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Biology 22: Genetics and Molecular Biology. Spring Semester 2008. End of lecture click to practice some problems. Practice Problems. Try your skills at the following problem. First some rules & concepts are reviewed. Problem Powerpoint will be updated with answers and posted on ecompanion.
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Biology 22: Genetics and Molecular Biology Spring Semester 2008
Practice Problems • Try your skills at the following problem. • First some rules & concepts are reviewed. • Problem • Powerpoint will be updated with answers and posted on ecompanion.
Using Probability in Genetic Analysis 1. Probability (P) of an event (E) occurring: P(E) = Number of ways that event E can occur Total number of possible outcomes Eg. P(Rr) from cross Rr x Rr 2 ways to get Rr genotype 4 possible outcomes P(Rr) = 2/4 = 1/2
Using Probability in Genetic Analysis 2. Addition Rule of Probability – used in an “either/or” situation P(E1 or E2) = P(E1) + P(E2) Eg. P(Rr or RR) from cross Rr x Rr 2 ways to get Rr genotype 1 way to get RR genotype 4 possible outcomes P(Rr or RR) = 2/4 + 1/4 = 3/4
Using Probability in Genetic Analysis 3. Multiplication Rule of Probability – used in an “and” situation P(E1 and E2) = P(E1) X P(E2) Eg. P(wrinkled, yellow) from cross RrYy x RrYy P(rr and Y_) = 1/4 x 3/4 = 3/16
Using Probability in Genetic Analysis 4. Conditional Probability: Calculating the probability that each individual has a particular genotype
INTRODUCTION TO PROBLEM Phenylketonuria (PKU) is a genetic disorder that is characterized by an inability of the body to utilize the essential amino acid, phenylalanine.
PROBLEM Jack and Jill do not have PKU, however each has a sibling with the disease. What is the probability that Jack and Jill will have a child with PKU?
P p P p PP Pp pp Pp Using Probability in Genetic Analysis 4. Conditional Probability Jack is P_, Jill is P_ Parents of Jack or Jill: Pp x Pp P(Pp) = 2 ways to get Pp 3 possible genotypes P(Jack is Pp) =2/3 P (Jill is Pp) = 2/3 X
Using Probability in Genetic Analysis 4. Conditional Probability P(child with PKU)= P(Jack is Pp) x P(Jill is Pp) x P(child is pp) = 2/3 x 2/3 x 1/4 = 1/9 P(child without PKU)= 1-1/9 = 8/9
Using Probability in Genetic Analysis To calculate probability of child without PKU, look at all possibilities for Jack and Jill.
Using Probability in Genetic Analysis 5. Ordered Events: use Multiplication Rule For Jack and Jill, what is the probability that the first child will have PKU, the second child will not have PKU and the third child will have PKU? P(pp) x P(P_) x P(pp) = 1/9 x 8/9 x 1/9 = 8/729
Using Probability in Genetic Analysis 6. Binomial Rule of Probability – used forunordered events P = n! (asbt)s! t! a = probability of event X (occurrence of one event) b = probability of event Y = 1-a (occurrence of alternate event) n = total s = number of times event X occurst = number of times event Y occurs (s + t = n)
Using Probability in Genetic Analysis 6. Binomial Rule of Probability! = factorial= number multiplied by each lower number until reaching 1 5! = 5 x 4 x 3 x 2 x 1 1! =1 3! = 3 x 2 x 1 = 3 x 2! 0! = 12! = 2 x 1
Using Probability in Genetic Analysis 6. Binomial Rule of Probability Out of 3 children born to Jack and Jill, what is the probability that 2 will have PKU? n=3, a=1/9, s=2, b=8/9, t=1 3! (1/9)2(8/9)1= 3 x 2! (1/81) (8/9)= 24 2! 1! 2! 1! 729
Using Probability in Genetic Analysis The same result can be obtained using the multiplicative rule if all possible birth orders for families of three are considered: 8/729 + 8/729 + 8/729 = 24/729
