problem. Here p = Prob(aabb) = (1/4)(1/4) = 1/16, while N = 20
     1. Prob(0) = (1-1/16)^20 = 0.2751
     2. Prob(at least one) = 1-Prob(0) = 1-.2751 = 0.7249