1. Define the following terms and utilize these terms in an appropriate
context:
chi-square (X2)
analysis probability
product rule (law)
summation rule (law)
conditional probability binomial
expansion Hardy-Weinberg
Equalibrium
2. Use the product rule of probability to determine the expected frequency of the occurrence of independent events.
3. Use a combination of the product rule and sum rule in applicable situations.
4. If the frequency of a homozygous recessive condition in a population is given, calculate the frequency of the recessive gene by using the inverse of the product rule; from this frequency, then calculate the frequency of the dominant gene.
5. For any number of cases or trials, select the correct term from the expansion of a binomial and determine the probability of a certain set of events occurring.
6. Use the chi-square method of determining the probability of an event occurring by chance alone, and interpret whether a deviation from the expected is acceptable, significant or highly significant.
Resources: Text pgs 54-58
Laws of Probability
A. Definition: of among n possible, equally likely outcomes there are m outcomes in which event A happens, then the probability of event A is: P(A) = m/n
EX. cards: 52 possible outcomes = n
4 are aces = m
Probability of drawing an ace = 4/52 or 1/13
EX. Aa + Aa there are four outcomes AA, Aa, Aa, aa all are equally likely therefore n = 4
Probability of a dominant phenotype (m=3) = 3/4
Probability of recessive phenotype (m=1) = 1/4
Probability of heterozygous genotype (m=2) = 2/4
or 1/2
B. Addition Rule (Summation Rule)
Probability of mutually exclusive events
-Probability of the occurrance of either A or B is the probability
of A plus the probability of B minus the probability
of the joint occurrance of A and B
P(AorB) = P(A) + P(B) - P(AB)
EX. the probability that a card drawn from a deck is either an ace or a spade
P= 4/52 + 13/52 - 1/52
= 16/52 or 4/13
ace spade ace/spade
*most events in Mendelian genetics are mutually exclusive, meaning the occurrance of one precludes the possibility of the other (EX. kings and aces are mutually exclusive)
*the probibility of some one set of mutually exclusive events is the sum of the probs of individual events
EX.
Aa X Aa probability of dominant phenotype = 1/4
(AA)
+2/4 (Aa)
3/4
EX.
Probability of a king, a queen or an ace
4/52 + 4/52 + 4/52 = 3/13
*conditional probability - the occurrance
of B is conditional on the occurrance of A P(B/A)
EX. card is drawn then without replacing it a second card is drawn
pull a spade 13/52 ---> pull a club 13/51
----> pull a spade 12/50
C. Multiplication rule - probability of the occurrance of A followed by B is the probability of A times the conditional probability of B given that A has occurred. P(A and B) = P(A) X P(B/A)
EX. pick an ace followed by a king (without replacement)
4/52 X 4/51 = 4/663 (figure your lottery odds!!!!)
EX.
pick an ace, an ace, a king
4/52 X 3/51 X 4/50 = 2/5525
Independent events- The probability that both of two independent events will occur is the product of the individual probabilities and like wise for more than two events
EX. what is prob that if a family has 5 children, all will be boys
1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32
EX. how many different gametes can be produced by AaBbCcDd?
1/2 x 1/2 x 1/2 x 1/2 = 1/16 or 16 different genotypes
EX. what is the probability that you will throw a head then a tail
1/2 x 1/2 = 1/4
EX. what is probability of a 5 then a 6 with the throw of a dice
1/6 x 1/6 = 1/36
EX. what is the probability of throwing a Yahtzee on the first toss?
Product and Summation
What is the probability of a 5 and a 6 without respect
to order?
5 then 6 P = 1/6 x 1/6 = 1/36
6 then 5 P = 1/6 x 1/6 = + 1/36
1/18
EX. probability of one boy and one girl
Girl
first P = 1/2 x 1/2 = 1/4
Boy first P = 1/2 x 1/2 = + 1/4
1/2
EX. probability of 2 boys and one girl
Girl,
Boy, Boy 1/2 x 1/2 x 1/2
Boy, Girl, Boy 1/2 x 1/2 x 1/2
Boy, Boy, Girl 1/2 x 1/2 x 1/2
*there are eight different ways to have three children, three of them are two boys and a girl
Binomial Expansion- used to determine the probability of a particular combination rather than going through all possibilities.
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
(a + b)6 = a6 + 6a5b + 15a4b2
+ 20a3b3 + 15a2b4 + 6ab5 + b6
3 children ----> (a + b)3
B G B 1/2 x 1/2 x 1/2
= 1/8
B B G 1/2 x 1/2 x
1/2 = 1/8
G B B 1/2 x 1/2 x
1/2 = 1/8
Expansion (a + b)3 = a3 + 3a2b + 3ab2 + b3
let
a = boys = 1/2 let b = girls = 1/2
then 3a2b = 2 boys and one girl therefore 3(1/2)2 (1/2)
= 1/8
Be careful: if the question says at least then you may have more possibilities to concider
Hardy Weinberg applications
given: Aa x Aa creates
| A | a | |
| A | AA | Aa |
| a | Aa | aa |
What is the probability of creating AA, or aa or Aa?
What if we are considering albanism and the frequence in the population of the aa individual is actually 1/20,000?
if the frequency of homozygous recessive is 1/20,000
then the probability of aa is 1/20,000 which is created by multiplying the prob of (a) with the prob of (a)
or a x a = 1/20,000 What is the frequency (probability) of a?
1/20,000 =
1/141 (prob of a) therefore prob of A = 140/141
Prob of AA = 140/141 x 140/141
Prob of Aa = (140/141 x 1/141) + (140/141 x 1/141)
Prob of aa =
1/141 x 1/141 = ~1/20,000