Transcript Ch 3 end with solution to end of lecture problem(s)

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.
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?
Using Probability in Genetic
Analysis
4. Conditional Probability
Jack is P_, Jill is P_
Parents of Jack or Jill: Pp x Pp
P
p
P
PP
Pp
p
Pp
pp
X
P(Pp) = 2 ways to get Pp
3 possible genotypes
P(Jack is Pp) =2/3
P (Jill is Pp) = 2/3
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.
Jack
Jill
P_ child
Probability
1/3 PP
1/3 PP
1
1/9
1/3 PP
2/3 Pp
2/3 Pp
1/3 PP
1
1
2/9
2/9
2/3 Pp
2/3 Pp
3/4
3/9
Total=8/9
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 for
unordered 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 occurs
t = 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
3! = 3 x 2 x 1 = 3 x 2!
2! = 2 x 1
1! =1
0! = 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:
1st child
2nd child
3rd child
Probability
PKU=1/9 No= 8/9 PKU=1/9
No=8/9
PKU=1/9 PKU=1/9
8/729
8/729
PKU=1/9 PKU=1/9 No=8/9
8/729
8/729 + 8/729 + 8/729 = 24/729
The End